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2025-03-21T14:48:29.699932
| 2020-01-20T21:36:45 |
350824
|
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|
Stack Exchange
|
Find closed-form expression to $f(n)$
For all $n \in \mathbb{N}$, let ${\mathcal A}_n := \left\{\lceil n/2\rceil, \lceil n/2\rceil+1,\dots, n-1 \right\}$ and
$$f(n) := \begin{cases} \min\limits_{a \in {\mathcal A}_n} \frac 1 4 \binom n a f(a) & \text{if $n\geq 4$}\\ \qquad 1 & \text{otherwise} \end{cases} $$
I am looking for a closed-form expression for $f(n)$, but I could not find it.
So far what I have done is to try to get it into Python, I got the first 100 values, and I thought I would be able to guess the closed-form expression, but it is too difficult to guess it.
The first 10 values are:
$$
\begin{array}{cl}
n & f(n) \\
\hline
0 & 1 \\
1 & 1 \\
2 & 1 \\
3 & 1 \\
4 & 1.0 \\
5 & 1.25 \\
6 & 1.875 \\
7 & 3.28125 \\
8 & 6.5625 \\
9 & 14.765625
\end{array}
$$
Putting it in MS Excel I found that it is NOT an exponential expression as I guessed, but I couldn't find any more.
Your $f(n)$ depends on $f(n)$, so you probably meant for $a$ to range from $n/2$ up to $n-1$, not up to $n$. Did you notice that $f(2n)=f(2n-1)$ for $n\ge4$? (At least, that's true for $4\le n\le 15$. I didn't check further.) Have you tried looking at the sequence of numerators of the $f(n)$, since the denominators are just powers of $2$.
It was a typo, I fixed it, thanx! But $\forall n\gneq m>3. f(n)\ne f(m)$ according to my Python program... And yes, even when omitting the $\frac 1 4 $ factor, it is still very hard to guess the genral closed form expression...
An experimental observation, if this helps - minimum is actually attained on $\lceil n/2\rceil$ for $n\geqslant14$. In particular, this implies that $f(2n)=2f(2n-1)$ for $2n\geqslant16$, and $f(2n+1)=\frac14\binom{2n+1}nf(n+1)$ for $2n+1\geqslant15$
@DudiFrid Sorry, my comment had a missing 2. It should say $f(2n)=2\cdot f(2n-1)$. And I just checked this up to $n\approx50$.
@DudiFrid Do you agree with my edits?
The following answer depends on what I mentioned in a comment — that the minimum is attained on $\lceil n/2\rceil$ for $n>13$; I don't know how to prove it. For $n\geqslant3$,
$$
f(n)=\frac1{3\times4^{\lceil n/2^\ell\rceil+\ell-5/2}}\frac{n!}{\operatorname{rni}(n/2)!\operatorname{rni}(n/4)!\operatorname{rni}(n/8)!\cdots\operatorname{rni}(n/2^\ell)!}
$$
where $\ell=\max(0,\lceil\log_2(n/13)\rceil)$ and $\operatorname{rni}(x)$ is the nearest integer to $x$, with the convention $\operatorname{rni}(n+\frac12)=n$ for integer $n$.
My experiments show that the minimum is unnecessarily attained in $\lceil n/2 \rceil$...:\
@DudiFrid I meant for $n\geqslant14$, let me add it. Did you find any counterexample? Sometimes it is also attained at $n-1$, but for all $n\geqslant14$ that I checked it is attained at $\lceil n/2\rceil$
I am now running another kind of check, currently it shows that $f(n)=\frac14\binom n{\lceil\frac n2\rceil}f(\lceil\frac n2\rceil)$ for $14\leqslant n\leqslant7500$
...up to 10000 by now
|
2025-03-21T14:48:29.700140
| 2020-01-20T22:47:54 |
350828
|
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"authors": [
"Carlo Beenakker",
"José Hdz. Stgo.",
"M. Khan",
"alpoge",
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"url": "https://mathoverflow.net/questions/350828"
}
|
Stack Exchange
|
Question about a lesser-known "class number formula" of Gauss
My question refers to article 301 of section 5 of Gauss's D. A. - there Gauss gives an asymptotic formula for the mean number of classes of forms with positive discriminant ($D>0$):
$$h(D) = \frac{4}{\pi^2}\log (D) + \delta$$
where $\delta$ is the following constant:
$$\delta = \frac{8}{\pi^2}C+\frac{48}{\pi^4}\sum_{n=2}^{\infty}\frac{\log (n)}{n^2} - \frac{2 \log 2}{3\pi^2}$$
and $C$ is the Euler-Mascheroni constant. This formula is noteworthy because Gauss also evaluates the error term $\delta$ in this class formula. I tried to search on the web for references to this analytic result of Gauss but all I found is articles about his asymptotic class number formula for forms with negative discriminant ($D<0$), which Gauss gives in the next article of D. A. (article 302).
My question is mainly about getting a reference for this result of Gauss. If such a reference doesn't exist, I'll be glad to hear an expert opinion on the historic significance of this formula (Gauss did publish this formula; thereby it's plausible that it did have influence).
He gives an asymptotic for the number of genera, which is akin to giving an asymptotic for $\sum_{n\leq X} 2^{\omega(n)}$ —- hence the log. [Note that he says genera around $+D$ or $-D$, and $h(-D)$ is way bigger.] Let me know if I’ve misunderstood!
Mmmh... Isn't art. 301 of the Disquisitiones a reference for this result of Gauß?
@JoséHdz.Stgo. --- I presume with "reference for this result of Gauss", the OP means "a reference which discusses this result of Gauss".
If I recall correctly there may be some information on this in Ayoub's book "Introduction to Analytic Number Theory". Having said this, I have not looked at this book for many years and I do not have easy access to a copy.
@CarloBeenakker: I wonder if the OP is aware of this previous question in MO: https://mathoverflow.net/a/109396/1593
Concerning the historic significance of Gauss's article 301: Marius Overholt traces back to this publication on the growth of class numbers the use of local averaging to study the growth of arithmetic functions.
I think the only detailed discussion of this particular result of Gauss is due to Dirichlet, and appears in his 1838 publication "on the use of infinite series in the theory of numbers", which is available on internet archive on p.376-391 of G. Lejeune Dirichlet's Werke. In fact, this publication of Dirichlet is discussed in a recent (2018) biography of Dirichlet (by Uta C. Merzbach), which cites the pages in Dirichlet's publication on Gauss's formula from D.A article 301:
Finally, Dirichlet observed that by the same kind of analysis he could find the formulas presented in article 301 of Gauss's beautiful work:
Suppose, for example, that it's a question of obtaining the mean number of genera for the determinant $-n$, a number which we shall denote by $F(n)$. If one compares art. 231 (of the D.A) where all the complete characters assignable a priori are enumerated, with art. 261 and 287, where the illustrius author showed that only half of these characters correspond to really existing genera, one could easily find (...) five equations (...) which (on appropriate summing and substituting), result in the asymptotic formula of the mean value of the number of genera for a determinant $-n$ as $$\frac{4}{\pi^2}(\mathbb{log (n)}+\frac{12C'}{\pi^2}+2C-\frac{1}{6}\mathbb{log (2)})$$
which coincides with the result of M. Gauss.
Looking at Dirichlet's relatively long reconstruction of Gauss's result, i gained the impression that this is one of those results where truly rigorous analytic methods were needed in it's derivation (this is also evident from the title of Dirichlet's memoir). I think this makes clear that, despite not giving a proof for this result (not even in his Nachlass), Gauss was fully aware of some equivalent form of Dirichlet's L-series technique; Gauss's formula is too precise, and therefore it cannot be assumed that he conjectured it on empirical basis. This might seem to contradict Gauss's own statement from art. 301, according to which he discovered this result after a long tables study - but i guess the "tables study" needs to be interpreted as a "semi-empirical derivation".
|
2025-03-21T14:48:29.700442
| 2020-01-20T23:12:21 |
350830
|
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|
Stack Exchange
|
Functions that are Khinchin integrable but not Henstock-Kurzweil integrable
I posed this question on Mathematics SE recently, though by the total lack of attention it has gotten, I do not anticipate an answer and bring it here.
What are some Khinchin integrable $f$ which are not Henstock-Kurzweil integrable?
I myself have just begun studying gauge integration for a course and am unable to answer this question$-$given we apply gauge integrals to deal with non-Lebesgue / non-Riemann integrable functions, I was surprised to learn there even was a more general integral, and am curious as to what non-gauge integrable functions necessitate it.
Please let me know by a comment if this question is not appropriate for MO. I will delete it accordingly.
Let $F$ be the function from example 6.20 c) in [1]. That is, fix a perfect nowhere dense subset $E$ of $[0, 1]$ such that $0, 1\in E$ such that $0 < |E| < 1$, for example the fat Cantor set. Write $(0, 1)\setminus E$ as a countable union of open intervals $(0, 1)\setminus E = \bigcup_{n=1}^\infty I_n$. Let $\rho_n$ be the length of a largest subinterval of $[0, 1]$ disjoint from $I_1, ..., I_n$. Let $A = E\cup \{c_n : n\in\mathbb{N}\}$ where $c_n$ is the middle point of $I_n$. Define $G:A\to\mathbb{R}$ as $$G(x) = \begin{cases} 0, & x\in E, \\ |I_n|+\rho_n, & x = c_n\end{cases}$$ and extend $G$ to a continuous function $F:[0, 1]\to\mathbb{R}$ by letting $F$ to be linear on every open interval $J\subseteq A^c$ with end-points in $A$.
The function $F$ is $ACG$ but it has no a.e. derivative so $F$ is not in $ACG_*$. Then $F_{ap}'$ exists a.e. and is Khintchine integrable. However it cannot be Henstock-Kurzweil integrable, since supposing otherwise, if $G$ is the indefinite integral of $F_{ap}'$, then $F-G$ is in $ACG$ and $$(F-G)_{ap}' = F_{ap}'-G_{ap}' = F_{ap}'-G' = F_{ap}'-F_{ap}' = 0\text{ a.e.}$$
so from uniqueness of Khintchine integral we would have $F-G = \text{const.}$ which implies that $F$ is $ACG_*$. $\unicode{x21af}$
Also see example 16.19 in [1] for a function which is in $ACG$ but not in $ACG_\Delta \supseteq ACG_*$. The same argument shows that this function is Khintchine integrable but not Henstock-Kurzweil integrable.
[1] "The Integrals of Lebesgue, Denjoy, Perron, and Henstock" by Russell A. Gordon
Huh, I didn't know MathJax obeyed \unicode!
|
2025-03-21T14:48:29.700606
| 2020-01-21T00:37:48 |
350833
|
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|
Stack Exchange
|
Dualizing module for $\operatorname{Aut}(F_n)$
In The complex of free factors of a free group (pdf at Hatcher's page), Hatcher and Vogtmann defined a simplicial complex $FC_n$ called the ``complex of free factors'' of the free group $F_n$. They proved that $FC_n$ is homotopy equivalent to a wedge of $(n-2)$-spheres. They call the top reduced homology group of this complex the Steinberg module of $\operatorname{Aut}(F_n)$ and ask if it is a rational dualizing module for $\operatorname{Aut}(F_n)$. That is, they ask if: $$H^i(\operatorname{Aut}(F_n);\mathbb Q) \cong H_{2n-2-i}(\operatorname{Aut}(F_n);\tilde H_{n-2}(FC_n;\mathbb Q)).$$ Has there been any progress on this question?
In our paper here, Himes, Miller, Nariman, and myself prove that Hatcher-Vogtmann’s question has a negative answer, at least for $n=5$. It also probably has a negative answer for larger $n$, but our ignorance of the unstable cohomology of $Aut(F_n)$ prevents us from proving this via our techniques. I do not know any reasonable conjecture as to what the dualizing module is.
|
2025-03-21T14:48:29.700715
| 2020-01-21T03:53:57 |
350842
|
{
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"authors": [
"Nik Weaver",
"Sanae Kochiya",
"Yemon Choi",
"erz",
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"url": "https://mathoverflow.net/questions/350842"
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|
Stack Exchange
|
Approximation of Inductive Tensor Product $C(X) \bar{\otimes} C(Y)$
The following question is from Banach Algebra Techniques in Operator Theory written by Ronald G. Douglas.
Assume both $X, Y$ are Banach spaces and $X \otimes Y$ is the algebraic tensor product. Let ${X^*}_{\leq1}$ be the closed unit ball in $X^*$. For $w \in X \otimes Y$, define $\|w\|_i = \sup\{\left|\sum_{k = 1}^n \phi(x_k) \psi(y_k)\right| : x_k \in X, y_k \in Y, w = \sum_{k = 1}^n x_k \otimes y_k\}$ (one of $w$'s expression in $X \otimes Y$), $\phi \in {X^*}_{\leq1}, \psi \in {Y^*}_{\leq1}$}. One can check this is a norm on $X \otimes Y$ and we let ($X \mathbin{\bar{\otimes}} Y, \| \cdot \|_i$) be the completion of the set ($X \otimes Y, \| \cdot \|_i$).
Now assume $X, Y$ are both compact Hausdorff topological spaces and hence ($C(X), \| \cdot \|_{\infty}$), ($C(Y), \| \cdot \|_{\infty}$) are Banach spaces. Show that ($C(X) \bar{\otimes} C(Y), \| \cdot \|_i$) is isometrically isomorphic to ($C(X \times Y), \| \cdot \|_{\infty}$). Here $X \times Y$ is equipped with the product topology.
Note that any norm $\|\cdot\|$ in $C(X) \oplus C(Y)$ (the direct sum of two Banach Spaces) is equivalent to $\|\cdot\|_1$ because both $C(X), C(Y)$ are Banach spaces equipped with $\|\cdot\|_{\infty}$ (hence $\|(f_x, f_y)\|_1 = \|f_x\|_{\infty} + \|f_y\|_{\infty}$. Meanwhile, one can find a homeomorphism between $C(X)\oplus C(Y)$ and $C(X\times Y)$ because $\|f\|_{\infty} \leq \|f_x\|_{\infty} + \|f_y\|_{\infty} \leq 2\|f\|_{\infty}$. Hence I directly start finding relation between ($C(X)\oplus C(Y), \|\cdot\|_1$) and ($C(X) \mathbin{\bar{\otimes}} C(Y), \|\cdot\|_i$)
$$\Large Question Part$$
Say $w \in C(X)\bar{\otimes} C(Y)$ and here I have difficulty finding upper bound of $\|w\|_i$ with respect to $\|\cdot\|_1$. Naively I consider partition of unity of $X$, say {$P_i, i \leq n$} and $\sum_{i \leq n}fP_i$ is one to break down $f$. Hence this could be one of the expression of $f$ part in $w$. I do not know if $n$ is the max number of pieces of $f$ I can break down.
According to hints in the book, by Krein-Milman, it suffices to consider extreme points in $X^*$ and $Y^*$. Before using this, I believe I need to collect enough information of $w$.
Regarding terminology: I think this is usually called the injective tensor product of Banach spaces, not the "inductive" tensor product
What is $C(X) \times C(Y)$? The cartesian product? If so, it can't be homeomorphic to $C(X \times Y)$. Look at the case where $X$ and $Y$ are finite, the dimensions don't match.
@NikWeaver I believe your concern is correct. Here I should change the $\times$ to $\oplus$ because I am taking direct sum of two Banach spaces (i,e, ($C(X), |\cdot|{\infty}$ and ($C(Y), |\cdot|{\infty}$). When $X$ and $Y$ are finite, dim$C(X)$ = $\vert X \vert$ because in this case $C(X)$ is the span of characteristic function of singleton. Hence dim[$C(X) \oplus C(Y)$] will be equal to dim$C(X \times Y)$.
@YemonChoi I am confused by the difference between the terminology used in my book and results from Google. These two terms mean the same but in case someone will check the book I will follow what the book is using.
The vector space dimension of $C(X) \oplus C(Y)$ is the sum of $|X|$ and $|Y|$, the dimension of $C(X\times Y)$ is their product.
For nice space $X$ and $Y$, $C(X\times Y)=C(X,C(Y))$.
@NikWeaver Assume both $X$ and $Y$ are finite sets. Let's fix $x_1 \in X$. For any two different $y_i, y_j \in Y$, ($\chi_{{x_1}}, \chi_{{y_i}}$) is linearly independent with ($\chi_{{x_1}}, \chi_{{y_j}}$).Then in a basis of $C(X) \oplus C(Y)$, at least we have $\vert Y \vert$ many linearly independent elements. For any two different $x_n, x_m \in X$ and a fixed $y \in Y$, also we have ($\chi_{{x_n}}, \chi_{{y}}$) is independent with ($\chi_{{x_m}}, \chi_{{y}}$). If there are no mistakes above, then the basis size of $C(X) \times C(Y)$ will be $\vert X \vert$*$\vert Y \vert$.
@NikWeaver Now for set bijiection between $C(X \times Y)$ and $C(X) \oplus C(Y)$, you can refer to this link: https://math.stackexchange.com/questions/3516784. In $X \times Y$ (equipped with product topology), let $P_x, P_y$(resp.) be projection on $X, Y$(resp.). For any $f \in C(X \times Y)$, I consider the mapping $f \mapsto (f \circ P_x, f \circ P_y)$. The other direction is similar (please check the links for more details).
You are wrong. Consider the case where $X$ and $Y$ are both singletons.
@NikWeaver I noticed where my mistake is .... In general your statement is right. When one of $X$ or $Y$ is infinite, $\vert C(X) \oplus C(Y) \vert$ is smaller than $\vert C(X \times Y) \vert$. I made a silly mistake here. $f_x \circ P_x$ is continuous in $X \times Y$ whenever $f_x \in C(X)$ but not the other way.
No ... no. I really think you should start by trying to understand the finite dimensional case.
Thank you for your advice, @NikWeaver. Now I finally realize when finding basis I should start from leaving all but one coordinate $0$ but I can not imagine I missed this ….
|
2025-03-21T14:48:29.701044
| 2020-01-21T05:19:57 |
350843
|
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"Ian Agol",
"Michael Albanese",
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|
Stack Exchange
|
Existence of fibered surfaces in arbitrary 4-manifolds?
It is apparently a result of F. González-Acuña that all closed orientable 3-manifolds contain a fibered knot. (I am not sure exactly where to find a published proof of this result and as an aside I would be interested in hearing about any proofs/references that anyone knows.)
I am wondering to what extent this is true in higher dimensions and specifically in dimension 4. In particular, is there always a fibered 2-sphere in a closed orientable 4-manifold? If not, what about arbitrary genus surfaces?
For the 3-manifold case, one could use that the figure 8 knot is universal. The pullback of the figure 8 will be a fibered link in the branched cover. One can plumb hopf bands to create a knot then.
Sorry if this is basic, but what is a fibered surface?
a surface whose complement is fibered
I suppose that, for a $n$-manifold $M$, containing a "fibered codimension-2 manifold" $N\subset M$ means that $N$ has trivial normal neighbourhood $\nu N = N\times D^2$ and its complement $M \setminus \nu N$ fibers over $S^1$ with a fibration that extends the projection $\partial (\nu N) = N \times S^1 \to S^1$ of its boundary. The fiber of the fibration is some $(n-1)$-manifold $Y$ with boundary $N$.
Any such manifold $M$ is obtained from a $(n-1)$-manifold with boundary $Y$ and a self-diffeomorphism $\varphi$ of it (that restricts to the identity on the boundary). You use $\varphi$ as the monodromy to construct a fibration over $S^1$ with fiber $Y$, and then cap the boundary component by gluing $\partial Y \times D^2$. Following Gordon, when $\varphi$ is trivial, we may say that $M$ is obtained by spinning $Y$. In this case we get $\pi_1(M) = \pi_1(Y)$.
The first thing to do is to calculate some topological invariants of such a manifold $M$ obtained in this way. We get $\chi(M) = \chi(N)$. So if a four-manifold $M$ contains a fibred 2-sphere it must have $\chi(M)=2$. If it contains a more general fibred connected surface $N$, we get $\chi(M) = \chi(N) \leq 2$. I would also bet that $M$ is even and has zero signature (I haven't checked this). One could look at the possible fundamental groups $\pi_1(M)$ one could get. Geometrisation of 3-manifolds gives some information on the possible $Y$ and the possible $\varphi$.
Thank you for this - I did not notice that about the Euler characteristic, and the signature does indeed vanish as @MarcKegel pointed out. How does geometrisation give info about $Y$ and $\phi$?
A fibered codimension $2$ link in a manifold $M$ can be identified with an open book decomposition of $M$ (the object that is described at the beginning of Bruno Martelli's answer).
The existence of open books was studied in Quinn's paper:
Open book decompositions, and the bordism of automorphisms.
In particular, any odd-dimensional (closed oriented) manifold admits the structure of an open book. In even dimensions there are obstructions. For example, the only surface admitting an open book is $S^2$.
A general obstruction is coming from the signature as Bruno Martelli was pointing out. Given an open book decomposition of $M$, it is relatively easy to construct an orientation reversing diffeomorphisms on the neighborhood of the codimension $2$ manifold and on its complement. Thus the signature has to vanish.
In the above paper, Quinn identifies another (more complicated) algebraic obstruction to the existence of open books in even dimensions larger than $4$ and shows on the other hand that these are the only obstructions.
In dimension $4$ the question about the existence of open book structures seems to be unsolved (apart from that the signature has to vanish). (Quinn's paper uses the $s$-cobordism theorem and thus the argument is probably not working in dimension $4$.)
|
2025-03-21T14:48:29.701422
| 2020-01-21T08:27:18 |
350846
|
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"Matthieu Romagny",
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|
Stack Exchange
|
Global functions on a product of schemes over artinian ring
For a morphism of schemes $f:X\to S$ with $S=\text{Spec}(R)$ affine, let's write $A(X)=H^0(X,\mathcal{O}_X)$. I'm interested in the morphism of $R$-algebras
$$
c:A(X)\otimes_R A(Y)\to A(X\times_SY)
$$
for some $X,Y$ over $S$ which I don't want to assume affine or proper. With no affineness condition, the classical case where $c$ is shown to be an isomorphism is when $X\to S$ is flat and $A(Y)$ is flat (SGA3 I, Lemme 11.1).
I want to prove that $c$ is an isomorphism in the following case: $R$ is an artinian ring with algebraically closed residue field and $X$ and $Y$ are smooth, of finite type, connected if that helps, over $R$.
In fact in my intended application I have $X=Y=G$ a smooth connected group scheme. Over the residue field, by Chevalley we know $G$ is an extension of an abelian variety by a an affine group scheme, or by the 'dual' Chevalley also an extension of an affine by an anti-affine. But I'm not sure this is relevant in any way.
My attempts to prove this so far were as follows. If $X,Y$ are both affine, or both proper, then the claim is true. I tried dévissage, that is, $X\to S$ factors as an affine morphism $X\to Z$ followed by a proper morphism $Z\to S$ (for schemes, Temkin 2011; for group schemes this is Chevalley) but was unable to really go further.
Another approach I tried was to use the Künneth formula (say in the form
stacks project Tag 0FLQ. Taking $H^0$ in the Künneth formula, in the RHS I get exactly $A(X\times_SY)$, while in the LHS the algebra $A(X)\otimes_R A(Y)$ should appear as a graded piece of the abutment of the second hypercohomology spectral sequence, if I'm right. The problem here is that I'm not sure how to write that spectral sequence, because instead of a functor, in this case I have a bifunctor (namely $\otimes$) and then I get lost.
Note that although I'd be quite surprised, it is still possible that my guess for the isomorphism is wrong, and I'd be happy (well... would I?) to be disproved.
Thank you for any help or suggestion!
Do you know an example of $Y/S$ smooth and of finite type (with $S=\operatorname{Spec} R$ as in your question) such that $A(Y)$ is not flat over $R$?
@Piotr: Excellent question. I must think about it.
@PiotrAchinger We probably can construct an example using deformation theory. Take $Y$ to be the deformation of $\mathbb{A}^2\setminus 0$ over $S=k[e]/e^2$ given by gluing the charts $k[x^{\pm 1},y][e]$ and $k[x,y^{\pm 1}][e]$ using the infinitesimal automorphism $id+e \frac{1}{xy}\frac{\partial}{\partial x}$ of the intersection $k[x^{\pm 1},y^{\pm 1}][e]$. Then no function of the form $x+e f,f\in k[x^{\pm},y]$ extends to a global function on $Y$ so $e x\in H^0(Y,\mathcal{O})$ is an elmement killed by $e$ that is not divisible by $e$, this module is thus not flat.
@SashaP: your example is nice, it's of course a very natural one. I'll try to compute the map $c$ in case $X=Y=$ your example, and see how it's like. Thanks!
|
2025-03-21T14:48:29.701650
| 2020-01-21T09:24:04 |
350849
|
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|
Stack Exchange
|
Random matrix properties
Let $\mathbf{H}_{N,K}$ be a random matrix whose entries are i.i.d complex Gaussian random variables with variance $1$. Then, we know from the law of large number that if $N,K\rightarrow\infty$, we have
$$\frac{1}{N}\mathbf{H}^\mathrm{H}\mathbf{H}\rightarrow\mathbf{I},$$
where $\mathbf{I}$ is the identity matrix. Now, I have some questions regarding this large dimension analysis:
Let define $\mathbf{A}=\mathbf{H}^\mathrm{H}\mathbf{H}$. If $f$ be a continues function, is it true to write?
$$f(\mathbf{A})\rightarrow f(\mathbf{\mathbb{E}[\mathbf{A}]})=f(N\mathbf{I}).$$
More over, can we write as follows?
$$\mathbf{H}^{\mathrm{H}}f(\mathbf{A})\mathbf{H}\rightarrow \mathbf{H}^{\mathrm{H}}f(N\mathbf{I})\mathbf{H},$$
and for special case $f(\mathbf{A})=\mathbf{A}$, we have?
$$\mathbf{H}^{\mathrm{H}}\mathbf{A}\mathbf{H}\rightarrow N\mathbf{H}^{\mathrm{H}}\mathbf{H}\rightarrow N^2\mathbf{I}.$$
I don't see how this would be consistent with Marchenko-Pastur
Your identity matrix $\mathbf I$ must be $K\times K$ and hence cannot be the limit of anything as $K\to\infty$.
Let me check this for real matrices and $f(A)={\rm tr}\,A^2$. The $N$ eigenvalues $\mu_n$ of $A$ have in the limit $N\rightarrow\infty$ at fixed $N/K=\lambda\leq 1$ the Marchenko-Pastur distribution
$$\rho(\mu)=\mathbb{E}\left[\sum_{n=1}^N\delta(\mu-\mu_n)\right]=N\frac{\sqrt{\lambda_+-\mu/N}\sqrt{\mu/N-\lambda_-}}{2\pi\lambda\mu},\;\;\lambda_-<\mu/N<\lambda_+,$$
with $\lambda_\pm=(1\pm\sqrt\lambda)^2$. The function $f(A)$ tends in this limit to
$$f(A)\rightarrow\int_{N\lambda_-}^{N\lambda_+}\mu^2\rho(\mu)\,d\mu=N^3(1+\lambda),$$
which differs from the answer $f(N I)=N^3$ conjectured in the OP.
|
2025-03-21T14:48:29.701779
| 2020-01-21T09:47:08 |
350850
|
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|
Stack Exchange
|
Derived category of a fiber product
Let $X = Y \times_Z W$, where $X,Y,Z,W$ are Noetherian schemes, and consider the pullback diagram associated to $X, Y, Z, W$. We have a diagram
$$
\require{AMScd}
\begin{CD}
D(Z) @>>> D(Y)\\
@VVV @VVV\\
D(W) @>>> D(X)
\end{CD},
$$
where $D(-)$ means either the bounded derived category of coherent sheaves, or the unbounded category of quasi coherent sheaves, and the maps are given by pullback of modules.
I was wondering whether this diagram is a push out diagram. I don’t expect this to be true in the world of standard algebraic geometry, i.e. if this holds I would expect one has to take the derived fiber product and consider the categories as $\infty$ or DG categories. However, not having much knowledge of derived algebraic geometry, I don’t know whether this result holds.
No, this is false. Consider what happens when the intersection is empty, for instance two distinct points in the line. If you reverse the roles of pushout and pullback, then the statement is often true-- though I'm not sure what the correct level of generality is.
@Phil Tosteson I see, thanks!
Empty intersection seems to work fine. For the case of dg categories of unbounded complexes of quasi-coherent sheaves, see Theorem 4.7 in https://arxiv.org/abs/0805.0157. The case of dg categories of coherent complexes is more complicated, see https://arxiv.org/abs/1312.7164.
@PavelSafronov yeah, actually I gave it another thought yesterday and I wasn’t able to spot why empty intersection should not work. Thank you very much for the references!
|
2025-03-21T14:48:29.701925
| 2020-01-21T10:36:48 |
350852
|
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|
Stack Exchange
|
A randomized central limit theorem
Let $X_k$, $k = 1, 2, \dots$, be a sequence of i.i.d. random variables with finite second moments. Also, let $N_k \geq 1$, $k = 1, 2, \dots$, be a sequence of random variables taking integral values, such that $\lim_k N_k = \infty$ a.s.. Furthermore, assume that each $N_k$ is independent of the $X_k$'s.
If $S_k := \sum_1^{N_k} X_k$, does it follow that $(S_k - \mu N_k)/\sigma\sqrt{N_k}$ converges in distribution to the standard normal variable
(where $\mu = \mathbb{E}[X_k]$ and $\sigma^2 = \mathbb{V}[X_k]$) as $k \to \infty$?
Yes: the distribution of $(S_k - \mu N_k) / (\sigma \sqrt{N_k})$ is a mixture of the distributions of standardised sums $Y_k = (\sum_{j=1}^k X_j - k\mu) / (\sigma \sqrt{k})$. Since the distribution of $Y_k$ converges to standard normal, the same is true for the mixture.
Otherwise you can redo the CLT proof with the characteristic function $\mathbb{E}(e^{i\alpha (S_k-\mu N_k)/\sigma\sqrt{N_k}})=\mathbb{E}(\mathbb{E}(e^{i\alpha (X-\mu )/\sigma\sqrt{N_k}})^{N_k})\approx \mathbb{E}((1-\frac{\alpha^2}{2N_k})^{N_k})\approx e^{-\alpha^2/2}$.
$\newcommand{\ep}{\varepsilon}
\newcommand{\Si}{\Sigma}$The answer is yes, and it is enough that $N_k\to\infty$ just in probability or, equivalently, in distribution (rather than almost surely), which means that for each real $b$
$$P(N_k\le b)\to0$$
as $k\to\infty$.
Indeed, let $F_n$ be the cdf of $\sum_1^n(X_i-\mu)/(\sigma\sqrt n)$ and let $G_k$ be the cdf of $S_k$.
Take any real $x$. By the central limit theorem, $F_n(x)\to\Phi(x)$ as $n\to\infty$, where $\Phi$ is the standard normal cdf.
Take now any real $\ep>0$. Then there is some natural $A$ such that $|F_n(x)-\Phi(x)|<\ep/2$ for all $n>A$. Further, there is some natural $K$ such that $P(N_k\le A)<\ep/2$ for all $k>K$.
Take now any such $A$ and $K$, and then take any natural $k>K$. Write
$$G_k(x)-\Phi(x)=\sum_{n=1}^\infty P(N_k=n)F_n(x)-\Phi(x)=\Si_1+\Si_2,
$$
where
$$|\Si_1|=\Big|\sum_{n=1}^A P(N_k=n)(F_n(x)-\Phi(x))\Big|
\le\sum_{n=1}^A P(N_k=n)=P(N_k\le A)<\ep/2,$$
$$|\Si_2|=\Big|\sum_{n>A} P(N_k=n)(F_n(x)-\Phi(x))\Big|
\le\sum_{n>A} P(N_k=n)\ep/2\le\ep/2,$$
whence
$$|G_k(x)-\Phi(x)|<\ep$$
for all $k>K$. That is, $G_k(x)\to\Phi(x)$ as $k\to\infty$, as desired.
The answers are correct!
|
2025-03-21T14:48:29.702080
| 2020-01-21T12:23:28 |
350857
|
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|
Stack Exchange
|
Constructive contents of “the support of a sheaf is closed” or “the flat locus is open”
Out of curiosity I started to go through an introductory course on schemes and to convince myself that I could prove every result (after an appropriate reformulation if necessary) constructively, i.e. without use of the law of excluded middle. See here for a nice explanation of how to avoid the use of the law of excluded middle (and hence the existence of prime ideals!) by defining schemes as certain locally ringed locales. This project is going quite well, but there is one thing that is leaving me a bit puzzled.
There are some well known technical statements in algebraic geometry of the form
$\newcommand{\sheaf}{\mathcal}\newcommand{\from}{\colon}\newcommand{\into}{\hookrightarrow}$
Proposition 1: Let $X$ be a scheme and let $\sheaf{F}$ be a quasi-coherent sheaf of modules on $X$. If $\sheaf{F}$ is finitely generated then the support $\{ x \in X \mid \sheaf{F}_x \neq 0\}$ is a closed subset of $X$.
Proposition 2: Let $S$ be a scheme. Let $f \from X \to S$ be a morphism which is locally of finite presentation. Let $\sheaf{F}$ be a quasi-coherent $\sheaf{O}_X$-module which is locally of finite presentation. Then $\{ x \in X \mid \sheaf{F}_x \text{ is flat over } \sheaf{O}_{S,f(x)}\}$ is an open subset of $X$.
There are also the corresponding affine statements
Proposition 1a: Let $A$ be a commutative ring and $M$ finitely generated $A$-module. Then the set of all prime ideals of $A$ such that $M_\mathfrak{p} \neq 0$ is closed in the prime spectrum of $A$ (and coincides with $V(\mathrm{Ann}(M))$).
Proposition 2a: Let $f \from A \to B$ be a ring homomorphism of finite presentation and let $M$ be a finitely presented $B$-module. Then the set of all prime ideals $\mathfrak{p}$ of $B$ such that $M_\mathfrak{p}$ is flat as an $A$-module is open in the prime spectrum of $B$.
There are of course other such statements concerning the smooth locus of a morphism etc. My problem is that I don't even know how to state these propositions constructively. Schemes as locales do not have underlying sets of points which one could test on whether the stalk of a sheaf at that point satisfies some condition.
Regarding the first proposition (closedness of the support), one could try to define the support of a finitely generated module as the closed subscheme $V(\mathrm{Ann}(M)) \into \mathrm{Spec}(A)$ and for the global case one can maybe show that these subschemes can be glued together sensibly.
However the radical ideal definig the open subset in Prop. 2a cannot be described as easily. The proof certainly makes the impression that something nontrivial is going on and my question is how to capture the constructive content of these arguments. I assume a constructive proof won't be easy (in some way or another some $\mathrm{Tor}$ groups will be involved and already the definition of these groups is a bit problematic in constructive mathematics (see here)). Therefore for now I will be happy with a formulation of the statement which is sensible in a constructive setting.
|
2025-03-21T14:48:29.702311
| 2020-01-21T12:28:39 |
350858
|
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|
Stack Exchange
|
Linearly independent support vectors of a convex set
Let $\Omega\subset\mathbb{R}^n$ a compact strictly convex set containing $0$ in its interior and let $k\leq n$.
Given a vector $x\neq 0$ in $\mathbb{R}^n$ a supporting vector $\xi_x$ in the direction of $x$ is a vector satifying $h(x)=\langle x,\xi_x\rangle$ where $h(x):=\sup\{\langle x,u\rangle|u\in\Omega\}$ is the supporting function of $\Omega$.
Given $k$ linearly independent directions $e_1,...,e_k$, on what condition on $\Omega$ can we show that the supporting vectors in the corresponding directions $\xi_{e_1},...,\xi_{e_k}$ are linearly independent?
More generally, what are the properties of the map associating to a unit vector $x$, the corresponding support direction $\xi_x$ ?
According to your definition, the set of supporting vectors in the direction of $x$ (for $x \ne 0$) is a hyperplane. If $n\ge 3$ and $e_1$ and $e_2$ are linearly independent, the hyperplanes corresponding to $e_1$ and $e_2$ will always intersect, i.e. it is possible to have $\xi_{e_1} = \xi_{e_2}$.
Thank you for your remark, I forgot to mention that I choose $\xi_x\in\partial\Omega$ (and $\langle x,\xi_x\rangle=h(x)$).
|
2025-03-21T14:48:29.702410
| 2020-01-21T13:44:42 |
350863
|
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|
Stack Exchange
|
Special element of a commutative ring
Let $R$ be a commutative ring with $1$ and $S $ be a multiplicative subset of $R $. I am looking for an equivalence condition for the following property in $R $:
Property: There exists a fixed element $s\in S $ such that for each idempotent element $e $ of $R $ we have either $se=0$ or $s (1-e)=0$.
Note: Above property has a geometric interpretation, actually, $s $ is an element that annihilates a set of idempotents whose closure is a connected component in the $Spec (R) $, when $s $ is not a unite element.
The collection of clopen subsets of $\operatorname{Spec}(R)$ on whose complement $s$ vanishes identically forms a filter (it is closed under intersections, enlarging the subset, and doesn't contain the empty set). Your property is equivalent to it being a maximal filter. The usual construction using Zorn's lemma should show that such an element always exists.
Thanks for your comment. I fixed the question.
Your comment is useful for me.
Can you explain a little more about filters?
|
2025-03-21T14:48:29.702509
| 2020-01-21T14:08:51 |
350866
|
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|
Stack Exchange
|
Computations using "Stover's spectral sequence"
In this article from 1990, Stover describes a specral sequence which converges to the higher homotopy groups of the homotopy colimit of a diagram $\underline{X}$ of topological spaces.
The second page terms $E^2_{p,*}$ are described as values of the $p^{th}$ derived functor of the colimit functor $\lim_p (\pi_*(\underline{X}))$ (where I use $\lim$ to mean the colimit functor) on the associated diagram of homotopy algebras $\pi_*(\underline{X})$. Here, for $p=0$ we take the normal colimit functor $\lim (\pi_*(\underline{X}))$.
This derived functor is defined as $\pi_p({\lim{F_*}})$ - that is, the $p^{th}$ homotopy group of the simplicial $\Pi$-algebra $\lim{F_*}$, where $F_*$ is a free simplicial resolution of the diagram $\pi_*(\underline{X})$.
In order to take the above $p^{th}$ homotopy group $\pi_p(\lim F_*)$, we need the face and degeneracy maps in the simplicial object $\lim F_*$. How are the face and degeneracy maps for $\lim F_*$ defined here?
More generally, are there examples in the literature of the usage of this spectral sequence in making concrete computations - or alternatively in aiding computations?
EDIT: Having offered a bounty on this question which has since expired, I would like to emphasise that a satisfactory answer would be one which answers my question about the face and degeneracy maps. An answer to the second, more general question, would be a bonus.
A simplicial resolution of a diagram is a simplicial object in the category of diagrams. In particular, if we take colimits dimension-wise, we obtain a simplicial object.
I am not aware of any computations using Stover's spectral sequence.
Something related which has been developed in the meantime is the calculus of homotopy functors. This starts with the Blakers-Massey theorem, which can be stated as a theorem about homotopy groups of a colimit, in a range. My instinct is that Stover's spectral sequence on its own may be hard to wield, but enhancing it with the Goodwillie calculus may yield calculations.
|
2025-03-21T14:48:29.702925
| 2020-01-21T14:10:41 |
350867
|
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|
Stack Exchange
|
How can you order a free group?
A left order on a (discrete) group $G$ is a total order on $G$ satisfying $\forall g,h,k \in G: g < h \implies kg < kh$. A right order is defined symmetrically, and a biorder is an order that is both at once. An order is scattered if it has no densely ordered subset of cardinality at least two, where a set $S$ is densely ordered if $a, c \in S \wedge a \neq c \implies \exists b \in S: a < b < c$.
I'm interested in order types of left orders (and biorders, why not). It's easy to show that left orders on the integers $\mathbb{Z}$ all have order type $\mathbb{Z}$, and on $\mathbb{Z}^2$ there are two constructions of left orders, and as far as I can tell one gives order type $\mathbb{Z}^2$ in lex order (so scattered), and the other orders are dense (so not scattered).
Let $F_2$ be the free nonabelian group on two generators.
Does $F_2$ admit a scattered left order?
Does $F_2$ admit a scattered biorder?
I had an awesome application for this, but I broke it already. Now I'm just curious. I don't actually know what you get from the Magnus embedding $a \mapsto 1+a$, I admit I was too lazy (or scared?) to even give it real thought, and I did not notice a statement in the literature.
More generally, one may ask:
What are the order-types of left orders on $F_2$?
What are the order-types of biorders on $F_2$?
More generally, I'm interested in information on order types of orders on any torsion-free groups, there's plenty of literature on orders but I haven't seen much about order types.
Weaker than your first question, but $F_2$ acts faithfully on a scattered left order. Indeed, for each quotient by a term of the lower central series one get such a (non-faithful) action, and then one can concatenate.
Does everything? Is there a simple torsion-free non-orderable group?
Erm wait, dyadic rationals.
Don't understand your last 2 questions. You're using "simple" in the meaning "non-complicated"? Anyway every torsion-free abelian group is orderable. Nevertheless I agree that for $p\ge 2$ $\mathbf{Z}[1/p]$ (and hence $BS(1,p)$) cannot act faithfully on any scattered totally ordered space.
I mean simple in the usual sense. With the first question I meant, does everything act on a scattered left order? If yes, then your observation is not very interesting. With the second question, I figured if you have a simple torsion-free non-orderable group, then that will be a counterexample. But then I realized $\mathbb{Z}[1/2]$ is an easier example. I agree you can also f.g.ify it.
Let us first clarify the relationship between scattered, discrete, and dense orders. The last two notions are standard in the theory of ordered groups.
An order (left or two-sided) is discrete if there exists a smallest (necessarily unique) positive element. If such an element does not exist, the order is called dense.
We claim that no dense order (left or two-sided) is scattered. Moreover, we claim that, if the order is dense, then the entire group is a densely ordered set. Indeed, if $a < b$ then $e < a^{-1}b$ and, since the order is dense, there exists $c$ such that $e < c < a^{-1}b$ and we obtain $a < ac < b$.
On the other hand, discrete orders are not necessarily scattered. For instance the lexicographic bi-order on $\mathbb{Z} \times F_2$, with any bi-order on $F_2$, is discrete with smallest element $(1,e)$, but there are densely ordered subsets, namely the copy of $F_2$.
Going back to the posed question, since free groups of rank > 1 do not admit discrete bi-orders (in fact, no centerless group admits a discrete bi-order) they do not admit scattered bi-orders.
On the other hand, there are discrete left-orders on free groups of rank > 1, and, therefore, the question is more interesting there.
The claim that centerless groups do not admit discrete bi-orders follows from Theorem 2.1 in the paper cited below, while the claim that free groups admit discrete orders (that are, moreover Conradian) is Corollary 3.6 in the same paper.
Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O., Discretely ordered groups, Algebra Number Theory 3, No. 7, 797-807 (2009). ZBL1229.06008.
Very nice! Can you give a reference for the last two paragraphs?
I believe you also give a complete answer to the fourth question, as there is a unique dense countable total order without endpoints.
Added a reference.
By the way, as I recall, the Magnus embedding order is one of those that can be obtained from the lower central series.
|
2025-03-21T14:48:29.703251
| 2020-01-21T14:27:04 |
350869
|
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Stack Exchange
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The symmetric monoidal closed structure on the category of $\mathcal{F}$-cocomplete categories and $\mathcal{F}$-cocontinuous functors
In 6.5 of the book by Kelly,
Basic concepts of enriched category theory. Reprints in Theory and Applications of Categories, No. 10, 2005.
the author claims that the $2$-category $\mathsf{Cat}_{\mathcal{F}}$ of $\mathcal{F}$-cocomplete categories and $\mathcal{F}$-cocontinuous functors is in some sense monoidal closed. Indeed $\mathsf{Cat}_{\mathcal{F}}$ has a very natural notion of internal hom, because $\mathsf{Cat}_{\mathcal{F}}(A,B)$ is $\mathcal{F}$-cocomplete. Kelly's result shows that one can define a tensor $\otimes: \mathsf{Cat}_{\mathcal{F}} \times \mathsf{Cat}_{\mathcal{F}} \to \mathsf{Cat}_{\mathcal{F}}$ having the usual property of a monoidal closed structure (up to replacing isomorphisms with euquivalences of categories.
Q1: Unfortunately, I do not understand where $A \otimes B$ is defined, to my understanding he starts mentioning it, but I do not get where the definition is given. Could someone help me to understand it?
After a while, I think I got the definition (even if I do not find it in the book, and I wish someone can tell me what the author is doing there), which is quite involved and comes from a $2$-dimensional adaptation of a classical result in $1$-dimensional category theory. The result is mostly due to Kock and Seal, but I shall mention the Chap. 6 of PhD thesis of Martin Brandenburg, Tensor categorical foundations of algebraic geometry, because he gathered the existing literature in a coherent way.
Thm. (Seal, 6.5.1 in Ref.) Let $T$ be a coherent (symmetric) monoidal monad on a (symmetric) monoidal category C. Then $\mathsf{Mod}(T)$ becomes a (symmetric) monoidal category.
Seal does not show that it is monoidal closed, because he does not assume closedness of the base, yet he proves monoidality of the Eilenberg-More category of algebras. Hopefully, if the base is closed, so is the EM-category.
Q2: Unfortunately, to my understanding, the literature contains a plethora of notions of nice monads: coherent, commutative, monoidal, strong, Hopf... can someone guide me among these options? Which kind of monads induces a monoidal closed structure on the category of algebras?
Q3: I wanted to use a Kock-like result in the case in which $T$ is the free completion under $\mathcal{F}$-colimits over $\mathsf{Cat}$ in order to show that $\mathsf{Cat}_{\mathcal{F}} \cong T\text{-}\mathsf{Alg}$ is monoidal closed. Indeed to me, this is what Kelly is secretly doing. Does anyone know a reference that uses this kind of argument? Indeed here many $2$-dimensional and size-related subtleties should be taken into account.
Could you give a reference for Kock's results ? he has several paper on strong & commutative monads and I can't find the one where this theorem is...
@SimonHenry, thanks. I edited.
I think the reference you gave only show that the category of algebra is a "closed category", it does not show it is monoidal. I will be tempted to think that to get a monoidal category of algebras you need some presentabitliy/accessibility assumption, or at least cocompletness. In the case of interest to you this result only gave you that functor categories are $\mathcal{F}$-cocomplete, which you already know.
@SimonHenry, I hope I finally managed to give a good reference. Seal does not show that it is monoidal closed, because he does not assume closedness of the base, yet he proves monoidality of the Eilenberg-More category of algebras. No assumption on presentability is needed. I believe that if the base is closed, so is the EM-category.
@IvanDiLiberti have you seen Bourke's paper on Skew structures in 2-category theory and homotopy theory? Section 6 discusses a 2-dimensional version of Kock's result due to Hyland and Power, which Bourke argues is nicely analyzed via a skew closed structure on the 2-category $T\text{-}\mathrm{Alg}_s$ of algebras and strict morphisms.
Thank you very much! I remember John mentioned me the paper of Hyland and Power, but he did not mention to me his own paper! I think this might really be an answer to my question!
Looking a bit closer to the paper, the fact that everything is strict might be a problem in this case. Maybe this is fixable, I will think about it.
One doesn't need monads to understand Kelly's tensor product. Q3 is probably answered in https://core.ac.uk/download/pdf/82548027.pdf
Let $\mathcal{V}$ be a complete and cocomplete closed symmetric monoidal category. Let $\mathcal{F}$ be a small set of indexing types aka weights (these are just $\mathcal{V}$-functors from a $\mathcal{V}$-category to $\mathcal{V}$) and $\mathsf{cat}_{\mathcal{F}}$ denote the $2$-category of essentially small $\mathcal{V}$-categories which are $\mathcal{F}$-cocomplete; the following construction does not work for large categories. I denote the Hom-categories of $\mathcal{F}$-cocontinuous functors by $\mathrm{Hom}_{\mathcal{F}}$; Kelly writes $\mathcal{F}\mathrm{-Cocts}$.
