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2025-03-21T14:48:30.932156
| 2020-05-10T07:25:24 |
359899
|
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"Brendan McKay",
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|
Stack Exchange
|
How many ways to cover a N×N chessboard with white and black boxes by some restrictions?
Suppose we have a N×N chessboard and the boxes ■, □.
We should cover the chessboard with those boxes but there can not have the 2×2 square $\scriptstyle{\begin{array}{cc}\square&\square\\
\square&\square\end{array}}$
on the chessboard.
Can we calculate the ways to cover the chessboard?
Furthermore,
(1) if we connect the up(left) boundary and the down(right) boundary together, which means the up(left) and down(right) boundary form a 2×N(N×2) rectangle and it should not have the 2×2 square
$\scriptstyle{\begin{array}{cc}\square&\square\\
\square&\square\end{array}}$
too.
Can we calculate the ways?
(2) if we have k black boxes, with the (1) chessboard, can we calculate the ways?
The first one is http://oeis.org/A139810 which has no formula or other information. I doubt that a simple formula exists, but it would be quite interesting to know what the asymptotic behaviour is.
a closed-form expression for case 1 is given by Simon Cowell https://arxiv.org/abs/1506.03580 ; references for upper and lower bounds are also given.
the Cowell expression is given as Mathematica code in http://oeis.org/A255936
@CarloBeenakker About $3^n$ operations if I understand it.
Thank you @CarloBeenakker for your edit and comments, it helps me understand the problem deeply.
Thank you @BrendanMcKay for your comment, it's kinda sad to know that there has no simple formula of the problem. the upper and lower bounds are given in KKP93 and SL90.
The link [KKP93] is wrong and I relink it here KKP93
|
2025-03-21T14:48:30.932297
| 2020-05-10T08:52:11 |
359904
|
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"Christoph Mark",
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|
Stack Exchange
|
Integrals and finite dimensionality in braided Hopf algebras
Let $H$ be a Hopf algebra with invertible antipode. Let $A$ be a braided Hopf algebra in the Yetter-Drinfeld category ${}_H^H\mathcal{YD}$ over $H$.
A nonzero left integral in $A$ is a nonzero element $x\in A$ such that $yx=\epsilon(y)x$ for all $y\in A$. A nonzero right integral in $A$ is a nonzero element $x\in A$ such that $xy=\epsilon(y)x$ for all $y\in A$.
If $A$ is a classical Hopf algebra with trivial braiding, it is known that the existence of a nonzero left integral or nonzero right integral in $A$ implies finite dimensionality of $A$. This is a result due to Sweedler, Integrals for Hopf algebras, Annals of Mathematics, 1969.
My question concerns the braided case: Is the analogous result true for every braided Hopf algebra $A$? Does the existence of a nonzero left integral or nonzero right integral in $A$ imply finite dimensionality?
A reference for this general result would be welcome. If it helps, you could assume additionally that $A$ is a $\mathbb{Z}_{\geq 0}$-graded braided Hopf algebra, and if it helps further, you could assume additionally that $A$ is a connected $\mathbb{Z}_{\geq 0}$-graded braided Hopf algebra (i.e. $A^0=\mathbf{k}1$ where $\mathbf{k}$ is the ground field).
Remark. The proof of Sweedler for classical Hopf algebras does not directly carry over to the braided case.
I think the answer to your question is affirmative.
Consider the Radford biproduct $A\rtimes H$ defined in [1], which is an ordinary Hopf algebra over $\Bbbk$ defined on the vector space $A\otimes_{\Bbbk} H$. This construction does not require $A$ to be finite-dimensional over $\Bbbk$. The following was shown in [2, Section 4.6]: If $x$ is a non-zero left integral for $A$ and $\Lambda$ is a non-zero left integral for $H$, then $$\Lambda_{(1)}\cdot x\otimes \Lambda_{(2)}\in A\rtimes H$$
is a non-zero left integral of $A\rtimes H$. Hence, $A\rtimes H$ is finite-dimensional by the classical Larson--Sweedler result. In particular, $A$ is finite-dimensional over $\Bbbk$.
[1] Radford, David E., The structure of Hopf algebras with a projection, J. Algebra 92, 322-347 (1985). ZBL0549.16003.
[2] Burciu, Sebastian, A class of Drinfeld doubles that are ribbon algebras., J. Algebra 320, No. 5, 2053-2078 (2008). ZBL1163.16025.
Thank you, I was thinking about the same solution but had no references. I am sure it is true like this, now.
|
2025-03-21T14:48:30.932572
| 2020-05-10T09:43:54 |
359907
|
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"Dmitri Pavlov",
"Maxime Ramzi",
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|
Stack Exchange
|
Symmetric monoidal structure of the heart of $S^1$-spectra
How to give a symmetric monoidal structure of $SH^{S^1}(k)^{\heartsuit}$ (after $\mathbb{A}^1$-localization)? The standard answer is
$$E_1\otimes E_2:=(E_1\wedge E_2)_{[0,0]}$$
but I don't see why truncation functors are compatible with smash products.
I don't know about the case you're interested in, but for usual spectra and usual chain complexes, you need the LHS to be a smash product over $H\mathbb Z$ ( or $Hk$ if your chain complexes are over $k$) - is that not something that would matter here ?
@MaximeRamzi Here the smash product is between spectra of simplicial sheaves (of sets), not simplicial abelian sheaves.
@NanjunYang: The smash product is defined for spectra of simplicial sheaves of sets.
For the new version of your question, you may reduce to connective objects (objects of $\mathscr C_{\geq 0}$, $\mathscr C$ being your stable $\infty$-category. So you only need to prove the following : if $E$ is connective, and $f$ is a $0$-equivalence between connective objects, then so i $E\otimes f$. Again, I don't know your specific situation so I don't know how easy that is, but it may be good to know that this suffices. If your smash product is sufficiently nice (which I assume it is), it suffices to show that connective objects are stable under tensor product
What I call $\otimes$ in my previous comment is what you call $\wedge$
@MaximeRamzi I've understood that $(HM\wedge HN)_{\leq 0}=H(M\otimes N)$.
|
2025-03-21T14:48:30.932698
| 2020-05-10T10:43:20 |
359911
|
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"Bill Johnson",
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|
Stack Exchange
|
Existence of injective compact operators
We know that if $X$ is a separable Banach space, then for every infinite dimensional Banach space $Y$, there exists an injective compact operator from $X$ to $Y$.
My query is for every Banach space $X$ (need not be separable ) do there exist a Banach space $Y$ and an injective compact operator $T:X\to Y$?
No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X = \ell_\infty^*$. Then you have no chance of building such operators.
Another possibility for counterexamples comes with non-separable reflexive spaces (or, more generally, WCG spaces), which contain non-separable weakly compact sets. Injective bounded linear operators are then homeomorphic embeddings of such sets with respect to the weak topology, so cannot be compact.
I was actually trying to check that it never exists if $X$ is non-separable.
@YCor, take $T\colon \ell_\infty\to c_0$ given by $T (\xi_k){k=1}^\infty = (\xi_k / k){k=1}^\infty$. It is compact and injective.
Thanks, I can see why I was stuck! In any case this provides useful context.
Perhaps it is worth pointing out that $X$ has the property if and only if $X^$ is weak$^$ separable, which is a somewhat weaker condition than that $X$ embeds isomorphically into $\ell_\infty$.
@BillJohnson, with the first known example being your space $J! L$ :-)
@TamaszKania: And a better one in our recent joint paper. :-)
|
2025-03-21T14:48:30.932829
| 2020-05-10T11:00:55 |
359913
|
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"cll",
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|
Stack Exchange
|
The state of art of the singular Levi problem -- and hyperkähler quotients
One of the versions of the classical Levi problem asks the following:
Let $X$ be a complex manifold. Is it true that $X$ is Stein iff
$X$ admits a smooth exhaustion strictly plurisubharmonic function?
The answer was proved to be affirmative by [Grauert68], I believe.
When we allow $X$ to have singularities we come straight away to a problem of defining a strictly plurisubharmonic function. Narasimhan uses the following definition in his proof of the singular version of Levi problem (see [FN80]).
Let $X$ be a reduced complex analytic space. A function $\rho\colon X\to \mathbb R$ is called plurisubharmonic if for every point $x\in X$ there exist an embedding of a neighbourhood $U$ of $x$ into $\mathbb C^N$ such that $\rho|_U$ comes as the restriction of a plurisubharmonic function $\rho'\colon \mathbb C^N\to \mathbb R$ to $U$. A plurisubharmonic function is called strictly plurisubharmonic if its small enough local perturbations are also plurisubharmonic functions.
Please check my words as I'm not an expert in complex analysis.
As far as I understand, Narasimhan does not impose any smoothness assumptions on plurisubharmonic functions in his proof. I wonder if a definition of a "smooth" strictly plurisubharmonic function can be given in easier terms, in terms of the $\partial\overline{\partial}$-operator for instance.
It seems too naïve to think that the following definition of a strictly plurisubharmonic function could allow one to make a conclusion about Steinity.
Let $X$ be a reduced complex analytic space. A function $\rho\colon X\to\mathbb R$ is called smooth strictly plurisubharmonic in a naïve sense if the following holds
The restriction of $\rho$ to the set of non-singular points of $X$ is smooth.
The form $\sqrt{-1}\partial\overline{\partial}\rho$ is a strictly positive $(1,1)$-form i.e. its restriction to the tangent space of any (maybe singular) point is strictly positive.
Now assume we have an exhaustion strictly plurisubharmonic function on $X$ in the sense of this naïve definition. What are we able to conclude about $X$? Which further assumptions on $\rho$ or on $X$ should one impose to conclude that $X$ is Stein?
My motivation for the question comes from the following observation made in [HKLR87].
Let $X$ by a hyperkähler manifold equipped with a HKLR-compatible $U(1)$ action i.e. the one that rotates complex structures. More precisely, for every $\lambda\in U(1)$
$$
\lambda^*\omega_I = \omega_I \:\:\:\:\: \lambda^*\Omega_I = \lambda\Omega_I
$$
where $\omega_I\in \Lambda^{1,1}_IX$ is the Kähler form and $\Omega_I\in\Lambda^{2,0}_I X$ is the holomorphic symplectic form. Let $\mu\colon X\to \mathbb R$ be a moment map for this action i.e.
$$
d\rho = \iota_\varphi\omega_I
$$
where $\varphi$ is the vector field tangent to the $U(1)$-action.
Choose a complex structure $J\in\mathbb H$ which anticommutes with $I$. Let $\overline{\partial}_J$ be the $\overline{\partial}$-operator on differential forms for the complex structure $J$. Then
$$
-\sqrt{-1}\partial_J\overline{\partial}_J \rho = \omega_J
$$
In particular, $-\rho$ is a smooth strictly plurisubharmonic function.
The problem with the observation above is that most of examples of HKLR-compatible $U(1)$-actions arise on hyperkähler quotients of finite or infinite dimensional affine spaces. They are quite rarely smooth. So if one wants to conculde anything about Steinity of $X_J$ one needs an understanding of the notion of plurisubharmonicity in the singular case.
(a little bit disappointing) postscriptum. I am able to prove that under mild assumptions hyperkähler quotients of affine spaces are affine varieties when regarded as complex varieties with the complex structure $J$. My proof uses another methods. So alas, these varieties do not give us any interesting examples of Stein non-affine varieties. However, it is very tempting for me to apply some singular Levi problem to conclude more or less straightforwardly that these varieties are Stein. The exhaustion condition is not too difficult to check.
Bibliography:
[Grauert68] Grauert, H.: On Levi's problem and the imbedding of real-analytic manifolds. Ann. of Math. 68, 460-472 (1968)
[FN80] Fornaess, J.E.; Narasimhan R.: The Levi Problem on Complex Spaces with Singularities. Math. Ann. 248, 47-72 (1980)
[HKLR87] Hitchin, N.J.; Karlhede, A.; Lindström, U.; Rocek, M.: Hyperkähler metrics and supersymmetry. Commun. Math. Phys. 108 (1987), 535 -- 589
Privet, Anya.
Your reference [FN80] actually seems to contain an answer to this problem! They state (in particular, see the question 1.5 in the introduction) that the class of weakly psh functions, i.e. functions which are psh being restricted on any holomorphic disk coincides with the class of psh functions in the restriction sense.
I think it is implied by your condition (I do not precisely understand the assumptions, but clearly if $\partial \bar \partial f$ is well defined and positive then its restriction on any curve is nonnegative too).
Privet, dorogoy. I mean the Zariski tangent space (https://en.wikipedia.org/wiki/Zariski_tangent_space). You made a nice remark about curves. I believe that the arguments with disk embeddings allow to give a definition of psh as "$\partial\overline{\partial}\rho$ is non-negative on the smooth part" plus the continuity of $\rho$ -- and it's equivalent to the one of Narasimhan (correct me if i'm wrong). But I'm not at all sure if the same argument should work for strictly psh functions.
As far as I remember, the definition of $\partial\overline{\partial}\rho$ exists but isn't really tame. I don't know any reference.
|
2025-03-21T14:48:30.933152
| 2020-05-10T11:14:43 |
359914
|
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|
Stack Exchange
|
Decomposition of cohomology via algebraic cycles (or correspondences)
In the paper " On the calculation of local terms in the Lefschetz-Verdier trace formula and its application to a conjecture of Deligne" by Richard Pink, he wrote in the first page that
To obtain finer information one would like to split up the total cohomology using algebraic cycles and to describe the Galois representations on the individual factors. The process of splitting can be described in terms of correspondences; hence one wants to use a Lefschetz trace formula for the twist of a correspondence by Frobenius.
I already see some sort of examples on modular curves (and Shimura varieties): One obtain the Galois representation corresponding to a cusp form by a "cut out" process from the cohomology of modular curves via the action of the Hecke algebra, which can be interpreted as Hecke correspondences. I am curious on the purely geometric picture: namely on a "good enough" (smooth, projective, etc) algebraic variety $X$ over a field $k$, we can define the $\mathbb{Q}$-algebra of correspondence $\mathrm{Corr}(X/k)$, generated by isomorphism classes of diagrams $$X\xleftarrow{c_1}Z\xrightarrow{c_2}X$$ where $c_1$ is proper and $c_2$ is finite etale. This algebra acts on the etale cohomology groups by $c_1^*c_{2!}$. So my question is
What can we say about the structure of the $\mathbb{Q}$-algebra of correspondences? Is there a general description of the decomposition of the etale cohomology of $X$ under the action of the algebra of correspondences? Is there a reference on this?
Intuitively, the action of the correspondence algebra commutes with the Galois action, so should give smaller Galois representations as in the modular curve case.
The keyword for this question is "motive".
@Sasha Thanks for this hint. I am not familiar at all to the theory of motives so I didn't realize they are related.
|
2025-03-21T14:48:30.933313
| 2020-05-10T11:24:30 |
359915
|
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|
Stack Exchange
|
Does every cofinal branch through Kleene's O compute true arithmetic?
My question concerns cofinal branches through Kleene's $O$, which is a set of natural numbers and a computably enumerable relation $<_O$ on this set that provides
ordinal denotations for any desired computable ordinal. For every number $n\in O$, the $<_O$-predecessors of $n$ in $O$ are a computably enumerable set of natural numbers that is well-ordered by $<_O$,
representing a computable ordinal, and every computable ordinal is
represented in this way. Meanwhile, the set $O$ itself is not computable nor even hyperarithmetic, for
it is $\Pi^1_1$-complete.
I am interested specifically in the complexity of cofinal branches through
Kleene's $O$. Let us say that $z$ is a cofinal branch through $O$
if $z\subset O$, the members of $z$ are linearly ordered by $<_O$,
and $z$ contains one index of every computable ordinal rank.
My intuition is that such branches must have high Turing degree,
but I haven't been able to prove this. For example, because of the
close connection between $O$ and computable ordinals, it would seem
reasonable to suppose that every cofinal branch can compute WO, the
set of Turing machine programs computing a well-ordered relation on $\mathbb{N}$.
Question 1. Does every cofinal branch through Kleene's $O$
compute a $\Pi^1_1$-complete set of natural numbers?
An affirmative answer would imply, in particular, that every cofinal branch $z$ could compute $O$ itself.
Failing this, perhaps every branch can at least compute the set TA
of true arithmetic assertions.
Question 2. Does every cofinal branch through Kleene's $O$
compute true arithmetic?
In other words, if I have a cofinal branch $z$ through Kleene's
$O$, and I use $z$ as an oracle, can I compute whether a given
arithmetic sentence is true in the standard model?
This question arose recently in the seminar I am running with Wesley Wrigley, in connection with some of his work, which concerns the Feferman-Spector theorem, asserting that there are some cofinal branches through $O$ for which the theory that arises by iteratively adding consistency statements is not complete, even for $\Pi^0_1$ arithmetic truth. Notice that the issue of whether this theory is incomplete, however, is not the same as the question whether the path itself, when used as an oracle, can compute true arithmetic.
An idea: fix a sequence $A=(\alpha_i){i\in\omega}$ cofinal in $\omega_1^{CK}$. Now consider the forcing notion $\mathbb{P}$ where conditions are finite sequences $(n_i){i<k}$ with $\vert n_i\vert_\mathcal{O}=\alpha_i$ and $n_j<\mathcal{O}n{i}$ for all $j<i<k$ (and conditions are ordered by extension as usual). How complicated are the correspondingly generic paths, that is, the $<_\mathcal{O}$-downwards closures of the set of notations occurring in the conditions in the generic filter? (And I think this is independent of choice of $A$.)
This forcing is isomorphic to the forcing to add a Cohen real. But the Turing degree of the branches will not generally be respected by the isomorphism. So I'm not sure how much we can get this way.
Yes, it's not enough to use the isomorphism type of the forcing. But there might still be something we can get from the fact that the set of notations of a given length extending a given notation isn't too complicated.
I see; you want to argue something like this: it is dense that a given TM program does not compute some instance of TA.
Perhaps we can use that all bounded initial segments of the branch are uniformly c.e. That is, for each n in z, we can enumerate $z\upharpoonright n$.
@NoahSchweber I can prove that the generic path you describe must at least compute the halting problem. The reason is that for any program p it is dense to include an index either of type 1 or type 2, where the type 1 index works only if p halts and the index of type 2 works only if it doesn't. We can parameterize those program types, and simply search for it in z, thereby solving the halting problem. I believe that this idea ramps up to get TA from any generic path.
More specifically, my idea about type 1 and type 2 is that we will stick on another $\omega$ many terms (by powers of $2$) and then include the index $3\cdot 5^e$, where $e$ is a program that is recognizably of the form (1) wait for $p$ to halt, then climb through that $\omega$ tower, or (2) climb through the tower as long as $p$ is not halting. Since it is dense to have this occur (with programs recognizably of this form), we can eventually find it in $z$. So we can decide the halting problem from $z$. By now doing this in a nested way, we can similarly decide TA.
A small observation: $O$ is hyperarithmetic in any of its paths $z$. The point is that $n\in O$ iff there's an $<_O$-preserving embedding of the $<_O$-predecessors of $n$ into $z$. That makes $O$ $\Sigma^1_1$ in $z$, and since it's $\Pi^1_1$ even without $z$, it's hyperarithmetic in $z$. There might be some hope for estimating the relevant level of the $z$-hyperarithmetic hierarchy and so dragging this observation down to a level near what you asked about.
Yes, Andreas @AndreasBlass, I agree completely. This was the main reason to think that the Turing degree of the cofinal paths should be high. But I don't see how to reduce the hyperarithmetic reduction to Turing reduction.
Goncharov, Harizanov, Knight and Shore investigated the Turing degrees of $\Pi^1_1$ cofinal branches (which they call "paths through $\mathcal{O}$"). They showed there is a $\Pi^1_1$ cofinal branch which does not compute $\emptyset'$, so certainly doesn't compute true arithmetic. On the other hand, H. Friedman showed there is a $\Pi^1_1$ cofinal branch which computes $\mathcal{O}$ (reference can be found in the GHKS paper).
That's a lovely result, I was totally unaware of that!
Thanks very much! Here is the DOI for the JSL version: https://doi.org/10.2178/jsl/1082418544
|
2025-03-21T14:48:30.933972
| 2020-05-10T12:05:29 |
359918
|
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|
Stack Exchange
|
Compact embedding of space of signed Radon measures into Sobolev space $W^{-1,q}$ from Evans paper; Does it work in one space dimension?
Background: I work on a PDE problem where I have some approximating sequence of measure-valued functions and I need to compactly embed it into some negative Sobolev space $W^{-m,q}$ on the bounded interval in $\mathbb{R}$. I am mostly interested in the spaces where $q=2$. I found only one such embedding in the one theorem from the paper:
Evans - Weak convergence methods for nonlinear partial differential equations, 1990.
Theorem 6 (Compactness for measures, page 7): Assume the sequence $\{\mu_k\}_{k=1}^{\infty}$ is bounded in $\mathcal{M}(U)$, $U \subset \mathbb{R}^n$. Then $\{\mu_k\}_{k=1}^{\infty}$ is precompact in $W^{-1,q}(U)$ for each $1 \leq q <1^*$.
Here $\mathcal{M}(U)$ represents space of signed Radon measures on $U$ with finite mass, $U \subset \mathbb{R}^n$ is an open, bounded, smooth subset of $\mathbb{R}^n, n \geq 2$ and $1^*=\frac{n}{n-1}$ represents a Sobolev conjugate.
The identical theorem (Lemma 2.55, page 38) is given in the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, with a difference that instead of $1 \leq q <1^*$, in there is written $1 \leq q <\frac{n}{n-1}$ (here it isn't written explicitly that $n\geq 2$).
My question: does the Theorem 6 works in one dimension ($n=1$)? That is do we have a compact embedding of space $\mathcal{M}(U)$ into the space $W^{-1,q}(U)$, where $U \subset \mathbb{R}$?
And additionaly:
I assume that if we have compact embedding into $W^{-1,q}(U)$, then we have it also in the $W^{-m,q}(U), m\geq 1$?
Are there any other measure spaces (e.g. space of finite positive measures $\mathcal{M}_+$, space of probability measures with finite first moment $Pr_1$, etc.) that are compactly embedded into some negative Sobolev spaces $W^{-m,q}(U)$?
I think that if we use definition of the Sobolev conjugate: $\frac{1}{p^*}=\frac{1}{p}-\frac{1}{n}$, we get for $p=1,n=1$ the $\frac{1}{1^*}=\frac{1}{1}-\frac{1}{1}\Rightarrow 1^*=\infty$. So we would have that theorem 6 (maybe) works for every $1 \leq q < \infty$ (and then for $q=2$ also)? If we use $p^*=\frac{np}{n-p}$ we would have for $n=1,$ $p^*=\frac{p}{1-p}$ and here we could not take $p=1$ and get $p^*$.
I usually do not deal with the measure-valued and negative Sobolev spaces, so I don't know much about them. Help with this would be great and I definitely need it. And any additional reference besides the two mentioned above would be nice. Thanks in advance.
yes, the statement is written in dimensions $n=1$ too, see my answer below. As a general rule of thumbs things are easier in lower dimensions, and clearly here the statement is distinguished because the threshold for Morrey's inequality $W^{1,p}\subset C^\alpha$ in dimension 1 is $p=1$, so in this particular case the statement of Theorem 6 holds verbatim for all $p\geq 1$.
as should be clear from my answer below, the key is to first obtain a continuous embedding $W^{1,p}\subset C^{\alpha}$, which is then automatically transferred to a compact embedding via Ascoli-Arzelà. And in dimension 1 this works for all $p\geq 1$.
Here is a partial answer, which has to do with dual compact embeddings: If the embedding between (resonable) Banach spaces $X\subset\subset Y$ is compact then the dual embedding is compact too, $Y^*\subset\subset X^*$.
This is useful here since the space of Radon measures is the dual of continuous bounded functions, $\mathcal M(U)=(C_b(U))^*$.
Now for $p>n$ we have that $W^{1,p}$ is continuously embedded into some Hölder space $C^\alpha$ (for some $\alpha\equiv \alpha(n,p)$). By the Arzelà-Ascoli theorem this shows that the embedding
$$
W^{1,p}(U)\subset\subset C_b(\bar U)
$$
is compact too.
As a consequence we have that the embedding
$$
\mathcal M(U)\subset\subset W^{-1,q}(U)
$$
provided that $q=p'$ is such that $p>n$, ie for all $q<1^*=\frac{n}{n-1}$ (this is exactly why the cirical $1^*$ exponent appears in your Theorem 6).
As for the second part of the question: since the embedding $W^{m,p}\subset W^{1,p}$ is trivially continuous for $m>1$, the reversed embedding $W^{-1,q}\subset W^{-m,q}$ is continuous. Then the composition of "compact$\circ$continuous = compact" $\mathcal M\subset\subset W^{-1,q}\subset W^{-m,q}$ also gives compactness for
$$
\mathcal M(U)\subset\subset W^{-m,q}(U).
$$
Thank you for the answer and the comments. It is nice to be sure that Theorem 6 works in one dimension also. In the case I am interested in $W^{1,2}$ is continuously embedded into $C^{0,\alpha}, 0<\alpha\leq \frac{1}{2}$. By Arzela-Ascoli it is compactly embedded in $C_b$ and by duality argument $\mathcal{M}$ is compactly embedded in $W^{-1,2}$. I wasn't sure in the Theorem 6 is it true that $1^* = \infty$? If it is true, then $\mathcal{M}$ is compactly embedded in $W^{-1,q},q<\infty$ based on your answer? And thanks again.
Yes, formally you can tink of $1^*=\infty$, and indeed $\mathcal M\subset\subset W^{-1,q}$ for all $q<\infty$ in dimension 1.
Thank you for the follow up. I think that this will help me a lot.
I am happy with the answer, but I always leave a week or two before I accept the answer. I take the time to try to use the answer I got on the problem I am trying to solve. And also I would like to see if there is an answer to the second part of my question (about other measure spaces). If not, this answer is quite nice and it will be accepted definitely. Thanks again.
Oh sure, sorry about that, for some reason I thought you were a new user here and that you didn't know about our MO system. I'm sorry. And I'm not chasing for points either, it just annoys me when there are nice answers given to a quiestion (I'm not saying mine is!) but users just leave them hanging out there. I'll remove my comment ASAP, now I feel terrible.
Everything is fine. I understand what you saying. And we are all here for solving problems I hope, and not for points. Thanks again.
|
2025-03-21T14:48:30.934383
| 2020-05-10T14:25:46 |
359926
|
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|
Stack Exchange
|
Is there a stacky definition of irreducible symplectic manifold?
I am now interested in studying symplectic structures in the field of stacks.
In particular, is there a stacky definition of irreducible symplectic manifold ?
I'm also interested in similar things in derived algebraic geometry.
Any comment welcome! Thank you!
Have a look at https://arxiv.org/pdf/1903.05632.pdf. In page number 7, they define Symplectic stacks. It might be of some use.. You can also look at http://marle.perso.math.cnrs.fr/publications/Liegroupoids.pdf This define/discuss symplectic Lie groupoids. It is not necessary to remind that Lie groupoids and differentible stacks are closely related.. :)
There is a relatively large body of recent literature on derived (shifted) sympectic structures - look in works of a subset of Calanque, Pantev, Toen, Vaquie, Vezzosi, etc. A review is in https://arxiv.org/pdf/1603.02753.pdf
Thank you very much! And what is the difference between the definition on the third page of http://arxiv.org/abs/1111.6294v1 and the shifted Symplectic structures?
I apologize for the many questions, but I would be grateful if you could also give me some good literature for learning derived algebraic geometry and shifted symplectic structures.
The definition you mention in arxiv.org/abs/1111.6294v1 looks like an attempt at a $0$-shifted symplectic structure, but seems to omit any sort of closure condition. Beware that the survey arxiv.org/pdf/1603.02753.pdf which Balazs mentions really just summarises CPTVV, neglecting to mention much related literature. For differentiable stacks, a summary on shifted structures is https://arxiv.org/pdf/1804.07622 , including references to related mathematical physics papers. Some of Safronov's papers are also a good place to look if you want to see where these structures arise in the wild.
Thank you very much! Does that mean that the result of TPVV can lead to the result of http://arxiv.org/abs/1111.6294v1 ?
yes, by throwing away the higher structure
Many times Thank you. And what are the open problems with shifted symplectic structure?
|
2025-03-21T14:48:30.934548
| 2020-05-10T14:44:54 |
359928
|
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|
Stack Exchange
|
Maximum partite subset of edges of a hypergraph
If $X\neq \varnothing$ is a set we say that ${\frak P} \subseteq {\cal P}(X)$ is a partition of $X$ if
$\bigcup{\frak P} = X$, and
$P\neq Q \in {\frak P} \implies P\cap Q = \varnothing$.
Let $H = (V,E)$ be a hypergraph with $V \neq \varnothing$ and $\bigcup E = V$. A partition ${\frak P}$ of $V$ is said to be splitting if for all $e\in E$ and $P \in {\frak P}$ we have $|e \cap P| = 1$, and we call such a hypergraph admitting a splitting partition partite. It is easy to see that the edges of a partite hypergraph all have the same cardinality - and every splitting partition also has that cardinality.
If $E_1 \subseteq E$ we call $H|_{E_1}:=(\bigcup E_1, E_1)$ the subhypergraph induced by $E_1$.
Question. If $H=(V,E)$ is a hypergraph with $\bigcup E = V$, is there $E_1\subseteq E$ such that $H|_{E_1}$ is partite, but whenever we have $E_2\subseteq E$ with $E_1\subseteq E_2$ and $E_1\neq E_2$, then $H|_{E_2}$ is no longer partite?
Note. A straightforward application of Zorn's Lemma seemed to bear no fruit; I wasn't able to prove that the union of a chain of partite edge sets needs to be partite again. Of course I will remove this question if I committed a stupid mistake and there is an easy argument showing that partiteness carries through unions of chains.
I don't understand the question. What's to stop $E$ itself being partite, in which case all its subsets will be partite?
Right @gowers, in that case $E$ is the (only) maximal partite subset of itself. But the question is, even if $H=(V,E)$ is not partite, can we find $E_1 \subseteq E$ such that $(\bigcup E_1, E_1)$ is partite, but no proper superset of $E_1$ has this property?
|
2025-03-21T14:48:30.934799
| 2020-05-10T14:59:14 |
359931
|
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|
Stack Exchange
|
Covering families of sets by small-measure partitions
Let $(X,\mathscr{A},\mu)$ be a probability space and let $\{A_1,\ldots,\}\subset\mathscr{A}$ be a countable family of sets with small measure: say $\mu(A_i)\le\epsilon$. I am trying to show that one can find a countable (disjoint!) partition $\{B_i\}$ of $X$ with the following property: Each $A_i$ is covered by some $(B_j)_{j\in J}$ such that
$|J|$ is small (say, $1/\epsilon$) and $\mu(\cup_{j\in J}B_j)$ is not too large (say, $O(\epsilon)$ or even $O(\sqrt\epsilon)$).
We can assume that $\mathscr{A}$ is a Borel $\sigma$-algebra induced by some metric, if it helps.
Edit. It was pointed out by Fedja and others that the previous formulation, which required a condition like $\mu(B_i)\le\epsilon^2$, has atomic counterexamples.
If $B_i$ are disjoint and cover $A_i$'s, then they have to be a refinement of ring generated by $A_i$'s, right?
Yes, I think that’s right.
but then isn't $A_n=[\frac{1}{n},\frac{1}{n}+\varepsilon]$ a counterexample? Among $B_i$ you will have $(\frac{1}{n+1},\frac{1}{n}]$, as well as $(\frac{1}{n+1}+\varepsilon,\frac{1}{n}+\varepsilon]$, and so you won't be able to cover $A_j$'s with a finite number of $B_j$, or am i misreading the question?
The answer is negative.
First we may always assume that there are only finitely many $B_i$ -s: The sum of the measures of the $B_i$-s converges so we may take the union of all but finitely many of them with measure of this union less that $\epsilon^2$ and replace this cofinite set of $B_i$-s by their union.
Now let the $A_i$ be independent sets of measure $\epsilon$ each (like in Andrey's comment) and $B_i$ a finite cover as required. Each $A_i$ is covered by a union of some $B_j$-s whose measure (of the union) is of order $\epsilon$ (or whatever the required bound is- as long as it is of order smaller than 1). There are only finitely many such unions. The indicator functions of the $A_i$-s tend weakly to the function which is constantly $\epsilon$. It follows that the measure of $A_i$ intersection with each of the finite unions above tend to something of order the measure of the union times $\epsilon$ which is of order smaller that $\epsilon$ so it is impossible that all the $A_i$ are covered by such unions.
Consider $X = [0, 1]^{\aleph_0}$ with cylindrical sigma algebra and product measure (of Lebesgue ones). Let $A_i = [0,1]\times\ldots \times [0, \varepsilon]_i \times [0,1]\times\ldots$ (cylinder with measure epsilon). What are $B_i$? It seems it is not possible.
Let $X=[0,1]$, let $\epsilon\gt0$, and let $\{A_i:i\in\mathbb N\}$ be the set of all $A\subseteq[0,1]$ such that $A$ is a finite union of rational intervals and $\mu(A)\lt\epsilon$. Let $\{B_i:i\in\mathbb N\}$ be any countable partition of $[0,1]$. Given any $n\in\mathbb N$, we can find $A_i$ which has nonempty intersection with each of the sets $B_1,B_2,\dots,B_n$, whence $\{B_j:j\in J\}$ covers $A_i$ only if $J\supseteq\{1,2,\dots,n\}$. Therefore $|J|$ may be required to be arbitrarily large. Moreover, if the sets $B_j$ are measurable, $\mu(\bigcup_{j\in J}B_j)$ may be required to be arbitrarily close to $1$.
|
2025-03-21T14:48:30.935053
| 2020-05-10T15:11:17 |
359932
|
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|
Stack Exchange
|
Is there a spherical analogue of polar duality for spherical complexes?
Let $P$ be a spherical complex, which essentially means a tiling of a sphere, let us say the $(d-1)$-dimensional sphere $\mathbb{S}^{d-1}$ in $\mathbb{R}^d$ to fix notation, where each cell is a spherical polytope. These are higher-dimensional analogues of spherical polyhedra.
My question is: is there a known construction of a combinatorially dual spherical complex, say $Q$ (assuming such a $Q$ always exists)?
So I am essentially asking for a spherical analogue of polar duality (polar duality assigns a dual convex polytope to a given convex polytope).
What is a "spherical polytope"? And how would it differ from a spherical complex?
@SamHopkins, yes I am not sure if this is standard terminology or not. I have read in a book that a spherical polytope is the intersection of some closed hemispheres which is non-empty and does not contain a pair of antipodal points. A spherical complex is a tiling of the whole $(d-1)$-dimensional sphere by spherical polytopes.
That being said, I have also seen the expression "spherical polytope" used to mean a spherical complex, in the sense that I described in my previous comment. I am not sure which one is more standard terminology.
Thanks, I understand now.
Naive question: can a spherical complex always be converted to a complete polyhedral fan with the same combinatorics?
If so you might be able to use duality of fans to get what you want.
@SamHopkins, I am new to this area. I don't know what is a "complete polyhedral fan" for instance.
A (convex) polyhedral cone is this: https://en.wikipedia.org/wiki/Convex_cone#Polyhedral_and_finitely_generated_cones. A fan is a collection of polyhedral cones which intersect properly (the intersection of any two cones in the fan is a cone in the fan which is a common face of both). "Complete" just means it fills up all of space. Intuitively if you intersect a polyhedral fan with a sphere centered at the origin you should get a spherical complex in your sense; and it would seem the reverse procedure is possible too.
@SamHopkins, I understand now. Do you have any reference to duality of complete polyhedral fans please? I think this may be just what I need (after intersecting with the sphere).
See the "dual cone" section in that Wikipedia page. The dual fan should just be the fan of all the dual cones (unless I'm making a bone-headed mistake).
@SamHopkins, yes wonderful. This is precisely the construction I was thinking about actually. Is it clear that combinatorially, it gives the dual of the original complex? This is the part I was not sure about.
@SamHopkins, solved, thanks to your comments, and this wikipedia page https://en.wikipedia.org/wiki/Dual_cone_and_polar_cone. Many thanks! Could you please write an answer?
see my edit. Now I'm less certain that this duality makes sense.
Since there seems to be no easy answer, I am very curious about the following experiment: take, e.g., one of the spherical triangulations (say, of the 3-sphere) that is known to not correspond to a 4-polytope. Can you explicitly construct a spherical complex dual to this one? Maybe this gives a hint how to do it in general. On the other hand, I can imagine that if we go to high enough dimensions, the vertex-figures of some spherical complexes are not polytopal, and so there is indeed no dual spherical complex in this sense.
As discussed in the comments, this stuff is more commonly described using the language of fans. A (convex, polyhedral) cone in $\mathbb{R}^{d}$ is the intersection of finitely-many half-spaces through the origin. A (polyhedral) fan is a collection of cones that intersect properly: the intersection of any two cones in the fan is again a cone in the fan which is a common face of both cones. A fan is complete if the union of all the cones in the fan is all of $\mathbb{R}^d$. By intersecting a complete fan with a sphere centered at the origin we obtain a spherical complex in the sense of the question-asker. This procedure should also be reversible by "coning over" a spherical complex. Cones have dual cones, and in this way we get dual fans. This gives the desired combinatorial duality for spherical complexes.
EDIT: Whoops, now I am actually less sure of exactly how duality of cones leads to duality of fans. The dual cones of a fan will not fit together into a fan. So this does not answer the question. (Of course, if our fan happens to be polytopal, then we can use polar duality of polytopes.)
EDIT 2: I asked Vic Reiner about this question, and he gave me a lot of good information. He pointed out that the question of the existence of "dual" CW complexes is a difficult and subtle point, as discussed for instance in this other MO question. However, for a PL cell decomposition of $\mathbb{S}^d$ there exists a dual PL cell decomposition of $\mathbb{S}^d$ with a dual face lattice, as proved in Proposition 4.7.26(iv) on pg. 214 of the "Oriented Matroids" book by Björner et al. The spherical complexes you describe (a.k.a. polyhedral fans) will certain be PL, but this result still does not quite answer your question because it is not clear that the dual PL cell decomposition will correspond to a polyhedral fan.
No, what's troubling me is that if I take all the cones in a fan and take their duals, they will not fit together into a fan.
Could you give an example where the dual cones of a fan do not fit together into a fan please?
Even the most basic example of say 6 equiangular cones in $\mathbb{R}^2$. These cones are acute so their duals will be too big.
What you kinda want to do is take the normal fan of a fan. But I'm not sure that makes sense.
I understand your objection. Can one perhaps construct a fan out of the duals of the polyhedral cones?
You can maybe mimic the construction of a normal cone/normal fan, but for the faces of your spherical complex instead of a convex polytope: https://en.wikipedia.org/wiki/Normal_fan.
At any rate you should unaccept this answer, because while it might be useful for you as a starting point, it doesn't answer your question. I'm now convinced your question is very interesting, though, and I'm surprised I've never heard discussion of "dual polyhedral fans" before. One possible answer is that only genuinely polytopal fans have duals: an analogous thing is that only planar graphs have duals.
By the way, in case the complete polyhedral fan $T$ is the normal fan of a convex polytope $P$, one may view the normal fan of $P^$, with $P^$ being the polar dual of $P$, as a kind of dual of $T$. So this seems to provide a partial answer to my question in this important special case.
Yes of course: in the polytopal case things work perfectly.
(But that case is probably not so interesting for you, because I think it means you could deform your spherical complex into a convex polytope.)
Yes, but this is good for me, because the fan I am interested in is a normal fan. However, it would be interesting to find a more general and more direct construction, should there exist one.
I never studied PL cell decompositions. Just out of curiosity, is there a PL cell decomposition of a sphere (represented as a higher dimensional cube), which is not equivalent as a CW complex to that of a spherical complex? I am not sure if my question makes sense (I don't know the definitions well).
I think everything I’m talking about are CW complexes (maybe even regular CW complexes). You’d have to look at the references to be sure.
Yes, but I was wondering essentially about the last sentence in your answer, when you wrote "it is not clear that the dual PL cell decomposition will correspond to a polyhedral fan". Can you explain a bit more please this statement?
The spherical complexes you define should be a very special class of PL decompositions of the sphere. I would recommend looking at that part of "Oriented matroids" to understand the definitions more precisely.
|
2025-03-21T14:48:30.935617
| 2020-05-10T15:53:16 |
359935
|
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|
Stack Exchange
|
Are there algorithms for deciding or solving conjugacy in integer quaternion rings?
I am doing some research on the quaternions and their role in Non-commutative cryptography. I have found a number of articles, but it is still unclear to me if there is a known solution to the Conjugacy Search Problem (CSP) or Conjugacy Decision Problem (CDP) in integer quaternion rings:
CSP: Determine whether there exists a $z$ such that $zy=xz$ for a given $(x, y) \in R$
CDP: Find a $z \in R$ such that $zy=xz$ for a given $(x, y) \in R$
Here $R$ is either the ring of Lipschitz quaternions ($\{ai+bj+ck+d\mid a,b,c,d\in\mathbb{Z}\}$) or the Hurwitz quaternions (which is the union of the Lipschitz quaternions with a second copy of the lattice shifted by $(\frac12,
\frac12, \frac12, \frac12)$; i.e., where all coordinates are either integers or integers $+\frac12$).
This paper is a PKC scheme based on the Conjugacy search problem and decision problem
Valluri and Narayan - Quaternion public-key cryptosystems
However, I am inclined to assume that the CSP has not been solved for the quaternions, based on the above. Does an algorithm for solving the CSP for quaternions exist?
I'm going to assume you are discussing the quaternions over $\mathbb{R}$. In this case what you are asking is essentially the conjugacy structure of $\mathrm{SO}(3)$, and it's easiest to see geometrically. Every rotation in three dimensions is a rotation by some angle around an axis, and two rotations by the same angle are conjugate via any rotation that takes one axis to the to other. The structure of the double cover doesn't change much about the conjugacy.
https://mathoverflow.net/questions/358830/conjugacy-in-the-quaternion-group/358837#358837
@JohnVoight Is that true for all quaternion groups, like the Lipschitz or Hurwitz?
I removed the irrelevant [algebraic-groups] and [free-groups] tags.
When you say 'the group' do you mean 'the ring'? Neither the Hurwitz nor Lipschitz quaternions form a field, so you have to be very careful using expressions like $z^{-1}$ in such a case. The question of whether there's a $z$ such that $zy=xz$ is very different from the question of whether there's an invertible such $z$ (which is nearly trivial; very few Hurwitz or Lipschitz quaternions are invertible multiplicatively, after all...)
(Also, if you do mean 'as a group' then the problem is trivial, because the group structure on Hurwitz or Lipschitz quaternions is the additive one, which is still commutative.)
Presuming that the question here is about conjugacy in these rings, then the problem is likely to be Complicated, by analogy with similar problems in $SL_2(\mathbb{Z})$ for instance. I think there's an interesting question here but you need to be more specific about it.
@user157838: yes, the linear algebra approach works in the unit group of any quaternion algebra over a (computable) field.
@StevenStadnicki I think you are right, I do mean conjugacy in rings, and I realize now that my question was not formulated correctly. I apologize, I am still very shaky with this kind of math. I will edit the question to be more specific
Excuse me, I would like you to clarify something. I've looked up few papers on this theme. Is it some non-commutative version of NTRU?
@LevSoukhanov Not exactly, I am just interested in the applicability of a quaternion ring to non-commutative crypto.
If I am not mistaken, the question as written now follows by doing the linear algebra approach over $\mathbb{Q}$ and then scaling $z$ to get integer coefficients.
I'm currently reading https://arxiv.org/pdf/1709.02079.pdf this, hopefully it is related.
As @AchimKrause points out, currently the problem is still trivial because considered over Q. You should probably work with polynomials over H (similar to standard NTRU) to obtain interesting lattice problem.
@user157838: Just a side note: you cite some authors, but their English is so bad, that I wouldn't even bother reading their papers. But then, they got published by Hindawi, so... this should tell a lot.
@AlexM. Is Hindawi considered a reputable publisher? I looked into it briefly and found they were generally considered to be ok. Furthermore, I didn't want to discard a paper solely because of the English.
@user157838: No, Hindawi is not ok. Their peer reviewing process is... "problematic", to say the least. I constantly find emails from them in my "spam" folder, in which they promise very quick acceptance for publication. Such aggressive self-promotion should be condemned; and the promise of such quick reviewing tells a lot about its quality. Regarding the authors' English, I simply cannot understand the quoted paragraph "Regarding this [...] predicted groups": it simply makes no sense in English. Anyway, your question was about something else, so let us end this off-topic discussion here.
@AlexM. I have removed that source
If you are talking about quaternions, then you can rewrite the equation $y = z^{-1}xz$ as $zy - xz = 0$ which is $\mathbb{R}$-linear in $z$. Therefore both problems can be solved quite easily using linear algebra. (I.e. introduce real indeterminates $z_1, \ldots, z_4$ via $z = z_1 + i z_2 + j z_3 + k z_4$, insert into the latter formula and obtain linear system for $z_1, \ldots, z_4.$
If you are talking about quaternion group, then it's even easier as you can just test all eight elements.
What about the Libschitz or Hurwitz groups?
I hope that you aren't missing the fact that the OP is working over $\mathbb Z$.
@AlexM.: The OP changed the question. It was not originally about the integral case.
|
2025-03-21T14:48:30.936045
| 2020-05-10T16:03:05 |
359936
|
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|
Stack Exchange
|
Do prime pairs inbetween and equidistant from adjacent integer powers cover all the prime numbers?
