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2025-03-21T14:48:29.859639
| 2020-02-12T20:11:36 |
352571
|
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|
Stack Exchange
|
When is the homotopy category of an accessible $\infty$-category accessible?
Let $\mathcal C$ be an accessible $\infty$-category, and let $ho(\mathcal C)$ be its homotopy category. I can think of two "trivial" reasons for $ho(\mathcal C)$ to be accessible:
$ho(\mathcal C) = \mathcal C$;
$ho(\mathcal C)$ is small with split idempotents.
Otherwise, I am aware of very few examples where $ho(\mathcal C)$ is accessible. Indeed, given that $ho(Spaces)$ is very far from accessible, I think I should expect that it is quite rare for $ho(\mathcal C)$ to be accessible.
However, I know of one interesting class of examples where $ho(\mathcal C)$ is "nontrivially" accessible. Let $k$ be a field.
Write $D(k)$ for "derived $\infty$-category of $k$", i.e. the category of chain complexes over $k$ localized ($\infty$-categorically) at the quasi-isomorphisms. This is a presentable $\infty$-category.
Then $ho(D(k))$ is the usual derived category of $k$, which is equivalent to the usual 1-category of graded $k$-vector spaces (though of course the triangulated structure is different), and so is obviously accessible (locally presentable, in fact).
This one class of examples has me second-guessing my expectation that taking homotopy categories rarely preserves accessibility.
Questions:
What are some other examples of accessible $\infty$-categories $\mathcal C$ such that $ho(\mathcal C)$ is also accessible (which do not satisfy (i) or (ii) above)?
Given such an example, is the functor $\mathcal C \to ho(\mathcal C)$ accessible (i.e. does it preserve $\kappa$-filtered colimits for some $\kappa$?)
Can we (partially) characterize this condition, giving necessary and / or sufficient conditions for $ho(\mathcal C)$ and $\mathcal C \to ho(\mathcal C)$ to be accessible?
Is there anything to be said in particular about the case where $\mathcal C$ is stable and presentable, or even more particularly the case where $\mathcal C$ is the derived $\infty$-category of a ring?
More broadly, at the moment it seems very mysterious to me that $ho(D(k))$ comes out to be accessible. I'd appreciate any perspective which makes this fact look less mysterious.
Give a look at Example 5.3 in Generalized Brown representability in Homotopy categories by Rosicky. The author shows that $\mathsf{Ho}(\text{SSet}_n)$ is accessible.
@IvanDiLiberti Thanks! Just to be clear, this is not the homotopy category of $n$-truncated spaces as one might naively guess (and the homotopy category of $n$-truncated spaces is not accessible for $n \geq 1$ because idempotents do not split -- the counterexample appearing in HTT <IP_ADDRESS>, which I think is due to Freyd, uses only 1-truncated spaces).
Not directly related to the question, but a paper you might be interested, with a similar line of inquiry (for combinatoriality) is: https://arxiv.org/abs/1702.00240v2
@DavidWhite Thanks!
@IvanDiLiberti Does your criterion here show, perhaps, that the derived category of a ring $R$ is not concrete (and hence not accessible) unless $R$ is a division ring? If so, that would be very interesting.
I honestly do not know, but the answer might be close to be yes. Whenever $R$-modules are as complex as $\mathbb{Z}$-modules, you have a high chance of building a weak classifying object (Def. 4.4). Moreover, since these categories are quasi-stable (in the language of the paper), one can even try and apply Thm. 4.13. In fact, the whole proof is somehow based on the fact that $\mathbb{Z}$-modules cannot be homotopy concrete.
Here's something we can say which addresses a large class of examples. Let us say that a (possibly noncommutative) ring $R$ is not zero-dimensional if there exists $r \in R$ which is neither a right unit nor a right zero-divisor.
Claim: Let $R$ be a ring which is not zero-dimensional. Then the derived category $ho(D(R))$ of left $R$-modules is not concrete (i.e. does not admit a faithful functor to $Set$) and in particular is not accessible.