If $\mathcal{C},\mathcal{D},\mathcal{E} \in \mathsf{cat}_{\mathcal{F}}$, let us call a $\mathcal{V}$-functor $\mathcal{C} \otimes \mathcal{D} \to \mathcal{E}$ bi-$\mathcal{F}$-cocontinuous if it preserves $\mathcal{F}$-colimits in each variable (see Section 1.4 in Kelly's book for the definition of $\otimes$ here). The tensor product $\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D}$ is defined by the universal property
$$\mathrm{Hom}_{\mathcal{F}}(\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D},\mathcal{E}) \simeq \{\text{bi-}\mathcal{F}\text{-cocontinuous functors } \mathcal{C} \otimes \mathcal{D} \to \mathcal{E}\}.$$
Of course, these equivalences should be natural in $\mathcal{E}$. We can also write
$$\mathrm{Hom}_{\mathcal{F}}(\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D},\mathcal{E}) \simeq \mathrm{Hom}_{\mathcal{F}}(\mathcal{C},\mathrm{Hom}_{\mathcal{F}}(\mathcal{D},\mathcal{E})).$$
The idea of the construction of this tensor product of categories is very similar to the construction of the tensor product of modules (take a free module on the product and then introduce the bilinear relations).
Here, we start with the free cocompletion $\widehat{\mathcal{C} \otimes \mathcal{D}}$ of $\mathcal{C} \otimes \mathcal{D}$, which is the $\mathcal{V}$-category of $ \mathcal{V}$-functors $(\mathcal{C} \otimes \mathcal{D})^{op} \to \mathcal{V}$. We have the Yoneda embedding $Y : \mathcal{C} \otimes \mathcal{D} \to \widehat{\mathcal{C} \otimes \mathcal{D}}$.
Let $\Phi$ denote the set of cylinders in $\mathcal{C} \otimes \mathcal{D}$ which are either of the form
$$F \xrightarrow{\lambda} \mathrm{Hom}_\mathcal{C}(G(-),A) \xrightarrow{\text{ can }} \mathrm{Hom}_{\mathcal{C} \otimes \mathcal{D}}(G(-) \otimes B, A \otimes B)$$
for some indexing type $F : \mathcal{J}^{op} \to \mathcal{V}$ in $\mathcal{F}$, some $\mathcal{V}$-functor $G : \mathcal{J} \to \mathcal{C}$, some colimit cylinder $\lambda : F \to \mathrm{Hom}_\mathcal{C}(G(-),A)$ and some $B \in \mathcal{D}$, or of the form
$$F \xrightarrow{\mu} \mathrm{Hom}_\mathcal{D}(H(-),B) \xrightarrow{\text{ can }} \mathrm{Hom}_{\mathcal{C} \otimes \mathcal{D}}(A \otimes H(-), A \otimes B)$$
for some indexing type $F : \mathcal{J}^{op} \to \mathcal{V}$ in $\mathcal{F}$, some $\mathcal{V}$-functor $H : \mathcal{J} \to \mathcal{D}$, some colimit cylinder $\mu: F \to \mathrm{Hom}_\mathcal{D}(H(-),B)$ and some $A \in \mathcal{C}$.
Now consider $\mathrm{Alg}(\Phi)$, the full subcategory of $\widehat{\mathcal{C} \otimes \mathcal{D}}$ which consists of those $\mathcal{V}$-functors $P : (\mathcal{C} \otimes \mathcal{D})^{op} \to \mathcal{V}$ which map all cylinders in $\Phi$ to limit cylinders in $\mathcal{V}$. It is a non-trivial result that this is a reflective subcategory (Theorem 6.5 in Kelly's book), in particular cocomplete. This also needs the extra assumption that $\mathcal{V}$ is locally bounded. Let $R : \widehat{\mathcal{C} \otimes \mathcal{D}} \to \mathrm{Alg}(\Phi)$ denote the reflector.
Define $\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D}$ as the smallest full subcategory of $\mathrm{Alg}(\Phi)$ which is closed under $\mathcal{F}$-colimits and contains the image of
$$\mathcal{C} \otimes \mathcal{D} \xrightarrow{Y} \widehat{\mathcal{C} \otimes \mathcal{D}} \xrightarrow{R} \mathrm{Alg}(\Phi).$$
It is clear that $\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D}$ is an object of $\mathsf{cat}_{\mathcal{F}}$. It has the required universal property: Theorem 6.23 in Kelly's book says that $\mathrm{Hom}_{\mathcal{F}}(\mathcal{C} \boxtimes_{\mathcal{F}} \mathcal{D},\mathcal{E})$ is equivalent to the category of $\Phi$-comodels in $\mathcal{E}$, which by definition are $\mathcal{V}$-functors $\mathcal{C} \otimes \mathcal{D} \to \mathcal{E}$ which map the cylinders in $\Phi$ to colimit cylinders. According to the definition of $\Phi$, these are exactly the bi-$\mathcal{F}$-cocontinuous functors $\mathcal{C} \otimes \mathcal{D} \to \mathcal{E}$.
Just a doubt. Should $\mathcal C \boxtimes_{\mathcal F}\mathcal D$ be a subcategory of $\text{Alg}(\Phi)$, rather than being a subcategory of $\widehat {\mathcal C \otimes \mathcal D}$?
@GiorgioMossa Yes, thanks! It was a typo.
I am not aware of Kock's works.
Nevertheless Kelly provides the definition of its tensor product in the next page: it defines its tensor product $\mathcal A \otimes_{\mathcal F} \mathcal B$ as the $\mathcal F$-theory generated by the sketch $((\mathcal A \otimes \mathcal B)^{op},\Phi)$ where $\Phi$ is made of the $\mathcal A \otimes \mathcal B$ cylinder of the form $\lambda \otimes B$ and $A \otimes \mu$ defined in the previous paragraph ($\lambda$ and $\mu$ range over the family of colimit cylinders of $\mathcal A$ and $\mathcal B$ respectively).
Edit: I see your problem was not with the tensor in $\mathcal F-\mathbf{Cat}$ the tensor $\mathcal A \otimes \mathcal B$. This is the tensor product of $\mathcal A$ and $\mathcal B$ as $\mathcal V$-categories, the definition can be found in section 1.4 page 12.
I think it is important to stress the fact $\mathcal A \otimes \mathcal B$ is not a $\mathcal F$-complete category.
I hope this helps.
This is more or less a copy-paste of Kelly's words, but I frankly do not get them. $A \otimes B$ should be a $V$-category with $\mathcal{F}$-colimits, not a theory, nor a sketch. I am probably missing something incredibly simple, but I do not understand Kelly's words.
I've made an edit. It should address your doubts.
I think I understood, but I will not accept this answer, hopefully will come with a more explanatory answer. This answers Q1, but does not make Kelly more readable at all. That was the main point of my question.
I think it is not fair to wait for an answer which answers all three questions. I suggest to open a new thread for the whole monad thing (which is, in fact, not necessary for Kelly's tensor product).
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2025-03-21T14:48:29.704021
| 2020-01-21T15:21:53 |
350872
|
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|
Stack Exchange
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Is there a ring for which the reducibility of a polynomial is undecidable?
Let $R$ be a ring such that all of its elements have a finite number of divisors, ie $\forall r\in R\, |\{x\in R: x|r\}|<\infty$.
Then we can decide whether a polynomial in $R[t]$ is reducible through Kronecker's method.
Even in the ring $\mathbb{Q}[x,\sqrt{x},\sqrt[4]{x},...]$, where $x$ has an infinite number of divisors, it is easy to list any polynomial's possible factors, and reducibility is decidable.
Is there a ring for which the reducibility of a polynomial is undecidable?
Is there a countable ring with computable ring operations and decidable equality for which the reducibility of a polynomial is undecidable?
Yes that is a good suggestion @Matt F.. Btw, where could I find an examplu of ring with uncomputable operations?
Yes, but the answer is a bit unsatisfying. This answer is a summary of the very nice paper Computable Fields and Galois Theory, Russel Miller, Notices of the AMS, 2008.
First of all, if one could not even compute with the elements of the ring $R$ at all, it would be unclear what it would mean for factorization to be computable. The usual solution is to talk about "computable rings", meaning a countable (or finite) ring $R$ where the elements are indexed by integers and the operations of addition and multiplication are given by computable functions.
The following is an example of a computable field in which factorization is not decidable: Let $p_n$ be the $n$-th prime and let $T_n$ be the $n$-th Turing machine and let $K = \mathbb{Q}(\sqrt{p_n} \ : \ T_n \ \mbox{halts})$. Given any element $\theta$ of $K$, there is a finite expression which witnesses that $\theta$ is in $K$: namely, the algebraic expression for $\theta$ in terms of finitely many $p_n$'s and a transcript of the ruunning of the corresponding Turing machines. This can be made into a proof that $K$ is computable, in particular (in response to Matt F.'s comment) there is no issue with testing equality. However, $x^2 - p_n$ is reducible iff $T_n$ halts, so we cannot test reducibility.
The fields one normally meets in life do not have this issue. Reducibility is computable over $\mathbb{Q}$, $\mathbb{R} \cap \mathbb{Q}^{\mathrm{alg}}$ and over finite fields. If reducibility is computable over $k$, then it is computable over $k(t)$ and over $k[t]/f(t)$ for $f$ irreducible. That already covers the most obvious fields you know.
Searching for "computable fields" turns up a fair bit of recent research.
+1. Let me add to this that there's a surprising (to me at least) computable algebra fact here: while this undecidability result prevents the usual construction of the algebraic closure from being computable in the appropriate sense it is still the case that every computable field has a computable algebraic closure (Frolich-Shepherdson if I recall correctly). The subtlety here is around the phrase "algebraic closure:" for every computable field $F$ there is a computable field $G$ such that $G$ is isomorphic to the algebraic closure of $F$, but no embedding $F\rightarrow G$ need be computable!
@MarkSapir It is computable, but factorization is not decidable. I think that is what I wrote?
This is going to be a mess to write out but: Each element of $K$ is writable as $\sum_{J \subseteq I} c(J) \prod_{i \in I} \sqrt{p_i}$ where $I$ is some finite index set and $c$ is a function from subsets of $I$ to $\mathbb{Q}$. It is easy to add, multiply and test equality of expressions in this form. (continued)
For each $N$, take the first $N$ Turing machines and find all those which halt in $\leq N$ steps; call this list $\mathcal{T}(N)$. Note that $\mathcal{T}(N)$ is finite. Consider all functions $c : 2^{\mathcal{T}(N)} \to \mathbb{Q}$ where the numerator and denominator of all numbers in the image is between $-N$ and $N$. This is still a finite list; call it $\mathcal{K}(N)$. (continued)
Every element of $K$ is described by an expression in some $\mathcal{K}(N)$. Breaking ties within $\mathcal{K}(N)$ in some manner, we indexed $\bigcup \mathcal{K}(N)$ by positive integers so that $+$, $\times$ and $=$ are computable (continued)
The definition of computable field in Miller's article requires that my elements elements be indexed by $\mathbb{Z}_{>0}$, not by equivalence classes for a computable equality relation, but that is easy to deal with: Modify the definition of our set to only include those expressions in $\bigcup \mathcal{K}(N)$ which are not equal to an expression which we listed sooner.
Pinging @MarkSapir , since I forgot to do it when I started writing this.
Miller sketches this argument in slightly less detail than I do (page 8).
Are you sure you want to claim computability over $\mathbb{R}$? From what I understand, $\mathbb{R}$ is not constructively a field; I doubt factoring even univariate degree-$2$ polynomials is decidable over it.
@darijgrinberg You are right, I should speak of the field of real algebraic numbers. Fixed now.
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2025-03-21T14:48:29.704383
| 2020-01-21T16:35:01 |
350878
|
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|
Stack Exchange
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which operators are "really truly positive"?
Let's say that an operator G on a Hilbert space $\mathcal{H}$ is "really truly positive" iff
$\Vert\exp(-tG) \exp(-tG^*)\Vert_{op}<1$ for all $t>0$
How can we characterize the set of operators G which are really truly positive? I conjecture that $G$ is really truly positive iff the eigenvalues of $(G+G^*)$ are all nonnegative. I have yet to find an exception to this rule. This is obvious if $G$ is normal because in that case $\exp(-tG) \exp(-tG^*) = \exp(-t(G+G^*))$ -- but if $G$ doesn't commute with its adjoint this formula does not hold and so it is harder to try to prove anything.
[[A word on notation used here... Let $\exp(M)$ denote the operator exponential of an operator $M$. Let $\Vert M\Vert_{op}$ denote the operator norm of $M$. Let $G^*$ denote the adjoint of an operator $G$.]]
Welcome to MathOverflow! I suggest to explicitly mention in the question that you consider an operator on a Hilbert space.
What is the BCH condition?
The BCH formula (https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula) gives an explicit formula for how to compute exp(X)exp(Y).
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2025-03-21T14:48:29.704499
| 2020-01-21T16:46:31 |
350880
|
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How bad a proper forcing of size $\aleph_1$ can be?
This question concerns proper forcings of size $\aleph_1$. In the context of
$\rm ZFC+\neg CH$, I couldn't find any counter example to the following property. Suppose $\mathbb P$ is a proper forcing of size $\aleph_1$ in $H_\theta$, and suppose $p$ is $(M,\mathbb P)$-generic, where $M\prec H_\theta$ is countable and contains $\mathbb P$. Then if $M^*\prec H_\theta$ is an end extension of $M$ in the sense that $M\cap\omega_1=M^*\cap\omega_1$, $p$ is then $(M^*,\mathbb P)$-generic.
I may notice that I am not looking for a artificial construction of a single condition, I want every generic condition $p\in\mathbb P$ have(morally) the above property.
Let me give you the motivation, and then I state my question in a formal way.
Definition(Aspero-Mota): A forcing notion $\mathbb P$ is finitely proper if whenever $\mathcal M$ is a finite set of countable models and $p$ is a condition in $\bigcap \mathcal M$, then there is a condition $q\leq p$ which is $(M,\mathbb P)$-generic for every $M\in\mathcal M$.
The following is a particular case of their theorem..
Theorem The forcing axiom for meeting $\aleph_2$-dense sets by finitely proper forcings of size $\aleph_1$ is consistent with continuum large.
But they give an example of a proper forcing of size $\aleph_1$ which is not finitely proper. I have a relaxed version of the above property(they call it $\aleph_{1.5}-cc$), let us call it $\aleph_{\sqrt{\pi}}-cc$. I think this notion is iterable using virtual models and we get a forcing axiom with $2^{\aleph_0}=\aleph_3$. This is slightly stronger than Aspero-Mota's axiom(in the sense of the above particular case), but in fact this is not important. I wonder if we can prove the proper forcing axiom for posets of size $\aleph_1$ does not imply $2^{\aleph_0}=\aleph_2$.
What is the issue? The issue is that in our construction using two types models we don't have $\in$-chains, so we can't extend our conditions inductively to be generic for all models in the side condition. But if every proper poset of size $\aleph_1$ is $\aleph_{\sqrt{\pi}}-cc$, then it seems everything is fine and we can show that proper forcing axiom for poset of size $\aleph_1$(meeting $\aleph_2$ many dense sets) is consistent with $2^{\aleph_0}=\aleph_3$.
I think for a proper poset $\mathbb P$ of size $\aleph_1$, the $M$-genericity of a condition $p$ should only depend on $M\cap\omega_1$ when $\rm CH$ fails. The following is a formal attempt to find a structure theory for posets of size $\aleph_1$.
Question Assume $\rm CH$ fails. Let $\langle p_\alpha:\alpha<\omega_1\rangle$ be an enumeration of $\mathbb P$. Can one find a function $\rho:\omega_1\rightarrow\omega_1$ such that
if $M\prec H_\theta$ is countable, then for every $p\in M$, there is some $\beta\leq \rho(M\cap\omega_1)$ such that $p_\beta\leq p$ is $(M,\mathbb P)$-generic?
Remarks:
If the answer is yes, then it means we don't have to go unboundedly may times for finding generics for various models with the same $\omega_1$, and by a suitable decoration of side condition, we get $\rm PFA_{\omega_2}[\aleph_1]$ is consistent with $2^{\aleph_0}=\aleph_3$
Maybe the answer has to do with Chang's conjecture!
We may assume $\mathbb P$ has the $\omega_1$-approximation property.
We are also interested in the following case. If $\mathbb P$ is a counterexample to the above question we may think about a forcing notion of size $\aleph_1$ which kills the properness of $\mathbb P$, though itself enjoys the above property.
finitely proper and $\aleph_{1.5}-cc$ are not necessarily the same, but for the forcing of size $\aleph_1$ they coincide.
I've told David about the question, he seemed interested. I'll let you know if he says something about it today.
Good, Thanks @AsafKaragila!
Any updates on this one? I’ve gotten really interested in these sorts of questions over the past few months.
What can you say about the role of the failure of CH in the context of your first paragraph?
@ToddEisworth, No updates so far! I have to get back to this question soon. I also like it , it's a kind of structural question. Our iteration is with finite conditions, thus CH fails, on the other hand if CH holds then a poset on the underling set $\omega_1$ whose conditions carry countable information can have size $\omega_1$, like adding a Cohen subset of $\omega_1$, that I think it does not the property mentioned above. Thus $\neg CH$ rules out such examples, in fact if CH fails we have to work with a poset of size $\aleph_1$ whose conditions are finite, like all examples we know
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2025-03-21T14:48:29.704832
| 2020-01-21T16:55:28 |
350881
|
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|
Stack Exchange
|
Associativity may fail by little?
It is a well-known result on group theory that if a group has many pairs of commuting elements then it is abelian.
This motivated the following pseudo-conjecture.
If a (possibly infinite) set $S$ with a binary operation $\cdot$ is
such that for many triples $a,b,c\in S$ it holds $(a\cdot b)\cdot c
= a \cdot (b\cdot c)$, then $(S,\cdot)$ is a semigroup.
Exploring a little bit on this, a colleague told me he has read somewhere (he can't remember where) that the statement above is false.
If $S$ is a finite set, then there exists a binary operation $\star$ that satisfies $(a\star b)\star c = a\star(b\star c)$ for all but just one ordered triple $(a,b,c)\in S\times S\times S$.
Such a result implies that an algorithm that checks if a certain operation is associative must check indeed all triples of elements, which is kind of funny and rather unintuitive (to me, at least).
We couldn't find a proof of this last result, although playing with small sets it seems to be true.
My questions are the following:
1) Does anybody here know a reference for this result?
2) Is there a constructive proof for such an example?
Related: https://mathoverflow.net/questions/311209/whats-the-maximum-probability-of-associativity-for-triples-in-a-nonassociative , and the answer https://mathoverflow.net/a/311213 .
Suppose you show it true for small almost semi group S. Then let S' have S as a proper substructure, let z in S' and z not in S, and declare x*y be z for x and y with at least one outside of S. Then S' fails associativity at exactly the same spot as does S. So you just need one example on three elements for S, or even on two elements if such exists. Gerhard "Small Failures Are Easily Propagated" Paseman, 2020.01.21.
https://math.stackexchange.com/a/2860039
Concerning randomized algorithms to test associativity, the basic idea of the Rajagopalan–Schulman algorithm is outlined in https://rjlipton.wordpress.com/2010/06/03/an-amplification-trick-and-stoc-2010/
The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28.
The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$.
Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $w$ are distinct members of $S$, define $a\cdot a = u$, $a\cdot u = v$ and $x\cdot y = w$, for any $(x,y)$ other than $(a,a)$ and $(a,u)$. Then a case by case verification shows that $(a\cdot a)\cdot a = w\neq v = a\cdot(a\cdot a)$ but that, for every other triple $(x,y,z)\in S^3$, we have $(x\cdot y)\cdot z = x\cdot(y\cdot z)$.
Direct enumeration shows that there are (up to isomorphism) $10$ magmas of order $3$ with exactly one non-associative triple. (There are $124$ of order $4$.)
FWIW, the construction in https://math.stackexchange.com/a/2860039 works directly for sets of size $\ge3$ (and it is simpler).
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2025-03-21T14:48:29.705112
| 2020-01-21T17:00:03 |
350882
|
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"Agno",
"Eigentime",
"Fedor Petrov",
"LSpice",
"Michael Anderson",
"Pietro Majer",
"T. Amdeberhan",
"Wolfgang",
"https://mathoverflow.net/users/106114",
"https://mathoverflow.net/users/12489",
"https://mathoverflow.net/users/151396",
"https://mathoverflow.net/users/20722",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/29783",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/66131"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:625797",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350882"
}
|
Stack Exchange
|
A special type of generating function for Fibonacci
Notation. Let $[x^n]G(x)$ be the coefficient of $x^n$ in the Taylor series of $G(x)$.
Consider the sequence of central binomial coefficients $\binom{2n}n$. Then there two ways to recover them:
$$\binom{2n}n=[x^n]\left(\frac1{\sqrt{1-4x}}\right) \tag{1-1}$$
and
$$\binom{2n}n=[x^n]\left((1+x)^2\right)^n. \tag{1-2}$$
This time, take the Fibonacci numbers $F_n$ then, similar to (1-1), we have
$$F_n=[x^n]\left(\frac1{1-x-x^2}\right). \tag{2-1}$$
I would like to ask:
QUESTION. Does there exist a function $F(x)$, similar to (1-2), such that
$$F_n=[x^n]\left(F(x)\right)^n? \tag{2-2}$$
Of course I agree with the stated equalities (1-1)–(2-1), but does "In view of this" really mean (as it sounds like to me) "as a consequence", or does it just mean "Analogously"? (Maybe "In view of this" was meant to go before "I would like to ask"?) Also, in what sense is (1-1) similar to (2-1), other than that both are extracting coefficients of (fixed) ogf's?
Does not Lagrange inversion give an equation for $F$?
@FedorPetrov: does it give something "explicit"?
Sure: choose any nonzero value for $a_0$, and write $F(x)=a_0+a_1x+a_2x^2+...$.
Expanding $F(x)^n$ gives you a LINEAR equation in $a_n$ as a function of the preceding ones, the coefficient of $a_n$ being $na_0^{n-1}$. In the special case of Fibonacci numbers, I do not know if $F(x)$ is "explicit", but formally it exists.
Added: choosing $a_0=1$ gives you
$$F(x)=1+x+x^2/2-x^3/3+x^4/8+x^5/15-25x^6/144+11x^7/70-209/5760x^8-319/2835
x^9 +...$$
and any other nonzero $a_0$ gives $a_0F(x/a_0)$.
Second addition: since some people seem interested in this expansion, two remarks. Call $a_n$ the coefficient of $x^n$. First, it must be easy to show
that the denominator of $a_n$ divides $n!$ (and even $(n-1)!$). Second, and much more interesting, is that the numerator of $a_n$ seems to be always smooth, more precisely its largest prime factor never exceeds something like $n^2$. This is much more surprising and may indeed indicate some kind of explicit expression.
Third addition: thanks to the answers of Fedor, Richard, and Ira, it is immediate to see that $F(x)$ is a solution of the differential equation
$y'-1=x/y$, giving the recurrence formula for the coefficients $c_n$ of $F$:
$$\sum_{0\le n\le N}(n+1)c_{n+1}c_{N-n}-c_N=\delta_{N,1}$$
Does this help ?
Thank you. I wish (although I did not state it) we could find something more "explicit".
Have you noticed that for $n=5k+1$, the prime factors of the numerators are overall even much smaller than for the other $n$'s (with the biggest one about 4 times smaller than "usual")? For $n=6,11,\dots,46,51$ the biggest prime factor is as small as $5,5,31,61,101,151,211,281,241,359$. Moreover for $n=5k+1$ those numerators seem to contain high powers of $5$, at least $5^{2k-1}$. In general, only certain primes occur, starting with $5,11,19,29,31,41,59,61,71,79,89,101,109,131,139,149,151,179,181,191,199$. Really intriguing - do those have some (class number or whatever) property in common?
And here are some patterns, inspired by the sequence $31,61,101,151,211,281$ in the above comment: the numerator for $n=5k+1$ is divisible by $5k(k-1)+1$. It looks like there are many other quadratic sequences among the divisors: If $n \ne 0\pmod {11}$, it seems like the numerator is divisible by $n^2-7n+11=(n-3)(n-4)-1$, while for $n = 0\pmod {11}$, just a factor $11$ is missing. For even $n=2k$, it is divisible by $ (n-4)(n-8)-1$ and most often (!) also by $k(k-1)-1$. And so on. Maybe, all prime factors can be captured by such quadratic sequences, which would explain why they are small!
Wolfram alpha does give a solution to that ODE: https://www.wolframalpha.com/input/?i=D%5By%2Cx%5D+-+1+%3D+x+%2F+y
Lagrange–Bürmann formula claims that $$[w^{n-1}]H'(w)(\varphi(w))^n=n[z^n]H(g(z)),$$
where $g(z)$ and $f(w)=w/\varphi(w)$ are compositionally inverse power series without constant term (that is $g(f(w))=w$, $f(g(z))=z$), $H$ arbitrary power series. So if you choose $H$ equal to the antiderivative of $1/\varphi$, we get $$[w^{n-1}]\varphi^{n-1}=n[z^n]H(g(z))=[z^{n-1}](H(g(z))'=[z^{n-1}]\frac{g'(z)}{\varphi(g(z))}=\\ [z^{n-1}]\frac{g'(z)f(g(z))}{g(z)}=[z^{n-1}]\frac{zg'(z)}{g(z)}.$$
So, if you look for $\varphi$ such that
$$\sum_{n=1}^
\infty ([w^{n-1}](\varphi(w))^{n-1})z^{n-1}=u(z)$$
is a fixed function (satisfying $u(0)=1$), you should at first solve $zg'(z)/g(z)=u(z)$ that reads as $$(\log g)'=u(z)/z,\quad\log g(z)=\log z+\int_0^z \frac{u(t)-1}tdt+{\rm const},\\ g(z)=C z\exp\left(\int_0^z \frac{u(t)-1}tdt\right),$$
after that solve $$g(z)/\varphi(g(z))=z,\quad\text{i.e.}\,\, \varphi(s)=s/g^{-1}(s).$$
The choice of $C$ is your freedom.
You may try Fibonacci numbers, that is, $u(z)=1/(1-z-z^2)$.
We get
$$
g(z)=Cz(1-\alpha z)^{-\alpha/\sqrt{5}}(1-\beta z)^{\beta/\sqrt{5}},\quad
\alpha=\frac{1+\sqrt{5}}2,\,\beta=\frac{1-\sqrt{5}}2.
$$
The inverse map $g^{-1}$ is not explicit of course. You may probably look at it as a Christoffel–Schwarz type map (inverse of antiderivative of a product $\prod (z-z_i)^{c_i}$, where $c_i$ are all equal to $-1$ in our case).
A little elementary algebra then gives a nice equation for F (your $\varphi$ ): $\left( F(x) - \alpha x \right)^{\alpha} = \left( F(x) - \beta x \right)^{\beta}$
@Eigentime hm, where is $C$ in this equation?
I forgot to mention that I put $C=1$ to get the equation. From Henri Cohen's answer we see that we get the other solutions as $a F(x / a)$
Another way to state Fedor's answer is Exercise 5.56(a) of Enumerative Combinatorics, vol. 2. Namely, if $G(x)=a_1x+a_2x^2+\cdots$ is a power series (say over $\mathbb{C}$) with $a_1\neq 0$ and $n>0$, then
$$ n[x^n]\log \frac{G^{\langle -1\rangle}(x)}{x} =
[x^n] \left(\frac{x}{G(x)}\right)^n, $$
where $^{\langle -1\rangle}$ denotes compositional inverse. Letting both sides equal the Fibonacci number $F_n$ (using the indexing $F_1=F_2=1$) gives
$$ F(x) =\frac{x}{\left( x\exp \sum_{n\geq 1}F_n\frac{x^n}{n}
\right)^{\langle -1\rangle}}. $$
One can find a closed expression for $\sum F_n\frac{x^n}{n}$ by integrating $\sum_{n\geq 1} F_nx^{n-1}=1/(1-x-x^2)$, but there is no simple formula for the resulting compositional inverse.
Perhaps interesting to note the following.
In the notation of Henri Cohen,
$$
F(x)=1 + x+\frac{x^2}{\color{red} 2}-\frac{x^3}{\color{red} 3}+\frac{x^4}{\color{red}8}+\frac{x^5}{15}-\frac{25 x^6}{\color{red}{144}}+\frac{11 x^7}{70}-\frac{209
x^8}{\color{red}{5760}}-\frac{319 x^9}{2835}+\frac{8569
x^{10}}{\color{red}{44800}}-\frac{625 x^{11}}{4536}-\frac{1212751
x^{12}}{\color{red}{43545600}}+\frac{2759 x^{13}}{13650}-\frac{155302219
x^{14}}{609638400}+\dots
$$
Compare this to the sequence
$$
\sum_{k=0}^n \frac{(-1)^k}{k!}=1,0,\frac{1}{\color{red}2},\frac{1}{\color{red}3},\frac{3}{\color{red}8},\frac{11}{30},\frac{53}{\color{red}{144}},\frac{103}{280},\frac{2119}{\color{red}{5760}},\frac{16687}{45360},\frac{16481}{\color{red}{44800}},\frac{1468457}{3991680},\frac{16019531}{\color{red}{43545600}},\frac{63633137}{172972800},\frac{2467007773}{6706022400},\dots
$$
Just a coincidence? Perhaps there is a closed-form after all...
FWIW: the coefficient of $x^{20}$ also has matching denominator with the sequence.
and in any case the denominators in the latter sequence are multiples of the denominators of the coefficients
If fitted to the denominators in the latter sequence, the numerators of $F$ become $1,1,1,-1,1,2,-25,44,-209,-5104,8569,-550000,-1212751$... maybe "smooth" in terms of prime factors, maybe growing absolute values, but with an intriguing irregularity.
@Wolfgang some of these are also in A258943.
@AccidentalFourierTransform Wow, and again those not in the "exponential reversion" (never heard that term before) seem to be divisors of the corresponding entries. Fascinating relationship...
In fact, after doing more computing, it turns out the numbers in my above comment are not always integers. This fails for some primes $p$ (with a factor $p$ too much) for certain arithmetic progressions with distance $p$, e.g. $n=13k+8, 37k+9,37k+26,47k+23,53k+6,...$ with each time $k\ge1$. Meaning that e.g. $c_{21}\cdot A053556[21]$ (where A053556 are the denominators of the "Subfactorial coefficients"), still has a $13$ as denominator. (cont'd)
(cont'd) And it seems to be exactly for the $n$ in those (listable) arithmetic progressions that the corresponding primes remain in the denominator of $c_{n}\cdot{A053556[n]}$, while for all other $n$, this is an integer. A different question would be whether $\frac{A258943[n]}{c_{n}\cdot{A053556[n]}}$ is always an integer...
@Wolfgang It looks like the n-th term in the series expansion of $F(x)$ equals: $$\frac{A258943(n-1)}{(n-1)! + (n-2)!}, x^n$$
@Agno yes exactly! That is another way of stating what I have said in my answer. Note that $(n-1)! + (n-2)!=n (n-2)!=\dfrac{ n!}{n-1}$.
@Wolfgang Aaah, yes. I see it now. Failed to make the final step! Still curious whether there is some sort of a pattern in 258943.
@Agno Yes there is. :)
The power series $F(x)$ is closely related to the series of the "exponential reversion of Fibonacci numbers" $$R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}$$ (the $r_n$ are A258943, quoted in a comment). In fact it appears that, again in the notation of Henri Cohen, $$a_{n+1}=nr_n,$$ equivalently $$F'(x)=xR'(x).$$
So if the Fibonacci numbers are encapsulated by $$x=\sum_{n\ge1}F_n\frac{y^n}{n!}=y+\frac{y^2}{2!}+2\frac{y^3}{3!}+3\frac{y^4}{4!}+5\frac{y^5}{5!}+\cdots,$$ the reverse series of this is
$$
y=R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}=x-\frac{x^2}{2!}+\frac{x^3}{3!}+\color{red}{2}\frac{x^4}{4!}-\color{red}{25}\frac{x^5}{5!}+-\cdots,$$
while $$\begin{align}F(x)=1+\sum_{n\ge1}a_n {x^n} &=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+8\frac{x^5}{5!}-125\frac{x^6}{6!}+-\cdots\\
&=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+4\cdot\color{red}{2}\frac{x^5}{5!}-5\cdot\color{red}{25}\frac{x^6}{6!}+-\cdots\end{align}.$$
Possibly this relationship is not even specific to the Fibonacci numbers.
EDIT: It looks like the sequence $\{a_n\}$ has finally yielded its secrets. Given the conjectured "smoothness" of these coefficients (i.e. all prime factors are relatively small) as mentioned in Henri Cohen's answer, I have looked again into the factors and those quadratic sequences mentioned in the comments, and fortunately there are enough primes in them, such that finally I was able to find the pattern! We have for the sequence $\{r_n\}$ $${ r_n=\begin{cases} {(-1)^k} \prod\limits_{j=1}^k(n^2-5nj+5j^2) \quad\text{for
}\ n=2k-1, \\ \\ {(-1)^kk\cdot} \prod\limits_{j=1}^{k-1}(n^2-5nj+5j^2) \quad\text{for }\ n=2k. \end{cases}}$$ Once found, it should not be hard to prove that rigorously.
As pointed out by Agno in a comment, we can reduce to linear factors and write the product in terms of the Gamma function simply as $$r_n= {\sqrt5^{ \,n-1 }\frac { \Gamma \left( \frac{5-\sqrt {5}}{10}n \right)}{ \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) }}.$$More generally, if we start with a Lucas sequence $$f_0=0,\ f_1=1,\ f_n=pf_{n-2}+qf_{n-1}\quad(n\ge2),$$ the reversed series has $$\boxed{r_n= {\sqrt{4p+q^2}^{ \,n-1 }\frac { \Gamma \left[\dfrac n2 \Bigl(1-\dfrac{q}{\sqrt{4p+q^2}} \Bigr)\right]}{ \Gamma \left[1-\dfrac n2 \Bigl(1+\dfrac{q}{\sqrt{4p+q^2}} \Bigr)\right]}}}.$$ Note that whenever the argument in the denominator is a negative integer, the coefficient $r_n$ vanishes, e.g. this happens when $p=3,q=2$ for all $n\equiv0\pmod4$.
As far as the sequence of the signs, it is after all quite regular and is in fact self-similar (that is, unless $\sqrt{4p+q^2}$ is rational). This self-similar behaviour of the signs can be seen by virtue of the (negative) argument of the Gamma function in the denominator, knowing that $\Gamma$ changes signs at each negative integer and the multiples of $\sqrt{4p+q^2}$ occurring in the argument do the rest. (Think e.g. of the self similarity features of the Wythoff sequence.)
For the reversion of the original Fibonacci sequence, I have displayed here the signs of the first $1500$ even and then the first $1500$ odd coefficients and found that their quasi periodicity comes out nicely when putting exactly $76$ in each row (writing "o" instead of "$-$" for better visibility). The "longest pairings" are colored:
all patterns "$++--++--++$" in yellow and
all patterns "$--++--++--$" in blue.
Very well done, Wolfgang :-) I had also started to spot a few patterns, but had not come even close to this. You should most certainly add this as a formula for the sequence A258943 on Sloane's!
@Agno indeed, I was already thinking about that
Just to share two observations:
You could rewrite $r_n$ for $n=odd$ as a "closed" form:
$${\frac {i{5}^{n/2-1/2}\pi, \left( -1 \right) ^{n/2}}{\sin \left( 1/10
,\pi, \left( \sqrt {5}n+15 \right) \right) \Gamma \left( 1-n/2+1/10
,\sqrt {5}n \right) \Gamma \left( 1-n/2-1/10,\sqrt {5}n \right) }}$$
When $n=$ a prime $>5$, then a unique property of $r_n$ is that all its prime factors are always congruent to {1, 4} mod 5 (A045468).
Note that in my previous comment I used $r_n$ as the $n$-th element of A258943, i.e it excludes the $\frac{1}{n!}$ factor.
@Agno Little typo (must be "5" instead of "15"). We can indeed write it for odd $n$ as $$r_n=\frac 1{n!}{\frac {(-5)^{(n-1)/2}\pi}{\cos\left(\frac{n\sqrt {5}}{10}\pi\right) \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) \Gamma \left( 1-\frac{5-\sqrt {5}}{10}n \right) }}$$ and for even $n$ $$r_n=\frac 1{n!}{\frac {-i\cdot (-5)^{(n-1)/2}\pi}{\sin \left(\frac{n\sqrt {5}}{10}\pi\right) \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) \Gamma \left( 1-\frac{5-\sqrt {5}}{10}n \right) }}$$
Nice! And taking it one step further:
$$A258943(n):= \displaystyle {\frac {2,\pi,(-5)^{(n-1)/2}}{\Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) \Gamma \left( 1-\frac{5-\sqrt {5}}{10}n \right) \left( {\rm e}^{i\pi n\left(\frac{1}{2\sqrt{5}}\right)}-, {\rm e}^{i\pi n\left(1-\frac{1}{2\sqrt{5}}\right)} \right ) }}$$
and its 'functional equation' is:
$$A258943(-n)\cdot A258943(n)=\frac{(-1)^n}{n^2} \qquad n \in \mathbb{N}$$
Or much simpler :-)
$$A258943(n):= \displaystyle \frac{\sqrt{5^n}}{\sqrt{5}}\cdot \frac{\Gamma\left(\frac{n}{2}-\frac{n}{2\sqrt{5}}\right)}{\Gamma\left(1-\frac{n}{2}-\frac{n}{2\sqrt{5}}\right)}$$
Wow...........!
Nice how you managed to expand the formula towards the Lucas sequences! A while ago I played with the "Imaginary Golden Ratio" that has the "Imaginary Fibonacci" sequence: $$f_0=0,\ f_1=1,\ f_n=f_{n-1}-f_{n-2}\quad(n\ge2) $$
(see here: https://mathforums.com/threads/imaginary-golden-ratio.17605/) This sequence could also be 'reversed' using your formula (with $=−1,=1$) and thereby yields an explicit expression for:$$A183611(n)=r_{n,-1,1}\cdot (-1)^{n+1}$$
And just for fun, two (extremely inefficient, I guess related to Wilson's theorem) primality tests using $r_{n,p,q}$
$$f(n)=n\cdot\left(\Big\lfloor\frac{1}{n^3}\frac{r_{n,1,1}}{r_{-n,1,1}}\Big\rceil-\frac{1}{n^3}\frac{r_{n,1,1}}{r_{-n,1,1}}\right)$$
$$g(n)=n\cdot\left(\frac{1}{n^3}\frac{r_{n,-1,1}}{r_{-n,-1,1}}-\Big\lfloor\frac{1}{n^3}\frac{r_{n,-1,1}}{r_{-n,-1,1}}\Big\rceil\right)$$
that both yield $0$ when $=$ composite en $1$ when $=$ prime, except for $=5$
and $=3$ respectively.
Added later. Simpler: $f(n)= (r_{n,1,1})^2 \pmod n$ and $g(n)= (r_{n,-1,1})^2 \pmod n$
I started writing this before Richard's answer appeared, with which it overlaps a lot, but I still have something to add.
Let us look at a more general problem: Suppose that $G(x) = 1+g_1x+g_2x^2+\cdots$ and that
$[x^m]G(x)^m = c_m$ for $m\ge1$. It is clear that the $g_i$ can be expressed uniquely in terms of the $c_i$, and we would like to find an explicit formula.
Let
$$
R(x) = \exp\biggl(-\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr)
$$
and let $f=f(x)$ satisfy $f = xR(f)$, so $f(x) =\bigl(x/R(x)\bigr)^{\langle -1\rangle}$.
By formula (2.2.7) of my survey paper on Lagrange inversion (a corollary of Lagrange inversion),
$[x^m](x/f)^m = c_m$, so by the uniqueness of $G(x)$, we have $G(x) = x/f(x)$. By formula (2.2.4) of my paper (a form of Lagrange inversion) we have for $\alpha\ne -m$,
$$[x^m] (f/x)^{\alpha}=\frac{\alpha}{m+\alpha} [x^m] R(x)^{m+\alpha}$$
so
$$
[x^m] G(x)^{-\alpha}=\frac{\alpha}{m+\alpha} [x^m] \exp\biggl(-(m+\alpha)\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr).
$$
In particular, taking $\alpha=-1$ gives the formula for the coefficients of $G(x)$ in terms of the $c_i$:
we have $g_1=c_1$ and for $m>1$,
$$
g_m=-\frac{1}{m-1} [x^m] \exp\biggl(-(m-1)\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr).
$$
We can use these formulas to find some nice examples of $[x^m]G(x)^m=c_m$.
First take $c_m$ to be the constant $C$. Then $R(x)=\exp(-\sum_{n=1}^\infty C x^n/n) = (1-x)^C$, $f$ satisfies $f=x(1-f)^C$, and
$$
\begin{aligned}G(x)^{-\alpha} &= \sum_{m=0}^\infty (-1)^m\frac{\alpha}{m+\alpha} \binom{C(m+\alpha)}{m} x^m\\
&=1+\sum_{m=1}^\infty (-1)^m \frac{\alpha C}{m}\binom{C(m+\alpha)-1}{m-1}x^m.
\end{aligned}
$$
In particular, if $C=1$ then $f=x/(1+x)$ and $G=1+x$. If $C=-1$ then $f$ is $xc(x)$ where $c(x)$ is the Catalan number generating function, $c(x) =(1-\sqrt{1-4x})/(2x)$, and $G(x) = 1/c(x)$. If $C=2$ then $f=xc(-x)^2$, so $G(x) = c(-x)^{-2}$.
For another example take $c_1=-1$ and $c_m=0$ for $m>1$. Then $R(x) = e^{x}$ so $f(x)$ is the "tree function" satisfying $f = xe^f$, $G = x/f = e^{-f}$ and
$$G(x)^{-\alpha} = \sum_{m=0}^\infty \alpha (m+\alpha)^{m-1}\frac{x^m}{m!}.$$
For any desired sequence $a_0,a_1,a_2,\cdots$ there is a unique function (a formal power series) $F(x)$ with $[x^n]F(x)=a_n$ namely $F(x)=\sum a_ix^i.$
Sometimes $F(x)$ is a polynomial. This happens exactly when the sequence is zero from some point on. Sometimes $F(x)$ is a rational function. This happens exactly when the sequence satisfies a linear homogeneous recurrence relation with constant coefficients. Then the denominator is determined by the recurrence and the numerator by the initial conditions. And $F(x)$ might or might not be expressible in a closed form of some other given type, such as $\frac{P(x)}{\sqrt[k]{Q(x)}}$ for $P,Q$ polynomials.
You desire that there is a function $G(x)$ with $[x^n](G(x)^n)=a_n$ for all $n$. Again there is a unique formal power series $G[x]=1+s_1x+s_2x^2+\cdots$ such that $$[x^n](G(x)^n)={\large \lbrace}\begin{array}{lr}
1 & \text{for } n=0\\
a_n & \text{for } n \geq 1\\
\end{array} $$
So you need to either have $a_0=1$ or restrict the requirement to $n \geq 1.$
The unique formal power series $G(x)$ is relatively easy to find term by term. It might or might not be a polynomial or expressible in a closed form of some other given type.
In the case of the central binomial coefficients, $F(x)$ is as you give in and $G(x)=1+2x+x^2.$
For the Fibonacci sequence the $F(x)$ is a rational function but the $G(x)$ doesn't appear at first glance to be anything nice.
|
2025-03-21T14:48:29.706287
| 2020-01-21T17:06:00 |
350883
|
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|
Stack Exchange
|
What (or how) are the new spaces of derived algebraic geometry?
I am a beginner in derived algebraic geometry and I am trying to develop some visual and geometrical intuition about derived schemes (and stacks), or more precisely about the new geometrical phenomena that they introduce.
It seems to me that (broadly speaking) the new spaces that derived geometry gives rise to are:
(Possibly higher dimensional) Loop spaces. They arise as self-intersections: e.g. see comments of J.Pridham and the answer of DamienC (1) below.
Derived infinitesimal disks/Formal neighbourhoods. Originated by nilpotent extensions. See for example the definition 1.1 in Vezzosi - A note on the cotangent complex in derived algebraic geometry.
QUESTION: What else? (See also an answer of DamienC below (2)). I think that while 1 and 2 are already present in derived schemes other phenomena require derived stacks.
I would like to see more examples that have some geometrical interpretation. There are cases of derived stacks for example in Toen - Higher and derived stacks: A global overview. Such examples include the derived stack of rank $n$ local systems over some topological space (and the derived moduli stack of vector bundles), derived linear stacks, and the derived stack of perfect complexes.
However I am unable to obtain a geometrical meaning for this examples.
EDIT: What is the geometrical interpretation of the higher homology groups in (for example) the derived stack of vector bundles over a projective variety $\mathbb{R} \underline{{Vect}_{n}}(X)$?
According to this paper from Toen-Vezzosi some of the motivation for this derived stack comes from the will to build a smooth moduli space (unlike the underived case).
When $X=S$, a smooth projective surface, they claim that the tangent space at a point $E$ is:
$T_{E} \mathbb{R} \underline{\operatorname{Vect}}_{n}(S) \simeq-H^{2}(S, \underline{E n d}(E))+H^{1}(S, \underline{E n d}(E))-H^{0}(S, \underline{E n d}(E))$.
However here the $H^{2}$ term (which is the derived part) seems to come from the fact of $S$ is $2$-dimensional and not from any singularity or self-intersection (which seems strange to me)
If you look at the example of (2), which is quite similar (I think it is the derived stack of local systems), the $H^{2}$ term appears when you take into account the self intersection of $0$ in $\mathbb{A}^{1}$, (i.e. a truly derived structure).
What I am misunderstanding here?
The difficulty in finding a visualisation is that whereas underived nilpotents come from truncating schemes with more points, the derived structure is entirely formal. One example you might like to mull is the derived fibre product ${0}\times^h_{\mathbb{A}^1}{0} = \mathrm{Spec } k[x_1]$, for $x_1$ in chain degree $1$ (cochain degree $-1$). The tangent space at the closed point is $k[-1]$, which you should think of as the homotopy kernel of $0 \to k$.
Thank you for the example! Is that the loop space at $0$, right? I am writing an answer/reflection that is related to this but is too long so I will add it in the main question.
Yes, in a sense that's the loop space at $0$.
I edited in a link to @DamienC's answers, but saw a comment, not an answer, from @JonPridham. I guessed that that link was the right one, and so edited it in, too. I hope that was correct.
Thanks, I have edited a bit the text cause the second answer of Damien was really to the third question "3-what else..." and not to the first one. Feel free to modify anything else for a better understanding.
The example of the derived stack $RVect$ of vector bundles on an algebraic surface $S$ is very similar to the one of the derived stack $RLoc$ (I usually drop the "R", but I keep it to be consistent with your notation) that I gave in on eof my answers. Indeed $RVect_n(S)=RMap(S,BG)$ ... tbc...
... continued ... just like in my answer, if you let $Y=Spec(k[\tau])$ (this is a derived affine scheme), and look at $Y$-points of $RVect_n(S)=RMap(S,BGL_n)$ you'll get the datum of a $k$-point, that is a vector bundle $E$ on $S$, together with a class in $H^2(S,End(E))$.
On a side note, for a later occasion, I think it'd be better to start a new question than significantly changing the question several times after answers have been given. The new question can refer to the previous one for background. But it seems to me that multiple editions makes the whole thread not easy to follow for people who may want to catch up.
Yes, that's true. I will consider it in the following
I'm not quite sure what kind of answer you're expecting, but here is a geometric example that may help to grasp some intuition.