Is it true that every odd prime number is a member of one or more pairs of primes $(p_1, p_2)$ such that $p_1$ and $p_2$ are inbetween and equidistant from two adjacent powers of some number $n$? i.e.
$0 < p_1 - n^m = n^{m+1} - p_2 $ for some positive integers $n, m$
For example the prime number $73$ is a member of the pair $(73, 179)$ where $73 - 6^2 = 6^3 - 179$
Your equations should be $|p_1-n^m|=d$ and $|p_2-n^{m+1}|=d$ to reflect the opening sentence.
Is $\ n\ $ fixed or variable?
@Wlod AA n,m, and d can be any positive integers, chosen for each p
Thank you (for being arbitrary; and is a user variable? :) ). #### What would be your numerical evidence?
You may say "odd prime".
This boils down to finding a primality radius $r$ of $q:=\frac{n^{m+1}+n^{m}}{2}$ less than $l:=n^{m+1}-q$, which should exist provided $n$ is large enough, under a sufficiently strong form of Goldbach's conjecture.
Where a primality radius of an integer $u$ is a non negative intever $v$ such that both $u+v$ and $u-v$ are primes.
However, this question is of a different character. Goldbach's conjecture asks for one prime pair for 2n for every n, whereas this asks if every prime is part of a special pair for certain n. It is unclear that one statement follows from the other. Gerhard "Making An Unusual Pair Indeed" Paseman, 2020.05.10.
One could study first $\ \rho(a\ b)\ :=\ {(p\ q)\in\Bbb P^2: p+q=a+b\ \text{and}\ a\le p\le\frac{a+b}2}\ $ for every $\ (a\ b)\in\Bbb Z\ $ such that $\ 0<a<b-1.$
For every odd prime $\ p,\ $ and for every integer $\ n>1\ $ there exists exactly one positive integer $\ m=\mu(p\ n)\ $ such that
$$ n^m\ <\ p\ \le\ n^{m+1}. $$
Odd prime $\ p\ $ is said to be $\ n$-doubtful (or $n$-questionable) $\,\ \Leftarrow:\Rightarrow\,\ q:=n^m+n^{m+1}-p\,\ $
is not a prime, where $\ m=\mu(p\ n).$
Odd prime $\ p\ $ is trustful $\,\ \Leftarrow:\Rightarrow\,\ $ there exists integer $\ n>1\ $ such that $\ p\ $ is not $n$-doubtful. Otherwise, $p$ mistrusted.
The OP's conjecture (actually a question) was that every odd prime is trusted.
==============================================
The first $2$-questionable prime is $\ p=23\ $ because
$$ 2^4 < p\ < 2^5 $$
and
$$ q\ :=\ 2^4+2^5\ -\ p\ =\ 25 = 5^2 $$
is not a prime. Thus, there remains to verify primality (or non-primality) of
$$ n^m + n^{m+1} - 23 $$
just for a very limited number of cases, for all $\ n>2\ $ such that $\ n<23$:
$\ 3^2<23<3^3\ $ -- well, $\ 13+23=36=3^2+3^3\ $ and $\ q:=23\ $ is a prime hence the conjecture holds.
The next $2$-questionable prime is $\ 41.\ $ However,
$$ 67\ =\ 3^3+3^4-41 $$
is a prime. The conjecture holds for $\ p:=41.$
There is just one more $2$-questionable prime in the
$\ [2^5;2^6]\ $ range, namely $\ p:=47.\ $ But $\ 47\ $ is not $3$-doubtful -- indeed:
$$ 61\ = 3^3+3^4-47\ $$
is a prime. Next, we get a $2$-doubtful prime $\ 127\ $ from the higher end of of $\ [2^6;2^7].\ $ It's not $3$-doubtful though:
$$ 197\ =\ 3^4+3^5-127 $$
is a prime.
Given a huge list of consecutive primes, a computer program can verify the conjecture quickly within the given range of primes.
I am still curious about the smallest prime that is both $2$- and $3$-doubtful, as well as about
the general question about prime $\ d_n\ $ which is the smallest among $k$-doubtful for every $\ k\le n.$
REMARK doubtful = questionable (but of course :) ).
Actually:
The smallest $2$- and $3$-questionable prime is $\ p:=73.$
Indeed,
$$ 119 = 2^6+2^7-73 $$
is not a prime ($119=7\cdot 17$) hence prime $73$ is $2$-questionable. Also,
$$ 35 = 3^3+3^4-73 $$
is not a prime hence prime $73$ is $3$-questionable. Great!*
Furthermore, the same prime $\ p:=73\ $ is also $4$-questionable
since
$$ 247 = 4^3+4^4-73 $$
is not a prime $\ (247=13\cdot 19),\ $ as well as $5$-questionable:
$$ 77 = 5^2+5^3-73 $$
is not prime. Thus,
Prime $\,73\ $ is the smallest that is $2$- and $3$- and $4$-questionable. Furthermore, prime $\,73\ $ is the smallest that is $2$- and $3$- and $4$- and $5$-questionable.
However, the conjecture holds for prime $73$ since it is not $6$-questionable; indeed:
$$ 179 = 6^2+6^3-73 $$
is a prime.
It's interesting that you're hitting with such small numbers - mostly $2$s and a few $3$s. The overall asymptotic situation is good but I wouldn't expect you to hit so fast so often.
|
2025-03-21T14:48:30.936341
| 2020-05-10T16:10:00 |
359937
|
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|
Stack Exchange
|
An orthogonal matrix that satisfies a property must be a permutation matrix
Let $A$ be an $n\times n$ orthogonal matrix such that $\sum_{k=1}^na_{ik}^3a_{jk}=\sum_{k=1}^na_{jk}^3a_{ik}$ for every $1\le i,j\le n$.
Original question which is solved by a counterexample given (For the new question see Edit 2):
I want to show that $A$ is either a permutation matrix (i.e., all but one entry in each row and column is zero, and that non-zero entry is $1$ or $-1$), or all entries of $A$ have absolute value $1/\sqrt n$.
This question happens in my calculation on some functions from the tangent space of $A$.
I will provide a proof of $n=2$ without using the $\sin$ & $\cos$ representation of $A$:
$a_{11}^3a_{21}+a_{12}^3a_{22}=a_{11}a_{21}^3+a_{12}a_{22}^3$, so $a_{11}a_{21}(a_{11}^2-a_{21}^2)=a_{12}a_{22}(a_{22}^2-a_{12}^2)$. Since they are unit vectors, $a_{11}^2-a_{21}^2=a_{22}^2-a_{12}^2$, so it's easy to deduce the desired result. However, I find it hard to generalize this method.
Any suggestions will be appreciated.
Edit: Ok let's add the condition that $\sum_{k=1}^na_{ki}^3a_{kj}=\sum_{k=1}^na_{kj}^3a_{ki}$ for every $1\le i,j\le n$. I have proved that this must be true from the assumptions.
Edit 2: the comment gives a counterexample of this. As Ilya Bogdanov suggested, it is true that if $A$ is irreducible, then each non-zero entry of $A$ has the same absolute value?
@LSpice The tangent space of $A$ in the manifold $O(n)$, which is $A$ times all skew symmetric matrices, and from that (and my original problem) I got this question.
I see. I believe that the usual grammar is "the tangent space at $A$ to $\operatorname O(n)$."
You say "I have proved that this must be true from the assumptions", which suggests that there's a real problem behind this one that you're not specifying. Why not say what that is?
@LSpice What I meant is from the assumption of $\sum_{k=1}^na_{ik}^3a_{jk}=\sum_{k=1}^na_{jk}^3a_{ik}$.
What if you take the direct sum of a (normalised) 2x2 Hadamard and a 2x2 identity matrix?
@PadraigÓCatháin Well you find a counterexample. Thanks!
So, @Padraig's comment siggests that any direct sum of normalized Hadamard matrices, with rows and columns being permuted arbitrarily, works. Perhaps, it is better to ask about irreducible matrices...
Check out weighing matrices -- they are nxn orthogonal matrices with k non-zero entries in each row and column. There should be also lots of irreducible examples of these.
Sorry, the matrix below is perhaps too large for a comment.
It seems that here is an irreducible counterexample:
$$
\frac12\begin{bmatrix}
1& 1& -1& 1& 0& 0& 0& \cdots& 0& 0& 0& 0& 0\\
1& 1& 1& -1& 0& 0& 0& \cdots& 0& 0& 0& 0& 0\\
0& 0& 1& 1& -1& 1& 0& \cdots& 0& 0& 0& 0& 0\\
0& 0& 1& 1& 1& -1& 0& \cdots& 0& 0& 0& 0& 0\\
\vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \ddots& \vdots& \vdots& \vdots& \vdots& \vdots\\
0& 0& 0& 0& 0& 0& 0& \cdots& 0& 1& 1& 1& -1\\
-1& 1& 0& 0& 0& 0& 0& \cdots& 0& 0& 0& 1& 1\\
1& -1& 0& 0& 0& 0& 0& \cdots& 0& 0& 0& 1& 1\\
\end{bmatrix}
$$
Surely, one can find many more similar examples.
So I would ask whether all nonzero elements of an irreducible orthogonal matrix satisfying the property have the same absolute value.
|
2025-03-21T14:48:30.936558
| 2020-05-10T16:20:50 |
359938
|
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|
Stack Exchange
|
Are twin primes the only solution to this equation?
Let $p,q_i, i \ge 1$ be primes, $m$ a positive integer.
The equation
$$
p.\prod_{i=1}^m(q_i-1)-(p-1).\prod_{i=1}^mq_i=2
$$
for $m=1$ has all twin primes $p,q_1=p+2$ as solution.
Are there solutions for all $m\ge2$ ?
The simplest heuristic suggests probably yes, but it seems unlikely to be possible to prove this.
I'm not sure what to expect. Without the primality condition,
we have for each $m$ an equation of the same degree as the number of variables
(namely $m+1$),
so we expect some power of $\log N$ solutions up to $N$,
unless there's a polynomial family. But primality costs
a factor of $(\log N)^{m+1}$, suggesting a delicate balance.
@NoamD.Elkies I think the leading term cancels, leaving us with a degree $m$ equation. This makes the probable situation more clear.
In the case $m=2$, you can take $q_1=p+2$ and $q_2=(p^2+p+2)/2$, so that your desired equation is just a polynomial identity. Standard conjectures then ensure that $p, q_1$ and $q_2$ are simultaneously prime, infinitely often
An exhaustive search up to $1000$ finds the solutions
$$
(p,q_1,q_2) = (3, 5, 7), (11, 13, 67), (59, 67, 487),
$$ $$
(191, 277, 613), (251, 463, 547), (347, 571, 883)
$$
for $m=2$ and
$$
(p,q_1,q_2,q_3) =
(3, 5, 7, 37), (3, 7, 7, 11), (23, 37, 107, 131), (83, 137, 283, 797)
$$
for $m=3$.
Given an odd prime $p$ and integer $m \geq 2$, we can set $q_1=p+2$ and then, for each $i$ with $2 \leq i \leq m$,
$$
q_i = \frac{ 4+ (p-1) \prod_{j=1}^{i-1} q_j}{2}.
$$
One can check that
$$
p \cdot \prod_{i=1}^m (q_i-1) - (p-1) \cdot \prod_{i=1}^m q_i =2,
$$
and each $q_i$ is a polynomial in $p$, of degree $2^{i-1}$ (with coefficients in $\mathbb{Z}[1/2^{i-1}]$).
It is plausible that this construction should produce infinitely many values of $p$ for which $p$ and each $q_i$ are simultaneously prime. That being said, I may be missing some local obstruction. While this very likely leads to infinitely many examples for $m \in \{ 2, 3, 4 \}$, the smallest with $m=4$ being
$$
(p,q_1,q_2,q_3,q_4)=(3,5,7,37,1297),
$$
examples with $m=5$ are rather thinner on the ground : the smallest is with $p=512351711$.
|
2025-03-21T14:48:30.936839
| 2020-05-10T16:31:07 |
359939
|
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|
Stack Exchange
|
Is there Hodge isomorphism between Dolbeault and Harmonic on noncompact manifold
As is well known , Hodge theorem tells us
Let $(X, g)$ be a compact hermitian manifold. Then the canoni.
cal projection $\mathcal{H}_{\bar{\partial}}^{p, q}(X, g) \rightarrow H^{p, q}(X)$ is an isomorphism.
i.e., Dolbeault cohomology groups is isomorphic with the corresponding harmonic spaces.
My question is:
1) Is there the Hodge decomposition on noncompact manifolds?
2) Are Dolbeault cohomology groups isomorphic with the corresponding harmonic spaces.
?
https://arxiv.org/abs/1812.11764
|
2025-03-21T14:48:30.936942
| 2020-05-10T16:45:52 |
359941
|
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|
Stack Exchange
|
Computation of fraction field of formal series over the integers
What is the fraction field $K$ of the domain $\mathbb Z[[X]]$?
It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in L\setminus K$.
Indeed a Laurent series $f(X)=\sum_{i\in \mathbb Z} q_iX^i \in L \; (q_i=0 \operatorname {for} i\lt\lt 0)$ must have as its coefficients rational numbers whose denominators involve only finitely many primes (which depend only on $f(0)$) in order to belong to $K$.
But is that necessary condition sufficient?
Edit
No, that necessary condition is not sufficient.
For example if $p$ is a prime the series $\sum_{i\geq 0}\frac {X^i}{p^{i^2}}$ does not belong to $K=\operatorname {Frac}\mathbb Z[[X]]$, even though the denominators in its coefficients only involve the single prime $p$.
This follows from Elad Paran's Example 2.3(c) quoted by Arno in his very pertinent answer below.
Here https://math.stackexchange.com/questions/140054/what-is-the-fraction-field-of-rx-the-power-series-over-some-integral-doma is an unanswered related question.
It might be tempting to first solve the probably easier question when $\mathbf{Z}$ is replaced with $\mathbf{Z}_p$ (or another local PID).
For an integral domain $A$ with fraction field $\operatorname{Frac}(A)=K$ we can define $H=\bigcup_{0\neq a\in A}\frac{1}{a} A[[\frac{x}{a}]]$. We have the series of inclusions
$$\operatorname{Frac}(A[[x]])\subseteq H[\frac{1}{x}]\subseteq K((x))$$
where $H[\frac{1}{x}]=\operatorname{Frac}(H)$ is the ring mentioned in the other answer. Moreover, we can describe explicitly when equality is achieved.
For instance a theorem of Gilmer [1] shows that $\operatorname{Frac}(A[[x]])=K((x))$ iff for any sequence $s_n$ of nonzero elements of $A$ we have $\bigcap s_nA\neq (0)$. A theorem of Benhissi [2] shows that $\operatorname{Frac}(A[[x]])=H[\frac{1}{x}]$ iff for any nonzero $a\in A$ we have $\bigcap a^nA\neq(0)$.
Everything is downhill from here. Obviously $A=\mathbb Z$ doesn't satisfy such a condition so unfortunately the inclusion $\operatorname{Frac}(\mathbb Z[[x]])\subset H[\frac{1}{x}]$ is strict. To make things worse, Rivet has a theorem [3] that shows that it is impossible to characterize which power series belong to $\operatorname{Frac}(A[[x]])$ when $A$ is a discrete valuation ring, using criteria that involve only knowing valuations of individual coefficients. So you can put conditions on which primes appear in denominators and to which multiplicities, yet this will not allow you to conclude whether your series belongs to the fraction field or not.
A textbook reference is the monograph [4], particularly chapter 2 on pathologies of $R[[x]]$. It goes over an exposition of constructions that are used in the literature to prove these sort of negative results. The main tool is the ability to explicitly construct power series in $H[1/x]$ that are so sparse that they can't be in $\operatorname{Frac}(A[[x]])$.
[1] Gilmer, R. "A note on quotient fields of $D[[X]]$" Proc. Amer. Math. Soc. 18, 1138-1140 (1967)
[2] Benhissi, A. "Corps des fractions de $A[[X]]$" Communications in algebra. 25 (9),
2861-2879 (1997)
[3] Rivet, R. "Sur le corps des fractions d’un anneau de séries formelles à coefficients dans un anneau de valuation discrete" C. R. Acad. Sci., Paris, Sér. A 264, 1047-1049 (1967)
[4] Brewer, J.W. "Power Series Over Commutative Rings", CRC press (1981)
Dear Gjergji, would you say that there is as of now no good description of the fraction field of $\mathbb Z[[X]]$ ? Is this known as an open question?
@GeorgesElencwajg I wouldn't call it an open problem, but yes, everyone seems in agreement that there is no simple description. See this older answer for another take at a not-so-explicit description of this field of fractions (invert all irreducibles): https://mathoverflow.net/questions/194056/some-questions-about-the-ring-zx/194074#194074
Thank you @Gjergji but indeed I find the recipe "invert the irreducibles" not useful at all. After all the required field is obtained by the recipe "invert all non-zero series" ...I would say a description is satisfactory only if it tells you whether a given series with coefficients in $\mathbb Q$ is in the fraction field or not. I hope the problem is not undecidable...
@GeorgesElencwajg One problem is that such an algorithm needs more than just accessing the coefficients individually: no matter how far we go, the answer can still be yes or no, as can be checked by truncating the series or putting very big denominators. You need more information on the series, but how to describe it?
There exists a relatively simple necessary and sufficient condition. A power series $ f\in \mathbb Q((X))$ lies in the field of fractions of $\mathbb Z[[X]]$ if and only if it satisfies the following two conditions
There exist only finitely many primes $p$ appearing in the denominators of coefficients of $f$.
For each $p$ appearing in the denominators, there exists a monic polynomial $Q \in \mathbb Z_p[X]$ such that $Q f \in \mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ]$, in other words the powers of $p$ appearing in the denominators of coefficients of $Q f$ are bounded.
Another way of saying the second condition is that $f$ is the sum of an element in $Z_p[[X]] [ \frac{1}{p} ]$ plus an element in $\mathbb Q_p(X)$.
Proof of "only if"
Given a ratio $\frac{A}{B}$ with $A, B \in \mathbb Z[[X]]$, the only primes appearing in the denominators of coefficients of $\frac{A}{B}$ are the primes dividing the leading coefficient of $B$, of which there are finitely many.
Now fix one of these primes. By the Weierstrass preparation theorem in $\mathbb Z_p[[X]]$, we can write $B$ as an invertible element $U$ of $\mathbb Z_p[[X]]$, times a power of $p$, times a monic polynomial $Q \in \mathbb Z_p[X]$. (Theorem 1.3 of the paper Jesse Elliott linked, after dividing by a suitable power of $p$ to make one of the coefficients zero. ) Writing $f =\frac{A}{B} = frac{A}{U Q p^n}$, we see that $Q f = \frac{A}{U p^n} \in \mathbb Z_p[[X]] [\frac{1}{p} ]$ because $U$ is invertible. This verifies the second condition.
Proof of "if"
Let $p_1,\dots, p_m$ be the primes appearing in the denominator of $m$ and let $Q_{p_1}, \dots, Q_{p_m}$ be the $p$-adic power series guaranteed by the condition. We will show the existence for each $p$ of a power series $U_{p} \in \mathbb Z_p[[X]]^\times $ such that $U_{p} Q_{p} \in \mathbb Z[[X]]$.
To do this, we can assume by removing a factor of $X^k$ from $Q_{p}$ that the leading coefficient of $Q_{p}$ is nonzero. Because it is a nonzero $p$-adic number, it has the form $p^{n_i}$ times a $p$-adic unit for some natural number $n_i$. Let the first coefficient of $U_{p_i}$ be the inverse $p$-adic unit, and then choose inductively for each $n$ the coefficient of $X^n$ in $U_{p}$ so that the coefficient of $X^n$ in $Q_{p} U_{p}$ is an integer in the range $\{0,\dots, p^{n_i}-1\}$. This is possible since that range hits every element of $\mathbb Z_p / p^{n_i} \mathbb Z_p$.
Now $\prod_{i=1}^m (Q_{p_i} U_{p_i} ) \in \mathbb Z[[X]]$, and $$f \prod_{i=1}^m (Q_{p_i} U_{p_i} ) = Q_{p_i} f U_{p_i} \prod_{j=1}^{i-1}(Q_{p_j} U_{p_j} )\prod_{j=i+1}^{m}(Q_{p_j} U_{p_j} ) $$ is the product of $Q_{p_i} f \in \mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ] $ with $U_{p_i} \in \mathbb Z_p[[X]]$ and several terms $Q_{p_j} U_{p_j}\in \mathbb Z[[X]] \subset \mathbb Z_p[[X]] $. Thus $f \prod_{i=1}^m (Q_{p_i} U_{p_i} )$ lies in $\mathbb Z_p[[X]] [\frac{1}{X}, \frac{1}{p} ] $, i.e. the powers of $p$ dividing the denominators of its coefficients are bounded. Since these primes $p$ are the only primes that divide the denominators of coefficients of $f$, and no primes divide the denominators of the coefficients of $(Q_{p_j} U_{p_j} )$, it follows that the coefficients of $f \prod_{i=1}^m (Q_{p_i} U_{p_i} )$ have bounded denominators, and thus we may clear denominators by multiplying by a natural number, exhibiting $f$ as a ratio in $\mathbb Z[[X]]$.
Proof that the alternative second condition is equivalent:
That such $f$ satisfy the second condition is clear. For the converse, given $A \in \mathbb Z_p[[X]]$ and $Q$ monic in $\mathbb Z_p[X]$ of degree $n$, to show that $\frac{A}{Q}$ has this form, we may assume by dividing $Q$ by any factors that are a unit in $\mathbb Z_p[[X]]$ that all non-leading coefficients of $Q$ are divisible by $p$. One then checks that $A= BQ +R $ for $B \in \mathbb Z_p[[X]]$ and $R \in \mathbb Z_p[X]$ of degree at most $n$, one checks that the natural map $\mathbb Z_p[X]/Q \to \mathbb Z_p[[x]]/Q$ is an isomorphism by subtracting a suitable multiple of $Q$ to cancel the degree $n$ term mod $p$, then the degree $n+1$ term mod $p$, and so on, and then the degree $n$ term mod $p^2$, and the degree $n+1$ term mod $p^2$, and so on, etc.
Thanks a lot Will. This looks extremely interesting and I'll try to understand the details . (As I already told Arno in a comment, this is foreign territory for me).
The description of the irreducible elements in $\mathbb{Z}_p[[X]]$ is often called the Weierstrass preparation theorem in this ring.
I think you meant $p_i \neq p$, not $\ell \neq p$ or $p_i \neq P$. I followed your argument until the last paragraph. What do you mean by "every ratio"? Can you explain the argument in the last paragraph in further detail?
@JesseElliott. You're right about $p_i$, sorry. Given a ratio $ \frac{A}{Q}$ with $A \in \mathbb Z_p[[X]]$ and $Q \in \mathbb Z_p[[X]]$ monic of degree $n$, with all the non-leading coefficients of $Q$ divisible by $P$, we can write $A = BQ + R$ where $B \in \mathbb Z_p[[X]]$ and $R \in \mathbb Z_p[X]$ has degree $<n$. To do this we subtract a suitable multiple of $Q$ from $A$ to cancel all the terms of degree at least $n$ modulo $p$, then cancel the terms of degree at least $n$ modulo $p^2$, and so on. Because we are adding higher powers of $p$, both the term we subtract and
@JesseElliott the remainder converge, and then we have $\frac{A}{Q} = B + \frac{R}{Q}$.
@GeorgesElencwajg Edited to give a clearer proof.
Thank you, Will.
The condition that only finitely many primes divide the denominators is not sufficient. One needs to add at least the additional condition that their growth is polynomial. I suspect there is a more standard reference, but see for Example 2.3(c) in Algebraic patching over complete domains by Elad Paran. It says there that ${\rm Quot}(\mathbb{Z}[[X]])$ is contained in the field
$$
\left\{\frac{1}{c}\sum_{i=m}^\infty\frac{b_i}{a^i}X^i:a,b_i,c,m\in\mathbb{Z},ac\neq 0\right\}.
$$
This does of course not yet answer the question how to describe ${\rm Quot}(\mathbb{Z}[[X]])$ more precisely.
How interesting! Many thanks since this is completely foreign territory to me and I would never have stumbled upon that article by myself.
Perhaps, it is more convenient to rephrase the condition in the exercise as follows: There exist $a$ and $c$ such that $cf(aX)$ has integer coefficients.
Great comment, @Ilya. Круто !
If $R$ is a UFD with field of quotients $K$, then the group of units $K^*$ is the direct product of $R^*$ with the free abelian group generated by the irreducible/prime elements of $R$ (one for each class of associates). This at least gives you a description of the group $K^* = K \backslash \{0\}$. In the case of $\mathbb{Z}[[X]]$, which is a UFD, its prime elements are described here (see Theorem 1.4): https://www.ams.org/journals/tran/2014-366-08/S0002-9947-2014-05903-5/S0002-9947-2014-05903-5.pdf . Moreover, $\mathbb{Z}[[X]]^*$ is isomorphic to $\{1,-1\} \times (1+X\mathbb{Z}[[X]])$, where also $1+X\mathbb{Z}[[X]]$ is the additive group of the universal lambda ring $\Lambda(\mathbb{Z})$ over $\mathbb{Z}$. Perhaps more is known about the structure of that group.
Thanks a lot for the answer and the link, Jesse. MathOverflow is so great because it allows you to interact with people, like you, who have actually done research in the question you are interested in.
This is actually a remark on Arno Fehm's answer, but too long for a comment (the remark is due to P. Samuel). Let $p$ be a prime, and let us look at power series $u(X)=\sum_{n\geq 1}a_nX^n$ in $\mathbb{Q}((X))$ satisfying $u^2-pu+X=0$. This is equivalent to $a_1=p^{-1}$, $a_n=a_1a_{n-1}+\ldots +a_{n-1}a_1$. So the $a_i$ are uniquely determined, and of the form $\dfrac{b_i}{p^i} $ with $b_i\in\mathbb{Z}$. Still $u$ does not belong to $K$, because it is integral over $\mathbb{Z}[[X]]$ which is integrally closed. If it were in $K$, it would belong to $\mathbb{Z}[[X]]$, and this is clearly not the case.
Thank you, abx.
|
2025-03-21T14:48:30.937732
| 2020-05-10T17:37:34 |
359947
|
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|
Stack Exchange
|
A lisse mixed sheaf as an extension of pure lisse sheaves
I am trying to understand Corollary 1.8.11 in Deligne's Weil II paper. The statement is that for a normal scheme $X_0$ that is of finite type over $\mathbb{F}_q$, every lisse $\ell$-adic $\iota$-mixed sheaf $\mathscr{F}_0$ on $X_0$ is an extension of pointwise $\iota$-pure lisse $\ell$-adic sheaves on $X_0$.
My question is rather basic: how is this statement not a tautology? By definition given in 1.2.6 of that article, a constructible $\ell$-adic sheaf is $\iota$-mixed if it is a successive extension of $\iota$-pure sheaves. Moreover, a subsheaf or a quotient of a lisse sheaf is again lisse (because lisse sheaves are the same as $\ell$-adic representations of $\pi_1$), so isn't 1.8.11 essentially a restatement of the definition? I am confused, any help would be appreciated.
Subqotients of lisse sheaves are not necessarily lisse: if $j:U\to X$ is an open immersion with complement $i:Z\to X$ then $j_{!}\mathbb{Q}{l,U}$ is a subsheaf of $\mathbb{Q}{l,X}$ and $i_*\mathbb{Q}_{l,Z}$ is its quotient.
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2025-03-21T14:48:30.937831
| 2020-05-10T18:53:30 |
359954
|
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|
Stack Exchange
|
Locally compact Polish groups acting on standard Lebesgue spaces
If $G$ is a countable discrete group, then one can consider the Bernoulli shift $2^G$. $G$ acts on $2^G$ via shift, and letting $\mu$ be the product of the $(1/2, 1/2)$-measure in each coordinate, then $(2^G, \mu)$ is a Borel, essentially free, probability measure preserving action of $G$ on a standard Lebesgue space.
My question is whether there is any analogue of this for locally compact Polish groups. More precisely, if $G$ is a locally compact Polish group, does $G$ admit a Borel, essentially free, probability measure preserving action on a standard Lebesgue space?
Yes - this is Proposition 1.2 of Adams, Elliott and Giordano (MathSciNet review), Amenable actions of groups, Trans. Amer. Math. Soc. 344 (1994), 803-822.
I think that the analogue of a Bernoulli shift is rather the translation action on a Poisson point process. It appears in the work of Ornstein and Weiss on entropy theory of amenable groups as the analogue of the Bernoulli shift for locally compact groups and a more general form of it appears in the book of Kornfeld Sinai and Fomin under the name of Poisson suspensions.
An informal description of the model is as follows. Let $\Pi_m$ a Poisson point process over $G$ with the Haar measure $m$ as the intensity measure.
This can be viewed as a measure on the countable discrete subsets of $G$ and the action of $G$ is by translating the whole configuration.That is for $\nu\subset G$, $g\nu=(g^{-1}h)_{h\in\nu}$. This action is free, ergodic and measure-preserving.
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2025-03-21T14:48:30.937976
| 2020-05-10T19:11:01 |
359956
|
{
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"Mare",
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|
Stack Exchange
|
Almost split sequences coming from bimodules
Let $A$ be a finite dimensional algebra with enveloping algebra $A^e$.
Auslander and Reiten proved in "On a theorem of E. Green on the dual of the transpose" that
$Hom_A(Tr_{A^e}(A),M) \cong \tau(M)$ for any non-projective indecomposable $A$-module $M$.
Question 1:We have $Hom_A(Tr_{A^e}(A),M) \cong D( Tr_{A^e}(A) \otimes_A D(M)) $, is this isomorphic to $M \otimes_A \tau_{A^e}(A)$?
Question 2: In case question 1 is true, when we have an almost split sequence of $A$-bimodules
$0 \rightarrow \tau_{A^e}(A) \rightarrow X \rightarrow A \rightarrow 0$ and we tensor it with an $A$-module $M$ over $A$, we get an exact sequence $0 \rightarrow \tau(M) \rightarrow M \otimes_A X \rightarrow M \rightarrow 0$. Is this exact sequence again almost split? If not, does this at least work when $A$ is a symmetric algebra where question 1 has indeed a positive answer?
Let $A=k[x]/(x^2)$ and $S$ the simple $A$-module for a field $k$. If $\eta\colon 0\to \tau_{A^e}(A) \to X \to A\to 0$ is the almost split sequence ending in $A$ over $A^e$, then $S\otimes_A X$ is a semisimple module. It follows that $S\otimes_A \eta$ is a split exact sequence. Hence Question 2 is not always true.
There is an analogue of this for group algebras $kG$ and Hopf algebras with involutive antipode, with $k$ an algebraically closed field of characteristic $p$. See https://www.sciencedirect.com/science/article/pii/0021869386901730 for the group algebra case. Then, if $0\to \tau(k) \to E\to k\to 0$ is the almost split sequence ending in the trivial module $k$, then tensoring with any module $M$, we obtain an exact sequence $0\to \tau(k)\otimes_k M \to E\otimes_k M \to k\otimes_k M\to 0$, where $k\otimes_k M \simeq M$. If $p \not\mid\dim_k M$, then this sequence is almost split. Hence the tensor product is not always almost split. I would guess the result would be something similar for bimodules. Henning Krause had a PhD-student who address this question at some point. I only remember that it is not true in general, but I don't remember for what reason.
Here is how one can analyze this in QPA2:
gap> Q := Quiver( RIGHT, "Q(1)[a:1->1]" );
Q(1)[a:1->1]
gap> KQ := PathAlgebra( GF( 2 ), Q );
GF(2) * Q
gap> rels := [ One( KQ ) * Q.a * Q.a ];
[ Z(2)^0*(a*a) ]
gap> A := KQ/rels;
(GF(2) * Q) / [ Z(2)^0*(a*a) ]
gap> M := AlgebraAsBimodule( A );
<2>
gap> R := UnderlyingRepresentation( M );
<2>
gap> U := AsModule( LEFT, R );
<2>
gap> TrU := TransposeOfModule( U );
<2>
gap> DTrU := DualOfModule( TrU );
<2>
gap> p := ProjectiveCover( U );
<(4)->(2)>
gap> q := KernelEmbedding( p );
<(2)->(4)>
gap> V := Source( q );
<2>
gap> homVDTrU := Hom( V, DTrU );
Hom(2, 2)
gap> f := BasisVectors( Basis( homVDTrU ) )[ 1 ];
<(2)->(2)>
gap> T := Pushout( f, q );
<4>
gap> RR := UnderlyingRepresentation( T );
<4>
gap> TT := AsBimodule( RR );
<4>
gap> S := SimpleModules( RIGHT, A );
[ <1> ]
gap> MM := TensorProductOfModules( S[ 1 ], TT );
<2>
gap> IsSemisimpleRepresentation( UnderlyingRepresentation( MM ) );
true
The module $T$ is the middle term in the almost split sequence ending in $A$ as a bimodule.
Thanks. Is it also possible to test question 1 with QPA 2? The "result" looks like it should naturally hold, but I can not see how an isomorphism might look in general. So maybe it is just wrong for some random algebra like a non-selfinjective Nakayama algebra.
I tried a bigger example, but QPA2 has gotten to be very slow compared to earlier, so it didn't seem to finish. For a Nakayama algebra with two vertices it looked like it was true (which one probably can check quite easily by hand).
|
2025-03-21T14:48:30.938212
| 2020-05-10T19:18:30 |
359958
|
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"Monroe Eskew",
"Noah Schweber",
"Walter Bruce Sinclair",
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|
Stack Exchange
|
Positive set theory and the "co-Russell" set
This is a more focused version of a question which was asked at MSE a couple years ago, but is still unanswered there. That question asks about a broad range of theories, whereas this version focuses on a single one.
Let $S=\{x: x\in x\}$ be the "dual" to Russell's paradoxical set $R$, and consider the question "Is $S\in S$?" If we try to work purely "naively" there does not seem to be an immediate argument one way or the other (dually to the situation for $R$). However, in light of Lob's theorem it's a bit premature to leap to the conclusion that there actually are no such arguments.
So it remains plausible that there is a "reasonable" set theory in which the question "Is $S\in S$?" is decided in a nontrivial way. Note that the nontriviality requirement rules out both $\mathsf{ZF}$- and $\mathsf{NF}$-style set theories: the former prove $S=\emptyset$, while the latter prove that $S$ is not a set in the first place.
The most appealing candidate to me appears to be $\mathsf{GPK}_\infty^+$. This theory proves that $S$ is a set and is nonempty. It also (in my opinion) has a very attractive intuition behind it, as well as an interesting model theory. So my question is:
Does $\mathsf{GPK}_\infty^+$ decide whether $S\in S$?
Could you say a bit more how NF proves that $S$ is not a set? (Or did you just mean that it doesn’t obviously prove it is a set?)
@EmilJeřábek $\mathsf{NF}$ proves that the complement of a set is a set, so if $S$ were a set $R$ would be as well.
there are ZF style theories which indeed can decide on that, take ZF with Ur-elements, now if we restrict the non-well founded parts of those theories to just the Quine atoms, and work with models in which there is a set of all quine atoms, then clearly the question is also trivially decidable?
@ZuhairAl-Johar I would still call that a trivial solution (although you're right, it's a different trivial solution to the one I stated). I'm really interested in $\mathsf{GPK}_\infty^+$.
@NoahSchweber, I've put the answer to the MSE version of this question that you've linked in the head post. A pure formula means just a positive formula (i.e. bounded universal quantifiers are not allowed). According to that answer the solution is simply found as: $$\exists A \forall y (y \in A \iff y \in y \lor y=A)$$.
How about set theories with anti-foundation? You'd certainly get $S \neq \emptyset$ in those, but I do not see off the top of my head what status $S \not\in S$ gets.
@AndrejBauer I'm also interested in those, but that would be more appropriate for the MSE version of this question (and I mentioned them there); I decided to focus on $\mathsf{GPK}_\infty^+$ here partly to keep the question more concrete and also because I've gradually come to like that theory in particular. I've edited to clarify this point.
$\mathsf{GPK}^{+}_{\infty}$ looks interesting indeed, especially to a topologically minded person like me.
Minor note: There seems to be a typo in the SEP article for the axiom schema of closure (the variable $y$ is not captured by its quantifier).
@AndrejBauer Re: antifoundation, over something $\mathsf{ZF}$-flavored Aczel's and Boffa's axioms each make $S$ a proper class. The only $\mathsf{ZF}$-flavored versions I'm aware of which make $S$ a nonempty set are fairly artificial, like "There is a Quine atom" + "There is a set of all Quine atoms."
@NoahSchweber, as regards your last comment, $ZFCU $ + hereditary size axiom do indeed prove $S \not \in S$. I personally don't think this theory is artificial. The hereditary size axiom can prove infinity by itself (this axiom is: for every set there is a set of all sets hereditarily smaller than it), this axiom is a non-trivial theorem of ZF, and its not directly related to the self membership problem, you can build hierarchies using this function. This works as long as all self membered sets in the theory are hereditarily smaller than some set (i.e., not necessarily quine atoms)
What’s this about Lob’s Theorem?
@MonroeEskew Probably https://math.stackexchange.com/questions/2135587/variant-of-g%C3%B6del-sentence/2135649.
@ZuhairAl-Johar Artificiality is subjective, but I do think that theory is artificial. I don't see any motivation for your hereditary size axiom; merely being a nontrivial theorem of $\mathsf{ZFC}$ isn't sufficient for me. Regardless, this question is again specifically about $\mathsf{GPK}_\infty^+$.
@MonroeEskew The point is that "This statement is unprovable"/"This statement is provable" may at first glance look parallel to $R$/$S$; Lob's theorem then vaguely suggests that the self-membership question for $S$ might not be undetermined. It's purely motivational, feel free to ignore it.
@NoahSchweber, a non-trivial theorem of ZF. (not just ZFC). I think the hereditary size axiom is quite natural!! But adding the restriction of self membered sets to be hereditarily of a size below some specific set size is what's artificial. I agree.
Well the question had been there for two years. Obviously no one knows the answer!?
@ZuhairAl-Johar "Well the question had been there for two years. Obviously no one knows the answer!?" Nobody on MSE knew the answer two years ago. That doesn't mean that nobody on MO knows the answer now.
@NoahSchweber, it looks hard. But as you said possibly someone knows the answer now. Usually in those theories (having big sets) its the small sets that really count, as it is the case with NF, we always at the end concentrate on ZF like sets that gives most of the power for the theories, and features of the big sets remains a mystery most of the times because we don't have much control over them, I think the matter is the same with positive set theory. The concentration is on hereditary isolated sets.
Just out of curiosity : in either case, would S have a non-null proper subset?
@WalterBruceSinclair This is equivalent to $S$ having at least two elements. The answer is yes; besides the universal set, one way of generating elements of $S$ is as follows. For each set $x$ let $C(x)={y: x=y\vee x\in y}$ be the "cone above $x$." This exists by positive comprehension, and since $x\in C(x)$ we have $C(x)\in C(x)$ and so $C(x)\in S$. So this lets us generate lots of elements of $S$.
This answer has been superseded by a more general argument (based on a result of Cantini's) that I have posted as an answer to Noah's original MSE question. In particular, a fairly weak fragment of $\mathsf{GPK}$ is enough to entail that $S \in S$.
I will show that $\mathsf{GPK}_\infty^+$ is consistent with $S \in S$, under the assumption of a weakly compact cardinal. Specifically, I'll show that the 'standard models' of $\mathsf{GPK}_\infty^+$ (which require weakly compact cardinals) satisfy $S \in S$. I've made very little progress with proving anything from $\mathsf{GPK}_\infty^+$ directly, although this is probably my failing.
The main idea of the argument is very simple: Show that these models satisfy that for every set $X$, there exists a set $Y$ satisfying $Y = X \cup \{Y\}$. You then note that $S$ must be a fixed point of this operation, so it must be the case that $S \in S$.
For the sake of completeness, I will give the construction of a model of $\mathsf{GPK}_\infty^+$ in a way that should feel familiar to classical set theorists. I'm also doing this because I couldn't find a presentation of this construction that I liked online, although I did use the Stanford Encyclopedia of Philosophy article Alternative Axiomatic Set Theories to get the general idea.
Construction
Recall that a forest is a partially ordered set in which the set of predecessors of any element is well-ordered. A tree is a forest with a unique minimal element. I'll use the word path to refer to a downwards-closed, linearly ordered set. (I can't remember if there's a standard term for this.) The height of a path is its order type as an ordinal. The height of a node $x \in T$ is the height of the path $\{ y \in T : y < x\}$. To make the presentation more uniform between successor and limit stages, we'll refer to paths as having children, rather than just nodes. A node $x$ is the child of a path $A$ if $\{ y \in T : y < x\} = A$. A path $A$ is the parent of a node $x$ if $x$ is the child of $A$. Finally, a branch is a path of maximal height. (The forest we will construct will not have any dead paths that are shorter than the forest itself.)
We will build a forest (although really it's just a tree and an isolated branch corresponding to the empty set) whose nodes of height $\alpha$ are labeled with sets of paths of height $\alpha$. (Note that this isn't circular, because the paths of height $\alpha$ do not contain nodes of height $\alpha$.) Every set of paths of height $\alpha$ will be attached to precisely one node of height $\alpha$. We will use the convention that lowercase letters refer to sets of paths and that uppercase letters refer to paths of sets.
At stage $0$, there is a unique path of length $0$, the empty path $\Lambda$, and there are two sets of such paths, the empty set and the singleton $\{\Lambda\}$, so these are the roots of the two trees in our forest.
At non-zero stage $\alpha$, given the forest built up to height $\alpha$, we add each set $x$ of paths of length $\alpha$ as a node. The parent of the node $x$ is the unique path $A$ with the property that for any node $y \in A$,
for every $B \in x$, there exists a $C \in y$ such that $B$ extends $C$ and
for every $C \in y$, there exists a $B \in x$ such that $B$ extends $C$.
It is not hard to show by induction that this is well defined.
Now we build this forest up to a height $\kappa$, where $\kappa$ is a weakly compact cardinal. The elements of the model are the branches of the forest, and the element of relation is defined by $A \in B$ iff for every $\alpha < \kappa$, $A \upharpoonright \alpha \in B(\alpha)$. This structure is typically referred to as the $\kappa$-hyperuniverse, but I'll just call it $M$.
Argument
I claim that $M$ satisfies the following:
For every $X \in M$, there exists $Y \in M$ such that $Z \in Y$ if and only if either $Z \in X$ or $Z = Y$.
It's easy to see how this resolves the status of $S \in S$. Let $S' = S \cup \{S'\}$. Clearly by construction, $S' \supseteq S$ and $S' \in S'$. Therefore we have $S' \in S$, but this implies that $S \supseteq S'$, so $S = S'$ and $S \in S$. I think the most reasonable approach to resolving your question would be to prove this claim directly from $\mathsf{GPK}_\infty^+$, but I can't even show that $\mathsf{GPK}_\infty^+$ entails the existence of a Quine atom. (Or that any two Quine atoms are equal, for that matter.) Someone probably knows how to do this, though, and I'd be interested to see it.
Proof of claim. To prove this, consider the path corresponding to $X$. Let $Y$ be the $\kappa$-indexed sequence of nodes on the forest defined inductively by $Y(\alpha) = X(\alpha) \cup \{ Y \upharpoonright \alpha\}$ for each $\alpha < \kappa$. We need to show that this is well defined.
For $\alpha = 0$, $Y \upharpoonright 0 = \Lambda$, so $Y(0) = X(0) \cup \{\Lambda\}$.
For a non-zero $\alpha$, assuming we've shown that $Y \upharpoonright \alpha$ is actually a path on the forest, we immediately get that $Y(\alpha) = X(\alpha) \cup \{ Y \upharpoonright \alpha\}$ is a node of height $\alpha$ in the forest. We just need to show that $Y(\alpha)$'s parent is $Y\upharpoonright \alpha$. Fix a node $Y(\beta) \in Y \upharpoonright \alpha$ (for some $\beta < \alpha$). For any $A \in Y(\alpha)$, either $A \in X(\alpha)$ or $A = Y\upharpoonright \alpha$. In the first case, by the fact that $X\upharpoonright \alpha$ is a path, there must exist a $B \in X(\alpha)$ such that $A$ extends $B$. In the second case, $Y \upharpoonright \beta \in Y(\beta)$ and is extended by $Y\upharpoonright \alpha$ (by the induction hypothesis). Essentially the same argument gives that for any $B \in Y(\beta)$, there is an $A \in Y(\alpha)$ extending $B$.
Therefore $Y$ is a branch of the forest and corresponds to an element of $M$. $Y$ is clearly a superset of $X$ and clearly contains $Y$ as an element. We need to show that $Y = X \cup \{Y\}$.
Suppose that $Z \in Y$. For each $\alpha < \kappa$, we get that either $Z \upharpoonright \alpha = Y \upharpoonright \alpha$ or $Z \upharpoonright \alpha \in X(\alpha)$. Because of the way forests/trees work, if there is an $\alpha$ such that $Z \upharpoonright \alpha \neq Y \upharpoonright \alpha$, then this must also be true for any $\beta \in (\alpha,\kappa)$. Therefore, if $Z \neq Y$, then for some $\alpha < \kappa$, we have that $Z \upharpoonright \beta \in X(\beta)$ for all $\beta \in (\alpha,\kappa)$. It's not hard to show that the same must be true for any $\beta <\kappa$, so we have that $Z \in X$, proving the claim.