The proof is a straightforward generalization of Freyd's argument that the homotopy category of spaces is not concrete, and in particular is robust enough that "$D(R)$" could be interpreted to have various "boundedness" conditions if one likes.
Beginning of Proof of Claim: Pick $r \in R$ which is neither a unit nor a zero-divisor. If $M$ is a left $R$-module, define the submodule $M r^\alpha$ for any ordinal $\alpha$ inductively by $M r^0 = M$, $M r^{\alpha+1} = M r^\alpha r$, and taking intersections at limit ordinals. Note that if $\phi: M \to N$ is any $R$-module map, then $\phi(M r^\alpha) \subseteq N r^\alpha$ for any $\alpha$.
Let $\alpha$ be an ordinal. For each $n \in \mathbb N$, let $W_\alpha^{(n)}$ be the set of strictly increasing sequences of ordinals $\alpha_0 < \alpha_1 < \dots < \alpha_m$ where $m \leq n$ and $\alpha_m \leq \alpha$. Let $F_\alpha^{(n)} = R\{W_\alpha^{(n)}\}$ be the free left $R$-module on generators given by $W_\alpha^{(n)}$, and define $M_\alpha^{(n)} = F_\alpha^{(n)} / K_\alpha^{(n)}$, where $K_\alpha^{(n)}$ is generated by those elements of the form $[\alpha_1,\dots,\alpha_n] - r[\alpha_0, \alpha_1, \dots, \alpha_n]$ (when $n = 0$ we interpret this to mean that $r[\alpha_0] \in K_\alpha^{(n)}$).
Lemma 1: For $n \in \mathbb N$, the map $M_\alpha^{(n)} \to M_\alpha^{(n+1)}$ is injective.
Proof: This is straightforward, using the fact that $r$ is not a right zero-divisor.
Define $M_\alpha = \cup_{n \in \mathbb N} M_\alpha^{(n)}$. We have a natural filtration $M_\alpha^{(0)} \subseteq \cdots \subseteq M_\alpha^{(n)} \subseteq \cdots M_\alpha$, and the associated graded is naturally identified with $\oplus_{n \in \mathbb N} \oplus^{ W_\alpha^{(n)} \setminus W_\alpha^{(n-1)}} (R/(Rr))$.
Lemma 2: Pick a set of "canonical" coset representatives in $R$ of the nonzero elements of $R/(Rr)$. Then every element of $M_\alpha$ may be written uniquely in the form $\sum_i r_i w_i$ where the $r_i$ are canonical coset representatives and the $w_i$ are distinct words of some $W_\alpha^{(n)}$.
Proof: The existence of such a representation is basically obvious. Suppose that two such representations denote the same element of $M_\alpha^{(n)}$, where $n$ is minimal: $\sum_i r_i w_i = \sum_j s_j v_j$. Then they have the same image in the associated graded, and so the $w_i$'s of length $n$ match up with the $v_j$'s of length $n$; since the choice of coset representatives has been normalized, their coefficients in fact coincide. Deleting these words, we get an identification of representations in $M_\alpha^{(n-1)}$, contradicting the minimality of $n$.
Lemma 3: For all ordinals $\beta$, $M_\alpha r^\beta$ consists of those elements whose canonical representation as chosen above contains only words $[\alpha_0,\dots,\alpha_n]$ where $\alpha_0 \geq \beta$.
Proof: Using the canonical representation from Lemma 2, this is now an easy transfinite induction.
Conclusion of Proof of Claim: Note that $M_\alpha r^\alpha = R/(Rr)$ canonically, and this module is nonzero because $r$ is not a right unit. Let $f_\alpha$ be the map $R/(Rr) = M_\alpha r^\alpha \rightarrowtail M_\alpha$. Then for $\alpha < \beta$, there is no factorization of $f_\beta$ through $f_\alpha$ by Lemma 3. Pick $d \in \mathbb Z$ such that $(R/r)[d]$, each $M_\alpha[d]$, and the fiber $(M_\alpha/f_\alpha)[d-1]$ of $f_\alpha[d]$ are all in whichever flavor of $ho(D(R))$ we are working with. Then the maps $f_\alpha[d]$ are a proper class of pairwise-inequivalent weak cokernels of their fibers in $ho(D(R))$. But as Freyd shows, a proper class of pairwise-inequivalent weak cokernels out of a fixed object in a pointed category imply that the category is not concrete.