In differential geometry, when an intersection is badly behaved (e.g. it doesn't have the expected dimension) one can geometrically perturbe one of the two factors. For instance, if you are intersecting tow submanifolds $X,Y\subset Z$, and if $X$ is locally given as the zero of some functions $X\overset{\text{loc}}{=}\{f_1=\dotsb=f_k=0\}$, you may want to introduce a deformation $X_{t_1,\dotsc,t_k}$ of $X$ defined as
$$
X_{t_1,\dotsc,t_k}\overset{\text{loc}}{=}\{f_1=t_1,\dotsc,f_k=t_k\}.
$$
One of the main idea of derived geometry is to replace these deformation/perturbation parameters $t_i$'s by a homological perturbation:
$$
X_{t_1,\dotsc,t_k}\overset{\text{loc}}{=}\{f_1=d\tau_1,\dotsc,f_k=d\tau_k\},
$$
with $\operatorname{deg}(\tau_i)=-1$ (my degree convention is cohomological).
Homological perturbation has two advantages above geometric perturbations:
it can be made functorial.
it exists even in the (quite non-flexible) algebraic setting, where geometric perturbation may not exist.
Let's try to apply informally the above reasoning to the case discussed in Jon Pridham's comment: consider $X=Y=\{x=0\}$ inside $Z=\mathbb{A}^1=\operatorname{Spec}(k[x])$. You deform $\{x=0\}$ to $\{x=d\tau\}$ and then proceed with the intersection of $\{x=d\tau\}$ with $\{x=0\}$, and get $\{d\tau=0\}$, which is just a ($k$-)point (it is $0$ in $\mathbb{A}^1$) together with a self-homotopy (given by $\tau$). This is indeed the "space" of derived loops in the affine line that are based at $0$.
I apologize for self-promoting, but you can read an informal account of how to view derived self-intersections as some kind of based loop spaces in the introduction of Calaque, Căldăraru, and Tu - On the Lie algebroid of a derived self-intersection.
Thank you for the answer. It is quite concrete an clarifying. The intro of the paper is useful too.
However, I am unsure about higher homotopical groups (i.e. $deg(\tau_{i})<-2$). In which situations you can have higher (or abritary high?) homotopical perturbations (you dont need higher dimensional singularities for that, right?)?
On the other side if are all the derived nilpotents (of any derived scheme/stack) of homotopical nature? If so, the relevant geometry/topology would be homotopy classes (i.e. maps from the topological space of our scheme/stack to some n-spheres, )
Also, thinking about my last sentence it seems to me that broadly speaking the new spaces that derived geometry gives rise to are self intersections spaces (i.e. formal loop and higher dimensional formal loop spaces) and "derived thickenings" by nilpotent extensions (yielding formal disks/higher dimensional formal disks). Does this make any sense?
In my first comment I meant "$deg(\tau_{i})\leq2$ " and "On the other side, are all the derived nilpotents (in any derived scheme/stack) of homotopical nature?"
$\DeclareMathOperator\Map{Map}\DeclareMathOperator\ad{ad}$This is an attempt to answer the third question: what else?
Let $X$ be a compact space and let $G$ be an affine algebraic group. One can contemplate the following (underived) higher stacks:
$BG$: the classifying stack of $G$-torsors.
$X_B$: the constant stack associated to $X$.
One can consider the higher underived mapping stack $\Map(X_B,BG)$, which is nothing but the ordinary (ie non-derived) stack of $G$-local systems on $X$. Its tangent complex has amplitude $[-1,0]$:
in degree $-1$, at a $k$-point $P$ ($P$ is a $G$-local system), its cohomology is $H^0(X,\ad(P))$, where $\ad(P)$ is the linear local system associated with $P$ and the adjoint $G$-representation $\mathfrak{g}$: $\ad(P)=P\times_G\mathfrak{g}$.
in degree $0$, at a $k$-point $P$, its cohomology is $H^1(X,\ad(P))$.
The infinitesimal theory $\Map(X_B,BG)$ doesn't capture anything about higher cohomology groups $H^{*\geq 2}(X,\ad(P))$.
If you're looking at the derived mapping stack $\mathbb{R}{\Map}(X_B,BG)$ instead, then its tangent complex at a $k$-point $P$ is the full de Rham cohomology $H^{*+1}(X,\ad(P))$.
Why is this so? The point is that the underived stack $\Map(X_B,BG)$ doesn't see anything else than the fundamental groupoid of $X$: this is because $BG$ is a $1$-truncated Artin stack. But if we allow families of $G$-local systems parametrized by geometric objects intrinsically carrying homotopical information (affine derived schemes), then we get back the missing information. For instance, it's a good exercise to check that if $Y$ is the derived self-intersection of $0$ in the affine line that was mentionned in previous answers (i.e. $Y=\operatorname{Spec}(k[\tau])$, $\operatorname{deg}(\tau)=-1$), then a $Y$-point in $\mathbb{R}{\Map}(X_B,BG)$ is the datum of a $k$-point $P$ and a class in $H^2(X,\ad(P))$.
Did you really mean $\mathbb R{\operatorname{Map}}$ for the derived functor? I thought it was usually denoted $\mathbf R{\operatorname{Map}}$.
Thanks for your answer Damien, I am still a bit unsure about how to intepret geometrically the class in $H^{2}(X,ad(P))$ (in the example of $Y=$self-intersection of $0$ in $\mathbb{A}^{1}$). While the $H^{0}$ and $H^{1}$ should correspond to the usual automorphisms and deformations of the underived stack I guess that the $H^{2}$ should take into account the derived deformations (coming from the deformations of the derived loop space at $0$?). However $H^{2}(X,ad(P))$ should correspond to a class of $2$-cycles/holes (dependent on sections of a bundle/local system/whatever) but (...)
seems strange to me that that set of $2$-dimensional objects parametrizes the deformations of the derived $1$-dimensional loop space. Does higher dimensional loop spaces work in the same way (i.e. if $Y$ was the self int. of 0 in $\mathrm{A}^{n}$ you would need classes in $\left.H^{i}(X, a d(P)), i=2, \ldots, n+1\right)$?
What happens if $Y$ is a more complicated object, like the derived self-intersection of a closed subvariety (say of $\operatorname{dim}=m$ ) of some other variety (say of $\operatorname{dim}=l$ )? We can know something about the $H^{i}$ 's needed in that case?
Another concern I have is that I cannot achieve an unified vision of what the higher homology groups reprsent in derived stacks. For example in a derived moduli stack of vector bundles $V$ over a projective surface $S$ I have read that the tangent space at some point $V$ is $(-1)^{i} \sum_{i=0}^{2} H^{i}(S, \operatorname{End}(V))$ where the $H^{2}$ term comes from the fact that $S$ is 2 -dimensional rather than any singularity or self-intersection. Being your example (the derived stack of local systems) so similar to that one the different origin of $H^{2}$ sounds strange to me (..)
The one I see strange is the derived stack of vector bundles case not your example, (as I would expect the higher $H^{i}$ 's were present in a stack only when singularities and self-intersection arise!). Btw the article when I saw that about the derived stack of vector bundles is https://arxiv.org/pdf/math/0210407.pdf
If you replace $S$ with a closed topological surface, and the stack of vector bundles with the one of local systems, things fit nicely together. Say $S$ is the gluing (push-out) of two surfaces $S_{\pm}$ with same boundary boundary $C$. Then $Loc(S)$ is equivalent to the derived intersectin of $Loc(S_+)$ and $Loc(S_-)$ in $Loc(C)$.
To answer the comments that starts as "seems strange to me". You'd rather want to look at $Y_n$-points for $Y_n=Spec(k[x_n])$ with $x_n$ sitting in degree $-n$. They'll give you a $G$-local system $P$ and a class in $H^{n+1}(X,ad(P))$. The degree is not related to the codimension, but rather to how many times you self-intersect $0$: $Y_1$ is the self-intersection of $0$ in the affine line $Y_0$, $Y_2$ is the self intesection of $0$ in $Y_1$, ... $Y_{n+1}$ is the self intesection of $0$ in $Y_n$.
The motivation with higher cohomology groups leads me to the following question; in the case $dim(X) = 1$ where the cohomologies $H^n(X, ad(P))$ trivially vanish for $n>1$, will it then be true that the derived and undervied stacks coincide?
@Robert Hanson: yes! For instance, both the derived and underived stack of local systems on $S^1$ are equivalent to the adjoint quotient [G/G].
Nice thank you!
|
2025-03-21T14:48:29.707039
| 2020-01-21T17:25:43 |
350885
|
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|
Stack Exchange
|
Definition of the derivative of a Poisson structure on a manifold given by bivector called a Poisson bivector
What is the derivative of a Poisson structure on a manifold given by a Poisson bivector?
Is this term used anywhere, or should an answer invent a definition? On an orientable manifold you have the modular vector field $\Phi$ = $\pm$ the divergence of the Poisson bivector field $\pi$ [Poisson Structures, §4.4.3].
In coordinates, applying \partial_k to the bivector \Theta^{ij}\partial_i\partial_j is no problem. How to say it coordinate free?
@JimStasheff: Hi Jim! Actually, the coordinate expression is unhealthy because it will not look like a partial derivative hitting the Poisson bivector in any other coordinate chart. You could put a connection $\nabla$ on the tangent bundle (e.g. a Levi-Civita one) and replace $\partial_k\to\nabla_k$. But it's unclear what problem that solves for you. (Perhaps a symplectic connection is more natural, if the Poisson structure is of maximal rank.)
Thank you - very very helpful comments
Since I first asked, I discovered the derived bracket version {f,g} = [[Theta,f],g]
|
2025-03-21T14:48:29.707149
| 2020-01-21T17:30:24 |
350886
|
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|
Stack Exchange
|
Representation of optimal controls as diffusions
In reading this post I couldn't help but wonder the following question:
Let $\sigma>0$ and suppose, as in the motivational post, we are given a stochastic optimal control problem:
$$
\begin{aligned}
v(t,x)= & sup_{a \in \mathcal{A}} \, \mathbb{E}^{t,x}\left[
\int_t^T f(X_s^a,a_s)ds + g(X_T^a)
\right]\qquad 0\leq t< T
\\
dX_t^a =& x_0 + \int_0^t b(X_s^a,a_s)ds + \int_0^t \sigma dW_s,
\end{aligned}
$$
optimized over the set of all predictably measurable controls. In this case, we know that (under some conditions so that BSDE methods apply) there exist exists an optimal control $\hat{a}_t$ and it is of the form:
$$
a_t = \alpha(X_t,V_t,\nabla V_t,\Delta V_t),
$$
where $V_t$ is the problem's value function (which is also a local-martingale).
My question is, under what (reasonable) conditions is:
$\alpha$ continuous,
$V_t,\nabla V_t$, and $\Delta V_t$ can all be expresses as strong solutions to some (finite-dimensional) time-homogeneous SDEs?
|
2025-03-21T14:48:29.707238
| 2020-01-21T18:05:11 |
350888
|
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|
Stack Exchange
|
Rational functions and power series
Let $P(z)/Q(z)$ be a rational function ($\mathbf{C}$ coefficients) and assume that its Taylor series $\alpha(z)=\sum_{n\geq 0} a_n z^n$ around $z=0$ has radius of convergence $1$. Consider the power series $|\alpha|(z)=\sum_{n\geq 0}|a_n|z^n$, which also has radius of convergence $1$. Is $|\alpha|(z)$ a rational function?
I suppose the generating function of $a_n=\cos nt$ is rational while the generating function of $|\cos nt|$ is not, for irrational $t$.
@AlexandreEremenko Of course you mean irrational multiples of $\pi$.
@Robert Israel: yes, this is what I mean.
Note that $\sum_n |a_n|^2 z^n$ is always a rational function when $\sum_n a_n z^n$ is a Laurent expansion of a rational function. @AlexandreEremenko
And more generally if $\sum_{n} a_n z^n$ and $\sum_{n} b_n z^n$ are rational functions, so is $\sum_n a_n b_n z^n$. But $\sum_n |a_n| z^n$ is not of this form.
Just to provide some context for this question. Under the hypothesis for the radius of convergence and the extra requirement that all $a_n$ are integers, a theorem of Fritz Carlson says that $\alpha(z)$ is either rational or it has the circle as a cut. I suspect that even in this this situation there are examples for which $\alpha$ is rational and $|\alpha|$ has the circle as a cut, but I am unable to find an explicit one.
|
2025-03-21T14:48:29.707484
| 2020-01-21T20:43:38 |
350896
|
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|
Stack Exchange
|
Visualizing hyperbolic metric of punctured sphere
Uniformization of the 3-punctured sphere generates a "pants" configuration with three legs narrowing down to cusps. This is supposed to have a metric of constant negative curvature, and I can see this in the cusps, and also in the saddle regions where they join, but I am at a loss to understand how the curvature can remain negative in the middle region. Simplistic depictions (e.g. on the Wikipedia pair-of-pants page) clearly do NOT show negative curvature throughout. As another example, Hubbard Fig. 3.5.1 (ref below) shows a surface which does have negative curvature throughout, and has two cusps coming in; however, it has a finite-size hole going out, and that hole is clearly going to have to grow, rather than shrink, to maintain negative curvature.
In short, I can't understand how a surface can start at zero radius, then traverse a section where size is growing, and then contract back to zero size, without encountering a section where curvature is positive. I can calculate that it is happens using the uniformizing metric, but I would really appreciate some guidance in how to visualize or understand it property. Even if I think of embeddings in 3 space with self-intersections, I can't seem to construct a visualization that works.
ref: Hubbard, John H., "Teichmuller Theory", volume 1, Matrix Editions, 2006.
Some intuitive understanding can be obtained from contemplating the universal covering, pictured by the infinite order triangular tiling of the hyperbolic plane
Would you have the same trouble with visualizing the Euclidean structure on a torus?
I think trying to embed such things curvature-preserving into 3-dimensional space is just the wrong way to think about them. For example iis reallyhard to visualize the flat metric on a torus like that. However, it is much easier by taking just a flat square and identifying opposing sides. This gives us the flat metric on a torus.
Let us apply the same idea to the three punctured sphere. Now we do not want curvature zero, but constant curvature minus 1. So we should not take a flat square, but some (ideal) polygon in two dimensional hyperbolic space. Take 6 ideal triangles with angles 60°,60° and 0° (so the vertices with 0°are ideal) glue them together to get an ideal hexagon with angles 0°,120°,0°,120°,0° and 120° and glue the edges next to one ideal vertex together.
To vertify that the result really has constant curvature minus -1, we have to show that everypoint has a small neighborhood that isometrically embeds into two dimensional hyperbolic space. Let us first go back to the toy example of a flat torus. Why does that argument work there. It is clear for the points in the interior of a square. For the points on the edges it also works, since we glue always glue two edge together. And for the cornerpoint it also works, since there 4 corners with 90° each meet.
And now the same argument works in our hyperbolic setting; in the interior everything is ok, we identify one edge with exactly one other edge and at the corner exactly three corners with 120° each meet.
It might be a nice exercise to relate this construction to the picture in the comment above by მამუკა ჯიბლაძე.
I agree with Henrik very much. Just a side comment: after drawing a lot of pictures about the local geometry in this style, you might also try to fashion a global topological picture of how these local pictures glue together. This can be done most easily with fabric, or some other very flexible material. Bill Thurston was a true genius at imagining these things in his mind, but a lot of the "large scale" understanding of the negative curvature of surfaces of negative euler characteristic can be gained in this way. I'm very far from the right person to ask about this, but here I am anyway..
But, you can embed a hyperbolic cylinder and visualize it pretty easily. And, the 3-punctured sphere is just two cylinders coming in, one going out, yet it's behaving very differently from a cylinder, because a hyperbolic cylinder can never grow and then shrink again. I guess I just have to dig into the model and see where this intuition from cylinders is going wrong.
|
2025-03-21T14:48:29.707807
| 2020-01-21T21:13:05 |
350898
|
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|
Stack Exchange
|
Finite energy solution for Allen -Cahn equation
I am interested in the Allen-Cahn equation in $ R^N$ and one can consider the related energy functional
$$ E(u):= \frac{1}{2}\int_{R^N}| \nabla u(x)|^2 dx + \frac{1}{4} \int_{R^N} (u^2-1)^2dx.$$ There has been a lot of work on this equation and in particular on the DeGiorgi conjecture. My question is related to whether any of the solutions have finite energy. So here is my exact question. Lets take $N=9$ and suppose $ x=(x',x_9)$.
Question. Does there exist a function $u$ with $ -1 <u<1$ with a $ u_{x_9}>0$ and $\lim_{x_9 \rightarrow \infty} u(x',x_9)=1$ and $ \lim_{x_9 \rightarrow -\infty} u(x',x_9)=-1$. Furthermore $ E(u)<\infty$.
The reason I ask this is I see some results about 'finite energy solutions' yet they just impose growth on the energy in terms of $B_R$ (ball radius $R$ centered at the origin) so I thought if the full energy is finite maybe something is trivial (but I just can't see it..)
Thanks for the comments.
Maybe lower bound |grad u| by |du/d_{x_9}| and then the integral over x_9 for any fixed x' is bounded away from zero.
@Willie Wong. I saw a comment that seems to have disappeared (or maybe I accidentily erased something). So at this point I am just looking for a function as described (it does not need to be a critical point)
@user36212 I will need to think a bit about what you are suggesting.
It's Willie's answer below, no need to think further...
Sketch of an argument:
For a fixed $x'$, let $f(x')$ denote the measure of the set $\{ u(x',x_9) \in (-1/2,1/2) \}$.
Note that for fixed $x'$ you have
$$ \int |\nabla u(x',x_9)|^2 d x_9 \geq \int |\partial_{x_9} u(x',x_9)|^2 d x_9 \geq 1 / f(x') $$
(on the ends of the interval (it is an interval by monotonicity) defining $f(x')$ the function takes values $-1/2$ and $1/2$ respectively, and so the minimizer is the linear function with slope $1/f(x')$.)
For fixed $x'$ you also have
$$ \int (u^2 - 1)^2 dx_9 \geq \frac9{16} f(x') $$
Finiteness of energy requires both be integrable in $x'$, which is not possible.
This argument doesn't seem to depend on the dimension, provided $N \geq 2$. Also, isn't it the case that the De Giorgi conjecture only asks for $C^2$ solutions to the Euler-Lagrange equation, and not actually solutions with finite energy?
That you very much for your answer. Regarding your question; you are correct, they don't ask for finite energy. I was just adding this for some other reasons.
|
2025-03-21T14:48:29.708010
| 2020-01-21T21:37:19 |
350900
|
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|
Stack Exchange
|
Orbits of tensor product $\operatorname{St}_2\otimes\operatorname{Sym}^2(\mathbb C ^3)$
Let $G_1=\operatorname{GL}_2(\mathbb C)$ act on $V_1=\mathbb C^2$ via the standard multiplication. Denote this representation by $\operatorname{St}_2$. Let $G_2=\operatorname{SL}_3(\mathbb C^3)$ act on $V_2=\mathbb C^6$ via $\operatorname{Sym}^2(\mathbb C^3)$. Then we have the representation of $G_1\times G_2$ on $V_1\otimes V_2$. I am curious about the orbits of this action. Do we know there are a finite number of orbits? If so, how many orbits are there and how do we describe them?
I appreciate any comments/references.
(This seems to be an exercise from representation theory, and thus I am not sure if it is appropriate to ask this question here.)
First consider the actions of $G_i$ on their respective representations. That's pretty much basic linear algebra. Then think about how these orbits combine in the tensor product.
@VítTuček Thanks for your comment. But that is basically the same as my question. It seems that one can not always build the orbits of the tensor product from each pieces in an easy way. One evidence of the last statement comes from the example of orbits of tensor products $\mathbb C^{n_1}\times \mathbb C^{n_2}\times \mathbb C^{n_3}$ of the group $GL_{n_1}\times GL_{n_2}\times GL_{n_3}$ https://mathoverflow.net/questions/67897/invariants-and-orbits-of-n-tensors. Even both $\mathbb C^{n_1}\times \mathbb C^{n_2} $ and $\mathbb C^{n_3}$ have finite number of orbits, the tensor does not ingeneral
You are right. I had a wrong mental picture of tensor product.
Did you try looking at the papers suggested in the answer to the question you linked? You can start with finitely many orbits of $GL_2 \times GL_3 \times GL_3$ on $\mathbb{C}^2\otimes \mathbb{C}^3\otimes\mathbb{C}^3$, intersect with $\mathbb{C}^2\otimes \mathbb{C}^3\odot\mathbb{C}^3$ and consider decomposition of these orbits under $GL_2\times SL_3$ embedded into $GL_2\times GL_3\times GL_3$ diagonally on the second factor.
Unless I missed something the question belongs to classical algebraic geometry (and also linear algebra) rather than representation theory. It pertains to classifying pencils of conics in the plane.
@AbdelmalekAbdesselam Thanks for your comment. Could you provide a reference?
@abdelmalekabdesselam Your comment made something click about both algebraic geometry and representation theory for me.
@PaulSiegel: Thank you :) I'm curious, what is it that clicked?
@QingZhang: I would have to search the literature for the precise answer to your question, but if you google search "pencil of conics" with quotes you will find useful references. Also, an excellent book on this general area is "Classical algebraic Geometry: aModern View" by Igor Dolgachev, at U. Michigan. You could also email him.
Sasha already gave a detailed answer, but this reference may be useful too: http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>.4948&rep=rep1&type=pdf
Yes, the number of orbits is finite. Indeed, as Abdelmalek mentioned, this is the question of classification of pencils of conics. The orbits are the following.
First, assume that two conics in the pencil are non-proportional.
1) Assume that at least one of the conics in the pencil is nondegenerate. Then this conic is isomorphic to $\mathbb{P}^1$ and the intersection points of the pencil is a subscheme of length 4.
1a) If the 4 points are distinct, this is a quadruple of points in general position on $\mathbb{P}^2$, all such quadruples are conjugate, and in appropriate coordinates this pencil can be written as
$$
\langle xy - xz, xz - yz \rangle.
$$
1b) If two points collide, this is a triple a points and a tangent vector at one of them, so the pencil can be written in the form
$$
\langle xy + xz, yz \rangle.
$$
1c) If two pairs of points collide, the pencil can be written in the form
$$
\langle x^2, yz \rangle.
$$
1d) If three points collide, the pencil can be written in the form
$$
\langle x^2 - yz, xy \rangle.
$$
1e) If all four points collide, the pencil can be written in the form
$$
\langle x^2 - yz, y^2 \rangle.
$$
2) Assume now that all conics in the pencil are degenerate. This is possible in either of two cases.
2a) All conics contain a given line. Then each conic is the union of this line and an extra line. Extra lines also form a pencil, and its intersection point can lie on the fixed line or away from it. This gives two more orbits:
2a') $\langle xy, xz \rangle$.
2a'') $\langle xy, x^2 \rangle$.
2b) All conics have a fixed singular point, but no common lines: $\langle x^2, y^2 \rangle$.
3) Next, assume that all conics in the pencil are proportional (equivalently, one of the conics is zero). These orbits are parameterized by the rank of the conic.
3a) $\langle x^2 - yz, 0 \rangle$.
3b) $\langle xy, 0 \rangle$.
3c) $\langle x^2, 0 \rangle$.
3d) $\langle 0, 0 \rangle$.
So, altogether there are 12 orbits. However, I could forget something.
EDIT. The orbit closure order, I think, is the following.
Orbits of type 1 are ordered as $(1e) < (1d), (1c) < (1b) < (1a)$; orbits $(1d)$ and $(1e)$ are incomparable and both sit between $(1e)$ and $(1b)$.
$(2a') < (1d)$, and incomparable with $(1c)$ and $(1e)$.
$(2a'') < (2a')$, $(2a'') < (1e)$.
$(2b) < (1e)$, and incomparable with $(2a')$, $(2a'')$.
$(3a) < (1e)$, and incomparable with type 2 orbits.
$(3b) < (3a)$, $(3b) < (2b)$, $(3b) < (2a'')$.
$(3d) < (3c) < (3b)$.
thank you very much for your excellent answer. For our purpose, we also need the closure relations of those orbits. Here is our guess: $(3d)\subset \overline{(3c)}\subset \overline{(3b)}\subset \overline{(3a)},$ $(3a)\subset \overline{(2a')}\subset \overline{(1c)}$, $(3a)\subset \overline{(2a'')}\subset \overline{(1c)}, (3a)\subset \overline{(2b)}\subset{\overline(1e)}\subset \overline{(1d)}\cap \overline{(1c)}$, $(1d)\subset \overline{(1b)}\subset \overline{(1a)}$, and $(1c)\subset \overline{(1b)}\subset \overline{1(a)}$. The notations are the same as in your answer.
Dear @Sasha, Could you confirm if the closure relations in the last comment. I am sorry for the messy notations. I know it is better to draw a tree about the closure relations. But I don't know how to do it in a comment. It will be great if you could provide some references. Thanks again.
I added a description of the order as I see it. It is not quite the same as you suggest: First $(3a)$ is not contained in the closures of type 2 orbits, because it contains a smooth conic, while those don't. Second, $(2a'') < (1e)$ because $x^2-yz$ in $(1e)$ can degenerate to $yz$. Third, $(2a') < (1d)$ because $x^2 - yz$ in $(1d)$ can degenerate to $yz$. Fourth, $(2a')$ is incomparable with $(1c)$, because $(1c)$ contains a double line, while $(2a')$ does not.
Thanks for your reply. I will think about it.
|
2025-03-21T14:48:29.708675
| 2020-01-21T22:21:16 |
350901
|
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|
Stack Exchange
|
Negation of cardinal characteristics
Looking over the various cardinal characteristics of the continuum all of them are defined by a sentence of the form:
"$\mathfrak{x}$ is the least cardinality of a subset of $\omega^\omega$ that $\ldots$"
Is there any meaningful cardinal characteristic whose definition can be given by
"$\mathfrak{x}$ is the supremum of the cardinalities of subsets of $\omega^\omega$ that $\ldots$"
I tried negating the definition of the usual cardinal characteristics, e.g. $\mathfrak{b},\mathfrak{d}$, but there are easy examples of size $2^{\aleph_0}$.
So, I am afraid the answer is "No", but I would be glad to hear otherwise.
What about the supremum of the cardinalities of strong-measure-zero subsets of $\mathbb R$? I don't know if you'd consider it a "cardinal characteristic" or not, especially since it is consistent that this cardinal is countable, but otherwise it fits the bill. There's a question about it here: https://mathoverflow.net/questions/281024/the-strong-measure-number/281041#281041.
@WillBrian: I have to see what is the strong-measure-zero subsets of R.
@WillBrian. This seems a good candidate, but is it consistent that $\mathfrak{s}_+$ is a cardinal between $\aleph_0$ and $2^{\aleph_0}$?
It is a theorem that $\mathfrak{s}_+$ is a cardinal between $\aleph_0$ and $2^{\aleph_0}$ (inclusive). It is consistent that it is $\aleph_0$ (this happens in Laver's model -- google "Borel conjecture" to read more about it), that it is uncountable but less than $2^{\aleph_0}$ (this happens in the random real model), and that it is equal to $2^{\aleph_0}$, even when CH fails (this happens in the Cohen model).
@WillBrian. Then this answers the question! Do you mind writhing it out so I can give you some credit :)
@WillBrian. Can you give me a reference for the consistency results? I am more interested in the random reals case, where $\mathfrak{s}_+$ is strictly between $\aleph_0$ and $2^{\aleph_0}$.
The results about the Cohen and random model can be found in chapter 8 of Bartoszynski and Judah's book Set Theory: on the structure of the real line. For the random real model, it follows from Theorem 8.2.8 that there are uncountable strong measure zero sets, and it is Theorem 8.2.11 that every strong measure zero set has size $\leq \aleph_1$.
So you don't consider the cardinal characteristic $\mathfrak c= 2^{\aleph_0}$ meaningful?
Is $\aleph_1$ a meaningful characteristic? This is the largest cardinal of a subset of the reals which carries a well-order all initial segments of which are Borel.
@Goldstern I like your characterization of $\aleph_1$, but you can not change the value of this "characteristic".
|
2025-03-21T14:48:29.708893
| 2020-01-21T23:14:52 |
350903
|
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"Mido",
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}
|
Stack Exchange
|
Eigenvalues of product of symmetric positive definite matrices
Let $T_1, \ldots, T_n$ by real symmetric positive definite matrices, with eigenvalues bounded below by $\mu > 0$.
Can I say
$$
\frac{x^T T_1 T_2 \ldots T_n x}{x^T x} \geq \mu^n
$$
If these matrices commute the result is straightforward, but I'm interested in the case where these matrices don't necessarily commute.
Edit: Not sure that you can say this for the $n=2$ case either.
Knowing the eigenvalues have a lower bound like that is not sufficient for what you want, because $x^T M x$ is unrelated to the eigenvalues of $M$ for $M$ not symmetric. In particular, can't the left side be negative already for $n=2$?
Oh good point. I'm not sure. In the $n=2$ case the eigenvalues are real and positive and so I thought that implied $x^T T_1 T_2 x > 0$, but I guess that may not be true.
Yes indeed, perhaps you can find a vector $x$ making the left-hand side negative.
It is true that the product $M=T_1T_2$ of two positive definite symmetric matrices has real and positive eigenvalues. And conversely, every matrix $M$ with real positive eigenvalues can be factored $M=T_1T_2$ as above. But $x^TMx$ does not need to be positive. Here is an example:
$$M=\begin{pmatrix} 3 & a \\ -a & -1 \end{pmatrix}, \qquad \sqrt3<a<2.$$
The eigenvalues, roots of $X^2-2X-3+a^2$, are real and positive, while
$x^TMx=3x_1^2-x_2^2$ is indefinite.
|
2025-03-21T14:48:29.709019
| 2020-01-21T23:34:38 |
350904
|
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|
Stack Exchange
|
Classifying space for Thompson's group F?
Let $\mathcal C$ be the free monoidal category generated by an object $X$, and a morphism $X \otimes X \to X$.
This category contains exactly two connected components: that of the monoidal unit $1\in \mathcal C$, and that of $X\in \mathcal C$. (In general, two object $A$ and $B$ of a category are said to be in the same connected component if they are related by a zig-zag of arrows $$A\to Y_1\leftarrow Y_2\to Y_3\leftarrow Y_4\to Y_5\leftarrow\ldots \to B.$$
In the case of $\mathcal C$, any two non-unit objects are related by a single morphism.)
Let $\mathcal C'\subset \mathcal C$ be the connected component of $X$ and
let $|\mathcal C'|$ denote the geometric realisation (of the simplicial nerve) of $\mathcal C'$.
The article
Marcelo Fiore, Tom Leinster, An abstract characterization of Thompson's group $F$, Semigroup Forum 80 (2010), 325-340, doi:10.1007/s00233-010-9209-2, arXiv:math/0508617.
proves that $\pi_1(|\mathcal C'|)$ is isomorphic to Thompson's group $F$.
Question: Is $|\mathcal C'|$ a classifying space for Thompson's group $F$?
Just to make sure I understand: Fiore and Leinster talk about a different category, namely, the monoidal category freely generated by an object $X$ and an isomorphism $X \otimes X \to X$. That category is a groupoid, so your question has a positive answer for their category. Now it easily follows from their result that for your non-groupoid category the fundamental group is $F$, and you are asking if you also get a $K(F,1)$ from your category. Right?
@Omar Antolin-Camarera: Yes, your understanding is correct.
|
2025-03-21T14:48:29.709159
| 2020-01-21T23:44:02 |
350906
|
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|
Stack Exchange
|
double shuffle lie algebra
I have a question about the definition of the double shuffle lie algebra discussed in section 1.3 of Sarah Carr's thesis (see https://www.imj-prg.fr/theses/pdf/sarah_carr.pdf)
Recall the definition of the double shuffle lie algebra:
Consider two coproducts, coshuffle and costuffle, that are equipped in two noncommutative polynomial algebras respectively, $\mathbb{Q}⟨⟨x,y⟩⟩$ and $\mathbb{Q}⟨⟨y_i; 1\leq i < \infty⟩⟩$. These coproducts are defined on the generators and one can extend them through multiplication (concatenation):
$$\begin{align}
\Delta_⧢: \mathbb{Q}⟨⟨x,y⟩⟩ &\to \mathbb{Q}⟨⟨x,y⟩⟩ \otimes \mathbb{Q}⟨⟨x,y⟩⟩ \\
x &\mapsto x \otimes 1 + 1 \otimes x \\
y &\mapsto y \otimes 1 + 1 \otimes y
\end{align}$$
$$\begin{align}
\Delta_*: \mathbb{Q}⟨⟨y_i⟩⟩ &\to \mathbb{Q}⟨⟨y_i⟩⟩ \otimes \mathbb{Q}⟨⟨y_i⟩⟩ \\
y_i &\mapsto \sum_{m+n=i} y_m \otimes y_n
\end{align}$$
Coshuffle can be considered as the dual of shuffle product of multiple zeta values, and costuffle as the dual of stuffle product of multiple zeta values.
Then the double shuffle lie algebra $\mathfrak{ds}$ is a vector subspace of $\mathbb{Q}⟨⟨x,y⟩⟩$ generated by elements $f$ such that $f$ are primitive for $\Delta_⧢$ and $\pi_y(f)$ are primitive for $\Delta_*$, where $\pi_y$ is a linear map that transforms $f$ to an element in $\mathbb{Q}⟨⟨y_i⟩⟩$ because costuffle is operated in $\mathbb{Q}⟨⟨y_i⟩⟩$. The definition of $\pi_y$ is
$$\pi_y: \mathbb{Q}⟨⟨x,y⟩⟩ \to \mathbb{Q}⟨⟨y_i⟩⟩ \\
\tilde{\pi_y}(x^{k_1-1}yx^{k_2-1}yx^{k_3-1}y\cdots x^{k_n-1}yx^{k_{n+1}})=
\begin{cases}
0, \; k_{n+1} \neq 0 \\
y_{k_1}y_{k_2}y_{k_3}\cdots y_{k_n}, \; k_{n+1} = 0
\end{cases} \\
\pi_y(f) = \tilde{\pi_y}(f) + \sum_{n\geq 2} (f|x^{n-1} y)\frac{(-1)^{n-1}}{n}y_1^n$$
My question is about this map $\pi_y$. The definition of $\tilde{\pi_y}$ is quite natural, but why do we need to add that summation term in the definition of $\pi_y$? Some other documentation tells me that the summation term comes from the extended double shuffle (EDS) relations of multiple zeta values, which is described in the first three sections of this paper. But I cannot figure out how EDS leads to the formula for $\pi_y$, nor can I find references that explain it. Very appreciated if anybody knows the references.
The double shuffle Lie algebra described relations among (formal) multiple zeta values, modulo products. However, the description in terms of primitivity with respect to the coshuffle and costuffle coproducts require us to assign finite values to divergent multiple zeta values.
For example, in weight three, we have two primitive elements with respect to the coshuffle coproduct:
$$x_0^2x_1 -2x_0x_1x_0 + x_1x_0^2$$
and
$$x_1^2x_0 -2x_1x_0x_1 + x_0x_1^2$$
both of which involve divergent integrals.
One can show that, for the shuffle product, there is a unique way to associate a value to divergent integrals such that $\zeta(1)=0$ and the shuffle product holds for products of not-necessarily-convergent integrals
$$\zeta(x_0^k)\zeta(w)=\text{Sum over shuffles of }x_0^k\text{ and }w.$$
This gives shuffle regularised MZVs.
Similarly, there is a unique way to associate a value to divergent sums such that $\zeta(1)=0$ and the stuffle product holds for products of not-necessarily-convergent sums, giving stuffle regularised MZVs.
However, the two regularisations do not agree: consider $\zeta(1,1)$. By the shuffle shuffle product, we get
$$\zeta(1,1)=\frac{1}{2}\zeta(1)^2=0,$$
while the stuffle product tells us that
$$\zeta(1,1)=\frac{1}{2}(\zeta(1)^2-\zeta(2))=-\frac{1}{2}\zeta(2).$$
As such, in order for the double shuffle Lie algebra to encapsulate all the double shuffle relations, we need to switch between the regularisations, which is where the extra term in $\pi_Y$ arises. One can compute directly that that additional term encodes the value of the stuffle regularised $\zeta(1,1,...,1)$. It is also easy enough to see that every divergent MZV can be written as a stuffle polynomial of convergent MZVs and $\zeta(1,1,...,1)$, so I'm going to hand-wave and say that this makes $\pi_Y$ sufficient to encode the change from shuffle regularisation to stuffle regularisation.
|
2025-03-21T14:48:29.709415
| 2020-01-21T23:49:38 |
350907
|
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|
Stack Exchange
|
Isomorphism classes of split extensions
Let $p$ be a prime number and $n$ an integer such that $p\geq n$. Let $P(n)$ denotes the number of partitions of $n$. Can we conclude from Theorem 1.1 and Theorem 1.3 in the reference FINITE_p-GROUPS_OF_MAXIMAL_CLASS_AND_EXPONENT_p, that the number of isomorphism classes of splits extensions of $( \mathbb{Z} / p \mathbb{Z} )^n$ by $\mathbb{Z} / p \mathbb{Z} $ with a non abelian middle groups is $P(n)-1$ ?. Does the subgroup $( \mathbb{Z} / p \mathbb{Z} )^n$ should be here characteristic in $(\mathbb{Z}/p\mathbb{Z})^{n}\rtimes \mathbb{Z}/p\mathbb{Z}$ ?.
Any help would be appreciated so much. Thank you all.
What's the question?
The formula you have in mind doesn't even work for $n=1$. $P(1)=1$ so $P(1)-1=0$ but there is a split extension. Note, the subgroup might not be characteristic, for example when it's a direct product.
Thank you verret but I m focus here on the case $n\geq 2$ and the midle groups of the splits extensions must be non abelian.
If I understand correctly, you are counting the conjugacy classes of elements of order $p$ in $\operatorname{GL}(n,\mathbb{F}_p) $ — or equivalently, of matrices $N$ in $\operatorname{M}_n(\mathbb{F}_p) $ with $N^p=0$, but $N\neq 0$. If $p\geq n$ you get all nilpotent nonzero matrices, hence indeed $P(n)-1$ conjugacy classes using Jordan normal form, but this is false for $p<n$: for $p=2$, for instance, you get $[n/2]$ (corresponding to partitions with only 1 and 2, and at least one 2).
Also, in the case when a subgroup $C_p^{n-1}$ is centralized by the $C_p$, the subgroup $C_p^n$ is not characteristic.
Thanks for noticing. What do you think of my answer?. Is there an error in one of the answer stages?
|
2025-03-21T14:48:29.709664
| 2020-01-22T00:16:55 |
350909
|
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"TOM",
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|
Stack Exchange
|
Concentration of maxima of a random polynomial with Rademacher coefficients
Let $X_1,\ldots, X_n$ be independent Rademacher random variables (i.e. $\mathbb{P}(X_i=\pm 1)=1/2$). Consider the random polynomial $$P_{n}(t)=c+X_{1}t+X_2t^2+\cdots+X_{n}t^n.$$
Is it well known how to get good upper bounds on probabilities of type
$$\mathbb{P}(|\max_{t\in [0,x]}|P_{n}(t)|-\mathbb{E}\max_{t\in [0,x]}|P_{n}(t)||>y)?$$
Theorem 5.3.2 in Talagrand's book states the following (using here somewhat different notations):
Let $X_1,X_2,\dots$ be independent Rademacher random variables. Let $U$ be a subset of the closed ball $B(u_0,s)$ in $\ell^2$ centered at some $u_0\in\ell^2$ and of some radius $s>0$. Let
$$S:=\sup_{u\in U}\Big(c(u)+\sum_{i\ge1}u_iX_i\Big),$$
where $c\colon U\to\mathbb R$. Then
$$P(|S-ES|\ge y)\le C\exp\Big(-\frac{y^2}{Cs^2}\Big) $$
for some universal real constant $C>0$ and all real $y>0$.
If $ES$ in the latter display is replaced by the median of $S$, then $C$ may be replaced by $4$.
Take now any real
$$s\ge\inf_{u\in\mathbb R^n}\max_{t\in[0,x]}\sqrt{\sum_{j=1}^n(u_j-t^j)^2}.$$
Then, by the above statement, the probability in question is bounded from above by $ C\exp\Big(-\dfrac{y^2}{Cs^2}\Big)$. In particular, here one may take $s=\sqrt{\sum_{j=1}^n x^{2j}}$.
Thank you for a quick answer!
|
2025-03-21T14:48:29.709776
| 2020-01-22T00:31:20 |
350911
|
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"Paolo Leonetti",
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|
Stack Exchange
|
Sign in Dirichlet's approximation theorem
Fix $\alpha \in \mathbf{R}$. The classical Dirichlet's approximation theorem states there exist infinitely many rationals $p/q$ such that
$$
\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^2}.
$$
Question. Fix $\alpha \in \mathbf{R}$. Is it true that there exist infinitely many rationals $p/q$ such that
$$
0\le \alpha- \frac{p}{q}\ll\frac{1}{q^2}\,\,?
$$
What does the $\ll$ mean in this context?
"There exists an absolute constant $c>0$ such that the inequality $0\le \alpha-p/q \le c/q^2$ holds for infinitely many rationals $p/q$"
Yes, this follows from considering the continued fraction of $\alpha$. If $p_n/q_n$ is the $n$th convergent to $\alpha$ and $n$ is odd then
$$ 0\leq \alpha - \frac{p_n}{q_n} \leq \frac{1}{q_n^2}.$$
|
2025-03-21T14:48:29.709862
| 2020-01-22T00:58:29 |
350913
|
{
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"Jeff Strom",
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|
Stack Exchange
|
When is a homotopy pushout contractible?
Let $B \leftarrow A \to C$ be a span of spaces, and consider the homotopy pushout $B \cup_A C$.
Question: When is $B \cup_A C$ contractible?
This is a pretty open-ended question. I'm interested in necessary conditions or sufficient conditions or interesting examples or special cases, etc.
Some additional conditions I'd be happy to take as blanket assumptions:
I don't think too much is lost if we assume that $A,B,C$ are connected.
I'm happy to assume that $A \to B$ is 1-connected (conventions vary about what this means; I mean that the homotopy fiber of $A \to B$ is 1-connected, i.e. that $A \to B$ is $\pi_1$-surjective).
For somewhat obscure reasons, I'm particularly interested in the case where $A \to C$ is the projection $A_0 \times C \to C$ out of a binary product.
For similarly obscure reasons, I'm particularly interested in the case where $C$ is a loopspace, or even where $C = \Omega \Sigma C_0$ is a free loopspace.
Observation:
By pasting on a few more pushout squares, we obtain the identity $\Sigma(B \vee C) = \Sigma A$.
I'm not sure what to make of this, though.
You've probably already thought of this, but you can get some algebraic necessary conditions by using Mayer-Vietoris in your favorite (co)homology theory.
@John I think this is covered already by Tim's observation, which is an almost unstable version of what MV gives you
@Tim Knots are one rich source of examples. If $K \subset S^3$ is a knot there is a pushout square with $C = S^3 - K$, $B$ a tubular neighborhood of the knot in $\mathbb R^3$ (contractible), and $A$ is $B - K$ (equivalent to $S^1$). However, this doesn't produce many examples which satisfy $4$, because $\pi_1 C$ is rarely abelian.
The paper Structure theorems for homotopy pushouts. I. Contractible pushouts. by John Klein (MR1490203) is all about contractible pushouts.
Assume all the spaces are connected, $\pi_1 A \to \pi_1 B$ is surjective, and $\pi_1 C$ is abelian.
The pushout, $P = C \sqcup_A B$, is contractible if and only if $\pi_1(P)$ and all the $H_i(P)$ are trivial. However, by Seifert-Van-Kampen, and the hypothesis on $\pi_1 A \to \pi_1 B$, we have that $\pi_1 C$ surjects onto $\pi_1 P$. Thus $\pi_1 P$ is abelian, so $P$ is contractible if and only $H_i(P) = 0$ for all $i$.
For this, as you and John observed, it is necessary and sufficent for $H_i(A) \to H_i(B) \oplus H_i(C)$ to be an isomorphism for all $i \geq 1$.
Let me add some additional remarks on the enumeration question:
How many such spaces $A$ are there sitting over $B\times C$ such that the homotopy pushout
$$
B \leftarrow A \to C
$$
is contractible?
If $B\cup_A C$ is contractible, then the sum of the maps $\Sigma A \to \Sigma B$ and $\Sigma A \to \Sigma C$ gives a homology isomorphism
$$
\Sigma A \to \Sigma B \vee \Sigma C\, .
$$
If $A$ is connected, then we conclude that the map is also a weak homotopy equivalence, so $\Sigma A$ is a wedge of $\Sigma B$ and $\Sigma C$ in this case. The converse to this statement is also true.
Hence, we are reduced to classifying those spaces $A$ whose suspension splits as a wedge of $\Sigma B$ and $\Sigma C$. Hence, this problem can be formulated as a kind desuspension problem.
There is a classifying space ${\cal D}(B,C)$ which is the realization of (the nerve of) a category whose objects are spaces $A$ with structure map $A \to B\times C$ such that $\text{hocolim}(B \leftarrow A \to C)$ is contractible. A morphism of this category is a weak homotopy equivalence of spaces over $B\times C$.
Classification Result: Assume $A,B,C$ are $1$-connected. There is a function
$$
\pi_0({\cal D}(B,C)) \to \{B\vee C, B\wedge C\}
$$
(the target is the abelian group of homotopy classes of stable maps) which is a bijection in a metastable range
$$
\max(b,c) \le 3\min(r,s)-1\, ,
$$
where $b$ is the homotopy dimension of $B$ (the CW complex of minimal dimension having the homotopy type of $B$), $c$ is the homotopy dimension of $C$, $r$ is the connectivity of $B$ and $s$ is the connectivity of $C$.
Here I have given $B$ and $C$ basepoints.
The function in (4) is easily described as the stable homotopy class of the weak map
$$
\Sigma B\vee \Sigma C \overset{\simeq}\leftarrow \Sigma A \to \Sigma (B \wedge C)
$$
where the right arrow is given by suspending the composition
$A \to B\times C \to B\wedge C$.
If we want to work beyond the metastable range, the homotopy type of ${\cal D}(B,C)$ can be determined "up to extensions" via the coefficients of the identity functor in homotopy functor calculus.
See the paper
Klein, John R.; Peter, John W. Fake wedges. Trans. Amer. Math. Soc. 366 (2014), no. 7, 3771–3786
which extends my earlier paper:
John R. Klein, Structure theorems for homotopy pushouts. I. Contractible pushouts, Math. Proc. Cambridge Philos. Soc. 123 (1998), no. 2, 301–324.
|
2025-03-21T14:48:29.710194
| 2020-01-22T04:22:53 |
350917
|
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|
Stack Exchange
|
Non-algebraic representations of $\text{SL}_n(\mathbb{R})$
My question is easily stated: are all continuous finite-dimensional real representations of $\text{SL}_n(\mathbb{R})$ algebraic representations?
This is false if you drop the word "continuous" (e.g. using noncontinuous field automorphisms of $\mathbb{R}$).
It is also false if you work with $\text{GL}_n(\mathbb{R})$ instead of $\text{SL}_n(\mathbb{R})$. For instance, there are many examples in the answers to this question. However, all of the continuous examples there use the determinant in some way, so they don't give examples of non-algebraic representations of $\text{SL}_n(\mathbb{R})$.
@MarkSapir: Margulis's super-rigidity implies that every representation of SL(n,Z) virtually extends to a continuous representation of SL(n,R), but as far as I know (I am not an expert) it just gives a continuous extension. I would also expect that if this is true, then invoking Margulis would be like killing a fly with a nuclear bomb.
Every field automorphism of $\mathbf{R}$ is continuous (however you get representation using non-continuous field embeddings into $\mathbf{C}$).