Very nice, I like this a lot! I'll accept it once I've had a chance to read it in detail and convince myself it works (granted it doesn't answer the full question, but based on how long this has been open I think this is a good stopping point for it).
|
2025-03-21T14:48:30.939224
| 2020-05-10T19:30:13 |
359961
|
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|
Stack Exchange
|
Calculation of $\pi_3^s$ via killing spaces and Steenrod squares
$\newcommand{\Sq}{\text{Sq}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Sf}{\mathbb{S}}$I am grateful to anyone who will help me, because I know it is not a short calculation and it requires some time. I hope that there is someone who enjoys, as me, these things.
I am (self) studying the book Fomenko and Fuchs - Homotopical Topology (MSN). In chapters 4 and 5 it discusses the Serre spectral sequence and cohomology operations and it shows how to compute $\pi_1^s, \pi_2^s$ using these tools. Then, in Lecture 31, it leaves $\pi_3^s$ as an exercise. Explicitly (p.417):
The $\Sf^n_{n+k}$ are the killing spaces of $\Sf^n$, i.e. the spaces with the same homotopy groups as $\Sf^n$ starting from $n+k$ and zero before. The $E_k$ are the underkilling spaces of $\Sf^n_{n+3}$, i.e. the spaces that at any stage quotient the $(n+3)$-homotopy group of $\Sf^n_{n+3}$ by $\mathbb{Z}_2$ . We want to prove that this chain ends at $E_3$, i.e. that $H^{n+3}(E_3) \cong 0$, and conclude that $\pi_3^s \cong \Z_8$ (at $2$).
If I understood properly, I have to set up spectral sequences for six fibration:
\begin{align*}
K(\Z, n-1) &\to \Sf^n_{n+1} \to \Sf^n \\
K(\Z_2, n) &\to \Sf^n_{n+2} \to \Sf^n_{n+1} \\
K(\Z_2, n+1) &\to \Sf^n_{n+3} \to \Sf^n_{n+2} \\
K(\Z_2, n+2) &\to E_1 \to \Sf^n_{n+3} \\
K(\Z_2, n+2) &\to E_2 \to E_1 \\
K(\Z_2, n+2) &\to E_3 \to E_2
\end{align*}
Here is the calculation I have done, but it is not correct because I obtain $H^{n+3}(E_3) \ncong 0$. Certainly it must be $\text{Sq}^1l_{n+3} \neq 0$, differently from what I wrote in the file.
I briefly explain the conventions in the pdf.
Coefficients ring is always $\Z_2$.
For every spectral sequences I show only three columns: the left column of the SS (cohomology of the fiber), the bottom row of the SS (cohomology of the base) (since we can choose $n$ so large to avoid terms in between) and the resulting cohomology of the total space. The differentials go up right by one position.
I marked in red the elements which do not survive in the SS.
I marked in blue the Steenrod-squares-relations which do not convince me: they do not convince me since I have the feeling that the Steenrod squares have the same problem of the cup product with spectral sequences, i.e. if they are 0 in $E_\infty$ maybe they are not really 0 in the total cohomology.
The point (4) is not discussed in the book, so probably I am not right and the error is somewhere else. If instead I am right, there would certainly be corrections that make things works (for example, if the last blue relation is wrong, then everything works). Anyway, it is an exercise from a book, so I suppose it must have a clear solution and the relations must not have the ambiguity that I am claiming.
EDIT I
I write here the four blue-marked relations
$\Sq^3 {i_{n+2}} = 0$. I think this is right, otherwise $\Sq^3 i_{n+2} = i_{n+5}' = \Sq^2 i_{n+3}'$. Then $i_{n+6}'' = \Sq^3 i_{n+3}' = \Sq^1\Sq^2 i_{n+3}' = \Sq^1\Sq^3 i_{n+2} = 0$, which is a contradiction.
$\Sq^4\Sq^1 i_{n+2} = i_{n+7}$. I think this is correct if 1) is correct. In fact $0 = \Sq^2\Sq^3 i_{n+2} = \Sq^4\Sq^1 i_{n+2} + \Sq^5 i_{n+2} = \Sq^4\Sq^1 i_{n+2} + i_{n+7}$.
$\Sq^1 k_{n+3} = 0$. This must be correct, otherwise the computation will end and we obtain that $\pi_3^s$ is $\Z_4$ at 2. By the way, I cannot prove it.
$\Sq^1 l_{n+3} = 0$. This must be wrong, but I don't know how to prove it.
(For the Adem relations I used Alex Kruckman's Adem relations code.)
EDIT II
Since nobody answers, I make the question a little more general. I don't know how to solve 3, but I can imagine it's the same as 1: suppose that it is non-zero in order to get a contradiction using Adem's relations. I worry about 4 instead, because there I have to prove that it is not zero. Then my general question could be
If $\Sq^1x = 0$ in $E_\infty$ which are the common arguments that can I try to prove that $\Sq^1x \neq 0$ in the target of the spectral sequence?
I have edited with a link to the book I think you mean and specific page and lecture references, but please correct if I got it wrong. Also, when you define your commands, make sure that there is no whitespace after the closing $; that whitespace will appear in the rendered result, making it start with spurious blank lines or such.
I suggest you look at Mosher-Tangora as well as McCleary’s book on spectral sequences. They both have computations similar to what you are looking for.
@user51223 M-T seems to use different fibrations, but I will look. Anyway, if you know these computations and have some hints, I will be happy to hear them!
|
2025-03-21T14:48:30.939921
| 2020-05-10T19:46:05 |
359963
|
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|
Stack Exchange
|
Number of real roots of irreducible polynomials that are solvable by radicals
Let $n \geq 3$ be a natural number. Define the set $X_n$ as the set of natural numbers that appear as the number of real roots an irreducible polynomial of degree $n$ over $\mathbb{Q}$ which is solvable by radicals can have.
Example: In case $n=p$ is a prime, we have $X_p= \{1,p \}$.
Is $X_n$ or the cardinality $|X_n|$ also known for other values?
It would be interesting to see the beginning of the sequence $a_n=|X_n|$ for small values of $n$, maybe it appears in the oeis.
what (trivial or nontrivial) things do you know about $X_n$ in general?
@ArnoFehm I dont really know much about $X_n$ except for the stated result for prime numbers.
You certainly knew $X_4={0,2,4}$, so is $X_6$ the first case you don't know?
@YCor Thanks, yes 6 is the first non-trivial case it seems.
GAP might provide for small $n$ the list of $k$ such that $S_n$ has a transitive solvable subgroup that contains a product of $k$ disjoint transpositions (which is a necessary condition for the existence of such a polynomial with $n-2k$ real roots). Nevertheless for $n=6,8$ this yields no obstruction.
Certainly $n\in X_n$, but probably that was clear to you already.
Klueners and Malle have a database of number fields of degree $\leq 19$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $n$ with $4 \leq n \leq 18$, $X_{n} = \{ k : 0 \leq k \leq n \text{ and } k \equiv n \pmod{2} \}$.
EDIT: If $n = 2k$ is even, then $\mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/k\mathbb{Z}$ contains elements of order $2$ with any desired even number of fixed points. It seems possible that one can obtain the same thing for $n$'s that are multiples of $3$. The smallest $n$ that does not fall into this category is $n = 25$, and using the classification of transitive subgroups of $S_{25}$, one finds that $X_{25} \subseteq \{ 1, 5, 9, 13, 17, 21, 25 \}$.
So quite surprisingly for primes it tends to be very small and for composite it's as large as possible... until 19. For every such $n$ and $k$ with $n-k$ even you manually checked inside the list the polynomials until finding one with $k$ real roots? or you found a quicker way?
Thanks, that motivates to experiment a bit also for some n>=20. I will try my luck a bit to see if the pattern continues.
Jensen On the number of real roots of a solvable polynomial
includes a proof of:
Loewy’s theorem. Let $K$ be a real number field and $f(X)$ an irreducible polynomial in $K[X]$ of odd degree $n$. If $p$ is the smallest prime divisor of $n$ and the Galois group of $f(X)$ over $K$ is solvable, then $r(f) = 1$ or $n$ or satisfies the inequalities $p ≤ r(f)≤ n−p +1.$
|
2025-03-21T14:48:30.940179
| 2020-05-10T19:55:39 |
359965
|
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|
Stack Exchange
|
Basis vs Schauder basis in normed spaces
Following the conventions from Heil: "A Basis Theory Primer" and Albiac, Kalton: "Topics in Banach Space Theory", we might define a basis of an (infinite-dimensional) normed space $V$ as a sequence $(e_n)$ in $V$, such that for any $x \in V$ there is a unique sequence of scalars $(\lambda_n)$, such that $x = \sum_n \lambda_n e_n$ (converging in norm), whereas for a Schauder basis we demand that these coefficients are produced by linear continuous functionals $(e^*_n)$, such that $e^*_m(e_n) = \delta_{mn}$ and $\lambda_n = e^*_n(x)$.
Now, if we work in a separable Banach space, these two notions coincide (theorem 4.13 in Heil and theorem 1.1.3 in Albiac, Kalton), but what if $V$ is a separable normed space which is not complete? Is there a simple, instructive example in which these linear functionals $(e^*_n)$ exist but fail to be continuous?
You ask for an instructive example, so I'll be long winded.
Suppose $(x_n)$ is basis for a normed space $(X,\|\cdot\|)$. The partial sum projections $S_n$ are well defined but might be discontinuous. Define a new, larger, norm on $X$ by $|x| := \sup_n \|S_n x\|$. Then $(x_n)$ is a Schauder basis for $(X,|\cdot|)$. Note that this is the first step in the proof that every basis for a Banach space is a Schauder basis. The harder part of the proof is to show that $(X,|\cdot|)$ is complete when $(X,\|\cdot\|$ is complete. When $(X,\|\cdot\|)$ is not complete, $(X,|\cdot|)$ may not be complete, but if you are building a basis for $(X,\|\cdot\|)$ that is not a Schauder basis, it would be nice to have one such that the basis is a Schauder basis for some natural complete norm on $X$. Also, you would like your non Schauder basis to be in a some natural normed space; let's say an inner product space that is a non closed subspace of $\ell_2$ under the usual $\ell_2$ norm, $\|\cdot\|_2$. Well, $\ell_2$ has lots of non closed subspaces that are complete under a natural norm, the most used being $\ell_1$ under its norm $\|\cdot\|_1$. It should not be too hard to find a Schauder basis for $\ell_1$ such that at least one of the biorthogonal functionals is not in $\ell_2$. Perhaps the simplest such example is $x_1 = e_1$ and $x_n = e_1+e_n$ when $n>1$. (Here $(e_n)$ is the standard unit vector basis.) Notice that the biorthogonal functionals in $\ell_\infty = \ell_1^*$ are given by $x_1^* =(1,-1,-1,\dots)$ and, for $n>1$, $x_n^* = (0,0,\dots,1,0,\dots)$, where the $1$ is in the nth coordinate. Only $x_1^*$ is not continuous in the $\ell_2$ norm--this makes it easy to verify that $(x_n)$ is a basis for $(\ell_1,\|\cdot\|_2$).
Great! Thank you for this very nice clarification and neat example! Just one small detail: should it be $x^*_n = (0,...,0,1,0,...)$ for $n>1$? (maybe I'm not reading the notation properly)
Yes; sorry for the typo. I'll fix it.
|
2025-03-21T14:48:30.940402
| 2020-05-10T19:57:54 |
359966
|
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|
Stack Exchange
|
Asymptotic behaviour of solutions to system of ODEs
Let $Y:(0,+\infty)\to\mathbb{R}^n$ be a solution to the system of ODEs
$$
L[Y]=0,
$$
where $L$ is a linear operator which behaves, in a neighbourhood of 0, as
$$
L[Y](r)\simeq-Y''(r)-\frac{1}{r}Y'(r)+\frac{1}{r^2}Y(r)+AY(r)
$$
where $A$ is a constant $n\times n$ real matrix. I'm interested in understanding the behaviour in a neighbourhood of 0 of the kernel of $L$.
If $A=0$, then it's clear that any element of the kernel behaves, at main order near 0, like a linear combination of
$$
re_j, r^{-1}e_j\quad{j=1,\dots,n}
$$
where $e_j$ is the $j$-th element of the standard basis of $\mathbb{R}^n$. I wonder if this is true as well when $A\ne0$.
In first place, we can surely approximate the behaviour of $L$ by setting all elements of the diagonal of $A$ to 0 (since it appears a term of size $1/r^2$ in the expression of $L$).
In general, if we assume that all the components of $Y$ have the same "size", then one may think that the conclusion still holds true, because formally
$$
\frac{1}{r^2}Y(r)+AY(r)\simeq \frac{1}{r^2}Y(r)
$$
but I don't think that the assumption here is justified. Without this assumption the same conclusion may fail, suppose indeed that the first equation is given in the following form
$$
-y_1''-\frac{1}{r}y_1'+\frac{1}{r^2}y_1+y_2=0,
$$
here I don't know if we have control on the size of $y_2$; if for instance $$y_2\simeq\frac{1}{r^3}y_1\qquad (\ast)$$ the behaviour near 0 of $y_1$ changes completely. On the other hand I don't know if it's possible to have a behaviour like $(\ast)$ in a system like this.
Is there any formal way to face this problem? Does it depend on the form of $A$? Any help is very appreciated.
I am no expert here, but I believe it might be worth giving a look into [Coddington & Levinson, Theory of ordinary differential equations] chapters 4 and 5. I suspect that you're tryinf to deal with a singularity of the second kind.
|
2025-03-21T14:48:30.940583
| 2020-05-10T20:40:20 |
359974
|
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|
Stack Exchange
|
In the two-person Killing the Hydra game, what is the winning strategy?
My question is which player has a winning strategy in the
two-player version of the Killing the Hydra game?
In their amazing paper,
Kirby, Laurie; Paris, Jeff, Accessible independence results for Peano arithmetic, Bull. Lond. Math. Soc. 14, 285-293 (1982), ZBL0501.03017,
Laurie Kirby and Jeff Paris introduced the Killing the Hydra game,
in which one attempts to kill the Hydra by cutting off its heads.
At stage $n$, when you make a cut, just below a head, the Hydra grows $n$ copies of
itself, copies of the position from one lower node (if any) up to the node preceding the neck that had been cut, and whatever is above that node. To illustrate, here are some initial moves in the Hydra
game:
The Hydra game involves some fascinating issues in mathematical
logic, because of its connection with Goodstein's
theorem.
Specifically, what Kirby and Paris proved is
Theorem.
Every strategy in the Killing the Hydra game will eventually succeed in
killing the Hydra; and
This fact is not provable in Peano Arithmetic (PA).
My question here, however, is concerned with the natural
two-player version of the game. Specifically, given a finite
Hydra tree, we play a two-player version of the Killing the Hydra
game, where each player makes a cut on their turn, and the Hydra
grows new heads according the original Hydra rules. The first
player without a move loses---you want to cut the very last head.
Question. Which player has a winning strategy? What is the
winning strategy?
Since every play of the game will lead eventually to a win for one
of the players, it follows by the fundamental theorem of finite
games that one of the players will have a winning strategy. Which
player is it that has the winning strategy? And what are the
winning moves?
The Kirby-Paris theorem is quite robust with respect to the game rules, for it works even when the Hydra grows many more than $n$ copies at stage $n$, or fewer; but I expect that the two-player version might be sensitive to such changes in the rules. Please provide an answer for any reasonable version of the game to which the Kirby-Paris theorem still applies.
I suspect that there's a fairly natural ruleset such that every clopen game of rank $<\epsilon_0$ can be represented as a two-person hydra game, possibly up to some fairly benign transformations; if so, for each ${\bf d}\le_T{\bf 0^{(\epsilon_0)}}$ there will be a game of that type all of whose winning strategies compute ${\bf d}$ (and since every clopen game of rank $<\epsilon_0$ has a winning strategy computable in ${\bf 0^{(\epsilon_0)}}$ that's optimal).
This is a wonderful question and I regret that I can only give it one upvote.
@NoahSchweber I'd love to hear more, if you can prove this. Meanwhile, I've described what I view as the standard Hydra rule set, and I don't know who wins.
Maybe the analysis of Hackenbush game might be useful, and equivalently there a might be a natural way of analysing this game as version of Nim that grows more heaps?
I suggest the following variant, which avoids making arbitrary assumptions about how the hydra grows heads, and which may be more amenable to an explicit description: first Alice plays as Hercules, then Bob plays as the Hydra (i.e., chooses the number of copies made), then Bob plays as Hercules, then Alice plays as the Hydra, and so on.
Is this necessarily a simple problem? Given an arbitrary non-drawn chess position, either White has a winning strategy from that position or Black does, but figuring out which one has it can be extremely difficult, and analysis must be specific to that position. These hydras can be much more complex than chess positions. You may have to just calculate it to the end. In fact maybe there could be a complexity result lurking in this question. I mean it takes stronger axioms than PA to prove that the game is guaranteed to end at all...
We can think of this as a game of "omega-nim;" to more precise since the game you are describing is impartial, operating under the normal play convention, and finite we have that the Sprague-Grundy Theorem applies. In other words, to every "hydra-ordinal" there is an "omega-nimber."
Already this suggests thinking of the plus signs in the hydra as being something roughly analogous to the heaps in Nim. Let me make this precise.
Definition The set of hydras (or hydrae?), $\mathcal{H}$, is defined recursively as
$0 \in \mathcal{H}$
$ n \in \mathbb{N} \implies \omega^n \in \mathcal{H}$
$ \kappa_0,...,\kappa_{n-1} \in \mathcal{H} \implies \sum_{i}\omega^{\kappa_i} \in \mathcal{H}$
We define the "winner function" as:
Definition We denote by $w(\kappa, n, i )$ the "winner of the game $\kappa$ at the $n^\text{th}$ step if it is currently $i^\text{th}$ player's ($i \in \{0,1\}$) turn to move," i.e. $n $ is the number of hydra heads that will grow out if player $i$ cuts a head off of $\kappa$ and $w(\kappa, n, i )$ is the winner under optimal play. Since $w(\kappa, n, i )= 1- w(\kappa, n, 1- i )$ we will sometimes just consider the case $w(\kappa, n) \overset{\text{def}}{=}w(\kappa, n, 0) $ for simplicity.
The answer to your question is to compute $w(\kappa,1,i)$; but of course, in order to answer it we are going to need to define the following
Definition A strategy is a $\sigma: (\mathcal{H} \setminus \{0\}) \times \mathbb{N} \longrightarrow \mathcal{H} $ such that $\sigma(\kappa,n)$ is a legal position at step $n+1$ that succeeds a legal position at $n$. Equivalently we could have allowed for a "virtual position/move" $-1$ and defined $\sigma': \mathcal{H} \times \mathbb{N} \longrightarrow \mathcal{H} \cup \{-1\}$ as $\sigma'(0,n) = -1$ and $\sigma'(\kappa,n) = \sigma(\kappa,n)$ otherwise. We let $\mathcal{S}$ be the set of all such strategies.
Let us try to define the winning $\sigma$ by taking cases on the different possible "heaps."
The heap has size 0 or 1
Since $w(0,n,i) = 1-i$ the strategy $\sigma(0,n)$ is meaningless (that's we defined $\sigma $ on $(\mathcal{H} \setminus \{0\})$); likewise we see that $w(1,n,i) = i$ so that $\sigma(1,n)$ is forced to be $0$. This easily generalizes to the following case
The hydra is a natural number
It is straightforward to prove by induction that $w(k,n,i) = 1- ((k+i) \% 2)$ where $\%$ is remainder after division since by the rules of the game no new hydras grow after cutting a head off a hydra of the form $\omega^0 + \omega^0 +... + \omega^0 = \omega \cdot k $. Likewise $\sigma(1,n)$
The hydra is of the form $\kappa + 1$
If $\kappa ' = \left( \sum_{i}\omega^{\kappa_i}\right) + 1 = \kappa + 1$ then $w(\kappa' ,n,i) = 1 - w(\kappa' - 1 ,n,i) = 1-w(\kappa ,n,i) $. This can proven by a "sum of games" style proof. The idea is the following: if it is $i$'s turn then either:
making a cut on $\kappa$ wins the game,
making a cut on $\omega^0 = 1$ wins the game,
or none of the above
but these are respectively true if and only if
The cut $\sigma(\kappa,n)$ loses the game (for the game $\kappa$ not $\kappa+1$) for all $\sigma$
proof by induction using the following two observations: 1) $\omega^0=1$ (or "parity") is a "loop-invariant" of the game and 2) $\sigma(\kappa,n) < \kappa$ (see the proof of thm 2 of [Kirby; Paris] for the proof of the inequality) 3) eventually we will hit $\kappa \in \omega $ for any strategy $\sigma$ ( once again see [Kirby; Paris]) which is the previous case
the position $\kappa$ is a losing position, but this is true iff the first case is true
or none of the above, but this is true iff the first case is false
Therefore the game is lost or won regardless of what move is made; it only depends on the "parity."
The hydra is of the form $\kappa + \lambda $
By a proof by induction and using the fact that $\sigma(\kappa,n) < \kappa$ and $\sigma(\lambda,n) < \lambda $ (see thm 2 of [Kirby; Paris]) we have that
\begin{equation}
w(\kappa + \lambda , n ) = w(\kappa , n ) \oplus w( \lambda , n ) \oplus 1 .
\end{equation}
Since we have already proven it for $\lambda =1$ and we can assume $\kappa > 2$ we can take the following cases in the induction
Player 1 cuts $\kappa$ then Player 2 cuts $\kappa$
Player 1 cuts $\kappa$ then Player 2 cuts $\lambda$
Player 1 cuts $\lambda$ then Player 2 cuts $\kappa$
Player 1 cuts $\lambda$ then Player 2 cuts $\lambda$
Which all lead to smaller cases to which we can apply the induction hypothesis.
Lets verify for simple examples: since $\sigma(\omega,n) = n $ we have that $w(\omega,n) = n \% 2 $ and also $w(0,n) =1$ which agrees with $w(\omega+ 0 ,n) = n \% 2 = w(\omega , n ) \oplus w( 0 , n ) \oplus 1$. Similarly $w(\omega+ 1 ,n) = n \oplus 1 = w(\omega , n ) \oplus w( 1 , n ) \oplus 1$ and more generally we have that $w(\omega+ k ,n) = n \oplus k = w(\omega , n ) \oplus w( k , n ) \oplus 1$. Likewise by a second induction we have that
\begin{equation}
w\left(\sum_i \kappa_i , n \right) = 1 \oplus \bigoplus_i w( \kappa_i , n ) .
\end{equation}
The hydra is of the form $\omega ^ \kappa $
The point here is that is $\omega ^ \kappa \neq \omega ^ {\lambda +1}$ for all $\lambda \in \mathcal{H}$ then all of the cuts are made inside of the $\kappa$. By induction, we also see that if $\omega ^ \kappa = \omega ^ {\lambda +1}$ then $w(\omega^\kappa,n)$ only depends on the parity of $n$ since if $\sigma$ is the "subtract 1" cut then $\sigma(\omega^\kappa,n) = \omega^\lambda \cdot n$ and $w(\omega^\lambda \cdot n, n+1, 1 ) = \bigoplus_i w( \omega^\lambda , n+1 ) $ by the previous section, so that $w(\omega^\kappa, n ) = \min\{1 \oplus \bigoplus_i w( \omega^\lambda , n + 1) ,1 \oplus w(\kappa, n )\}$ by a similar proof to the last section. The second term corresponds to the following: if player 0 can lose/win the game $\kappa$ then $\kappa$ corresponds to an odd/even game, but when the time finally comes to split $\omega^1$ player 1 will be left with an even/odd game.
Great! This is excellent. For those readers who might benefit, here is an accessible account of transfinite Nim that I wrote on my blog a few years back: http://jdh.hamkins.org/transfinite-nim/.
Thanks! I will be sure to check it out; it might make the problem easier to attack.
|
2025-03-21T14:48:30.941285
| 2020-05-10T21:51:13 |
359981
|
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|
Stack Exchange
|
Principal ideal of a non-associative magma
The definitions of an ideal of an algebraic structure $A$ (as a substructure $I$ such that the product of $A$ and $I$ is a subset of $I$) do not involve associativity.
However, the definitions of a principal ideal I know (for a semigroup or a ring) assume associativity.
For example, the left principal ideal $S^1a$ of a semigroup $S$ is an ideal because $S^1(S^1a) = (S^1S^1)a$.
The two-sided principal ideal of a semigroup is the set $S^1aS^1$ which is defined due to associativity.
https://en.wikipedia.org/wiki/Green%27s_relations
I am trying to generalize the definition of a principal ideal to non-associative structures.
Would it be a correct generalization to say that a principal ideal is an ideal that can be obtained by taking a single element and all finite products with the element as one of the oprands?
The left (right) principal ideal of a non-associative magma $M$ generated by an element $a$ is the set that includes $a$ and all finite products of elements of $M$ where $a$ is the rightmost (resp. leftmost) operand.
The two-sided principal ideal of $M$ generated by $a$ is the set that includes $a$ and all finite products of elements of $M$ that contain $a$ as an operand.
I am wondering if there is a notation for the sets of finite products for a non-associative structure.
Instead of $S^1a$ or $S^1aS^1$ it must include all possible combinations of $...S^1S^1a$ or $...S^1S^1aS^1S^1...$.
Are there better approaches or formulations of a principal ideal of a non-associative magma?
Are there generalizations of a principal ideal for non-associative rings, algebras, etc.?
Yes and no. As in any structure, you have the notion of substructure generated by a single element. In associative structures, this behaves well. Actually, say in a group, you have two kinds of substructures: subgroups, and normal subgroups (= $G$-subgroups, $G$ being viewed endowed with $G$-action). In the latter case, there is no particular good behavior of single-generated substructures. In semigroups, you in addition have ideals, and single-generated behave well. In magmas, single-generated substructures or left/two-sided ideals don't behave well. Not simpler than 2-generated ones, say.
@YCor The purpose of the question is mostly methodological. It may be helpful to understand a principle on the simplest structure, and then to find that it behaves well in a more complex structure.
I guess that the notion of principal ideal is not likely to be very helpful in magmas or non-associative algebras in general, and doesn't look simpler than the notion of ideal generated by 2 elements.
@YCor A definition of a non-associative ideal generated by a set of elements would be even better. Are there any references on the item?
I expect you'd get something such as an inductive definition (as you'd give to a computer), and nothing simpler. Do it inside a magma $M$ is enough (then extend by linearity): the 2-sided ideal generated by $Y\subset M$ is the union over all $n$ of all possible $n$-fold products involving at least one element of $Y$. Recursively, define $M_1=M$, $Y_1=Y$, $M_n=\bigcup_{p+q=n}M_pM_q$, $Y_n=\bigcup_{p+q=n}(M_pY_q\cup Y_pM_q)$ with $p,q>0$ in unions. Then the ideal generated by $Y$ is $\bigcup_{n\ge 1} Y_n$.
@YCor Thank you. It works for me. Would you like to delete, close, or answer the question?
In a magma $M$, one can describe the 2-sided ideal generated by a subset $Y$ as follows: define by induction
$$M_1=M,\;Y_1=Y,\; M_n=\bigcup_{p,q\ge 1,p+q=n}M_pM_q,\;Y_n=\bigcup_{p,q\ge 1,p+q=n}(M_pY_q\cup Y_pM_q).$$
Then the 2-sided ideal generated by $Y$ is $Y_\infty=\bigcup_{n\ge 1} Y_n$.
An alternative definition, is to define $Y'_1=1$, $Y'_{n+1}=Y'_nM\cup MY'_n\cup Y'_nY'_n$; then $\bigcup_{n\ge 1} Y'_n=Y_\infty$.
If $R$ is a scalar ring (=commutative associative unital) and $A$ is an $R$-algebra (not assumed associative), if $Y$ is a subset of $A$, one can define $Y_\infty$ as previously (using only multiplication). Then the $R$-submodule generated by $Y_\infty$ equals the 2-sided ideal generated by $Y$. In case $A$ is unital, this is also the additive subgroup generated by $Y_\infty$.
|
2025-03-21T14:48:30.941681
| 2020-05-10T22:12:26 |
359983
|
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|
Stack Exchange
|
Discriminant ideal in a member of Barsotti-Tate Group
Let $S = \operatorname{Spec} R$ an affine scheme (in our case latter a complete dvr) and $p$ a prime. Then Barsotti-Tate group or $p$-divisible group $G$ of height $h$ over $S$
is an inductive system
$$(G_n, i_n:G_n \to G_{n+1})$$
where $G_n$ are finite, locally free
group schemes of order $p^{nh}$, such that all $i_n$ are closed embeddings with $G_n=G_{n+1}[p^n]$ (equality of sub-groupschemes).
I have a question about a proposition I found in my seminar notes (more precisely a strange reduction step there):
Assume, that for our BT group the base scheme $S= \operatorname{Spec} R$ is spectrum $R$ is a complete, discrete valuation ring. with alg. closed residue field $k=R/ \mathfrak{m}_R$. That's a meaningful reduction in light of the fpqc-descent. Denote it's fraction field by $F:=Frac(R)$.
Let $G_{n}$ an arbitrary member of the inductive system of the Barsotti-Tate group. Denote by $G_{\nu, F}:= G_{n} \times_R \operatorname{Spec} F$ the generic fiber. The proposition tells:
Proposition: The discriminant ideal $\mathcal{D}_{G_n/R} \subset \mathcal{O}_{G_n}$ of the $G_{n}$ depends only on $G_{n, F}$. (in the sense that all parameters determine the ideal sheaf can be extracted from $G_{n, F}$)
The proof begins with a statement I not understand:
It says that the claim above is trivial if $p \in R^{\times}$, ie $p$ invertible in $R$. So assume $p \in \mathfrak{m}_R$ [...]
That's exacly the question: Why is the statement of the Proposition in the case case $p \in R^{\times}$ trivial? Any idea why?
An alternative (or better "classical") proof can be found in J. Tate's paper $p$-Divisible Groups. But the concern of this question is only to understand the "triviality" above.
If $p \in R^\times$ then the discriminant ideal is trivial because all $G_n$ are etale.
@DavidHansen: but why $p \in R^\times$ imply that $G_n$ is etale?
In Tate's paper I mentioned above I found indeed that the discriminant ideal is generated by a power of $p$ (page 164, Prop. 2), but it's a non trivial result. SoI not see why the implication $p \in R^\times \Rightarrow G_n$ etale is "elementary". Could you sketch the proof or give a reference for the argument?
@MortyPB. The following is a standard fact (found in most introductory references on group schemes): Any finite flat group scheme of $p$-power order is 'etale if $p$ is invertible on the base. To prove this, reduce to checking this over a field, and then use that multiplication by $p^N$ is simultaneously $0$ for $N \gg 0$ (by assumption) and unramified (as the map on tangent spaces is given by $p^N$, which is invertible), so the group scheme is geometrically reduced (and thus 'etale).
Ok, so over a field the story follows from a decomposition into etale and connected part. Structure theorem tells about the connected part that it's of order coprime to $p$, so it's trivial by Lagrange. So over field it's fine.
The last part of your argument I not understand: Why the fact that multiplication by $p^N$ for $N$ big enough ($N$ bigger then the group order) acts as zero and induce simultaneously an unramified morphism, imply that $G$ is geometrically reduced?
Unramified finite type schemes over a field are 'etale and hence geometrically reduced.
yes, I see. Last remark: how do you perform the reduction of the problem from "over $R$" to "over a field"?
A finitely presented flat map is 'etale if it is fibrewise 'etale
@Anonymous: Sorry for digging out this thread but there is still a detail in your proof of the claim that Any finite flat group scheme over $k$ of $p$-power order is 'etale if $p$ is invertible on the base. We By Deligne theorem the multiplication by $p^N$ on $G$ indeed kills $G$; in other words the map $p^N: G \to G$ factorize over $Spec(k)$. Then you claim that the map on tangent spaces of $G$ induced by $p^N$ is also given by multiplication by $p^N$? Question: why the induced maps on tangent spaces are given by multiplications by $p^N$?
More precisely I understand that the induced map over tangent spaces over neutral element $e \in G$ is the multiplication by $p^N$-map. But over other points of $G$ I'm not sure why this is true. Could you give a reference for the proof you gave above. Meanwhile I found another proof of this claim but I would like to understand yours.
Sorry, I missed this until now. Looks like you got an answer elsewhere
@Anonymous: No problem, yes meanwhile I got an answer. But could you tell me:
was this exactly the idea you had in mind? namely we want to deduce
that the multiplication map $f_{p^n}: G \to G$ (which is in our case
the zero map as we already know) is unramified. Therefore
we have to analyse the induced map
on differentials $f_{p^N}^\Omega_{G} \to \Omega_{G}$.
What we know that as you said the induced map
$f_{p^N}^: T_e G \to T_e G$ on the tangent space at the stalk
on the neutral element $e \in G$ is liteally given by multiplication
with $p^n$.
And this is an isomorphism (since $p^n \in k^$) and
therefore the dual map on differentials at stalk of $e$
$f_{p^N}^\Omega_{G,e} \to \Omega_{G,e}$ is isomorphism as well,
especially $f_{p^N}$ is unramified at $e$.
The crucial point that confused me before was why in order to check
that $f_{p^N}^*\Omega_{G} \to \Omega_{G}$ is unramified it was
sufficient to check it on $e$-stalk $\Omega_{G,e}$.
As Remy pointed out with reference to Bhatt's notes it's indeed
a fact that that a map on differentials of group schemes are
completly determined on how the map act on the stalk
$\Omega_{G,e}$. For sake of completness: was this exactly
the argument you implicitly used in your proof in the comment
to show that $f_{p^N}$ is unramified?
Yup, this is indeed what I had mind
|
2025-03-21T14:48:30.942102
| 2020-05-10T22:48:55 |
359988
|
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|
Stack Exchange
|
Existence theory with an integral equation
I am reading a paper in which it is proposed that one can solve a problem from mathematical physics by establishing an existence theory for a system of equations. One of the equations in the system is not a PDE and so the system is not local: it's not written in the paper, but it should be an integral of $t$ taken from $0$ to $\infty$ (hence it does not satisfy a local PDE at every point). To clarify, the other two equations in the system are elliptic PDEs.
In order to have an existence theory for the system, one would have to show that the theory works when $\phi$ takes a 'wide range of possible values', but I find it unclear what it would mean for $\phi$ to take a sufficiently wide range of values.
Edit: System of equations added (first equation is the Dirac equation).
$$
\begin{cases}
D \tilde{\psi_t} = 0,\\
\phi = 2 \displaystyle\int\limits_0^{\infty}\! \chi_t \bigg((w_t^{-})^2|\psi_t^{-}|^2 + (w_t^{+})^2|\psi_t^{+}|^2 \bigg) u_t^2 \: \text{d} t \\
H_{\Sigma} - Tr_{\Sigma} K = 0 \\
\end{cases}
$$
It has been requested that I define all the symbols being used: this is a bit lengthy but geometrically interesting so here goes. Start with a Riemannian $3$-manifold equipped with a metric $g$ and a symmetric $2$-tensor field $k$. We will now define the metric given by conformal flow of metrics: this is used in papers by Bray and Agol, Storm and Thurston. This flow is given by
$$
\begin{cases}
g_t = u_t^4 g,\\
\frac{d}{dt} u_t = v_t u_t\\
\Delta_{g_t} v_t =0 \:\: \text{on} \: M_t \\
\end{cases}
$$
where $v_t =0$ on $\partial M_t$ and $v_t(x)$ tends to $-1$ as $|x|$ tends to $\infty$. Assuming $(M,g_t)$ contains an outermost minimal surface, $\partial M_t$ is the region outside of that surface. Take two copies of $M_t$ denoted by $M_t^-$ and $M_t^+$, where the copy is equipped with a metric $g_t^+ = (w_t^+)^4 g_t$ and $w_t^+ = (1 + v_t)/2$ (swap plus with minus for the other metric).
Now take what Bray refers to as the doubled manifold $\tilde{M}_t$, which is the union of $M_t^-$ and $M_t^+$. The associated metric $\tilde{g}_t$ is defined piecewise as the union of $g_t^-$ and $g_t^+$. The spinors in the equations are defined on the manifolds you would expect.
$\Sigma$ is a hypersurface which is given by a graph $t=f(x)$ which sits inside a general warped product space equipped with a warped product metric $(M \times \mathbf{R}, g + \phi^2 dt^2)$, for some function $\phi$ defined on $M$. $\chi_t(x)$ is defined to be $1$ when $x$ is in $\Sigma_t$ and $0$ when $x$ is in $\Sigma_0 - \Sigma_t$. The notation denotes that the conformal flow is now being performed on $\Sigma_0$. The third equation can be written in local coordinates to see the dependence on $g$, $k$ and the warping factor $\phi$, but essentially it is a second-order quasi-linear elliptic PDE. $K$ is just the extension of $k$ on $M$ to the product space and $H$ is the mean curvature.
So this is the issue really, what kind of existence theory will one have for the system to deal with the second integral equation. The coupling between $g$ and $\phi$ is highly non-trivial.
There are entire branches of "integro-differential equations", "pseudo-differential operators", "non-local PDEs" etc. Without knowing what the equations are, however, it is difficult to tell which one may be appropriate here.
I've added the system if that helps. The first and third equations are elliptic PDE and the third equation can be written explicitly in local coordinates to see the dependence on the metric $g$ and on $\phi$. The trouble is the equation in the middle, which is simply an integral, so I'm not exactly sure what this now implies for trying to find an existence theory for the whole system. Obviously the second equation cannot satisfy a PDE locally at each point though.
You should define your symbols, and their corresponding domains. if the symbols are all independent from each other, then your equation is linear and can be solved without doing any work. So there must be some non-trivial coupling between the three displayed equations for this to be an interesting system.
@Willie Wong: Thanks for your response, I have defined the quantities, let me know if I missed anything. Essentially the coupling between $\phi$ and $g$ is highly non-trivial, because at each time step there is a new $g$ and a new $\phi$ and it's hard to know what $\phi$ is doing. I am basically trying to find out how a theory of existence for the system might deal with the second integral equation (probably this existence theory will need some geometric intuition).
I don't quite understand the paragraph that is second from the bottom. What is $\Sigma_t$ and $\Sigma_0$? Also, even though you write $M_t$, am I right in understanding that all of the $M_t$ are diffeo to $M$, with all of the boundaries $\partial M_t$ identified as $\partial M$? (I am trying to understand how you can integrate over $t$ in the second equation when the functions $\psi_t$ may have different domains.)
Also, it looks like $\psi, w, u$ etc can be treated essentially as just given functions (they can be solved independently of $\phi$ and $\Sigma$). Is that correct?
Basically the idea is to couple the conformal flow of metrics to the equation for the manifold $\Sigma$ (third equation), so you do that by performing the conformal flow on the manifold $\Sigma_0$. $\Sigma_0$ is a manifold solving the third equation such that its boundary has zero mean curvature. Following the previous notation, $g_t = (u_t^4 (g + \phi^2 df^2), \Sigma_t)$ is a conformal flow, where $\Sigma_t$ is the region outside the surface $\partial \Sigma_t$.
Yes I believe all the $M_t$ must be diffeomorphic to $M$ because the only thing which is being evolved is the metric. $\psi$ is a harmonic spinor satisfying the Dirac equation, so it must depend on the metric, same with $u$ and $w$ which depend on the metric at each time step as they satisfy the equations of Bray's conformal flow of metrics.
|
2025-03-21T14:48:30.942533
| 2020-05-11T00:01:51 |
359992
|
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|
Stack Exchange
|
Can you give me an example of a totally disconnected subgroup of a topological group that is not a topological group?
In the book The topology of fiber bundles, Steenrod characterize bundle over a base $X$ and totally disconnected structural group $G$ as follows.
Theorem: Let $X$ be arcwise connected, arcwise locally connected,
and semi locally 1-connected. Let $G$ be totally disconnected topological group. Then the equivalence classes of principal bundles over $X$ with group $G$ are in 1-1 correspondence with the equivalence classes (under inner automorphisms of
$G$) of homomorphisms of $\pi_1(X)$ into $G$.
In the differentiable context (every manifold here is closed and connected), examples of Ehresmann-Feldbbau fiber bundles (fiber bundles where the group be not, necessarily, a topological group) with totally disconnected groups occur in foliated bundles:
Proof: A foliated bundle is characterizazed by a homomorphisms $\varphi:\pi_1(M)\longrightarrow \text{Diff}(F)$, where $F$ is the fiber. With the $C^\infty$-topology, the group $\text{Diff}(F)$ is a metrizable topological group that act continuously in $F$. Since that the fundamental group of a compact manifold is finitely genereted (a discussion can be found here), the group $G=\text{Image}(\varphi) $ is an enumareable set. Now, given $f,g\in G$, $f\neq g$, denoting by $d$ a metric that generates the topology of $\text{Diff}(F)$, the function $F(h)=\frac{d(f,h)}{d(f,h)+d(f,g)}$, $F:\text{Diff}(F)\rightarrow \mathbb{R}$ , is a continuous function satisying $F(f)=0$ and $F(g)=1$. Since G is an enumerable set, there exists $\alpha$, $0<\alpha<1$, such that $F(h)\neq \alpha $ for all $h\in G$. Thus, $G$ is equal the disjoint union $(G\cap F^{-1}(-\infty,\alpha))\cup (G\cap F^{-1}(\alpha,\infty))$, $f\in \cap F^{-1}(-\infty,\alpha)$ and $g\in \cap F^{-1}(\alpha,\infty)$, therefore $G$ is totally disconnected.
I want to show that a foliated fiber bundle is also a fiber bundle with a group $G$. This motives the question below:
Question: The group $G$ is a topological group? More generally, every totally disconnected subgroup of a topological group is, itself, a topological group?
Where you say more generally, I assume you mean a subgroup equipped with the subspace topology? Because it seems to me the answer is yes, and it has nothing to do with total disconnectedness: a subgroup of a topological group, when given the subspace topology, is a topological group. Mostly it comes down to checking that if $H \hookrightarrow G$ is a subspace, then the product topology on $H \times H$ coincides with the subspace topology on $H \times H$ inherited from the product space $G \times G$.
Yes, the subgroup $H$ has the subspace topology. The group operations $p:H\times H \rightarrow H$ and $i:H\rightarrow H$, $p(g,h)=gh, i(g)=g^{-1}$ are continuous? This is true when $H$ is closed subgroup. In general not. Therefore, a subgroup is not, in general, a topological group with the subspace topology.
You say in general not. I don't think I follow. What example do you have in mind? (FWIW, Wikipedia seems to agree with me, https://en.wikipedia.org/wiki/Topological_group#Properties, paragraph 5, but it seems to me a straightforward argument.)
Elizeu, that doesn't answer my question. If you have an example, please say what it is.
I don't have in mind an example. But, the continuity of $p:G\times G\rightarrow G$ not implies the continuity of $p:H\times H\rightarrow H$.
If $H$ is a subgroup of a topological group $G$ and is equipped with the subspace topology, then $H$ is a topological group.
It's mostly a matter of seeing that the product topology $Top_{prod}$ on $H \times H$ matches the subspace topology $Top_{sub}$ on $H \times H$ coming from $G \times G$. The product topology on $H \times H$ is the smallest topology such that the two projections $\pi_1, \pi_2: H \times H \to H$ are continuous. An open in $H$ is of the form $U \cap H$ for some open $U$ of $G$, and $\pi_1^{-1}(U \cap H) = (U \cap H) \times H = (U \times G) \cap (H \times H)$ is open in the subspace topology on $H \times H$. A similar statement holds for $\pi_2$ in place of $\pi_1$. Hence $Top_{prod} \subseteq Top_{sub}$. But also, the subspace topology on $H \times H$ is the smallest topology such that the inclusion $i: H \times H \to G \times G$ is continuous. For opens $U, V$ of $G$ we have $i^{-1}(U \times V) = (U \times V) \cap (H \times H) = (U \cap H) \times (V \cap H)$; the latter is open in the product topology. So $Top_{sub} \subseteq Top_{prod}$.
If $m_H: H \times H \to H$ is the restriction of multiplication $m_G: G \times G \to G$, and $U \cap H$ is a typical open in $H$, then, $m_H^{-1}(U \cap H) = m_G^{-1}(U) \cap (H \times H)$ is open in $H \times H$ under the subspace topology, hence under the product topology by the above paragraph.
If $i_G: G \to G$ denotes inversion and $i_H: H \to H$ is its restriction, then $i_H^{-1}(U \cap H) = i_G^{-1}(U) \cap H$ is open in $H$ for a typical open $U \cap H$.
Your answer is very clear. I have made confusion with the notion of a Lie subgroup. Thanks.
|
2025-03-21T14:48:30.942914
| 2020-05-11T00:04:22 |
359993
|
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|
Stack Exchange
|
Analogy between Stiefel-Whitney and Chern classes
There is a clear similarity between Stiefel-Whitney and Chern classes, if one replaces base field $\mathbb R$ with $\mathbb C$, coefficient ring $\mathbb Z/2$ with $\mathbb Z$ and scales the grading by a factor of $2$. For instance, both can be defined by the same axioms (functoriality, dimension, Whitney sum, value on the tautological bundle over $\mathbb P^1$).
Is there a deep reason behind this correspondence? The best explanation I have so far is the structure of classifying spaces. Being Grassmanians, they admit Schubert cell decomposition (which is essentially an algebraic fact). For cohomology of complex Grassmanians, differentials vanish for dimensionality reasons, and for real ones they vanish when reduced mod $2$.
There is a number of similar phenomena, for instance, $BO(1,\mathbb R) = K(\mathbb Z/2, 1)$ and $BU(1) = K(\mathbb Z, 2)$ which says that in both cases, topological line bundles are completely determined by their first characteristic class.