So the outline, as in Freyd's proof, is to use the fact that in $ho(D(R))$, basically every map is a weak cokernel (in fact, all maps are if we're working in the unbounded derived category).
Here's a further generalization:
Claim: Let $\mathcal T$ be a triangulated category and with a $t$-structure such that $\mathcal T^\heartsuit$ is has coproducts, which are exact. Suppose there exists $M \in \mathcal T^\heartsuit$ and a monic endomorphism $M \overset r \rightarrowtail M$ in $\mathcal T^\heartsuit$ which is not an isomorphism, and such that $Hom(M,M/r)$ is an $r$-torsion $End(M)$-module. Then $\mathcal T$ is not concrete, and in particular not accessible.
Proof: For most of the proof, we work in $\mathcal T^\heartsuit$. Similar to before, for every ordinal $\alpha$, we define $W_\alpha$ to be the set of finite subsets of $\alpha+1$, and set $M_\alpha$ to be the cokernel of the map $\oplus^{W_\alpha} M \xrightarrow{1-s} \oplus^{W_\alpha} M$, where $s$ carries the $[\alpha_0<\dots< \alpha_n]$ -copy of $M$ to the $[\alpha_1<\dots<\alpha_n]$ copy of $M$ via the map $r$. A straightforward transfinite induction shows that the $[\alpha_0<\dots<\alpha_n]$'th structure map $M \to M_\alpha$ is in $r^{\alpha_0} Hom(M,M_\alpha)$, and in particular the $[\alpha]$th structure map is in $r^\alpha Hom(M,M_\alpha)$.
It is straightforward to see that the natural map $M_\alpha \to M_{\alpha+1}$ actually splits, and its cokernel $Q$ sits in an exact sequence $M/r \to Q \to M_\alpha \to 0$, where the map to $M_\alpha$ is obtained by deleting the last letter of each word (which is always $\alpha$ here), and the copy of $M/r$ corresponds to the generator $[\alpha]$ (moreover, the $\alpha$th structure map $M \to M_\alpha$ is nonzero and factors through $M/r$). We claim now that $r^{\alpha+1}Hom(M,M_\alpha) = 0$. This is proved by induction on $\alpha$ using the following
Lemma: Let $M$ be an object of an abelian category and $r: M \to M$ a map. If $0 \to A \to B \to C$ is exact, and if $Hom(M,C)$ is $r$-torsion, then any map $\phi \in r^{\alpha+1} Hom(M,B)$ factors (necessarily uniquely) through $A$, and the factored map lies in $r^\alpha Hom(M,A)$. Dually, if $A \to B \to C \to 0$ is exact and $Hom(M,A)$ is $r$-torsion, then if $r^\alpha Hom(M,C) = 0$ it follows that $r^\alpha Hom(M,B) = 0$, for $\alpha \geq 1$.
We apply the lemma to the exact sequences $0 \to M_\alpha \to M_{\alpha+1} \to Q \to 0$ and $M/r \to Q \to M_\alpha \to 0$ at the successor steps of a transfinite induction to conclude that indeed $r^{\alpha+1} Hom(M,M_\alpha) = 0$.
Now we conclude as before: the $\alpha$th structure map $M \to M_\alpha$ doesn't factor through the $\beta$th structure map $M \to M_\beta$ for $\beta > \alpha$ because then it would lie in $r^\beta Hom(M,M_\alpha) = 0$, and it is in fact nonzero. Since $M\to M_\alpha$ are weak cokernels in $\mathcal T$, they form a proper class of pairwise-inequivalent weak cokernels out of $M$, so that $\mathcal T$ is not concrete.
|
2025-03-21T14:48:29.860357
| 2020-02-12T20:26:44 |
352572
|
{
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"LSpice",
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|
Stack Exchange
|
Is it possible to express the functional square root of the sine as an infinite product?
Cross-post from MSE.