The answer to your question is yes, and more generally true for every connected semisimple Lie group, and even perfect is enough. This is due to the fact that every perfect subalgebra of matrices is the Lie algebra of an algebraic subgroup. This can be applied to the (Lie algebra of the) graph of your representation.
@YCor: Maybe one should mention that this graph is a closed subgroup of a Lie group, hence a Lie subgroup, and therefore algebraic by your argument.
@YCor: "Every field automorphism of R is continuous": Even more is true: The identity is the only field automorphism of $\mathbb{R}$.
@YCor, the implication "perfect matrix Lie algebra $\implies$ Lie algebra of algebraic group" doesn't seem to be easily googlable. Could convert your comment into a short answer with a reference?
I found a reference for the result YCor quoted: it is Corollary 7.9 in Borel's book "Linear Algebraic Groups".
@YCor: Thanks for the comment. Your answer is convincing; however, I am puzzled by one thing. In the paper "Rationality of representations of linear Lie groups" by D. H. Lee and T. S. Wu (Proc. Amer. Math. Soc. 114 (1992), no. 3, 847-855), the authors claim that there are counterexamples to this among connected semisimple real algebraic groups. Their paper uses too much Lie theory for me to easily follow it. How can I reconcile what they do with what you do?
(this paper is what originally led me to worry that there might be counterexamples for $\text{SL}_n(\mathbb{R})$)
Thanks for the reference! Indeed my argument was too sloppy. One more problem with dealing with algebraic geometry while working with real points! Indeed the idea is to say that the graph of the representation is an algebraic subgroup... But at the level of complex points it's no longer the graph of a map in their example. Nevertheless, this seems to still hold for simply connected $G$ (in the algebraic sense), and in particular for $\mathrm{SL}_n(\mathbf{R})$. Indeed Theorem 1 (equivalence between (iii) and (iv)) of your linked paper answers your question.
PS: link to the Proc AMS paper: https://www.ams.org/journals/proc/1992-114-03/S0002-9939-1992-1072344-X/S0002-9939-1992-1072344-X.pdf
To summarize the counterexample: consider the homomorphism $\mathrm{SL}_3\to\mathrm{PGL}_3$, whose kernel has order 3. At the level of real points, it induces a homomorphism $f$ between unit components. Then $f^{-1}$ is an isomorphism of the underlying Lie groups that is not algebraic.
@YCor: Thanks! Can you say a bit more about why (iv) in the linked paper has something to do with the algebraic group being simply-connected? It seems to be a rather different condition (but as I work in a different area, I might be missing some equivalences that are obvious to specialists).
I didn't mean this (although it should be true): I'm just saying that $G=\mathrm{SL}_n(\mathbf{R})$ satisfies (iv) (namely that the inclusion of $G$ in its (complex) Zariski closure is its complexification).
@YCor: Got it, thanks!
|
2025-03-21T14:48:29.710588
| 2020-01-22T07:21:47 |
350927
|
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|
Stack Exchange
|
Continuum percolation in 1d
What is known about continuum percolation in 1d?
By this, I mean, for $d \in \mathbb{N}$, the Poisson-Boolean model of disks of radius $r_0 \in \mathbb{R}$ with centres arranged randomly in $[0,1]^{d}$, with a Poisson number of disks at density $\lambda$ per unit volume, taking the case $d=1$. A pair of disks is connected if they overlap.
In site percolation on the integer lattice in 1d, the critical site occupation probability $p_c=1$, otherwise the chain of sites will break somewhere.
But in 1d continuum percolation, can $r_0(\lambda)$ scale slowly enough to zero as $\lambda \to \infty$ for the probability of a break goes to zero?
Whatever $r_0$ and $\lambda$ are, there will be infinitely many gaps of size bigger than $2r_0$, so that there is no percolation.
So basically, only if $2r_0$ is the width of the domain do you get percolation.
OK - sorry - I hadn't read the question properly. So you are looking at a finite segment of the line, namely [0,1] rather than all of $\mathbb R$. My answer applies to the infinite case. So in the finite case, there should be some sort of phase transition, I think. I can't tell you exactly what it is immediately though.
You're answer is correct. So, if the ratio $r_0/w$, where $w$ is the domain width, goes to a constant, is that may be sufficient for percolation?
Can you see any literature on this? Or has it not been studied?
I'm not understanding here. I thought $w$ was fixed and you are looking at how $r$ should scale with $\lambda$. (Certainly $w$ may as well be fixed because changing $w$ by a factor of $t$ is equivalent to changing $r$ by a factor of $t$ and $\lambda$ by a factor of $1/t$).
Yes sorry, just set $w$ to 1. I mean simply that as $\lambda \to \infty$, you at least need $r_0$ a strictly non-zero fraction of the domain width, i.e. does not go to zero.
I'm really not expert in this field. But I'm sure there is literature if you look for it. Probably the 1D case is such that you can derive a pretty reliable estimate by simple calculations (this is much harder in 2D and higher).
So set $w=1$. I prefer to think about $\lambda(r)$ rather than $r(\lambda)$. Of course one is just the inverse of the other. So given $r$, you want it to be very likely there is another centre at a distance in the range $[r,2r]$. How likely? Likely enough that this can be continued for $1/r$ steps. So: you want $(1-e^{-r\lambda})^{1/r}$ to be close to 1.
See for example this paper.
Covering by random intervals and one-dimensional continuum percolation
C. Domb, J. Stat. Phys. Vol. 55(1-2), 1989
and also
Exact solution of a one-dimensional continuum percolation model
A. Drory, Phys. Rev. E Vol. 55(4), 1997
among other works, including many by the second author above.
|
2025-03-21T14:48:29.710832
| 2020-01-22T09:33:39 |
350931
|
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|
Stack Exchange
|
Tiling rectangle with trominoes — an invariant
There are two types of trominoes, straight shapes and L-shaped. Suppose a rectangle $R$ admits at least one tiling using trominoes, with an even number of L-trominoes.
EDIT: we do not admit ALL orientations, but only those which are constructed by starting in a box, and next box is placed to the left, or below previous box.
Hence, only 2 out of the 4 possible orientations of the L-tromino is allowed. Hence, these are the four trominoes that we are allowed to use:
$$\begin{align}\newcommand{\x}{\large\blacksquare}\newcommand{\o}{\large\phantom{\square}}
&\o\o\o\o\o\o\o\o\o\o\x\\[-5pt]
&\x\x\o\o\o\o\x\o\o\o\x\\[-5pt]
&\x\o\o\o\o\x\x\o\o\o\x\o\o\o\x\x\x
\end{align}$$
Prove that every tiling of $R$ must use an even number of L-trominoes.
This smells a lot like a classical tiling problem, but unlike the classical cases (dominoes and missing corners), the situation is not to show that a tiling is impossible…. Some nice invariant is perhaps what I am looking for.
I actually know how to prove the above statement, but it requires a lot more machinery than I would like. Moreover, I am interested in a more general question, (regarding representation theory and cylindrical Schur functions, and the proof I know does not generalize to this situation), but I hope that a good proof of the above problem generalizes.
For the interested: The general setting supposes that the rectangle is a torus, and that we use $k$-ribbons, where $k$ is odd. We then want to show that the number of $k$-ribbons which occupies an even number of rows, occur an even number of times. Hence, a proof not relying on the fact that $R$ has boundaries, or heavily uses the fact that each shape has three squares, is of extra interest.
EDIT2: 2022-12-09 So, I am in the process of writing up a proof of this (and some other stuff).
Do you know whether the statement can be strengthened to: every tiling of $R$ must use an even number of "north/east plus south/west L-trominoes" and an even number of "north/west plus south/east L-trominoes"? Or any other statement about the orientation of the L-trominoes?
Isn't the result false on a torus? Take rows $(112, 133, 232)$ and numbers indicate where the tiles are.
hm really? http://www.matematika-shkolnikam.ru/43.jpg
Fedor's theorem is stronger than mine.
@FedorPetrov Ah, sh*t, i need one extra condition - I want to show that no rectangle admits both odd and even number of L-trominoes then (the special case I am interested in seem to only admit even number of L-trominoes)...
For torus, you can also use the rectangle tiling $(111,223,233)$ which has two L's.
Also, since $3$ is prime, the sizes are $m\times n$ where one is divisible by $3$, so you always have a tiling with no L's which is even.
@VilleSalo Well, I need to add some extra conditions for the torus - the case I have in mind does not simply glue the edges together, but there is also a shift...
But it is nice to know that the shift might be relevant.
I guess the tag should be [tag:tiling], not [tag:tilting].
To make the edit clearer: the permitted orientations are "2 by 2 block missing northwest/top-left square" and "2 by 2 block missing southeast/bottom-right square", right?
@user44191 yes, that is correct.
A more general theory is due to Igor Pak, https://www.ams.org/journals/tran/2000-352-12/S0002-9947-00-02666-0/S0002-9947-00-02666-0.pdf.
@RichardStanley yes, I am aware of that, but his machinery does not generalize to the torus setting, so I was hoping for an alternative approach...
Have you looked at diagonals skewed to your pieces? (Two of the pieces occupy 2 squares of a certain diagonal direction, while the others occupy only one;alternatively,two pieces lie within two diagonals, whereas the others cross three diagonals.) Gerhard "Hoping For A New Slant" Paseman, 2020.01.22.
It may be worth noting that any invariant that is linear in the "filled" squares (more formally, that can be extended to a linear function in the "filled" squares) will not work. A 2 by 6 rectangle can be filled by 3 non-overlapping same-orientation L-trominos with 3 squares blank; those squares can then be filled by 2 I-trominoes minus another one.
A possible simplification: is there some invariant (besides "total number of squares modulo 3") on Young tableaux under 1) lengthening any row by 3, 2) turning a row into 4 rows (of the same length), and 3) lengthening any 3 rows of the same length by 1? This should correspond to invariants under addition of arbitrary I-trominos.
@user44191 well, the invariant is the sign of a Sn-character... as far as I know, this is quite complicated...
Do you have any examples of tilings with different numbers of the two orientations of the L-trominos?
@user44191 No, and for rectangles, they have to be equal in numbers..
@PerAlexandersson: your last comment on the two L-shapes having equal multiplicity, does it follow from the results in Pak's paper? From just a cursory reading I couldn't figure it out.
@YaakovBaruch I believe that it does, but I do not remember exactly from where
@user44191 - adding to your comment about linear invariants - it's even worse. A disjoint union of a 1x3 horizontal rectangle, a horizontal cylinder of circumference 9 and length 2, and a 3x3 torus is indistinguishable from the 10x3 rectangle in terms of: area=30, left/right border length=10, top/bottom length=3, number of corners=4, absence of concavity along the border! Yet the disjoint figure can be tiled with unequal numbers of the 2 L-shapes, due to the torus. I can't think of any equation (linear or not) that could capture the connectedness and thus distinguish between the 2 cases.
@YaakovBaruch Feel free to edit the question, I did not manage to get that command to work (there is some issue with the -5pt command, i think.
@FedorPetrov, the link from your comment now goes to a generic landing page, and the Wayback Machine doesn't remember the image. Do you know where to find it now?
Prior comment clarified: any invariant must separate the 2 cases $\quad\newcommand{\x}{\large\blacksquare}\begin{align}\x\[-5pt]\x\[-5pt]\x\end{align},\cup,$ $\big\uparrow\begin{align}&\x\x\x\x\x\x\x\x\x\[-5pt]&\x\x\x\x\x\x\x\x\x\end{align}\big\uparrow,\cup,$ $\big\uparrow\begin{align}&,,\longrightarrow\[-5pt]&\x\x\x\[-5pt]&\x\x\x\[-5pt]&\x\x\x\[-5pt]&,,\longrightarrow\end{align}\big\uparrow\quad$ and $\quad\begin{align}&\x\x\x\x\x\x\x\x\x\x\[-5pt]&\x\x\x\x\x\x\x\x\x\x\[-5pt]&\x\x\x\x\x\x\x\x\x\x \end{align},\quad$ since the 1st can be tiled with different numbers of L-shapes.
@LSpice I think, it was 5*9 partitioned onto corners. Like this:112233455/162773445
6688799AA/BCC8DE9AF/BBCDDEEFF
@YaakovBaruch I know it's a three year old thread but, yes, "number of northwest L's" - "number of southeast L's" is Pak's invariant $f_1$. See the discussion after Theorem 1.4,
@PerAlexandersson Are you still interested in this? I thought about it a bit this morning, and I think the following are probably right: (1) Given a ribbon tiling of the torus, it is always possible to cut it open to make a make a ribbon tiling of a cylinder (see https://www.symmetricfunctions.com/cylindricSchur.htm ). (2) It should be pretty straightforward to work out a version of Pak's theory for cylindrical ribbon tableaux. However, (3) the same size torus can give rise to many different cylindrical shapes, and it isn't clear how to relate them.
Also (4) going from "ribbon tiling" to "ribbon tableaux" (the analogue of Pak's Lemma 3.4) isn't so easy in the cylindrical case. In short, I put some thought into this, figured out a few things, but there is a lot which can go wrong and I'm not sure it is worth writing down the details.
@DavidESpeyer Yes, so I am still interested in this, and I hope to have a preprint out soon (within 3-4 weeks). It has some applications to cylindric Schur functions as you noted which was one of the inspirations of this problem.
So, one can find an invariant that works, but it is not the most natural as it requires some non-commutative thinking.
Here is a start on a proof. I ask Per to finish it up.
The constraints allow me to color diagonals of the array in three distinct colors. (I choose a periodic pattern of red, green, and blue.) Two of Per's pieces remove one of each color, while the other two (because of how they orient and how I choose my diagonal coloring) cover two of one color and one of another. This imbalance should allow a parity argument for the L shaped pieces. Note this doesn't work if any of the other orientations are allowed.
Gerhard "Only Giving Half A Proof" Paseman, 2020.01.22.
I'm still seeing issues with this because I think of a strip of nested L pieces of one orientation spanning part of the torus. Perhaps Per has more of the picture and can use the idea of solving with this invariant anyway. Gerhard "Maybe He Can Force Herringbone" Paseman, 2020.01.22.
Indeed, if I use a three by three torus, I can tile it with three L shapes of the same orientation. Hopefully he can rule that and similar out. Gerhard "If Not One, The Other" Paseman, 2020.01.22.
Well, forget about the torus for the moment, but I am not convinced about the diagonal argument. You can have one type of L-piece, three copies with the 3 possible colorings. This is indistinguishable from three of the straight shapes.
That is true, but if two L's are oriented in opposite fashion, there is only one configuration (up to color) that maintains equicolor balance. Gerhard "If You Look For Balance..." Paseman, 2020.01.22.
@GerhardPaseman. Your argument is not completely fruitless. With simple vector arithmetic it can be used to show that the numbers of the two L-shapes are congruent (mod 3). But nothing more than that.
|
2025-03-21T14:48:29.711465
| 2020-01-22T11:04:36 |
350934
|
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|
Stack Exchange
|
Non-trivial examples of regular Lagrangian flow in BV case
What is a concrete example of BV vector field $v$ with $\mathrm{div}\, v = 0$ that makes Ambrosio's theory of regular Lagrangian flow relevant?
With concrete I mean that we can compute the flow explicitly. An example I provided in a related question seems to have much stronger solutions.
Another related question is this: Relationship between three different definitions of solutions for ODE with irregular coefficient.
Here is what I believe is a relevant example: consider the vector field $X$ in $\mathbb R^3$,
$$
X=a_1(x_2, x_3)\frac{\partial}{\partial x_1}+a_2(x_1, x_3)\frac{\partial}{\partial x_2}+a_3(x_1,x_2)\frac{\partial}{\partial x_3}, \quad\text{so that div $X=0$.}
$$
Assume that $a_j\in L^\infty$, $\frac{\partial a_j}{\partial x_1}, \frac{\partial a_j}{\partial x_2}\in L^1$, and
$\frac{\partial a_1}{\partial x_3}, \frac{\partial a_2}{\partial x_3}$ are Radon measures. Then bounded measurable solutions of the Cauchy problem for the PDE, $Xu=F$, $u=g$ given on $∑$ a transversal hypersurface to $X$ are locally uniquely determined by $F, g$.
In fact, it is even possible to say that the "generic" example in 3D is
$$
X=a_1(x_1,x_2, x_3)\frac{\partial}{\partial x_1}+a_2(x_1,x_2, x_3)\frac{\partial}{\partial x_2}+a_3(x_1,x_2,x_3)\frac{\partial}{\partial x_3},
\text{with $a_j,\ $div $X\in L^\infty$},
$$
and
$$
\frac{\partial a_j}{\partial x_1}, \frac{\partial a_j}{\partial x_2}\in L^1, \quad
\frac{\partial a_1}{\partial x_3} \text{ Radon measure},
\frac{\partial a_2}{\partial x_3}, \frac{\partial a_3}{\partial x_3}\in L^1.
$$
|
2025-03-21T14:48:29.711684
| 2020-01-22T11:12:23 |
350935
|
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|
Stack Exchange
|
Radon transform on complex Grassmannians
Let $Gr_{i,n}$ denote the Grassmannian of complex linear $i$-dimensional subspaces in the Hermitian space $\mathbb{C}^n$. Let $1\leq i<n/2$. Consider the Radon transform between space of functions on two different Grassmannians:
$$R\colon C^\infty(Gr_{i,n})\to C^\infty(Gr_{n-i,n})$$
given by $(Rf)(E)=\int_{F\subset E} f(F) dF$ where the integration is with respect to the Haar measure on the Grassmannian of complex $i$-subspaces of $E$.
Is it true that $R$ is an isomorphism? A reference would be most helpful.
Remark. For real Grassmannians the corresponding question has positive answer, see Gelʹfand, I. M.; Graev, M. I.; Roşu, R. The problem of integral geometry and intertwining operators for a pair of real Grassmannian manifolds. J. Operator Theory 12 (1984), no. 2, 359–383.
|
2025-03-21T14:48:29.711762
| 2020-01-22T11:54:37 |
350937
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350937"
}
|
Stack Exchange
|
Suggestions for two-choice game played in ladder graph
I was just working on counting all the possible Nash Equilibrium solutions for a two-choice game played on a ladder graph (I got my results and all that for a generic number of players).
And I was wondering if you'd have any suggestions for related papers/books to see if I can find a practical application or any ideas to generalise it in graph theory. Thanks!
|
2025-03-21T14:48:29.711810
| 2020-01-22T12:17:22 |
350939
|
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"site": "mathoverflow.net",
"sort": "votes",
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|
Stack Exchange
|
Abelian lie algebra homology
Let $\mathfrak g$ be an abelian Lie algebra over $\mathbb Z.$ We can consider its Lie-algebra homology, say as $\mathrm{Tor}^{U(\mathfrak g)}_*(\mathbb Z,\mathbb Z)$ and its group homology as $\mathrm{Tor}^{\mathbb Z[\mathfrak g]}_*(\mathbb Z,\mathbb Z).$ Is it true that these two functors coincide ? If so, I believe it should be known, maybe somebody know a reference?
|
2025-03-21T14:48:29.711858
| 2020-01-22T12:22:44 |
350940
|
{
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"authors": [
"Bort",
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350940"
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|
Stack Exchange
|
Process in 1D, conclusions from gap distributions
preface: sorry if I am using the wrong nomenclature, part of my problem is that I don't know enough to google what I am looking for.
Let's say I have a large sorted set (or an ensemble of sets) of events $x_i$ and I know that the "gaps" $\Delta = x_i-x_{i-1}$ are distributed in a way that $P(\Delta)\sim \Delta^\alpha$ for $\Delta \ll 1$. Is there anything to learn from just knowing $\alpha$? Or are there bounds to which values $\alpha$ can have under certain typical assumptions?
The process as a whole should be stationary, by which I mean that there is no explicit dependence on time.
The only thing I can come up with is that for $\alpha>0$ the events are repulsive in some sense whereas for $\alpha<0$ they are attractive. $\alpha=0$ should include the standard Poisson-process and correspond to independent events.
if the "events" are eigenvalues of a random matrix, then $\alpha$ will generically take on one of the three values $1,2,4$.
Thanks @CarloBeenakker , yeah I remember that. They are not directly, but maybe related? Say I have a pos-def matrix and am updating it randomly (in some fashion) and everytime the lowest eigenvalue crosses zero there is some renewal event (an instability in the system). Is there any statement about the zero-crossings?
all of this will depend on the type of process you are considering; a study of the spacing distribution of zero-crossings in a particular physics context is given in https://arxiv.org/abs/1305.2924
|
2025-03-21T14:48:29.712001
| 2020-01-22T13:55:23 |
350947
|
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|
Stack Exchange
|
Do these surfaces intersect?
For any real numbers $a_{1},a_{2},\cdots a_{6}$ and $b_{1},b_{2},\cdots b_{6}$
with $\sum_{i=1}^{6}a_{i}^{2}=1$ and $\sum_{i=1}^{6}b_{i}^{2}=1$,
does the equation $$ x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}\left(\sum_{i=1}^{6}b_{i}x_{i}\right)^{2}=1 $$ always have a solution $x_{1},x_{2},\cdots x_{6}$ in $\mathbb{R}$ with $\sum_{i=1}^{6}x_{i}^{2}=6$? Thanks.
what have you tried? what is your background? this may be suitable in math stack exchange if you can edit this mentioning what all you know
@PraphullaKoushik I have tried special cases and the answer is yes for the special cases; and on the other hand, I could not find any counter-example. Best regards.
which special cases have you tried? add that in the question.. what techniques you know about intersection of surfaces add that in the question
@PraphullaKoushik I tried the special cases of choosing some of them to be 0 's or 1's. I am not in the area of algebraic surface, but this question is about intersection of surfaces, which I believe the researchers in the area of surfaces or related can answer. Thanks.
There have been a relatively large number of edits in a short space of time. It would be preferable to work out before hand what one wants to write, and then stick with it
The editing or rephrasing was due to no answer received for the original question. Best regards.
@PraphullaKoushik Just wondering if you know the answer to the question or have any helpful idea that could lead an answer to the question. Thanks a lot.
@mathers1 sorry, can not think of any way to proceed
This should be possible to answer using cylindrical algebraic decomposition.
@gyashfe Thank you for your comment. Could you please let us know what cylindrical algebraic decomposition is exactly and show us how should it possible to use it to get the correct answer to this question? Thanks.
@mathers1 it is an algorithm for quantifier elimination over a real closed field (see also the Tarski-Seidenberg theorem). You can find it, together with some computer implementations, on the internet.
@gyashfe But this question is about the existence of solutions and the equation is generic. How could the algorithm that you referred to be used to get the correct answer to this question? Thanks.
@mathers1 The algorithm can directly take as input a quantified formula of the form: $\forall a_1,\ldots,a_6,b_1,\ldots,b_6 \exists x_1,\ldots,x_6: \lnot(\sum a_i ^2 = \sum b_i ^2 = 1) \vee (\prod_{i=1}^4 x_i ^2 (\sum a_i x_i)^2 (\sum b_i x_i)^2 = 1)$ and return a truth value. If there are free variables, it gives as output the precise conditions on them which make the formula true (these conditions are given by polynomial equations and inequalities).
The answer is "yes" though the argument is rather ad hoc and doesn't generalize to vectors in more general positions.
We have $6$ unit vectors $v_j$ out of which the first $4$ are pairwise orthogonal and want to show that there exists a vector of length $\sqrt 6$ such that $\prod_{j=1}^6 |\langle x,v_j\rangle|\ge 1$ (to get below $1$ is trivial). Consider all sums $y=\sum_{j=1}^6\varepsilon_j v_j$ where $\varepsilon_j=\pm 1$ and choose the one with the largest length. Replacing some $v_j$ with $-v_j$, if necessary, we can assume WLOG that it is $y=\sum_j v_j$. Comparing $y$ with $y-2v_j$ (one sign flip), we see that $\langle y,v_j\rangle\ge 1$ for all $j$. Unfortunately, $y$ is a bit long, but it cannot get the length greater than $4$ (the $4$ pairwise orthogonal vectors produce length $2$) and we have
$$
\|y\|^2=\sum_j \langle y,v_j\rangle=:\sum_j (1+u_j), \quad 0\le u_j\le 3
$$
Reducing the length to $\sqrt 6$ means that we have to multiply $y$ by $\left(1+\frac 16\sum_j u_j\right)^{-1/2}$, so it suffices to show that
$$
\prod_j(1+u_j)\ge \left(1+\frac 16\sum_j u_j\right)^3
$$
i.e.
$$
\prod_j(1+u_j)^{1/3}\ge 1+\frac 16\sum_j u_j.
$$
However, on $[0,3]$, we have $(1+u)^{1/3}\ge 1+\frac u6$ (the LHS is concave, so it is enough to check the endpoints) and Bernoulli finishes the story.
Thank you for your answer. When you multiply $y$ by $\left(1+\frac 16\sum_j u_j\right)^{-1/2}$, how would you have $\langle y,v_j\rangle\ge 1$ for all $j$? Thanks.
@mathers1 I would not. I would just have the product $\prod_j \langle x,v_j\rangle$ equal to $\left(1+\frac 16\sum_j u_j\right)^{-3}\prod_j(1+u_j)$ and the rest of the post is about showing that it is still $\ge 1$.
|
2025-03-21T14:48:29.712295
| 2020-01-22T14:58:23 |
350950
|
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|
Stack Exchange
|
Index of the Fredholm operator
I have two vector bundles $E_1$, $E_2$ over $M$ and an embedding of the smooth sections $\lambda : \Gamma(M, E_1) \rightarrow \Gamma(M, E_1 \oplus E_2)$. I consider a Fredholm differential operator $D_1 : \Gamma(M, E_1) \rightarrow \Gamma(M, E_1)$ which can easily be lifted to the Fredholm differential operator $D_2 : \Gamma(M, E_1 \oplus E_2) \rightarrow \Gamma(M, E_1 \oplus E_2)$. What can be said about their indices $\mathrm{Ind}(D_1)$ and $\mathrm{Ind}(D_2)$?
could you please more explain how do you deduce $D_2$ from $D_1$? Moreover is $\lambda$ the obvious embedding?Or you fix an arbitrary embedding?Finally can you explain what is the motivation for this question?
|
2025-03-21T14:48:29.712370
| 2020-01-22T16:57:51 |
350952
|
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|
Stack Exchange
|
Fubini's theorem on arbitrary foliations
In what follows $ \mathbb{R}^{n+m} = \{(x,y): x \in \mathbb{R}^n, \ y \in \mathbb{R}^m \} \ .$
Suppose $G: U \to V $ is a $C^1$-diffeomorphism from an open subset of a manifold to an open subset of $\mathbb{R}^{n+m}$. We write
$$ (\xi,\eta)= G^{-1}(x,y) \ . $$
This is a local parameterization of $U$. We will think of
$$ U_\eta: = G^{-1}(\mathbb{R}^n \times \{y\} \cap V) $$
as the "horizontal" fibers, which are $n$-dimensional $C^1$-manifolds. Similarly, we refer to
$$ U_\xi = G^{-1}( \{x\} \times \mathbb{R}^m \cap V) $$
as the ``vertical" fibers, which are $m$-dimensional $C^1$-manifolds.
Question: Is there a version of the Fubini's theorem that equates the integral over $U$ to a double integral over fibers?
I am wondering if such a result exists in any textbook. In any other publication? It must be, because it is so natural to desire an integration over the original foliations.
This is a paraphrase of the question asked previously here.
I came up with the following:
Let$ DG_{|U_\xi} (\xi,\eta)$ be the derivative, at point $(\xi,\eta)$, of the map $G_{|U_\xi}$, i.e. the restriction of $G$ to $U_\xi$. This restriction is from an $m$-dimensional $C^1$-manifold to a subset of $\mathbb{R}^m$.} So its derivative and its Jacobian are defined.
Suppose for some fixed $\eta_0$, the union of vertical fibers along $U_{\eta_0}$ covers the set $U$.
Lemma Under the assumptions above, for any integrable function $f: U \to \mathbb{R} $,
$$ \int_U f = \int_{U_{\eta_0}}\left(\int_{U_\xi} f(\xi,\eta) \frac{|\det DG_{U_\xi} (\xi,\eta)| \cdot |\det DG_{U_{\eta_0}} (\xi,\eta_0)|}{|\det DG(\xi,\eta)|} \ d\mathcal{H}^m(\eta)\right) d\mathcal{H}^n(\xi) \cdot $$
That is saying that to integrate over $U$ simply integrate along the fibers and then sum over the "base", which is what Fubini's theorem says in the standard orthogonal coordinates of the Euclidean space. The correcting factor is can be memorized as
$$ \frac{\text{ Jacobian along fibers} \times \text {Jacobian along base}}{\text{full Jacobian}} \ .$$
Corollary: Integration in spherical coordinates. There, the fibers are orthogonal to the base, and so, the full Jacobian also factors, cancelling the numerator. Therefore, the correcting factor disappears (=1), and we get the familiar
$$ \int_U f \ d \mathcal{L}^n = \int_0^\infty \left( \int_{\mathbb{S}^{n-1}(r)\cap U} f \ d \sigma \right) dr \ . $$
Example: Let $P$ be a parallelepiped in the plane, resulting from tilting a rectangle so that the acute angle between its edges $A$ and $B$ is $\theta$. Then, foliating $P$ by parallel copies of the edges $B$, indexed $B_x$, leads to
$$ \int_P f \ d \mathcal{L}^2 = \int_A \left( \int_{B_x} f \sin(\theta) \ d \mathcal{H}^1 \right) dx \ = \sin(\theta) \int_A \left( \int_{B_x} f \ d \mathcal{H}^1 \right) dx \ . $$
Again my questions are:
1) Is there some alternative formula known out there?
2) Are there references giving specifically this? (I am not interested in "it can follow"s and "it can be done"s!)
Is this a question?
I'm voting to close this question as off-topic because it is not a question.
Are there references in a textbook to this? That is the question.
By the way, a recent edit fixed a number of small typos, but one of them wasn't actually: parallelopiped is an acceptable alternate spelling of parallelepiped.
A nice version of the manifold version of Fubini's theorem is in Differentialgeometrie und Fasserbündel by Rolf Sulanke and Peter Wintgen:
Let $\phi \in C^1(M,N)$, where $M,N$ are smooth manifolds of dimensions $m,n$, respectively, with $m \ge n$. Let $\omega \in \Omega^{m-n}(M)$ and $\eta \in \Omega^n(N)$, and let $f : M \to {\bf R}$ be measurable (meaning, its superposition with any map is Lebesgue measurable). Assume the set of critical values of $\phi$ has measure zero in $N$ (again, this means the image under any map has Lebesgue measure zero).
If the $m$-form $f\omega\wedge\phi^*\eta$ is integrable on $M$, then for almost all $x \in N$ the integral:
$$
\int\limits_{\phi^{-1}(x)} f\omega
$$
is well-defined and, moreover, when treated as a function of $x$ and multiplied by $\eta$, it is integrable on $N$ and:
$$
\int\limits_M f\omega\wedge\phi^*\eta =
\int\limits_N \bigg( \int\limits_{\phi^{-1}(x)} f\omega \bigg)\, \eta
$$
|
2025-03-21T14:48:29.712653
| 2020-01-24T21:48:25 |
351088
|
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|
Stack Exchange
|
Regularity of conformal maps
In order to define what it means for a map $f \colon \Omega \subseteq \mathbb R^n \to \mathbb R^n$ to be conformal, it is sufficient to require that $f$ is everywhere differentiable. Does conformality automatically implies that $f$ is $\mathcal C^1$ (hence real-analytic, see below)?
By complex analysis, we know the answer is positive when $n=2$.
In higher dimensions, Liouville's theorem characterizes conformal maps as Möbius transformations, but it is stated for $f \in W^{1,n}$ in Wikipedia. Is it known whether it also holds when $f$ is assumed everywhere differentiable?
Can one retinterprete the concept of conformality in terms of certain elliptic PDE? hence regularity would imply real analyticity? In dimension 2 it is CR equation, an elliptic PDE.
Can one retinterprete the concept of conformality in terms of certain elliptic PDE? hence regularity would imply real analyticity? In dimension 2 it is CR equation, an elliptic PDE.
Let $n\geq 3.$ Let $\Omega$ be an open connected subset of $\mathbb R^n,$ and let $f:\Omega\to\mathbb R^n$ be a function having a pointwise derivative $Df(x)$ everywhere satisfying $(Df)^T(Df)=g(x)I$ with $g(x)>0.$ Then $f$ is continuously differentiable.
By the inverse function theorem, $f$ is a local homeomorphism. By shrinking $\Omega$ we can assume $f$ maps $\Omega$ homeomorphically to $f(\Omega),$ and that $\Omega$ and $f(\Omega)$ are bounded.
Note $$\|Df\|^2=c_ng(x)=c_n\det(g(x)I)^{1/n}=c_n|\det Df(x)|^{2/n}$$
where $\|\cdot\|$ is Frobenius norm, and $c_n$ is a constant depending on $n.$
By the change of variables formula (proof sketch), $$\int_\Omega |\det Df(x)|\;dx=\mu(f(\Omega))<\infty.$$
So $f\in W^{1,n}(\Omega,\mathbb R^n)$ and you can use the $W^{1,n}$ result you mentioned.
Beautiful. Just to be sure, the point is that the integral is also equal to the $L^n$ norm of $Df$ (up to a constant)?
@seub: yes, $|Df|^n$ is integrable, so $Df$ is in $L^n.$
@seub Note that this answer assumes that $df$ is invertible at every point. Some people allow the differential to degenerate in their definition of conformal maps.
@AsafShachar I am not aware of this convention. As far as I am concerned, a conformal map is an angle-preserving map. In other words it is a map whose derivative at any point is a linear similarity. It does not make sense to me to include maps having critical points.
|
2025-03-21T14:48:29.712831
| 2020-01-24T21:54:10 |
351089
|
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|
Stack Exchange
|
Functional equation of twisted triple product L-function
Let $\mathbb{E}=E_1\times E_2\times E_3$ denote the product of three elliptic curves over $\mathbb{Q}$ of prime level $p$ and consider the $p$-adic Galois representation $$V_p(\mathbb{E})=H^1_{et}(E_{1/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{2/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{3/\bar{\mathbb{Q}}}, \mathbb{Q}_p).$$ Denote by $L(\mathbb{E}, s)=L(V_p(\mathbb{E}),s)$ the associated triple product $L$-function. It has a functional equation centered at $s=2$ with global sign equal to $a_p(E_1)a_p(E_2)a_p(E_2)\in \{ \pm 1 \}$ (cf. Gross-Kudla '92). Here, $a_p(E_i)$ denotes the $p$-th Fourier coefficient of the weight 2 normalized newform of level $\Gamma_0(p)$ associated to $E_i$ by modularity.
Let $\chi$ be a Dirichlet character modulo $p$ and denote by $L(\mathbb{E}\otimes \chi, s)$ the $L$-function attached to the Galois representation $V_p(\mathbb{E})\otimes \chi$. My question is: what is the functional equation of $L(\mathbb{E}\otimes \chi, s)$ and what is its global sign?
Thank you in advance for any help.
this is a great question, i've often wondered this myself!
There is a functional equation for $L(\mathbb{E} \otimes \chi, s)$, but it relates $L(\mathbb{E} \otimes \chi, s)$ to $L(\mathbb{E} \otimes \bar\chi, 4-s)$. If $\chi$ is not trivial or quadratic, then $L(\mathbb{E} \otimes \chi, s)$ and $L(\mathbb{E} \otimes \bar\chi, s)$ are different functions, so you cannot use this to deduce anything in particular about vanishing at the central value. The Langlands $\varepsilon$-factor is still defined, and it is a complex number of absolute value 1, but it isn't $\pm 1$, so you can't reasonably call it a "sign"; at a guess it is probably something like $\tau(\chi)^4 / p^{2n}$, where $p^n$ is the conductor of $\chi$ and $\tau(\chi)$ is the Gauss sum.
(You can try to cheat by considering the function $M(s) = L(E \otimes \chi, s) L(E \otimes \bar\chi, s)$, which does satisfy a functional equation relating $M(s)$ and $M(4-s)$, but unfortunately the order of vanishing of $M(s)$ at $s = 2$ is automatically even anyway, since $L(E \otimes \bar\chi, s) = \overline{L(E \otimes \chi, \overline{s})}$, so both factors have the same order of vanishing. So although $M(s)$ does have a functional equation, you can't get any nontrivial vanishing information out of it.)
The moral here is that the whole story of L-values vanishing because of "sign" phenomena is specific to self-dual settings.
Thank you very much for your clear answer. I do have some interest in the case where the character is quadratic. How would I go about computing the epsilon factor as you did? Do you have a reference that could be useful? Thank you.
I would do this by computing the Deligne $\varepsilon$-factor of the associated Weil-Deligne representation at $p$ (which might sound scary, but is actually rather explicit and simple since all the WD reps have a very simple shape).
Alright, thank you for your help, I will try this.
If you get stuck, you might like to consult Ramakrishnan's paper "Modularity of the Rankin--Selberg L-series", where he proves the existence of the Rankin--Selberg lifting $GL_2 \times GL_2 \to GL_4$; he does this using the converse theorem, so the core of the proof is a very careful analysis of the analytic continuation and functional equation of the $GL_2 \times GL_2 \times GL_2$ triple product.
|
2025-03-21T14:48:29.713080
| 2020-01-25T00:25:48 |
351093
|
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|
Stack Exchange
|
A countable set theory providing choice?
Instead of Zermelo set theory $Z$ take $Y$ = $Z$ minus the power set axiom plus
Enumerability: $\forall x(x\neq \emptyset \to\exists f[f:\mathbb{N}\overset{onto}{\frown}x ])$
$\imath$ is the definite description operator so that $\imath$x(x is the number of acting US presidents) = 1. $\mu$ is the least operator, so that 0=$\mu$(+<1) in number theory.
Let M be an infinite set in Y of pairwise disjoint non-empty sets. Let $f$ be a surjection from $\mathbb{N}$ to M. Let $h$ be a surjection from $\mathbb{N}$ to $\bigcup M$. Define:
$\forall i(g(i)\overset{def}{=} \imath x\exists m(\langle m,x\rangle\in h\wedge m=\mu n(\langle n,x\rangle\in h\wedge x\in f(i)))$
I state without giving a proof:
$\mathbf{Theorem} \ G=\{x|\exists i(i\in \mathbb{N}\wedge \langle i,x\rangle\in g\}$ contains precisely one member from each of the members of $M$.
Has this been looked at, and is it otherwise worthwhile pursuing if one as me has reservations towards the power set?
In definition of $x$, maybe $x \in f(i)$ should be out of $\mu$ operator? Maybe lack a parenthesis after $h$, inside $\mu$ operator.
Frankly I found the notation made this question much harder to follow. But let me see if I've understood correctly. Let $M$ be a countable collection of pairwise disjoint sets, which by your axiom has an enumeration $M={M_1, M_2,\dots}$. Likewise, there is an enumeration of $\bigcup M = {x_1, x_2, \dots}$. Your claim is we can choose one element from each $M_i$, simply by choosing the element which occurs first in the enumeration ${x_1, x_2, \dots}$. Is that right?
It is well-known that AC is equivalent to the well-ordering principle over ZF. You just stated a more specific version of the well-ordering principle in ZF - powerset. More generally one could assert that every set is the image of a function on an ordinal.
@Nate Eldrigde I believe what you state is a fair rendering in natural language. Each g(i) is the x in f(i) which occurs first in h; isn't that right?
@MonroeEskew Yes, there are these equivalences of the AC, and I am glad you point to well-ordering as a near version. My main concern here is that Enumerability is a very reasonable alternative to the power set axiom which postulates inscrutable entities, given the LS-theorem and large degree of non-absoluteness. I was surprised that choice may be a bonus. I am also interested in the strength of these theories with Enumerability instead of power set, especially when related to the Reverse Mathematics Program.
@FrodeAlfsonBjørdal Well you can just work in 2nd order arithmetic, where everything is countable. On the other hand, it is a very mainstream position to treat the collection of all real numbers as an object capable of being used in constructions just like anything else.
@MonroeEskew I like to be able to offer new ways of looking at things, and am skeptical towards the existence of a set of all real numbers. Perhaps $ZF-Power+Enumerability$ is sufficient to capture enough of second order number theory to be a foundation for mathematics.
I don’t see how a position can be a foundation for mathematics if it excludes large areas of modern mathematics.
@MonroeEskew There are many points of view on what mathamatics is and should be. Confer the Foundations of mathematics email list for many recent discussions. Soloman Feferman's article A Little Bit Goes a Long Way is good on this. For a long time many have opted for weak systems, also in light of the Reverse Mathematics Program. Skolem, Weyl and others were skeptical towards classical set theory and Fraenkel suggested Second Order Arithmetics as sufficient for Analysis. Nelson opted for a weak predicative system on a par with Robinson's Q. Computable mathematics is at the level of RCA(0).
@FrodeAlfsonBjørdal You can have a philosophical view on what mathematics should be. But what mathematics is is a more factual question. The goal of “providing a foundation for mathematics” is a utilitarian one, and we should be able to check whether your preferred system succeeds at formalizing mathematics as it actually is.
@MonroeEskew What is a philosophical view? There is no such thing as what mathematics really is. All those I listed are influential mathematicians. The utility of mathematics lies in its ability to support computations. I suggest we discuss this further elsewhere, if at all.
@FrodeAlfsonBjørdal I just mean look at the actual practice of mathematics.
@MonroeEskew As I have pointed out and as you may see if you will look, there are many mathematical practices. I am surprised that you seem to think otherwise.
@FrodeAlfsonBjørdal But you are choosing to privilege one kind above others-- those that deny uncountable sets. So ZF-powerset+countability could be a foundation for "a kind of mathematics", not simply a foundation for mathematics--unless you insist on interpreting all the uncountable mathematics as people talking about countable models of ZFC or similar theories.
@MonroeEskew I am not seeking to privilege anyone, and believers in uncountable sets have been privileged for too long. Beyond RCA(0) all things mathematical are becoming unclear. My theory W is probably close to PI(1-2)-comprehension, and perhaps above, which is more than sufficient for science and speculative purposes. In general it is a good precept to be satisfied with as weak mathematical systems as possible, as they are safer.
@MattF. I agree, but only up to a point. Simpson e.g. recently pointed out in lecture: "Our message is that, if mathematicians exercise some care and work within the strictures of the known “weak” systems or other systems which are finitistically reducible, then all is well. The finitistic core is protected from contamination. Hilbert’s program lives!"
I forgot to include Per Martin Löf and Vladimir Voevodsky in my miniscule survey, by the was. Many in the proof theoretic traditions have been skeptical towards uncountability, by the way.
@FrodeAlfsonBjørdal So what is the source of your skepticism? Is it the consistency strength of the powerset axiom, or just some metaphysical qualm? Because your system does not preclude accommodating very high strength assumptions like, "there is a countable transitive model of ZFC + a huge cardinal."
I already adduced LS, and this was also the reason for Skolem's skepticism. Of course, if ZFC + a measurable cardinal has a model then it can be accommodated in my system. There are other problems as with what Feferman called "monsters", e.g. the Banach-Tarski "paradox".
Yet Skolem’s paradox lives on in your system! Your system may countenance countable models that believe they are uncountable. Also your system may countenance models that believe in paradoxical decompositions of the unit ball. How are these things even possible?! I don’t see how you have alleviated the headache.
@MonroeEskew No, the system does not say that there are uncountable infinities! As I stated, if the systems are consistent they have countable models in my system. There is no problem, and systems can have all kinds of countable models if consistent.
@FrodeAlfsonBjørdal I’m not sure you can claim the Skolem paradox as a motivation for your system, when you don’t seem to find it paradoxical at all. The reason it is supposed to be perplexing is that there can be a model of classical set theory which is wrong about countability. I don’t see how you make Banach-Tarski any less perplexing either.
For example: I do not believe that $ZF$ is inconsistent, but I believe that $ZF$ has no standard model.
|
2025-03-21T14:48:29.713734
| 2020-01-25T01:28:19 |
351095
|
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|
Stack Exchange
|
Is a representation of $\operatorname{SL}_n$ defined over a field $k$ if its image is contained in $\operatorname{GL}_n(k)$?
Let $k$ be a subfield of $\mathbb{C}$ and let $f\colon \operatorname{SL}_n(\mathbb{C}) \rightarrow \operatorname{GL}_m(\mathbb{C})$ be an algebraic homomorphism such that $f(\operatorname{SL}_n(k)) \subset \operatorname{GL}_m(k)$. Question: must $f$ be defined over $k$? In other words, are the matrix entries of $f(x)$ polynomials in the entries of $x \in \operatorname{SL}_n(\mathbb{C})$ whose coefficients lie in $k$?
I'd also be interested in the question with $\operatorname{SL}_n$ replaced by other semisimple groups, but the above is the most important special case for what I am doing.
It suffices to prove the following, right? If $p \in \mathbf{C}[x_1,\dotsc,x_d]$ with $p(k^d) \subseteq k$, then $p \in k[x_1,\dotsc,x_d]$. For $d=1$ it is true. I don't know about the general case, though.
@MartinBrandenburg: doesn't the general case then follow from induction on $d$?
@MartinBrandenburg: While that is a true fact about polynomials, it doesn’t imply the question since SL_n is not isomorphic to affine space.
@AndyPutman but $SL_n$ is not too far from being an affine space: the Bruhat decomposition produces a Zariski open subset isomorphic to $\mathbb{A}^{n^2-1}$.
Sure SL_n is not affine space, but isn't it sufficient to consider the polynomials in the entries of $f$?
@MartinBrandenburg - yes, but you don’t know about what you get when you plug arbitrary values into them, just matrices with det 1.
@PiotrAchinger - that’s a good point, and allows an argument in the style of what Martin suggests.
@PiotrAchinger: I would say that the Bruhat decomposition produces a Zariski-open subset in ${\rm SL}_n$ that is isomorphic to ${\Bbb A}^{n^2-n}\times ({\Bbb G}_m)^{n-1}$, which is a Zariski-open subset in ${\Bbb A}^{n^2-1}$.
Yes, and in fact something much more general is true Let $X$ and $Y$ be affine varieties defined over a field $k$. If the $k$-points of $X$ are Zariski dense, $X$ is reduced, and $f: X_{\mathbb C} \to Y_{\mathbb C}$ sends $X(k)$ to $Y(k)$, then $f$ is defined over $k$.
This was inspired by comments of Martin Brandenburg, Jef L, Andy Putman, and Piotr Achinger.
Proof: First note that by embedding $Y$ in affine space, we may assume $Y = \mathbb A^n$. Second note that by viewing a map to $\mathbb A^n$ as an $n$-tuple of maps to $\mathbb A^1$, we may assume $Y =\mathbb A^1$. Thus $f$ is a polynomial function in $\mathbb C[X]$, and we want to check it lies in $k[X]$.
Because $\mathbb C[x]= k[X] \otimes_k \mathbb C$, we may write $f = \sum_{i=1}^n \alpha_i f_i$ where $\alpha_i \in \mathbb C$ and $f_i \in k[X]$. Without loss of generality, we may assume that $\alpha_1=1$ and that the $\alpha_i$ are $k$-linearly independent. (Add $1$ to the list of $\alpha_i$s, then for any linear relation, use that relation to remove whichever $\alpha_i$ is not $1$ and adjust the $f_i$s appropriately.)
Now for $x \in X(k)$, we have $f(x) = \sum_{i=1}^n \alpha_i f_i(x)$. Because the $\alpha_i$ are $k$-linearly independent, and $f(x)\in k$, this implies $f_i(x)=0$ for $i>1$. Then because $X(k)$ is Zariski dense, this implies $f_i=0$ for $i>1$, so $f=f_1 \in k[X]$. QED
The Zariski density can be checked for $SL_n$ using the birational map to affine space obtained by forgetting one entry, and for other semisimple groups using the Bruhat decomposition as suggested by Mikhail Borovoi.