Also, I have been told that Thom polynomials for Thom-Boardman singularities of maps between real or complex manifolds have the same coefficients when expressed in $w_i$ and $c_i$. Can these facts be explained in a similar way?
что касается предпоследнего абзаца -- может быть, просто повезло, что у $S^0$ и $S^1$ нет высших гомотопических групп?..
Here is one way I like to think of the analogy.
The maximal torus of diagonal matrices $T^{n} \subset U(n)$ gives a map $BT^n \to BU(n)$ which on integral cohomology gives an isomorphism from $H^{\ast}(BU(n))$, which is a polynomial algebra on $n$ generators of degrees $2i$, $1\leq i\leq n$, to $H{\ast}(BT^n)^{\Sigma_n}$, which is the polynomial algebra on the symmetric polynomials $\sigma_i$ in the $n$ standard degree 2 generators. The Chern class $c_i$ is the element of the domain that maps to $\sigma_i$.
The maximal $2$-torus of diagonal matrices $(C_2)^{n} \subset O(n)$ gives a map $BC_2^{n} \to BO(n)$ which on mod $2$ cohomology gives an isomorphism from $H^{\ast}(BO(n))$, which is a polynomial algebra on $n$ generators of degrees $i$, $1\leq i\leq n$, to $H^{\ast}(BC_2^n)^{\Sigma_n}$, which is the polynomial algebra on the symmetric polynomials $\sigma_i$ in the $n$ standard degree $1$ generators. The Stiefel-Whitney class $w_i$ is the element of the domain that maps to $\sigma_i$.
Thinking of $BT^{n}$ as $(CP^{\infty})^{n}$ and $BC_2^{n}$ as $(RP^{\infty})^{n}$ may help.
I like to imagine that these maps are part of the explanation of splitting principle, since $BT^n$ and $B(C_2)^n$ are classifying spaces for split bundles. This makes generators of their cohomology a sort of "universal Chern roots".
I agree. I gave a new proof of the splitting priniciple in http://www.math.uchicago.edu/~may/PAPERS/Split.pdf that emphasizes that point of view.
Any rank $n$ real vector bundle $E\to X$, $X$ compact $CW$-complex is $\newcommand{\bZ}{\mathbb{Z}}$ is $\bZ/2$-oriented and, as such it has a $\bZ/2$-Thom class $\tau_E\in H^n_{cpt}(E,\bZ/2)$. Then
$$w_n(E)=\zeta^*\tau_E\in H^n(X,\bZ/2),$$ where $\zeta:X\to E$ is the zero-section
Any complex vector bundle $E\to X$, $X$ compact $CW$-complex of complex rank $n$ is $\bZ$-oriented and, as such it has a $\bZ$-Thom class $\tau_E\in H^{2n}_{cpt}(E,\bZ)$. (Note that $2n$ is the real rank of $E$.) Then
$$c_n(E)=\zeta^*\tau_E\in H^{2n}(X,\bZ).$$
Thus in both cases the top Stieffel-Whitney class and the top Chern class are Euler classes, with different choices of coefficients. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bP}{\mathbb{P}}$
To get the rest of the Stieffel-Whitney/Chern classes on then needs to rely on some basic facts
$$ H^\bullet(\bR\bP^n,\bZ/2)\cong\bZ/2[w]/(w^n+1),\;\; H^\bullet(\bC\bP^n,\bZ)\cong\bZ[c]/(c^n+1),\tag{1}
$$
where $w\in H^1(\bR\bP^n,\bZ/2)$, $c\in H^2(\bC\bP^n,\bZ)$ are the Euler classes of the (duals) of the tautological line bundles.
These results suffice to construct the Stieffel-Whitney/Chern classes. This is the approach pioneered by Gronthendieck for the construction of Chern classes. For details see Chapter 5 of these notes.. As explained in Example 4.3.5. of these notes duality, under the guise of Thom isomorphism is also responsible for the isomorphisms (1).
One could claim that duality or Thom isomorphism is what makes things work. What is behind Thom isomorphism? As described in Bott-Tu, this follows from two basic facts about cohomology. The first is the Poincare lemma with compact supports
$$H^k_{cpt}(\bR^n, G)=\begin{cases} 0, & k\neq n,\\
G, &k=n.
\end{cases}
$$
and the second is the Mayer-Vietoris principle which, roughly specking says that one can recover the cohomology of an union from the cohomology of its parts. View this as a local-to-global principle, a way a patching local data to obtain global information. The orientability condition is the one that allows the local-to-global transition.
Fantastic course, many thanks for sharing it!
Let me try a high-brow answer using equivariant stable homotopy theory.
By the stable Thom isomorphism, the integral (co)homology of $BU$ agrees with that of $MU$; likewise the $\mathbb{Z}/2$-(co)homology of $BO$ agrees with that of $MO$.
Let $H\underline{\mathbb{Z}}$ be the $C_2$-equivariant Eilenberg--Mac Lane spectrum for the constant Mackey functor and let $M\mathbb{R}$ be the Real cobordism spectrum. Using the theory of Hu and Kriz, we see that $H\underline{\mathbb{Z}}$ is Real oriented and thus the $RO(C_2)$-graded groups $H\underline{\mathbb{Z}}^{\bigstar}M\mathbb{R}$ agree with $H\underline{\mathbb{Z}}^{\bigstar}[[ \overline{c}_1, \overline{c}_2, \dots ]]$, where $\overline{c}_i \in H\underline{\mathbb{Z}}^{i+i\sigma}M\mathbb{R}$. The $\overline{c}_i$ define thus maps $M\mathbb{R} \to \Sigma^{i+i\sigma}H\underline{\mathbb{Z}}$, which forget to the usual Chern classes $c_i\colon MU \to H\mathbb{Z}$.
The geometric fixed points of $M\mathbb{R}$ are $MO$, while those of $H\underline{\mathbb{Z}}$ are $\prod_{k\geq 0}\Sigma^{2k}H\mathbb{Z}/2$ and thus come with a projection $p\colon \Phi^{C_2}H\underline{\mathbb{Z}} \to H\mathbb{Z}/2$. Thus combining $\Phi^{C_2}$ with $p$, the $\overline{c}_i$ define maps $MO \to \Sigma^iH\mathbb{Z}/2$, which we claim to be the $w_i$.
I have not thought through this, but I guess one way to show a thing like this is to use homology instead. We have $\pi_{\bigstar}H\underline{\mathbb{Z}} \otimes M\mathbb{R} \cong \pi_{\bigstar}H\underline{\mathbb{Z}}[\overline{b}_1, \dots]$. Write $\pi_*\Phi^{C_2}H\underline{\mathbb{Z}} = \mathbb{Z}/2[u]$. Then
$$H_*(MO; \mathbb{Z}/2)[u]\cong (\Phi^{C_2}H\mathbb{Z}/2)_*(MO) \cong \pi_* \Phi^{C_2}(H\underline{\mathbb{Z}} \otimes M\mathbb{R}) \cong \mathbb{Z}/2[u][\Phi^{C_2}\overline{b}_1, \dots] $$
Here, we use that we can get the geometric fixed points by inverting one element $a_{\sigma}$ and that $M\mathbb{R}$ is of finite type. Killing the $u$ on both sides, gives that we obtain indeed the whole homology of $MO$ by this geometric fixed points construction. Thus, I expect that by playing around with dual (co)homology classes, one should get indeed that one obtains the $w_i$ from the $\overline{c}_i$.
(Why was it natural to use the projection $p$? Indeed, when we apply geometric fixed points to the $\overline{c}_i$ all other components are $0$. This we can see by lifting the $\overline{c}_i$ to $BP\mathbb{R}^{\bigstar}M\mathbb{R}$ and observing that the geometric fixed points of $BP\mathbb{R}$ are just $H\mathbb{Z}/2$.)
Could you please add few words to mention what $\Phi$ is? Hu and Kriz do explain it referring to Lewis-May-Steinberger, but...
Sure. $\Phi = \Phi^{C_2}: \mathrm{Sp}^{C_2} \to \mathrm{Sp}$ is the geometric fixed point functor. It is characterized by the following properties: 1) $\Phi^{C_2}\Sigma^{\infty}X = \Sigma^{\infty}X^{C_2}$; 2) $\Phi^{C_2}$ is symmetric monoidal; 3) $\Phi^{C_2}$ commutes with homotopy colimits. If you want to compute the geometric fixed points of $M\mathbb{R}$ you can argue like follows: $M\mathbb{R}_{k+k\sigma}$ is the Thom space of the universal bundle on $BU(n)$ with the complex conjugation action. Its fixed points are the corresponding Thom space over $BO(n)$...
...further $M\mathbb{R} \simeq \mathrm{hocolim} \Sigma^{-k-k\sigma} M\mathbb{R}_{k+k\sigma}$ (this is a general fact, see e.g. Section 2 of Hill--Hopkins--Ravenel's Kervaire paper). As $\Phi^{C_2}$ is symmetric monoidal, it also preserves duals and thus takes $\Sigma^{-k-k\sigma}(-) = D(S^{k+k\sigma}) \otimes -$ to $\Sigma^{-k}$ as the $C_2$-fixed points of $S^{k+k\sigma}$ are $S^k$. Thus, we see that the geometric fixed points of $M\mathbb{R}$ are the homotopy colimit over $k$-fold desuspensions of Thom spaces over $BO(n)$, i.e. exactly $MO$.
Is there a particular reason why you use $M\mathbb{R}$ instead of $\Sigma^\infty BU_{\mathbb{R}}$ or is that only for convenience of computation?
@DenisNardin You're right that it would be actually more natural to use the latter because one doesn't have to use the Thom isomorphism. It was more out of habit.
$\newcommand{\Z}{\mathbf{Z}}$The Chern classes may be viewed as a map $c: \mathrm{BU}(n) \to \prod_{i=1}^n K(\Z,2i)$. Call the target $X_n$. Complex conjugation defines a $\Z/2$-action on $\mathrm{BU}(n)$, whose fixed points are $\mathrm{BO}(n)$. Give $X_n$ the $\Z/2$-action defined by viewing $K(\Z, 2n)$ as $\Omega^\infty(\Sigma^{2n,n} \mathrm{H}\Z)$, where $S^{2n,n} = S^{n + n\sigma}$ (here, $S^\sigma$ is the one-point compactification of the sign representation of $\Z/2$), and $\mathrm{H}\Z$ is the Eilenberg-Maclane spectrum associated to the constant Mackey functor. I forgot to write this in the previous version of this answer, but (as in Lennart's answer) there is a projection map $\Phi^{C_2} \mathrm{H}\Z\to \mathrm{H}\mathbf{F}_2$. The map $c$ is $\Z/2$-equivariant, and taking $\Z/2$-fixed points and composing with the above projection yields the map $\mathrm{BO}(n) \to \prod_{i=1}^n K(\Z/2, i)$ given by the Stiefel-Whitney classes. I haven't done this explicitly, but to check that these are in fact the Stiefel-Whitney classes, one reduces to the case of line bundles (by the splitting principle); then, it follows from the fact that $\mathrm{BU}(1)$ is equivariantly equivalent to $\Omega^\infty \Sigma^{2,1} \mathrm{H}\Z$ (and this identification is given by the first Chern class), and taking fixed points and projecting gives the identification of $\mathrm{BO}(1)$ with $\Omega^\infty \Sigma \mathrm{H}\mathbf{F}_2$ (where this identification is given by the first Stiefel-Whitney class). [This is essentially the same as Lennart's answer.]
It's not true that $\Sigma^nH\mathbb{Z}/2$ is the homotopy fixed points nor the genuine fixed points, nor the geometric fixed points of $\Sigma^{(1+\sigma)n}H\mathbb{Z}$. In which sense did you want to take the fixed points?
@DenisNardin You're right. Edited. (Had excluded the step made explicit in Lennart's reply.)
|
2025-03-21T14:48:30.943732
| 2020-05-11T03:32:03 |
360003
|
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|
Stack Exchange
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Existence of an injective unbounded below operator
Given an infinite dimensional Banach space, can we always find an infinite dimensional Banach space $Y$ and an injective bounded operator $T:X\to Y$ such that $T$ is not bounded below?
If $X^{*}$ is $w^*$-separable, then for every Banach space $Y$, there exists an injective compact operator $T: X\to Y$ (see Goldberg and A.H. Kruse, The Existence of Compact Linear Maps Between Banach Spaces. Proc. A.M.S. 13 (1962), 808-811) and we know that a compact operator is not bounded below.
We also know that for $X$ nonseparable we may not find any injective compact operator into any Banach space (Existence of injective compact operators).
Yes. Using the injective property of $\ell_\infty$, get an operator $S:X\to \ell_\infty$ that is compact on some infinite dimensional subspace $X_0$ of $X$. Let $Q: X \to X/X_0$ be the quotient map. Define $T:X\to \ell_\infty \oplus X/X_0$ by $Tx = (Sx, Qx)$.
With a slightly different argument you can replace $\ell_\infty$ by any infinite dimensional Banach space with $S$ being even nuclear.
A more challenging problem is whether there is always an injective strictly singular operator from $X$ into some Banach space. (Hint: The answer is negative.)
How about inessential instead of strictly singular? (https://link.springer.com/content/pdf/10.1007/BF01682943.pdf) My intuition would be that the answer is negative too.
Right, Tomek. Same example I had in mind for strictly singular.
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2025-03-21T14:48:30.943848
| 2020-05-11T04:03:14 |
360004
|
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|
Stack Exchange
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Open problems for shifted symplectic structures
I am now interested in shifted symplectic structures.
What are the open problems of shifted symplectic structures regarding the moduli space of sheaves ?
Especially now I am interested in moduli space of sheaves on K3 surface.
Any comment welcome! Thank you!
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2025-03-21T14:48:30.943905
| 2020-05-11T04:19:40 |
360006
|
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Stack Exchange
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zero extension of positive currents are always positive
In Demailly's Complex Analytic and Differential Geometry page 139:
He said the trivial (zero) extension of the positive current $T$ (on $X\setminus E$), which denoted by $\tilde T$ is always positive on $X$. It seems like quite obvious, but I came across some obstacle when verifying that claim.
Notice that a current $T\in\mathcal D^{'}_{p,p}(X)$ is said to be positive if $\langle T, u\rangle\geqslant 0$ for all test forms $u\in\mathcal D_{p,p}(X)$ that are strongly positive at each point.Another way of stating the definition is:
$T$ is positive if and only if $T \wedge u \in \mathcal D_{0,0}^{'}(X)$ is a positive measure for all strongly positive forms $u \in\mathcal C_{p,p} ^{\infty}(X)$.
This is so because a distribution $S\in\mathcal D^{'}(X)$ such that $S(f)\geqslant 0$ for every non-negtive function $f\in\mathcal D (X)$ is a positive measure.
Here is my thought:
First, select arbitrary $u \in\mathcal C_{p,p} ^{\infty}(X)$, we have $u \in\mathcal C_{p,p} ^{\infty}(X\setminus E)$, then we would like to show that
$$\langle \tilde T\wedge u,f\rangle=\langle T\wedge u,f\rangle\geqslant 0, \qquad f\in\mathcal D(X),$$
due to the construction of $\tilde T$. However, supp $(f)\cap (X\setminus E)$ isn't compact in $X\setminus E$. Then, how can I infer LHS is non-negative?
Any help and suggestion are appreciated. Thanks a lot!
This is where the finite mass condition comes in. It ensures that, with your notations, $\langle T \wedge u ,f \rangle$ is convergent. To simplify notation, say $T= \sum \lambda_I dz_I \wedge d \bar {z_I}$ and $u=dz_J\wedge d \bar{z_J}$. Then $\langle T\wedge u,f\rangle = \int_{X\setminus E}\lambda_{J^c} fdV_{\mathbb C^n}$ while we know that the functions $\lambda_I$ are non-negative on $X\setminus E$ and locally integrable near $E$.
I meant that the $\lambda_I$'s are positive measures (they don't necessarily have a density wrt the Lebesgue measure) with locally finite mass near $E$, hence $\langle \lambda_I, f\rangle$ is well-defined.
@Henri, thanks for your help. One more question maybe sounds stupid, why we need $\langle T\wedge u,f \rangle $ is convergent, as when consdering the local representation, it seems like non-negative already?
The quantity $\langle T \wedge u ,f \rangle $ has to be a number (not $+\infty$) if you want to make $\widetilde T$ into a current.
@Henri, ok, now I get it, thanks for your patience and assistance again.
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2025-03-21T14:48:30.944076
| 2020-05-11T06:45:26 |
360013
|
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Stack Exchange
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Kähler differential of completion of algebra
Let $(R, \mathfrak{m}) $ be a local $k$- algebra and $\Omega^{1}_{R}$ denote the module of Kahler differential. Does the canonical map $ \Omega^{1}_{R} \otimes_{R} \hat{R} \rightarrow \Omega^{1}_{\hat{R}}$ is an isomorphism?, where $\hat{R}$ is the completion of $R$ with respect to $\mathfrak{m}$ and $k$ is any field of characteristics $0$.
If not, under what condition on $R$ turn this map to be isomorphism?. Is smoothness of $R$ will suffices to conclude this?
No, this is already false for $R = k[t]_{(t)}$, $\widehat{R} = k[[t]]$, because the equation $d(\sum a_n t^n) = \sum n a_n t^{n-1}$ holds in the target but not in general in the source (N.B. your map goes in the wrong way). That's why for complete rings one usually works with continuous differentials rather than Kahler differentials.
Ok got it. Can you tell me what is $\Omega^{1}_{k[[x]]}$?
nothing particularly useful. We know that $k[[x]]$ is the increasing union of smooth $k[x]$-algebras (by Neron desingularization) which seems to show that $\Omega^1_{k[[x]]}$ is free of infinite rank. I don't know what you want to do with it but it's probably the wrong object to look at.
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2025-03-21T14:48:30.944188
| 2020-05-11T07:59:04 |
360017
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Stack Exchange
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Removing half of H_1 of a 2-manifold through a 3-dimensional bounding manifold
Let $M$ be a 2-manifold. If $X$ is a 3-manifold such that $\partial X = M$, we know that the image of the boundary map $H_2(X, M) \rightarrow H_1(M)$ has rank equal to one half the rank of $H_1(M)$.
My question is that given any half rank subspace/subgroup $C$ of $H_1(M)$, can I always find a 3-manifold $X_C$ such that $\partial X_C = M$ and the image of the boundary map $H_2(X_C, M) \rightarrow H_1(M)$ is exactly $C$? If the answer is not, what is the constraint on $C$ for such a 3-manifold $X_C$ to exist?
$M$ definitely needs to be oriented, otherwise you won't even find any $X$ bounded by $M$. Also, a necessary condition is that your subspace forms a Lagrangian subspace with respect to the intersection pairing on $H_1(M)$ (or is orthogonal to a Lagrangian subspace of $H^1(M)$ with respect to the cup product). I have the feeling that that might be sufficient, using a surgery argument, but I am not sure (hence a comment instead of an answer).
@AchimKrause Thanks for the comment. I am indeed mostly interested in the case of orientable 2-manifold $M$. Could you elaborate on the surgery argument that yields the conclusion that the subspace $C$ has to form a Lagrangian subspace of $H^1(M)$ ?
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2025-03-21T14:48:30.944299
| 2020-05-11T08:45:01 |
360018
|
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"url": "https://mathoverflow.net/questions/360018"
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|
Stack Exchange
|
Uniform Lipschitz function approximation by shallow neural networks
Fix $d\in \mathbb{N}$. Let $F_1$ be the set of all 1-Lipschitz functions mapping $[0, 1]^d$ to $\mathbb{R}$.
For $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ and $m \in \mathbb{N}$, let $N_\varphi^m$ be the set of feed-forward neural network functions with input dimension $d$, output dimension 1, hidden dimension m, two layers and activation function $\varphi$. That means $N_\varphi^m$ is the set of functions $h: \mathbb{R}^d \rightarrow \mathbb{R}$ such that $h(x) = b_0 + \sum_{i=1}^m b_i \varphi(a_0 + \sum_{j=1}^d a_j x_j)$ for $x\in \mathbb{R}^d$, where $a_0, ..., a_d \in \mathbb{R}$ and $b_0, ..., b_m \in \mathbb{R}$ are the weights of the network.
I am looking for the following result, which I expect to exist somewhere in the literature (for a suitable activation function $\varphi$):
Does the following hold?
For any $\varepsilon > 0$, there exists some $m \in \mathbb{N}$, such that for any $f \in F_1$ there exists $f^m \in N_\varphi^m$ so that for all $x \in [0, 1]^d$ it holds $|f(x)-f^m(x)|<\varepsilon$.
More abstractly, I am looking for a standard universal approximation result for neural networks, but the necessary hidden dimension $m$ should only depend on the function class (Lipschitz functions), not on the specific function.
In this paper the authors achieve this kind of result (Theorem 1), but they require deep neural networks instead of shallow ones.
Maybe you can check Theorem 4 in Poggio et al. "Why and When Can Deep-but Not Shallow-Networks Avoid the Curse of Dimensionality: A Review"
Thanks, this indeed looks like the kind of result I need. However, I'll leave the check mark open for now, as there might be results out there with a more direct proof or more general activation functions!
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2025-03-21T14:48:30.944433
| 2020-05-11T09:52:18 |
360021
|
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"url": "https://mathoverflow.net/questions/360021"
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|
Stack Exchange
|
Support of random closed walk in arbitrary graph
Researching a question related to closed walks on graphs, I have come across the following problem. Let $G$ be a connected graph on $n$ vertices and $k=O(\log(n))$. Pick a random closed walk on $G$ as follows: Pick a uniformly random vertex $v$ of $G$ and then pick a uniformly random closed walk $w$ of length $2k$ starting and ending at $v$. I am interested in showing that with high probability the support of $w$, i.e., the number of vertices it passes through, is $\Omega(k^{\alpha})$ for some constant $\alpha>0$. More formally, is it true that
$$\Pr[\text{support}(w)<k^{\alpha}]\to 0 \text{ as } n\to \infty?$$
It is not very important to me that $k=O(\log(n))$, so anything without this restriction would also be of interest!
Generally, I have had a hard time finding litterature dealing with properties of random closed walks. Is there any I have overlooked?
Such a bound for regular graphs valid for $\alpha<1/4$ was recently obtained in "Support of Closed Walks and Second Eigenvalue Multiplicity of Regular Graphs"
https://arxiv.org/pdf/2007.12819.pdf Theorem 1.3
Indeed, thank you! I coauthored the paper :)
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2025-03-21T14:48:30.944532
| 2020-05-11T09:53:08 |
360022
|
{
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}
|
Stack Exchange
|
Reference for limits of schemes with non-affine transitions?
Inverse systems of projective schemes appear in several contexts, for example:
in constructing the Zariski-Riemann space of a projective variety,
in studying subvarieties of a projective variety passing through a given point by the crutch of infinitesimal thickenings (e.g. the formal group of an abelian variety),
etc.
Almost everywhere in the literature, inverse systems are studied under the assumption that the transition morphisms in the system are affine (mostly because in this case the limit is representable).
Do you know some references (articles, books...) where inverse systems are studied without the affineness assumption?
The natural object one wants to study is the limit of the system as an fppf sheaf, and one wants to do some geometry with it, that is, attach geometric quantities to it like a topological space, a dimension, define regularity properties etc.
|
2025-03-21T14:48:30.944622
| 2020-05-11T10:26:05 |
360024
|
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|
Stack Exchange
|
Separability of Minkowski Sum of well-behaved sets
Let $A$ and $C$ be non-empty simply connected and connected subsets of $\mathbb{R}^k$ and suppose that $C$ is convex. Then is the Minkowski sum $A+C$ separable?
What do you mean by "separable" ?
There's a dense separable subspace, since both $A$ and $C$ are subspace of a Euclidean space
Then it's true, because any subset of a separable metric space is also separable.
Maybe @FlabbyTheKatsu had in mind that $A+C$ is simply connected? In this case the answer is negative: take for $A$ the letter $C$ and for $B$ the letter $I$ on the plane.
Yes, I meant to ask as Taras said. Thank you for the anwser.
|
2025-03-21T14:48:30.944702
| 2020-05-11T11:53:38 |
360029
|
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|
Stack Exchange
|
Closed form for Jacobi sum $\sum_{a\in \mathbb{F}_{p^2}}\chi(a)\chi(1-a)$
Let $\mathbb{F}_{p^2}$ be a field with $p^2$ elements and $\chi:\mathbb{F}_{p^2}^*\to\mathbb{C}^{*}$
be a multiplicative and non-trivial character on the multiplicative group $\mathbb{F}_{p^2}^*$ (extended by $\chi(0)=0$). Let $$J(\chi)=\sum_{a\in \mathbb{F}_{p^2}}\chi(a)\chi(1-a).$$ Is it possible to find a closed form for $J(\chi)$?
Writing that whole sum makes sense and is very similar to what you do in the $\mathbb{F}_p$ case.
However, I doubt that this generator approach will be useful here. As you already noted, this completely takes into account the multiplicative structure of our field, but none of the additive structure. And the appearance of $1-a$ really makes this an additive problem. So essentially you are summing $\chi(ab)$ over $a,b$ with $a+b=1$. And it is completely hopeless to characterise the solutions to such an equation in terms of a multiplicative generator.
Compare to how primitive roots are useful to find the number of quadratic residues but completely useless to find e.g. the number of adjacent quadratic residues i.e. numbers where $a$ and $a+1$ are quadratic residues.
Is $p$ prime? or just a prime power? Also there is clearly no such character, unless you mean a character $\mathbb{F}_{p^2}^\to\mathbb{C}^$, or mean it to be valued in $\mathbb{C}$.
I don't think there is any closed-form formula, no.
This is a Jacobi sum $J(\chi,\chi)$ (for the field $\mathbb{F}_{p^2}$). One reference is http://www.math.mcgill.ca/goren/SeminarOnCohomology/mycohomologytalk.pdf.
|
2025-03-21T14:48:30.944845
| 2020-05-11T12:02:39 |
360030
|
{
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|
Stack Exchange
|
$\mathbb Z$-graded manifold is isomorphic to the structure sheaf of supermanifold locally
In this paper, definition 4.4.1 about supermanifold and definition 4.6.1 about graded manifold:
Definition 4.4.1: An supermanifold $\mathcal{M}$ is a locally ringed space $(M,\mathcal O_M)$ which is locally isomorphic to $(U, C^\infty(U)\otimes \wedge^\bullet V^*)$ where $U\subset \mathbb R^n$ is open and $V$ is some finite-dimensional real vector space.
Definition 4.6.1: A graded manifold $\mathcal M$ is a manifold $M$ which locally looks like $(U, C^\infty(U)\otimes \text{Sym} (V^*))$, where $U ⊂ \mathbb R^n$ is open and $V$ is a graded vector space.
Then it claims without proof in Remark 4.6.1 that:
Remark 4.6.1:
One can construct an isomorphism between the the structure sheaf of a supermanifold and the local model of a graded manifold, which will be in the category of $\mathbb Z$-graded algebras.
My question:
Using definition 4.4.1 for supermanifold and definition 4.6.1 for graded manifold above, how can we construct such isomorphism?
My trial:
In the second definition, write graded vector space $V = \bigoplus_{i \in \mathbb Z} V_{i}$.
Let $V_{\bar{0}} = \bigoplus_{k \in 2 \mathbb Z} V_k$ and $V_{\bar{1}} = \bigoplus_{k \in 2 \mathbb Z+1} V_k$, then $V = V_{\bar{0}}\oplus V_{\bar{1}}$ and $V^{*} = V_{\bar{0}}^{*} \oplus V_{\bar{1}}^{*}$.
$$\text{Sym}^{n}(V^{*}) = \text{Sym}^{n}(V_{\bar{0}}^{*} \oplus V_{\bar{1}}^{*}) = \bigoplus_{0 \le k \le n}(\text{Sym}^{k}V_{\bar{0}}^{*}\otimes\bigwedge^{n-k}V_{\bar{1}}^{*}).$$
$$\text{Sym}(V^{*})
= \bigoplus_{n} \text{Sym}^{n}(V_{\bar{0}}^{*} \oplus V_{\bar{1}}^{*}) \\
= \bigoplus_{n} \bigoplus_{0 \le k \le n}(\text{Sym}^{k}V_{\bar{0}}^{*}\otimes\bigwedge^{n-k}V_{\bar{1}}^{*})
= \text{Sym} V_{\bar{0}}^{*} \otimes \bigwedge^{\bullet} V_{\bar{1}}.$$
Therefore, locally a graded manifold $\mathcal M$ looks like $(U, C^{\infty}(U) \otimes \text{Sym} V_{\bar{0}}^{*} \otimes \bigwedge^{\bullet} V_{\bar{1}})$.
Compared to $(U, C^\infty(U)\otimes \wedge^\bullet V^*)$ in definition 4.4.1, I got an extra term $\text{Sym} V_{\bar{0}}^{*}$.
The only way to eliminate this term is to require $V_{\bar{0}} = 0$, i.e. $V = V_{\bar{1}} = \bigoplus_{k \in 2 \mathbb Z+1} V_k$ is made of odd components.
However, in definition of graded manifold, there is no such requirement on the graded vector space $V$, so Definition 4.6.1 and Remark 4.6.1 are not consistent, and this is where I got puzzled.
A way to fix this might be like this, use a slightly different definition of $\mathbb Z$-supermanifold:
For example, in Mnev's paper, definition 4.22 and the following, it requires the open set $U$ belongs to a graded vector space $W$(the target of even characters, as the $V_\bar{1}$ we defined here), and $V$ the odd fiber.
Thanks for your time and effort.
I think it is a typo: the second local model has infinite real dimensional fibers over every point of $M$, while the first doesn't. Maybe the vector space in the second example is odd?
@BenMcKay I'm sorry, I missed the condition that $V$ in the second definition is a graded vector space.
If you set $V_{\bar{0}} = 0$ and $V_{\bar{1}} = V$, then a supermanifold is straight forwardly a graded manifold, as basically you've already shown. A graded manifold with bosonic directions is not a supermanifold, so there cannot be a functor going in the other direction. What do you think is missing?
@IgorKhavkine so you mean the graded vector space $V$ in definition of $\mathbb Z$-graded manifold is required to be odd?
The odd components are odd.
@Andrews could you perhaps make your question more precise? As is, the only way I can answer it is by repeating what I already wrote. Please voice what you think is missing.
In the updated question you write "The only way to eliminate this term is to require $V_{\bar{0}} = 0$ [...]" That is correct. A supermanifold is a graded manifold. But not every graded manifold is a supermanifold, as you've noticed. The Remark is talking about an isomorphism between the structure sheaves of any one supermanifold and a corresponding graded manifold. It cannot be referring to an isomorphism between the two categories, because such an isomorphism does not exist.
@IgorKhavkine Thank you for your kindly explanation. Would you mind to make this into an answer? So I can accept the answer and this question will be marked as "answered".
A supermanifold is a graded manifold with only odd components, more precisely the equivalence is seen by setting $V_{\bar{0}}=0$ and $V_{\bar{1}} = V$ in your notation. In the opposite direction, a graded manifold is a supermanifold only if it has no even components, that is $V_{\bar{0}} = 0$.
|
2025-03-21T14:48:30.945114
| 2020-05-11T13:03:24 |
360035
|
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|
Stack Exchange
|
Sendov's conjecture
It has been more than fifty years for famous Sendov's conjecture which states that if $p(z)$ is a polynomial of degree $n$ having all its zeros in the unit disc $|z|\leq 1$ then each of the n roots is at a distance no more than 1 from at least one critical point. It has also been more than twenty years for its proof for polynomials of degree $n=8$. I believe, after that no progress on greater values of $n.$ Some claims are there on arxiv platform, which are not yet validated by the experts I feel. There is a review article http://www.indianmathsociety.org.in/mathstudent-part-1-2019.pdf#page=101
It is very sad and unfortunate to hear that Sendov passed away on January 19, 2020 and a nice tribute article is published in Journal of Approximation Theory, see https://www.sciencedirect.com/science/article/pii/S0021904520300423?via%3Dihub
There is another paper on the conjecture for high degree of polynomials, see https://www.ams.org/journals/proc/2014-142-04/S0002-9939-2014-11888-0/home.html
A new approach of solving this problem using Grace Theorem by constructing apolar polynomials is given in the article 'A remark on Sendov Conjecture' published in Comptes rendus de l’Acade'mie bulgare des Sciences, Vol 71 (6), 2018, pp.731-734.
Let me know if anything more significant latest contributions?
https://arxiv.org/abs/2012.04125
For a contribution with simple tools as continued fraction see https://math.stackexchange.com/questions/4791581/sendovs-conjecture-and-squaring-polynomial
A 2010 status report is given by D. Khavinson et al. in Borcea's variance conjectures on the critical points of polynomials. A 2019 update is in A note on a recent attempt to prove Sendov's conjecture, by N.A. Rather and Suhail Gulzar.
Concerning the contents of the status report, I have placed a question here:https://mathoverflow.net/q/402980/37855, which contains more recent info.
|
2025-03-21T14:48:30.945268
| 2020-05-11T13:14:46 |
360037
|
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"Pol van Hoften",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360037"
}
|
Stack Exchange
|
The category of finite flat group schemes
Let $\varphi: G \to G^\prime$ be a homomorphism of finite flat group schemes over a locally noetherian base scheme $S$, and let $h_\varphi: h_G \to h_{G^\prime}$ be the induced homomorphism of sheaves of abelian groups on $(\mathfrak{Sch}/S)_{\mathrm{fppf}}$. I wonder what are the equivalent conditions for $h_{\varphi}$ to be injective (surjective).
More specifically, are the following equivalent:
$\varphi$ is a closed immersion;
$h_\varphi$ is injective.
And are the following equivalent:
$\varphi$ is faithfully flat;
$h_\varphi$ is surjective.
Furthermore, is the category of finite flat group schemes over $S$ an abelian category?
The first is true by (https://stacks.math.columbia.edu/tag/04XV) which tells you that closed immersions are the same thing as proper monomorphisms. For your last question, see https://mathoverflow.net/questions/7688/the-category-of-finite-locally-free-commutative-group-schemes
Finite flat group schemes cannot be an abelian category because finite groups is not an abelian category. Finite flat commutative group schemes on the other hand has a chance of being abelian, and in the post cited by @user45878 it is explained that this is true when $S$ is the spectrum of a field and false for some other bases (e.g. a mixed characteristic DVR of large ramification index).
|
2025-03-21T14:48:30.945638
| 2020-05-11T13:22:25 |
360038
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360038"
}
|
Stack Exchange
|
On $s$-harmonic functions
Is this statement true?
A bounded positive function $v\in C^{2}(\mathbb R^N)$ decaying to zero which is $s$-harmonic function ($s\in (0, 1)$) outside a ball behaves like $|x|^{2s-N}$ near infinity. That is, if $N>2s, $ then $|x|^{N-2s} v(x) \to l$ whenever $|x|\to \infty$ for some $l>0.$
Yes, it is. Although there are more elementary ways to prove this fact, I suppose the following argument is the shortest.
Every positive $s$-harmonic function in an arbitrary open set $D$ can be represented in terms of the Poisson kernel $P_D(x, y)$ (with $y \in \mathbb{R}^N \setminus D_M$) and the Martin kernel $M_D(x, z)$ (with $z \in D_M \setminus D$). Here $D_M \subseteq \overline{D}$ is the Martin compactification of $D$.
For an exterior domain, these kernels decay as $|x|^{2s-N}$ as $|x| \to \infty$, except the Martin kernel $M(x, \infty)$ with pole at infinity, which is asymptotically constant (at least if $2s < N$; this is true for any exterior domain, and for the complement of a ball we even have explicit expressions at hand). If your function decays at infinity, the contribution of $M(x, \infty)$ is necessarily zero, and hence it is asymptotically equal to $|x|^{2s-N}$.
For details, see my article with Krzysztof Bogdan and Tadeusz Kulczycki:
K. Bogdan, T. Kulczycki, M. Kwaśnicki, Estimates and structure of -harmonic functions, Prob. Theory Rel. Fields 140(3–4) (2008): 345–381, DOI:10.1007/s00440-007-0067-0
The article is unfortunately not very reader-friendly; if you have any questions, feel free to ask.
|
2025-03-21T14:48:30.945884
| 2020-05-11T13:44:52 |
360040
|
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|
Stack Exchange
|
Extension of Dedekind domains and their codifferent
Let $A\subset B$ be a finite extension of Dedekind domains. Let $0\neq b\in B$ and $0\neq a\in A$ such that $(a)=(b)\cap A$. In particular, we have $a=b\cdot c$ for some $c\in B$. Now for any $A$-linear map $\varphi:B\to A$ the map $x\mapsto\varphi(c\cdot x)$ maps the ideal $(b)$ into $(a)$. Thus we obtain an $A$-linear map $\overline{\varphi}:B/(b)\to A/(a)$. This defines a homomorphism of $A$-modules (even $B$-modules):$$\textrm{Hom}_A(B,A)\to\textrm{Hom}_A(B/(b),A/(a)),\, \varphi\mapsto \overline{\varphi}.$$
I have checked several examples and this map was always surjective. Is this always the case?
Moreover, if we relax the condition of $A$ and $B$ being Dedekind domains to just being some integral regular domains, can we still say something on the image of this map? Can we describe it just in terms of the finite ring extension $A/(a)\hookrightarrow B/(b)$?
Any $A$-module homomorphism $B/b\to A/a$ gives, by composing with $B\to B/b$, a homomorphism $B\to A/a$. Using the surjection $A\to A/a$ and the fact that $B$ is $A$-projective, the map lifts to give a homomorphism $B\to A$.
So, the only real assumption needed is that $B$ is $A$-projective.
Thanks! So this shows that any homomorphism $B/b\to A/a$ comes from a homomorphismm $B\to A$ which maps $(b)$ to $(a)$. Is it clear that every such can be obtained by multiplication with $c$ as above?
okay, I got it!
|
2025-03-21T14:48:30.946016
| 2020-05-11T14:24:52 |
360043
|
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|
Stack Exchange
|
Distribution boundary value of analytic function and wave front sets
Assume $f(z)$ is analytic in the tube domain $\mathbb R^n\oplus iC$, where $C\subset \mathbb R^n$ is a convex cone. Under the assumption $|f(x+iy)|\leq 1/|y|^k$, we know by a Theorem of Martineau (see also Hormander, volume 1, Theorem 3.1.15) that the limit $\lim_{y\to 0, y\in C} f(x+iy)$ exists as a tempered distribution $f(x)$ on $\mathbb R^n$, uniformly in proper cones $y\in C'\subset C$. The convergence is in the weak topology, and in fact in the strong topology on the space of tempered distributions of fixed order $k$.
Question: Let $\Gamma\subset T^*\mathbb R^n$ be the wave front set of $f(x)$. Is it true that $f(x+iy)\to f(x)$ also in the Hormander topology $C^{-\infty}_\Gamma$? If not true in general, can some conditions be given on $f$ that would ensure such convergence?
You need to check Condition (ii) in Definition 8.2.2 in the first volume of Hörmander's ALPDO. Let us note $f(x+i0)$ the limit-distribution of your question and let $\Gamma$ be its wave-front-set. Let $K\times V$ be a compact-conic subset of the phase space with empty intersection with $\Gamma$, let $\phi$ a smooth compactly supported function of $x$, supported in $K$. We must take a look at
$$
\mathcal F\bigl(f(x+iy)\phi(x)-f(x+i0)\phi(x)\bigr)(\xi)\quad \text{for $\xi\in V$}.
$$
With $T_y$ standing for the Fourier transform of $f(x+iy)-f(x+i0)$, we define for $\alpha\in \mathbb N^n$,
$$
J_\alpha(\xi,y)= \xi^\alpha\int T_y(\eta) \hat \phi(\xi-\eta) d\eta=
\int T_y(\eta)(\xi-\eta+\eta)^\alpha \hat \phi(\xi-\eta) d\eta.
$$
We define for $\beta, \gamma\in \mathbb N^n$,
$$
J_{\beta,\gamma}(\xi,y)
=\int T_y(\eta)\eta^\beta(\xi-\eta)^\gamma \hat \phi(\xi-\eta) d\eta
=\int T_y(\eta)\eta^\beta\ \widehat{D^\gamma \phi}(\xi-\eta) d\eta,
$$
and we have for $1=\chi_0+\chi_1$, $\chi_1$ supported near $\Gamma$, $\chi_0$ supported in $V$,
$$
J_{\beta,\gamma}(\xi,y)=\int\chi_{0}(\eta)
T_y(\eta)\eta^\beta\ \hat{\phi}_\gamma(\xi-\eta) d\eta
+
\int \chi_1(\eta)T_y(\eta)\eta^\beta\ \hat{\phi}_\gamma(\xi-\eta) d\eta.
$$
Let us verify the bounds and note that $\phi_\gamma$ is in the Schwartz space.
The first integral ($\eta$ is there a fast-decreasing direction for $T_y(\eta)$ since we are away from the bad directions of the wave-front-set) is uniformly rapidly decreasing in $\xi$:
we have
$$
\vert\int\chi_{0}(\eta)
T_y(\eta)\eta^\beta\ \hat{\phi}_\gamma(\xi-\eta) d\eta\vert
\lesssim
\int\chi_{0}(\eta)
(1+\vert\eta\vert)^{-N} (1+\vert\xi-\eta\vert)^{-N} d\eta\lesssim
(1+\vert\xi\vert)^{-N+n+1}
$$
We check the second integral for $\xi \in V$: there we have from the empty-intersection above
$$
1+\vert\xi-\eta\vert\gtrsim 1+\vert \xi\vert +\vert \eta\vert
$$
and this gives the uniform fast decay since $T_y(\eta)$ is bounded above by $(1+\vert\eta\vert)^{N_0}$:
we have
$$
\vert\int \chi_1(\eta)T_y(\eta)\eta^\beta\ \hat{\phi}_\gamma(\xi-\eta) d\eta\vert\lesssim
\int (1+\vert\eta\vert)^{N_0+\vert \beta\vert}(1+\vert\xi-\eta\vert)^{-N} d\eta\lesssim
\int (1+\vert\eta\vert)^{N_0+\vert \beta\vert}(1+\vert\xi\vert+\eta\vert)^{-N} d\eta
\lesssim
(1+\vert\xi\vert)^{-N+N_0+\vert \beta\vert+n+1}.
$$
To complete the proof it is needed to check the convergence to 0 with $y$ of the above bounds.
Thanks for the answer! I don't quite understand how to think about the first summand, the one with $\chi_0$. We have the Schwartz distribution $T_y(\eta)$, which is paired with the Schwartz function $\chi_0(\eta)\hat{\phi_\gamma}(\xi-\eta)$. I don't see how to use the knowledge $\eta\in V$ to be able to write $T_y(\eta)\leq C(1+|\eta|)^{-N}$, let alone have $C$ independent of $y$. In fact, everything you write applies to any weakly converging sequence, which need not also be in Hormander topology; surely somehow the particular case of holomorphic boundary value should be used?
@Dima I should have written this as a claim: for the first integral, since $\eta\in V$, thus away from the bad directions for $f(x+i0)$ you have fast decay. For instance, if you take a look at the one-dimensional homogeneous $(x+i0)^a$ with wave-front-set ${0}\times\mathbb R_+^*$, then I think that a direct calculation of the Fourier transform of $(x+iy)^a$ shows a uniform fast decay for $y\ge 0$ for $\xi$ negative. I believe indeed that this is the place where the assumption is used.
Do you think Hormander convergence $f(x+iy)\to f(x+i0)$ holds in general? Computing explicitly is only possible in very few cases.
Yes, I think so, at least in the framework of Theorem 3.1.11 in the Hörmander's ALPDO. I guess that it is indeed possible to have an explicit expression for the Fourier transform using Formula (3.1.13) there; also the homogeneous example $(x+i0)^a$ has a Fourier transform $\hat T$ supported in $\mathbb R_+$ and the Fourier transform of $(x+iy)^a$ is $e^{-y\xi} \hat T(\xi)$, so it vanishes as well on $\xi<0$. To go back to that Formula (3.1.13), it is quite likely that there is an analogous device in several dimensions.
One can use the argument in Hormander's theorem 8.1.6 to show there is convergence in $C^{-\infty}_{C^o}$, where $C^o$ is the dual cone - that's what happens in the example you consider. However it could happen that $\Gamma$ is strictly smaller than $C^o$, and then I don't see any way to approach the convergence question in the general case.
Then it is true in one dimension, since you have only two directions for the variable $\xi$ in the wave-front-set and one direction is ruled out by the inclusion of Theorem 8.1.6.
Yes, that's true. But in higher dimension it can happen that $\Gamma$ is smaller than $C^o$, and then it's not clear how to proceed.
|
2025-03-21T14:48:30.946353
| 2020-05-11T16:19:32 |
360050
|
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"F.Abellan",
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|
Stack Exchange
|
The $\infty$-category of natural transformations as an end
Let $\mathcal{C}$ be an $\infty$-category viewed as a fibrant scaled simplicial set with all 2-simplices thin and let $\mathfrak{C}\!at_{\infty}$ be the $\infty$-bicategory of $\infty$-categories. A model for $\mathfrak{C}\!at_{\infty}$ is given by applying the marked version of the homotopy coherent nerve to $\operatorname{Set}_{\Delta}^{+}$.
We define the $\infty$-bicategory of functors $\operatorname{Fun}(\mathcal{C},\mathfrak{C}\!at_{\infty})$ as in the $(\infty,1)$-case. Thin 2-simplices are given by maps of simplicial sets $\mathcal{C}\times \Delta^2_{\sharp} \to \mathfrak{C}\!at_{\infty}$ where $\Delta^2_{\sharp} $ denotes the maximally scaled 2-simplex.