It seems that your proof works without the assumption that $X$ is affine.
Why is $k[X]$ a free vector space over $k$?
Why does every ${\rm Aut}({\Bbb C}/k)$-invariant element of $\Bbb C$ lie in $k$?
@Mikhail Borovoi 1. Every vector space is free. 2. Because $\mathbb C$ is algebraically closed, every automorphism of a sub field extends to an automorphism of the field. Apply this to $k(x)$ if $x$ is transcendental or its Galois closure if not.
@MikhailBorovoi New, slicker, proof avoids using any facts about automorphisms and thus avoids arguing with algebraic closures, transcendence bases, etc.
You write "a field $k$", but it should be "a subfield $k$ of $\mathbb{C}$", right?
At some point do we need the fact that $X$ is reduced (of course $SL(n)$ is, so this would only be relevant for the more general statement). I am thinking of $X= \mathrm{Spec} k[\epsilon]/\epsilon^2$, $Y = \mathbb{A}^1$, and $f(\epsilon) = \alpha \epsilon$, where we could have $\alpha \in \mathbb{C} \setminus k$.
@Sarah: Any connected linear algebraic group $G$ over a field $k$ of characteristic 0 is unirational, and therefore, the set of $k$-points in $G$ is Zariski-dense. See Borel, Linear Algebraic Groups, 2nd ed., Corollary 18.3. By Will Sawin's proof, the answer to you question is Yes for any connected linear algebraic group $G$ over any subfield $k$ of $\Bbb C$.
Replacing $\mathbf{C}$ with any algebraically closed field $C$ containing a given field $k$, this seems to work for $k$ an arbitrary perfect field.
@YCor Good point. I think my automorphism-free proof works even for imperfect fields.
This follows from Theorem 6 in Steinberg's "Some consequences of the elementary relations in $SL_n$" in "Finite groups — coming of age", Contemporary Math. 45 Amer. Math. Soc. (sorry, I could only find a Google books link), together with the remarks at the end of the proof. Restricting $f$ to $\Gamma = SL_n(\mathbb{Z})$, the theorem yields a polynomial map $g$ (which is equal to $f$ for us) and a map $h$ that is trivial for us (since it is trivial on a finite index subgroup of $\Gamma$). As remark (b) mentions, $g$ is defined over $\mathbb{Q}(f(\Gamma))$, which is what you want.
In retrospect, this may have been a bit overpowered, since it does not use the algebraicity assumption.
This is really a special case of Margulis’s superrigidity theorem (though it takes some work to get from superrigidity to this, and possibly that work might require an anawer to the OP’s question). For $\mathbb{Q}$, Steinberg’s theorem was first proved by Bass-Milnor-Serre, and for general fields it was proved by Ragunathan.
(all of these references only give the rational part, but that is the most decisive part — the finite order bit can be gotten from the rational bit in an elementary way).
@AndyPutman Thank you for these very interesting comments!
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2025-03-21T14:48:29.714524
| 2020-01-25T01:30:52 |
351096
|
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|
Stack Exchange
|
multiplicative structure of Ext
Basically, I am trying to compute something with the Adams spectral sequence (as a toy example). The $E^2$ page reduced to computing $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$, where $\Gamma = \mathbb{F}_2 x/ x^2$ and $deg(x) =1$. Keeping very close track of degree, I found that $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$ is only nonzero in degree $s=t$. In this case, we have $\mathbb{F}_2 \{ \phi_t \} $ where $\phi_t \in Hom^t_{\mathbb{F}_2 x/x^2} (\Sigma^t \mathbb{F}_2 x/x^2, \mathbb{F}_2)$ sends $1$ to $1$ and $x$ to $0$. (the superscript $t$ denotes morphisms that lower degree by $t$.)
So I have come to the point that I need to understand the multiplicative structure of $Ext$. I know I am supposed to get that $\phi_1^2 = \phi_2$ (because I know what the spectral sequence is supposed to converge to--the 0th column should have a copy of $\mathbb{F}_2$ in each degree, but all of it generated by $\phi_1$ ). However, I am not really sure how to make sense of $\phi_1^2$. So my questions are
1) how to make sense of $\phi_1^2$? or have I confused things somewhere along the way
2) in order to work with the multiplicative structure in the second page of Adams spectral sequence, I need to understand the multiplicative structure of $Ext$. I would like to work directly with objects in $Ext$, but how does one multiply? and is it easier to do it this way or does one typically use the isomorphism between objects of $Ext$ and extensions, multiply the extensions (which is simple), and then translate back? But the isomorphism for higher Ext does not seem easy to work with.
Note: I specifically am trying to do with without the cobar complex.
Edit: I should add that the resolution I took was $\ldots \Sigma^2 \mathbb{F}_2 x/x^2 \xrightarrow{d} \Sigma \mathbb{F}_2 x/x^2 \xrightarrow{d} \mathbb{F}_2 x/x^2 \to \mathbb{F}_2$, where $d$ sends $\Sigma^{t} 1$ to $\Sigma^{t-1} x$ and everything else to $0$. I'm adding the $\Sigma$ to help keep track of degree.
You can actually pick any projective resolution, so what is probably optimal in this case is to take $... \rightarrow x^2\mathbb{F}_2[x]/x^4 \rightarrow x\mathbb{F}_2[x]/x^3 \rightarrow \mathbb{F}_2[x]/x^2 \rightarrow 0$. Then it boils down to calculating the endomorphism complex of this resolution.
so your $\phi_t$ is a map which sends $x^t \in x^t\mathbb{F}_2[x] / x^{t+2}$ to $1$ in $\mathbb{F}_2$, and we need to somehow find its lift to the map of complexes, which shouldn't be too hard (it exists and unique up to homotopy due to rhs being projective resolution). I presume we can take $\phi_t$ to be the map which sends $\alpha \in x^k \mathbb{F}_2[x]/x^{k+2}$ to the $x^{-t}\alpha \in x^{k-t} \mathbb{F}_2[x]/x^{k-t+2}$ (so basically just shifts the complex by $t$ and maps the terms isomorphically where possible and to zero otherwise).
That allows you to take compositions, and now the fact that $\phi_t \phi_s = \phi_{t+s}$ is clear. so if you want to compose exts, you need to resolve both modules and lift your maps to the maps of resolutions, and it is typically easy to do.
(if my notation is unclear $x^k \mathbb{F}_2 / x^{k+2}$ is $\Sigma^k \mathbb{F}_2[x]/x^2$ in your notation)
So, maybe I should format this as an answer.
Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(Hom(A^{\bullet}, B^{\bullet}))$, where $A^{\bullet}, B^{\bullet}$ are resolutions of objects $A, B$ respectively, and $A^{\bullet}$ needs to be projective, and no requirement on $B^{\bullet}$ but we will take it to be projective too to define composition.
Composition for Exts is then induced by composition $Hom(A^{\bullet}, B^{\bullet}) \otimes Hom(B^{\bullet}, C^{\bullet}) \rightarrow Hom(A^{\bullet}, C^{\bullet})$.
Let us denote $R = \mathbb{F}_2[x]/x^2$. Then, $\mathbb{F}_2$ admits the following resolution: $A^{\bullet} = ... \rightarrow \Sigma^2 R \rightarrow \Sigma R \rightarrow R \rightarrow 0$, the maps being multiplication by $x$.
The hom-complex $Hom(A^{\bullet},A^{\bullet})$ is then quasiisomorphic to $Hom(A^{\bullet}, \mathbb{F}_2)$ (it is always true for any pair of complexes such that left one is a complex of projective modules and the right one is switched to the quasiisomorphic one). Concretely, it means that any closed map $\phi: A^{\bullet} \rightarrow \mathbb{F}_2$ can be lifted to the map $\phi: A^{\bullet} \rightarrow A^{\bullet}$ uniquely up to homotopy. So, the map we lift is your $\phi_t$, and the lift is defined as $\phi_t: A^k \rightarrow A^{k-t} = Id$ for $k \geq t$ and $0$ otherwise.
It is now clear that $\phi_t \phi_s = \phi_{t+s}$
$\blacksquare$
the rule is basically "switch everything to resolutions and then map complexes instead of objects". it works very well if you have a category with enough projective or injective objects
Thank you! In order to determine what the lift of $\phi_t$ is, I need to know what the quasi-isomorphism is, correct? Is it given by composing on the left with the projection $A^{\bullet} \to \mathbb{F}_2$?
Yes, indeed (this is essentialy an unique choice because resolution is always quasiisomorphic to what it resolves, namely, its 0-th cohomology). You should also check another answer, it gives not only a recipe but puts into a broader context of the theory of quadratic algebras and Koszul duality (which I completely forgot about when writing this answer).
The composition product $Ext(N,P) \otimes Ext(M,N) \to Ext(M,P)$ can be computed as follows. Given cocycles $x : N_{s_1} \to \Sigma^{t_1} P$ and $y : M_{s_2} \to \Sigma^{t_2} N$, let $\{ y_s : M_{s+s_2} \to \Sigma^{t_2} N_s\}$ be a chain map lifting $y$. Then $xy$ is represented by the cocycle $x y_{s_1} : M_{s_1+s_2} \to \Sigma^{t_2} N_{s_1} \to \Sigma^{t_1+t_2} P $.
The point is that one needs (a finite bit of) the chain map lifting $y$, but only the cocycle $x$.
Saves effort.
(The $M_s$ and $N_s$ are the terms in projective resolutions of $M$ and $N$, of course.)
In this case, you're dealing with a Koszul algebra ---this is precisely your observation that the spectral sequence is concentrated in the diagonal. Its Koszul dual is a polynomial algebra, and this gives you the multiplicative structure on $\mathsf{Ext}$: each diagonal entry is one dimensional with generator $x_s$, and you have that $x_sx_t=x_{t+s}$, so $\mathsf{Ext}$ is a polynonimal algebra generated by $x_1$.
This can be seen at the level of the (reduced) bar construction $B\Gamma$, whose dual computes your $E_2$-page. It has zero differential, and it is generated in each homological degree $s$ by $x_s = [x\vert\cdots\vert x]$ with $x$ appearing $s$ times. The product is dual to the coproduct of $B\Gamma$, which is simply $\Delta(x_s) = \sum x_i\otimes x_j$.
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2025-03-21T14:48:29.714962
| 2020-01-25T03:55:14 |
351100
|
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Stack Exchange
|
Coronas of Spaces and the Borsuk Shape Category
I was thinking about some topological ideas, especially those relating to shape theory, and came across some interesting constructions that seem to relate to shape theory, but that I can't quite situate within my own knowledge - so was wondering if there is any related ideas out there in math.
Let $\beta$ be the Stone-Cech compactification functor. For any space $S$, let us define a new space $CS$ as follows: Consider the space $\beta(S\times [0,1))$ and the projection $\beta \pi_2:\beta(S\times [0,1))\rightarrow [0,1]$ given by extending the projection to the second coordinate. Define $CS$ to be the fiber of $1$ under $\beta \pi_2$ - note that if $S$ is itself compact Hausdorff, this is just $CS=\beta(S\times [0,1))\setminus S\times [0,1)$. This space somehow "expands" $S$ to be large enough that any map $S\times [0,1)$, when thought of as a homotopy missing its endpoint, will in some sense have a limit.
More precisely, if $Q=\prod_J[-1,1]$ for some indexing set $J$ with projections $\pi_j:Q\rightarrow[-1,1]$ for each $j\in J$ and $B$ is some compact subset of $Q$, one may explicitly describe the maps $CA\rightarrow B$: they are continuous functions $f:S\times [0,1)\rightarrow Q$ such that for any open set $N$ containing $B$, there is some $\alpha$ such that if $t>\alpha$ then $f(a,t)\in N$ under the equivalence relation that $f\sim g$ if for every $j\in J$ and $\varepsilon>0$ there is some $\alpha$ such that if $t>\alpha$ then $\pi_j(f(a,t))-\pi_j(g(a,t))<\varepsilon$. Note that every compact Hausdorff space $B$ can be embedded in some such $Q$.
This characterization seems very similar to the definition of morphisms in Borsuk's Shape Category except that morphisms in that category are taken up to homotopies, whereas these continuous functions $CA\rightarrow B$ are more strictly continuous maps - I haven't worked out all the details, but I think that under an appropriate notion of equivalence (e.g. some sort of homotopy in the corona of $A\times [0,1)\times [0,1]$ using the two embedding of $CA$ into this with last coordinate $0$ and $1$), one can get "morphism" in the shape category as an equivalence class of maps $CA\rightarrow B$.
This leaves me with a natural question: can one use this construction to handle some of the nasty spaces (e.g. the Warsaw circle) that shape theory handles without working in a category defined only up to homotopy? More explicitly, suppose we have some set of reasonably nice spaces (maybe something like compact Hausdorff) - can we define a category whose morphisms from $A$ to $B$ are continuous maps $CA\rightarrow B$? The stumbling block is that there's no obvious composition law to use - but maybe there is some clever composition law, or perhaps $C$ can be given the structure of a comonad via some map $CA\rightarrow CCA$ (although no such maps come to mind!), or maybe the explicit description of the maps $CA\rightarrow B$ can somehow be used (although it seems to become difficult to show well-definedness when one tries to do this).
Is there a way to make a category of "shapes" not up to homotopy along these lines? Is there any literature related to this idea?
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2025-03-21T14:48:29.715209
| 2020-01-25T04:57:07 |
351103
|
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Stack Exchange
|
A question in paper " A note on Odd zeta values " by Tanguy Rivoal and Wadim Zudilin on page 6
I am studying research paper " A note on odd zeta values " by Tanguy Rivoal and Wadim Zudilin .
Note-> This question has been closed 2 times on math.stackexchange . Earlier it was posted on MathOverflow but people here said it should be posted on math stackexchange. But they have closed it 2 times consecutively.
Please help I am badly struck on it.
Notations:
I couldn't think about how authors write the equation - we obtain $\phi_{(n) }^{-1} ( F(t) (t+m)^5 )_{ t=-m} $ $ \in \mathbb{Z} $ for m= 0 ,1,...,n by defininig $\phi_{(n)} = \prod_{2√n<p\leq n } p^{\rho_0(n/p) } $ .
Kindly see last 2 lines of image posted below , I have underlined(in black) the part in which i have question .
I have understood everything in research paper till this argument but I don't have a clue on how authors derives the underlined part.
I think they are using [12, lemma1] (Paper-" One of odd zeta values from $\zeta(5)$ to $\zeta(25) $ is irrational by elementary means " By Wadim Zudilin )whose image I am posting below but unfortunately I am not able to obtain this relation from the lemma.
Image of lemma ->
It would be really helpful for me if anyone can tell me how to deduce this equation $\phi_{(n) }^{-1} ( F(t) (t+m)^5 )_{ t=-m} $ $\in \mathbb{Z}$?
first of all, did you try to ask the authors?
Authors are almost always happy to answer questions from people interested in their work – the whole idea that someone is actually reading their paper is immensely satisfying.
Simulposted to m.se, https://math.stackexchange.com/questions/3521840/an-argument-in-lemma-of-research-paper-irrationallity-of-values-of-riemann-zet without notice to either site. That's an abuse – please don't do that.
@Gerry Myerson questions are different, please don't downvote. This question is different and harder than question asked on stackexchange. I request you to take your downvote back
@Dxdxdade I downvoted because that's also a straightforward question -- just a matter of working out the notations. Please don't post on MO unless you have a serious doubt about an argument in a paper. Please consider using MSE instead. By the way, I don't like how you post the questions. It's better if you rewrite things in usual LaTeX without images. It will also benefit your understanding.
@Dxdxdade By "serious doubt" I mean that you suspect that there is something wrong/incomplete in the argument, and you can substantiate that. So simply "not having any clue" is not enough to ask here. My advice would be that you rewrite the arguments (maybe taking particular cases and introducing your own notation) until you recognise which basic properties or theory you can apply.
You have your answer now. Please don't use the bounty system to evade normal community moderation. As has been pointed out, the question probably won't be considered appropriate for this site, and so if the community wants to close it, they should be allowed to.
If a question is closed on MSE that is not in itself a justification for asking it on here, especially if the people on MSE have the view that the question is unclear or too basic
Not that I care (I use MO less and less), but the bounty for this question disappeared. Funny.
@DamienC see comment of Todd Trimble
You had an answer in January, Arnold, which you accepted. What purpose does it serve to introduce edits to the question now?
The edits you have made to this and other old questions are mostly cosmetic and do not appreciably improve those posts. I don't see how that will get you out of a ban. It might be worth contacting the moderators for advice.
It's very easy (and this is probably why people are telling you that the question is not appropriate for MO).
You have a certain integer $X$, and you'd like to prove that $X$ is divisible by $\Phi_n$.
You know that $v_p(X)\geq\rho_0(n/p)$ for every odd prime $2\sqrt n\leq p\leq n$ (from the "Notations" section), meaning that $X$ is divisible by the $\rho_0(n/p)$-th power of $p$.
Hence you're done.
here $\psi(t) $ = $\frac {\Gamma'(t) } { \Gamma(t) } $ and $\rho_0(t) $ is as defined in original question
|
2025-03-21T14:48:29.715545
| 2020-01-25T08:33:32 |
351108
|
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|
Stack Exchange
|
Constructible étale sheaves on X are étale algebraic spaces over X
I saw the following statement in a paper of Bhatt-Mathew:
Let $X$ be a quasicompact quasiseparated scheme. Then there is an equivalence of categories between constructible étale sheaves (of sets) on $X$ and algebraic spaces $\mathcal{Y}\to X$ over $X$ whose structure map is étale.
I'm familiar with the proof that lcc étale sheaves of sets $F$ are equivalent to finite étale schemes $Y\to X$, which works by showing that constant finite sheaves of sets on $X$ correspond to trivial finite étale surjections over $X$ then using a descent argument (namely, if $F$ is the constant étale sheaf with value $S$ on $X$, it is representable by the étale $X$-scheme $X\times S$. Then the locally constant constructible case follows by unwinding the sheaf data to glue the $U_i \times S_i$ where the $U_i$ form a cover of $X$ such that the restriction of $F$ to $U_i$ is the constant sheaf valued in the set $S_i$.
For the case of an algebraic space $\mathcal{Y}\to X$ étale over $X$, I don't understand why there is a stratification $\{X_i\}$ of $X$ such that the pullback $Y_i:=\mathcal{Y}\times_X X_i \to X_i$ to each stratum is finite étale, and conversely, given a sheaf that is constructible with respect to a particular stratification $\{X_i\}$ of $X$, I don't see why this helps us build an algebraic space from the corresponding family of finite étale schemes $Y_i \to X_i$.
Locally constant constructible sheaves are not finite étale over X. They are just étale schemes of finite presentation. What is true is that any algebraic space has a finite stratification by locally closed subspaces which are genuine schemes. Such stratifications induce stratifications on the base scheme by étaleness. Conversely, for any locally constant constructible sheaf, there is a finite stratification of the base scheme such that the sheaf becomes locally constant over each stratum.
@Denis-CharlesCisinski Thanks! I wonder why finite étale things show up then. Is this just because étale-locally, every finite presentation étale map is finite étale (since étale maps are locally quasifinite)?
If the base scheme is of dimension zero, then all étale coverings of finite presentation are finite. This implies that any étale map of finite presentation is finite over a dense open subset of the base. Therefore, we may find stratifications over which your étale maps becomes finite.
@Denis-CharlesCisinski Sorry, one last question: Once you find a stratification of the base scheme in the converse such that the sheaf becomes locally constant over each stratum, this gives an étale scheme over each stratum. How do you show that these paste back together to an algebraic space?
I am confused by the Bhatt-Mathew statement, can you give a reference? What is the etale algebraic space corresponding to a skyscraper sheaf?
@PiotrAchinger It's mentioned in their paper on the arc topology, but I've provided a reference to the proof in SGA below.
To complete the argument: SGA1 shows descent for the v-topology of Bhatt and Mathew (descent along universal epimophisms). Both étale algebraic spaces and sheaves on the small étale sites form sheaves of categories (stacks) for the v-topplogy. Moreover both stacks take filtered limits of schemes with affine transition maps to filtered colimits of categories. To establish the equivalence, we may assume that the base scheme if affine with separably closed residue fields. My previous comments are then a proof that the equivalence holds.
In case anyone wants to know a reference, I found it now:
SGA4, Exp. IX, Prop. 2.7
Statement (Translated):
Proposition 2.7 Let $X$ be a quasicompact and quasiseparated scheme, and let $F$ be a sheaf of sets. For $F$ to be constructible, it is necessary and sufficient that it be isomorphic to the coequalizer of a pair of morphisms $H\rightrightarrows G$, where $H$ and $G$ are sheaves of sets representable by by étale schemes of finite presentation over $X$.
How does it answer the question?
@PiotrAchinger Sorry, my mistake, it's prop 2.7. Going to fix it now.
Ok, I see now! So e.g. for a closed immersion $i\colon Y\to X$ and $F$ the constant sheaf of two-element sets on $Y$ (representable by two copies of $Y$), $i_* F$ is representable by two copies of $X$ with two copies of $X\setminus Y$ identified...
|
2025-03-21T14:48:29.715848
| 2020-01-25T08:48:55 |
351109
|
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|
Stack Exchange
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Reasons for inapplicability of complete induction to tour expansion
It is known that tour expansion is a rather poor heuristic for generating short Hamilton cycles even in the planar Euclidean case. That comes as a surprise when learning of that for the first time.
The surprise is due to the fact that the heuristic seems to resemble complete induction if one starts with the convex hull of the pointset as the base case and then inserts as the next point the one that incurs the least length-increase of the tour:
base case:
the relative order of the points on the convex hull is the same as in the shortest tour through all points.
induction step:
if the the tour $T_n$ through the first $\left|CH\right|+n$ points is optimal, then the shortest tour $T_{n+1}$ through one additional point is again optimal ($\left|CH\right|$ shall be the number of points on the convex hull and $n\geq 0)$.
Question:
What are the reasons that prove that greedy tour expansion is not a variant of complete induction?
What I am looking for is a proof that clearly indicates, which of the conditions for the applicability of complete induction are violated by greedy tour expansion.
"then inserts as the next point the one that incurs the least length-increase of the tour:" Meaning "inserts between 2 adjacent points in the existing tour" or something else?
@fedja yes, that is the tour expansion heuristic, so that the relative order of points isn't changed by later insertions.
the greedy tour expansion corresponds to th CHCI heuristic described and analyzed in Investigating TSP Heuristics for Location-Based Services
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2025-03-21T14:48:29.716013
| 2020-01-25T11:26:41 |
351112
|
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|
Stack Exchange
|
Poset filtrations
Consider a chain complex $C$ and a poset $P$ so that there is a filtration by subcomplexes $C^p$ of $C$ where $p\in P$ in such a way that $p<q$ implies $C^p \leqslant C^q$.
As a second option, consider the situation when $C$ is $P$-graded. Then one can set $C^p = \{ x : |x| < p\}$ and $\overline{C}^p = \{ x : |x|\leqslant p\}$. These two filtrations are related, but would give different sub-quotients. If the poset $P$ is discrete, there's no problem, however.
Is there any literature on the possibility of producing spectral sequences out of such data, or doing homological algebra with this data?
There is recent work on the homotopical algebra of the simplicial analoque of your situation. Lurie defined a P-stratified space as being a space over the Alexandroff space corresponding to P. Very recently in his thesis,
https://arxiv.org/abs/1908.01366,
Sylvain Douteau has looked at this from the point of view of simplicial sets over the nerve of P. This was for fixed $P$ but he then freed things up in a subsequent article on the arxiv. His abstract is:
"In this article, we construct a cofibrantly generated model structure on the category of spaces stratified over a fixed poset, and show that it is Quillen-equivalent to a category of diagrams of simplicial sets. Then, considering all those model structures together, we construct a cofibrantly generated model structure on the category of all stratified spaces. In both model categories, weak-equivalences are characterized by stratified homotopy groups." (see https://arxiv.org/abs/1911.04921).
There is related material in another recent thesis:
Nand-Lal.A simplicial appraoch to stratified homotopy theory. (https://livrepository.liverpool.ac.uk/3036209/)
These approaches may provide ideas for handling your situation. For instance by converting the chain compolexes into simplicial abelian groups (provided the chain complexes are zero in negative dimensions) using Dold-Kan then adapting Douteau's arguments to get them to be Abelian group objects in a category of presheaves on the category of elements of $Ner(P)$. That should give a good context to study their homological algebra, e.g. by resolving by fibrant objects.
I hope this helps.
This is great, thank you!
I can give you other references on P-stratified spaces, but I think that Douteau and Nand-Lal are a good place to start. Interpretation of the invariants gives a link with infinity categories using the idea of exit paths which dates back to D. Treumann, Exit paths and constructible stacks, Compositio Mathematica, 145, (2009), 1504–1532.
To be completely transparent, what I mostly need is to produce a spectral sequence out of a complex filtered by subcomplexes indexded by a poset. I more-or-less know how to do this, with ideas dating back to Eilenberg and Cartan, but I thought maybe a more modern approach would be nice to cite.
Would thinking of the $P$-filtered complexes as being analogues of a fibre `space' and then using an analogue of the Serre spectral sequence give you what you need? There would be a possible map to the nerve of $P$ and then using Dold-Kan you might be able to get an interpretation of what Eilenberg and Cartan did (which I do not have available to me).
|
2025-03-21T14:48:29.716367
| 2020-01-25T12:55:32 |
351116
|
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|
Stack Exchange
|
About means of positive definite matrices
This question is related to this recent one.
Recall that on the cone $SPD_n$ of $n\times n$ symmetric positive definite matrices, there are various notions of means, which extend the same notions already known for positive scalars: arithmetic, geometric, harmonic. In addition, they satisfy the same order:
$$m_H(A,B)\le m_G(A,B)\le m_A(A,B).$$
I wander whether the following kind of mean is known. I call it the cofactor-mean, denoted $m_C(A,B)$, but shall be happy to adopt a well-established terminology.
Let $A\mapsto \hat A$ denote the cofactor map, and $B\mapsto \check B$ its inverse. For positive definite symmetric matrices, $\hat A=(\det A)A^{-1}$ and $\check B=(\det B)^{\frac1{n-1}}B^{-1}$. The cofactor-mean is
$$m_C(A,B)=\widehat{\frac12(\check A+\check B)}.$$
(Apology: the widehat is not wide enough, it should cover the fraction $\frac12\,$.)
It satisfies the following inequality
$$m_H(A,B)\le m_C(A,B).$$
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2025-03-21T14:48:29.716474
| 2020-01-25T14:03:01 |
351118
|
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|
Stack Exchange
|
Large Fourier submatrices with small operator norm
Consider a finite abelian group $G$ (I'm mostly interested in $\mathbb{Z}_2^n$).
For two subsets $A$ and $B$ of $G$, one can form a submatrix of the Fourier transform matrix on $G$ by keeping only the rows corresponding to $A$ and the columns corresponding to $B$.
Assuming $A$ and $B$ have some fixed cardinality, how small can the operator norm of the submatrix be? Can we find explicit large $A$ and $B$ such that the norm is small?
Any relevant pointer appreciated.
Let's normalise the Fourier matrix ${\mathcal F}$ to have entries of magnitude 1. Then the Fourier submatrix ${\mathcal F}_{A,B}$ corresponding to two subsets $A,B$ has Frobenius norm $(|A| |B|)^{1/2}$ and rank at most $\min(|A|, |B|)$, hence must have operator norm at least $(|A| |B|)^{1/2} / \min(|A|,|B|)^{1/2} = \max(|A|,|B|)^{1/2}$.
This bound is pretty much sharp. For instance, if $A$ is a subgroup of $G$, and $B$ is formed by selecting $k$ elements from each coset of the orthogonal complement $A^\perp$, then $|B| = k |A|$ and a routine application of Plancherel's theorem shows that ${\mathcal F}_{A,B}$ is equal to $|B|^{1/2}$ times an isometry, so the operator norm is exactly $\max(|A|,|B|)^{1/2}$. Even when $G$ has very few subgroups (e.g., $G$ is a cyclic group of prime order), one should be able to concoct similar examples (losing some constants) by using approximate groups in place of groups.
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2025-03-21T14:48:29.716602
| 2020-01-25T14:06:29 |
351120
|
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|
Stack Exchange
|
Is finding positive integer solutions of $\zeta(a/b) = c$ equivalent to deciding the rationality of $\gamma$?
This question requires little bit of explanation of the background hence it is a bit lengthy. Note: The question was initially posted in MSE but did not get answers hence posting in MO.
For every integer $n \ge 2$ there is a unique real $1< x_n <1.73$ such that the Riemann nzeta function $\zeta(x_n) = n$. Example
$$\zeta(1.72865..) = 2, \text{ } \zeta(1.41785..) = 3, \text{ } \zeta(1.29396..) = 4, ...
$$
It is likely that all $x_n$ are irrational but this is an open problem. I have devised a methodology to get some partial results in this direction.
If a rational $x_n > 1$ exists then, WLOG, let $x_n = 1 + \dfrac{a_n}{b_n} = 1 + \dfrac{1}{i_n + f_n}$ where $i_n$ is the integer part of the irreducible fraction $b_n/a_n$ and $f_n$ is its fractional part. From the Stieltjes series expansion of the Riemann zeta function, we obtain
$$
f_2 \approx 0.37241 \le f_n \le f_{n+1}
< 1 - \gamma \approx 0.422785
$$
where $\gamma$ is the Euler-Mascheroni constant. $f_n$ is strictly increasing follows from the fact that $\zeta(x)$ is strictly decreasing on our region of interest $x \ge x_2$.
Thus if $\zeta(1 + a_n/b_n) = n$ then we can write $b_n = ka_n + r_n$ for some integers $k$ and $r_n$ such that $\gcd(r_n,a_n) = 1$ and $0.37241 < \dfrac{r_n}{a_n} < 0.422789$.
Now for each $a_n \ge 1$ we can eliminate the possible values of $r_n$ by computer verification to increase the lower bound on $a_n$ as shown below.
Example: Proving $a_n \ge 6$ for $n \ge 2$.
For $a_n = 1,2,3,4$, there is no $r_n$ satisfying $0.37241 < \dfrac{r_n}{a_n} < 0.422789$ so these values of $a_n$ are eliminated. For $a_n = 5$ there is one possible value, $r_n = 2$ which satisfies both $\gcd(r_n,a_n) = 1$ and $0.37241 < \dfrac{r_n}{a_n} < 0.422789$. So we take $r_n = 2$ for further test. We observe that
$$f_3 \approx 0.3932265 < \frac{2}{5} < f_4 \approx 0.4018059$$
Since $\dfrac{2}{5}$ lies between two consecutive values of $f_3$ and $f_4$ is strictly increasing, we will never have a situation where $f_n = \dfrac{2}{5}$ for any $n$. Hence $r_n = 2$ is ruled out for $a_n = 5$. The integers $3$ and $4$ are witnesses for the elimination of $\dfrac{2}{5}$. With this we have exhausted all possibilities for $a_n = 5$ and so we can conclude if $\zeta(x_n)\in N$ then $a_n \ge 6$. Similarly we can eliminate $a_n = 6,7,\ldots$ and so on.
Note: As $a_n$ increases the number of possible $r_n$ increases. Each $r_n$ must be eliminated individual with witness to completely eliminate $a_n$. So after a point, $a_n$ will have multiple witnesses.
Definition: A witness for a given integer pair $(a,r), a > r$ is an integer $w_{a,r}$ such that $f_{w_{a,r}} < \dfrac{r}{a} <f_{1+w_{a,r}}$
The existence of a witness $w_{a,r}$ is therefore sufficient to prove that $\zeta\Big(1 +\dfrac{a}{x+r}\Big)$ is not integer for any integer $x > a > r$. By actual computation, I found the witness for every permissible pair $(a,r), a \le 10^4$ which implies that
If $x_n > 1$ is rational and $\zeta(x_n) \in N$ then the difference between
the numerator and denominator of $x_n$ is greater than $10^4$.
Size of the witness
I noticed that most integers have small witnesses but a few of them have large witnesses. I looked at the list of integers $a$ whose largest witness for some $r$ is larger than all the witness of all the integers less than $a$. Let us call these the maximal witnesses. Given below are the first few values of the maximal witnesses:
a Max of w_{a,r}
5 3
12 12
19 42
45 130
30 292
97 701
123 3621
518 16503
913 31689
1308 49858
1703 71984
2098 99521
2493 134724
2888 181317
3283 245888
3678 341249
4073 496221
4468 790559
4863 1598102
5258 11274164
Some of the values $a = 5,19,123,5258$ are the non-successive denominators in the continued fraction expansion of $\gamma$ but in general each of these values of $a$ are the denominators in increasing order for the best rational approximation of $\gamma$. In other words, the closer is $\dfrac{r}{a}$ to $1-\gamma$, the larger is its witness $w_{a,r}$ which makes computationally eliminating $a$ more difficult.
This finally bring me to my question.
Question: Is the problem of finding positive integer solutions $\zeta(p/q) = u$ equivalent to deciding the rationality of $\gamma$.
Related question
|
2025-03-21T14:48:29.716906
| 2020-01-25T14:20:34 |
351121
|
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|
Stack Exchange
|
Variants of Nicholson's inequalities for prime numbers, involving the Lambert $W$ function
The purpose of this post is ask about two closely related/inspired conjectures from inequalities due to Nicholson (see [1]) and Visser [2]. If my reasonings are right should be stronger versions of Nicholson's inequalities that you can see from the linked Wikipedia. My doubt is if there are some chances to disprove the following conjectures. The Wikipedia related to the Lambert $W$ function is this, I add it as reference to recall the definition and asymptotics. Here, we are denoting the $k$-th prime number as $p_k$, for integers $k\geq 1$.
Conjecture 1. For all integer $n\geq 5$ the inequality $$\left(\frac{p_{n+1}}{p_{n}}\right)^n<n W(n)$$
holds, where $W(x)$ is the main/principal branch of the Lambert $W$ function.
Conjecture 2. For all integer $n\geq 1$ the inequality
$$\left(\frac{\log p_{n+1}}{\log p_{n}}\right)^n<e^{-\gamma}\frac{e}{\Omega}$$
holds, where $\Omega=W(1)$ is the omega constant, thus here also $W(x)$ is the main/principal branch of the Lambert $W$ function, and $\gamma$ is the Euler-Mascheroni constant (as remark here, we add that $\frac{e^{-\gamma}}{\Omega}=\frac{e^{\Omega}}{e^{\gamma}}<1$).
Question. I would like to know if using the theory about prime numbers or prime gaps we can prove that some of previous conjectures (Conjecture 1 or Conjecture 2) is false/wrong. Many thanks.
I tried to write these in some way that these conjectures in the post harmonize well. I wrote some scripts in Pari/GP that can be evaluated in the web Sage Cell Server, feel free to ask about it in comments.
References:
[1] John W. Nicholson, see the section Comments for the sequence A182514
in The On-Line Encyclopedia of Integer Sequences (2013), or the Wikipedia Firoozbakht's conjecture.
[2] Matt Visser, Primes and the Lambert $W$ function, Cornell University Library (November 2013).
I know that the article [2] is available also as the preprint on arXiv with identificator 1311.2324
Many thanks for the upvote.
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2025-03-21T14:48:29.717092
| 2020-01-25T14:51:28 |
351122
|
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Stack Exchange
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Are open $\mathbb{G}_m$-invariant subschemes of an affine scheme precisely the homogeneous radical ideals of its coordinate ring?
This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: math.stackexchange. However it doesn't seem to get much attention there and I thought I would try it here. My questions are regarding actions of a group scheme $G$ on a scheme $X$. I'm fine with assuming $G$ affine.
$\newcommand{\IG}{\mathbb{G}}
\newcommand{\pmo}{{\pm 1}}
\newcommand{\IZ}{\mathbb{Z}}
\newcommand{\Spec}{\mathrm{Spec}}
\newcommand{\tensor}{\otimes}
\newcommand{\into}{\hookrightarrow}
\newcommand{\iso}{\cong}
\newcommand{\onto}{\twoheadrightarrow}
\newcommand{\sheaf}{\mathcal}
\newcommand{\inv}{{-1}}$
Originally, I was thinking about actions of the multiplicative group $\IG_m = \Spec(\IZ[T^\pmo])$ on affine schemes $X = \Spec(R)$. The category of affine schemes with a $\IG_m$-action is equivalent to the category of $\IZ$-graded rings as follows:
To a $\IZ$-graded ring $R$ associate the action $\IG_m \times \Spec(R) \to \Spec(R)$ which is given by the map of rings $R \to \IZ[T^\pmo] \tensor R$, $f = \sum f_d \mapsto \sum f_d \tensor T^d$, where $f_d$ are the homogenous components of $f$. Conversely, if the action $\IG_m \times \Spec(R) \to \Spec(R)$ is given, call $f_d \in R$ homogenous of degree $d$ if it is sent to a homogenous element of rank $d$ by the morphism $R \to \IZ[T^\pmo] \tensor R \cong R[T^\pmo]$, where we regard the latter ring as graded by the powers of $T$.
If $Y \subseteq X$ is a closed subscheme, I call $Y$ invariant under the action if the morphism $\IG_m \times Y \into \IG_m \times X \to X$ factors over $Y \into X$.
Proposition: The correspondence
$$\{\text{Closed subschemes of }X\} \iso \{\text{Ideals of }R\}$$
restricts to a bijective correspondence
$$\{\IG_m\text{-invariant closed subschemes of }X\} \iso \{\text{Homogeneous ideals of }R\}.$$
For example if $Y$ is $\IG_m$-invariant, then by definition the given action restricts to an action of $\IG_m$ on $Y$ and the inclusion $Y \into X$ becomes a morphism of affine schemes with a $\IG_m$-action. Hence the surjective map $R=\sheaf{O}(X) \onto \sheaf{O}(Y)$ is a morphism of graded rings. But its kernel is $I$ which is, hence, homogeneous.
I was trying to prove the corresponding result for open subschemes:
Conjecture 1: The correspondence
$$\{\text{Open subschemes of }X\} \iso \{\text{Radical ideals of }R\}$$
restricts to a bijective correspondence
$$\{\IG_m\text{-invariant open subschemes of }X\} \iso \{\text{Homogeneous radical ideals of }R\}.$$
It is easy to see that a homogeneous radical ideal defines an invariant open subscheme. First note that the union of invariant open subschemes is again invariant. If the open subschemes $U_j$ are defined by $I_j$, then $\bigcup U_j$ is defined by $\sqrt{\sum I_j}$. Hence it suffices to consider $U = D(f)$ for $f$ a homogeneous element. But then it is easy to see that $U$ is invariant. In fact, the restricted action $\IG_m \times U \to U$ makes $U = \Spec(R[f^\inv])$ into an affine schemes with a $\IG_m$-action. The associated grading on $R[f^\inv]$ is the natural grading that you would expect on a localisation at a homogeneous element.
However I cannot manage to prove the converse, i.e. if $I$ is a radical ideal of $R$ such that the open subscheme $X_I \into X$ defined by $I$ is $\IG_m$-invariant then $I$ is homogeneous. By the proposition it would suffice to prove
Conjecture 2: If $G$ is an (affine) group scheme acting on a scheme $X$ then the bijection
$$\{\text{Open subschemes of }X\} \iso \{\text{Reduced closed subschemes of }X\}$$
given by “reduced closed complement” and “open complement” restricts to a bijection
$$\{G\text{-invariant open subschemes of }X\} \iso \{G\text{-invariant reduced closed subschemes of }X\}.$$
This certainly sounds reasonable if one thinks for example of an action of a topological group on a topological space, where the complement of an invariant subset is clearly invariant again. However I cannot prove it in the context of schemes and I'm not quite sure that it is correct. If $X = \Spec(R)$ is reduced and of finite type over an algebraically closed field and $G = \IG_m$ (so that $\IG_m \times X$ is again reduced and of finite type), then it suffices to consider $k$-valued points, so that the reduced complement of an invariant open subscheme is indeed invariant again.
If $G = \IG_m$ and $X$ is affine then this conjecture is equivalent to the first one.
My conjecture is also equivalent to the claim that if $U \subseteq X$ is an open $\IG_m$-invariant subscheme then $U$ is a union of subschemes of the form $D(f) \subseteq X$ where $f$ in $R$ is a homogeneous element. In particular, this would imply (a special case of) the following conjecture:
Conjecture 3: If $G$ is an (affine) group scheme acting on a scheme $X$ and if $U \subseteq X$ is an open $G$-invariant subscheme then $U$ is a union of $G$-invariant affine open subschemes.
Unfortunately I really don't have a good intuition for actions of groups schemes and I would be glad about some clarification.
In Conj. 3 you mean that $U$ is such a union?
@MartinBrandenburg Yes, sure.
Your definition
$\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X \to X$ factors over $Y \hookrightarrow X$
can be simplified a bit. A better one for our purposes is
The two subschemes of $\mathbb G_m \times X$ defined as the inverse image of $Y$ under the right projection $\mathbb G_m \times X \to X$ and the action map $\mathbb G_m \times X \to X$ are equal.
To see they are equivalent, note that the universal property of a fiber product implies your factorization is equivalent to the claim that the immersion $\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X $ factors through the map from $\left(\mathbb G_m \times X \right) \times_X Y $ to $\mathbb G_m \times X $. That fiber product is the inverse image of $Y$ under the action map, and $\mathbb G_m \times Y$ is the inverse image of $Y$ under the right projection. So your factorization is equivalent to $p^* Y \subseteq a^* Y$, where $p$ is projection and $a$ is action. But $p^* Y \subseteq a^* Y$ is equivalent to $a^* Y \subseteq p^* Y$, and therefore equivalent to $p^* Y =a^* Y$, because we can swap the projection and action maps using the automorphism of $\mathbb G_m \times X$ that sends $(g,x)$ to $(g^{-1}, gx)$.
Now this one is equivalent for open subschemes and their closed complements because the operation of taking closed complement is compatible with pullback under smooth morphisms.
Where does this use the special case $\mathbb{G}_m$? Angelo wrote that it is false for general group schemes.
@MartinBrandenburg Then the maps $\mathbb G_m \times X\to X$ are not smooth, and in particular the inverse image of a reduced closed subscheme can clearly be non-reduced. I think this approach does work for general smooth group schemes.
Your conjecture 2 is false if you don't assume that $G$ is reduced (in positive characteristic there are affine group schemes that are not reduced).
As to conjecture 3, it is hopelessly wrong (think of the action of $\mathrm{GL}_n$ on $\mathbb P^{n-1}$). The only non-trivial case I know is Sumihiro's theorem: a normal algebraic variety with an action of a torus can be covered by invariant affine open subsets. Being normal is essential: consider the standard action of $\mathbb{G}_{\mathrm m}$ on $\mathbb P^1$, and glue together the origin and the point at infinity.
Intermediate cases
While the smooth group case has been affirmed by Will Sawin and the most general case has been refuted by Angelo, there is quite some space in between where we can still find more affirmative cases.
1 A Simple Example
You may skip this example and read the further sections if you are only interested in the results. Otherwise, it serves two purposes:
A more gentle start than abstract schemes (after all, it is not quite a reserach-level question and I'd like to make some allowances for a broader range of readers accordingly)
A motivating example showing some (but not all) interesting behaviour investigated in general
1.1 Algebraic Setting Consider the ring $R = \mathbb{F}_2[T]/(T^2-1)$ as a $\mathbb{Z}/2\mathbb{Z}$-graded ring with the grading $\mathrm{deg}(T) = 1 + 2\mathbb{Z}$, inducing the ring homomorphism
$$\begin{split}\alpha\colon R &\to R[S]/(S^2-1)\\T &\mapsto TS.\end{split}$$.
1.2 Geometric Setting Consider the group scheme
$$G = \mathrm{Spec}(\mathbb{Z}[S]/(S^2-1)) = \mathrm{Spec}(\mathbb{Z}[\mathbb{Z}/2]),$$ the affine scheme $X = \mathrm{Spec}(R)$ and the group action $$G \times X = \mathrm{Spec}(R[S]/(S^2-1)) \to \mathrm{Spec}(R) = X$$ given by $\alpha$.
1.3 Classical Subobjects When working in the unmodified setting of the question, a somewhat unclean picture emerges:
Open subschemes of $X$:
$\emptyset$ (invariant)
$X$ (invariant)
Closed subschemes of $X$:
$X$ (invariant, non-reduced)
$\emptyset$ (invariant, reduced)
$\mathrm{Spec}(\mathbb{F}_2)$ (non-invariant, reduced)
Ideals of $R$:
$(0)$ (homogeneous, non-radical)
$(1)$ (homogeneous, radical)
$(T+1)$ (non-homogeneous, radical)
So far, no real surprises. There are two ideals (resp. closed subschemes) corresponding to the empty open subscheme, but neither of them is homogeneous (resp. invariant) and radical (resp. reduced). However, as indicated by my choice of numbering, there still appears to be a nice correspondence between invariant open subschemes, invariant closed subschemes and homogeneous ideals, except that for a perfect correspondence we'd like to ignore the existence of the nilpotent non-homogeneous element $X+1$ so that $(0)$ would be a radical. So, we'll do just that.
1.4 Better Subojects When the closed subscheme is only required to be as reduced as an invariant subscheme can be and the ideal is considered to have the radical property only with respect to the homogeneous elements, we instead obtain the following picture:
Open subschemes of $X$:
$\emptyset$ (invariant)
$X$ (invariant)
Closed subschemes of $X$:
$X$ (invariant, maximally reduced)
$\emptyset$ (invariant, maximally reduced)
$\mathrm{Spec}(\mathbb{F}_2)$ (non-invariant, whatever)
Ideals of $R$:
$(0)$ (homogeneous, homogeneously radical)
$(1)$ (homogeneous, homogeneously radical)
$(T+1)$ (non-homogeneous, whatever)
2 Diagonalizable and Affine
Consider the diagonalizable group schemes $\mathrm{D}(M) = \mathrm{Spec}(\mathbb{Z}[M])$ for Abelian groups $M$ (in particular, $\mathbb{G}_{\mathrm{m}} = \mathrm{D}(\mathbb{Z})$). Then, a $\mathrm{D}(M)$-action on an affine scheme $X$ is the same as an $M$-graded ring. Restricting ourselves to these, i.e. diagonalizable groups and affine schemes, your first two conjectures are nearly correct and the last one may even be strengthened slightly.
2.1 Definition A homogeneous ideal is called homogeneously radical iff it contains all homogeneous elements of its radical. An invariant closed subscheme $C$ is called maximally reduced if any of its invariant closed subschemes on the same points is already identical to $C$.
2.2 Lemma When a diagonalizable group acts on an affine scheme, there is a bijective three-way correspondence working exactly like you'd expect between
homogeneously radical homogeneous ideals,
maximally reduced invariant closed subschemes and
invariant open subschemes.
Proof. The equivalence of the first two points is immediate from the correspondence between homogeneous ideals and invariant closed subschemes. The equivalence of the latter two points is a special case of the more general treatment in the next section.
2.3 Corollary From this, we get the following alterations of your three conjectures:
2.4 Fixed Conjecture 1 The assignment $$\{\textrm{ideals of \(R\)}\} \to \{\textrm{open subschemes of \(X\)}\}$$ restricts to a bijective correspondence $$\{\textrm{homogeneously radical homogeneous ideals of \(R\)}\} \cong \{\textrm{invariant open subschemes of \(X\)}\}.$$
2.5 Fixed Conjecture 2 The open-complement assignment $$\{\textrm{closed subschemes of \(X\)}\} \to \{\textrm{open subschemes of \(X\)}\}$$ restricts to a bijective correspondence $$\{\textrm{maximally reduced invariant closed subschemes of \(X\)}\} \cong \{\textrm{invariant open subschemes of \(X\)}\}.$$
2.6 Fixed Conjecture 3 Any invariant open subscheme is the union of invariant principal opens (and thus in particular of invariant affine open subschemes).
3 General Case
I have not yet thought about the adaptations which would enable Conjecture 1 to be considered for general schemes and Conjecture 3 has been refuted by Angelo. However, in light of the diagonalizable and affine case it seems straightforward to salvage Conjecture 2. Throughout this section, let $G$ be a group scheme acting on a scheme $X$ via $\alpha\colon G \times X \to X$.
3.1 Lemma When a quasi-compact flat group scheme acts on a scheme, the “open complement” assigment restricts to a bijection of
maximally reduced invariant closed subschemes and
invariant open subschemes.
Proof. The remainder of this section.
3.2 Definition Given a closed subscheme $C$ of $X$, its invariant hull is the scheme-theoretic image of $\alpha^*C$ under the projection $\pi\colon G \times X \to X$.
3.3 Lemma Let $C \subseteq X$ be a closed subscheme and $C'$ its invariant hull. Then $C \subseteq C'$.
Proof. Using the unit of $G$, the inclusion of $C$ into $X$ factors through both the projection and the action with a single morphism $f\colon C \to G \times X$. Consequently, seen as $\alpha\circ f$ the inclusion of $C$ into $X$ factors through $\alpha^*C$ and seen as $\pi\circ f$ it factors through the image $C'$ of $\alpha^*C$ under $\pi$.
3.4 Lemma Let $C \subseteq X$ be a closed subscheme, $C'$ its invariant hull and $D \subseteq X$ a $G$-invariant closed subscheme containing $C$. Then $C' \subseteq D$.
Proof. We have $\alpha^*C \subseteq \alpha^*D = \pi^*D$ by monotonicity and invariance, yielding the result.
3.5 Lemma Let $C \subseteq X$ be a closed subscheme and $C'$ its invariant hull. If $G$ is flat and qc, $C'$ is invariant.
Proof. Since scheme-theoretic images of qc morphisms commute with flat base change, $\pi^*C'$ is the image of $D = G \times \alpha^*C \subseteq G \times G \times C$ under the projection $\pi_{\blacktriangle\!\triangledown\!\blacktriangle}$ eliding the middle component and $\alpha^*C'$ is the image of $$E = {\underbrace{\langle\pi_{\vartriangle\!\blacktriangledown\!\vartriangle}, \alpha\pi_{\blacktriangle\!\triangledown\!\blacktriangle}\rangle}_g}^*\alpha^*C$$ under $\pi_{\blacktriangle\!\triangledown\!\blacktriangle}$. As $$f = \langle\pi_{\blacktriangle\!\triangledown\!\vartriangle}, \mu\langle\pi_{\vartriangle\!\blacktriangledown\!\vartriangle}, \iota\pi_{\blacktriangle\!\triangledown\!\vartriangle}\rangle, \pi_{\vartriangle\!\triangledown\!\blacktriangle}\rangle$$ restricts to an isomorphism $D \to E$ over $G \times X$, they agree. (This boils down to a calculation that $$\alpha\pi_{\blacktriangle\!\triangledown\!\vartriangle} = \alpha gf$$ and thus $$D = \pi_{\blacktriangle\!\triangledown\!\vartriangle}^*\alpha^*C = (\alpha\pi_{\blacktriangle\!\triangledown\!\vartriangle})^*C = (\alpha gf)^*C = f^*E,$$ which is tedious in this formalism but obvious when spelled out on points.)
3.6 Lemma Let $C \subseteq X$ be a closed subscheme, $C'$ its invariant hull and $U \subseteq X$ a $G$-invariant open subscheme not meeting $C$. If $G$ is quasi-compact, then $U$ doesn't meet $C'$, either.
Proof. We show the contrapositive. As $G$ is qc, so is $\pi$, meaning that by [02JQ] for any field-valued point $p$ in $C'$ there is a point $q$ of $\alpha^*C$ such that $p$ is a specialization of $\pi(q)$. We note that if $p$ lies in both $C'$ and $U$ then so does $\pi(q)$, meaning $q$ lies in both $\alpha^*C$ and $\pi^*U = \alpha^*U$. Therefore, $\alpha(q)$ is a common point of $C$ and $U$.
3.7 Corollary The claim of 3.1 Lemma holds.
Proof. Given an invariant open subscheme, consider its reduced closed complement $C$ and its invariant hull $C'$. Now, $C'$ is the maximally reduced (3.4 Lemma) invariant (3.5 Lemma) closed subscheme on the same support as $C$ (3.3 Lemma and 3.6 Lemma), providing the inverse assignment.
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2025-03-21T14:48:29.718038
| 2020-01-25T14:54:55 |
351123
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Stack Exchange
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A diagonal generating function for Fibonacci: Part II
In my earlier MO question, I mentioned although we have for the Fibonacci numbers that
$$F_n=[x^n]\left(\frac1{1-x-x^2}\right),$$
is there a function $F(x)$ such that $F_n=[x^n]\left(F(x)\right)^n$?
Several people offered their responses and I thank Henri Cohen, AccidentalFourierTransform, Fedor Petrov, Richard Stanley, and Ira Gessel, and Aaron Meyerowitz for that. The general agreement is that there seems to be "no closed formula" or "explicit formula".
Notation. Let $[x^ny^n]G(x,y)$ be the coefficient of $x^ny^n$ in the Taylor series of $G(x,y)$.
This time, I would like to increase the number of variables and ask:
QUESTION. Is this true? There is an explicit generating function such that
$$F_n=[x^ny^n](G(x,y))^n.$$
Define $G(x,y) := F(xy)$?
I will add "explicit".
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2025-03-21T14:48:29.718247
| 2020-01-25T15:04:53 |
351124
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Stack Exchange
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Limited sum for whole sum approximation
Let $d_n, n\in\{1,2,\cdots,N\}$ be $N$ realizations drawn independent and identically from uniform distribution on $(0,L)$ where $L=\gamma\sqrt{N}$ with constant $\gamma$. Suppose that we need to approximate the sum
$$\alpha=\sum_{n=1}^{N}d_n^{-3},$$
with the restricted sum
$$\hat{\alpha}=\sum_{n\in\mathcal I}d_n^{-3},$$
where $\mathcal I\subset N$. We define the set $\mathcal I$ as the largest $|\mathcal I|$ element of random variables $d_1^{-3},d_2^{-3},\cdots,d_N^{-3}$. Now, the question is to find the order of size of subset $\mathcal{I}$ which produce a good approximation ($\hat{\alpha}\simeq\alpha$), i.e.,
$$\mathbb{P}[|\alpha-\hat{\alpha}|\leq\epsilon]\geq 1 -\beta.$$
Is $|\mathcal I|$ in order of sub-linear respect to $N$?
If you miss even a single realization you may have an error as large as $1/\epsilon^3$ with probabiity $\epsilon/L$ that vanishes only linearly in $\epsilon$, so this will never provide a good approximation.
But, this occurs with a very small probability. By choosing $|\mathcal{I}|$ largest value of $d_n^{-3}$, it seems that by properly choosing $|\mathcal{I}|$ we have a good approximation.
But the probability that all of $N$ points be in $(0,\epsilon)$ is $(\epsilon/L)^N$ which is so small.
For example, assume that $N$ points are $L/N,2L/N,\cdots,L$. Then, the question is that is it possible to truncate the series $\sum_{i=1}^{N}i^-3$ to have a good approximation?
It is computationally efficient for me to choose only a small fraction of them.
Did you perhaps mean the relative error? Or perhaps $L$ varies with $N$? Otherwise each $d_n^{-3}$ is greater than $L^{-3}$, so the error is at least $L^{-3} (N - |\mathcal{I}|)$. Thus, the complement of $\mathcal{I}$ necessarily has bounded size if $\mathbb{P}[|\alpha-\hat\alpha|\leqslant \epsilon]\geqslant 1 - \beta$ and $N \to \infty$.
You are correct. $L$ is of order of $\sqrt{N}$.
(Edited after noticing an error which completely changes the answer.)
This is not a complete solution; however, it strongly suggests that the answer is positive.
Let $U$ be a random variable with uniform distribution on $[0, 1]$. The random variable $U^{-3}$ has tail $\mathbb{P}[U^{-3} > x] = x^{-1/3}$, and hence it is in the domain of attraction of a stable law with index $\alpha = \tfrac{1}{3}$.
Observe that $d_n = L U_n = \gamma \sqrt{N} U_n$ for an i.i.d. sequence $U_n$ with uniform distribution on $[0, 1]$. By the invariance principle, the processes
$$ X^N_t = \frac{1}{N^3} \sum_{n = 1}^{\lfloor N t \rfloor} U_n^{-3} $$
converge (in the appropriate Skorokhod topology) to the increasing $\tfrac{1}{3}$-stable Lévy process (i.e. the $\tfrac{1}{3}$-stable subordinator) $X_t$. Clearly,
$$ \sum_{n = 1}^{\lfloor N t \rfloor} d_n^{-3} = \frac{N^{3/2} X^N_t}{\gamma^3} \, . $$
Denote $J = |\mathcal{I}|$, and let $\hat\alpha_N$ be the sum of $J$ largest variables among $d_n^{-3}$, $n = 1, 2, \ldots, N$. Then
$$ \hat\alpha_N \approx \frac{N^{3/2}}{\gamma} \times (\text{sum of $J$ largest jumps of $X_t$, $t \in [0, 1]$}) . $$
(I am not rigorous here. However, I am rather convinced one can turn this into a completely argument.)
The question thus turns into the following problem, with $\alpha = \tfrac{1}{3}$ and $\delta = \tfrac{3}{2}$:
How many largest jumps of the $\alpha$-stable subordinator $X_t$, $t \in [0, 1]$, one has to add in order to get an approximation which is within $\pm\gamma N^{-\delta} \times \epsilon$ from $X_1$ with probability $1 - \beta$.
This has been studied a lot: it is the question how fast does the Ferguson–Klass–LePage series of a stable subordinator converge. In particular, since the $n$-th largest jump of $X_t$ is comparable with $n^{-1/\alpha}$, the error should be of the order $J^{1 - 1/\alpha}$ when $J$ largest jumps are taken into account. This suggests that $J \approx N^{\delta / (1/\alpha - 1)}$ should do the job. In our case, this gives $J \approx N^{(3/2) / (3 - 1)} = N^{3/4}$, and thus it suggests that it is sufficient to take roughly $|\mathcal{I}| = N^{3/4}$ largest values of $d_n^{-3}$.
I bet one can find a reference for what is written above (and in fact at least some authors do work with Pareto-distributed jumps $U_n^{-1/\alpha}$ rather than the ordered jumps of $X_t$). I am not an expert in this area, though, and I failed to find a reference in a (very) quick Internet search. The closest one that I encountered is:
Bentkus, V., Juozulynas, A. & Paulauskas, V. Lévy–LePage Series Representation of Stable Vectors: Convergence in Variation. Journal of Theoretical Probability 14, 949–978 (2001)
One may also search in the standard reference for simulation of stable random variables:
Janicki, A., and Weron, A. (1994). Simulation and Chaotic Behaviour of $\alpha$-stable Stochastic Processes. Marcel Dekker, New York
If we first choose $d_n=\frac{L}{nN}$ and include the $M$ smallest $d_n$'s in the restricted sum, then the relative error is
$$E=\frac{\sum_{n=M+1}^N 1/n^3}{\sum_{n=1}^N 1/n^3}=\frac{\psi ^{(2)}(M+1)}{\psi ^{(2)}(1)}+{\cal O}(N^{-2}).$$
This ratio of polygamma functions amounts to less than a 1% error for $M=6$, independent of $N$.
For a statistical test I compared $N=50$ and $N=500$ at a fixed $M=10$: shown below are the two histograms of the cumulative distribution of the error $E$, obtained from $10^3$ realizations of the set of random variables $d_1,d_2,\ldots d_N$. As you can see, the two histograms ($N=50$ on the left, $N=500$ on the right) are nearly the same, with $E<1\%$ happening with probability 0.9, so you can keep $M$ fixed as you scale up $N$.
Thank you so much. I only said $nL/N$ as an example. Maybe, it is better to write the problem in probabilistic way. I mean that $\mathbb{P}(|\alpha-\bar{\alpha}|\leq\beta)$.
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2025-03-21T14:48:29.718654
| 2020-01-25T15:40:02 |
351125
|
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|
Stack Exchange
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Packing circles with radii 1, 2, 3, ..., n in a rectangle
For each positive integer n, let $a_n$ be the area of the smallest rectangle whose area is a whole number, and inside which it is possible to pack all n circles of radii 1, 2, 3, ..., n respectively (with no overlaps).
Is it possible to determine $a_n$ precisely?
For example $a_{12}$ is at most 2466 (https://puzzling.stackexchange.com/questions/92949/my-mothers-dish-collection), and can perhaps be proved to be precisely that.
It's certainly not precisely that. One of the comments links a configuration in a square of area ~2518.16. There has been a lot of work into packing circles into squares, with very few bounds actually proven optimal.
@Wojowu You are right. Have edited accordingly.
On the same website (last updated in 2015) are the best known results up to N = 72.
Here's a better solution for $n=12$, with area approximately 2496:
Even better, with area approximately 2463:
Here's @MattF's suggestion, but it's worse in both dimensions:
@GerhardPaseman, if I consider only circles 6 through 12, this is the best solution I have found:
I added your suggested modification, but I guess the 4 is bigger than it looks.
Suppose you ignore plates one through five temporarily. How small a rectangle do you get for the other seven? (and does it involve a four grouping of 9 12 10 11 with 8 6 7 in a column?) Gerhard "Looking For A Dish Pattern" Paseman, 2020.02.01.
Thanks. I was expecting something slightly different, which would lead to an easy insertion of the remaining disks. However, I am not seeing how to improve the arrangement you provided. Gerhard "Now Looking For Place Settings" Paseman, 2020.02.01.
Also, this suggests improving lower bounds by packing the largest plates first. Maybe you can provide some suggestions for those? Gerhard "Save Bigger Parties For Later" Paseman, 2020.02.01.
None of my solutions provide lower bounds. This problem is highly nonconvex, and the solver yields optimality proofs only for convex problems.
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2025-03-21T14:48:29.718825
| 2020-01-25T17:13:16 |
351132
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|
Stack Exchange
|
Table of products for Lie algebra inner product of roots and weights
For a simple Lie algebra $\frak{g}$, it is usual to scale the inner product so that the shortest simple root has length $2$. With this conventions, where can I find a table (online) of the following values for a general simple $\frak{g}$:
i) $(\alpha_i,\alpha_j)$, for all $i,j=1,\dots, r:=\mathrm{rank}(\frak{g})$
ii) $(\alpha_i,\pi_j)$, for $\pi_j$ a fundamental weight
iii) $(\pi_i,\pi_j)$?
For individual simple types, look at the planches at the end of Bourbaki, Chap; 4-6 (or English translation published by Springer). Google Scholar permits an online search.
This information is available using the Atlas of Lie Groups and Representations software http://www.liegroups.org/software.
Here is an example, comments are in braces.
atlas> set G=Sp(4)
Variable G: RootDatum
atlas> set f=G.invariant_form {W-invariant bilinear form}
Added definition [3] of f: (ratvec,ratvec->rat)
atlas> set alpha=highest_short_root(G)
Variable alpha: vec
atlas> f(alpha,alpha)
Value: 6/1 {need to normalize it differently}
atlas> set g(ratvec v,ratvec w)=f(v,w)/3
Added definition [3] of g: (ratvec,ratvec->rat)
atlas> g(alpha,alpha) {normalized invariant form}
Value: 2/1
atlas> void:for alpha in G.posroots do
for beta in G.posroots do
prints(alpha, " ", beta, " ", g(alpha,beta)) od od
[ 1, -1 ] [ 1, -1 ] 2/1
[ 1, -1 ] [ 0, 2 ] -2/1
[ 1, -1 ] [ 1, 1 ] 0/1
[ 1, -1 ] [ 2, 0 ] 2/1
[ 0, 2 ] [ 1, -1 ] -2/1
[ 0, 2 ] [ 0, 2 ] 4/1
[ 0, 2 ] [ 1, 1 ] 2/1
[ 0, 2 ] [ 2, 0 ] 0/1
[ 1, 1 ] [ 1, -1 ] 0/1
[ 1, 1 ] [ 0, 2 ] 2/1
[ 1, 1 ] [ 1, 1 ] 2/1
[ 1, 1 ] [ 2, 0 ] 2/1
[ 2, 0 ] [ 1, -1 ] 2/1
[ 2, 0 ] [ 0, 2 ] 0/1
[ 2, 0 ] [ 1, 1 ] 2/1
[ 2, 0 ] [ 2, 0 ] 4/1
atlas> void:for alpha in G.posroots do for beta in
G.fundamental_weights do
prints(alpha, " ", beta, " ", g(alpha,beta)) od od
[ 1, -1 ] [ 1, 0 ]/1 1/1
[ 1, -1 ] [ 1, 1 ]/1 0/1
[ 0, 2 ] [ 1, 0 ]/1 0/1
[ 0, 2 ] [ 1, 1 ]/1 2/1
[ 1, 1 ] [ 1, 0 ]/1 1/1
[ 1, 1 ] [ 1, 1 ]/1 2/1
[ 2, 0 ] [ 1, 0 ]/1 2/1
[ 2, 0 ] [ 1, 1 ]/1 2/1
atlas> void:for alpha in G.fundamental_weights do
for beta in G.fundamental_weights do
prints(alpha, " ", beta, " ", g(alpha,beta)) od od
[ 1, 0 ]/1 [ 1, 0 ]/1 1/1
[ 1, 0 ]/1 [ 1, 1 ]/1 1/1
[ 1, 1 ]/1 [ 1, 0 ]/1 1/1
[ 1, 1 ]/1 [ 1, 1 ]/1 2/1
|
2025-03-21T14:48:29.718947
| 2020-01-25T17:13:26 |
351133
|
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"Tom Copeland",
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|
Stack Exchange
|
Getzler's stable graphs for modular operads
In The semi-classical approximation for modular operads, Getzler displays a table at the bottom of page two enumerating certain stable graphs. (This is related to the MO-Q "Stable graphs: Feynman diagrams and Deligne Mumford space.")
The expression in parentheses in the table can be identified with OEIS A263634, the logarithmic polynomials with important applications to the operator calculus of Appell sequences, to moment-cumulant theory, to isolating indeterminates in the construction of power series compositions, and to the related algebra of symmetric polynomials/functions as an avatar (e.g.f. vs. o.g.f. rep) of the Faber polynomials for transforming the symmetric power sum polynomials into either the complete symmetric polynomials or the elementry symmetric polynomials, i.e., the Newton-Waring identities.
The partition polynomials preceding the logarithmic polynomials remain relatively mysterious to me. I would like to check if these polynomials or a reduced form of them are already in the OEIS. Does anyone have an extension to higher orders of this part of the table, or can someone present a simple method of extending it?
(The table in the upper part of the page contains essentially the first few partition polynomials for Lagrange compositional inversion of formal Taylor series / e.g.f.s, OEIS A134685).
There is a possibility that these polynomisls are related to a transform of the Catalan numbers,
|
2025-03-21T14:48:29.719081
| 2020-01-25T17:32:04 |
351134
|
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"Dmytro Taranovsky",
"Fedor Pakhomov",
"Trevor Wilson",
"https://mathoverflow.net/users/113213",
"https://mathoverflow.net/users/1682",
"https://mathoverflow.net/users/36385"
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"sort": "votes",
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}
|
Stack Exchange
|
$0^\#$ in weak theories vs large cardinals in $L$
To better understand the transition from large cardinal axioms consistent with the constructible universe $L$ to large cardinal axioms transcending $L$, I am looking for natural equiconsistencies between "strong large cardinal in $L$" and "weak set theory + $0^\#$ exists". As a result, I formed the following question/conjecture, but other such equiconsistencies would also work as an answer.
Question: Is $\mathrm{ATR}_0$ + $Π^1_1$ transfinite induction + "$0^\#$ exists" equiconsistent with ZFC + {there is a virtually $n$-Ramsey cardinal}$_{n∈ℕ}$?
Here (i.e. for the purpose of this question), an infinite cardinal $κ$ is $n$-Ramsey if for every $f:κ^{<ω}→\{0,1\}$, there is $s$ homogeneous for $f$ such that $κ∈h^n(s∪\{κ\})$ where $h(x)=\{α∈x: \mathrm{ordertype}(x∩α)=α\}$. 'virtually' means that $s$ need only exist in a generic extension of $V$.
$\mathrm{ATR}_0$ is a commonly used subsystem of second order arithmetic consisting of basic axioms, induction for sets, and the ability to iterate arithmetic comprehension along a well-ordering. Over $\mathrm{RCA}_0$ (a weak base theory), it is equivalent to a variety of propositions, including open determinacy, clopen determinacy, analytic perfect subset property, $Σ^1_1$ separation, comparability of well-orderings, and so on.
The real number $0^\#$ (zero sharp) is the theory of finite tuples of Silver indiscernibles for $L$. We can formalize it in $\mathrm{ATR}_0$ + $Π^1_1$ transfinite induction as follows: $0^\#$ is a complete theory satisfying certain basic axioms such that for every ordinal $α$, the Skolem hull using $α$ indiscernibles is well-founded.
We have:
* "$0^\#$ exists" is a $Σ^1_3$ statement; the above Skolem hull coded as a real number is computable from $0^\#$ and a real number coding $α$.
* $0^\#$ is unique: $\mathrm{ATR}_0$ gives uniqueness of $L$, and given two real numbers leading to closed classes of indiscernibles $I$ and $J$, using induction we can iterate $α→I_{J_α}$ to get a common set of indiscernibles.
* The proof of $Σ^1_1$ determinacy (accessed Jan 25, 2020) goes through in this setting (even without $Π^1_1$ transfinite induction); I do not know whether the reversal holds.
* KP + "$0^\#$ exist" with transfinite induction restricted to $Σ_1$ formulas appears to have the same strength (with well-foundedness of the Skolem hull interpreted through existence of the transitive collapse); however, the proof of $Σ^1_1$ determinacy does not appear to go through there.
For every $n∈ℕ$, in $\mathrm{ATR}_0$ + $Π^1_1$ transfinite induction + "$0^\#$ exists", each Silver indiscernible is, in $L$, totally indescribable, completely ineffable, and virtually $n$-Ramsey. To prove the Ramseyness, one uses $Π^1_1$ transfinite induction to get an indiscernible with enough indiscernibles leading to it, which for $f∈L$ will give a homogeneous set in $V$. Now, given $f$ and a real number coding $L_κ$, existence of a desired homogeneous set is $Σ^1_1$, so it also exists $L[G]$ if $G$ is generic for $\mathrm{Coll}(ω,κ)$ (note that $L[G]$ is closed under hyperjumps).
Stronger theories: Adding more induction gives stronger forms of Ramseyness, and $Π^1_1-\mathrm{CA}_0$ + "$0^\#$ exists", through existence of the hyperjump of $0^\#$ (and $Σ^1_2$ correctness of $L$ here), implies homogeneous sets with (within limits) any desired amount of diagonalization.
Weaker theories: $Π^1_1$ transfinite induction is important since in $\mathrm{ATR}_0$ without extra induction, the above definition of $0^\#$ can be split into existence of a set $M$ with basic properties of $0^\#$ such that every constructible set is inside a well-founded iterate of $M$ (let us call it weak $0^\#$ here) and a transfinite induction instance guaranteeing that for all $α∈\mathrm{Ord}$, $α$th iterate of $M$ is well-founded; and for both the weak and the strong version, we appear to need some induction (ordinary $Σ^1_1$ (equivalently, $Π^1_1$) induction suffices) to guarantee uniqueness of $0^\#$. Here are some conjectured correspondences for provable $Σ_2^L$ statements (and if proved, potential answers):
$\mathrm{ATR}_0$ + weak $0^\#$ — ZFC + {there is $n$-iterable cardinal}$_{n∈ℕ}$
$\mathrm{ATR}_0$ + weak $0^\#$ + $Σ^1_1$ induction — ZFC + {there is $ω^n$-Erdos cardinal}$_{n∈ℕ}$
$\mathrm{ATR}_0$ + weak $0^\#$ + full induction (on $ω$) — ZFC + {there is $α$-Erdos cardinal}$_{α<ε_0}$
$\mathrm{ATR}_0$ + strong $0^\#$ implies that Silver indiscernibles are virtually almost Ramsey in $L$ and more; I conjecture correspondence with ZFC + (schema over $n$) $∃κ∈\mathrm{Ord} \, ∀f:κ^{<ω}→\{0,1\}$ "there is $s$ in a generic extension of $V$ with $s$ homogeneous for $f$ such that $s^n(0)$ exists", where $s(α)$ is the $(1+α)$th element of $s$.
The trace of large cardinal axioms in L: As an aside, related to their relativization to $L$, virtual large cardinal axioms (even virtual rank-into-rank) tend to be weaker than $0^\#$. However, we can define analogs of transitive models to get the trace of strong large cardinal axioms in $L$. For example, ZFC + measurable cardinal proves the same $Σ^L_2$ statements as ZFC (or ZFC + $V=L$) augmented with (schema, $T$ is a finite fragment of ZFC + measurable): There is a generic extension of $V$ with a transitive model $M$ of $T$ such that every $M$-cardinal is an $L$-cardinal in $V$. More generally, this also works for $Σ^L_{n+1}$ statements using $\{α: L_α ≺_{Σ_n} L\}$ in place of $L$ cardinals, and also for $Π^L_{n+2}$ statements if, in addition, by varying the extension, the height of $M$ can be arbitrarily large.
This isn't quite what you're looking for, and doesn't answer your specific question, but you may be interested to look at the following paper if you haven't already: Cheng and Schindler, Harrington's principle over higher order arithmetic (https://arxiv.org/abs/1503.04000)
@DmytroTaranovsky Great and well-researched question! Unfortunately I don't know how to construct a model with a virtually Ramsey cardinal from $0^#$. Let me comment, what could be done in the other direction, i.e. with regard to the reduction of a weak base theory enhanced with $0^#$ to $\mathsf{ZFC}+{\text{"there is virtually n-Ramsey cardinal"\mid n\in \mathbb{N}}$. By "reduction" I mean some apropritate partial conservation result.
@DmytroTaranovsky A weak base theories for which this works well are weak $\Pi_2$-axiomatizable set theories. For example a theory $\mathsf{BST}$ that is Extensionality + Regularity + Infinity + Axioms of Rudimentary Closure. In this setting $0^#$ is defined to be a set of naturals (encoding a complete extension of $\mathsf{ZFC}+V=L$ with indecernibles) such that for any ordinal $\alpha$ the model with $\alpha$-many indecernibles generated from $0^#$ is isomorphic to a transitive model (I will not be concerned with uniqueness).
@DmytroTaranovsky The theory $\mathsf{BST}+\text{"$0^#$ exists"}$ could be easily reduced to the theory $\mathsf{BST}+$ {there is a complete extension $X$ of $\mathsf{ZFC}+V=L$ with indecernibles and ordinals $\alpha_0<\alpha_1<\ldots<\alpha_{n}$ such that model with $\alpha_i$-many indecernibles generated from $X$ is isomorphic to $L_{\alpha_{i+1}}$ | $n\in \mathbb{N}$}. The latter theory in turn could be easily reduced to $\mathsf{ZFC}+{\text{there is a virtually $n$-Ramsey cardinal} \mid n\in\mathbb{N}}$.
@DmytroTaranovsky However, $\mathsf{ATR}_0^{\mathsf{set}}$ (which is the direct set-theoretic counterpart of $\mathsf{ATR}_0$) contains Axiom Beta, which might be problematic for the mentioned reductions.
@FedorPakhomov With respect to strength using $0^#$, ATR$_0$ without extra induction is likely similar to BST. However, using Axiom Beta (or for KP, bounded collection), we appear to have a hierarchy with a four-way correspondence: more induction $⇔$ more diagonalization of indiscernibles $⇔$ stronger partition properties in $L$ $⇔$ (at some points) fuller meaningfulness of $0^#$.
|
2025-03-21T14:48:29.719554
| 2020-01-25T23:26:40 |
351141
|
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|
Stack Exchange
|
Number of sequences satisfying termination conditions
This question was originally posed on math.SE but seems to require research-level mathematics expertise:
Two players play each other in a match of games of chess where the match winner is the first to achieve either $m>0$ total game wins or $0<n<m$ game wins in a row. (There are no ties.) How many distinct possible sequences of game results exist as a function of $n$ and $m$?
For small values, one can write out the sequences (as in the linked question).
Clearly the shortest sequence of games is of length $n$ (of which there are two) and longest sequence of games is of length $2 m - 1$, but we seek the total number of distinct sequences.
First steps
Assume, without loss of generality, that the winning player is $A$ and the loser is $B$.
Let $0<k \leq 2m-1$ be the length of a sequence. If $k<n$ there is no match winner; we can ignore such cases.
There are, then, two ranges of interest:
$n \leq k < m$
$m \leq k \leq 2m -1$
In the first case, the only way $A$ wins the match is to win $n$ games in a row. Of course the last game (position $k$) $A$ wins, terminating the match. Thus the final $n$ games $A$ must win. That leaves $k-n$ preceding "other" slots. The last one of these must be won by $B$. Thus there are $k - n - 1$ remaining "unassigned" slots which can be won in any order so long as $A$ does not win $n$ in a row.
The number of ways this can be achieved is:
$$\sum\limits_{j=0}^{k-n-1} {k-n-1 \choose j}$$
but again, this over counts in certain cases because it includes cases in which $A$ wins $n$ consecutive games.
In the second case there are two categories of ways $A$ wins: by winning $m$ games or by winning $n$ in a row. In either of these cases, the last game (slot $k$) $A$ must win. If $A$ wins $m$ total games (on the $k$th game) then $A$ has won $m-1$ games in the $2m - 2$, and that means that $B$ has won the remaining $m-1$ games (but neither have won $n$ in a row).
The total number of ways this can be done is:
$$\sum\limits_{j=0}^{m-1} {2m - 2 \choose j}$$
but note that some of these cases may have $A$ win $n$ in a row. We must subtract those cases.
Another way to view part of the problem
Perhaps we can view this problem as a path through a lattice graph. Consider the number of sequences in which $A$ wins the match by winning $m$ games (and not by winning $n$ in a row). Each game won by $A$ is a lattice step to the right and each game won by $B$ is a step upward. Let's assume $A$ will be the winner. Thus the terminal nodes are those on the right-most column of the graph, $m$ steps horizontally.
Each path with be an alternating sequence of "streaks" of winning alternating by $A$ then $B$ then $A$... However, each of these streaks can be only of length $1, 2, \ldots, n-1$.
So here we have $m=15$ and $n=4$. Each streak is of length less than $4$.
Each set of streaks of the winner $A$ represent an integer partition of $m$ restricted to elements $1, 2, \ldots, n-1$. (This is not the case for $B$, whose sum must be less than $m$ because $B$ loses.)
So the question can be restated as finding the number of distinct sequences of integer partitions of $m$, and for each such distinct sets of "streaks." Then, we find the number of integer partitions for $B$ that have the same number of streaks, or one greater or one less. Then we can alternate a streak for $A$ with a streak for $B$, and so on.
Counting this total number should give the number of sequences (paths) in which $A$ wins by reaching $m$ won games (but not having a streak of $n$).
Of course, this is not the full solution to the problem... but perhaps a casting that will help.
Let $g(i,j,k)$ be the number of sequences if the winner of the most recent game has won $i$ games, the opponent has won $j$ games, and the current win streak is $k$ games. Then $g$ satisfies the following recurrence:
$$
g(i,j,k) =
\begin{cases}
1 &\text{if $i=m$ or $k=n$},\\
g(i+1,j,k+1) + g(j+1,i,1) &\text{otherwise}.
\end{cases}
$$
The idea of the recurrence is to condition on the winner of the next game, as follows. If the most recent winner wins the next game, then both the win count $i$ and the consecutive win count $k$ increase by 1, and the opponent win count $j$ does not change, yielding $g(i+1,j,k+1)$. If instead the most recent winner loses the next game, then the win count $i$ does not change, the win count $j$ of the opponent increases by 1, the consecutive win count $k$ starts again at 1, and the opponent becomes the most recent winner, yielding $g(j+1,i,1)$.
We want to compute $g(0,0,0)/2$. For small $m$ and $n$, I get the following results:
\begin{matrix}
m\backslash n&1&2&3&4&5&6&7&8&9&10\\
1&1\\
2&1&3\\
3&1&5&10\\
4&1&7&26&35\\
5&1&9&68&112&126\\
6&1&11&174&362&442&462\\
7&1&13&445&1176&1560&1689&1716\\
8&1&15&1137&3827&5535&6207&6400&6435\\
9&1&17&2908&12491&19722&22915&23992&24266&24310\\
10&1&19&7445&40874&70503&84930&90323&91950&92324&92378
\end{matrix}
I may misread your notation by 1. For $(n,m) = (2,3)$ I get ten: $AA, BB, ABB, BAA, ABAA, BABB, ABABBk BABAA, ABABA, BABAB$, which your table has for $(3,3)$. Is there a slight convention misalignment?
$g(0,0,0)$ counts wins by either player. The table shows $g(0,0,0)/2$, which counts wins for player A.
Ah... That's the source of the misalignment (+1). Thanks. If you could give just a bit more of the derivation of the recursion relation, I'll accept.
|
2025-03-21T14:48:29.720172
| 2020-01-26T00:06:00 |
351143
|
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"Arshak Aivazian",
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|
Stack Exchange
|
Classification of associative polynomial functions
What is known about a classification of associative (binary) polynomial functions? First of all, it is interesting in two cases: over Integral domain (or even over field) and over ring of integers modulo $n$.
(A polynomial binary function is function $R \times R \to R$ induced by a polynomial in two variables $P$ over a ring $R$.)
If you find an associative polynomial P(x,y) where the term Q(x,y) = P(x,P(y,x)) is also associative, then the structure with underlying set R and binary operation P generates a locally finite variety; further, you just have to check three more terms in P to see if they and all other terms in P are associative. In case you needed lots of associative polynomials. Gerhard "Search On Hyperassociativity In MathOverflow" Paseman, 2020.01.25.
Over an infinite integral domain you can classify all polynomials that satisfy the associativity functional equation. A quick degree consideration of both sides tells you that the polynomial is at most degree $1$ in each variable, so the only answers are $x,y, c+x+y$ and $c_1(x+c_2)(y+c_2)-c_2$. In fact you can classify all $n$-variable polynomials which satisfy the $n$-ary version of associativity, as done in the paper "A description of n-ary semigroups polynomial-derived from integral domains".
As far as finite rings, I doubt you can say anything meaningful. For example, already over $\mathbb Z/p\mathbb Z$, any binary function can be written as a polynomial map, so you would be asking for a classification of all associative operations on this set (this number increases very fast).
Many thanks !
(Note: in your answer there are no polynomials x and y)
|
2025-03-21T14:48:29.720432
| 2020-01-26T02:03:30 |
351148
|
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|
Stack Exchange
|
Sum of multi-index factorials
Fix $d\in\mathbb{N}\setminus\{0\}$. For $j\in\mathbb{N}\setminus\{0\}$, let
\begin{align*}
[j] = \Big\{\alpha\in \mathbb{N}^d: \sum^d_{i=1}\alpha_i=j\Big\}.
\end{align*}
For $\alpha\in[j]$, define the multi-index factorial $\alpha! =\prod_{i=1}^d (\alpha_i !)$. What is a good bound for
\begin{align*}
\sum_{\alpha \in [j]} (\alpha !)
\end{align*}
in terms of $j$, when $d$ is fixed and $j$ is large?
You'll get different answers depending on j bigger than d or d bigger than j. Do you know which you need? Gerhard "Answer Depends On The Both" Paseman, 2020.01.25.
Updated the question. Please consider fixed d while j is large.
So I believe the argument in my post actually works when d^2 is less than j and gives an upper bound of (d-1)j(j!) on your sum when d is greater than 1. It would be nice to know for what pairs (d,j) we have that the subsums involving tuples with largest element (j-k) is greater than the subsum with tuples having largest element (j-k-1). Maybe you can find the general argument. Gerhard "Staying With Specialization For Now" Paseman, 2020.01.25.
The logarithm of the sum in question is $\sim j\ln j$ if $d=o(j)$.
Indeed,
assume that $j\ge d$ and, moreover, $d=o(j)$.
By Stirling's formula (see e.g. formula (26)),
$$(2\pi n)^{1/2}(n/e)^n<n!<(2\pi(n+1))^{1/2}(n/e)^n \tag{1}$$
for integers $n\ge0$ (with $0^0:=1$).
Let $S$ denote the sum in question. Let $a:=\alpha$ and $a_i:=\alpha_i$.
Note that $S\ge j!$, whence, by (1),
$$\ln S\gtrsim j\ln j.\tag{2}$$
On the other hand, again by (1), the arithmetic-geometric-mean inequality, and the convexity of $u\ln u$ in $u\ge0$ (with $0\ln0:=0$), for any $a\in[j]$
$$a!<(2\pi)^{d/2}e^{-j}\prod_1^d(a_i+1)^{1/2}\exp\sum_1^d a_i\ln a_i \\
\le (j/d+1)^{d/2}\exp(j\ln j) =\exp(j\ln j+o(j\ln j)).
$$
Also, the cardinality of $[j]$ is $\le j^d=\exp o(j\ln j)$ (actually, the cardinality of $[j]$ is $\binom{j+d-1}{d-1}$). So,
$$\ln S\lesssim j\ln j. \tag{3}$$
Thus, by (2) and (3),
$$\ln S\sim j\ln j,$$
as claimed.
Here is an approach that suggests (for $d$ smaller than $j$, say $d^2$ less than $j$) the sum is not larger than $dj(j!)$. Unfortunately, we are only bounding part of the sum by $d(j/2 - d)(j!)$. These are the parts which have an $\alpha_i$ term at least as large as $(j/2+d+1)$.
Note that there are $d$ terms whose product is $j!$: these are the $d$ tuples where all but one of the $\alpha_i$ are zero.
Now let's set $k=0$, and look at a single tuple whose largest $\alpha_i$ is $(j-k)$. By replacing this $\alpha_i$ by $(j-k-1)$ and adding $1$ to one of the other $d-1$ places, we get $d-1$ distinct tuples whose largest term is $(j-k-1)$ and whose sum of (the products derived from each of) these $d-1$ tuples is at most $(k+d-1)/(j-k)$ times the single tuple, meaning the sum of all (products derived from) tuples with largest element $(j-k-1)$ is less than the sum of all (products derived from) tuples with largest element $(j-k)$. So when $2k$ is less than $j+1-d$, we get the sum of tuples with largest element $(j-k)$ is less than $d(j!)$. So we can bound a large part of the sum by $(j+1-d)d(j!)/2$. If we could extend this argument down to $k=j/d$, we would have the sum bounded above by $(j-j/d)d(j!)$.
Gerhard "Turning Multiplication Back Into Addition" Paseman, 2020.01.25.
Actually, the sums may be bounded by a unimodal sequence which would lead to each subsum of terms derived from tuples with largest a_i being j-k is still less than d(j!). So I think we can establish a bound of dj(j!) even in the case d is near j in size. Gerhard "Ever Hopeful For A Proof" Paseman, 2020.01.25.
Let us also show for each natural $d$
$$S_{d,j}\sim j!d \tag{1}$$ (as $j\to\infty$),
where $S_{d,j}$ is the sum in question.
The key here is the recursion
$$S_{d,j}=\sum_{b=0}^j b!S_{d-1,j-b} \tag{2}$$
for $d=2,3,\dots$ and $j=0,1,\dots$, with the initial conditions $S_{d,0}=1$ and $S_{1,j}=j!$. The latter initial condition obviously implies (1) for $d=1$.
Proceed by induction on $d$. Take any $d=2,3,\dots$ and $j=0,1,\dots$. Then, by (2) and induction, for any fixed natural $B$
$$S_{d,j}=\sum_{b=0}^j (j-b)!S_{d-1,b} \\
=j!S_{d-1,0}+(j-1)!S_{d-1,1}+S_{d-1,j}+S_{d-1,j-1} \\
+\sum_{b=2}^B (j-b)!S_{d-1,b} +\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b} \\
=j!+O((j-1)!)+(d-1+o(1))j!+O((j-1)!) \\
+O((j-2)!) +\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b} \\
=(d+o(1))j!+\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b}.
$$
Let now $B\ge1$ be large enough so that $S_{d-1,b}\le b!d$ for all $b>B$; such $B$ exists by induction. Then, noting that $(j-b)!b!$ is log-convex in $b\in\{0,\dots,B\}$, we see that for $j\ge3$
$$\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b}\le \sum_{b=B+1}^{j-2} (j-b)!b!\,d \\
\le \sum_{b=2}^{j-2} (j-b)!b!\,d\le(j-2-1)(j-2)!2!\,d=o(j!).$$
Now (1) follows by the multiline display.
|
2025-03-21T14:48:29.720725
| 2020-01-26T05:35:01 |
351154
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351154"
}
|
Stack Exchange
|
A trapping set with finite measure
Does there exist a measurable subset $T$ of $[0, \infty)$ with finite measure and some $\epsilon > 0$ such that for every $r$ with $0 < r < \epsilon$, $nr$ is in $T$ for infinitely many positive integers $n$?
Note: The integers $n$ such that $nr$ lie in $T$ can depend on $r$.
No. Denote $T_k=T\cap [k,k+1)$. Then $\sum |T_k|<\infty$ (where $|X|$ stands for the measure of $X\subset \mathbb{R}$). Choose a segment $[a,b]\subset (0,\epsilon)$. Note that if $r\in [a,b]$ and $nr\in T_k$, then $na\leqslant nr< k+1$ and $nb\geqslant nr\geqslant k$, thus $n\in [k/b,(k+1)/a]$. The union of $n^{-1}T_k$ over all positive integers $n\in [k/b,(k+1)/a]$ has measure at most $$|T_k|\cdot \sum_{n\in [k/b,(k+1)/a]} n^{-1}\leqslant C|T_k|,$$
where $C$ depends only on $a$ and $b$. Now choose $k_0$ so that $C\sum_{k>k_0} |T_k|<b-a$. By the pigeonhole principle there exists a point $r\in [a,b]$ not covered by $n^{-1} T_k$ with $k>k_0$ and positive integer $n$.
|
2025-03-21T14:48:29.720820
| 2020-01-26T08:42:16 |
351160
|
{
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"authors": [
"Dieter Kadelka",
"Hanul Jeon",
"High GPA",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/109527",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351160"
}
|
Stack Exchange
|
$X$ is Polish and $N$ is countable. Is $N^X$ Polish?
$X$ is a separable, completely metrizable topological space equipped with its sigma algebra of Borel sets. $N$ is a countable space.
$X^N$ is the collection of all mappings from $N$ to $X$. It is equipped with product topology.
Is $N^X$ a Polish space?
Is $X^N$ a Polish space?
Is $X^X$ a Polish space?
It is obvious that $(2)$ is a Polish space, (3) is not in most cases. I guess $(1)$ is also Polish because $X^N$ and $N^X$ usually have the same structure?
It could depend on which topology is bestowed on the collection of mappings. (The usual choice is to give a product topology, though.)
@HanulJeon Yes you are right!
What do you mean by "$X^N$ and $N^X$ have the same structure"? In most cases these are completely different objects. What is the topology of $N^X$? A "product topology" does not make much sense for this case.
If you really mean these spaces $A^B$ to consist of all functions $B\to A$, then (1) and (3) are too big to be Polish. They have cardinality $2^{\mathfrak c}$ where $\mathfrak c$ is the cardinal of the continuum. Separable metric spaces have cardinality at most $\mathfrak c$.
On the other hand, (2) is, like any product of countably many Polish spaces, well known to be Polish.
Let us work in a convenient category of topological spaces, ie assume that we have function spaces and sufficiently many nice spaces around. My favourite such category is the category of $\mathrm{QCB}_0$-spaces, which are $T_0$ quotients of countably-based spaces [1].
Inside the category of $\mathrm{QCB}_0$-spaces, Schröder has obtained a characterization of the Hausdorff spaces $\mathbf{X}$ such that $\mathbb{R}^\mathbf{X}$ is Polish [2]. These spaces are now called coPolish spaces. A Polish space is coPolish iff it is locally compact.
A space such as $2^{(\mathbb{N}^\mathbb{N})}$ even fails to be countably based.
[1] https://www.sciencedirect.com/science/article/pii/S0304397501001098
[2] https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.200310111
|
2025-03-21T14:48:29.720971
| 2020-01-26T09:35:16 |
351162
|
{
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"authors": [
"Jianrong Li",
"LeechLattice",
"https://mathoverflow.net/users/11877",
"https://mathoverflow.net/users/125498"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351162"
}
|
Stack Exchange
|
How to show that $x_{k+1}+x_{k+2} + \cdots + x_n < 2m$?
Let $k \le n$ be positive integers and let $m$ be a positive integer. Assume that $x_1, \ldots, x_n$ are non-negative integers and
\begin{align}
& x_1^2 + x_2^2 + \cdots + x_n^2 - (k-2) m^2=2, \\
& x_1 + \cdots + x_n = k m, \\
& x_1 \ge x_2 \ge \cdots \ge x_n.
\end{align}
How to show that $x_{k+1}+x_{k+2} + \cdots + x_n < 2m$?
It is easy to see that the result is true for $k=1,2$.
In the case of $k=3$, we have
\begin{align}
& x_1^2 + x_2^2 + \cdots + x_n^2 = m^2 + 2, \\
& x_1 + \cdots + x_n = 3 m, \\
& x_1 \ge x_2 \ge \cdots \ge x_n.
\end{align}
We have to estimate the solutions of the above equations. Are there some method to do this? Thank you very much.
I believe it's possible to assume $m\geq 1$ is a real number, and apply Lagrange multipliers.
@LeechLattice, thank you very much. I tried to use your method to prove the case of $k=3$, $n=7$, and it works. But when I tried the case of $k=4$, $n=7$, the critical values are $$ [x_1 = 0, x_2 = 0, x_3 = 0, x_4 = 1, x_5 = 1, x_6 = 1, x_7 = 1, l_1 = 1/2, l_2 = 0, m = 1], \ [x_1 = 0, x_2 = 0, x_3 = 0, x_4 = -1, x_5 = -1, x_6 = -1, x_7 = -1, l_1 = -1/2, l_2 = 0, m = -1].$$ They don't satisfy the constrains $x_1 \ge x_2 \ge \cdots \ge x_n$. In some other cases, it is also possible that the critical values are not integers. Do you know how to solve these problems?
I am afraid it is not true. Test the situation when $x_2=x_3=\ldots=1$, equations read as $x_1^2+(n-1)=(k-2)m^2+2$, $x_1+(n-1)=km$, that gives $x_1^2-x_1=(k-2)m^2-km+2$. Let's think that both $k$ and $m$ are large (say greater than 1000). Then $x_1$ is something like $\sqrt{k}m$, $x_{k+1}+\ldots+x_n=n-k=km+1-k-x_1$ is something like $(k+o(k))m\gg 2m$.
It remains to find the solution of $x_1^2-x_1=(k-2)m^2-km+2$ with large $k$ and $m$. It reads as $x_1^2-x_1+m^2-2=k(m^2-m)$. For fixed $m$ it is solvable if $x^2-x+m^2-2$ may be divisible by $m^2-m$ (then it may be divisible for large $x$ that makes $k$ also large). Modulo $m$ we may choose $x\equiv 2\pmod 2$. Modulo $m-1$ we need $x^2-x-1\equiv 0$, so just take $m-1=a^2-a-1$ for some large $a$. Now combine solutions modulo $m$ and $m-1$ by Chinese remainders theorem.
|
2025-03-21T14:48:29.721130
| 2020-01-26T09:46:59 |
351163
|
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"authors": [
"Derek Holt",
"Francesco Polizzi",
"Geoff Robinson",
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|
Stack Exchange
|
Direct proof (or reference) that a given $p$-group is extra-special
Writing a paper on algebraic surfaces, I was led to consider the finite group $\mathsf{H}(A)$ whose presentation is the following.