Given two functors $F,G: \mathcal{C} \to \mathfrak{C}\!at_{\infty}$ we can use Proposition 2.4 in here to compute a model for $\operatorname{Nat}(F,G)$ in terms of a suitably lax slice construction.
I would like to show that $\operatorname{Nat}(F,G)\cong \lim_\limits{\operatorname{Tw}(C)}\operatorname{Fun}(F(-),G(-))$, which should obviously be true but I am having trouble constructing the universal cone for this functor.
The version of this statement for functors between two $(\infty,1)$-categories can be found as proposition 2.3 of https://arxiv.org/abs/1408.3065, or in section 5 of https://arxiv.org/abs/1501.02161 , this is not exactly what you are asking, but maybe this is close enough ?
@SimonHenry I am aware of those two references. I want to relate Example 6.8 in the second paper your mentioned with the model of Nat(F,G) above.
Let $\mathcal{C}$ be an $\infty$-category and $\mathbb{D}$ be an $\infty$-bicategory. In joint work with W. Stern (Enhaced twisted arrow categories ) we show that there is a natural equivalence of $\infty$-categories
$$\operatorname{Nat}(F,G) \simeq \lim_{\operatorname{Tw}(\mathcal{C})^{\operatorname{op}}} \operatorname{Map}_{\mathbb{D}}(F(-),G(-))$$
The "op" comes from the fact that we are using right fibration $\operatorname{Tw}(\mathcal{C}) \to \mathcal{C} \times \mathcal{C}^{\operatorname{op}}$.
|
2025-03-21T14:48:30.946515
| 2020-05-11T16:58:03 |
360051
|
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|
Stack Exchange
|
Conditions for optimal stationary strategies in MDPs
I have a specific Markov decision process (MDP) which is generated from a problem in another domain. What I would like to show is that under the limit of means criterion (no discounting) the optimal strategy depends only on the state and not the history of actions. The problem is the literature on MDPs is very large and each proof of optimal stationary strategies make different assumptions about the structure of the chain. Since my chain is generated from a problem outside of this domain it doesn't exactly fit any of the sets of assumptions I have seen so far. So I am looking for the weakest set of assumptions necessary for proving the existence of an optimal stationary strategy. Also, a resource for conditions for stationarity that doesn't require extensive knowledge of MDPs would be greatly appreciated.
I will now detail both the problem which generates my MDP and the MDP itself.
The problem is as follows:
An agent is playing a game with the following steps of play
An agent draws a card from a deck.
The reward for each possible action is determined by the card the agent draws. The action space is some bounded continuous interval i.e. ($[0,M]$ for some constant $M$). The reward is also from some bounded continuous interval i.e. ($[0,D]$ for some constant $D$). The reward is monotone increasing in the action.
Based on the action the agent plays he is forced to sit out a (potentially random) number of rounds
Card is returned to the deck, the agent draws another card (Back to step 1.)
Some additional facts there can be an uncountably infinite number of cards, the set of actions and punishment for each action is the same regardless of the card drawn , all that changes is the value of the reward for each action. Now I want to show that the agents action depends only on their card.
This can be modeled as the following MDP. There are an uncountable number of states (corresponding to drawing a card) each with an uncountable number of possible actions. There is also a finite length chain of states with a single action that transitions to the next element in the chain (corresponding to waiting), the last element transitions randomly to one of the uncountable states . When in one of states with many actions an agent picks an action and gets placed in the appropriate spot in the single action chain to induce the waiting period until he is returned randomly to a many action state.
Intuitively one can think of this as an infinitely wide chain connected to a path which leads back to the infinitely wide chain.
Things I have tried:
The biggest roadblock is the uncountable action and state space which is usually assumed finite or at least countable. However, the reason this assumption is needed (as far as I am aware) is to prevent infinitely long chains with certain undesirable properties. I feel like my infinite width chain should satisfy some different property that makes up for this.
At some point I came across a sufficient condition that only needed a specific state to be visited infinitely often. This is very close to being satisfied in my chain except for a degenerate strategy which plays an action which never gets any reward or punishment regardless of which card they draw. This causes the chain to always be in the infinitely wide section and hence never revisit the same state infinitely often.
Could you add a bit more structure? It is not even clear whether the rewards are bounded.
As Michael suggests, in this generality there need not be a stationary optimal strategy, nor, in fact, an optimal strategy. If the set of actions is not compact (or the payoff function is not continuous over the set of actions), then even if there is only one card in the deck and no punishment at all, there need not be an optimal action, since there might not be an action that yields the supremum of the payoffs.
Thank you both I missed this subtlety. Indeed there is nice structure on the reward. Here is the relevant portion I added to the main text. The reward for each possible action is determined by the card the agent draws. The action space is some bounded continous interval i.e. ($[0,M]$ for some constant $M$). The reward is also from some bounded continuous interval i.e. ($[0,D]$ for some constant $D$). The reward is monotone increasing in the action.
I think your problem is that you have not specified your model. In your verbal formulation it does not seem even to be a Markov decision model. My proposal is the following formulation: Let
$S := \mathbb{N}_0$ be the state space,
$A := [0,M]$ be the action space, endowed with Borel-$\sigma$-algebra $\cal{A}$, $q \colon S \times A \to P(S)$ be a Markov kernel with $P(S)$ be the space of all probabability measures on $S$ with the property that $p(s,a) = \delta_{s-1}$ for $s > 0$. The reward is a bounded (measurable) function $r \colon S \times A \to \mathbb{R}$ (you have punishment) with $r(s,a) \equiv 0$ for all $s > 0$. Then you are in the usual framework of MDP's. The only problem here is that $A$ is not countable. I think this is artificial generality.
Further you always return infinitely often to the state $s = 0$, so your second remark does not apply. Have a look into the book "Controlled Markov Processes" from E.B. Dynkin, A.A. Yushkevich, Springer (1979), ch. 7 or "Markov Decision Processes" of M.L. Puterman, Wiley & Sons (1994), ch.8.
Edit: As usual the problem with such models is: Describes it correctly the original problem? This usually can only be solved together with the user. If my formulation is correct it can be easily solved:
Let $X_a$ be any random variable with $\mathbb{P}(X_a = s) = q(0,a)(\{s\}), s \in S$ for $a \in A$. ($X_a+1$ is the random time until we reach $s = 0$ again, if we choose action $a \in A$). If there is any $a_0 \in A$ with $$\frac{r(0,a)}{\mathbb{E}(X_a+1)} \leq \frac{r(0,a_0)}{\mathbb{E}(X_{a_0}+1)}$$ for any $a \in A$, then it is optimal to choose always action $a_0$ if we are in state $s = 0$. We tentatively assume $\mathbb{E}X_a < \infty$ for all $a \in A$. Of course $a_0$ exists if $A$ is finite. (States $s \not= 0$ here are irrelevant.)
Your formulation is very close to correct. There is one problem I am seeing in your formulation you have $s=0$ as a single state. In reality there are many $s_0$ (corresponding to cards in the deck) that alter the reward function. Trying to use your formulation here, there is an uncountable set of actionable states which i will call $S_0$ and a distribution $D$ over this set. Now $p(s,a) = \delta_{s-1}$ for $s \notin S_0 \cup s_1$ and $p(s_1,a) = D$. The reward $r(s,a) = 0$ for all $s \notin S_0$. Now there is no state you return to infinitely often however you return to the set $S_0$.
If I understand you right (original descrition of the problem), $A$ and not $S$ should be interpreted as the deck of cards. Then an action $a \in A$ is just the card chosen. Since you put the chosen card back this simple interpretation is possible. Unfortunately in the original problem it is not clear what is action and what is state.
The card drawn is the state not the action. The cards are always drawn from the same distribution but you do not get to pick a specific card. Based on which card is drawn you have access to a different set of actions.
@TPaul Sorry, without more information about the "game" I do not know which model is appropriate. The essentially point is (and without further information only you can decide): Before (!) you draw a card is there any information which changes from draw to draw. For model building it is irrelevant what you call state and what action. These are words we use for convenience. We can make $A$ as complicated as we will, of course with the possible consequence that nothing can be efficiently computed. In your case $A$ may even be a set of functions or probability measures.
Thanks for all of your help. To answer your question there is no information which changes from draw to draw before you draw. To elaborate further, imagine every action results in a number of tokens being recieved in exchange for being placed in a certain place in the unactionable chain. Now imagine every card has a number which represents how much you value tokens in that round. So your action will depend on which card is drawn but the deck of cards does not change over time.
|
2025-03-21T14:48:30.947056
| 2020-05-11T18:07:44 |
360054
|
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|
Stack Exchange
|
Concentration (or two sided tail bounds around expectations) of maximum and minimum of $n$ iid, subgaussian random variables
I asked this on MSE, but got no answer, hence asking here now. Help appreciated!
My question is motivated by this question and this question, where the first was aimed for giving a one sided tail bound for maximum of subgaussians, and the second one was for two sided tail bounds for gaussians. I'm also motivated by questions like this one.
Let $\{X_1 \dots X_n\}$ be $n$ iid, subgaussian random variables so that $P[|X_i| \ge t] \le 2 exp (- \frac{ct^2}{ ||X_1||_{\psi_2}^2 }) \forall i, ||*||_{\psi_2}$ denoting the Orlicz norm. I'm looking for concentration inequalities for :
$$ X_{max} := max_{1 \le i \le n} X_i, \hspace{1mm} X_{min} := min_{1 \le i \le n} X_i $$
So to be more precise, I'm looking for tail bound functions $\alpha(t), \beta(t)$ of the form:
$$ P[ | X_{max} - \mathbb{E}X_{max} | \ge t ] \le \alpha(t), P[ | X_{min} - \mathbb{E}X_{min} | \ge t ] \le \beta(t) $$
where the following are decreasing functions of $t$:
$$ 0 \le \alpha(t), \beta(t) \to 0, t \to \infty.$$
Out of curiosity, I've been asking myself: why do I keep seeing various results on tail probabilities of only maximum of a set of random variables, but not minimum of them?
If you don't take absolute values, the max and min of symmetric random variables (such as Gaussians) have the same distribution, up to sign.
Regarding concentration of the max of Gaussians: See Theorem 5.2.2 in Vershynin's High Dimensional Probability book. Maximum is a Lipschitz function.
I think the sub-gaussianity condition by itself is inadequate for the bounds you want. Indeed, this condition implies upper bounds on the tails of $X_{\max}$ and hence on $EX_{\max}$, but leaves the difference $X_{\max}-EX_{\max}$ without proper control.
@IosifPinelis thanks for your comment! I think I still need to do some thinking to convince myself, but the reason I was thinking of such a two-sided tail estimate was I was hoping for as we increase $n, X_{max}$ should move further away from $0,$ and since each $X_i$ has a tail probability estimate around $EX_i,$ I was hoping there has to be some kind of "pivot" around which $X_{max}$ should concentrate, and I was hoping this "pivot" should be $EX_{max},$ but from what you wrote, we can't control $|X_{max} - EX_{max}|.$ I was hoping for a lower bound on $EX_{max}$ too (contd.)
@IosifPinelis (contd.) ...as intuitively, with large $n, EX_{max}$ should be far from $EX_1 $($X_i'$ s are iid). But then I wonder: is $X_{max}$ even a subgaussian? I guess I'll do some calculation of the tail estimate to get something like $P[ | X_{max} | \ge t] \le g(t),$ though I'd be more happy with something like $P[ | X_{max} -c | \ge t] \le g(t)...$ (but maybe it's impossible...)
@AryehKontorovich Thanks for the pointer to the theorem in Vershynin's book: yes max is a Lipschitz function, with the Lipschitz constant depending on $n,$ the number of $X_i$'s considered (or the dimension). But the theorem should still apply - I'll do the necessary calculation.
@IosifPinelis My mistake -- I misread the OP as "gaussian" rather than "subgaussian".
|
2025-03-21T14:48:30.947305
| 2020-05-11T18:16:41 |
360058
|
{
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"authors": [
"Nick Gill",
"Stefan Kohl",
"https://mathoverflow.net/users/28104",
"https://mathoverflow.net/users/801"
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"url": "https://mathoverflow.net/questions/360058"
}
|
Stack Exchange
|
Are there perfect DTI-groups which are not simple?
Recall that a subgroup $H$ of a group $G$ is called a TI-subgroup
if for every $g \in G$ it is $H \cap H^g \in \{1,H\}$.
We say that a group $G$ is a DTI-group if the derived subgroups of all
of its subgroups are TI-subgroups.
Question: Is it true that a non-trivial finite perfect DTI-group
is necessarily simple?
Remark: I know that -- provided that it exists -- a non-trivial finite
non-simple perfect DTI-group necessarily is minimal non-solvable.
Related question: References on a certain generalization of Dedekind groups
I was wondering whether the non-split version of $2^3.GL_3(2)$ might be a potential candidate. do you have an easy way to test this?
@NickGill Running a check over the perfect groups of order less than about 50000 with GAP did not reveal a non-simple perfect DTI-group. -- So any counterexample (if it exists!) must be bigger.
|
2025-03-21T14:48:30.947396
| 2020-05-11T18:24:07 |
360060
|
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|
Stack Exchange
|
Existence of nontrivial transfinite divisibility in $R$-modules
Let $R$ be a unital, possibly noncommutative ring and $s \in R$. For a right $R$-module $M$, define $Ms = \{ms \mid m \in M\}$; this is an additive subgroup of $M$, which is a module over the centralizer of $s$. For an ordinal $\alpha$, define $Ms^\alpha$ by transfinite induction: $Ms^0 = M$, $Ms^{\alpha+1} = (Ms^\alpha)s$, and take intersections at limit ordinals. Then the the groups $Ms^\alpha$ must stabilize eventually; say that the $s$-divisibility rank of $M$ is the minimal $\alpha$ such that $Ms^\alpha = Ms^{\alpha+1}$. Consider the following conditions:
For some $s \in R$, the $s$-divisibility rank of $R \in Mod_R$ is infinite (i.e. $R$ is not strongly $\pi$-regular).
For some $s \in R$ and some $M \in Mod_R$, the $s$-divisibility rank of $M$ is infinite.
For some $s \in R$, there exist right $R$-modules of arbitrarily large $s$-divisibility rank.
Obviously, $3 \Rightarrow 2 \Leftrightarrow 1$. I suspect that 1,2,3 are equivalent, and I think I can show this when $R$ is commutative, via a sort of "generalized Prufer module" construction. But I'm not sure when $R$ is noncommutative.
Question 1: Does $1 \Rightarrow 3$ when $R$ is noncommutative?
There is another condition which these appear to be related to. Let $m \in M$ and $n \in N$ be two elements of right $R$-modules. Say that $m \in M, n \in N$ are weakly equivalent if there exist maps of right $R$-modules $f: M \to N$ and $g: N \to M$ such that $f(m) = n$ and $g(n) = m$. This is an equivalence relation on elements of right $R$-modules. Consider the condition:
There exists a proper class of elements of right $R$-modules, no two of which are weakly equivalent.
Because $f(Ms^\alpha) \subseteq N s^\alpha$ for every map of right $R$-modules $f: M \to N$, we have that $3 \Rightarrow 4$.
Question 2: Does $4 \Rightarrow 3$?
The related question of whether $4 \Rightarrow 1$, can be stated contrapositively as: if $R$ is strongly $\pi$-regular, then does $R$ have only a set of weak equivalence classes of elements of $R$-modules?
|
2025-03-21T14:48:30.947556
| 2020-05-11T18:45:01 |
360061
|
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"url": "https://mathoverflow.net/questions/360061"
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|
Stack Exchange
|
very ample line bundle on singular curve
Let $X$ be a singular reduced irreducible projective curve over $k$(algebraically closed).
How to show that if $X$ has arithmetic genus 1,then for any smooth (closed) point $p\in X$,$\mathcal{ O}(3p)$ is very ample?
(If $X$ is nonsingular,it is well-known.)
You may refer to this article: https://www.sciencedirect.com/science/article/pii/S0022404903001014
|
2025-03-21T14:48:30.947618
| 2020-05-11T18:46:56 |
360062
|
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"JoshuaZ",
"Joël",
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"url": "https://mathoverflow.net/questions/360062"
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|
Stack Exchange
|
p2 - p1 = 2n for every 2n
Quote:
"Chen's work mentioned in the discussion of the Goldbach conjecture also showed that every even number is the difference between a prime and a P2."
from: link
However I can't get this verified or find different sources stating the same.
so question is: is this true? where can i find proof?
thanks.
I am not sure why there ate three votes to close. Someone cares to explain?
@Joël My guess is that the perception was that this is a simple enough question that a few minutes of research could have found an answer. In this case, I double checked my own answer by typing in "Chen's theorem" into Wikipedia.
The two relevant papers for Chen's original proof are Chen, J.R. (1973). "On the representation of a larger even integer as the sum of a prime and the product of at most two primes". Sci. Sinica. 16: 157–176. and Chen, J.R. (1966). "On the representation of a large even integer as the sum of a prime and the product of at most two primes". Kexue Tongbao. 11 (9): 385–386. See here for more details along with related strengthened results.
|
2025-03-21T14:48:30.947873
| 2020-05-11T18:59:49 |
360063
|
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"authors": [
"Gerhard Paseman",
"LSpice",
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|
Stack Exchange
|
Slicing up $\mathbb{N}\setminus\{1\}$
Let $\mathbb{N}$ denote the set of positive integers. For any prime $p\in\mathbb{N}$ let $p\mathbb{N} = \{np: n\in \mathbb{N}\}$. Is there a partition ${\cal P}$ of $\mathbb{N}\setminus\{1\}$ such that for all $B \in {\cal P}$ and every prime $p\in\mathbb{N}$ we have $|B \cap p\mathbb{N}|=1$?
Oops, I missed the every prime condition earlier. Start assembling sets of coprime integers. If you can't add a number to one of the existing sets, start a new set. In particular, each set will have at most one power of a given prime. Gerhard "Thinks Jumping Primes Are Cooler" Paseman, 2020.05.11.
@GerhardPaseman, if each member of $B$ contains only one element, then, for each $B$, $\lvert B \cap p\mathbb N\rvert$ will equal $0$ for most primes $p$. On edit: it seems that your revised construction is just what I described, isn't it?
Yes, I misread, and so seriously edited my comment above. It may be possible to give a slicker proof than what you have posted as an answer, but I'm not seeing it. Gerhard "More Prime Powers To You" Paseman, 2020.05.11.
Recursively define a sequence of $B$'s as follows. Initially, each is empty. At each step $n > 1$, place $n$ in the first $B$ that contains only elements coprime to $n$. Clearly, for each prime $p$, there is no $B$ that contains two distinct multiples of $p$. Now fix a prime $p$ and a natural number $N > 1$, and consider the first $B$ that contains no multiple of $p$ after step $N$ has completed. The first power $p^k$ of $p$ that is larger than $N$ cannot be placed in any earlier $B$ (since all have a multiple of $p$), so it will be placed in $B$ if no multiple $p d$ of $p$ with $N < p d < p^k$ has been.
As @StevenStadnicki points out, it's interesting to investigate the structure of these sets. (I started to do it by hand, and found it sort of addictive.) Here's some Haskell code to allocate the first $N$ numbers (doubtless both inefficient and unidiomatic, but it seems to work):
insert N [] = [(N, [N])]
insert N ((c,bs):bss) = if gcd c N == 1 then (N*c,N:bs):bss else (c,bs):(insert N bss)
insertTo 1 = []
insertTo N = insert N $ insertTo (N - 1)
One runs it as
map snd $ insertTo 1000
(for example), whose output starts
[[997,991,983,977,971,967,953,947,941,937,929,919,911,907,887,883,881,877,863,859,857,853,839,829,827,823,821,811,809,797,787,773,769,761,757,751,743,739,733,727,719,709,701,691,683,677,673,661,659,653,647,643,641,631,619,617,613,607,601,599,593,587,577,571,569,563,557,547,541,523,521,509,503,499,491,487,479,467,463,461,457,449,443,439,433,431,421,419,409,401,397,389,383,379,373,367,359,353,349,347,337,331,317,313,311,307,293,283,281,277,271,269,263,257,251,241,239,233,229,227,223,211,199,197,193,191,181,179,173,167,163,157,151,149,139,137,131,127,113,109,107,103,101,97,89,83,79,73,71,67,61,59,53,47,43,41,37,31,29,23,19,17,13,11,7,5,3,2],
[961,841,529,361,289,169,121,49,25,9,4],
[667,323,143,35,6],[899,437,221,77,15,8],
[713,247,187,21,10],[551,391,91,55,12],
[851,493,209,65,27,14],[299,133,85,33,16],
[377,253,119,95,18],[703,527,319,161,39,20],
[989,779,629,403,203,45,22],
[893,731,533,407,217,115,24],
[943,817,341,259,125,51,26],
[799,481,451,145,57,28],
[901,611,589,473,287,30] …
Running through this construction by hand, $B_1$ is (of course) the primes; $B_2$ appears to be the squares of the primes (and I suspect a proof would be pretty straightforward), and of course $2n\in B_n$. The structure of the $B$s after that point appears to get much more complicated, although for instance $m=12n+3\in B_{(m+1)/2}$ and $m=12n+9\in B_{(m-1)/2}$...
Indeed, I found it sort of fun to do it by hand, although it gets boring allocating each even number its own set. I'll put the Haskell code I used to generate examples in the post.
Place each $n$ that is not a prime power into its own $B=B(n)$. Then fill the rest of $B(n)$ with powers of primes that do not divide $n$.
|
2025-03-21T14:48:30.948150
| 2020-05-11T19:13:33 |
360066
|
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"Anton Petrunin",
"Ian Agol",
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|
Stack Exchange
|
Which convex bodies can be captured in a knot?
Which convex bodies can be captured in a knot?
This question is based on the discussion in "Is it possible to capture a sphere in a knot?".
We assume that the knot is made from an unstretchable, infinitely thin rope.
Comments:
By the construction of Anton Geraschenko, the question is equivalent to the existence of a graph embedded in the surface of the convex body that locally minimizes the total length. Such an embedding exists on some convex bodies; for example, on an equilateral triangle shown in the diagram (and on anything sufficiently close).
According to the original question a ball cannot be captured (in fact it cannot be captured in a link with 3 components). Moreover, a circular disk cannot be captured see my answer (thanks to Wlodek Kuperberg for asking). Likely, the same idea works for all convex bodies of revolution. Maybe all convex bodies of general position can be captured.
Can a circular disk be captured? Does it make a difference whether we use a knot or an unknot?
Thanks. Is there any convex body that can be captured by an unknot? The diagram with the equilateral triangle is unclear: knot? not?
@WlodekKuperberg about unknot, an unknot can be tangled at the intersection point; so that the same argument shows that it cannot be removed. But if you allow the thread to go thru it self, then no --- it cannot be captured --- the equidistant move reduces the length.
The regular tetrahedron and the cube seem like a natural pair of test cases (both can be nicely unrolled onto the plane, as long as you avoid the vertices).
@zeb yes, it seems that one can modify the example for triangle to show that tetrahedron can be captured. Most likely any convex polyhedron can be captured, but that will require a new idea.
That loop on an equilateral triangle doesn’t work: think of it as a billiard path in a triangle. Move path parallel, its length doesn’t change. When you approach the boundary, you can slip the knot off without increasing its length.
@IanAgol it works --- it has a knot at the crossing point.
What’s the proof?
@IanAgol the thread must be covered by a closed geodesic that is parallel to sides and it must have a triple point. Such a geodesic is unique, so you cannot move it.
@AntonPetrunin why must it be covered by a closed geodesic parallel to the sides with a triple point? And what thread are you talking about? The picture in the linked answer shows a trivalent graph, but this one has a degree six vertex
@IanAgol this graph is covered by a periodic geodesic that is locally minimizing. The triple point of this geodesic is the vertex of degree 6. As you claimed, the geodesic is a part of one-parameter family of geodesics (with the same length) but the near by geodesics in this family do not have triple points.
Okay, I guess I wasn’t clear what I am asking. This shows that the graph is locally minimal. But you claim that Geraschenko showed that this implies that there is a knot that is locally minimal. The proof he gives is incomplete, plus doesn’t apply to your graph since the vertex is degree 6, but he only shows a degree 3 vertex. In any case I am unconvinced.
@IanAgol Yes, I wanted to say "Geraschenko-type argument shows..."
A circular disc cannot be captured.
Likely the same idea can be used to show that any convex body of revolution cannot be captured.
It can be proved using a small variation of the idea as in the original answer.
It is sufficient to show that infinitesimal Möbius tranform $m$ of the disc can shorten the wrapping length while preserving the crossing pattern.
Once it is proved, moving in this direction will eventually allow the disc to escape.
Choose a Möbius tranform $m$ of the unit disc that is close to the identity map.
Denote by $u$ its conformal factor.
Denote by $U(r)$ the average value of $u(x)$ for $|x|=r$.
It is sufficient to show that $U(r)\le 1$ and the inequality is strict for $r<1$.
In this case by rotating the knot and applying the Möbius tranform will shorten the length.
Consider the circle $C_r$ of radius $r$ centered at the origin.
Denote by $r'$ the radius of $m(C_r)$.
Observe that $U(r)=r'/r$.
Evidently $r'\le r$ if $r\le 1$ and $r'< r$ if $r< 1$, whence the result.
|
2025-03-21T14:48:30.948499
| 2020-05-11T19:19:50 |
360068
|
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|
Stack Exchange
|
Quotients of subspaces of $C(\alpha)$
A well known problem, attributed to H. P. Rosenthal, asks whether or not every quotient of $C(\alpha)$, $\alpha$ countable ordinal, is $c_0$-saturated. As it is known, $C(\alpha)$ are $c_0$-saturated and also every quotient of $C(\alpha)$ contains $c_0$. Since a closed subspace of a quotient is a quotient of a closed subspace one may ask the following.
Question I. Let $X$ be a closed subspace of $C(\alpha)$. Is every quotient map $Q:X\to X/Y$ not strictly singular ? In particular what is the answer when $X/Y\cong c_0$ ?
This is formally a stronger question than the initial problem and has a positive answer when $X=C(\alpha)$. A possible negative answer indicates that the problem probably also has a negative answer. Moreover, the above question has a negative answer if the following has a positive one.
Question II. Does there exist a universal constant $C$ such that for every equivalent norm $|\|\cdot\||$ on $c_0$, there exists a countable ordinal $\alpha$ such that $(c_0,|\|\cdot\||)$ is $C$-isomoprhic to a subspace of $C(\alpha)$ ?
Do you know the answer to Question 1 when $X$ is the Schreier space (or, more generally, when $X$ has an unconditional basis)?
No I do not know. But I really concern for Question II. Is it possible Zippin's work to be related to that question?
Nice question. My initial thoughts. The answer to Q2 should be negative. The collection of renormings of $c_0$ is analytic, and so you can cook up a universal space for the property in the question that doesn't contain $C[0,1]$. In some sense (this isn't precise, since the universal space need not be $C(\alpha)$) this implies there is a countable $\alpha$ for all renormings, which shouldn't be true.
Is it possible to show that if for every specific equivalent norm there exists an $\alpha$ such that $C_0$ with this norm embeds into $C(\alpha)$ with a universal constant $C$ then there exists a countable ordinal $\beta$ such that $C(\beta)$ is universal for all these embeddings?
@ Spyridon Argyros I don't know. This goes back to your work with Dodos. But actually we may not need it. For all $n$ renorm $c_0$ so that every $n$ subset of the basis is, say, 2-equivalent to $\ell_1^n$. Now take an $\alpha$ where all of these $C$-embed into $C(\alpha)$. Continue this process (need to think more how to do this) to increase the $\ell_1$-index of some $C(\alpha)$ spaces. Then take the universal space $X$. Since $\ell_1$ does not embeds into any $C(\alpha)$'s, by the universal space construction it doesn't embed into $X$ either but the $\ell_1$ index of $X$ is $\omega_1$.
@Bunyamin Sari Just to understand. You claim that every universal space should contain $\ell_1$ ?
@ Spyridon Argyros If the above construction works we get: for all $\beta$ there exists a $(c_0,|.|{\beta})$ for which $\ell_1$-index is greater than $\beta$ and $C$-embeds into some $C(\alpha{\beta})$. So the universal space $X$ (constructed by [AD]) for the collection $C(\alpha_{\beta})$ should not contain $\ell_1$ but $X$ contain $\ell_1$ since the $\ell_1$ index is uncountable.
Actually, just realized the above construction is not possible. You can't get a renorming of $c_0$ with large $\ell_1$-index.
I agree with you. According to Dodos - Ferenczi (Fundamenta Mathematicae. 2007. , v.193, p.171 - 179) the spaces with separable dual are strongly bounded
@ Bill Johnson I. Gasparis informed me that Ted Odell ( arXiv:math/9201219) has proved that the quotients of the Schreier space are $c_0$ saturated and also the quotient map is not strictly singular.
|
2025-03-21T14:48:30.948762
| 2020-05-11T20:10:15 |
360071
|
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"Gerhard Paseman",
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|
Stack Exchange
|
Is asymptotic growth bound on a sequence equivalent to an asymptotic growth bound on its partial sum?
The following question was asked at https://mathoverflow.net/questions/360053/asymptotic-growth-bound-on-a-sequence-equivalent-to-an-asymptotic-growth-bound-o, but then deleted by the user:
I posted the following question on [Math Stack Exchange][1] on April 2, where it's been unanswered for over a month. I hope the good denizens of Overflow will allow me to migrate it here in hope of an answer or some advice on where to look further.
Question:
Let $(a_n)$ and $(b_n)$ be sequences of positive real numbers. Denote by $o$ the "little-oh" Landau symbol. Is it possible, in general, to find a third sequence $(c_n)$ such that $\sum_{k=0}^n a_k = o(b_n)$ if and only if $a_n = o(c_n)$? Is there a formula for such a $(c_n)$ in terms of $(a_n)$ and $(b_n)$?
It's also possible that some generalization of classical asymptotic analysis can be used to obtain a result in the same spirit, and I'm interested in hearing if this is so.
[1]: https://math.stackexchange.com/questions/3606416/asymptotic-growth-bound-on-a-sequence-equivalent-to-an-asymptotic-growth-bound-o
I think this question may make sense. Below is the answer to it.
The answer is yes or no, depending on how the quantifiers are placed.
If the question is this: Is it true that
$\forall (a_n)\ \forall (b_n)\ \exists (c_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n = o(c_n)\big]$?
then the answer is yes. Everywhere here, $(a_n),(b_n),(c_n)$ are sequences of positive real numbers.
Indeed, for any such $(a_n)$ and $(b_n)$, for all $n$ just let $c_n:=na_n$ if $\sum_{k=0}^n a_k=o(b_n)$ and $c_n:=a_n$ otherwise.
More interesting is this question: Is it true that
$\forall (b_n)\ \exists (c_n)\ \forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big]$?
Here the answer is no. Indeed, suppose that, to the contrary, the latter highlighted statement holds. For all natural $n$ let
$$b_n:=n\ln n.$$
Then I claim that for any $(c_n)$ such that
\begin{equation*}
\forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big] \tag{0}
\end{equation*}
we have
\begin{equation*}
c_n>\sqrt n
\end{equation*}
eventually, that is, for all large enough $n$.
Indeed, suppose otherwise: that for some natural $n_1<n_2<\cdots$ and all natural $j$
\begin{equation*}
c_{n_j}\le\sqrt{n_j}. \tag{1}
\end{equation*}
For all natural $k$, let now
\begin{equation*}
a_k:=\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{k=n_{i^2}\},
\end{equation*}
where $I\{\cdot\}$ denotes the indicator. That is, $a_k=\sqrt{n_{i^2}}$ if $k=n_{i^2}$ for some natural $i$, and $a_k=0$ otherwise. Then
\begin{equation*}
\sum_{k=0}^n a_k
=\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{n_{i^2}\le n\}
\le\sum_{i=1}^\infty\sqrt n\,I\{i^2\le n\}\le n=o(b_n).
\end{equation*}
So, by (0), $a_n=o(c_n)$, whence
\begin{equation*}
\sqrt{n_{i^2}}=a_{n_{i^2}}=o(c_{n_{i^2}})=o(\sqrt{n_{i^2}}),
\end{equation*}
by (1). This contradiction proves that indeed $c_n>\sqrt n$ eventually.
So, letting now $a_n:=\sqrt n/\ln(n+1)$, we have the condition $a_n=o(c_n)$ satisfied. However, here
\begin{equation*}
\sum_{k=0}^n a_k\sim\tfrac23\, n^{3/2}/\ln(n+1),
\end{equation*}
which is not $o(b_n)$. Thus, the second highlighted statement is false.
For the yes version above, it looks like cn is defined to give the complementary result. Gerhard "Would Like Some Further Explanation" Paseman, 2020.05.11.
@GerhardPaseman : It seems to me everything is correct there. E.g., if $a_k\equiv1$ and $b_n\equiv n^2$, then we take $c_n:=na_n=n$, and the equivalence $\iff$ holds. If $a_k\equiv1$ and $b_n\equiv n$, then we take $c_n:=a_n=1$, and the equivalence $\iff$ again holds. How do you get a complementary result?
It looks like I was thinking upper bounds, not asymptotics. You are right about the definition of cn. I am wondering though if you need a quantification over all n in your statement. Gerhard "Thanks For Your Checking In". Paseman, 2020.05.11.
Instead of the plain quantification over all $n$, the original question had $o(\cdot)$ asymptotics, which of course involves a $\forall\exists\forall$-type quantification.
|
2025-03-21T14:48:30.949032
| 2020-05-11T21:01:00 |
360074
|
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|
Stack Exchange
|
Is regularization of infinite sums by analytic continuation unique?
There are ill-posed summations that we can assign values to, take for concreteness,
$$ S = \sum_{k=0}^\infty k $$
to which we can assign $-1/12$ by several methods. Is there a fundamental and rigorous reason why these methods have to agree on the same value?
Perhaps the most powerful means we have available is the zeta function regularization, where we attach $\zeta(-1)$ (Riemann zeta function), which formally represents $S$, to the summation. If we have another analytic function that formally represents $S$ and analytically continues to assign a value to $S$, must the value always be the same, $-1/12$?
More concretely, suppose I have a function $\xi(s) = \sum_{k=0}^\infty f_s(k)$ such that $f_1(k) = k$ and $f_s(k)$ is a "nice" function (for the case of zeta function, $f_s(k)=k^{-s}$ is the exponential function in $s$). Suppose that $\xi(s)$ can be analytically continued to a meromorphic function with a value at $s=-1$. Must it be the case that $\xi(-1)=-1/12$?
More generally, what can be said about other divergent sums? If there is an analytic continuation that defines the value of such sums, must it be unique? Where can I look up known uniqueness results and techniques?
For your example $f_s(k)=k^{-s}$ we have $f_1(k)=1/k$, not $k$.
Let $\xi(s)=\sum_{k\geq 1}ke^{-(s+1)k}$. This extends to a meromorphic function on $\mathbb{C}$ for which $s=-1$ is pole, rather than $\xi(-1)=-1/12$.
Well, we can add the requirement that the function is defined to have a finite value (no poles) at s=-1.
Let $$f_k(s) = k^{-s}+(s+1)k^{-s-2},\qquad f_k(-1)=k$$ then $$F(s)=\sum_k f_k(s) = \zeta(s)+(s+1)\zeta(s+2), \qquad F(-1)=-1/12+1$$
It seems to me that $F(-1)=-1/12$.
$\lim_{s\to 1}(s-1)\zeta(s)=1$
Oops, you are right.
Mathematica tells me that Richard Stanley is right...
@MCH No. $\zeta(s)$ has a pole at $s=1$.
It has a single pole and thus a limit but the function is undefined at exactly $s=1$. For example, the sum $\sum_k s/k^{1+s}$ is of course equal to 0 when $s=0$, but that is not the same as the limit $s \zeta(1+s)$ at $s \to 0$, which is 1.
@MCH No Nobody is looking at $\sum_k 0 . k^{-1}$, why would you do so. $g(s)=(s-1)\zeta(s)$ is analytic at $s=1$ and $g(1)=1$, that's how complex analysis works.
But this is exactly what happens in your example, that's where the additional term 1 comes from. If you wish, amend the question to require a meromorphic function that has no poles at -1.
No.. Do you realize the title of your question is 'analytic continuation' ? Without analytic continuation you get $\sum_k f_k(-1)=\sum_k (k + 0.k^{-1})=\infty$ not $F(-1)=-1/12+1$..
What I'm saying is that with this argument you can attach contradictory values to any (even absolutely convergent, say a geometric etc) sum, because you can add each term with $s/k^{1+s}$ and claim you have increased the sum by $s \zeta(1+s)$ which is $1$ around $s\to 0$.
@MCH Interesting point. Reminds me of https://physics.stackexchange.com/a/492793/35699. Perhaps a stronger condition is implicitly lurking here.
@MCH $g(s)=(s-1)\zeta(s)$ gives $g(1)=1$ and $\sum_k 0.k^{-1}=0$. Both are different things. $g(s)$ is the analytic continuation of the function $(s-1)\sum_k k^{-s},\Re(s) > 1$. There is no contradictory stuff in there. It shows that analytic continuation gives any value to any series and yes even to the zero series. If your problem is to find which kind of analytic continuation summation gives $0$ to the zero series then this is another question. In every cases it won't help for $\sum_k k$ where you'll never find any condition saying that $k^{-s}$ is good and $k^{-s}+(s+1)k^{-s-2}$ is bad.
As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it is the unique constant term of the asymptotic expansion of the smoothed partial sums, which perhaps explains why it is the most "natural" value.
Let $\eta$ be any Schwartz function such that $\eta(0) = 1$. Then
\begin{align}
\sum_{n=1}^\infty n^s \eta(n \varepsilon)
&= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx
\end{align}
Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get
\begin{align}
\sum_{n=1}^\infty n^s
&= \sum_{n=1}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\
&\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=1}^\infty n^s \eta(n \varepsilon) \\
&= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\
&= \zeta(-s)
\end{align}
Is it obvious an $\eta(x)$ that makes that integral vanish exists for all $s$ independently of $s$?
@AlexArvanitakis I found a class of functions here that seem to cover all $s \neq -1$, namely $\eta_s(x) = \exp(-x)\left(1 - \frac{x}{s+1}\right)$.
That won't work for you; the $\eta$ on the LHS is $s$-independent.
@AlexArvanitakis The series is solved independently for each $s$. I've edited the post to clarify.
Thanks, that helped me
As the other answer showed, the regularization by analytic continuation is not unique.
The problem is mostly in that you choose this method. If you took other methods, like Dirichlet or Borel regularization, in most cases you would obtain the same result.
Analytic continuation in general does not belong here. Maybe there can be some constraints on this method that would restrict its results to those which are obtained by other methods, but so far it seems nobody proposed any.
What is 'here' in "Analytic continuation in general does not belong here"?
@LSpice analytic continuation does not belong to the set of the mutually-compatible methods that produce the same result when applicable (Abel, Cesaro, Borel, Dirichlet, etc). Analytic continuation without additional restrictions can yeld, in general, any result.
Thanks. I understand that different transformations that assign values to infinite sums (Borel sums etc) may result in different values. But I'm curious to know if, for example for the sum S above, there are possible functions (other than Riemann zeta) that can be analytically continued to a value different from -1/12.
@The thing is, the most of regularization methods AGREE on the same value. Alanytic continuation is less rigorous in this respect and I am sure you can find such function that it will give any value you want.
@Anixx At first the idea is that $\sum^L (-1)^n = \lim_{z\to 1} \sum_{n\ge 0} (-1)^n z^n$ is not the same as $\sum^A (-2)^n = \frac1{1+2z}|_{z=1}$. The former is a limit regularization and has good chances to be related from one method to the other while the latter is analytic continuation and usually is not related (except for zeta/power series with finitely many poles because one is the Mellin transform of the other).
|
2025-03-21T14:48:30.949481
| 2020-05-11T21:16:04 |
360075
|
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|
Stack Exchange
|
Solutions of PDE under changing topology
Let suppose we have a PDE on a manifold. I'm interested in the following question. How does the space of solutions of this PDE change when the topology of the manifold changes? For example in 2D we consider how the kernel of some partial operator changes under handle attachments. Do people study this type of problems?
Thank you!
After changing the topology, what is the new PDE you consider? How do you construct it from the original one?
Yes, this has been studied intensively for quite a while. In particular, people who work in gauge theory (as applied to 4-manifolds) have studied the effect of surgeries on the solutions to the ASD Yang-Mills equations and Seiberg-Witten equations using essentially analytic techniques. An early result of this kind by Donaldson shows that taking connected sum with ${\mathbb C}P^2$ (with its standard complex orientation) can change the number of solutions (counted with signs) from something non-zero to $0$. In other words, the moduli space (as an oriented $0$-manifold) can change under a simple topological operation on the manifold.
There is a vast literature on the subject, where one can detect much more subtle changes in the smooth topology of $4$-manifolds. You might look up the Fintushel-Stern knot surgery formula. There are also many results in other contexts such as moduli spaces of holomorphic curves.
Of course the most basic example is perhaps what you were alluding to in 2D. The dimension of the linear equations for harmonic forms (aka homology groups, by Hodge theory) depend on the topology of the underlying manifold.
|
2025-03-21T14:48:30.949628
| 2020-05-11T21:22:02 |
360076
|
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|
Stack Exchange
|
Effective semi-group of a singular abelian surface
Let $A$ be a singular abelian surface over $\mathbb{C}$; that is, an abelian surface of maximal Picard rank $\rho(A)=4$. By Shioda-Mitani we know $A \cong E \times E'$ where $E,E'$ are isogenous elliptic curves with CM in an imaginary quadratic field $\mathbb{Q}(\sqrt{-d})$. I'm not sure if this is standard terminology, but by the effective semi-group, I mean the semi-group $\text{NS}^{+}(A) \subset \text{NS}(A)$ of integral points in the effective cone of $A$.
We can take as a basis of $\text{NS}(A)$ the four classes $v, h, \Gamma, \Gamma_{\text{CM}}$, where $v,h$ are the vertical and horizontal classes in $E \times E'$, $\Gamma$ is the graph of an isogeny between $E, E'$, and $\Gamma_{\text{CM}}$ is the graph of the CM map. Obviously we get effective classes by taking non-negative integer linear combinations of these basis elements. However, $\text{NS}^{+}(A)$ is not finitely generated (see, page 1 of https://arxiv.org/pdf/alg-geom/9712019.pdf). So my questions are:
Do we have any understanding of the lattice points in $\text{NS}^{+}(A)$ which are not non-negative linear combinations of $v, h, \Gamma, \Gamma_{\text{CM}}$? Has this been studied anywhere? There are infinitely many such points, but I'm really lacking intuition for these.
Given an explicit class in $\text{NS}(A)$, is there any useful way of determining when it is effective? Other than the fact that it must intersect positively with an ample class. I haven't heard of such a condition in general, but I'm hoping maybe this particular case is easier.
Related: the situation for $\operatorname{NS}(E \times E)$ for a non-CM elliptic curve $E$ is worked out in Lazarsfeld's Positivity in algebraic geomtery I, Lemma 1.5.4. This should at least provide some intuition, even if the CM case might be a bit harder.
Below is a summary of the discussion in Lazarsfeld's Positivity in algebraic geometry I, Ex. 1.4.7, Lem. 1.5.4, and Rmk. 1.5.6.
Lemma. Let $D$ be an $\mathbf R$-divisor on an abelian surface $A$. Then the following are equivalent:
$D$ is nef;
$D$ is pseudo-effective;
$D^2 \geq 0$ and $D \cdot H \geq 0$ for any ample divisor $H$.
Proof. Implication 1 $\Rightarrow$ 3 is clear. For 2 $\Rightarrow$ 1, it suffices to treat the case where $D$ is effective and irreducible. Any translate $D + a$ for $a \in A$ is algebraically equivalent to $D$, so $D^2 = D(D + a) \geq 0$ as $D \neq D + a$ for $a \in A$ general. Finally, for 3 $\Rightarrow$ 2 it suffices to show that if $D$ is an integral divisor with $D^2 > 0$ and $D \cdot H > 0$, then some multiple of $D$ is linearly equivalent to an effective divisor. This follows from Riemann–Roch for abelian surfaces. (In fact $D$ is ample; see e.g. this post, or Prop. 1.5.17 in Lazarsfeld.) $\square$
Example. For example, if $E$ is an elliptic curve with CM in $\mathbf Z[\sqrt{-n}]$ for $n > 0$, and $\Delta, \Gamma \subseteq E \times E$ are the diagonal and the graph of "multiplication by $\sqrt{-n}$" respectively, then the matrix of the intersection form with respect to the basis $(h,v,\Delta,\Gamma)$ is
$$\begin{pmatrix}0 & 1 & 1 & n \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1+n \\ n & 1 & 1+n & 0 \end{pmatrix}.$$
A better basis is $(h+v,h-v,\Delta-h-v,\Gamma-h-nv)$, which gives the matrix
$$\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2n \end{pmatrix}.$$
With respect to this basis, the equations become
\begin{align*}
a^2 \geq b^2 + c^2 + nd^2, & & a \geq 0
\end{align*}
for a divisor $D = a(h+v) + b(h-v) + c(\Delta-h-v) + d(\Gamma-h-nv)$. These describe a circular cone in $\operatorname{NS}(A)_{\mathbf R} \cong \mathbf R^4$, so you can produce many effective classes close to the boundary with negative $\Delta$ or $\Gamma$ components. For example $(a,b,c,d) = (2m^2+1,2m^2,-2m,0)$ gives a divisor $D$ with
$$D^2 = 2\Big((2m^2+1)^2 - (2m^2)^2 - (-2m)^2\Big) = 2\Big( 4m^4 + 4m^2 + 1 - 4m^4 - 4m^2 \Big) = 2,$$
so $D$ is effective (even ample). Its coefficient in $\Delta$ is $-2m$.