I start with an anti-symmetric matrix $A=(a_{ij})$ of order $2n$ over $\mathbb{Z}_p$ (the finite group/field with $p$ elements), $p \geq 3$, and I consider the following system of generators and relations:
Generators:
$x_1, \ldots, x_{2n}, \; z$
Relations:
\begin{equation}
\begin{split}
x_1^p &= \ldots=x_{2n}^p=z^p=1 \\
[x_1, \, z] &= \ldots = [x_{2n}, \, z]=1\\
[x_i, \, x_j]& =z^{a_{ij}}
\end{split}
\end{equation}
where, by slight abuse of notation, the exponent $a_{ij}$ stands for any lifting in $\mathbb{Z}$ of $a_{ij} \in \mathbb{Z_p}$.
I can prove the following
Proposition 1. If $\det A \neq 0$, then the group $\mathsf{H}(A)$ presented as above is extra-special of order $p^{2n+1}$ and exponent $p$.
Proof. Calling $\omega$ the symplectic form on the $\mathbb{Z}_p$-vector space $V:=(\mathbb{Z}_p)^{2n}$ whose matrix (with respect to the standard basis) is $A$, one checks that $\mathsf{H}(A)$ is isomorphic to the Heisenberg group $\mathsf{H}(V, \, \omega)$. By standard linear algebra, up to a change of coordinates the symplectic form $\omega$ can be transformed into the standard symplectic form $\omega_{\mathrm{st}}$, so our group is also isomorphic to $\mathsf{H}(V, \, \omega_{\mathrm{st}})$, that has the desired properties, being the central product of $n$ copies of the $p$-group of order $p^3$ and exponent $p$. $\square$
Now, I was asked by the referee to give a direct proof of Proposition 1, namely a proof not involving symplectic forms. The motivation for this request is that the present proof is a bit far from the spirit of the paper, whose emphasis is mainly on group presentations.
Of course, I can give an equivalent proof involving nonsingular anti-symmetric matrices instead of symplectic forms, but I suspect that this is not what he/she/other pronoun was asking for. So let me ask the following two questions.
Question 1. Is there a proof of Proposition 1 essentially different from the one I gave?
Question 2. Is there any precise reference for Proposition 1?
I suspect that all of this is very well-known to the experts. Any hint to orientate myself in the extensive literature on $p$-groups would be greatly appreciated.
It is clear from standard properties on commutators that $H(A)$ is nilpotent of class at most $2$ with derived group and Frattini subgroup contained in $\langle z \rangle$
(just consider $H(A)/\langle z \rangle$, which is elementary Abelian of order $p^{2n})$.
The general property of commutators that you need is that we always have $[a,bc] = [a,b]^{c}[a,c]$. It is this, combined with the last relation(s), that ensures that the derived group of $H(A)$ is containedd in $\langle z \rangle$. It is necessary $A$ is antisymmetric, since we have $[x_{j},x_{i}] = [x_{i},x_{j}]^{-1}.$
The only remaining issue is check that $H(A)$ has center $\langle z \rangle$ (but no larger). This requires that $x_{1}^{b_{1}} x_{2}^{b_{2}} \ldots x_{2n}^{b_{2n}}$ is not central when $0 \leq b_{i} \leq p-1$ and not all $b_{i}$ are zero.
Using the above commutator relation $[a,bc] = [a,b]^{c}[b,c]$ repeatedly, and taking the commutator of the above element with each $x_{k}$, we see that this is equivalent to the non-singularity of $A$.
Ah, of course, fixed. Thank you very much for the answer, I will check the details.
In fact, since the group defined by the presentation clearly has nilpotency class at most 2 (it would be abelian if all of the $a_{ij}$ were 0), the commutator relation simplifies to $[a,bc] = [a,b][a,c]$, which make checking easier.
It suffices to check the commutator relation for $a, , b, , c, ,$ of the form $$x_1^{t_1}x_2^{t_2}\ldots x_{2n}^{t_{2n}},$$
right?
@DerekHolt : Yes, I was implicitly using the fact that $[x_{i},x_{j}]$ is central in this case (from the given relations), but perhaps I should have made this explicit.
@FrancescoPolizzi : You only need to check $[x_{k},x_{1}^{b_{1}} \ldots x_{2n}^{b_{2n}}]$ for each $k$, and this turns out to be $z^{b_{1}a_{k1} + \ldots + b_{2n}a_{k 2n}}$ in each case.
Isn't the condition required equivalent to the antisymmetric bilinear form defined by $A$ being non-degenerate, which is euqivalent to $A$ being non-singular?
@DerekHolt : I am not sure whether this was addressed to Francesco or me (or both). The question was to give a "direct" proof without using symplectic forms. But of course, extraspeciaal groups and symplecctic forms are inseparable, and it is certainly the case that the symplectic form proof and the above "direct" group-theoretic proof are equivalent, just using a different language.
Geoff and Derek: yes, for the reasons your are giving I suspect that every proof will be equivalent to the original one. However, I would like to comply with the referee's request, possibly learning some math during the process. Thanks again for your help.
Geoff Robinson and @DerekHolt: I have checked the details and everything works, thank you again for the clear and elegant proof. It seems to me that, exactly with the same argument, one can show that, if $A=(a_{ij})$ is
non-singular, then the group
\begin{equation}
\begin{split}
x_1^p &= \ldots=x_{2n-2}^p=z^p=1 \
x_{2n-1}^p & =x_{2n}^p=z \
[x_1, , z] &= \ldots = [x_{2n}, , z]=1\
[x_i, , x_j]& =z^{a_{ij}}
\end{split}
\end{equation}
is extra-special of order $p^{2n+1}$ and exponent $p^2$, right?
In fact, if I am not mistaken, just one relation of type $x_j^p=z$ is sufficient to force the exponent to be $p^2$...
For the first question, I think everything does go through with hardly any modification. Calling $G(A)$, the new group, it is clear that $G(A)/\langle z \rangle$ is elementary Abelian of order $p^{2n}$, so $G(A)$ has class at most $2$ and all its commutators are central (and, from general properties of class $2$ groups, the exponent is $p^{2}$). The linear independence of the rows of $A$ is equivalent to the fact that $\langle z \rangle$ is precisely the center of $G(A)$, just as before. As for the last question, yes, one such relation forces the group to have exponent $p^{2}$.
Thanks again! So the dichotomy is
(1) all the relations in the first row are of type $x_j^p=1$: the exponent is $p$;
(2) there is at least one relation of type $x_j^p=z$: the exponent is $p^2$.
By the classification of extra-special $p$-groups, we know that two such groups with the same order and exponent are isomorphic (at least if $p$ is odd), so the isomorphism class in the second case does not depend on the number of relations of type $x_j^p=z$.
And the isomorphism class of the group does not depend on $m$, as soon as $m >0$, right?
I think that's right, since there are only two isomorphism types of extraspecial group of order $p^{2n+1}$, so I was overcomplicating things.
|
2025-03-21T14:48:29.721564
| 2020-01-26T10:09:42 |
351166
|
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"Igor Rivin",
"Will Chen",
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|
Stack Exchange
|
Essential simple closed curves on a punctured torus vs those in the torus
Let $T$ be a compact oriented torus, let $p\in T$ be a point, and let $T^*$ be $T - \{p\}$.
In Farb-Margalit's Primer on mapping class groups, in the discussion after Proposition 1.5 they say that "the homotopy classes of simple closed curves in $T^*$ are in bijective correspondence with those in $T$", but they give no hint of a proof.
I presume they left out the word "essential"? Even if we're talking about essential simple closed curves, what is this bijection? The only natural thing is to look at the map induced by the inclusion $T^*\hookrightarrow T$, but then this would mean that in each coset of the commutator subgroup of $\pi_1(T^*,q)$ (for some base point $q\in T^*$), there is a distinguished class which can be represented by an essential simple closed curve?
My questions are: What is the right statement (and how do you prove it)? Also, given an element $w\in\pi_1(T^*,q)$ written as a word in $x,y$ for some nice choice of basis $x,y\in\pi_1(T^*,q)$, is it possible to recognize when $w$ has a representative which is a simple closed curve? an essential simple closed curve?
I'm a novice in this area, so references would be appreciated as well!
There is exactly one simple closed curve in each primitive homology class in $T^\ast$ and that curve is the shortest one in the homology class. Conversely, the simple closed curves on $T$ correspond exactly to the primitive homology classes (primitive = class $(a, b)$ where gcd of $a, b$ (sadly, also denoted by $(a, b)$ usually) equals $1$). For references, see the paper
McShane, Greg; Rivin, Igor, A norm on homology of surfaces and counting simple geodesics, Int. Math. Res. Not. 1995, No. 2, 61-69 (1995). ZBL0828.30023.
and references therein.
(the existence of a bijective correspondence in and of itself is not interesting since both are countable sets).
No, $W_{m, n}$ are the so called Sturmian words. Draw the line from $0$ to $(m, n)$ and look at the sequence of intersection with horizontal and vertical grid lines...
For a small simple closed curve around the puncture, is there no geodesic representative? (Theorem 1.1, 1.2 in your paper don't quite imply this, at least as far as I can see)
|
2025-03-21T14:48:29.721741
| 2020-01-26T10:14:25 |
351167
|
{
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"authors": [
"Dirk Werner",
"https://mathoverflow.net/users/127871"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351167"
}
|
Stack Exchange
|
$L_p(I,Y)^\perp=L_q(I,Y^\perp)$?
Let $X$ be a Banach space and $Y$ be a closed subspace of $X$. For $1<p<\infty$ consider the $p$-th power Bochner Integrable functions which takes values in $X$ and defined on the unit interval $[0,1]$, denoted by $L_p(I,X)$. It is clear that $L_p(I,Y)$ is a closed subspace of $L_p(I,X)$. It is well-known that if $X^*$ has RNP then $L_p(I,X)^*=L_q(I,X^*)$. My question is whether $L_p(I,Y)^\perp=L_q(I,Y^\perp)$?
My observations are the following:
Let $X$ be an Asplund space then for any $Y\subseteq X$, is also Asplund. Now since $(X/Y)^* = Y^\perp$, $Y^\perp$ has the Radon Nikodym Property and hence $L_p(I,X/Y)^* = L_q(I,Y^\bot)$. On the other hand $(L_p(I,X)/L_p(I,Y))^* = (L_p(I,Y))^\perp (\subseteq L_q(I, X^*))$.
Now suppose further, $X$ is reflexive, hence all the spaces involved will be reflexive.
So if the entities in the LHS quantities are the same, by uniqueness of preduels of reflexive spaces, we get:$L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$.
In order to prove the desired result it is enough to prove: $L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$. It is still unknown to me whether $L_p(I,X/Y) = L_p(I,X)/L_p(I,Y)$ is true even if $X$ is reflexive.
One always has $L_q(I, Y^\bot) \subset L_p(I, Y)^\bot$. If $X$ is reflexive and separable, you might use a Hahn-Banach argument to show equality.
Let's consider the following situation: $E$ and $F$ are Banach spaces, $D\subset E$ is a dense subspace, $Q: E \to F$ is a continuous linear operator, and its restriction $Q_0$ to $D$ is a quotient map onto its range $R$, i.e., $Q_0$ takes the open unit ball of $D$ to the open unit ball of $R$. Suppose $R$ is dense in $F$.
If the kernel of $Q_0$ is dense in the kernel of $Q$, then $x+\ker Q_0 \mapsto x+\ker Q$ is an isometry. Hence the canonical isometry $D/\ker Q_0 \cong R$ extends to an isometry $E/\ker Q\cong F$.
This applies in the notation of the question to $E=L_p(X)$, $F=L_p(X/Y)$, $Q(f)=q\circ f$ with $q:X\to X/Y$ the canonical map, and $D=$ the space of $X$-valued step functions.
|
2025-03-21T14:48:29.721899
| 2020-01-26T11:20:42 |
351169
|
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}
|
Stack Exchange
|
Is this invariant of an exotic 4-sphere trivial?
This is a rather naive attempt to construct an invariant of an exotic
4-sphere. Apparently, the lack of useful invariants in this context is a well known issue. This particular invariant is somewhat obvious which probably means it is useless - most likely, always zero. However, I would like to know for sure.
Consider a smooth manifold $M$ together with a homeomorphism
$f: \mathbb{S}^n\to M.$ There is a natural map (the suspension)
$$\phi: \mathbb{S}^{n-1}\times I\to \mathbb{S}^n$$
collapsing two $n-1$ - spheres at the ends to the poles of the $n$ - sphere, which we may interpret as a homotopy $\phi_t: \mathbb{S}^{n-1}\to \mathbb{S}^n$. For each $t\in I$ we have an oriented vector bundle
$\phi_t ^* f^* TM$ on $\mathbb{S}^{n-1}$. This bundle is obviously trivial on the ends, so we may turn it into a bundle $E\to \mathbb{S}^{n-1}\times \mathbb{S}^{1}$ simply by gluing this ends.
It is easy to see that if $M$ is the standard sphere then $E$ is trivial. For an exotic sphere this depends on the homotopy class $\pi_{n-1}(SO(n))$ of the corresponding clutching function which may be trivial or not. If I am not mistaken, this construction (unlike the common one using twists) works in all dimensions, including $n=4$. In particular, it gives an invariant of an exotic 4-sphere in the group $$\pi_3(SO(4))\cong \mathbb{Z}\oplus \mathbb{Z}.$$
([M. Kervaire. Some nonstable homotopy groups of Lie groups.(1960)].) The question is, is this invariant trivial?
Can not we obtain non-trivial invariant from a standard $S^4$ by choosing some elaborate homeomorphism: $S^4 \rightarrow S^4$?
Since the map $Top(4)/O(4) \to Top/O$ is 5-connected by Freedman-Quinn, the map $\pi_3(SO(4)) \to \pi_3(STOP(4))$ is injective. So this invariant only depends on $M$ as a topological manifold.
@skupers. Can you elaborate on the conclusion? We need (1) An analogous invariant in $\pi_3(STOP(4))$ (even though it would be trivial in the end) and (2) a commutative diagram. I am not sure about either.
Replace "vector bundle" by "topological microbundle" in your construction. You get an $n$-dimensional oriented topological microbundle on $S^{n-1} \times S^1$, and for $n=4$ this is depends on an element of $\pi_3(STop(4))$ (giving (1), only depending $M$ as a topological manifold).This is the same microbundle as obtained from your vector bundle on $S^{n-1} \times S^1$ by only remembering it is a microbundle (so (2) holds).
|
2025-03-21T14:48:29.722103
| 2020-01-26T11:49:52 |
351171
|
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|
Stack Exchange
|
Mysior's example of not completely Hausdorff space
https://www.ams.org/journals/proc/1981-081-04/S0002-9939-1981-0601748-4/S0002-9939-1981-0601748-4.pdf
In this link, there is the example of regular space, that is not completely regular. This space is also completely Hausdorff (https://math.stackexchange.com/questions/3516497/is-mysiors-example-completely-hausdorff/3516547#3516547). But in the article, the second page is remark, where is written, that if we add one point b to the space, with its local neighborhoods, that the space is also regular and not completely regular, but also not completely Hausdorff, because there is not the continuous function f, for which f(a)=f(b). I am interested in, how to prove that (that this space is not completely Hausdorff).
The statement is that for every continuous real $f$ on $X$ we have $f(a)=f(b)$. Hence the conclusion that $X$ is not functionally Hausdorff.
Its clear, but I cant prove that f(a)=f(b) for every realvalued function. I tried to prove as there is proved that A and a cant be separae by continuous function, that I cant do
The extended Mysior space is not completely Hausdorff because for any sequence of open sets $(U_n){n\in\omega}$ with $+\infty\in U_n\subset\overline U_n\subset U{n+1}$, the union $\bigcup_{n\in\omega}U_n$ contains the point $-\infty$ in its closure.
Writing $+\infty$ and $-\infty$ I had in mind the points denoted by $a$ and $b$ in the Mysior's paper. But it it is better to imagine those points $a,b$ as $+\infty$ and $-\infty$.
Thank a lot. But I cant understant how that implies that f(a)=f(b) for every realvalued function
If two points $a,b$ can be separated by a continuous function $f:X\to\mathbb R$ so that $f(a)=1$ and $f(b)=0$, the the sequence of open sets $U_n={x\in X:f(x)<\frac23-\frac1{2^{n+1}}}$ has the required property: $b\in U_n\subset \overline{U}{n+1}$ and the closure of the set $\bigcup{n\in\omega}U_n$ is contained in $f^{-1}((-\infty,\frac23])$ and hence does not contain the point $a$.
So, closure of the union of Un-s contained in f^(−1)((−∞,2/3]) and so f(a) is not 1 because, f(a) is in the closure of union of Un-s. But, sorry, I cant understand why the closure of union of Un-s contains "a". I am in the shame :(
Just analyze what happens with closures and neighborhoods in the Mysior space: the closure of the strip $[n,+\infty)\times[0,2)$ in the Mysior space adds the set $(n-1,+\infty)\times{0}$ and the open neighborhood of $(n-1,+\infty)\times{0}$ is the strip $(n-1,+\infty)\times[0,2)$ with finitely many points removed from any vertical segment. Then the closure of the latter set attachs $(n-2,+\infty)\times{0}$ minus finitely many points and this process continues to $-\infty$.
I am thankful very much
You are welcome.
Modify the proof in the paper: if $f(b)<p$ then, by continuity there is an $n$ such that $f(x,0)<p$ for all $x<-n$. The same argument as in the paper now works to establish that $f(a)\le p$. Likewise if $f(b)>p$ then $f(a)\ge p$.
|
2025-03-21T14:48:29.722442
| 2020-01-26T11:52:32 |
351172
|
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|
Stack Exchange
|
A determinantal mixture of probability densities
I came up with this operation after playing with determinantal point processes:
Given two probability densities $f,g$ defined on some measurable space with reference measure $\mu$, set
$$
f\star g(x):=\frac1{2(1-c^2)}\Big(f(x)+g(x)-2c\sqrt{f(x)g(x)}\Big)
$$
where $c:=\int \sqrt{fg}\,d\mu\in[0,1]$; note that the Cauchy-Schwarz (in)equality tels you that $c=1$ $\Leftrightarrow$ $f=g$, in which case we let $f\star f:=f$. It is easy to check that $f\star g$ is a probability density.
A few comments and first observations:
Where does this come from? $f \star g$ is the mean density (a.k.a one-point correlation function) of the determinantal point process associated with the projection kernel onto $\text{Span}(\sqrt{f},\sqrt{g})\subset L^2(\mu)$, whence the choice $f\star f=f$.
$c$ is known as the Bhattacharyya coefficient, i.e. $1-c$ is the squared Hellinger distance between $f$ and $g$.
$f$ and $g$ have disjoint supports $\Leftrightarrow$ $c=0$ $\Leftrightarrow$ $f\star g=\frac12(f+g).$
$Support(f\star g) = Support(f)\cup Support(g)$
For measurable sets $A,B$ with finite measure, we have:
$$
\frac{\mathbb 1_{A}}{\mu(A)}\star \frac{\mathbb 1_{B}}{\mu(B)}(x) =
\frac1Z
\begin{cases}
\mu(B) & \text{ if } x\in A\setminus B \\
\mu(A) & \text{ if } x\in B\setminus \\
\mu(A \bigtriangleup B)& \text{ if } x\in A\cap B
\end{cases}
$$
with normalization constant $Z:=2(\mu(A)\mu(B)-\mu(A\cap B)^2).$
Thus, if you partition $A\cup B$ into the three regions $A\setminus B$, $A \bigtriangleup B$, $B\setminus A$, the probability that the random variable with density $f\star g$ lies in one region is proportional to the surface of the two others.
Now, the questions: Can you find any other interesting properties for this operation? Is this worth further digging?
I'm pretty sure there is some more fun hidden with this determinantal mixture...
|
2025-03-21T14:48:29.722581
| 2020-01-26T12:09:03 |
351173
|
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|
Stack Exchange
|
Coefficients of factors in terms of those of the product
I have asked this question on Math Stackexchange but received no reply, so I am trying it here.
If we have two polynomials $p(x)=\sum_{i=0}^n a_i x^i$, $q(x)=\sum_{j=0}^mb_j x^j$, then the coefficients of the product $p(x)q(x)=\sum_{k=0}^{n+m}c_kx^k$ are given by
$c_k=\sum_{i=0}^ka_ib_{k-i}$ (where we take $a_i=0=b_j$ for $i,j<0$, $i>n$, $j>m$). Consider the expression for the $c_k$ as a system of $n+m+1$ (non-linear) equations in the $n+m+2$ variables $a_0,a_1,\ldots,a_n,b_0,b_1,\ldots,b_m$. What is known about solving this system? In other words, can we express the coefficients $a_i, b_j$ in terms of the $c_k$'s? I think this is not possible in full generality. I could not find anything useful except for the simplest cases, such as $n=1=m$. But I would be interested in special cases, partial results, etc.
Hi, the question is not of the research level, but considering you didn't get an answer on stackexchange I gonna answer it reel quick before it is closed.
So, the questions could be (1) are there any solutions to this system and (2) how does one find these solutions;
(1) is easy over $\mathbb{C}$: any polynomial factorizes into the product over its roots $a_n \prod_{i=1}^n (x-x_i)$, and your factors are some products of this decomposition. Over $\mathbb{Q}$, for example, this problem is much harder - polynomials which do not admit non-trivial decompositions are called irreducible.
(2) Is harder (and it actually depends on what do you expect as expression). Closed formula for this kind of things is more or less as hard as closed formula for roots, and as you probably know there is no closed formula in radicals for deg > 5.
Checking irreducibility over $\mathbb{Q}$ is possible, you can google it if this is the thing you were interested.
|
2025-03-21T14:48:29.722732
| 2020-01-26T15:04:58 |
351187
|
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|
Stack Exchange
|
Alexander duality and homology equivalence
While reading the paper of Kauffman and Taylor "Signature of links" I found the following situation.
In the proof of Theorem 2.6 they suppose that two links $L_1, L_2\in \mathbb{S}^3$ are topologically concordant, i.e. that there exists a proper embedding (not necessarily locally flat) of a finite number of cylinders $F\subset \mathbb{S}^3\times I$ such that $F\cap \partial({\mathbb{S}^3}\times I)$ is $L_1 \sqcup L_2$.
They want to study the manifold $M=\mathbb{S}^3\times I\setminus F$ and they claim that the inclusion in $M$ of each of its two boundary components (that are $\mathbb{S}^3\setminus{L_1}$ and $\mathbb{S}^3\setminus{L_2}$) is a homology equivalence mod $2$.
They say that this is a corollary of the Alexander Duality, which in this context states that $H_i(\mathbb{S}^3\times I, M)\cong H^{4-i}(F, \partial F)$, but I have no idea of how to show this implication.
The assertion also holds with $\mathbb{Z}$ coefficients and the groups of interest are i=1,2. Here is an alternative to Alexander duality. Using the long exact sequence of the pair and excision, $H_1(S^3 \times I \setminus \nu F)=H_{2}(S^3 \times I,S^3 \times I \setminus \nu F)=H_{2}(\nu F,\partial \nu F)$ is generated by the meridians of the cylinders. These meridians are images of the meridians of the link under the inclusion $S^3\setminus \nu L_k \hookrightarrow S^3 \times I \setminus \nu F$. For $i=2$, the same type of argument works.
Thank you for your answer. One question: are you supposing that $F$ admits a tubular neighbourhood inside $S^3\times I$? Because I do not know whether it is true, since $F$ could be embedded in a non-locally flat way.
I found a solution for my problem. I post here a brief proof, in case anyone needs it.
I use the version of Alexander Duality, as stated in Bredon's book "Topology and Geometry" that states that if $M$ is a closed oriented $n$-manifold and $K\subset L$ are nice compact subspaces, then there is an isomorphism $$H_i(M\setminus{K}, M\setminus{L})\cong H^{n-i}(L, K)$$
Step 1: If $F\subset D^4$ is a proper (not necessarily locally flat) surface in the $4$-disc, then $H_i(D^4\setminus{F}, S^3\setminus{\partial F})\cong H^{4-i}(D^4, F)$.
Coning off $(S^3, \partial F)$ we obtain $S^4\supset X$, with $X=F\cup c(\partial F)$, where $c(\partial F)$ denotes the cone on $\partial F$. At this point we can use Alexander Duality to get
$$
H_i(S^4\setminus{X}, S^4\setminus{(D \cup X)})\cong H^{4-i}(D\cup X, X)
$$
where $D$ is the complement of a small open collar of $(S^3, \partial F)$ inside $D^4$.
To prove Step 1 one simply notes that the inclusion
$$
(D^4\setminus F, S^3\setminus \partial F)\hookrightarrow (S^4\setminus{X}, S^4\setminus{(X\cup D)})
$$
induces isomorphisms in homology, and that by excision we have
$$
H^{4-i}(D\cup X, X)\cong H^{4-i}(D^4, F)
$$
Step 2: If $L_1, L_2$ are two links in $S^3$ and $F\subset S^3\times I$ defines a (non necessarily locally flat) concordance between $L_1$ and $L_2$, then the inclusion of $\partial_-M=S^3\setminus L_1$ in $M=(S^3\times I)\setminus F$ induces a homology equivalence.
Consider the cone on $(S^3, L_2)$ so to obtain $D^4\supset X$, where $X=F\cup c(\partial_+F)$. Notice that $\partial D^4\setminus \partial X=\partial_-M$.
Now apply step 1 to obtain
$$
H_i(D^4\setminus X, \partial_-M)\cong H^{4-i}(D^4,X).
$$
One obtains the thesis observing that $D^4\setminus X$ is homotopy equivalent to M, and that since both $D^4$ and $X$ are contractibles, the homology groups $H^{4-i}(D^4, X)$ all vanish.
Alexander Duality says that if $X$ is a nice enough subspace of $S^n$ then $\tilde H^i(X)\cong \tilde H_{n-i-1}(S^n-X)$. In this case, the top and bottom of the cobordism, $(S^3,L_1)$ and $(S^3,L_2)$ can be coned off, giving $S^4$ as the ambient space and $X=F\cup c(L_1)\cup c(L_2)$ as the subspace, with $c(Z)$ indicating the cone on $Z$. Since the cones are contractible, $\tilde H^i(X)\cong H^i(F,\partial F)$ from the long exact sequence of the pair. Putting together the pieces so far gives $$H^i(F,\partial F)\cong \tilde H_{4-i-1}(S^4-X).$$ Next, since $\tilde H_{j}(S^4)=0$ for all $i<4$, the long exact sequence of the pair gives $\tilde H_j(S^4,S^4-X)\cong \tilde H_{j-1}(S^4-X)$ for $j\leq 3$. So $\tilde H_{4-i-1}(S^4-X)\cong \tilde H_{4-i}(S^4,S^4-X)$ for $4-i\leq 3$, and $S^4-X$ is homotopy equivalent to your $M$. This gives the isomorphism you asked about since I believe the Kauffman-Taylor paper only needs it in this range of dimensions.
Thank you for your answer. I have a doubt and a comment.
it seems to me that the isomorphism $\tilde{H}^i(X)\cong \tilde{H}^i(F,\partial F)$ holds when $i\ne 1$. In fact, a posteriori, I know that $\tilde{H}_{2}(S^4-X)$ has rank equal to $\mu-1$, where $\mu$ is the number of components of the link, but from your isomorphism I get that it has rank equal to $\mu$.
It is not clear to me of how from this argument I can show that the inclusion in $M$ of each of the two boundary components induces the required homology equivalence.
Oops, you're right about that isomorphism being off for the i=1 case, but looking at the Kauffman-Taylor paper, I think you only need it for i=2, at least if your question is about the isomorphism in the diagram at the top of page 353. I don't see where they talk about a homology equivalence, though. Can you provide a more precise reference?
Of course, I am referring to the proof of Theorem 2.6 at page 355. In any case, your idea of the cones is very useful and I managed to prove the homology equivalence with it. I will probably post an answer with the solution in a while. Thank you very much :)
|
2025-03-21T14:48:29.723087
| 2020-01-26T15:17:27 |
351188
|
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|
Stack Exchange
|
The diagonal of the Weil restriction
Let $Y\to X$ be a finite surjective morphism of smooth projective geometrically connected varieties over $\mathbb{Q}$. Let $k$ be a number field and consider the induced morphism
$$f:Res_{k/\mathbb{Q}} Y_k\to Res_{k/\mathbb{Q}} X_k.$$
Let $\Delta\to Res_{k/\mathbb{Q}}X$ be the diagonal embedding of $X$ into the Weil restriction $Res_{k/\mathbb{Q}} X_k$ of $X_k$ to $\mathbb{Q}$.
What are the irreducible components of the scheme-theoretic inverse image $f^{-1}(\Delta)$? Is $f^{-1}(\Delta)$ equal to $Y$, and is the morphism $f^{-1}(\Delta)\to X$ the morphism $Y\to X$ we started with?
Here $f^{-1}(\Delta) $ is defined to be the fibre product of $\Delta$ with $Res_{k/\mathbb{Q}} Y_k$ over $Res_{k/\mathbb{Q}} X_k$.
Welcome to mathoverflow, Pat!
I believe that $f^{-1}(\Delta) \neq Y$. Let $Y$ be a torsor for an elliptic curve $E$ over $\mathbb Q$ and choose $k$ so that $Y_k \cong E_k$ or in other words $Y(k) \neq \emptyset$. Choose the torsor so that the diagram $C_k\cong E_k\to \mathbb P^1_k$ commutes. I believe this can be done by letting the torsor correspond to a cocycle in $H^1(G_\mathbb Q, E[2]) \subset H^1(G_\mathbb Q, E(\overline{\mathbb Q}))$.
In general, for a $\mathbb Q$ scheme $S$, from the functor of points persepective, $$f^{-1}(\Delta)(S) = \{\lambda: S_k \to Y_k : S_k\to Y_k\to X_k \text{ is defined over } \mathbb Q\}.$$
In our particular case, let's take $S = \mathbb Q$. Then, there are no maps from $S \to Y$ but there are maps from $S \to f^{-1}(\Delta)$: Simply take the origin for instance.
From the functor of points, it is at least clear that $Y$ sits inside $f^{-1}(\Delta)$. I am not sure what else is there.
Many thanks for this!
As with pretty much everything to do with Weil restrictions, it becomes more clear if we base change everything to an algebraically closed field $\overline{\mathbb Q}$. The base change of $k$ is just $\overline{\mathbb Q}^n$ for $n$ the degree of $k$ over $\mathbb Q$. Then the base change of the Weil restriction from $k$ to $\mathbb Q$ $Y$ is the Weil restriction of $Y$ from $\overline{\mathbb Q}^n$ to $\overline{\mathbb Q}$, in other words, $Y^n$. Similarly the base change of the Weil restriction of $X$ is $X^n$. So the inverse image is the inverse image of the diagonal $X$ inside $Y^n$, which is $Y \times_X Y \dots \times_X Y$.
So the geometric irreducible components will be the geometric irreducible components of that fiber product, and the arithmetic irreducible components will be the orbits on the geometric irreducible component of $\operatorname{Gal}(\overline{\mathbb Q} / \mathbb Q)$, which is the obvious action twisted by the action permuting the $n$ embeddings of $k$ into $\overline{\mathbb Q}$.
Thank you very much for your answer. It fully clarifies things up. I hope it's ok that I accepted Asvin's answer below. Both answers are however really useful to me!
|
2025-03-21T14:48:29.723297
| 2020-01-26T16:45:41 |
351194
|
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|
Stack Exchange
|
Example of nef divisor with Iitaka dimension 0
Does anyone know examples of a numerically non trivial nef divisor with Iitaka dimension 0? (Unfortunately, this question might be trivial as right now I can't think of any effective divisors with Iitaka dim 0 other than exceptional divisors)
Here's an example that works over an uncountable field.
Let $E$ be an elliptic curve in $\mathbf P^2$ and let $p_1, \ldots, p_9$ be 9 points on $E$. Blowing up these points, the proper transform $C$ of $E$ is an irreducible curve with self-intersection 0, hence nef.
Notice that if some multiple $mC$ of $C$ moves, it must move to a curve disjoint from $C$. But that implies that the corresponding power $N_C^m$ of the normal bundle of $C$ is trivial.
Now by choosing the points $p_1, \ldots, p_9$ on $E$ appropriately, we can arrange that $N_C$ is any desired element of $\operatorname{Pic}^0(C)$. In particular since our field is uncountable we can choose it to be a non-torsion element, so no power $N_C^m$ is trivial, and therefore no multiple $mC$ moves.
This is not exactly what was asked, but I think it is close to the spirit of the question:
Mumford's example of a non-ample divisor that is positive on every curve gives an example of a numerically non-trivial nef divisor with Iitaka dimension $-1$.
Here is a sketch, you can find a complete proof in many places, or fill in the gaps yourself:
Let $B$ be a curve of genus at least $2$. Then there exists $\mathscr E$, a locally free sheaf on $B$ of rank $2$ and degree $0$ such that $\operatorname{Sym}^m(\mathscr E)$ is stable for all $m>0$. (Proof: HW).
Let $X=\mathbb P(\mathscr E)$ and $D$ the divisor corresponding to $\mathscr O_{\mathbb P(\mathscr E)}(1)$. Then $D\cdot C>0$ for any effective curve $C\subseteq X$. (Proof: HW. Hint: use the stability assumption).
So, $D$ is nef, in fact as nef as it can be $\overset{..}\smile$.
However, $D$ is not ample, because $D^2=0$. (Proof: $\deg\mathscr E=0$).
This is the usual reason this example is mentioned, that is, that being positive on curves is not enough for being ample. But it actually gives an example of a numerically non-trivial nef divisor that does not have any effective representative, because if $D$ was represented by an effective divisor, then it would be positive on it, which would contradict that $D^2=0$. (And, in fact in that case, $D$ would be ample. In other words, any non-ample divisor on a surface that is positive on every curve must have an empty linear system).
|
2025-03-21T14:48:29.723466
| 2020-01-26T17:55:20 |
351199
|
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|
Stack Exchange
|
Explicit equations for rational elliptic surfaces (Halphen surfaces)
I am looking for explicit equations for rational elliptic surfaces in characteristic $2$. For me, a rational elliptic surface $X$ is a smooth projective surface $X$ which is rational and equipped with a morphism $X\to \mathbb{P}^1$ whose general fibres are elliptic curves. I work over an algebraically closed field $k$ of characteristic $2$.
In the article "Configurations of singular fibres on rational elliptic surfaces in characteristic two", William E. Lang says that he assumes the surface to be relatively minimal (so exceptional curves in the fibres) and then obtains an equation of the form
$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
where $a_i\in k[t]$ is a polynomial of degree $\le i$ in $t$. This is then an affine version of the surface, that we can see in $\mathbb{A}^3$ with coordinates $x,y,t$.
Do we always have this kind of equations? Why? And are all such equations giving rational elliptic surfaces?
Note that the pencil of elliptic curves can be put up to birational maps into a pencil of curves of degree $3m$ with multiplicity $m$ at $9$ points (Halphen pencil). As the integer $m$ is uniquely determined by the pencil, it seems strange to me that we can bound the degree of the equations of the $a_i$ when $m$ is large enough. Or is the equation only OK for $m=1$ ? If yes, then what would be equations for $m=2$, $m=3$ or $m=4$ ?
I'm just guessing, but could that be the Weierstrass normal form of an elliptic curve over $\mathbb{A}^1$?
Do you mean the Weierstrass form of the generic fibre, which is an elliptic curve over $k(t)$? Firstly, the surface is not uniquely determined by the generic fibre, so why can we have this form? In particular why are the coefficients polynomials? Also, why can we bound the degree of the coefficients and how can we decide that the surface is rational by doing this?
|
2025-03-21T14:48:29.723613
| 2020-01-26T18:16:46 |
351201
|
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|
Stack Exchange
|
A weak definition of multiple differentiability of a function of several variables
Several times I faced the following
Definition 1: A function $f: \mathbb{R^n}\mapsto \mathbb{R}$ is called $k$ times differentiable at $x_0$ iff all the partial derivatives of $f$ of order $k-1$ are differentiable at $x_0$.
Besides this definition, more popular is the other
Definition 2: A function $f: \mathbb{R^n}\mapsto \mathbb{R}$ is called $k$ times differentiable at $x_0$ iff $f$ is $k-1$ times differentiable in a neighborhood of $x_0$ and all the partial derivatives of $f$ of order $k-1$ are differentiable at $x_0$.
It's obvious, that in case $n=1$ these two definitions coincide. Besides that, it can be shown that Def. 2 coincides with general Frechet $k$-times differentiability and obviously it's not weaker than Def 1. On the other hand, it can be shown that Def. 1 is sufficient to prove general Young's theorem.
In this regard, the following question arose:
Does anyone know an example of a function (say 2 times differentiable of 2 variables) which satisfies the Def. 1, but not the Def. 2? I have some ideas of constructing such an example with the use of Sobolev mollifications, but it's quite complicated and ugly ….
If such an example exists, is Taylor theorem with Peano remainder valid for the functions which satisfy the Def. 1?
In MathJax, one should use *stars* stars, not $\textit{math-mode fakery}$ $\textit{math-mode fakery}$, for italics. I have edited accordingly.
And one should also pay due attention to Markdown. I have edited accordingly.
Here is an example (with $k=2$) of a function satisfying Def. 1 but not Def. 2. With $r:=\sqrt{x^2+y^2}$, let
$g(x,y):=xy/r\in[-r,r]$ if $r\ne0$, with $g(0,0):=0$. Everywhere here, $x$ and $y$ are any real numbers. Then $g$ is differentiable everywhere except at $(0,0)$ (and continuous everywhere), but $g$ has both partial derivatives everywhere:
$$\text{$g'_x(x,y)=y^3/r^3\in[-1,1]$ and $g'_y(x,y)=x^3/r^3\in[-1,1]$ if $r\ne0$, }$$
and $g'_x(0,0)=g'_y(0,0)=0$.
Let
$$f(x,y):=\sum_1^\infty\frac{r^3}{j^2}\,g(x-1/j,y).$$
Then $f$ is not differentiable at any of the points $(1/j,0)$ (and hence is not differentiable in any neighborhood of $(0,0)$), so that $f$ does not satisfy Def. 2.
However, by dominated convergence, $f$ has both partial derivatives everywhere, and (as $r\to0$)
$$f'_x(x,y)=\sum_1^\infty\frac{r^3}{j^2}\,g'_x(x-1/j,y)+\sum_1^\infty\frac{3rx}{j^2}\,g(x-1/j,y)\\
=O(r^2)=o(r),$$
$$f'_y(x,y)=\sum_1^\infty\frac{r^3}{j^2}\,g'_y(x-1/j,y)+\sum_1^\infty\frac{3ry}{j^2}\,g(x-1/j,y)\\
=O(r^2)=o(r).$$
So, $f'_x$ and $f'_y$ are both differentiable at $(0,0)$. Thus, $f$ satisfies Def. 1.
Ahaha, this is the same example I was thinking of, but with $x-1/2^j$ :) First I wanted to set these functions in balls of radius $1/2^{j+3}$, extend by 0 out of these sequence of balls, then mollify this function in some kind of Sobolev manner, then multiply by $x^3+y^3$ instead of $r^3$.) But then I quickly realized, that Sobolev mollification gives me a discontinuous function, so I started thinking about the series with $1/j^2$. Guess you were faster.) Thanks, Iosif!! Remark: a typo - $3r^2x$ and $3r^2y$...
@Alexander : Yes, this kind of reasoning seems a natural one to try. As for the possible typos, I think they are not typos: $(r^3)'_x=3r^2(x/r)=3rx$. (You can also check the dimensions: if $x$ and $y$ are measured in inches, then $r^3$ is measured in inches cubed, and hence $(r^3)'_x$ must be in inches squared.
Yep, sure, sorry, here come x and y instead of what I wanted to be r .)
@AlexanderKuleshov : Do you have a further response to this answer?
@IosifPinelis, there is a small paper by Alexander Kuleshov. It is shown there that Def. 1 is sufficient for Taylor theorem with Peano reminder, as well as for Young theorem (equality of mixed partial derivatives with different order). Your example is also cited in that paper, by the way.
@Skeeve : Thank you for your comment.
Now I got the answer to the second part of my own question: let $f(x,y)$ be twice differentiable at $(0,0)$ by Definition 1. Then $f(x,y)=P(x,y)+o(r^2)$, where $P(x,y)=f(0,0)+f_x(0,0)x+f_y(0,0)y+\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy+\frac{1}{2}f_{yy}(0,0)y^2$.
Really, denote $g(x,y):=f(x,y)-P(x,y)$. Then $g(x,y)=[g(x,y)-g(0,y)]+[g(0,y)-g(0,0)]$. We have $$g(x,y)-g(0,y) = g_x(\xi_1,y)x=[f_x(\xi_1,y)-f_x(0,0)-f_{xx}(0,0)\xi_1-f_{xy}(0,0)y]x=o(\|(\xi_1,y)\|)x=o(r^2),$$ where $\xi_1=\xi_1(x,y)\in(0,x)$. Similarly, $$g(0,y)-g(0,0)=g_y(0,\xi_2)=o(\|(0,\xi_2)\|)y=o(r^2).$$
So the classical Taylor theorem assumptions can be weakened in finite-dimensional case!?? Anyone saw this somewhere??
|
2025-03-21T14:48:29.723908
| 2020-01-26T19:09:15 |
351204
|
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|
Stack Exchange
|
Commutator algorithm
Let $M \in \mathrm{SL}(2, \mathbb{Z}).$ Is there an efficient algorithm to write $M$ as a commutator (group commutator, not algebra commutator) [or fail if this is impossible]?
Addendum: answering Mark Sapir's comments (every element in $SL(2, \mathbb{R})$ is a commutator EXCEPT $-I.$:
Thompson, R. C., Commutators in the special and general linear groups, Trans. Am. Math. Soc. 101, 16-33 (1961). ZBL0109.26002.
"Polynomial time" is not very satisfying :) I would assume that for an actual free group there should be a close-to-linear-time algorithm.
@MarkSapir You might be right. The fact that the equations are diophantine (we want integer solutions) is a bit problematic, of course (we are looking at $\mathbb{Z}$ points on some variety).
@MarkSapir And this is the ONLY exception. See Robert Thompson's 1961 Transactions paper.
@MarkSapir see the OP for reference (added)
For free groups, look at the linear-programming problem studied by Calegari in his proof that scl is rational. I think that writing w as a commutator is equivalent to finding an integer point in the slice with Euler characteristic equal to -1.
|
2025-03-21T14:48:29.724025
| 2020-01-26T19:36:26 |
351205
|
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|
Stack Exchange
|
Is the space of continuous functions from Polish space to Polish space Polish?
Theorem 4.19 in Kechris' Classical Descriptive Set Theory says that the space of continuous functions from a compact metric space to a Polish space is Polish. It is therefore obvious that the space of continuous functions from a compact Polish space to a Polish space is Polish.
The space of continuous function is equipped with a product metric, for example, the Chebyshev metric.
My question is, can we drop the "compactness"?
What anomaly would arise?
If we drop the "continuity", then some problems would arise for cardinality. However, with continuity, we don't have this problem.
Consider $C(\mathbb R,\mathbb R)$.
$K$ compact, $X$ metric: For space $C(K,X)$ use uniform convergence. (I assume Kechris says this somewhere before 4.19.) With other domains, for $C(Y,X)$ try the compact-open topology https://en.wikipedia.org/wiki/Compact-open_topology When $Y$ is locally compact Hausdorff, this still has some good properties.
I suppose that if this is the question you wanted to ask over there (https://mathoverflow.net/questions/351160/x-is-polish-and-n-is-countable-is-nx-polish/351182#351182), you should edit the other question and delete this one.
@Arno. Re: "you should edit the other question and delete this one"... Please keep in mind that High GPA is a new MathOverflow poster. Let's try to be gentle and welcoming.
Chebyshev? -- r'u sure?
@Arno They are different questions, in my humble opinion, because the other question does not really "work" but this one would possibly "work" in my guesses because continuity is imposed.
I'm still confused as to what topology you have in mind to put on the space of continuous functions. You say "a product metric" but as far as I can tell, the "Chebyshev metric" would induce the uniform topology, not the product topology. Can you please specify the topology precisely?
As already answered here ($X$ is Polish and $N$ is countable. Is $N^X$ Polish?):
Let us work in a convenient category of topological spaces, ie assume that we have function spaces and sufficiently many nice spaces around. My favourite such category is the category of $\mathrm{QCB}_0$-spaces, which are $T_0$ quotients of countably-based spaces [1].
Inside the category of $\mathrm{QCB}_0$-spaces, Schröder has obtained a characterization of the Hausdorff spaces $\mathbf{X}$ such that $\mathbb{R}^\mathbf{X}$ is Polish [2]. These spaces are now called coPolish spaces. A Polish space is coPolish iff it is locally compact. So we can weaken compactness to local compactness, but not further.
A space such as $2^{(\mathbb{N}^\mathbb{N})}$ even fails to be countably based.
[1] https://www.sciencedirect.com/science/article/pii/S0304397501001098
[2] https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.200310111
The other question does not involve continuity, though, so in my humble opinion, they are quite different
|
2025-03-21T14:48:29.724252
| 2020-01-26T21:13:43 |
351211
|
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|
Stack Exchange
|
Constant Martin kernel and amenability
Consider a finitely supported random walk on a discrete group G such that the support generates $G$ as a semigroup. The Martin kernels are then non-negative functions on the product $G \times M$ where $M$ denotes the Martin boundary of the random walk. Does anyone know an example with $G$ amenable such that there does not exist a point m in the Martin boundary for which the function $K( . , m)$ is constant $1$? And does anyone know an example of a non-amenable group for which there is an element $m$ in the Martin boundary such that the corresponding Martin kernel $K( . ,m)$ is constant $1$?
The Martin boundary of any amenable group determined by a symmetric probability measure always contains the function 1 - this is a result of Northshield. I am not aware of any answers to your other questions which seem to be quite interesting.
EDIT This is an observation (too long for a comment) concerning the answer of M. Dus below. If the minimal Martin boundary (i.e., the subset of the Martin boundary which consists of minimal harmonic functions) is closed, then the group action on the minimal Martin boundary is topologically amenable. The groups that admit topologically amenable actions are called amenable at infinity. Therefore, for any group which is not amenable at infinity the minimal Martin boundary is not closed (in particular, does not coincide with the whole Martin boundary) for any random walk on the group. See Theorem 6.5 and Section 6.E in this paper by Kaimanovich.
This is a partial answer to your second question. If the function 1 is minimal, then every bounded harmonic function is constant. To reformulate, if there exists a point $m$ in the minimal Martin boundary such that $K(\cdot, m)$ is constant 1,then the Poisson boundary is trivial, which cannot happen if the group is non-amenable. So you have to look at non-minimal points.
EDIT (taking into account the comments of R W)
However, for finitely supported measures, in several classes of groups (such as hyperbolic groups and relatively hyperbolic groups with respect to virtually abelian subgroups), the Martin boundary is minimal.
So in particular, for such examples of groups, the answer is no. You thus have to look for groups such that the minimal boundary is not the whole Martin boundary (such as the examples given by R W).
Many thanks for the replies! It seems to me that Northshields arguments show that there is a point with constant Martin kernel when the spectral radius - in the random walk sense- is 1. It is therefore relevant to ask if there is a finitely supported random walk on an amenable group for which the spectral radius is smaller than 1. Despite the fame of Kestens theorem it seems that the litterature does not contain such an example.
The question in the last comment should be ignored. There are examples of finitely supported random walks on amenable wreath products with spectral radius smaller than 1 in the 2005 and 2006 papers by Brofferio and Woess. In their examples there are constant Martin kernels and hence my original questions remain.