Remark. I don't know if every pseudo-effective class is algebraically equivalent to an effective one. (This is certainly false for "linearly equivalent", as can be seen with $p \times E - q \times E$ for different points $p, q \in E$.) On a general abelian surface I don't expect this to be true, because every effective class is ample if $A$ is simple, but I imagine there might be classes on the boundary of the nef cone (if $A$ has complex multiplication).
For a product of isogenous CM elliptic curves, there is a little more hope.
Thanks for your wonderful answer! It clears part of my confusion up. I suppose another complication is that certain classes on the boundary of the circular nef cone might actually be effective right? For example, the classes $v, h, \Delta, \Gamma$ are all on the boundary. And presumably any effective curve of square zero lies on the boundary.
The boundary is exactly given by $D^2 = 0$, since $D^2 > 0$ and $D \cdot H \geq 0$ force $D \cdot H > 0$. So effective classes on the boundary correspond to square $0$ curves, which have arithmetic genus $1$ by Riemann–Roch/adjunction, hence are elliptic since $A$ has no rational curves. Every such elliptic curve is isogenous to $E$ because it is an isogeny factor of $A = E \times E$. I don't know a way to predict whether a given class on the boundary comes from an elliptic curve, but you should be able to write down many examples by thinking of étale correspondences.
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2025-03-21T14:48:30.950123
| 2020-05-11T23:43:47 |
360080
|
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|
Stack Exchange
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Constructing non-trivial epimorphisms from commutative rings
As the title suggests, I wonder if anyone can share some techniques or references for constructing interesting epimorphisms from generic commutative rings. Generic is largely open to interpretation, but my thinking is that the main point is that the constructions shouldn't depend on global properties of the base ring, and relatedly that the discussion doesn't become trivial when specialized to quasi-local rings.
Although this question has its roots in commutative ring theory, scheme theoretic answers would also be welcome if that provides a more natural framework.
Why I'm asking this: For some important classes of rings, the behavior of epimorphisms is highly constrained. A couple of well-known examples: (i) if $R$ is reduced or Noetherian, then epimorphisms from $R$ are surjective if and only if $R$ has Krull dimension $0$; (ii) epimorphisms from Dedekind domains must be injective or surjective and they (more or less) decompose into a flat part and a torsion part;
My take away from these examples is that it might be interesting to study, generally, the definitional power of "epimorphisms from $R$ have property $(X)$" where $(X)$ is some intelligently chosen property of morphisms.
Of course, a vital prerequisite to this is the ability to construct interesting test epimorphisms. I quickly realized that, for all the papers on epimorphisms I've read, I have basically no idea how to produce them in a general setting.
The purpose of this post is to seek out knowledge of how to construct interesting epimorphisms of rings in situations as general as possible.
One situation I keep coming back to is the apparent need to understand how to construct test epimorphisms from quasi-local nonreduced rings of Krull dimension $0$ and quasi-local domains of dimension $1$.
Following is some thought vomit, please don't feel compelled to read this, especially if you are familiar with epimorphisms: The simplest epimorphisms are of course quotients by ideals and localizations at multiplicative subsets. These are in fact "good enough" epimorphisms to mostly understand what happens in the case $(X) = surjective$, as mentioned above.
Another classic source of epimorphisms is to adjoin 'pointwise' inverses to a ring, as $R \rightarrow R[x]/(ax^2-x, xa^2-a)$. The result of adjoining a pointwise inverse to every element of a ring is a Von Neumann Regular ring, sometimes called the universally absolutely flat $R$-algebra for obvious reasons. In general, the epimorphisms resulting from these constructions are not flat without the assumption that the minimal prime spectrum of $R$ is compact.
More sophisticated epimorphisms can be constructed by universally inverting maps between finite projective modules (so-called Cohn or universal localizations). If you can construct interesting projective modules over $R$, then I guess theoretically you can produce interesting (necessarily flat) epimorphisms by this method. Unfortunately this is already a dead-end if $R$ is quasi-local, because then universal localizations will just coincide with localizations at multiplicative sets.
More on flat epimorphisms.... when a ring has weak global dimension $1$ (is locally a valuation domain), then the flat epimorphisms coincide with "overrings," and are easy to understand completely in terms of intersections of "localizations" at primes. (For domains, "overring" of $R$ means a ring intermediate between $R$ and its field of fractions. The appropriate generalization to rings of weak global dimension $1$ would be the "ring of finite fractions," which, loosely speaking, has elements homomorphisms $I \rightarrow R$ for finitely generated faithful ideals of $R$, and then the meaning of "localization" needs to be slightly adapted too).
If we drop the wk. gl. dim. $1$ assumption, then we can still understand flat epimorphisms as Gabriel localizations (and even as Kaplansky ideal transforms in case they are finite type), but I find it becomes difficult to actually construct them with confidence due to the reflexive nature of the defining localizing system, i.e. the filter $\mathfrak{F}$ associated to a flat epimorphism $A \rightarrow B$ is $\mathfrak{F} = \{ I \subseteq A \mid IB = B\}$. Unless a localizing filter is a special instance of a universal localization, it hard hard to predict whether it will produce a flat epimorphism.
Another hope, I thought, might be to employ dominions to construct interesting epimorphisms. Recall that if $A \rightarrow B$ is a morphism of rings, the dominion of $A$ in $B$ is defined as $\operatorname{Dom(A,B)} = \{b \in B \mid b \otimes_A 1 - 1 \otimes_A b = 0 \}$. A morphism $A \rightarrow B$ is epi iff $\operatorname{Dom(A,B)} = B$. In general, the dominion of $A$ in $B$ does not induce an epimorphism $A \rightarrow \operatorname{Dom(A,B)}$ but if we keep taking successive dominions and limit intersections, transfinitely, then eventually the corresponding sequence of rings has to stabilize by cardinality considerations, and we get an epimorphism from $A$ as a result. The issue here is that the resulting epimorphism will usually be trivial unless $B$ is chosen carefully (but how??), and even so I'm not confident that the result of the transfinite construction would be workably transparent.
https://www.math.nagoya-u.ac.jp/~takahashi/epi18.pdf
@Hailong I'm familiar with this. It's a really interesting paper but doesn't strike me as relevant. Though I'd be willing to concede to focusing on Noetherian rings for now, the point I made in the post is important: the constructions should still be interesting for quasi-local rings and shouldn't depend heavily on global properties of the ring. Universal localizations are the focus of that paper, and they are trivial for quasi-local rings; moreover, I'm still interested in the construction of epimorphisms from rings that don't have interesting generalization-closed spectral subsets.
Section 6, in particular Theorem 6.10 did give way to construct examples of flat epi. that is not a localization. But now I see that you want more, just did not realize what you meant by "global properties".
@Hailong Dao Thanks for your comments. Right, I'm interested in test epimorphisms for the purpose of proving statements like "If $R$ is a ring such that all epis from $R$ are $(X)$, then $R$ is ......" . Unfortunately I don't think theorems like 6.10 with such geometric assumptions will help me.
(cont.) A lot of the time, as in the two examples I gave in the post, $(X)$ will force finite Krull dimension and localize well. It seems like an important starting point would be to build interesting epis from Noetherian local one-dimensional domains. Such epis would necessarily not be flat.
|
2025-03-21T14:48:30.950614
| 2020-05-12T00:01:25 |
360082
|
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|
Stack Exchange
|
Fun examples relating to Hopf surfaces
A Hopf surface is a compact complex surface whose universal cover is complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$. I would like to know whether anyone has any of the following examples:
(i) A non-compact complex surface whose universal cover is complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$.
(ii) A compact complex surface whose universal cover is smoothly isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$, but not complex analytically isomorphic to $\mathbb{C}^2 \setminus \{ 0 \}$.
Isn't $\mathbb C^2 \setminus {0}$ a silly example of (i)? (Also, \backslash doesn't space well: $\mathbb C^2 \backslash {0}$. Prefer \setminus: $\mathbb C^2 \setminus {0}$. I have edited accordingly.)
Only marginally less silly: take a cyclic group acting linearly and freely on $C^2 \setminus {0}$.
@DannyRuberman Thank you, but now the question remains: Can we find non-silly examples?
Look at Di Scala, Kasuya, Zuddas, Non-Kahler complex structures on $R^4$. These complex structures have embedded elliptic curves, so they won't be isomorphic to $\Bbb C^2 \setminus 0$ after removing a point. However, I doubt that they have compact quotients.
(ii) I would like to prove that there are no complex surfaces that satisfy (ii).
Indeed, suppose that the universal cover $\widetilde X$ of a complex surface $X$ is diffemorphic to $\mathbb C^2\setminus 0$. Let's prove that $\widetilde X$ is biholomorphic to $\mathbb C^2\setminus 0$.
First, we note that $X$ has a finite cover that is diffeomorphic to $S^3\times S^1$. Indeed, take any
$$S^3\subset \widetilde X=\mathbb C^2\setminus 0$$
that encircles $0$. Then, since the action of $\pi_1(X)$ on $\mathbb C^2\setminus X$ is discreet, there exists only finite number of elements $g\in \pi_1(X)$ such that $g\cdot S^3$ intersects $S^3$ in $\mathbb C^2\setminus 0$. Let's take such $g_1\in\pi_1(X)$ that $g_1\cdot S^3$ is disjoint from $S^3$. Let $\mathbb Z=\langle g_1\rangle $ be the group generated by $g_1$. Then it is not hard to see that $(\mathbb C^2\setminus 0)/\langle g_1\rangle$ is diffeomorphic to $S^1\times S^3$. Clearly $S^1\times S^3$ is a finite cover of $X$.
It remains now to apply the result of Bogomolov that complex surfaces with $b_2=0$ are either Hopf surfaces or Inoue surfaces https://en.wikipedia.org/wiki/Surface_of_class_VII . Since $\pi_1(S^1\times S^3)=\mathbb Z$, the complex structure on $S^1\times S^3$ is that of a Hopf surface. We conclude that $\widetilde X$ is byholomorphic to $\mathbb C^2\setminus 0$.
(i) As for (i), you can take $(\mathbb C^2\setminus 0)/\Gamma$, where $\Gamma$ is any finite subgroup of $SU(2)$.
Instead of using Bogomolov's result, you can use the fact that any complex surface homeomorphic to $S^1\times S^3$ is biholomorphic to a Hopf surface. This was proved by Kodaira in Complex Structures on $S^1\times S^3$, Proc. Nat. Acad. Sci. U.S.A. 55 (1966), 240–243.
Thanks Michael for giving this reference! I was not aware of this paper (regretfully)
|
2025-03-21T14:48:30.950835
| 2020-05-12T03:23:48 |
360089
|
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|
Stack Exchange
|
Cutting up the Bring surface into six pairs of pants
The Bring sextic, with 120 automorphisms, is the numerically most symmetric compact Riemann surface of genus 4. To cut it up into six pairs of pants, we need to cut along nine disjoint geodesic loops. How short can those loops be, and how symmetric can we make the decomposition?
I am studying the Bring sextic, by the way, because one can elegantly map all of the possible shapes of equilateral pentagons in the Euclidean plane to the points of the Bring sextic. (Calling it "the Bring sextic", by analogy with "the Klein quartic", obviates the clumsy need to choose between "the Bring surface" over $\mathbb{R}$ and "the Bring curve" over $\mathbb{C}$.)
For comparison, the Bolza quintic is the numerically most symmetric compact Riemann surface back in genus 2, with 48 automorphisms. The systole of the Bolza quintic is $2\operatorname{arccosh}(1+\sqrt{2})\approx 3.057$, and there are $12$ systolic loops (that is, loops of that length). There are triples of systolic loops that are disjoint, and such a triple cuts up the Bolza quintic into two isometric pairs of pants. In that decomposition, the three cuffs of one pair of pants are sewn to the three cuffs of the other, and all three twists, as a fraction of the systole, are $\operatorname{arccosh}((5+4\sqrt{2})/7)/(2\operatorname{arccosh}(1+\sqrt{2}))\approx 0.3213$.
Genus 3 is even prettier. The famous Klein quartic is the symmetry champion, with 168 automorphisms. It has $21$ systolic loops, each of length $8\operatorname{arccosh}(\frac{1}{2}+\cos(2\pi/7))\approx 3.936$. Some sextuples of them are disjoint, and such a sextuple cuts up the Klein quartic into four isometric pairs of pants. The 3-regular graph giving the sewing of the cuffs is $K_4$, the edge-graph of a tetrahedron, and all six twists are $1/8$.
With those cases as context, what happens in genus 4? The Bring sextic has $20$ systolic loops, each of length $2\operatorname{arccosh}((9+5\sqrt{5})/4)\approx 4.603$. Since our loops must be disjoint, however, we can take at most six of those $20$. Cutting along those six breaks the Bring sextic into three pieces, each with four loops of boundary. We need to cut each of those three pieces along another geodesic loop, to split it into two pairs of pants.
The Bring sextic has $30$ loops of length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$ (only slightly longer than systolic). For each of our three current pieces, there is one of those $30$ loops that splits it into two pairs of pants, leaving us with six isometric pairs of pants overall. The 3-regular graph giving the sewing of the cuffs in the resulting decomposition is $K_{3,3}$, a graph famous for its nonplanarity. The twists of the six systolic loops are $1/6$, while the twists of the three longer loops are $1/4$.
Is this the most symmetric way to cut up the Bring sextic? Or are there other decompositions that can compete with it for simplicity and symmetry?
Could you clarify your statement that “one can elegantly map all of the possible shapes of equilateral pentagons in the Euclidean plane to the points of the Bring sextic”? What is this map? (I just happened to be reading Matthias Weber's paper “Kepler's small stellated dodecahedron as a Riemann surface”, which is also about Bring's sextic, so I'm curious.)
The polygon space for equilateral pentagons in the plane and the Bring sextic are both compact, orientable, smooth surfaces of genus 4, so there are lots of maps between them. I am drafting a 35-page paper about such a map, and I hope that people who are curious about this area will be interested in reading it.
I am certainly curious to see this paper. But could you already explain what the automorphisms of order $2,4,5$ defining the triangle group $\Delta^0(2,4,5) \cong \mathfrak{S}_5$ acting on Bring's surface correspond to when acting on equilateral pentagons? (Or at least what are the fixed pentagon shapes under them?) Because the only action of $\mathfrak{S}_5$ on pentagons that I can think of is by permuting the vertices, and this doesn't preserve equilaterality, so I'm confused.
Rather than permuting the vertices of the equilateral pentagon, permute the order in which its edge vectors get assembled, tip to tail. The 24 $\frac{\pi}{5}$ vertices of the Bring sextic then correspond to the 24 pentagons whose five edges are evenly distributed around the unit circle, while the 30 $\frac{\pi}{4}$ vertices correspond to the 30 pentagons that have two pairs of coincident edges.
@Gro-Tsen, my paper about viewing the polygon space of equilateral pentagons as the Bring sextic is now available on arXiv: link. See, in particular, pages 10 and 11.
There is an alternative strategy for cutting up the Bring sextic into pairs of pants that might compete with the decomposition above in some ways.
Of the $30$ loops of the second-shortest length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$, there are sets of six that are disjoint. Cutting along such a set of six breaks the Bring sextic up into three pieces, each of which has four loops of boundary --- as happened also in the decomposition above.
The Bring sextic has $10$ loops of the third-shortest length, which is $2\operatorname{arccosh}((26+10\sqrt{5})/4)\approx 6.368$. For each of our current three pieces, there are two of those $10$ third-shortest loops, either of which can cut up that piece into two pairs of pants. With three binary choices, we get eight decompositions into isometric pairs of pants. The 3-regular graph that gives the sewing of the cuffs is $K_{3,3}$ in four of those eight decompositions, but is the edge graph of a triangular prism in the other four. The twists are somewhat simpler in all eight of those decompositions, however --- which might make them more attractive than the decomposition above for some purposes: The twists along the six shorter loops are $1/4$, as for the three loops of that same length in the decomposition above, but the twists along the three longer loops are $0$.
Not sure if this might be of any help, but have a look at the question about Cayley graph of $A_5$, that Cayley graph should be readily apparent on Bring's sextic given its $S_5$ symmetry.
|
2025-03-21T14:48:30.951259
| 2020-05-12T03:32:23 |
360090
|
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|
Stack Exchange
|
Splitting of short exact sequence in the category of finitely generated modules over a commutative Noetherian ring
In the category of finitely generated modules over a commutative Noetherian ring, the splitting of a short exact sequence can be checked locally at the maximal ideals of the ring. One reference for this is contained in the answer by Jeremy Rickard here Local property of split exact sequence .
My question is: Is there a reference of book or paper or some monograph where this fact (possibly along with a proof) appears ?
I would like to use this in a presentation and this result seems too natural to not have appeared anywhere before in literature.
Thanks.
Reiner's "Maximal Orders", Theorem 3.20, for instance.
@UriyaFirst Shouldn't that be an answer?
@PedroTamaroff I guess it should.
Reiner's "Maximal Orders", Theorem 3.20, for instance.
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2025-03-21T14:48:30.951458
| 2020-04-17T22:33:19 |
360093
|
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|
Stack Exchange
|
Jacobi fields on non-geodesic curves
The point of Jacobi fields is to study variations of geodesics through geodesics, but the Jacobi equation $D_t^2 J + R(J,\dot\gamma)\dot\gamma=0$ makes sense for any curve $\gamma$, not just for geodesics.
Are solutions to this equation still interesting in other situations? For instance, does a variation of a curve along a "Jacobi field" yield curves with related geodesic curvatures? Maybe this is related to e.g. curve shortening flow or some other already studied notions.
Interesting... One would expect the Jacobi equation to make no sense for non-geodesic $\gamma$ since it comes from the second variation of Energy and usually it doesn't make any sense to consider second derivatives at non-critical points.
@CarlosEsparza: Of course there is a metric running around on the path space (say of paths of Sobolev class $H^1$), hence one can define the second derivative everywhere.
In the two-dimensional case, variation of a curve along such a Jacobi field yield curves with the same geodesic curvatures.
(It is an application of Gauss--Bonnet.)
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2025-03-21T14:48:30.951902
| 2020-05-12T04:35:46 |
360095
|
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|
Stack Exchange
|
Spectral representation of closed operators with finite spectral bound
Assume $A$ is a closed linear operator on a Banach space $X$ and is densely defined. Assume the spectral bound $s(A) = \sup\{Re\lambda: \lambda\in \sigma(A)\}$ is finite. For example, if $A$ is the generator of a semigroup $T(t)$ with growth bound $\omega$, i.e., $\|T(t)\|\le Ce^{\omega t}$, we have $s(A) \le \omega$. Let $\alpha > s(A)$ be a real numer. Let $R(\lambda, A) = (\lambda - A)^{-1}$ be the resolvent. Do we have the following spectral representation of the semigroup $e^{tA}$?
$$e^{tA} = \frac{1}{2\pi i}\int_{\Gamma}e^{tz}R(z,A)dz,$$
where $\Gamma \subset \mathbb{C}$ is the vertical line $\{z\in\mathbb{C}, Re(z) = \alpha\}$.
In which sense would you expect the integral to converge? (Absolute convergence does certainly not hold.)
Yes, you are right, the integral cannot be absolutely convergent. My question is whether the integral is well-defined (convergent in the usual functional calculus sense) and equals the semigroup generated by $A$.
I guess there is no way since the resolvent cannot decay faster than $1/|z|$. The above integral does not converge even for multiplication operators.
@GiorgioMetafune: Well, since the exponential function oscillates along the imaginary axis, there might be a chance for this to converge as an improper integral. I wouldn't be too surprised if it holds, say, in finite dimensions (or even for analytic semigroups). But there are a lot of details to check...
Alright, I've done the calculation in the one-dimensional case. First note that there's the factor $\frac{1}{2\pi i}$ missing in front of the integral. But this little issue aside, we can use integration by parts and Cauchy's integral formula to see that the formula does indeed hold in the one-dimensional case (and thus also for diagonalizable matrices). I'd suspect that the same works for bounded generators $A$, but I have not checked this in detail. For analytic semigroups, things seem to be more involved since we cannot simply add a large circle on the left of the imaginary axis.
You are right. I just considered norm convergence.
I've looked it up now. The formula in question does indeed hold in the following sense:
Theorem. Let $(e^{tA})_{t \in [0,\infty)}$ be a $C_0$-semigroup on a complex Banach space $X$. Let $\omega \in \mathbb{R}$ be a real number that is strictly larger than the growth bound of our semigroup and let, for each $k > 0$, $\Gamma_k$ denote the complex line from $\omega - ik$ to $\omega +ik$. Then:
(i) The formula
$$
(*) \qquad e^{tA}x = \lim_{k \to \infty} \frac{1}{2\pi i} \int_{\Gamma_k} e^{zt} R(z,A)x \, dz
$$
holds for each $x$ in the domain of $A$.
(ii) If $X$ is an UMD space, then $(*)$ does even hold for each $x \in X$.
Reference: Proposition 3.12.1 and Theorem 3.12.2 in "Arendt, Batty, Hieber, Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)".
The same reference contains a counterexample that shows that one cannot drop the assumption in (ii) that $X$ be a UMD space, in general (this is Example 3.12.3 there; hardly surprising, the counterexample is a shift semigroup...).
You are right, nice to have this.
Many thanks for the reference, this is very helpful！
|
2025-03-21T14:48:30.952189
| 2020-05-12T04:52:12 |
360097
|
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|
Stack Exchange
|
Simple way to calculate the eigenvalues of a $2 \times 2 \times 2$ tensor
I am working with hypergraphs. The various matrices associated with hypergraphs are hypermatrix or tensors. I am interested in spectral aspects. In particular, I want to find all the eigenvalues explicitly for a class of hypergraphs. To start with, we can consider all the $0-1$ hypermatrices of order $2 \times 2 \times 2$. Then the characteristic polynomial is defined in terms of resultants of a certain system of homogeneous equations.
I want to know, some simple or at least a concrete way of calculating these eigenvalues. I have understood the method given in this paper. But this method was applicable only to special hypermatrices. Kindly share some references. Thank you.
Edit: I am interested in general uniform hypergraph then its associated matrices can be of any order and dimension. I thought, to start with, I will look at the simplest $2 \times 2 \times 2$ case.
See https://mathoverflow.net/questions/319644/eigenvalue-and-eigenmatrix-of-a-3d-tensor-how-to-calculate-it. See also https://www.stat.uchicago.edu/~lekheng/work/mcsc2.pdf, discusses eigenvalues of hypergraphs starting on slide 22. See also https://mathscinet.ams.org/mathscinet-getitem?mr=1325271 (but I don't have easy access to the actual article, so I'm not sure how relevant it really is).
Despite the above comment and already accepted answer, I think there might still be something to say here, and if some expert would care to step in, it would be most welcome. I'm a little fuzzy on this but it seems to me that a symmetric $2 \times 2 \times \dotsm 2$ tensor ($d$ factors) corresponds to a binary $d$-form, and the eigenvectors correspond to that form's critical points (on the projective line), ie., roots of the derivative (of a dehomogenization). The eigenvalues would presumably be the critical values, but I might be off by a scalar factor (???)...
... A scalar factor is not an issue at multiple roots of the original form (eigenvalue $0$) but for the others I would have to think about it. Well, I don't know how interesting hypergraphs on $2$ vertices are, but as far as eigenvalues of $2 \times 2 \times \dotsm \times 2$ tensors, I think that it should fall out easily from simple things with binary forms (even if I have some of the details wrong).
@ZachTeitler It is very interesting thank you. Can you share some references? Also, how to see that the eigenvectors correspond to that form's critical points (on the projective line) and eigenvalues are critical values? Can you share some examples as an answer also? I am interested in general uniform hypergraph whose associated matrices can be of any order and dimension. I thought, to start with, I will look at the simplest $2 \times 2 \times 2$ case.
As explained in a previous MO question, there is no unique generalization of the eigenvalue of an $n\times n$ matrix to an $n\times n\times n$ tensor. One approach is to construct a higher-order singular value decomposition. This has been worked out for the specific case of $2\times 2\times 2$ tensors by Ana Rovi in a M.Sc. thesis. Even for this simple case, there is no simple closed-form expression. A MatLab toolbox may be useful to implement the procedure.
Thank you for your answer and wonderful references.
|
2025-03-21T14:48:30.952543
| 2020-05-12T07:20:00 |
360103
|
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|
Stack Exchange
|
Eigenvalues of a matrix sum
I have a control system problem, which ends up in that the eigenvalues of the system matrix should have a negative real part, then the system is stable.
The system matrix is real but not symmetric. The values depend on the system parameters, but zeros will remain zeros and equal value will be equal.
M =
-126.49961 41.887902 1120.2778 13417.173 - 175.32356
-41.887902 - 126.49961 - 13417.173 1120.2778 - 28.050364
0.1282520 0. - 3.4974646 0. 0.5473535
0. 0.1282520 0. - 3.4974646 0.0875722
5.63488 0.9015356 - 153.66458 - 24.585101 0.
This one is fine, all eigenvalues have negative real part.
However, there is a disturbance that adds up:
Me =
- 308.54625 - 49.364927 0. 0. 0.
- 49.364927 - 7.8979926 0. 0. 0.
0. 0. 0. 0. 0.
0. 0. 0. 0. 0.
0. 0. 0. 0. 0.
One pair of eigenvalues of (M + Me) has positive real part.
Now, I can add a compensation like
Mc =
- r - x 0. 0. 0.
x - r 0. 0. 0.
0. 0. 0. 0. 0.
0. 0. 0. 0. 0.
0. 0. 0. 0. 0.
here with r = 85.41 and x = 41.88
and (M + Me + Mc) again has eigenvalues with negative real part if r and x are high enough.
The question is: How can I find sufficient values for r and x without numerically calculating the matrix eigenvalues? Is there some approximation?
Thanks for any hint
Alexander
You could use the Gershgorin circle theorem. For an $n\times n$ matrix $M$, consider $n$ circles $C_i$, $i=1,2,\ldots n$ in the complex plane centered at $M_{ii}$, with radius $\sum_{j\neq i}|M_{ij}|$. Each eigenvalue of $M$ lies in at least one of these circles. So you want to add to $M$ a correction that shifts all circles such that they lie in the left-half of the complex plane.
This is a constrained nearest stable matrix problem. You can try using the technique in Guglielmi, Lubich Matrix Stabilization Using Differential Equations https://doi.org/10.1137/16M1105840 .
|
2025-03-21T14:48:30.952718
| 2020-05-12T07:55:34 |
360105
|
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|
Stack Exchange
|
A soft question on the ABC conjecture
In Nature Vol 580, in an article about Shinichi Mochizuki's proposed proof of the abc-conjecture, there is a formulation saying:
The conjecture roughly states that if a lot of small primes divide two numbers $a$ and $b$, then only a few, large ones divide their sum, $c$.
Is that a relevant description, and if so, how to see the connection?
Posted on [math.se] about a month ago: Soft question about the ABC conjecture. (So far, no answers, only a few comments.)
This is not quite about the size of the primes, but rather about exponents in whuch they appear. If one of the numbers is divisible by a large power, that decreases its contribution to the radical. The ABC conjecture essentially states this can't happen for all of $a,b,c$ at the same time.
The more the word "roughly" contributes (and it does a lot here), the less the description is relevant!
@Wojowu - In a very big and very smooth number there have to be a lot of repetitions of small primes in the decomposition.
@Lehs That's indeed the case, but this should be seen as a special case of ABC, not the entire content of it. For instance, when proving Fermat's Last Theorem as a corollary of ABC, we also consider numbers like $a=x^4,b=y^4,c=z^4$. These needn't be "very smooth", instead we have all prime factors appearing in exponents larger than $1$. This is enough to use ABC.
If you are interested in the largest prime factor of $ab(a+b)$,
there is xyz conjecture.
Smooth solutions to the abc equation: the xyz Conjecture
This paper studies integer solutions to the ABC equation A+B+C=0 in which none of A, B, C has a large prime factor. Set H(A,B, C)= max(|A|,|B|,|C|) and set the smoothness S(A, B, C) to be the largest prime factor of ABC. We consider primitive solutions (gcd(A, B, C)=1) having smoothness no larger than a fixed power p of log H. Assuming the abc Conjecture we show that there are finitely many solutions if p<1.
|
2025-03-21T14:48:30.952915
| 2020-05-12T09:44:00 |
360112
|
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|
Stack Exchange
|
On the logarithmic derivative of an analytic function
I have reached upto this situation while exploring some properties of analytic functions which are bounded on the unit circle.
Suppose $f(z)$ be such an analytic function in the closed unit disc with $f(z)\neq 0$ in $|z|\leq 1$ and $|f(z)|$ attains its maximum at $z_0=e^{i\theta_0}$ on the unit circle $|z|=1. $ $f(z)$ also satisfies the conditions that
$$e^{i\theta_0}\frac{f'(e^{i\theta_0})}{f(e^{i\theta_0})}\geq 0$$
and $$\Re \left\{ e^{i\theta}\frac{f'(e^{i\theta})}{f(e^{i\theta})}\right\}\leq M,$$ for any real $\theta$ and for some fixed $M>0.$ I have been trying relating $|f'(e^{i\theta})|$ with $|f(e^{i\theta_0})|$ with inequality. Would I get an inequality of type $|f'(e^{i\theta})|\leq M|f(e^{i\theta_0})|$ under any circumstances? Suggestions will help me in going ahead.
Would it be easier to proceed if I take $ f(z)$ a polynomial?
What does your first condition mean? The left hand side is a complex number right?
@user159888 Yes, for some examples it will be a real number, but for most example it will not be right?
Could you explain the role of the first condition? Is the conclusione false, without?
@user250938 let $|f(z)|=R$; then the maximum condition for $R$ at some $|w|=r$ on $|z|=r$ implies $\partial R /(\partial r) (w)=0$ which translates into $wf'(w)/f(w)$ real by taking $\log f=\log R+i \arg f$ near $w$; the above is obviously true for non-zero minima or local maxima too, but now using specifically the global maximum property on the circle we get that $wf'(w)/f(w) \ge 0$ by a little proof involving Schwarz lemma
Great approach. That is right.
@user159888 I see, the first condition is always satisfied. By the way, once the quantity is real I see it is nonegative since it is a maximum (the modulus must be non decreasing in the radial direction), without using the Schwartz Lemma.
The function $f(z)=\exp\left ({\frac{1}{z} \log (1-z)}\right )$ is bounded in D and $zf'/f=\log (1-z)+i \arg (1-z)$. Its real part is upper bounded (but not lower bounded), the imaginary part is bounded. However $f'$ is unbounded and the estimate fails. This seems to be a counterexample. If it works, an approximation argument should show that any estimate as above fails on functions holomorphic up to the boundary.
|
2025-03-21T14:48:30.953091
| 2020-05-12T10:18:55 |
360113
|
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|
Stack Exchange
|
Do you recognise this setup of structure on a poset?
The setup is that we have a finite poset $P$, with a multiplicative rank function $r_{xy}:P\times P\rightarrow \mathbb{N}$, and a symmetric pairing $\langle\ ,\ \rangle:P\times P\rightarrow\mathbb{N}$. By multiplicative rank function we mean $r_{xy}r_{yz}=r_{xz}$ for $x\leq y \leq z$, but no other conditions on $r$, it need not vanish when $x$ and $y$ are incomparable, so I dont think $P$ is ranked in the usual sense. Our poset has a unique minimal element $\hat{0}$, and a distinguished maximal element $1$, but $1$ doesn't necessarily cover every element of $P$. From this, we get an associated automorphism $L:\mathbb{Q}[P]\rightarrow \mathbb{Q}[P]$ given by $$L(y)=\sum_{x\leq y}\frac{r_{\hat{0}x}}{r_{\hat{0}1}}\mu_P(x,y)x.$$
Extending our form to $\mathbb{Q}[P]$ by linearity, we are interested in the functions $f:P\rightarrow \mathbb{Z}$ that satisfy the functional equation: $$f(x)=\sum_{y\in P}f(y)\langle x,L(y)\rangle.$$
My question is, have you seen this bunch of structure before in any other posets? I was told that this might resemble Khazdan Lustzig type recurrences, but I couldn't see how to relate this to the general Khazdan-Luztig-Stanley polynomial of a poset. If you've seen this kind of thing in a different context, that would also be very useful to hear, I don't have much familiarity with these things, and any references that exhibit this setup would be helpful for me.
What does a "multiplicative rank function" mean? (Relatedly: is $P$ graded?)
Edited, I don’t believe $P$ is necessarily graded, since it can have saturated chains of different lengths. However if you know this setup in the graded case, I would be interested.
|
2025-03-21T14:48:30.953223
| 2020-05-12T10:32:14 |
360114
|
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|
Stack Exchange
|
Reference request: Name or use of this group of diffeomorphisms of the disc
Let $k \in \{0,\infty\}$, $G\subseteq \operatorname{Diff}^k(D^n)$ be the set of diffeomorphisms $\phi:D^n\to D^n$ of the closed $n$-disc $D^n$ (with its boundary) satisfying the following:
$
\phi(S_r^{n-1})=S_{r}^{n-1},
$
where for every $0\leq r\leq 1$, let $S_r^{n-1}:= \left\{
x \in D^n:\, \|x\|=r
\right\}$.
Is $G$ studied? If so what are some important papers on it?
It seems like a very intuitive object to be considered.
Note, when $k=\infty$ this is a subgroup of $\operatorname{Diff}_{\partial}^{\infty}(D^n)$, the group of diffeomorphisms of $D^n$ fixing both the origin and the boundary (well-studied by Hatcher for example).
Details:
I'm interested in either extreme, that is $C^{\infty}$ diffeomorphisms of $C^0$ (homeomorphisms).
Vague comment: It seems like your $
G \cong \sqcup_{0\leq r\leq 1} Diff(S^{n-1})$ or something along those lines.
Do you mean $C^\infty$-diffeomorphisms? the study of diffeomorphism groups of the circle is highly sensitive to the differentiability class.
I'm most interested in the $C^{\infty}$ or $C^0$ cases, more-so the former.
It seems that in the $C^\infty$ setting, the elements are (smooth) paths of diffeomorphisms of $S^{n-1}$, such that the endpoint at $0$ is in $O(n)$. In other words, this should be homotopy equivalent to $O(n)$. The reason is that we want $\phi$ to be a diffeo at $0\in D^n$ whose differential preserves the norm. On the other hand, for homeos, there is no boundary condition at $0$ by the Alexander trick.
|
2025-03-21T14:48:30.953369
| 2020-05-12T11:14:15 |
360117
|
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|
Stack Exchange
|
Isometric stratification preserves volume?
Let $K\subset \mathbb{R}^k$ be a non-empty compact subset let $f:K \to K$ be Lipschitz and surjective. If, moreover, $f$ is an isometry then clearly $f$ preserves the Lebesgue measure of $K$.
I consider the following generalization:
Let, $\{K_i\}_i$ be a collection of mutually-disjoint compact subsets of $K$ containing at-least two points, and for which $\bigcup_i K_i =K$ as a set, and let $f:K\to K$ be a Lipschitz function satisfying
$$
f|_{K_i} \mbox{ is an isometry}
$$
Intuition: So the isometry property of $f$ is stratified to each $K_i$
Does $f$ preserve the Lebesgue measure of $K$?
Obviously not, e.g., take $K_i$ to be singletons.
Thanks for pointing that out, $K_i$ should have at-least two points so that the isometry property means something..
You now excluded singletons, but there are still trivial counterexamples of the same flavour with $K_i$ not singletons (and infinite). [BTW you removed the assumption "is an isometry"after "$f|_{K_i}$" so the sentence is incomplete]
You still haven't addressed this. Take in $\mathbf{R}^2$ the diagonal matrix $(1,1/2)$ and take $K_i$ to be contained in horizontal lines...
Here is a result that may come close to what you want. As you know, if $(M_i,g_i)$, $i=0,1$ are two Riemann manifolds and $F:M_0\to M_1$ is an isometry, i.e., diffeomorphism such that $F^*g_1=g_0$, then the two manifolds have the same metric volumes, same curvatures etc.
A similar result holds for singular spaces. To better appreciate the following abstract definition consider the following simple experiment. Take two sheets of paper, of identical sizes. Then crumple one of them.
The resulting crumpled sheet is isometric with the unfolded sheet in the sense that there exists an obvious piecewise linear homeomorphism between them that sends a curve one one sheet to a curve one the other sheet of equal length.
You can imagine more complicated situations by opening a book to see its pages fanning out, glued along the book's spine. This is obviously not a manifold and if you fold some sheets you obtain a new space isometric to the original.
Now for the abstract definition. The appropriate category of spaces to work with are the compact subanalytic subsets of $\newcommand{\bR}{\mathbb{R}}$ of some $\bR^n$; see Sec. 2.1 of these notes for the exact meaning of subanalytic.
A subanalytic isometry between two compact subanalytic sets $X_0, X_1$ is a subanalytic homeormorphism $F:X_0\to X_1$ such that, for any subanalytic curve $\gamma:[0,1]\to X_0$ ($\gamma$ subanalytic, continuous injective map) the curve $F\circ \gamma$ has the same length as $\gamma$.
We say that two compact subanalytic sets $X_i\subset \bR^{n_i}$, $i=0,1$ are intrinsically isometric if there exists a subanalytic isometry between them.
Two intrinsically isometric subanalytic sets $X_0,X_1$ have the same dimension and
$$
\mu_m(X_0= \mu_m(X_1),\;\;m=\dim X_0=\dim X_1.
$$
Above, $\mu_m$ denotes the $m$-dimensional Hausdorff measure.
This is a special case of a theorem of J. Fu that shows much more. More precisely
$$\mu_k(X_0)=\mu_k(X_1),\;\;\forall k=0,1,\dotsc, m,
$$
where $\mu_k$ denotes the $k$-dimensional curvature measure of a compact subnalytic subsets. For example $\mu_0$ is the Euler characteristic. If $X\subset \bR^n$ is an $m$-dimensional smooth subanalytic manifold, then $\mu_{m-2}(X)$ m is, up to a universal multiplicative constant, the integral over $X$ of the scalar curvature of the induced metric. In this case these curvature measures appear in the tube formula $c=n-m$, $\newcommand{\bT}{\mathbb{T}}$
$$
{\rm vol}_n\big(\bT_r(X)\big)= \mu_m(X)\omega_c r^c+ \mu_{m-2}(X)\omega_{c+2}r^{c+2}+ \cdots,\;\;r\ll 1,
$$
where $\omega_d$ denotes the Euclidean volume of the
unit $d$-dimensional ball and $\bT_r(X)$ is tube of radius $r$ around $X$
$$
\bT_r(X)=\big\{ x\in \bR^n;\;\;{\rm dist}(x,X)\leq r\big\}.
$$
For more details and a proof of Fu's result see the same notes as above. Fu's theorem is in Section 4.5
|
2025-03-21T14:48:30.953680
| 2020-05-12T11:22:27 |
360118
|
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|
Stack Exchange
|
What is the geometric realization of the the nerve of a fundamental groupoid of a space?
It can be easily seen that there exists a functor $F:Top \rightarrow Grpd$ from the category of topological spaces to the category of groupoids defined as follows:
Obj: $X \mapsto \pi_{\leq 1}(X)$, where $\pi_{\leq 1}(X)$ is the fundamental groupoid of $X$.
Mor: ($f:X \rightarrow Y) \mapsto F(f):\pi_{\leq 1}(X) \rightarrow \pi_{\leq 1}(Y)$ where the functor $F(f)$ is defined as follows:
Obj: $x \mapsto f(x)$
Mor: $([\gamma]:x \rightarrow y) \mapsto [f(\gamma)]:f(x) \rightarrow f(y) $ where $[\gamma]$ is the homotopy class of path $\gamma$ in $X$ and $[f(\gamma)]$ is the homotopy class of path $f (\gamma)$ in $Y$.
Also it is not difficult to see that $F$ is well behaved with homotopy (for example in the chapter 6 of http://www.groupoids.org.uk/pdffiles/topgrpds-e.pdf)) that is in the sense that if $f,g: X \rightarrow Y$ are homotopic then the induced functors $F(f)$ and $F(g)$ are naturally isomorphic.
Also using this functor $F$ one can construct a 2-funntor $\tilde{F}: 1Type \mapsto Gpd$ where $1Type$ is the 2-category consisting of homotopy 1-types, maps and homotopy class of homotopies between maps and $Gpd$ is the 2-category consist of Groupoids, functors and natural transformations. Now according to Homotopy hypothesis of dimension 1 as mentioned in http://math.ucr.edu/home/baez/homotopy/homotopy.pdf this $\tilde{F}$ is an equivalence of 2-categories.
So from the above mentioned observations I felt that the functor $F$ is an interesting object of study.
Now if we consider the following sequence of functors:
$$
X \stackrel{F}{\mapsto} \pi_{\le 1}(X) \stackrel{N}{\mapsto} N(\pi_{\le 1}(X)) \stackrel{r}{\mapsto} r(N(\pi_{\le 1}(X)))
$$
where $N$ is the nerve functor and $r$ is the geometric realization functor.
My question is the following:
How the topological spaces $X$ and $r(N(\pi_{\leq 1}(X)))$ are related? It may be possible that my question does not make much sense when $X$ is a general topological space but then, does there exist any specific class of topological spaces $X$ which has a "good relation" with $r(N(\pi_{\leq 1}(X)))$?
I would be also very grateful if someone can refer some literature in this direction.
Thank you!
https://en.wikipedia.org/wiki/Postnikov_system
@DenisNardin Thanks for the comment. But can you please explain in little details how this helps in answering my question. Prior apologies if I sound stupid!
@AdittyaChaudhuri $|N \pi X|$ (using more usual notations...) is the $X_1$ from the Wikipedia article. It's a space satisfying $\pi_k(X_1, x) = 0$ for all $x \in X_1$ and $k \ge 2$, equipped with a map $X \to X_1$ inducing a bijection on $\pi_0$ and an isomorphism on all $\pi_1$'s.
PS: Note that this is essentially the statement of the dimension 1 homotopy hypothesis you mention (not exactly the same strictly speaking, but I think they're pretty much equivalent).
@NajibIdrissi Thank you for the valuable comment. I am trying to understand it. I did not know that this is equivalent to Homotopy Hypothesis in one dimension. This question just came to me out of only curiosity!
As far as I know, geometric realization of a simplicial set has structure of a CW complex... May be you can ask if you assume $X$ to be a CW complex, do you get an isomorphism of CW complexes between $X$ and $r(N(\pi_1(X)))$?
@PraphullaKoushik Yes that is true. But I did not mention about isomorphism anywhere in the question. But in this case I already got my answer from Najib Idrissi's comment.
So, you do know that there is an isomorphism when X is a CW complex?
@PraphullaKoushik No according to Najib Idrissi's comment it is $X_1$ coming from Postnikov system of $X$. But I am yet to find out how. I am trying.
@PraphullaKoushik $X$ and $r(N(\pi_{\le 1}(X)))$ are not isomorphic even when $X$ is a CW-complex. Consider the case of $S^2$. Every loop is trivial, and $\pi_{\le 1}(X)$ is equivalent to the trivial category. So its geometric realization is contractible.
@AdittyaChaudhuri No problem at all. Could you say what is not so clear? I would suggest the following steps: show that its path components are the path components of $X$ so that you can restrict yourself to a path connected $X$; show that the higher homotopy groups of $|N \pi X|$ vanish (there is a standard construction to build a contractible covering space, it is somewhere in Hatcher, look for "classifying space"); and show that the map $X \to |N \pi X|$ (obtained by playing with adjunctions) is an isomorphism on $\pi_1$. (This can also all be deduced from Ronnie Brown's answer below.)
@NajibIdrissi Thank you very much for the comment. I got what you said. I will check the details rigorously. According to your comment I guess if in particular $X$ is a homotopy 1-type then then the map $X \rightarrow |N\piX|$ is a weak homotopy equivalence . Now I feel that if $X$ is an $n-$ type and instead of fundamental groupoid functor if we start with some "appropriately generalised" n-functor then we will get a weak equivalence from $X$ to its " appropriately generalised" geometric realisation. (appropriately generalised as "the usual realisation" is a functor , not a n-functor. )
@NajibIdrissi I also guess if we move up the ladder i.e if $X$ is an arbitrary $C W$ complex then if we start with "appropriately generalised notion of infinity functor" instead of fundamental groupoid functor then we will get a weak equivalence from $X$ to it's appropriately generalised notion of " geometric realisation " of the "appropriately generalised nerve" of the image of $X$ under that "infinity functor". "I think this is what you meant to say about it's relation with Homotopy Hypothesis". Please let me know if my guess is wrong. I have very little background in these topics.
@AdittyaChaudhuri If you look at my answer, you generally do not have a canonical map $X\to \lvert N (\Pi_1) \rvert$. If $X$ is a CW-complex, you can use Whitehead's theorem to construct a homotopy inverse to the map $\lvert \operatorname{Sing}(X)\rvert \xrightarrow{\simeq} X$ then compose with the natural map on the left from my answer, but such a homotopy inverse is not uniquely defined (it is defined only up to a contractible ambiguity). If $X$ is not a CW-complex, such a map might not even exist (take a finite topological model of $S^1$, for example)..
@HarryGindi So are you trying to say that $\lvert N(\prod_1) \rvert$ is not exactly $X_1$ in the Najib Idrissi's comment because of that "contractible ambiguity" you mentioned in your comment above?
@AdittyaChaudhuri I'm sorry but you have to ask Najib what he meant. It's not the case that you can always build a literally continuous map (my example of the finite model for S^1 is the counterexample).