@KlausThomsen Very interesting example. Thank you ! I wanted to look for one, but did not find time to do so.
|
2025-03-21T14:48:29.724598
| 2020-01-26T22:02:59 |
351214
|
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|
Stack Exchange
|
Composition of couplings as a pullback construction
A metric measure space $(X,d,\mu)$ consists of a metric space $(X,d)$ together with a Borel measure $\mu$. A coupling between metric measure spaces $(X_1,d_1,\mu_1)$ and $(X_2,d_2,\mu_2)$ consists of a Borel measure $\nu$ on the product metric space $X_1\times X_2$ such that $(\pi_1)_*(\nu) = \mu_1$ and $(\pi_2)_*(\nu) = \mu_2$.
I would like to see composition of couplings between metric measure spaces (explained below) as a pullback construction, in the same way that composition of correspondences between sets can be seen as a pullback.
Classical set up for composition of couplings. I follow Section 6.2 of [1]. There, it is stated that a coupling $\nu$ between measures $\mu,\mu'$ on Polish metric spaces $X$ and $X'$ can be written in the form $\mu(\mathbb{d} x)Q_\nu(x,\mathbb{d}x)$, where $Q_\nu$ is a Markov kernel from $X$ to $X'$. In particular, if $\mu_1,\mu_2,\mu_3$ are probability measures on compact spaces $X_1$, $X_2$, and $X_3$, and $\nu_1$ and $\nu_2$ are couplings between $\mu_1,\mu_2$ and $\mu_2,\mu_3$ respectively, one can define a Markov chain $(A_1,A_2,A_3)$ where $A_i$ is a random variable with distribution $\mu_i$ on $X_i$, with initial distribution $\mu_1$, and with transition kernel $Q_{\nu_i}$ at step $i$. Then, the joint law of $(A_1,A_3)$ gives a coupling between $\mu_1$ and $\mu_3$, denoted by $\nu_1\nu_2$.
My hope is that the above situation gives a diagram (in some suitable category of metric measure spaces)
\begin{array}{ccccccccc}
& & (X_1 \times X_2,\nu_1) & & & & (X_2 \times X_3,\nu_2) \\
& \stackrel{\pi_1}{\swarrow} & & \stackrel{\pi_2}{\searrow} & & \stackrel{\pi_2}{\swarrow} & & \stackrel{\pi_3}{\searrow} \\
(X_1,\mu_1) & & & & (X_2,\mu_2) & & & & (X_3,\mu_3)
\end{array}
such that the pullback of the middle cospan exist, and is given by say $(X_1\times X_2\times X_3, \nu')$, which in turn induces a coupling
$(X_1\times X_3, (\pi_1,\pi_3)_*(\nu'))$ between $\mu_1$ and $\mu_3$, that coincides with the composite coupling $\nu_1\nu_2$.
Question. Is there a reasonable category of metric measure spaces that admits at least certain kinds of pullbacks and realizes my hope?
The problems that I have encountered are similar to the problems pointed out in Is there a category structure one can place on measure spaces so that category-theoretic products exist?. I would like the morphisms to be measure-preserving, or measure non-compressing (and distance non-increasing). The projection from the product measure is of those forms, if we restrict attention to metric probability spaces, which I am willing to do. But it is not clear to me what the measure on an arbitrary pullback should be. There is a measure on an arbitrary pullback, by seeing it as a subspace of the product. But it is not clear to me that the projections are measure preserving (or non-compressing) in that case.
[1] Tessellations of random maps of arbitrary genus, Grégory Miermont.
I think the relatively independent joining is a construction in ergodic theory somewhat similar to what you’re looking for.
@AnthonyQuas I see the analogy. Are you aware of any categorical interpretation of the relatively independent joining construction?
Sorry - I can't help you there
|
2025-03-21T14:48:29.724815
| 2020-01-27T02:44:50 |
351221
|
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|
Stack Exchange
|
A question on the problem of Dirichlet
Suppose $U$ is an open set in $\mathbb{R}^{n}$ ($n\geq2$) whose complementary is not polar, and $f$ is a real-valued function defined at least on the boundary of $U$. We know that the generalized Dirichlet problem has a solution, i.e., if $f$ is continuous and all boundary points of $U$ are regular, there is a function $H_{f}^{U}(x)$ that is harmonic on $U$ and tends to $f(y)$, as $x\to y$, for all $y$ in the boundary of $U$. The same is also true if $f$ is superharmonic, but not necessarily continuous, on a neighborhood of the boundary of $U$.
Suppose now that $U$ is an open set in $\mathbb{R}^{n}\cup \{\infty\}$, the one point compacification of $\mathbb{R}^{n}$. Do the same results hold for $f$ continue , or $f$ superharmonic (not necessarily continuous)? Do you know reference for that stuff?
Not sure if I understand correctly, but I bet the answer is yes: if one looks at harmonic functions, the one-point compactification of $\mathbb{R}^n$ is just the unit sphere in $\mathbb{R}^{n+1}$ (via stereographic projection), and so $\infty$ is no different from any other point of $\mathbb{R}^n$.
@Mateusz Kwasinski: $\infty$ is different when $n\geq 3$.
@AlexandreEremenko: In what sense? Kelvin transform allows one to transform "problems at infinity" into "problems at the origin".
@Mateusz Kwasnicki: Yes, but Kelvin's transform looks somewhat different in $n=2$ and $n\geq 3$.
I found a book that talks about the problem of Dirichlet for unbounded regions. This is Lester L. Helms' book on " potential theory", Springer, 2009, Chapter 5. According to this book, the answer to the first question is yes; i.e. if $f$ is continuous on the boundary of $U$ then $H^{U}_{f}(x)$ is harmonic and tends to $f(y)$, as $x\to y$, for all regular point $y$ in the boundary of $U$. But steel I do not know the answer to the second question.
|
2025-03-21T14:48:29.724983
| 2020-01-27T04:13:47 |
351224
|
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|
Stack Exchange
|
Lower bounding decoding error in a noisy adversarial channel
Problem description
Suppose we have a finite alphabet $\mathcal{X}$, where each letter $X \in \mathcal{X}$ indexes into some fixed set of distributions, $\{P_{1},\ldots,P_{|\mathcal{X}|}\}$. For example, you might like to think of each $P_{X}$ as Gaussian with different means.
A letter $X \in \mathcal{X}$ is picked uniformly at random and sent across a channel, in which an adversary picks $k$ random letters in $\mathcal{X}$ (excluding the true $X$) to form a set $U = \{ X_1, \ldots, X_k\}$, and sends on $N$ random samples, drawn i.i.d. from the mixture distribution $\bar{P}_U = \frac{1}{K}(P_{X_1} + \ldots + P_{X_k})$, denoted $Y_1,\ldots,Y_N$ (or $Y^N$ collectively). What is the minimum probability of error that the decoding receiver can achieve?
Target solution
As $n \rightarrow \infty$, we would expect the probability of error to be lower-bounded by $1 - \frac{1}{|\mathcal{X}| - K}$, but I'm having a hard time proving that this is the case. Ideally, we would find a tight non-asymptotic bound.
My attempt at a proof
It feels like this should be a simple application of Fano's inequality. Given an estimator $\hat{X}$ for $X$, we have,
$$P(\hat{X} \neq X) \geq 1 - \frac{I(X;Y^N) + 1}{\log |\mathcal{X}|},$$
where $I(X;Y^N)$ denotes the mutual information between $X$ and $Y^N$. I haven't been able to do better than using $I(X;Y^N) \leq nI(X; Y)$, and bounding the mutual information in terms of the maximum pair-wise KL divergence, $\max_{X,X' \in \mathcal{X}}\mathrm{KL}(P_{X}\Vert P_{X'})$. Unfortunately, this lower-bounds the error by zero as $n$ grows sufficiently large.
What does for each letter we have a corresponding distribution mean? A single letter has a probability, not a full distribution. Are you talking about multiple alphabets? What is the alphabet? Is it finite?
Sorry, it is a little unclear. I'll try to explain here and then improve the wording in the question. Roughly, the adversary produces some random samples that depend on the letters they chose.
There is a single finite alphabet, $\mathcal{X}$. You can think of each $x \in \mathcal{X}$ as parameterizing a distribution $P(\cdot|x)$. The distributions can be arbitrary, though we will likely need to assume that all pairs of distributions have bounded KL divergence. You might like to think of the $P$ distributions as Gaussians with different means.
OK, and it also seems that you have $P(x|x)=0,$ for each of the conditional distributions, am I right?
I'm not sure I understand, but I think perhaps I didn't explain it clearly enough. The $P(\cdot|x)$ distributions could be defined as a density on the reals, while the alphabet itself is finite. The alphabet can be thought of as indexing into some fixed set of distributions ${P_1,\ldots,P_{|\mathcal{X}|}}$.
I was confused, I understand now, I think.
I think you'll need some dependence on support sizes of the $P$'s or total variation distance between them (not KL). If each $P_x$ is almost uniform on the same very large set, perhaps with a hole (making pairwise KL divergence infinite), you'll need a number of samples growing with the support sizes.
Yes, I think you're right. In my setting, I'm happy to assume bounded pairwise KL divergence but we can probably get some more general results by making weaker assumptions on the support size.
Consider the special case where $|\mathcal X| = 2$ and $k=1$, that is, if you pick $P_1$ the adversary picks $P_2$ and generates a sample $Y^n = (Y_1,\dots,Y_n)$ drawn i.i.d. from $P_2$, and vice versa. Then your question turns into the minimal error in a (Bayesian) binary hypothesis test. By Le Cam's lemma:
$$
\inf_{\psi} \mathbb P\big( \psi(Y^n) \neq X\big) = \frac12\big( 1 - \| P_1^{\otimes n} - P_2^{\otimes n}\|_{\text{TV}}\big)
$$
If $P_1 \neq P_2$, we have $\| P_1^{\otimes n} - P_2^{\otimes n}\|_{\text{TV}} \to 1$ as $n \to \infty$, i.e., the two product distributions eventually separate and the minimum error probability goes to zero, i.e., you can do consistent hypothesis test. (You shouldn't expect a nonzero lower bound in the limit).
For the general case, it is enough to consider how small $\mathbb P(\psi(U) \neq X)$ can be made, where $\psi : \binom{[m]}{k} \to [m]$. That is, we want the optimal decision rule
$$
\psi^* = \arg\min_\psi \mathbb P(\psi(U) \neq X).
$$
Here $\binom{[m]}{k}$ is the set of all subsets of $[m] = \{1,2,\dots,m\}$ of size $k$.
This is a Bayesian decision problem, with a 0-1 loss. It is well-known that the optimal rule minimizes the posterior risk (see Lehaman and Casella, Theorem 1.1. in Chap 4, p.228), i.e.
\begin{align}
\psi^*(u) &= \arg \min_{j \in [m]} \mathbb P( j \neq X\mid U= u) \\
&= \arg \max_{j \in [m]} \mathbb P( j = X\mid U= u) \\
\end{align}
We have
\begin{align*}
\mathbb P( j = X\mid U= u) =
\begin{cases}
\frac{1}{m-k} & j \notin u \\
0 & j \in u.
\end{cases}
\end{align*}
For any $u$, an optimal decision rule can pick any $j \notin u$ to maximize the posterior risk (the rule is not unique). Let $j(u)$ be the smallest element not in $u$. Then $\psi^*(u) = j(u)$ is an optimal rule. For this rule, we have
$$
\mathbb P(\psi^*(u) = X \mid U = u) = \frac{1}{m-k}.
$$
Taking the expectation (and using smoothing), it follows that
$$
\mathbb P(\psi^*(U) = X) = \frac{1}{m-k}.
$$
Thus,
$$
\mathbb P(\psi(U) \neq X) \ge \mathbb P(\psi^*(U) \neq X) = 1-\frac1{m-k}.
$$
To see why the above is enough, note that the error based on $Y^n$ is bounded by the problem where we have access to both $Y^n$ and $U$:
$$
\min_\phi \mathbb P(\phi(Y^n) \neq X) \ge \min_{\widetilde\psi} \mathbb P(\widetilde\psi(Y^n,U) \neq X).
$$
The optimal solution of the latter problem is again the minimizer of the posterior loss
\begin{align*}
\widetilde\psi^*(Y^n,U) &= \arg\min_{j \in [m]} \mathbb P( j \neq X \mid Y^n, U) \\
&=\arg\min_{j \in [m]} \mathbb P( j \neq X \mid U)
\end{align*}
where the second line is by the Markov property. It follows that the problem with access to $Y^n$ and $U$ has the same solution as the case where we only observe $U$ and
$$
\min_{\widetilde\psi} \mathbb P\big(\widetilde \psi(Y^n,U) \neq X\big) = \min_\psi \mathbb P\big( \psi(U) \neq X\big) =1-\frac{1}{m-k}.
$$
The above argument in fact shows that whenever $X \to U \to Y^n$ (i.e., $X$ is independent of $Y^n$ given $U$), then
$$
\min_{\phi} \mathbb P \big(\phi(Y^n) \neq X\big) \ge \min_{\psi} \mathbb P \big(\psi(U) \neq X\big),
$$
which seems to be Bayesian decision-theoretic version of the data processing inequality.
Thanks! It is nice to see that you can get asymptotic results with total variation here. However, I want to emphasize that we should not consider this special case to be a general counter example. This same argument can be applied to the general $k = |\mathcal{X}| - 1$ setting, and lines up with my predicted error bound in the question. I am especially interested in settings where $k < |\mathcal{X}| - 1$ and a nonzero lower bound is expected (as proved with the data processing inequality).
@user124784, I have added some thoughts on the general case. I think in most cases you should expect a vanishing minimum probability of error.
Actually, I made some more comments. The general case seems to exactly reduce to the case $k = |\mathcal X|-1$. See my calculation before the example.
I'm pretty sure there is a mistake here. You have assumed that the $Y_i$'s are conditionally independent given X, but this is not the case. The adversary samples only a single set $S$ which is used for all $Y_i$ samples. My partial answer uses the data processing inequality to show that there will be a non-zero error --- no matter how large $n$ is we cannot exceed the information stored in $S$ on $X$. I will try to clarify this today by adopting your improved notation into the main question.
@user124784, OK, I removed my solution for the general case. What I had was for the case where the adversary picks a new $U$ for every draw of $Y$ which is not what your problem is asking for.
@user124784, I added a solution with your conjectured bound.
This is great! The observation you gave at the end that this can be interpreted as a Bayesian decision-theoretic version of the data processing inequality is very interesting. This answers my question almost entirely! I would still like to see a way to derive a non-asymptotic bound on the error with $n$, but I consider this to be sufficiently complete to accept. I will take a look at Lehamann and Casella, and other references, to see if I can find a way to find the relevant convergence rates.
@user124784, yes, that observation seems interesting to me as well! I have not noticed this before. The lower bound holds for any $n$, i.e., it is nonasymptotic. I guess you want to see how much above the lower bound you are with a given $n$? That needs more thinking. You might be able to get something using the Chernoff bound for hypothesis testing, etc... Intuitively, you shouldn't be able to do better than a hypothesis test for determining $U$ given $Y^n$.
Yes, sorry I think my use of terminology was slightly off. I want to quantify the closeness to the bound for a given $n$. I'll have to do some related reading and see where I end up. Thanks for the help! If you have any other thoughts, please do let me know!
I managed to get a partial solution to this.
Note that the described procedure forms a Markov chain $X \rightarrow U \rightarrow Y^n$. Thus, by the data processing inequality, we must have,
$$I(X;Y^n) \leq I(X; U).$$
The latter has no dependence on $n$, and can be computed in closed form. I believe we have,
$$I(X; U) = \log \frac{|\mathcal{X}|}{|\mathcal{X}| - k}.$$
This doesn't quite match what I expected in the lower bound, due to the log-factors.
In any case, I would like to capture this behavior through a dependence on $n$. I would expect the data processing inequality to be tight in the limit of $n \rightarrow \infty$ (under some reasonable assumption on the $P$s).
|
2025-03-21T14:48:29.725849
| 2020-01-27T08:25:50 |
351227
|
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}
|
Stack Exchange
|
Constraint from level sets of an analytic function
Let $\theta$ and $f$ be two real analytic non-constant functions defined on $[0,2\pi]$. For simplicity we assume $f$ has just two critical values $m<M$ (in the picture $-1$ and $1$); we index as $\{f_j\}_{j \in\{1,2,...,N\}}$ its invertible branches. Our hypotesis is that $\theta$ and $f$ satisfy $\forall \;a\in(m,M)$
$$
\sum_j \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{|f'(f_j^{-1}(a))|}=0.
$$
This is a balancing condition for 2-dimensional vectors represented with complex numbers and involving all points in a given level set (which is finite since $f$ is analytic).
EDIT. As pointed out in the comments it was probably unclear that the hypotesis is about a pair of functions $\theta$ and $f$ that satisfy the condition above. I am not claiming this relation holds for any $\theta$. Such pair of functions can for example be constructed by considering a closed regular curve on $[0,2\pi]$ of the form
$$
\gamma(t)=\sum_{j=1}^N a_j \sin(j\cdot t)+i\sum_{j=1}^N b_j \cos(j\cdot t),
$$
and observing that for example $\int_0^{2\pi}\gamma(t) \Bigl(\cos((N+1)\cdot t)\Bigl)^n=0$, $\forall n$. Reparametrizing by arc-length $t=v(s)$ and denoting with $\theta(s)$ the turning angle of $\gamma$ we get
$$
\int \text{e}^{i\theta(s)} \Bigl(\cos((N+1)\cdot v(s))\Bigl)^n=0.
$$
From here one can show that if such property holds for $\cos((N+1)\cdot v(s))$ then it must hold for any composition of this function with $g \in L^1$. Picking $g$ as the characteristic function of $[a,a+\delta]$ and derivating in $\delta$, we get the equality above for $f=\cos((N+1)\cdot v(s))$. You can read more about that in this other question of mine:
Given $\theta$, find $f$ such that $\int_{\mathbb{T}} \text{e}^{i\theta} \cos(h \cdot f) = 0,$ for all $h \in \mathbb{N}$.
We observe
$$
0=\lim_{a \to m^+} \sum_j \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{|f'(f_j^{-1}(a))|} = - 2 \lim_{a \to m^+} \sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))},
$$
where the second equality follows from the analiticity of $f$ at inner points. Similarly, derivating in $a$, one obtains $\forall \; n \in \mathbb{N}$
$$
\lim_{a \to m^+} \frac{d^n}{da^n} \sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}=0.
$$
I would like to prove from these hypoteses that also the sum over odd indices is constantly $0$, that is $\forall \;a\in(m,M)$.
$$
\sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}=0.
$$
Note that the fact that all the derivatives are $0$ in the limit is not enough. We could theoretically have something like $\text{e}^{-1/x}$, whose derivatives go to $0$ when $x$ approches $0$, but the function is not constant. Nevertheless, I have experimental evidence that this does not happen in our setting. I tried to exploit the fact that here we are dealing with a single analytic function and that its global behaviour is determined by its Taylor expansion at any point but I could not conclude the implication this question is about.
EDIT 2. Calling our function $F(a):=\sum_{j \text{ odd }} \frac{\text{e}^{i \theta(f_j^{-1}(a))}}{f'(f_j^{-1}(a))}$, proving that it is constantly $0$ is equivalent to prove the existence of $C_\varepsilon>0$ such that
$$
\| F^{(n)}(a) \| \leq C_\varepsilon^{n+1} n!, \; \forall n \in \mathbb{N}, \; a \in (m,m+\varepsilon).
$$
In this case, the series $0 \equiv \sum F^{(n)}(a) \frac{a^n}{n!}$ would converge uniformly to $F(s)$ in a neighbohood of $0$, providing the desired identity. This implication is explained for example in One-Sided Analyticity Condition Guarantees Analytic Function?.
Although this approach looks reasonable I am still stuck. Does anyone have some hints?
The question is pretty specific but I guess an answer in the positive could connect to properties of analytic functions interesting also from a more general perspective. Thanks.
Something is crazy as written: if $\theta$ is an arbitrary analytic function, you can just interpolate it to be anything you want at the evaluation points, which will definitely destroy the suggested identity.
@fedja $\theta$ and $f$ are arbitrary as long as they satisfy the condition on the level sets on top, which must hold for all the a. They are a specific pair of functions, which satisfy the given constraint. Is that unclear?
Question has been edited. Hope now it makes more sense. Thanks for pointing that out.
Something is still strange. Let's take $f=\cos(3x)$. Then the "odd" level set is $t,t+\frac{2\pi}3,t+\frac{4\pi}3$ and the "even" one is the same with $2\pi-$ everywhere for not too large $t$. The derivatives at the pre-images are all the same in absolute value, so we are just asking if the function $g(t)=e^{i\theta(t)}+e^{i\theta(t+\frac{2\pi}3)}+e^{i\theta(t+\frac{2\pi}3)}$ can satisfy $g(t)+g(2\pi-t)=0$ without being identically $0$. Let $\theta(t)=t+\psi(t)$ where $\psi$ is real-analytic, $2\pi$-periodic. We are interested in the Fourier coefficients of $e^{i\theta(t)}$ with indices divisible by $3$. It is enough to obtain some non-trivial sequence satisfying $a_{-k}=-a_k$ for $3\mid k$. This means that we should get some prescribed sequence for the Fourier coefficients of $e^{i\psi(t)}$ with indices $k\equiv -1\mod 3$. The restriction that $\psi$ must be real relates only the pairs of indices $m$ and $-m$, so we have no relation for the indices in our set and the differential (in any reasonable Banach algebra of real-analytic $2\pi$-periodic functions) of the mapping $\psi(\cdot)\mapsto e^{i\psi(\cdot)}$ is just $i$ times the identity, so as a mapping from real functions to the sequences of Fourier coefficients with indices congruent to $-1$ modulo 3 is onto with bounds, which means that we can get any sequence that is sufficiently small, thus refuting your conjecture. Am I missing anything?
Thanks for your answer. I guess you are on point. It is just not completely apparent to me how one handles the condition $\widehat{\psi}(m)=\overline{\widehat{\psi}(-m)}$ when it comes to construct $\text{e}^{i\psi}$ with the desired coefficients using your invertibility result.
Your answer in the negative is a bit sad for me ;) since I hoped to use my equality to prove certain periodicity properties of $\theta$ and $f$, in particular that, under the sum condition on level sets, if $\text{e}^{i\theta}$ is $2\pi$-periodic then also $f$ must be. Do you have maybe an intuition about that?
@Leonardo "It is just not completely apparent to me how one handles the condition..." You just force it at every iteration step. If you want to solve $\widehat{e^{i\psi}}(k)=a_k$ for $k\equiv -1\mod 3$ and $\psi_m$ is an approximate solution giving you discrepancies $b_k$, then you just add a real-valued function whose Fourier coefficients with the indices $k$ you are interested in are $-ib_k$ and the coefficients with indices $-k$ are $i\bar b_k$. This function adds just a little to the norm but decreases the discrepancy fixed number of times, so you have a geometric convergence.
|
2025-03-21T14:48:29.726320
| 2020-01-28T20:57:42 |
351376
|
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|
Stack Exchange
|
Plane partitions with equal margins
A plane partition of $n$ is an table of integers $A=(a_{ij})$ which add up to $n$ and non-increase in rows and columns. For example,
$$A= \begin{matrix} 331 \\
32 \ \ \\
11 \ \
\end{matrix}
$$
is a plane partition of $(3+3+1)+(3+2)+(1+1)=14$.
One can view plane partitions as an arrangement of cubes stacked in the corner, see more on Wikipedia.
Define margins to be 1-dim projections of cubes on all coordinate axis. These margins are triples $(\lambda,\mu,\nu)$ of partitions of $n$. For example, for $A$ as above we have $\lambda = (7,5,2)$, $\mu=(7,6,1)$ and $\nu=(7,4,3)$.
Question: What is the smallest $n$ for which there exist two different plane partitions of $n$ with the same margins $(\lambda,\mu,\nu)$?
Note: I know there are two different plane partitions of $2100$ with equal margins. This is a consequence of $p_2(n)<p(n)^3$ for $n\ge 2100$, where $p_2(n)$ is the number of plane partitions. Unfortunately, 2100 is way too large, and $p_2(2100)\approx 1.47\cdot 10^{141}$. So it would be nice to find a small explicit example or an argument why e.g. there is none, say, for $n\le 200$.
UPDATE (Jan 29): Thank you, John Machacek, Gjergji Zaimi and Brian Hopkins, and apologies for being unclear. No, I don't worry for the symmetries, so John's first answer is the correct one. It's so nice and clean and Gjergji's argument is so simple and convincing, there is not much to add other than the motivation.
This is related to the study of Kronecker coefficients and a bound by Ernesto Vallejo (here) which should be better known:
$$g(\lambda,\mu,\nu)\ge p(\lambda,\mu,\nu)$$
where $g(\lambda,\mu,\nu)$ is the Kronecker coefficient and $p(\lambda,\mu,\nu)$ is the number of (labeled) plane partitions with margins $\lambda,\mu,\nu$. So I became interested if this bound is ever effective. From asymptotic considerations I can see that $g(\lambda,\mu,\nu)\ge p_2(n)/p(n)^3 = \exp\Theta(n^{2/3})$, but what are those partitions? I then found an easy construction of such $\lambda,\mu,\nu$ and this many plane partitions as long as there is one example with $p(\lambda,\mu,\nu)\ge 2$. So now my pathetic 2100 can be replaces with 13. Nice. I will link my paper here discussing these and other bounds once we put it on the arXiv (joint with Greta Panova).
This reminds me of an old answer of mine https://mathoverflow.net/a/293624/297
@DavidESpeyer - Indeed. This is striking. Thank you.
Hopefully I understood the definitions and computed correctly. How about $n = 13$ with
$$A = \begin{matrix} 3 3 1 \\ 2 1 1 \\ 2 \ \ \end{matrix}$$
and
$$B = \begin{matrix} 3 2 2 \\ 3 1 \ \\ 1 1 \ \end{matrix}$$
where all margins are $(7,4,2)$.
I believe your $A$ and $B$ are equivalent under the $S_3$-symmetry of plane partitions, which maybe Igor wanted to disallow.
@SamHopkins Yes they are "transpose" of each other. Because of the equal margins, the margins still coincide as ordered tuples (which is how I interpreted the question). But I will wait for clarification.
@JohnMachacek, no symmetry needed to be avoided, this is exactly the example I wanted. Thanks! See the explanation added to the question in case you are curious.
@IgorPak, great my understanding of the question was correct. I am glad the example was helpful. I look forward to seeing the paper, it sounds very interesting!
I don't know if this is optimal, but here is a pair of plane partitions of $n=53$ with no symmetries that have equal projections on each individual axis:
$$A=\begin{matrix} 5 4 4 4 4 \\ 5 4 1 1 \ \ \\ 5 2 1 1 \ \ \\ 5 1 1 1 \ \ \\ 1 1 1 1\ \ \end{matrix}$$
and
$$B=\begin{matrix} 5 5 5 5 1 \\ 4 4 1 1 1 \\ 4 2 1 1 1 \\ 4 1 1 1 1 \\ 4 \ \ \ \ \ \ \ \ \end{matrix}$$
This is, of course, a bit like cheating because we just overlapped two cyclically symmetric partitions that are transpose with an L shape to break any symmetries.
If the question doesn't care about symmetries, and is simply asking for the smallest $n$ for which some plane partition of $n$ is not uniquely determined by the projection triple $(\lambda,\mu,\nu)$ then John Machacek's example is the minimal one.
Proposition: For $n\le 12$ a plane partition of $n$ is uniquely determined by the triple $(\lambda, \mu, \nu)$.
Proof: It is simple to check that for a classical partition $\alpha$, if $\alpha$ is not determined by the length of its longest row and longest column, $(\alpha_1, \alpha'_1)$, then we must have $\min(\alpha_1,\alpha'_1)\geq 3$ and $|\alpha|-\alpha_1-\alpha'_1\geq 1$.
Now let's denote the rows of our plane partition as $\beta^1, \dots, \beta^r$ where each $\beta^i$ is a partition. If $\beta^1$ doesn't satisfy the conditions of the previous paragraph then we could reconstruct $\beta^1$ from the knowledge of its longest row and column. Both of these can be found from the information of the projections of the plane partition. Once we reconstruct $\beta^1$ we can repeat the process and reduce the problem to reconstructing $\beta^2, \dots, \beta^r$. Eventually we recover the whole partition. Moreover we can read the plane partition using a different coordinate plane and obtain the same information for $\gamma^1$ and $\delta^1$ which denote the slices of our partition on the planes $y=0$ and $z=0$.
Since each of $\beta^1, \gamma^1, \delta^1$ satisfy the conditions of the first paragraph we have that the size of our plane partition is at least
$$|\beta^1|+|\gamma^1|+|\delta^1|-\frac{1}{2}((\beta^1)_1+(\beta^1)'_1+(\gamma^1)_1+(\gamma^1)'_1+(\delta^1)_1+(\delta^1)'_1)+1$$
$$\geq |\beta^1|+|\gamma^1|+|\delta^1|-((\beta^1)_1+(\beta^1)'_1+(\gamma^1)_1+(\gamma^1)'_1+(\delta^1)_1+(\delta^1)'_1)+10\geq 13$$
as desired.
this is a very nice construction and the argument is very clean. I wish I could accept multiple answers.
Here is a smaller example than Gjergji Zaimi's that has at least less symmetry than John Machacek's. Among the plane partitions of 14,
$$ A = \; \begin{matrix} 3 & 2 & 2 & 1 \\ 3 & 1 \\ 1 & 1 \end{matrix} $$
and
$$ B = \; \begin{matrix} 3 & 3 & 1 & 1 \\ 2 & 1 & 1 \\ 2 \end{matrix} $$
both have margins $((8,4,2), (7,4,2,1), (8,4,2))$.
The smallest pairs with equal distinct margins occur with $n = 16$, e.g.,
$$ C = \; \begin{matrix} 5 & 2 & 2 & 1 \\ 3 & 1 \\ 1 & 1 \end{matrix} $$
and
$$ D = \; \begin{matrix} 5 & 3 & 1 & 1 \\ 2 & 1 & 1 \\ 2 \end{matrix} $$
both have margins $((10, 4, 2), (9, 4, 2, 1), (8, 4, 2, 1, 1))$.
(I was able to get Mathematica to check for equal margins among all plane partitions through $n = 16$ and might be able to go a little farther. Luckily these examples came up within the bounds of my programming ability and computer power.)
|
2025-03-21T14:48:29.726870
| 2020-01-28T21:02:15 |
351377
|
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|
Stack Exchange
|
When does group cohomology $H^1(G,M)$ depend only on the image of $G$ in Aut($M$)?
To motivate the question (and narrow it down if the one I asked is too broad), I'm doing readings from Manin's cubic forms book. A while back I was asked to compute the Galois cohomology $H^1(G, Pic(X_{\overline k}))$ of various Del Pezzo surfaces. My calculations seem to be fine, except I made the assumption that this cohomology group depends only on the image of $G$ in $Aut(Pic(X_{\overline k}))$. I know this isn't true in general. Manin indicates it holds in my case by Proposition 31.3 in his book, on page 179 - however, I've been informed that this proposition is probably not correct as stated, so I'd like some other reference, argument, general principle, or even heuristic that tells me when I'm on safe ground making this assumption (hopefully one as elementary as possible, since my understanding isn't extensive).
I have also posted this question here: https://math.stackexchange.com/questions/3516911/when-does-group-cohomology-h1g-m-depend-only-on-the-image-of-g-in-autm
Could you include the statement of the proposition in Manin's book?
By inf-res, you computed a subgroup and its quotient is the kernel of $\operatorname{Hom}_{\bar G}(H,M) \to H^2(\bar G, M)$ where $\bar G$ is your image and $H$ the kernel. So for instance if $H$ is torsion and $M$ is torsionfree, you have everything.
|
2025-03-21T14:48:29.727015
| 2020-01-28T21:02:25 |
351378
|
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"url": "https://mathoverflow.net/questions/351378"
}
|
Stack Exchange
|
Are there always 1-dimensional projective representations
Let $G$ be a finite $p$-group and $(\mathbb{F},\mathcal{O},\mathbb{K})$ a sufficiently large $p$-modular system. Furthermore let $\mu_{p^\infty}\leq \mathcal{O}^\times$ the group of roots of unity of $p$-power order.
If $\alpha$ is a 2-cocyle with values in $\mu_{p^\infty}$ does there always exist a one-dimensional projective representation $G\to\mathcal{O}^\times$ with cocylce $\alpha$ ? In other words: Does the twisted group algebra $\mathcal{O}_\alpha[G]$ always have a one-dimensional representation $\mathcal{O}_\alpha[G]\to\mathcal{O}$?
I might be confused, but isn't a one-dimensional representation equivalent to a trivialization of $\alpha$ (i.e. a realization of $\alpha$ as a coboundary)? Without thinking more carefully, I guess there could be kernel in $H^2(G; \mu_{p^\infty}) \to H^2(G; \mathcal{O}^\times)$. But then I think you are asking if this is the zero map, which seems unlikely IMHO.
I don't think so. For example cyclic groups fit into a non-splitting extension $1\to C_p \to C_{p^2} \to C_p \to 1$ which defines a non-trivial cocycle $\alpha$, but there is a morphism $\mathcal{O}_\alpha[C_p] \to \mathcal{O}$ which maps a generator of $C_p$ to a primitive $p^2$th root.
And if $\mathbb{K}$ is algebraically closed for example, then $H^2(G,\mu_{p^\infty}) \to H^2(G,\mathcal{O}^\times)$ is even an isomorphism. First $H^\ast(G,\mu) \to H^\ast(G,\mathcal{O}^\times)$ ($\mu$ being the subgroup of all roots of unity) follows from the fact that $\mathcal{O}^\times / \mu$ is uniquely divisible, $\hat{H}^\ast(G,uniquely divisible)=0$ and the long exact sequence. And $H^2(G,\mu) = H^2(G,\mu_{p^\infty})$ follows because $G$ is a $p$-group.
Ah, dammit. The answer is no and the first group I would have looked at next is the counterexample... the extra special group $G=\langle x,y,z \mid x^p=y^p=[x,y]=z, z^p=1\rangle$ fits into an non-trivial extension $1\to C_p \to G \to C_p\times C_p\to 1$ and the corresponding 2-cocycle $\alpha$ gives us the counterexample, because $\mathcal{O}_\alpha[C_p\times C_p] = \mathcal{O}\langle X,Y \mid X^p=Y^p=XYX^{-1}Y^{-1}=\zeta_p\rangle$ does not have one-dimensional representations into something of characteristic zero. If $\rho$ was such a thing, $\rho(X)$ and $\rho(Y)$ would commute so that $\rho(X)\rho(Y)\rho(X)^{-1}\rho(Y)^{-1}$ would have to be both equal to $1$ and to $\zeta_p$.
While I sympathise with your frustration, you could probably delete the first sentence, and maybe even the second sentence up to the ellipsis.
By the way, for people like me who aren't finite-group theorists, it may be helpful to remark that this extra-special group is perhaps also known as the 3-dimensional Heisenberg group over $\mathbb F_p$.
|
2025-03-21T14:48:29.727250
| 2020-01-28T22:35:30 |
351382
|
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|
Stack Exchange
|
Reflecting Boundary conditions for advection-diffusion equations
I am trying to model the dynamics of phytoplankton in a water column using one-dimensional advection-diffusion partial differential equations.
$$\frac{\partial P}{\partial t}= D\frac{\partial^2 P}{\partial x^2}-v\frac{\partial P}{\partial x}+rP, $$
where D is the diffusion coefficient and $v$ is advection. Both $D$ and $v$ are constant.
The topwater surface is at $x=0$ and the bottom is at $x=L$. The advection velocity is positive downward. I am using the zero-flux boundary condition on the top surface, which is
$$ D\frac{dP(0)}{dx}-v P(0)=0. $$
My question is about the zero-flux boundary condition at the bottom of the water column. In most papers, the zero-flux boundary condition at the bottom is written also
$$ D\frac{dP(L)}{dx} - vP(L)=0. $$
However, since phytoplankton can not escape the boundary (no-flux), so is not it should be
$$ D\frac{dP(L)}{dx} + vP(L)=0? $$
That is phytoplankton should bounce back at the bottom, so advection velocity will be in the upward direction so + sign with advection term instead of negative at the bottom boundary at $x=L$?
Any thoughts on it, please?
Please define P and $v$. If $v$ a constant ?
Yes, both $D$ and $v$ are constant.
On the one hand, you say that the advection velocity $v$ is constant. On the other hand, you say it is downward at the top edge and upward at the bottom edge. How are these two statements reconciled?
The usual boundary condition
$$
D\frac{dP}{dx} -vP = 0
$$
seems correct regardless of whether one is considering $x=0$ or $x=L$. Note that the relative sign between the two terms has nothing to do with the orientation of the boundary surface. Instead, it encodes the relative orientation of the diffusion and advection effects, and that relative orientation is the same everywhere. Diffusion is always in direction of decreasing density $P$, and advection is always in direction of $v$.
For example, consider $P$ decreasing as $x$ approaches $L$. This means there will be diffusion towards the boundary. That has to be compensated by an advection velocity $v$ away from the boundary at $L$, which in your convention means negative $v$. Thus, $D \cdot dP/dx < 0$, and $vP<0$; to cancel, these have to subtracted from one another, corroborating the validity of the usual boundary condition.
Thank you so much and it makes much more sense. So just want to make sure I understood it correctly, advection velocity can not be constant than since it changes direction due to this relative orientation of diffusion and advection?
It seems to me that a solution with constant $v$ should be possible - it just implies that diffusion at one boundary is towards, at the other away from the boundary. I.e., $dP/dx$ has the same sign at the two boundaries.
Thanks again. I really appreciate your help.
|
2025-03-21T14:48:29.727460
| 2020-01-28T23:07:00 |
351384
|
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"Martin Brandenburg",
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|
Stack Exchange
|
Characterization of geometric morphisms without referring explicitly to the left adjoint?
Recall that a functor $f_\ast : \mathcal E \to \mathcal F$ between toposes is called a geometric morphism if it has a left exact left adjoint $f^\ast$. Is there an intrinsic characterization of such functors which doesn't refer explicitly to the left adjoint? Let's assume our toposes are Grothendieck.
Of course, the condition that $f_\ast: \mathcal E \to \mathcal F$ admit a left adjoint $f^\ast$ is equivalent, by the adjoint functor theorem, to the condition that $f_\ast$ preserves limits and sufficiently-filtered colimits. But I don't know how to encode the condition that $f^\ast$ preserves finite limits purely in terms of $f_\ast$.
Again by the adjoint functor theorem, one can say when $f^\ast: \mathcal F \to \mathcal E$ is the left adjoint of a geometric morphism $f_\ast$ without referring to $f_\ast$ directly -- it means that $f^\ast$ preserves colimits and finite limits. I'm looking for something similar which refers only to $f_\ast$ and not to $f^\ast$.
I'm also interested in the $\infty$-topos version of this question. It would also be interesting to know the answer when $\mathcal E, \mathcal F$ are Grothendieck abelian categories rather than toposes.
Is there some type of object $X$ so that $\hom(-,X)$ takes finite limits to colimits? And are there enough of them to characterize the finite limits? If so, then you are asking that $f_*$ preserves that class of objects.
Some sort of ind- or pro-thing?
@TheoJohnson-Freyd Unfortunately no. Suppose that $Hom(\emptyset \times \emptyset, X) = Hom(\emptyset, X) \amalg Hom(\emptyset, X)$. Then since $\emptyset \times \emptyset = \emptyset$ (this is true in any cartesian closed category, for example), we get $2=1$, a contradiction. Here $\emptyset$ is the initial object. There's some sliver of chance such things might exist in the abelian context...
Small remark: $f^(1)=1$ means that $\Gamma \circ f_= \Gamma$, where $\Gamma = \mathrm{Hom}(1,-) : \mathcal{F} \to \mathsf{Set}$ are global sections.
Also, $f^$ preserves binary products iff the canonical morphisms $f_[f^(B),A] \to [B,f_(A)]$ are isomorphisms (but here still $f^*$ appears, but at least no products).
|
2025-03-21T14:48:29.727650
| 2020-01-28T23:20:20 |
351385
|
{
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"authors": [
"Ian Agol",
"Laiyuan Gao",
"Robert Bryant",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351385"
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|
Stack Exchange
|
The isometric immersion of a positively curved projective plane in 3-dimensional Euclidean space?
In 1903, W. Boy showed that the real projective plane $\mathbb{R}P^2$ can be immersed in the Euclidean space $\mathbb{E}^3$ (see Werner Boy, Math. Ann. 57 (1903), no. 2, 151-184.). Suppose a Riemannian surface ($\mathbb{R}P^2$, g) has positive sectional curvature $K(x)$ everywhere. $K(x)$ needs not to be a constant. The question is: can this surface isometrically immersed in $\mathbb{E}^3$. I guess that there is no such an immersion. But I did not find a relative reference or a direct proof by myself.
As is well known that D. Hilbert proved that a hyperbolic plane can not be isometrically immersed in $\mathbb{E}^3$ (see D. Hilbert, Trans. Amer. Math. Soc. 2 (1901), no. 1, 87-99.). This is the motivation of the above problem.
I don’t think so: the immersion of the double cover of the sphere would be locally convex, hence globally convex, hence embedded. Or more easily, the projective plane would have a well-defined normal from mean-curvature, giving a section of the normal bundle. But since the pullback of the tangent bundle to R^3 is trivial, this would give a contradiction to the tangent bundle to RP^2 being nontrivial.
Ian is correct. In fact, the open problem is whether $\mathbb{RP}^2$ can be immersed in $\mathbb{R}^4$ so that the induced metric has non-negative curvature. (It can be embedded in $\mathbb{R}^4$ smoothly, and it can be embedded in $\mathbb{R}^5$ with a metric of positive curvature, in fact with constant positive curvature.)
If you allow immersions which are only C^1, you can use the Nash-Kuiper theorem to deform Boy's embedding to an isometric one.
To Robert: Are there any references related to your statements? Thank you.
To Thomas: I am not very clear how the curvature becomes positive as the surface deforms?
@LaiyuanGao: I don't know any recent references. There is an embedding of $\mathbb{RP}^2$ into $\mathbb{R}^5$, equivariant under an irreducible representation of $\mathrm{SO}(3)$ on $\mathbb{R}^5$, that has constant (positive) Gauss curvature. Gromov, in his 1986 book, Partial Differential Relations, asks whether $\mathbb{RP}^2$ can be embedded into $\mathbb{R}^4$ with positive Gauss curvature (p. 279, Question (e'')). I saw a preprint in 2005 that claimed to construct such an immersion, but it had a serious error and was never published, as far as I know. That's all I know.
@IanAgol: Is the 2-fold sphere you mentioned is an immersion of the projective plane into $R^3$?
@RobertBryant: Thank you for your references. If $RP^2$ can be immersed into $R^3$ with positive Gauss curvature then it seems that it can be embedded into $R^4$ with positive Gauss curvature by a deformation of that immersion in $R^3$ to an embeding in $R^4$ via removing self-intersections? I am not very sure.
@LaiyuanGao: Are you interested in $C^1$ isometric immersions or only interested in isometric immersions that are at least $C^2$ (or some other regularity). If you are satisfied with a $C^1$ isometric immersion, then, yes, by Nash-Kuiper, there is a $C^1$-isometric immersion of $(\mathbb{RP}^2,g)$ into $\mathbb{E}^3$ (as Thomas Richard said). However, if the immersion is to be everywhere $C^2$, then the metric $g$ cannot have positive Gausß curvature everywhere, as Ian Agol's argument shows. About the $C^1$ deformation to an embedding in $\mathbb{R}^4$ question, I am not sure.
@IanAgol: Now I understand your comment. Forget my above question. Thank you.
@RobertBryant: Thank you for your clarification. I considered the immersion which is everywhere $C^2$. Should a locally convex surface be globally convex? In fact, we have lots of closed and locally convex curves in the plane not embedded.
@RobertBryant: Understand. Due to Hadamard's theorem. Thanks.
@LaiyuanGao: That's true, but it can't happen for a compact surface in $3$-space if the Gauß curvature is positive everywhere. As Ian pointed out, the surface would have to be orientable, and then the Gauss map will make the surface into a covering space of the sphere, which ensures (if the surface is connected) that the surface is a $2$-sphere and that the Gauß map is 1-to-1. (This is what is different from the locally-convex-curves-in-the-plane case.)
Ian's argument of mean curvature is wonderfully simple. Here is another one. Rotate your surface to put it in generic position with respect to the heigth function z; then, the preimage of z is a Morse function f on RP^2, which has no critical point of index 1 (saddle point) since the surface is locally convex. Hence, every critical point of f has index 0 or 2 (local extremum). By Morse theory (or by the Poincaré-Hopf formula applied to a gradient-like vector field), the number of local extrema is the Euler characteristic of RP^2, hence 1. This is a contradiction, since there are at least one minimum and one maximum.
|
2025-03-21T14:48:29.727983
| 2020-01-28T23:20:23 |
351386
|
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"Johannes Hahn",
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|
Stack Exchange
|
Convolution algebra associated to a finite dimensional algebra
Given a finite dimensional $k$-algebra $A$ (we can assume it is given by a connected quiver with relations). One can form its trivial extension $T(A)$ (see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra ), which is a Frobenius algebra and thus it has a coalgebra structure, see for example the book by Kock on Frobenius algebras and 2D topological quantum field theories.
Now given any $k$-coalgebra $C$ and $k$-algebra $A$, we can form the convolution algebra $W_{C,A}={\rm Hom}_k(C,A)$.
This gives rise to (at least) two new algebras from a given algebra $A$, namely $W_{T(A),A}$ and $W_{T(A),T(A)}$ using that $T(A)$ has a coalgebra structure as a Frobenius algebra. I have no real experience with this topic, so sorry in case my questions are stupid.
Question 1: Have either of the algebras $W_{T(A),A}$ or $W_{T(A),T(A)}$ been considered/described before? Are they quiver algebras and if so, can their quivers be described?
Question 2: What do those two algebras look like in the concrete example where $A=kQ$ is a path algebra (maybe of Dynkin type)?
Question 3: In case those questions are non-trivial, can those algebras be calculated by a computer algebra system like GAP/QPA?
For the benefit of anyone reading who, like me, did not know the terminology/definition of $T(A)$: it is what a Banach algebraist might write as $A\ltimes A^$ where $A^={\rm Hom}_k(A,k)$, and the crossed-product notation indicates that we think of $A^$ as equipped with trivial multiplication with $A$ acting on $A^$ from left and right in the canonical way. Alternatively, elements of $T(A)$ can be viewed as "matrices" of the form $\begin{pmatrix} a & b^* \ 0 & a \end{pmatrix}$ with $a\in A$ and $b^\in A^$; the multiplication on $T(A)$ is then given by "matrix multiplication"
Just so I understand the concept of convolution algebra correctly: The convolution algebra is what we get when we use $Hom_k(C,A)\otimes Hom(C,A) = (C^\ast\otimes A) \otimes (C^\ast\otimes A) = (C\otimes C)^\ast\otimes (A\otimes A)$ and then apply multiplication and the dual of the comultiplication. Am getting this right? If so, I think $W_{C,A}$ is isomorphic to $W_{C,k} \otimes A$, correct?
@JohannesHahn I would think there is only one canonical way to get an algebra structure on this and it should be what you said. It looks like you are correct and so the convolution algebra might be rather boring, at least as a quiver algebra.
Indeed. If $C$ is a Frobenius algebra, then the convolution algebra $W_{C,k}$ is just $C$ itself via the isomorphismus $C\cong C^\ast$ that comes from the Frobenius structure.
well, I made a small mistake. I think, $C^{op} \cong W_{C,k}$ is the right answer.
Question 3: The algebra constructions $W_{C,A}$ or even $W_{T(A),A}$ are not available in QPA. There are no structures made for co-algebras in QPA.
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