@HarryGindi But if $X$ is a CW complex then will $\lvert N(\pi_{\leq X}(X)) \rvert$ be always $X_1$?
What do you mean X_1? Do you mean the first postnikov stage? Then yes. If you mean the 1-skeleton of X, the answer is no.
@HarryGindi I meant first postnikov stage. Thanks
The inclusion of groupoids into simplicial sets is fully faithful. Its left adjoint, $\Pi_1$ is given by left Kan extension of the functor $\Delta\to \mathcal{Gpd}$ sending the n-simplex to the contractible groupoid with objects $\{0,...,n\}$.
The entirety of the data of the homotopy type of the space $X$ is contained in its singular simplicial set, which is canonically a Kan complex. In particular, the fundamental groupoid functor you've written above is canonically isomorphic to the composite $\Pi_1 \circ \operatorname{Sing}$. Then we have a universal natural transformation $$\operatorname{Sing} \to N\circ \Pi_1\circ \operatorname{Sing}$$ given by the unit of the adjunction $Π_1\dashv N$. Taking geometric realizations, we obtain a span
$$\lvert \Pi_1 \rvert \cong \lvert N\circ \Pi_1\circ \operatorname{Sing}\rvert \leftarrow \lvert \operatorname{Sing}\rvert \xrightarrow{\simeq} \operatorname{id_{\mathbf{Top}}},$$
where the righthand map is the counit of the adjunction between simplicial sets and topological spaces $\lvert \bullet \rvert \dashv \operatorname{Sing}$, and is a natural weak homotopy equivalence by a theorem of Quillen.
So the lefthand map exhibits the nerve of the fundamental groupoid as stage 1 of the Postnikov System as mentioned by Denis in the first comment.
Thanks for the answer. Here in the first line by "inclusion of groupoids into simplicial sets" do you mean to say nerve functor? Also you mentioned "Universal natural transformation"... I came across that term in Mac lane (Working Mathematician -Pg 39 2nd edition).. did you mean to say in that way? But in that case a universal natural transformation is defined between 2 functors $T_0,T_1: C \rightarrow C \times 2$ where $2$ is the category $0 \rightarrow 1$ and $C$ is any category. I am not getting how we use this notion here. Also are we using the fact it is Universal?
1.) Yes, I mean along the nerve. 2.) By universal natural transformation, I mean 'Universal Morphism' as in this wikipedia article https://en.wikipedia.org/wiki/Adjoint_functors . The unit and counit of an adjunction are always universal in this sense.
Thanks for the explanation.
@AdittyaChaudhuri I just wanted to mention: This is a general pattern. At least when we have a Kan complex K, we can write down its map to its n+2-coskeleton, which is also a Kan complex.Then the unit map for the coskeleton adjunction, $K\to \operatorname{Cosk}{n+2} K$ exhibits $\operatorname{Cosk}{n+2}$ as the nth truncation, that is, the nth stage of the postnikov system. But note that when K is a Kan complex, the $1+2=3$-coskeleton for is canonically isomorphic to $NΠ_1(K)$.Taking the case $K=\operatorname{Sing}(X)$ gives the direct relation to topology by the same span as above.
Thank you very much for this remark on this "general pattern". I am not much well versed with these notions. I will take some time to completely understand what you are trying to say in the above comment.
In the answer you took geometric realisation of $\prod_1$ but $\prod_1$ is category valued . So did you mean to say $\lvert N \circ \prod_1 \rvert$ instead of $\lvert \prod_1 \rvert$ ?
yes. that's right.
Just for confirmation! In that case $\lvert N \circ \pi_1 \rvert$ will be same as $\lvert N \circ \pi_1 \circ Sing \rvert$? Is it?
@AdittyaChaudhuri yeah, implicitly they are two different constructions that are canonically isomorphic.
A very rough argument that can be (easily) formalized is as follows:
We have a notion of $\infty$-groupoids. These are like groupoids, but they have homotopies between morphisms, homotopies between homotopies, and so on. Every topological space presents an infinity groupoid by taking the objects to be points, morphisms to be paths, morphisms between morphisms to be homotopies of paths, etc.
If one takes the connected components of this we get the path components of our space. If one takes the connected components of the automorphism group of a point, we get the fundamental group. If one takes the connected components of the morphisms from a constant path to itself we get the second homotopy group, and so on.
Hence this $\infty$-groupoid can be seen as presenting all the homotopical information of our space. Now the fundamental groupoid is given by taking this $\infty$-groupoid and taking connected components of the morphisms between points to get an actual groupoid. Now we just remarked that the higher homotopy information (homotopy groups after the first) are all contained in the connected components of the morphism sets. By discretizing these sets we are removing all higher homotopical information.
So what should we expect when we realize it? Well we should expect that $\pi_0 , \pi_1$ are that of are spaces, but $\pi_n$ for $n>1$ is trivial. This is exactly what happens, we get the first Postnikov space for $X$, i.e. $K(\pi_1(X),1)$ (or really a disjoint union of these for each path component).
Thanks for the answer. Can you please explain in little detail about what do you mean by Connected components of Morphisms? Also I did not get what is meant by "discretizing the morphism sets"? Did you mean to say all higher $k-$ morphisms are identity?
Connected components of a groupoid mean the isomorphism classes. You are correct in your guess of what "discretizing" meant.
I have now (May 13) partitioned the answer into the blocks 1,2, as I think 2 is the simpler answer!
1 I hope the book Nonabelian Algebraic Topology will answer the question for you.
A groupoid is level one of a structure called a crossed complex which is a kind of nonabelian chain complex but also with the groupoid structure in dimensions $\leqslant 1$, which operates on the higher dimensional stuff. There is a homotopically defined functor $\Pi$ from the category of filtered spaces to crossed complexes, using the fundamental groupoid and relative homotopy groups and also a functor $\mathbb B$ from crossed complexes to filtered spaces such that $\Pi \mathbb B$ is naturally equivalent to the identity. This setup is particularly useful for CW-complexes with their standard cellular filtration.
Part of the thesis of the book is to use structured spaces, in this case filtered spaces, to get to link various dimensions, and in this way to use strict algebraic structures. Also the proofs use higher cubical homotopy groupoids, and are non trivial, but can involve the intuitive idea of allowing "algebraic inverses to subdivision", that is generalising to dimension $n$ the usual composition of paths. This is more difficult to do simplicially.
Part I of the book deals with dimensions $0,1,2$ where it is easier to explain the intuitions, and history. Section 2.4 discusses the classifying space of a group and of a crossed module, but the groupoid case comes in Chapter 11.
2 But an answer can easily be put: a groupoid $G$ has a set of objects say $G_0$ and its classifying space $BG $
also contains the set $G_0$.The fundamental groupoid $\pi_1(BG, G_0)$ is naturally isomorphic to $G$! That is, you need the concept of the fundamental groupoid $\pi_1(X,S) $ on a set $S$ of base points, which is formed of homotopy classes rel end points of paths in $X$ with end points in $S$. You can find this developed in the book "Topology and Groupoids". The notion itself was published in my paper
``Groupoids and Van Kampen's theorem'', Proc. London Math. Soc. (3) 17 (1967) 385-40.
The use of this Van Kampen Theorem involving a set of base points was to allow a theorem which could compute fundamental groups of spaces, such as the circle, where the traditional theorem did not apply.
See also this mathoverflow link.
Thank you Sir very much. I will definitely read the book you mentioned & will try to connect it to answer my question. But geometrically what is the meaning of geometrical realisation of the nerve of a fundamental groupoid? . So according to Najib Idrissi's comment it is $X_1$ which is coming from postkinov system of $X$ i.e sequence of spaces$X_n$ and maps $\phi_n:X \rightarrow X_n$ which induce isomorphisms from $\pi_i(X)$ to $ \pi_i(X_n)$ for $i \leq n$, bijection on $\pi_0$ and $\pi(X_n)=0$ for $i > n$. But how this is related to geometric realisation I am not getting properly!
Sir, in the comment above I assumed $X$ is a C W complex and "by sequence of spaces" I mean an inverse system of spaces.
Thank you Sir for the edit. I am trying to understand it.
|
2025-03-21T14:48:30.954792
| 2020-05-12T13:17:10 |
360127
|
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|
Stack Exchange
|
Branching from GL(a+b) to GL(a) x GL(b)$ using Gel'fand-Cetlin patterns
If one iterates the multiplicity-free branching rule from $\operatorname{GL}(n)$ representations (finite-dim, over $\mathbb C$) to $\operatorname{GL}(n-1)$ all the way down to $\operatorname{GL}(0)$, one obtains triangular "Gel'fand-Cetlin (or Zeitlin or Tseytlin or...) patterns", indexing a basis of the representation. The differences in the row sums indicates the weight of the basis element.
If one only iterates $b$ times (for $n=a+b$), this gives a branching law from $\operatorname{GL}(a+b)$ to $\operatorname{GL}(a) \times \operatorname{GL}(1)^b$, in terms of Gel'fand-Cetlin trapezoids of height $b$. Using the crystal structure on SSYT, in bijection with GC patterns, one should be able to make these trapezoids into a $\operatorname{GL}(b)$ crystal.
What are the highest-weight Gel'fand-Cetlin trapezoids, giving a branching law from $\operatorname{GL}(a+b)$ to $\operatorname{GL}(a)\times\operatorname{GL}(b)$?
As the above suggests, I think answering this is a matter of following bijections -- I just would rather quote a reference than reinvent the wheel. (Probably p385 of Zhelobenko's book? I don't have access to the whole thing, and the snippet Google provides has too much undefined notation for me to readily compare to GC patterns. If I'm reading correctly, he only gives a positive formula for fundamental representations, and uses a determinant to give a nonpositive formula in general.)
For a background and definition of GT-patterns, see https://www.math.upenn.edu/~peal/polynomials/gtpatterns.htm
Gelfand-Tsetlin trapetzoids are in bijection with skew SSYT, which do have a crystal structure. Is this what you are asking for?
Probably, though I suppose it's possible that there are multiple crystal structures and I want a different one. Anyway I want the unraisable ones.
|
2025-03-21T14:48:30.954948
| 2020-05-12T13:49:20 |
360130
|
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"Tomo",
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|
Stack Exchange
|
Local acyclicity with respect to a sheaf
I am trying to understand the definition 2.12 in SGA 4.5, chapter 7. The multitute of localizations confuses me, as I still do not understand very basic algebraic geometric notions.
To get some understanding, I tried to prove the following:
If $X\xrightarrow[]{id} X$ is locally acyclic with respect to $\mathcal{F}$, then $\mathcal{F}$ is locally constant.
Is this true? If yes, could you provide an explicit proof to help me?
When $\mathcal F$ is a constructible sheaf, this follows by SGA 4 IX 2.11 as the property that the identity map is locally acyclic implies that all the specialization morphisms are isomorphisms.
$\DeclareMathOperator\Spec{Spec}$This is an expansion of the comment given by Tomo: the locally acyclic condition is equivalent to that every specialization of geometric points induces isomorphism of stalks of étale sheaf $F$, and in the noetherian case, the latter is equivalent to $F$ being locally constant when $F$ is constructible.
Recall the definition 2.12 given in SGA 4.5 (also see the stacks), for a geometric point $\overline{x}:\Spec(k(x)^\text{sp})\rightarrow X$ lying over $x\in X$, consider the embedding $j$ of variety of vanishing cycle of identity map at $\overline{x}$ into the étale local scheme at $\overline{x}$, which is exactly $\overline{t}\rightarrow \Spec(O^\text{sh}_{\bar{x},X})$ where $\overline{t}$ is a geometric point of $\Spec(O^\text{sh}_{\bar{x},X})$ and $O^\text{sh}_{\bar{x},X}$ is the strict henselization of $O_{x,X}$ with respect to the geometric point $\overline{x}$. Such $\overline{t}$, if we understand it as a geometric point at $X$, is exactly the geometric points which are specialization of $\overline{x}$ in the sense that $\overline{x}$ factors through every étale neighborhood of these points.
For the identity morphism to be locally acyclic, it's required that the pullback of étale sheaf $F$ along $j$ is isomorphism, this is translated into the condition that restriction morphism of (étale) stalks form 'generalization' point to the 'specialization point' is isomorphism: $F_{\overline{x}}\cong F_{\overline{t}}$.
If $F$ is constructible, for any geometric point $\overline{s}$ on $X$, the stalk $F_{\overline{s}}$ is finite, denoted as $M=\{\sigma_{1},\dotsc,\sigma_{n}\}$, and consider the locally constant sheaf $\underline{M}$ on $X$, lift $\{\sigma_i\}$ to sections on an arbitrary étale neighborhood $U$, then these sections determine a sheaf morphism on $U$: $f:\underline{M}\rightarrow F|_{U}$. Here the condition of 'constructible' is important because the finiteness of stalk makes it possible to lift to a morphism between constant sheaf and $F$.
By descent, $U$ has only finite irreducible components, if we assume $U$ is affine without loss of generality. And after removing irreducible components not containing $\overline{s}$, we have $\underline{M}\cong F|_{U}$ because the stalks at generic points of irreducible components will be isomorphic to $F_{\overline{s}}$.
Here we don't distinguish between the 'real point' $s\in X$ and the geometric point $\overline{s}$ lying over $s$ because $\overline{s}$ is specialization of $\overline{t}$ iff $s$ is specialization of $t$.
|
2025-03-21T14:48:30.955171
| 2020-05-12T14:20:51 |
360131
|
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|
Stack Exchange
|
May-McClure "A reduction of Segal conjecture"
I am looking for a digitalized version of paper by J.P. May and J. McClure A reduction of Segal conjecture, as I need it to understand some lemma from Kuhn's Tate Cohomology and Periodic Localization of Polynomial Functors. The paper was published in Current Trends in Algebraic Topology, Canadian Mathematical Society Conference Proceedings, 1982.
I hope this is not breach of any rules - given the current situation, I cannot go to University's library and check the hard copy, and I cannot find any version of this article on the internet. Any help would be highly appreciated!
(If nobody has an access to a digitalized version, could sbdy quote what Corollary 4 is saying?)
I found it on professor May's web site at http://math.uchicago.edu/~may/PAPERS/42.pdf
Since this link might disappear, it is also archived at the Wayback Machine.
It works, thanks!
|
2025-03-21T14:48:30.955267
| 2020-05-12T14:58:11 |
360134
|
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|
Stack Exchange
|
abc triples with a symmetric condition
Recently, I have asked a question about the balance of abc triples. Since then I have come up with a different idea of a new criterion that somewhat combines balance and magnitude and has two advantages:
it does not imply a (somewhat arbitrarily chosen) threshold, like e.g. the idea of "good abc triples" or the question when to consider a triple "sufficiently balanced".
it is symmetric in $a,b,c$, meaning that it can be possibly modelized in terms of (hyper-?)elliptic curves, with things happening in $\mathbb Z$ rather than in $\mathbb N$.
If we denote the usual abc triples by "c-abc triples", my idea would be to introduce a subset called "a-abc triples" or for short, "a-triples" (maintaining the hyphen to avoid grammatical ambiguities), defined as follows:
A triple $(a,b,c)$ with $a<b$ and $a+b=c$ is an a-triple iff $a>\text{rad}(abc)$.
It is natural to define the a-quality of such a triple as $\frac{\log a}{\log\text{rad}(abc)}>1$ .
Since we have automatically $b,c>\text{rad}(abc)$ as well, we could consider equivalently
triples $(a,b,c)\in\mathbb Z^3$ with $a+b+c=0$ and $|a|,|b|,|c|>\text{rad}|abc|$.
It turns out that $95$ of the $241$ known "good" abc triples (i.e. with quality $\geqslant1.4$) are a-triples. The 10 ones with best a-quality are the following:
rk quality size merit a/b a-quality
66 1.4420 15.51 15.53 0.6363 1.4038
95 1.4316 13.28 12.18 0.8366 1.3948
151 1.4158 23.92 24.63 0.5997 1.3906
173 1.4121 29.38 31.48 0.3006 1.3815
9 1.5270 9.78 11.02 0.1139 1.3723
105 1.4290 10.44 8.74 0.6055 1.3710
240 1.4003 16.79 14.68 0.6427 1.3662
43 1.4526 9.43 8.28 0.3550 1.3629
28 1.4646 21.58 25.80 0.0302 1.3605
72 1.4403 16.98 17.38 0.1058 1.3538
Note that the penultimate one of those is quite imbalanced, but still has a good a-quality. As the size grows, the contribution of the imbalance is mitigated by taking the logs. Or look at the third one in the list (rank 151): "big" in size, very balanced, thus the a-quality is "hardly" smaller than the (c-)quality.
Looking at a-triples might shed some new light on the abc conjecture. My first question:
Are there still infinitely many a-triples?
We believe there are infinitely many a-triples and here are
two partial proofs.
The elliptic curve $x^3+y^3=6z^3$ has infinitely many coprime
integer solutions. Take $a=x^3,b=y^3,c=6z^3$.
We have $\log a \approx \log b \approx \log c$ by
the theory of elliptic curves. In addition abc implies
$\log a < (1-C) \log c$ can't happen infinitely often for fixed
$C > 0$, since this will give infinitely many abc triples violating
the abc conjecture.
Since infinitely many prime squares divide $abc$, we have a-triple.
There are infinitely many coprime
integer solutions to $y^2=x^3 + 2 z^6$ and take $a=x^3,b=2z^6,c=y^2$.
Again we have $a,b,c$ with roughly equal logarithms.
Here is sage session without taking the radical, just log(min(x^3,2z^6,y^2)/log(xyz):
sage: E=EllipticCurve(QQ,[0,2]);P=E.gens()[0]
sage: for k in [ 2 .. 20]:
....: x1,y1=(k*P).xy()
....: a=numerator(x1);b=denominator(x1).isqrt();c=a^3+2*b^6;
....: c=c.isqrt()
....: A=min(abs(_) for _ in [a^3,2*b^6,-c^2]);ra=prod(u for u,_ in factor(ZZ
....: (2*a*b*c),limit=10^6))
....: print k,RR(abs(A)).log()/RR(abs(ra)).log(),RR(ra).log(10)
2 0.622930427076373 3.38273726576133
3 0.882982381624213 7.14783363090242
4 0.994840130747121 14.5358637500049
5 0.962868637411342 24.2686482149114
6 1.00233452517062 33.6432311660788
7 0.921720461403240 47.6950749995341
8 0.999678413471366 60.8697877606209
9 0.982053057678340 78.1486267664457
10 0.984848581223119 96.8940057210651
11 0.995456196322965<PHONE_NUMBER>81424
12 1.00640878471394<PHONE_NUMBER>69602
13 0.983087805421523<PHONE_NUMBER>47524
14 0.989966424580658<PHONE_NUMBER>41615
15 1.00341504891259<PHONE_NUMBER>08444
16 0.986803892310106<PHONE_NUMBER>25422
17 0.997502907017965<PHONE_NUMBER>71681
18 1.00125521214337<PHONE_NUMBER>09346
19 0.997604588937612<PHONE_NUMBER>03754
20 0.998545019903716<PHONE_NUMBER>79189
So do I understand correctly: for your second curve, you have found numerically that for many of the integer solutions, the ratio "without taking the radical" approaches $1$ quite closely, and so you are hoping that more often than not, the radical is small enough that the ratio, if taken "with taking the radical", exceeds $1$? If so, it seems a bit fishy to me... Could you also display the magnitude of each $xyz$, in order to give an idea by which kind of factor the radicals need to be smaller for the respective triple to be indeed an a-triple?
Also, I don't seem to see a reason why the ratio even "without taking the radical" should not be able to exceed $1$ in theory. But curiously enough, it does not seem to happen. May you try to compute the radical for #12, or are the numbers way too big for that?
@Wolfgang I take partial radical: the radical of (x^3,2z^6,-y^2) is at most xyz, but in addition rad(xyz) might be smaller and I don't take this into account. You can run the sage session in a browser, check the web for that.
Yes I know. The idea is very clever, maybe there are other elliptic curves with even higher powers and still infinitely many coprime integer solutions, doing an even better job. Hopefully Noam Elkies comes along here...
@Wolfgang I edited with radical partial factorization up to 10^6 and log_10(xyz).
|
2025-03-21T14:48:30.955615
| 2020-05-12T15:23:54 |
360135
|
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|
Stack Exchange
|
Rank of a block of an invertible matrix
Let $A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$ be an invertible matrix where $A_{11}$ is square. Let $A^{-1} =: B = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix}$ and shape of $B_{11}$ is same as $A_{11}$.
What is known about the relationships between the ranks of $A_{21}$ and $B_{21}$? For example, if either $A_{11}$ or $A_{22}$ is invertible, then it can be easily shown using the block inverse formula that the ranks are equal. I presume when this is not true, the equality does not hold in general. Is this correct? In that case, what bounds are known?
If the diagonal blocks are not invertible, the other blocks should be of full rank, otherwise it is hard to see the whole matrix being invertible. Do you know of counterexamples? Gerhard "Full Rank Comes From Somewhere" Paseman, 2020.05.12.
Ah: Take some permutation matrices. Perhaps this will suggest some conditions the ranks can satisfy. Gerhard "I See Some Counterexamples Now" Paseman, 2020.05.12.
That paper from Strang and Nguyen gives the answer: https://epubs.siam.org/doi/abs/10.1137/S0036144503434381
See "theorem 2.1: (nullity theorem). Complementary submatrices of a square matrix and its inverse have the same nullity". In your case the complementary submatrices are exactly $A_{21}$ and $B_{21}$, so their ranks are equal.
|
2025-03-21T14:48:30.955736
| 2020-05-12T15:51:19 |
360138
|
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|
Stack Exchange
|
$ \overline{(A-A)}\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0 $?
Let $X$ be a separable Banach space and $A$ is a subset of $X$ such that
$$
A\cap\overline{B}(0,r) \text{ is weakly compact, } \forall r>0.
$$
Can we say that :
$$
\overline{(A-A)}\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0
$$
with $A-A=\{a-b:a,b\in A\}$.
Why are you asking this question again? Nik Weaver gave you a counterexample in dimension $2$.
No I made a mistake in the question I want the closure of $(A-A)$
OK; I'll answer when I have time.
The answer is negative. In $X :=\ell_2 \oplus_\infty \ell_1$ let
$x_n = (M_n \delta_n, 2 e_n)$ and $y_n = (M_n \delta_n, e_n)$, where $(\delta_n)$ is the unit vector basis for $\ell_2$ and $(e_n)$ is the unit vector basis for $\ell_1$ and $M_n \uparrow \infty$ fast. Let $B$ be the convex hull of the $x_n$ and $y_n$ and let $A$ be the closure of $B$. Then $A$ fails the conclusion since $A-A$ contains the unit vector basis of $\ell_1$. It remains to check that $A$ satisfies the hypothesis if $M_n \uparrow \infty$ sufficiently quickly. Suppose that $x$ is in $B\cap\overline{B}(0,r)$. Write $x$ as a convex combination of the $x_n$ and $y_n$ and let $a_n$, respectively, $b_n$ be the coefficient of $x_n$, respectively, $y_n$. Since $\|x\| \le r$, $a_n + b_n \le r/M_n$. Let $Q_N$ be the natural projection from $X$ onto the closed span of $\{(0,e_n) : n\ge N\}$. Thus
$$\|P_nx\| = \sum_{n=N}^\infty 2a_n+b_n \le 2r\sum_{n=N}^\infty 1/M_n,$$
which we can assume goes to zero as $N\to \infty$. By density of $B$ in $A$, this upper estimate is valid also for all $x\in A$. This implies that the projection of $A\cap\overline{B}(0,r)$ into $\ell_1$ is (norm) totally bounded, and thus $A\cap\overline{B}(0,r)$ is contained in $E+F$ with $E$ weakly compact and $F$ norm compact (where $E$ is the ball of radius $r$ in $\ell_2 \oplus \{0\}$), and hence $A\cap\overline{B}(0,r)$ is relatively weakly compact. Since it is clearly weakly closed, this shows that $A$ satisfies your hypothesis.
EDIT: After writing this, I realized that it was really, really silly to introduce $\ell_2$; I could just as well have used the scalar field and replace $M_n \delta_n$ with $M_n$. This gives an example in $\ell_1$, where weak compactness and norm compactness are the same. A similar argument works to give a negative answer to the OP's question in every non reflexive Banach space (one needs to know that a non reflexive space contains a normalized basic sequence that is not relatively weakly compact).
Thank you very much, sir, for the time you have given me. The reason why I asked this question, I will ask it as another question. I couldn't answer him
|
2025-03-21T14:48:30.955924
| 2020-05-12T16:01:04 |
360140
|
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|
Stack Exchange
|
What do you call $\operatorname{diam} (A)^d/\mathcal{L}^d (A)$ for $A \subseteq \mathbb{R}^d$ convex?
If $A \subseteq \mathbb{R}^d$ is convex, is there a more or less established name for the quantity $$\operatorname{diam} (A)^d/\mathcal{L}^d (A),$$
where $\mathcal{L}^d (A)$ is the Lebesgue measure of $A$?
The quantity is invariant under isometries and scaling.
For $d=2$ and $A$ a triangle, it is is an extension from the condition preventing angles from being too acute.
isn't this some kind of measurement of the isotropy/ellipticity? I don't lknow of any name for that quantity, though...
|
2025-03-21T14:48:30.956002
| 2020-05-12T16:29:26 |
360141
|
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|
Stack Exchange
|
Subschemes of the affine line over a domain
Let $R$ be a domain with affine spectrum $S$ and consider the scheme $X=\mathbb A^1_R=\operatorname {Spec}R[T] $ over $S$.
Let $P\subset R[T]$ be an ideal with $P\cap R=0$ and let $Y\subset X$ be the associated subscheme. The extension of $P$ to $(\operatorname {Quot}(R))[T])$ is principal, generated by some polynomial $f$ which we may assume has coefficients in $R[T]$. What is the relationship between $Y$, the subscheme $V(f)\subset X$ and (probably) inverse images of subschemes in $S$ ?
(I have asked this question three days ago on math stackexchange but got no answer nor comment)
Edit: I'm also interested in knowing if one can say more under the assumption that $P$ is prime.
$V(f)$ is the union of $Y$ and a subscheme of $X$ which is the pull back of some subscheme from $S$.
@Mohan: Well of course, that's the obvious guess, as I hinted in my question. I'm asking for a detailed, rigorous proof.
Are you assuming $R$ is Noetherian? I was.
No, I was not assuming $R$ noetherian. But I would be happy to read a proof of your claim assuming noetherianness.
Then, use Krull's principal ideal theroem and look at the prime decomposition of $(f)$. Set theoretically this is the union of $Y$ and the pull back of a closed subset of $S$, since all minimal primes containing $f$ have height one.
@mohan I can't make head or tail of what you write. Of course the minimal primes containing $f$ have height one but what has this got to do with the morphism to $S$ , the pull back of closed subsets of $S$ and the hypothesis that $P\cap R=0$?
The hypothesis says, $(f)=P\cap I$ where $I\cap R\neq 0$ and then $\sqrt{I\cap R} R[T]=\sqrt{I}$, since $I$ has height one.
"The hypothesis says, (f)=P∩I where I∩R≠0" No hypothesis says that for the good reason that $I$ is never mentioned in my question and you don't say what it represents. In the most charitable interpretation I guess $I$ means the ideal defining $Y$ and then your inequality is not always true. Anyway I'm getting a bit tired of trying to understand unjustified statements that I find quite enigmatic. this is not the way this site works: please write a complete answer in the answer box if you are capable and willing. Else, no hard feelings and I wish you all the best .
Here maybe is a helpful example: $R=k[a,b,c,d], P=(aT^2-b, cT^2-d)$. Then $f=1$ (so the generic fiber of $Y \rightarrow S$ is empty), $Y$ is finite degree $2$ over ${ad-bc=0} - {a=c=0}$ $\cup$ line over ${a=b=c=d=0}$.
@David Your example is very interesting but it doesn't quite satisfy my hypotheses. Indeed you have $P\cap R\neq 0$, contrary to what I suppose, since $c(aT^2-b)-a(cT^2-d)= -bc+ad\neq 0$ is in $P\cap R$. Nevertheless I'm sure one could extract a nice example from your situation and I thank you for this comment.
@lefuneste Sorry I forgot about $P \cap R = 0$. Instead $P=(aT^3-bT, cT^3-dT)$ and $f=T, Y$ as above $\cup {T=0}$.
Perfect, @David and thanks a lot. I would be very happy if you wrote down the answer in general, guided by your edifying example. (I tried to upvote your comment, but unfortunately couldn't, probably because I am a new user on this site)
@lefuneste I think I don't have much in general to add to what's already formulated in your question. I merely gave an example to show that special fibers over the base can have various different qualities from the generic fiber. Think of more examples with special fibers where the leading coefficient of $f$ vanishes but not all coefficients.
|
2025-03-21T14:48:30.956278
| 2020-05-12T17:02:30 |
360144
|
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|
Stack Exchange
|
Analog of the Birkhoff's ergodic theorem for the sequence of squares
Consider a dynamical system $(X, \mathcal{B}(X), \mu, T)$ where $(X, \mathcal{B}(X), \mu)$ is a measure space and $T$ is a measure-preserving, invertible transformation.
Then by the classical Birkhoff's ergodic theorem if $p\ge 1$, then for any $f\in L^p(X, \mu)$ the sequence
$$
\mathcal{M}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(T^n x)
$$
converges for almost every $x\in X$.
$\textbf{Question:}$ Is it true that the sequence
$$
\mathcal{A}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(T^{n^2} x)
$$
is convergent a.e. for $f\in L^p(X,\mu)$ and $p\ge 1$?
I'll be more than happy to see the answer to my naive question for a particular case when: $X=\mathbb{Z}$, $\mu$ is a counting measure and $T$ being a regular shift operator $Tf(x)=f(x+1)$. In this case
$$
\mathcal{A}_N f(x):=\frac{1}{N}\sum_{n=0}^N f(x+n^2), \qquad x\in\mathbb{Z}
$$
for $f\in \ell^p(\mathbb{Z})$.
It seems that the problem reduces to the study of the boundedness of the maximal function:
$$
f\mapsto\sup_N \mathcal{A}_N |f|.
$$
Is there a smart way of getting this boundedness from the corresponding result in the continuous setting? I tried to apply some known transference principles, but it seems to me, that the fact that there are large gaps between squares, namely $(n+1)^2-n^2\simeq n$, causes some trouble. Please excuse me if I'm overlooking something obvious here.
No - the sequence of squares is universally bad which was proved by Buczolich and Mauldin. I will quote from Tom Ward's review of their paper Divergent square averages, Ann. of Math. (2) 171 (2010), no. 3, 1479–1530.
A consequence of J. Bourgain's work [Inst. Hautes Études Sci. Publ. Math. No. 69 (1989), 5–45; MR1019960] is an ergodic theorem along squares, answering earlier questions of Bellow and Furstenberg: If $(X,\mathcal B,T,\mu)$ is a measure-preserving system, then the non-conventional ergodic averages
$$
\frac1{N} \sum_{n=0}^{N-1} f(T^{n^2} x)
$$
converge almost everywhere for $f\in L^p$ with $p>1$. Here a comprehensive - and negative - answer is given to his question of whether the result extends to all of $L^1$. The authors show that the sequence $(n^2)$ is universally bad: for any ergodic measure-preserving system there is a function $f\in L^1$ for which the above averages fail to converge as $N\to\infty$ for $x$ in a set of positive measure.
PS The Birkhoff theorem does not apply to your ``particular case'' as it requires the presence of a finite invariant measure.
Thanks for the comprehensive answer!
If $X=\mathbb{Z}$, $\mu$ is the counting measure, and $T$ is the shift operator given by $Tf(x)=f(x+1)$, then for all real $p\ge1$, $f\in\ell^p(\mathbb{Z})$, and $x\in\mathbb{Z}$, by Hölder's inequality,
$$
|\mathcal{A}_N f(x)|\le \frac1N\,\sum_{n=0}^N|f(x+n^2)|
\le\frac1N\,\|f\|_p\,(N+1)^{1-1/p}\to0
$$
and hence $\mathcal{A}_N f(x)\to0$ as $N\to\infty$.
Oh, that's right! So the problem of everywhere convergence trivializes on $\mathbb{Z}$. The question about the boundedness of the related maximal function seems to remain valid in this case though and it's answered for $p>1$ in Bourgain's paper.
No need for anything so sophisticated as $\ell^p\subset c_0$ :)
@RW : Do you think this one-line application of Hölder's inequality is sophisticated? Of course, as you suggest, you can also use $\ell^p\subset c_0$, but then you will also need to use the Cesàro mean theorem https://math.stackexchange.com/questions/155839/on-ces%C3%A0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn . I am not sure which way is simpler.
|
2025-03-21T14:48:30.956548
| 2020-05-12T17:51:02 |
360147
|
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|
Stack Exchange
|
Spherical harmonics expansion
In the context of $L^2$ space, it is usually stated that any square-integrable function can be expanded as a linear combination of Spherical Harmonics:
$$
f(\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, Y_\ell^m(\theta,\varphi)\tag 2
$$
where $Y_\ell^m( \theta , \varphi )$ are the Laplace spherical harmonics.
The context here is important because this equality holds only in the sense of the $L^2$-norm:
$$\lim_{N\to\infty} \int_0^{2\pi}\int_0^\pi \left|f(\theta,\varphi)-\sum_{\ell=0}^N \sum_{m=- \ell}^\ell f_\ell^m Y_\ell^m(\theta,\varphi)\right|^2\sin\theta\, d\theta \,d\varphi = 0.$$
Do we also have pointwise convergence almost everywhere?
the convergence is pointwise and uniform, see for example https://www.emis.de/journals/BBMS/Bulletin/bul954/KALF.PDF
@CarloBeenakker: But of course the convergence can't be uniform for a general $f\in L^2(S)$.
@Coltrane8: Your question has generated some negative reactions, and I suspect this may be due to the somewhat emotional and excited sounding tone you employed (this may not have been your intention, but it probably came across that way). I have taken the liberty of giving a more neutral sounding version of your question, but feel free to go back to the previous version if desired.
@Coltrane8: The question itself looks very reasonable to me, though I suspect that this is answered somewhere in the literature (or recognized as an open problem).
This 2002 paper indicates that the question was open at the time: https://projecteuclid.org/download/pdf_1/euclid.pcma/1416322431
@ChristianRemling I feel that your last comment should be the answer (non-pdf link). (Also, I believe there are counterexamples to simpler generalizations of Carleson’s theorem?)
@FrancoisZiegler: My guess would also be that the question is still open, but maybe one of the experts (Terry Tao, ...) can weigh in. And indeed, a.e. convergence need not hold for general ONBs (is this what you're alluding to in the second part of your comment?).
@ChristianRemling I had in mind Fourier series on $(\mathbf R/\mathbf Z)^n$ for $n>1$ where I believe subtle “summation method” problems already arise (but I am far from an expert).
A fundamental issue in appraising (uniform!?! especially...) pointwise convergence of eigenfunction expansions is having a reasonable estimate on the sup-norms of the eigenfunctions, as a function of the eigenvalue.
For Fourier series, the sup-norms are all $1$, and we can easily overlook this convenience.
For spherical harmonics, it is "well known" (see Stein-Weiss, and also some of my own notes...) that there is a polynomial bound for the sup-norm versus $L^2$ norm in terms of the eigenvalue. This does admit some abstraction in situations where a compact group acts transitively on the physical space on which we consider the functions.
Even for mildly general Sturm-Liouville situations, I myself do not know of a general approach to (uniform?) pointwise convergence, ... but I wish I did, and my ignorance surely indicates very little about the state of the art. :)
EDIT: that polynomial bound does also enable a reasonable "Sobolev space" set-up here, which gives tighter estimates than just saying "oh, it converges distributionally" (which, indeed, is a good remark).
Suppose, for a moment, that the spherical harmonics expansion for a function $g\in L^2$ is everywhere convergent to $g$.
Now change the function from $g$ to $f$ by changing the values of $g$ on some nonempty set of measure $0$. Then $f\in L^2$ and the Fourier coefficients $f^m_l$ of $f$ will be the same as $g^m_l$, and hence the expansion for $f$ will be the same as that for $g$. However, now the expansion for $f\in L^2$ will not be everywhere convergent to $f$.
Members of $L^2$ are actually, not functions, but classes of functions differing only on a set of measure $0$. So, it makes no sense to even talk about everywhere convergence to a member of $L^2$ in general.
However, if $f$ is smooth enough, then the corresponding expansion will converge to $f$ everywhere. For instance, by Theorem 1 on page 9 in the paper Kalf - On the Expansion of a Function in
Terms of Spherical Harmonics in Arbitrary Dimensions referenced in the comment by Carlo Beenakker, if $f\in C^1$, then the corresponding expansion will converge to $f$ uniformly and hence everywhere.
So how to deal with series sums term by term or derivation, for series expansions of classes of functions? I mean what is the formal development for the properties in analogy with usual functions?
Since $\lim_n(a_n+b_n)=\lim_n a_n+\lim_n b_n$ (provided that the latter two limits exist and, say, are finite) and $(f+g)^m_l=f^m_l+g^m_l$, the expansion for $f+g$ can be obtained by the term-wise addition of the expansions for $f$ and $g$ and, moreover, the expansion for $f+g$ will converge to $f+g$ if (and in the same sense as) the expansions for $f$ and $g$ converge to $f$ and $g$, respectively.
Previous comment continued: In particular, if $f$ and $g$ are in $C^1$, then the expansion for $f+g$ can be obtained by the term-wise addition of the expansions for $f$ and $g$ and, moreover, the expansion for $f+g$ will converge to $f+g$ uniformly and hence everywhere.
Ok, and the limit is taken in the norm topology of the Hilbert space in general? And the latter result is an application of the Thm in the link above, and in general not applicable for any other orthonormal set (say Bessel functions or Hermite polinomials). I mean there is no generalization and one must treat expansions on a particular set of functions, case to case.
Moreover I'm now still a bit confused, how can I now talk about convergence uniformly if the orthogonal projection doesn't distinguish by function differing for a set of measure $0$ ? Is that the case that the functions in the class of $f$ are not in $C^1$?
@Coltrane8 : I don't understand the point of your penultimate comment. Perhaps, you can rephrase/clarify it. As for your latest comment, members of $C^1$ (in contrast with $L^2$) are true functions (rather than equivalence classes). Moreover, if $f\in C^1$ and $g$ differs from $f$ on a nonempty set of measure $0$, then $g$ is not in $C^1$ (and not even in $C$).
I'm sorry. I'll rephrase: 1) The result stated above, where the expansion converge uniformly is valid for spherical harmonics. There is a more general theorem that says something like that for any expansion on any complete orthonormal set of $L^2$? (Say if I have another function defined on $R$ bounded and continuous, or $C^1$, and I expand it on the set of Hermite polinomials. Can I say it converges uniformly. I'm looking for this kind of generalization)
when you say $\lim_n(a_n+b_n)=\lim_n a_n+\lim_n b_n$, the definition of limit depends on the topology I suppose. So in this case all is good because we are using the topology inducted by the Hilbert norm on $L^2$ spaces?
@Coltrane8 : (i) I am not aware about general results on the uniform convergence of expansions in any orthonormal basis of $L^2$. (ii) For any two sequences $(a_n)$ and $(b_n)$ in any topological vector space, $\lim_n(a_n+b_n)=\lim_n a_n+\lim_n b_n$ if the latter two limits exist; this is just by the definition https://en.wikipedia.org/wiki/Topological_vector_space of a topological vector space. Of course, $C^1$ and $L^2$ are topological vector spaces.
@Coltrane8 You do have the (perhaps weak) consolation that if you work in the generalised context of the space of distributions on the sphere then you have everything you might desire (except for pointwise convergence)--the series always convergences in the distributional sense, you can freely manipulate the terms, you can differentiate term by term, .... The price you have to pay is that the corresponding convergence notion is much weaker than any of the classical ones for functions. Only you can know if it is too high.
|
2025-03-21T14:48:30.957187
| 2020-05-12T18:42:33 |
360153
|
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"url": "https://mathoverflow.net/questions/360153"
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|
Stack Exchange
|
Integrals of products of fractional parts
Let $((x)):=x-\lfloor x \rfloor -1/2$, where $\lfloor x \rfloor $ denotes the greatest integer $\le x$.
Let $a,b,c,...$ denote arbitrary natural numbers. It is clear that
$$ \int_0^a ((x/a)) dx =0.$$
A little exercise shows that
$$ \int_0^{ab} ((x/a))((x/b)) dx =\frac{\gcd(a,b)^2}{12}.$$
It appears that
$$ \int_0^{abc} ((x/a))((x/b))((x/c)) dx =0,$$
but I have not been able to determine a general formula for
$$ \int_0^{abcd} ((x/a))((x/b))((x/c))((x/d))dx.$$
I would be interested in a general formula for the last integral, and for similar integrals with a larger number of factors. It seems that these integrals, or the corresponding sums, such as
$$ \sum_{k=0}^{ab-1} ((k/a))((k/b)) = \frac{\gcd(a,b)^2-1}{12},$$
should be in the literature, but I have not been able to find a reference.
One place where I have seen this come up is in a paper of Lemke Oliver and Soundararajan. Here such quantities appear naturally in computing moments of some remainder terms, and Proposition 3.1 gives some basic properties. The sums are very closely related to the integrals (see Proposition 3.2 there). These are not too hard to establish, and I don't know really anything non-trivial about these integrals. I think it would be quite interesting to understand even the four fold integrals better.
In particular, a closed-form for the four-fold integral would lead to an asymptotic for the fourth moment of the error term for partial sums of the Euler totient function. (Eric Naslund, Maksym Radziwiłł, and I all looked at this a long time ago but didn't go anywhere.)
These quantities are studied under the name Franel Integrals in the literature. If we change notation so that $\psi(x) = x−⌊x⌋−1/2 $ is the saw tooth function, we may define the "fourth order" Franel integral as
$I(a,b,c,d)=\int_{0}^{1} \psi(ax) \psi(bx) \psi(cx) \psi(dx) dx.$
Expanding $\psi(x)$ in a Fourier series we see that:
$I(a,b,c,d) = \frac{1}{16 \pi^4} \sum_{\substack{s,t,u,v \in Z - \{0\}\\ sa+tb+uc+ue=0}} \frac{1}{stuv} =: \frac{1}{16 \pi^4} L(a,b,c,d)$
There is no explicit general closed form evaluation of this quantity but there are many special cases known, including
$L(1,1,1,1) = \frac{\pi^4}{5},$
$L(a,1,1,1) = \left(\frac{1}{3a} - \frac{2}{15a^3} \right) \pi^4,$
$L(a,a,b,b) = \frac{\pi^4}{9} + \frac{8 (a,b)^4 \zeta(4)}{a^2b^2}.$
Do similar identities to your final identity hold for other numbers of variables?
The "general identity" is, as far as I can see, essentially a consequence of expanding the function into a Fourier series and using orthogonality of the trigonometric system. This is how the integral is transformed into a counting problem for solutions of linear Diophantine equations. You can certainly do a similar thing for any other number of variables, but the formula clearly becomes a bit more complicated.
Thanks for the pointer to Franel integrals. That's very helpful. But there is something wrong with your formulas, because $I(1,1,1,1)=1/80$.
Andreas: indeed I was off by a factor of $\pi^4/16$ in my identities. I've corrected this now.
Mark: Thanks for the correction. But $I(2,1,1,1)=3/320$, while your formula says that it equals $1/120$. Do you have a reference for these formulas?
Andreas: indeed that was another typo, $L(a,1,1,1) = \left(\frac{1}{3a} - \frac{2}{15a^3} \right) \pi^4$. I've updated the posted. See: R. McIntosh's paper Franel Integrals of Order Four.
|
2025-03-21T14:48:30.957695
| 2020-05-12T19:32:48 |
360159
|
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|
Stack Exchange
|
Lower bounds on the length of circuits, depending on the number of times it crosses itself
I have this problem that I have been stuck on for months, and would like to know if somebody can tell me a way to attack the problem. Let me ask the problem in a simple example below.
Let $G(V,E)$ be a graph in which all vertices are degree 4. To simplify suppose there are no loops, and no repeated edges between vertices. An example would be to take a $d \times d$ square lattice and identify vertices and edges along opposite sides giving you a graph with $|V| = (d-1)^2$ on the torus (incidentally this graph has the property that it ``looks the same'' from any vertex). Now take any $v \in V$, and let $C$ be a circuit (no repeated edges) that starts and ends at $v$. Let $\text{len}(C) = L$, and suppose $C$ crosses itself $r$ times (i.e. there are $r$ vertices for which all 4 edges incident at the vertex are in $C$).
What is the minimum value of $L$?
If anyone can tell me if this problem is solved somewhere, or what general techniques are known to attack these kinds of problems, I'd really appreciate it. The actual graph that I'm interested in is more complicated than the one I mention here, but it is not completely random and has some symmetry. But all vertices are either degree 3 or 4, and some repeated edges are allowed (but no loops).
Edit: In view of the first comment below, let me make the question slightly interesting by requiring that the circuit is not a contractible loop on the torus.
Also let me provide another example in which I can also ask the same question. Let $G(V,E)$ be constructed by drawing $m$ latitudes and $n$ longitudes on the sphere. Then $|V| = mn + 2$ counting the north and south pole as vertices. Then let $C$ be a path (no repeated edges, and cannot visit the north or south pole in between) whose end points are the north and south pole. In this graph, all vertices except the north and south pole are degree 4. If $C$ crosses itself $r$ times, what is the minimum length of $L$ that makes it possible?
Without further conditions, you cannot say more than the trivial lower bound $L \geq 2r$ since all vertices in $C$ could have degree $4$.
@TonyHuynh I made some edits to the question.
|
2025-03-21T14:48:30.957892
| 2020-05-12T19:41:49 |
360161
|
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|
Certain morphism between graded modules
Say we have a morphism $f=(f_{i,j}) : \oplus_{i}M_{i} \rightarrow \oplus_{j}N_{j}$ (The direct sum of graded modules is finite) and $f_{i,j}$ are morphisms of graded modules but not necessary with the same degree.
Is it possible for the kernel to be graded?
Thanks
Modules over what? Associative ring? Commutative ring? Graded in what? positive integers? non-negative integers? an arbitrary abelian group?
Say positively graded ring and of course associative.
Are you sure this is what you intend to ask? It’s certainly possible for the kernel to be graded. For example, the kernel could be zero, or the $M_i$ could all be concentrated in the same degree, or the $f_{i,j}$ could all have the same degree.
Of course the kernel is not zero, the $f_{i,j}$ does not have the same degree and the $M_{i}$ are not concentrated in the same degree. Sorry I did not mention this hypothesis.
Jeremy Rickard, what about if all these graded modules are generated in the same degree?
There are still easy examples. For example, take $A$-modules for $A=k[x,y]$, graded with $|x|=|y|=1$. $M_1=A$, $N_1=N_2=A/(y)$, all generated in degree zero. $f_{1,1}$ and $f_{1,2}$ induced by multiplication by $1$ and $x$ respectively. Or all sorts of similar examples based on combining graded morphisms of different degrees but with the same kernel.
Here the situation is different. I understand this type of examples. You can always shift the left modules and then all the morphisms will have the same degree, namely zero degree.
|
2025-03-21T14:48:30.958031
| 2020-05-14T08:01:36 |
360317
|
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|
Stack Exchange
|
Improved upper bound for second moment of reduced residues modulo $q$?
The question asked here is a follow up to this question, which was answered by user GH from MO. [EDIT: The question is edited for clarification after receiving a comment on the original posting.]
As there, let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q. Then, following Montgomery and Vaughan in On the distribution of reduced residues,
the second moment of the number of reduced residues modulo $q$ in an
interval of length $h$ about its mean, $hP$, can be stated as
$$
M_2(q;h) =
qP^2
\sum_{\substack{{r \mid q }\\{r > 1}}}
\mu(r)^2
\left(
\prod_{\substack{ {p \mid q }\\{p \nmid r} }}
\frac{p(p-2)}{(p-1)^2}
\right)
r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right).
$$
Applying the inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ then gives the upper bound
$$\tag{1}
M_2(q;h)\leq qhP,
$$
while a lower bound is given by
$$
M_2(q; h) \geq qhP - qhPQ + O(qhP^2)
$$
where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.
Since $\{\alpha\}\leq 1$, we also have that $\{\alpha\}(1 - \{\alpha\}) \leq 0.25$. This allows for a second upper bound,
$$\tag{2}
M_2(q;h) \leq 0.25 \,
qP^2
\sum_{\substack{{r \mid q }\\{r > 1}}}
\mu(r)^2
\left(
\prod_{\substack{ {p \mid q }\\{p \nmid r} }}
\frac{p(p-2)}{(p-1)^2}
\right)
r^2 \phi(r)^{-2}.
$$
As pointed out by user Ofir Gorodetsky in the comment below, the upper bound in (1) is already optimal for those choices of $q$ such that the upper and lower bounds are of same order. For example, taking $q=\prod_{p\leq h^C} p$ for large enough $C$, we have that $M_2(q;h)\asymp qhP$.
Here I'd like to consider the rather opposite situation, when $C$ is small. Specifically, take $q=\prod_{p\leq h^C} p$, where $C<1$. Then we have that $Q=1$ and the lower bound takes the form
$$
M_2(q; h) \geq O(qhP^2),
$$
which leaves some more wiggle room between the upper and lower bounds, and I'd like to know whether the upper bound in (2) this case provides some improvement on (1).
Question
Given $q=\prod_{p\leq h^C} p$, where $C<1$. Does the upper bound for $M_2(q;h)$ stated in (2) provide a sharper bound compared to (1) in this case? And is it possible to express the upper bound in (2) a bit more simplified or asymptotically?
At least for some families of $q$, the upper bound is already optimal in the sense that there is a matching lower bound of the same order magnitude; this follows from the general lower bound given at the end of your linked question. To be concrete, taking $q = \prod_{p \le h^C} p$ for sufficiently large $C$, we have $M_2(q;h) \asymp qhP$.
Thanks for pointing to an important clarification! I've edited the question to focus on the case when $q=\prod_{p\leq h^C} p$ and $C<1$, as in this case the lower bound should be $M_2(q;h)\geq O(qhP^2)$ (is it a better or more correct way to write this inequality?).
Unfortunately, if you just use the fact that ${ \alpha}(1-{\alpha})$ is bounded, you do not obtain a superior upper bound. Indeed, assuming $q$ is squarefree (as we may), both $\mu(r)^2$ and $\prod_{p\mid q, , p \nmid r} p(p-2)/(p-1)^2$ are $\asymp 1$, and your idea leads to an upper bound of order $qP^2 \sum_{r \mid q, , r> 1} r^2/\phi(r)^2$, which is much larger qualitatively than $qPh$ for $q=\prod_{p \le x} p$ with $x \ge \log^2 h$...
A heuristic for using ${ \alpha } (1-{\alpha}) \le \alpha$ is that $r^2/\phi(r)^2$ being large is correlated with $r$ being large which is correlated with ${ h/r }(1-{h/r })$ being of order $h/r$.
Thanks for the effort! I'd be happy to accept that as an answer if you'd want to write it up as such.
|
2025-03-21T14:48:30.958271
| 2020-05-14T08:32:02 |
360318
|
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|
Stack Exchange
|
Convergence of a nonlinear iterative sequence
I have the following iterative sequence:
\begin{eqnarray*}
a_{t+1} &=& (1+\alpha-\beta)^2a_{t} - 2\alpha(1+\alpha-\beta)b_{t} +\alpha^2a_{t-1}+\frac{L}{a_{t}}, \\
b_{t+1} &=& (1+\alpha-\beta)a_{t} - \alpha b_{t},
\end{eqnarray*}
where $\alpha, \beta \in (0,1)$, $L>0$, $a_0=b_0=a_1>0$. The question is if $\{(a_t, b_t)\}_{t=1}^{\infty}$ will converge to the fixed point.
By the way, my numerical simulation suggests even the initial value of $a_0, b_0, a_1$ are not equal or positive, the sequence still can converge, so I suspect the initial value doesn't affect the convergence of the sequence.
|
2025-03-21T14:48:30.958349
| 2020-05-14T08:40:42 |
360319
|
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"url": "https://mathoverflow.net/questions/360319"
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|
More important or relevant progress in discretizing hard problems in physics in last decade
This is a reference request, and soft question as companion.
I'm curious to ask, from an informative point of view, what about the more important progress in the goal to discretize hard problems in physics and that were in the literature recently, let's say in this decade (2009-2020), as remarkable advances.
For the following question I was inspired in the problems explained in [1] (if I understand/interpret well the words of the professor in slides of his section What About Future Laws of Physics?, my understanding is that these discussions are related to hard problems in physics).
Question. I would like to ask as reference request, or soft question, for the more important or relevant recent progress in discretizing hard problems in physics.
I'm asking it as a reference request for the more recent advances, then I'm going to try to search and read those references from the literature. If in your discussion as soft question you want to refer about other past articles feel free to do it.
My knowledges in physics or mathematical physics and discretization methods aren't the best, but I think that this could be an interesting post for your colleagues, and reference for all us. Feel free to add your feedback about the post in comments.
References:
[1] David Tong, Physics and the Integers, University of Cambridge, Trinity Maths Society (2010).
I can't add more to my post, I'm going to read and appreciate the comments and answers
Many thanks for previous edits.
I understand the question as a request for pointers in the literature to research in the discretization of spacetime. There are two reasons why this is an active research topic, a fundamental and a practical reason: Fundamentally, spacetime might be discrete at the smallest levels (Planck scale); practically, to simulate relativistic quantum field theories (either on a classical computer, or, eventually, on a quantum computer), we need to discretize continuous degrees of freedom.
For an overview that I found instructive, and you might also like, I point to Causal Fermions in Discrete Spacetime by Farrelly and Short:
We consider fermionic systems in discrete spacetime evolving with a
strict notion of causality, meaning they evolve unitarily and with a
bounded propagation speed. First, we show that the evolution of these
systems has a natural decomposition into a product of local unitaries,
which also holds if we include bosons. Next, we show that causal
evolution of fermions in discrete spacetime can also be viewed as the
causal evolution of a lattice of qubits, meaning these systems can be
viewed as quantum cellular automata. Following this, we discuss some
examples of causal fermionic models in discrete spacetime that become
interesting physical systems in the continuum limit: Dirac fermions in
one and three spatial dimensions, Dirac fields and briefly the
Thirring model. Finally, we show that the dynamics of causal fermions
in discrete spacetime can be efficiently simulated on a quantum
computer.
Many thanks for your excellent answer.
After two months that was asked and answered the question I've considered to accept your excellent answer. Many thanks for share your knowledge in this site, is the best reference for many persons.
|
2025-03-21T14:48:30.958608
| 2020-05-14T09:45:33 |
360323
|
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"Daniel W.",
"Gerhard Paseman",
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|
Strengthening Sylvester's theorem
I am working on a problem in commutative ring theory, that deals with $p$-adic valuations. This leads to a number theoretical question that I want to explain in the following.
Let $n \in \mathbb{N}$ and $k$ an integer $\leq n/2$. Then, by the well-known result of Sylvester, there is an integer in $\{n, n-1,..., n-k+1\}$ that has a prime factor $>k$.
There is a very practical survey by Shorey and Tijdeman on generalizations of this theorem. For instance, it is known that the maximal distance of two positive integers having a prime factor $>k$ is $\leq \Big(\frac{1}{2} + o(1)\Big)\frac{k}{\log(k)}$. In particular, for large enough $k$ this difference is less than $\pi(k) < k$.
There are some similar results, but none of them helped me with my actual question, although they point in the direction of a positive answer to it:
Question: Can one explicitely give a constant $C$ with the following property?: For each positive integer $n \geq C$ that is not a prime power and for each integer $k$ with $1<k\leq n/2$, there are two different $a,b \in \{n,n-1,...,n-k+1\}$ having a prime factor $>k$.
Thank you in advance for any help!
I suspect the answer is C=13. (Still checking small cases.) For large k, you can use results like those of Nagura and extensions (there is a prime in (n, 6n/5) for n bigger than 24), and combinatorics for very small k. There is a region of k which aren't small but less than n/log n which makes this a challenge, if not an open problem. Gerhard "Will Try Jumping On It" Paseman, 2020.05.14.
By hand, I get n=18 as an exception. I suspect there are for each k greater than 2 only finitely many exceptional n, and for k=2 the only exceptions involve n or n-1 being a power of two. What computations have you done regarding this? Gerhard "Wants To See More Motivation" Paseman, 2020.05.14.
Actually, regarding my loss of knowledge concerning number theory, I hoped the result is a simple corollary of existing theory and therefore just asked. Thank you for your time and comments!
It turns out there are some simple arguments (Langevin especially) around Grimm's conjecture which don't give your result but bring you closer to it. I'm preparing an answer to make these more known. I'm also seeing if my recent work can apply to this question. Gerhard "Check Out 243490 And Sequel" Paseman, 2020.05.14.
I think we can prove it by understanding, refining, and reworking Sylvester's original proof. If you are willing to give it a go, I can provide a strategy and we can try it out, if you are willing to do a lot of the work of checking. I invite you to email me using the address hiding on one of my user pages. Gerhard "Up For Combinatorial Number Theory" Paseman, 2020.05.25.
OK. A simple strategy exists. At https://mathoverflow.net/a/361843/ is an observation which, coupled with standard theory and results on primes in large intervals, gives a way to compute C given not just two but d numbers in the interval with large prime factors. I think W should still collaborate on a formal write up. Gerhard "Considers This Problem As Solved" Paseman, 2020.05.31.
With motivation from the original poster, a key idea from Sylvester, and technical inspiration from Iosif Pinelis, I contribute an observation that helps toward an answer.
I use m instead of n and n instead of k. I start with the inequality that p! is strictly less than 3^p for p less than 7, and less than (p/2)^p for all larger integers p. We will set p =$\pi$(n).
Consider the product of integers in (m,m+n], and write it as W(n!)B, where W are all the prime factors (with multiplicity) at most n dividing (m+n)!/(m!n!), leaving B as the product of the remaining prime factors which are all larger than n, and B=1 if there are no such large prime factors.
Sylvester's observation is that W is at most (m+n-p+1)...(m+n). If B=1 the interval (m,m+n] has all numbers being n-smooth. The extended observation (which I think is new and hopefully orignal) is that WB is at most (m+n-p-d+1)...(m+n) if there are at most d many numbers in (m,m+n] which are not n-smooth. We fix d and observe that the original problem relates to d=1 in what follows.
Under supposition that there are not d+1 many non smooth numbers in (m,m+n], we now have (n!) Is at least (m+1)...(m+n-p-d). Write m as kn + i for positive integer k and non negative integer i (choose i less than n for less confusion). We now have (p+d)! Is at least (and for large enough n strictly greater than) k^(n-p-d).
So if (m,m+n] has at most d numbers which are not n-smooth, then we use the inequality above to note that when p+d is greater than 6, k is strictly less than ((p+d)/2)^((p+d)/(n-(p+d))). To save on plus signs, write q=p+d.
By the above, when q is at most 6 and n is at least 2q, then k is at most 2. (I leave the case n smaller than 12 and arbitrary d to the reader.) As n grows, q(1+ log (q/2)) will be less than n (because d is fixed), and one can use current literature or supercomputers to compute for which n this holds, in which case k is strictly less than e.
So given d, one can compute n0 without much challenge to find that (m,m+n] has d+1 non smooth numbers for n greater than n0 and for m at least as large as 3n.
To handle the remaining case for small d (d less than 6), use Nagura or similar as outlined in the other answer of mine to find d+1 non smooth integers in the interval for when m is in [n,3n). This should hold for m at least 150, giving C is less than 150.
Gerhard "Would James Joseph Be Approving?" Paseman, 2020.06.01.
For n less than 2q, we can still find a small bound on k. For the Nagura argument, it helps if k is less than 5; this should hold already for most practical applications of d and n. Gerhard "Leaving Small Computations To Reader" Paseman, 2020.06.01.
Also, the observation is new to me because I have not scoured the literature. I (and Daniel) would appreciate any reference having something close to the bound observed above on WB. Gerhard "Going Back To Modesty Now" Paseman, 2020.06.01.
By analyzing the product more carefully, I get 3k/2 is less than e, so k must be 1. This means we just have m in (n,2n) to worry about for n greater than n0. Further, for n less than n0 we can increase the denominator significantly to arrive at a smaller value of k. I think this will be an improvement on the Erdos paper. Gerhard "Is Still Working The Algebra" Paseman, 2020.06.01.
At Who first proved the generalization of Bertrand's postulate to (2n,3n) and (3n,4n)? are references to work establishing the existence of more than one prime in intervals that aren't too short, including work of Nagura. You can use many of these in a way similar to how I am about to show, which reveals that the main problem is for k small (but not too small).
We will pick n sufficiently large, and try to find a C using this. Pick a real x, say x is between n and n+1, and use Nagura's result to find a prime in (5x/6, x) whenever x is bigger than 30. So when k is bigger than n/6, we are already half way to our goal.
Now scale down by a factor of 2. When x is bigger than 60, there is a different prime in (5x/12, x/2), giving a number 2p which is less than n/6 below n, and has a factor larger than n/3. So for k bigger than n/6 and n at least 60, we have achieved our goal of finding two distinct numbers with prime factors bigger than k. For numbers n less than 60, one finds that primes and twice primes are close together so that this holds for k at least n/6 and n smaller than 60 and bigger than 36.
However, we need not stop there. We can scale down by 3,4,5 and larger to find numbers in (5x/6,x) which are three times a prime (or four times, or five or larger), getting at least five distinct numbers close to n .
In general, if you have a parameter $C_k$ so that for every $x \gt C_k$ there is a prime in $(x - x/k, x)$, you can then exhibit $k$ many distinct numbers below $n$ and greater than $n - n/k$ for $n \gt kC_k$ with prime factors greater than $n/k$. This gives more than you need for large values of your $k$ (different from the $k$ in $C_k$).
There is an argument from Langevin that goes like this: pick an arithmetic progression of $k$ terms, each term coprime to the common difference $d$. Define a map from each term to that prime $p$ such that the largest prime power which is a factor of that term is a power of $p$. Much of the time, this map is injective, so each term gets a different prime divisor. When it is not, then two terms are both divisible by a power of the same prime, say $p^e$. Since the terms are coprime to the difference, $p^e$ is less than $k$. Since $p^e$ is the largest prime power of one of the terms, that term must be no larger than lcm(1...p^e), so the term is less than lcm(1..k). So if n is large enough, k terms around n have large enough prime divisors, especially when $k \gt 4$. Unfortunately the lower bound grows with $k$.
It may be possible to push the lower bound (indeed, I have unpublished work which takes it down to about sqrt(lcm...)), but your condition is weaker. It may be possible to modify the term to largest prime power map to exhibit a logarithmic if not constant lower bound.
Gerhard "And Then There's Jumping Primes" Paseman, 2020.05.14.
Now you invested quite some work into this. Thank you very much! It seems to be a hard problem. Do you think it helps to restrict $k \leq n-P$, where $P$ is the biggest prime number $\leq n$? Maybe this is a distance short enough to overcome the hard cases in the middle...
If gaps between primes are as small as expected (like (log n)^2), we might get away with modified Langevin. Grimm's conjecture implies your result for k greater than 3. It's possible that my current work may help with values of k that small in spite of the current lack of knowledge of prime gaps. Gerhard "Hasn't Made The Leap Yet" Paseman, 2020.05.15.
One can extend both these arguments, but with less certainty and still not close enough to realizing the desired C. There is Harman-Baker-Pintz which allows your k to range from n^{0.525.} plus epsilon on up, and there is modified Langevin which holds for a given k and all n above lcm(1.. sqrt(k)), roughly. If you ask, I will post details. Gerhard "Mathematics Remodeled Almost To Order" Paseman, 2020.05.16.
I already noticed that my original ring theoretical question is related in many ways to prime gaps. Maybe the problem is equivalent to some conjecture on prime gaps. What do you think?
Equivalence to something directly related to prime gaps will be a challenge. To wit, a counterexample of significant size would say something about a gap below n, and also below n/2, n/3, and so on. It might tie in with studies on smooth numbers though. Grimm's conjecture should imply yours, but there may be weaker ones still to which yours might be equivalent. I would like to see the ring theoretic version of your conjecture. Gerhard "If It's Ready For PrimeTime" Paseman, 2020.05.17.
If we only could get a constant lower bound in the argument of Langevin in your answer. It already helped me in some partial cases of my problem with the growing lower bound. The ring theoretical question is about factorizations in the ring of integer-valued polynomials $\text{Int}(\mathbb{Z}) = { f \in \mathbb{Q}[x] \mid f(\mathbb{Z}) \subseteq \mathbb{Z} }$. It is already known that this ring has very wild factorization-theoretical properties, now we want to see how a special class of irreducibles behaves, namely $\binom{x}{n}$ for $n \in \mathbb{N}$.
It is a conjecture that all powers of $\binom{x}{n} = \frac{x(x-1)...(x-k+1)}{n!}$ have only the trivial factorization $\binom{x}{n}...\binom{x}{n}$. A positive answer to the question above leads to a proof for this conjecture.
So please let me know, if one day you are able to get down to a constant bound in the Langevin argument. Thank you for you very motivated answers and comments!
|
2025-03-21T14:48:30.959501
| 2020-05-14T10:47:48 |
360327
|
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|
Stack Exchange
|
Decomposition of the homogeneous polynomial ring $\{\mathbb R[x_{ij}]_{1\le i,j\le n}\}$ of degree 2 into Specht modules
I have tried to decompose this as following spans over the real field.
$V_1=\operatorname{span} \langle x_{ij}^2\rangle$
$V_2=\operatorname{span} \langle x_{ij}x_{jk}\rangle$
$V_3=\operatorname{span} \langle x_{ij}x_{kl}\rangle$
where the index $ij$ stands for the subset $\{i,j\}$ of size 2, thus the ordering within does not matter and there are no variables of the form $x_{ii}$. Also the indices $i,j,k,l$ are distinct.
Now, one determines
$V_1\cong M_{(n-2,2)}$
$V_2\cong M_{(n-3,2,1)}$
Next I tried for $V_3$ the following:
$E_{il,jk}=x_{ij}x_{kl}+x_{ik}x_{jl}$ is fixed precisely by the group $S_{i,l}\times S_{j,k}\times S_{n-4}$.
Moreover, $E_{il,jk}-E_{ij,kl}+E_{ik,jl}=2x_{ij}x_{kl}$, which shows that all of $V_3$ is generated. One can define a map $\phi:M_{(n-4,2,2)}\rightarrow V_3$ by sending the tabloid with last two rows $i,j$ followed by $k,l$, to the element $E_{ij,kl}$. One can get deceived into concluding that this is an isomorphism.
Unfortunately since the tabloid with rows $k,l$ and $i,j$ above swapped also maps to $x_{ij}x_{kl}$, this map is not an isomorphism.
However $x_{ij}x_{kl}+x_{ik}x_{jl}+x_{il}x_{jk}$ has fixed group $S_{(n-4,4)}$ and is isomorphic to $M_{(n-4,4)}$, so I know that this makes up ${n\choose 4}$ dimensions in $V_3$, but a dimension count shows that the remaining $2{n\choose 4}$ dimensions has many possible decompositions into Specht modules.
Could someone help with zeroing in on one decomposition?
Write $W_4$ for the $S_4$ representation $$({\rm Ind}_{S_2 \times S_2}^{S_4} ({\rm triv} \otimes {\rm triv}))_{S_2}.$$
Then similarly to the answer to your previous question, you are considering $$Ind^{S_{n}}_{S_4 \times S_{n-4}} (W_4 \otimes {\rm triv}),$$ which can be computed using the Pieri rule as soon as you know the decomposition of $W_4$ into irreducibles.
By definition $W_4$ corresponds to the plethysm $(2) \circ (2)$. In general, plethysm is hard to compute but any specific case can be done, and in fact the plethysm $(n) \circ (2)$ is known in general. This should allow you to compute $(x_{i_1 j_1}) \dots (x_{i_n j_n})$ where all indices are distinct.
We have $(2) \circ (2) = (4) + (2,2)$. So the answer is $$((4) + (2,2))*(n-4)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-3,3) + (n-4,4)$$ $$+ (n-2,2) + (n-3,2,1) + (n-4,2,2).$$
I think we should have $(n-1,1)$ and $(n-3,3)$ as everything adds to $n$. I wonder if we can write an explicit basis for each irreducible? I can do the trivial :). You can get sub representations doing things like like fix $i,j$ then look at $\sum_{k, \ell} x_{ij}x_{k\ell}$.
@John Thanks-- fixed the typo. It should be possible, since you can decompose $W_4$ easily (but I don't know the answer).
@PhilTosteson, what is a good reference article/book for the Plethysm computation? Googling doesn't seem to do much good. Also Sagan's book's index does not have the word Plethysm.
@PhilTosteson, does the subscript $_{S_2}$ at the end of the second line mean something? Otherwise I see no reason why $(3,1)$ doesn't appear in the expansion of the Plethysm $(2)\circ(2)$.
@Karthik Yes, it means take co-invariants with respect to the action of $S_2$. In other words $W_4$ is the action of $S_4$ on partitions into two blocks of size two. For this specific case, there is no need to learn about plethysm-- the decomposition is easy. If you're interested in learning more, I would reccomend Stanley's enumerative combinatorics volume 2.
Muchos Gracias senor @Phil ! Much appreciated. This problem or class of problems was one roadblock I'd been trying to clear but hadn't had even an elementary understanding of symmetry group reps. I look forward to more interactions in future!
|
2025-03-21T14:48:30.959774
| 2020-05-14T12:32:34 |
360332
|
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|
Stack Exchange
|
Complexity of set-partition problems
given a universe $\mathcal{U}$ of elements and a system $\mathcal{S}$ of weighted subsets of $\mathcal{U}$ whose union covers $\mathcal{U}$.
Assuming the existence of at least one subsytem $S\subseteq\mathcal{S}$ such that the disjoint union of its elements $\lbrace s_1,\,\dots,\,s_k\rbrace$ covers $\mathcal{U}$, what is the complexity of finding $S_{opt}\subseteq\mathcal{S}$, that resembles the disjoint set-cover of minimal weightsum, i.e. does the restriction to covers with disjoint sets suffice to make the problem polynomial?
The exact cover problem being one of Karp's
21 NP-complete problems, your optimization problem is NP-hard.
|
2025-03-21T14:48:30.959850
| 2020-05-14T14:17:12 |
360339
|
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|
Stack Exchange
|
Is there a non-convex function with non-decreasing average rate of change?
$\newcommand{\R}{\mathbb R}$
Let $f$ be a function from $\R$ to $\R$. It is said that $f$ is midpoint-convex if for any real $x$ and $y$ we have $f(x-y)+f(x+y)\ge2f(x)$ or, equivalently, $f(x+y)-f(x)\ge f(x)-f(x-y)$. So, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$
$$\frac{f(x+y)-f(x)}y\ge\frac{f(x)-f(x-y)}y.$$
One may call the ratio $\frac{f(x+y)-f(x)}y$ the average rate of change of the function $f$ over the interval $[x,x+y]$. Thus, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$ the average rate of change of $f$ over $[x,x+y]$ is no less than the average rate of change of $f$ over $[x-y,x]$.
By Sierpiński's theorem, every Lebesgue measurable midpoint-convex function from $\R$ to $\R$ is convex. However (assuming the axiom of choice), one can easily construct a (necessarily non-Lebesgue measurable) midpoint-convex function from $\R$ to $\R$ which is not convex -- see e.g. this, where $f(i_1)$ should be replaced by $f(b_{i_1})$.
Let us now say that $f$ has non-decreasing average rate of change if
$$\frac{f(x_2+y)-f(x_2)}y\ge\frac{f(x_1+y)-f(x_1)}y$$
for any real $x_1,x_2,y$ such that $x_1\le x_2$ and $y>0$. This is equivalent to the following:
$$\frac{f(x-y_2)+f(x+y_2)}2\ge\frac{f(x-y_1)+f(x+y_1)}2$$
for any real $x,y_1,y_2$ such that $0\le y_1\le y_2$.
One may note here that every function with non-decreasing average rate of change is obviously midpoint convex.
The question is then the one in the title: Is there a (necessarily non-Lebesgue measurable) non-convex function with non-decreasing average rate of change?
The standard example of a non-convex but midpoint-convex function is additive:
given a basis $B$ of $\mathbb R$ over the rationals $\mathbb Q$, choose one member $\alpha$ of $B$ and
if $x = \sum_{\beta \in B} c_\beta(x) \beta$ with $c_\beta(x) \in \mathbb Q$ (only finitely many nonzero), take $f(x) = c_\alpha(x)$. This has non-decreasing rate of change, indeed
$$\frac{f(x_1 + y) - f(x_1)}{y} = \frac{f(x_2 + y) - f(x_2)}{y} = \frac{f(y)}{y}$$
|
2025-03-21T14:48:30.959990
| 2020-05-14T14:33:59 |
360341
|
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"Martha Łącka",
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|
Stack Exchange
|
Properties of the spectrum of the Koopman representation
Let $G$ be a discrete countable infinite group acting on a compact metric space $X$ via homeomorphisms preserving a probability measure $\mu$.
A function $\lambda\colon G\to \mathbb C$ is an eigenvalue of the action of $G$ if there exists a function $f\in L^2(X,\mu)$ such that for every $g\in G$ one has $\lambda(g)\cdot f=f\circ g$.
In this paper: Ergodicity of the Cartesian product by E. Flytzanis, Trans. Amer. Math. Soc. 186 (1973), 171-176 (freely available link at AMS site), there are two results concering the product of two dynamical systems given by $\mathbf Z$-action:
a sufficient condition for the ergodicity of the product,
a description of the spectrum of their cartesian product.
It is also written there that the above results hold also for $G$-action, but it is not specified what does it exactly mean. Am I right that:
A product of $(X,G)$ and $(Y,G)$ is ergodic if the function
constantly equal to 1 is the only common eigenvalue for $(X,G)$ and
$(Y,G)$?
The set of eigenvalues of $X\times Y$ equals the set of all functions
of the form $f\cdot g$ (pointwise multiplication), where $f$ is an
eigenvalue for $X$ and $g$ is an eigenvalue for $Y$?
I need also two other properties, but I could not find the appropriate references for them (maybe I am wrong that they are true?):
The set of eigenvalues of a factor of a dynamical system $(X,G, \mu)$
is contained in the set of eigenvalues of $(X,G,\mu)$?
Every eigenvalue of an ergodic dynamical system is simple (beware: I
do not assume that $G$ is abelian).
For your first question the answer is "no". This is already seen when considering the regular representation of a finite group with trivial abelinzation.
As a general rule of thumb, it is convenient to replace "eigenvalues" with "finite dimensional subrepresentations" when considering non-commutative groups. Upon such a replacement you'd get a positive answer. Your second question is badly formulated. Are you looking for eigenvalues or eigenvectors?
The last two properties that you ask about are correct. For the first, a representation of a factor is a subrepresentation, and for the second, in the particular case of a compact group, every ergodic representation is a factor of the regular. Note that when asking about eigenvalues, the question always reduces to the case of a compact group.
Thank you, Uri Bader! Your answer is really helpful. For the first question - could you recommend some reference for this topic? In the second question I meant eigenvalues. If I understand correctly, an eigenvalue is a function from $G$ to $\mathbb C$ so I can consider pointwise multiplication (that is $f\cdot g(x)=f(x)\cdot g(x)$).
I am sorry for the confusion - I used $f$ and $g$ for eigenvalues which is probably not standard.
oh... this is a horrible notation you adopted there. The answer is negative here as well. I'll write it in an answer format.
The answer to both question is negative.
Take $G=S_3$, the symmetric group of the set $X=\{1,2,3\}$.
Then $L^2(X)$ is decomposed to the trivial representation and a another two dimensional irreducible representation.
As a $G$-space, $X\times X\simeq X \cup G$ where $X\subset X\times X$ is the diagonal and $G$ corresponds to the rest.
This shows that $X\times X$ is not ergodic, providing a counter example to the first question.
It also follows that $L^2(X\times X)$ contains a sub-representation isomorphic to the regular representation $L^2(G)$ and in particular it contains the one dimensional sign representation. This provides a counter example to the second question.
Thank you for the answer :)
|
2025-03-21T14:48:30.960263
| 2020-05-14T15:31:05 |
360344
|
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"Aaron",
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|
Stack Exchange
|
Solution for $Xa + X^Tb = c$ where $X^TX = I$?
There are three known $n\times1$ vectors: $a, b, c$, along with one unknown $n\times n$ matrix: $X$. I am only interested in the $n={2,3}$ cases.
$X$ is $2\times 2$ or $3\times 3$ rotation matrix with an unusual domain specific constriant:
$X^TX = XX^T = I$
$Xa + X^Tb = c$
Is there a solution for $X$ in terms of $a,b,c$? Based on where the problem came from, I know there isn't always a solution, but I have stumped myself trying to figure out how to solve it when there is one.
I have tried working out the $2\times 2$ case element-wise, and arrived at the following, equally(?) difficult problem:
$X = \begin{bmatrix}x_{11} & x_{12} \\ -x_{12} & x_{11}\end{bmatrix}$
$\begin{bmatrix}a_1+b_1 & b_2-a_2 \\ a_2+b_2 & a_1-b_1\end{bmatrix}\begin{bmatrix}x_{11}\\x_{12}\end{bmatrix}=c$
$Ax = c$ where $x^Tx=1$
This reduces to a system of polynomial equations in matrix elements, for which there exist standard methods of solving and even software solvers.
MSE https://math.stackexchange.com/ is a right forum for such type questions.
As a beginning of a search for solutions, we can take norms of each side. We then get
$$\|a\|^2 + \|b\|^2 + 2 b\cdot (X^2 a)=\|c\|^2,$$
which is enough to fix the angle between $b$ and $X^2 a$.
Similarly, multiplying through by $X$ and then dotting with $b$, we can conclude that
$$b\cdot (Xc)=\|b\|^2+b\cdot (X^2 a)=\|b\|^2+\frac{\|c\|^2-\|a\|^2-\|b\|^2}{2}=\frac{\|b\|^2+\|c\|^2-\|a\|^2}{2}.$$
Therefore, we know the angle that $b$ makes with $Xc$. Similar calculations show us the angle that $a$ makes with $X^T c$. In two dimensions, this is enough to find $X$ geometrically if it exists in most cases, and otherwise to say that there is no such solution. The geometry is slightly more involved in 3 dimensions, and I'm not immediately sure if there is more useful information to be extracted through dot products to help.
Here is an approach for $\mathbb R^3$, inspired by Michael Renardy's answer.
Let us temporarily expand the problem to $Aa+Bb=c$, where $A,B\in O_n(\mathbb R)$. Assuming that $|a|,|b|$, and $|c|$ satisfy the triangle inequality, we can find a solution, $(A_0,B_0)$. However, the space of all solutions is $(MA_0,MB_0)$ where $Mc=c$ and $M\in O_n(\mathbb R)$. Thus, we've reduced the problem to:
Given $A,B\in O_n(\mathbb R)$ and $c\in \mathbb R^n$, does there exist $X\in O_n(\mathbb R)$ such that $$Xc=c \quad \text{and} \quad I=AXBX.$$
Since $AX$ and $BX$ are inverses, they commute, and so $BXAX=I$ too. Evaluating at $c$, we get 3 equations,
$$ Xc=c, \quad X(Bc)=A^Tc, \quad X(Ac)=B^Tc.$$
Assuming that $c, Ac, Bc$ span your space (which happens in $\mathbb R^3$ for most $(A,B,c)$-triples), this specifies a unique candidate $X$ to test to see if it actually satisfies the problem.
Explicitly, if $P$ is the matrix whose columns are $c,Ac, Bc$ respectively, and $Q$ is the matrix whose columns are $c, B^Tc, A^Tc$ respectively, then $X=QP^{-1}$. We just need to check that $XX^T=I$ and $AXBX=I$. or see that these equations are violated.
\begin{equation}
Xa = \begin{bmatrix} a_1x_{1,1} + a_2x_{1,2}\\ a_1x_{2,1} + a_2x_{2,2}\\
\end{bmatrix}
\end{equation}
and
\begin{equation}
X^Tb = \begin{bmatrix} b_1x_{1,1} + b_2x_{2,1}\\ b_1x_{1,2} + b_2x_{1,2}\\
\end{bmatrix}
\end{equation}
and
\begin{equation}
Xa+X^Tb = \begin{bmatrix}a_1x_{1,1} + a_2x_{1,2} + b_1x_{1,1} + b_2x_{2,1}\\ a_1x_{2,1} + a_2x_{2,2}+ b_1x_{1,2} + b_2x_{2,2}\\
\end{bmatrix} = \begin{bmatrix}(a_1+b_1)x_{1,1} + a_2x_{1,2} + b_2x_{2,1}\\ (a_2+b_2)x_{2,2} + a_1x_{2,1} + b_1x_{1,2}\\
\end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\
\end{bmatrix}
\end{equation}
but $X$ is a rotation matrix gives us that
\begin{equation}
Xa+X^Tb = \begin{bmatrix}(a_1+b_1)\cos\theta - a_2\sin\theta + b_2\sin \theta \\ (a_2+b_2)\cos\theta + a_1\sin\theta - b_1\sin\theta\\
\end{bmatrix} = \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta \\ (a_2+b_2)\cos\theta + (a_1 - b_1)\sin\theta\\
\end{bmatrix}
\end{equation}
(if you only have $X^TX= I$ then you have to consider the extra case there $\sin \theta \to -\sin \theta $; i.e. rotation composed with reflection. Notice that $X$ is a rotation implies $X^TX= I$!) and therefore that
\begin{equation}
Xa+X^Tb = \begin{bmatrix}(a_1+b_1)\cos\theta +(b_2- a_2)\sin\theta \\ (a_2+b_2)\cos\theta + (a_1 - b_1)\sin\theta\\
\end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\\
\end{bmatrix}
\end{equation}
and therefore (by using the triangle inequality) you don't have a solution if for example
\begin{equation}
\frac{|(a_1+b_1)|}{\sqrt{2}} +\frac{|(b_2- a_2)|}{\sqrt{2}} < |c_1|
\end{equation}
and like wise for the second condition
\begin{equation}
\frac{|(a_2+b_2)|}{\sqrt{2}} +\frac{|(a_1- b_1)|}{\sqrt{2}} < |c_2|
\end{equation}
and you can probably come up with all kinds of other tests for failure, but here is the most general one:
An alternative/equivalent way to look at it is that you have an overdetermined system of 3 equations and 2 unknowns of the form
\begin{equation}
\begin{array}
& ax & + & by & =& c_1\\
cx & + & dy & =& c_2 \\
x^2 & + & y^2 & =& 1 \\
\end{array}
\end{equation}
where $a = a_1+b_1$ , $b=a_2- b_2$, $c = a_2+b_2$, and $d = a_1- b_1$; which is highly unlikely to have solutions.
Therefore you have solutions iff the solution to the system of
equations \begin{equation} \begin{array} & ax & + & by & =& c_1 \\ cx
& + & dy & =& c_2 \\ \end{array} \end{equation} also satisfies the
condition $ x^2 + y^2 = 1 $.
If you want to work out the $n=3$ case you can do the same exact thing but use the Euler angles; it will be long and tedious but you can probably get some kind of condition on the solutions.
You can possibly try this as well. Note that the orthogonality requirement can be relaxed using Schur complement as $\begin{bmatrix}I&X\\X^\top & I \end{bmatrix} \succeq 0$. So, we have the following:
$$
\max_{X\in R^{n\times n}} ~~ \|c-Xa-X^\top b\|_2\\
\hspace{-3cm}\mbox{subject to}\\
~~~~~~~~~~~~\begin{bmatrix}I&X\\X^\top & I \end{bmatrix} \succeq 0.
$$
Note that if the solution to this convex problem $X^*$ does not satisfy $\|c-Xa-X^\top b\|^2_2=0$, there does not exist a solution to the original problem.
Hope this partial treatment helps. And, thanks to Aaron for pointing out an error earlier.
Note that $(Xa)^T(X^Tb)\neq a^Tb$.You don't end up with $X^TX$, you get $(X^T)^2$.
Yikes! Thanks for that. My bad. Will think about getting around it.
An obvious necessary condition is that |a|, |b| and |c| can be sides of a triangle. The two-dimensional case can be analyzed further geometrically. If the triangle inequalities are satisfied, there are vectors congruent to a and b which form a triangle with c. The condition you require is that these vectors can be obtained either by rotating a and b in opposite directions (if X is proper orthogonal, and in this case the bisector of the angle between a and b stays the same), or by reflection across some axis (if X is improper orthogonal).
You can solve the $n=2$ case relatively easily with complex numbers:
$a$, $b$ and $c$ can be represented by complex numbers and the rotation matrix by a complex number z with modulus 1.
So the equations are:
$$za+z^{-1}b=c$$ with $|z|=1$.
The equation is equivalent to $$az^2-cz+b=0$$
which you can solve immediately using the quadratic formula:
$$z=\frac{c\pm\sqrt{c^2-4ab}}{2a}.$$
Clearly we only have solution to the problem if $$|\frac{c+\sqrt{c^2-4ab}}{2a}|=1$$ or $$|\frac{c-\sqrt{c^2-4ab}}{2a}|=1.$$
We can analyse further to obtain simpler conditions for a solution to exist by taking the modulus which gives $(za+b/z)(\bar{a}/z+\bar{b}z)=|c|^2$ or $2\Re(\bar{a}b/z^2)=|c|^2-|a|^2-|b|^2$. The inequality $|\Re(z)|\leq|z|$ implies $||c|^2-|a|^2-|b|^2|\leq 2|a||b|$ which is equivalent to the triangle inequality holding for $|a|$, $|b|$ and $|c|$ which as Michael Renardy noted is obvious geometrically and is clearly a necessary condition for a solution to exist. However it is also a sufficient condition.
In fact if $\bar{a}b=re^{\mu}$ and we set $z=e^{i\theta}$ the equation reduces to $2\Re(re^{\mu-2\theta})=|c|^2-|a|^2-|b|^2$ or $2r\cos(\mu-2\theta)=|c|^2-|a|^2-|b|^2.$ or $\cos(\mu-2\theta)=(|c|^2-|a|^2-|b|^2)/(2|a||b|).$ and hence if the triangle inequality condition is satisfied we have $|\cos(\mu-2\theta)|\leq 1$ which gives us a solution for $\theta \in \mathbb{R}$ and hence the rotation $z=e^{i\theta}$.
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2025-03-21T14:48:30.960768
| 2020-05-14T15:35:54 |
360346
|
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|
Stack Exchange
|
English reference for the Brauer-Kuroda formula
I'm currently trying to understand the Brauer-Kuroda formula.
Although there are many recent papers on the formula but they seem to be purely algebraic.
They say that original analytic approach is much easier and short.
So I hope to find a reference on the original proof of Brauer-Kuroda formula using the formula for the residue of Dedekind zeta function.
But the papers of Kuroda, Brauer are in German.
So is there any good reference for the formula in 'English' that just explains the papers of Kuroda, Brauer? In other words, the shortest reference in English.
Thank you.
See Section VIII.7 in Fröhlich-Taylor, Algebraic Number Theory, Cambridge University Press.
|
2025-03-21T14:48:30.960854
| 2020-05-14T16:04:29 |
360349
|
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"authors": [
"Iosif Pinelis",
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"user64494"
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|
Stack Exchange
|
Expected value of a truncated binomial
Let $X\sim B(n,p)$ be a binomial random variable and fix $0<k<n$. Are there any well-known bounds for $\mathbb{E} (X-k)^+$, where $(X-k)^+ =\max\{0,X-k\}$? I am particularly interested in interpreting this as a function of $p\in (0,1)$.
$\newcommand\vpi{\varphi}$Let
$$Z_n:=\frac{X_n-np}{\sqrt{npq}},$$
where $X_n:=X$ and $q:=1-p\in(0,1)$. Then $Z_n\to Z\sim N(0,1)$ in distribution (as $n\to\infty$). Also, $EZ_n^2=1$. So,
assuming that
$$z_{n,k}:=\frac{k-np}{\sqrt{npq}}\to z\in\mathbb R, \tag{0}$$
by the uniform integrability we have
$$\frac{E(X-k)^+}{\sqrt{npq}}=E(Z_n-z_{n,k})^+
\to E(Z-z)^+,$$
whence
$$E(X-k)^+\sim\sqrt{npq}\, E(Z-z)^+. \tag{1}$$
Also, note that for real $z$
$$\psi(z):=E(Z-z)^+=\int_z^\infty(u-z)\vpi(u)\,du=\vpi(z)-z(1-\Phi(z)),$$
where $\vpi$ and $\Phi$ denote, respectively, the pdf and cdf of $N(0,1)$. Here is the graph $\{(z,\psi(z))\colon|z|<3\}$:
Using more advanced tools than the central limit theorem for the binomial distribution, one can greatly improve estimate (1) (derived assuming (0)). Indeed, by the nonuniform Berry--Esseen bound (see e.g. formula (4.3.1)), for all real $t$
$$P(Z_n>t)-P(Z>t)=O\Big(\frac1{(1+|t|^3)\sqrt{npq}}\Big). \tag{2}$$
Everywhere here the constants in $O(\cdot)$ are universal. Also, for any real-valued random variable $Y$ and any real $y$,
$$E(Y-y)^+=E\int_y^\infty du\, 1_{Y>u}
=\int_y^\infty du\, P(Y>u).$$
So, by (2),
$$E(X-k)^+=\sqrt{npq}\,E(Z_n-z_{n,k})^+
=\sqrt{npq}\,E(Z-z_{n,k})^+ + O(1). \tag{3}$$
Of course, (3) implies (1) (when (0) holds), but (3) is much stronger and more general than (1).
That is perfect! I guess the natural question then becomes "what does $E(Z-z)^+$ look like", although that sounds much more Google-able.
I have added details on $E(Z-z)^+$ and also added a stronger estimate.
The command of Mathematica 12.0
Mean[TransformedDistribution[Max[x - k, 0],x\[Distributed] BinomialDistribution[n,p]]]
produces $${p^{\lfloor k\rfloor +1} (1-p)^{n-\lfloor k\rfloor } \left(-\binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)+k \binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)-k p \binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)+p \binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)+p \lfloor k\rfloor \binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)-\lfloor k\rfloor \binom{n}{\lfloor k\rfloor +1} \, _2F_1\left(1,-n+\lfloor k\rfloor +1;\lfloor k\rfloor +2;\frac{p}{p-1}\right)-p \binom{n}{\lfloor k\rfloor +2} \, _2F_1\left(2,-n+\lfloor k\rfloor +2;\lfloor k\rfloor +3;\frac{p}{p-1}\right)\right)}{(p-1)^{-2}} $$ for $k\ge 0,\, k <n$.
Mathematica also finds its variance through Variance[TransformedDistribution[Max[x - k, 0], x \[Distributed] BinomialDistribution[n, p]]].
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