It is known that the sine can be expressed as an infinite product: $$\sin(x) = x \prod_{n=1}^{\infty} \Big{(} 1 - \frac{x^{2}}{n^{2}{\pi}^{2}} \Big{)} .$$ We can define that functional square root of a function $g(\cdot)$ to be the function $f(\cdot)$ that satisfies $f(f(x)) = g(x)$. The square root of the sine function with respect to function composition has been discussed previously on MO on a number of occasions. For instance, here the formal power series is considered.
I wonder whether the functional square root of the sine also has an infinite product representation. If not, has any research been done on this question?
You refer to the functional square root. Is it unique?
Functional square root of sine is not unique, and it cannot be defined in the whole complex plane (according to a theorem of I. N. Baker). You can define it on $(0,\pi)$ with $f(0)=0$, for example. The resulting function is analytic in the component of Fatou set of sine adjacent to zero on the right (see, for example the first figure
here). But this function $f$ has no zeros in its domain, therefore a representation as infinite product does not make much sense. I mean that $\log f$ can be defined and any breaking of this log into summands gives you an infinite product representation. Of course this answer is related to a true product,
it does not exclude that there is some "formal" product representing this
square root, whatever a "formal product" may mean.
|
2025-03-21T14:48:29.860482
| 2020-02-12T20:43:25 |
352574
|
{
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|
Stack Exchange
|
Is integration by parts useful for obtaining global cancellation?
If $f$ and $g$ are functions on the real line, and $f$ is oscillatory, then an important technique for bounding the integral $\int fg$ is applying an integration by parts, writing
$$ \int_{-\infty}^\infty f(x) g(x) = - \int_{-\infty}^\infty F(x) g'(x) $$
where $F'(x) = f(x)$. If $f$ has rapid oscillation at a point, then this oscillation will cancel out in the integral formula for $F$ in terms of $f$, and thus have neglible impact on the values of $F$, leading to a useful bound. Thus it seems integration by parts is only useful if one wants to cancel out local cancellation in a particular integral / sum. For instance, one can use integration by parts to obtain strong bounds for oscillatory integrals of the form
$$ \left| \int \psi(x) e^{\lambda i \phi(x)} \right| $$
where $\lambda$ is a large parameter, by applying in integration by parts and then applying the triangle inequality to take in the absolute value sign into the integral. On the other hand, if we consider a bump function $\eta$ supported on $[N-1,N]$ and define $\psi(x) = \eta(|x|)$, then we have
$$ \int_{-\infty}^\infty x \psi(x)\; dx = 0. $$
Now integration by parts on $[-N,N]$ expresses this integral as
$$ 2N \int \eta(x)\; dx - \int_{-N}^N \Psi(x)\; dx $$
where $\Psi'(x) = \psi(x)$. The integral here is thus $\approx N$, and so it is not clear that integration by parts has expressed the cancellation in the integral.
My question is whether integration by parts is useful only for obtaining bounds via 'local cancellation', or whether it also captures 'global' cancellation that can occur on $f$ on different parts of the domain of $f$. If integration by parts cannot capture this bound, what techniques are available to study 'global cancellation' when estimating oscillatory quantities. Moreover, are there operators $T: L^p(X) \to L^q(Y)$ where global cancellation is useful when obtaining boundedness of these operators, or is local cancellation the only obstruction to boundedness? If local cancellation is only useful to $L^p$ estimates, does global cancellation become more relevant when proving the boundedness of operators in other norm spaces?
The short answer (in such generality as you ask about) is "it all depends on the particulars of the problem and your inventiveness in non-trivial cases". I guess some people will post some examples but there seems to be no general rule or method that always works or always does not. At least, that is what my personal experience suggests :-)
|
2025-03-21T14:48:29.860676
| 2020-02-12T22:05:43 |
352578
|
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"LSpice",
"Siddharth Bhat",
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|
Stack Exchange
|
Topology is to semi-decidability, coarse structures are to what?
There is a folklore correspondence between topology as semi-decidability amongst computer scientists, which is explained in places like:
The monograph Synthetic Topology: of Data Types and Classical Spaces
An answer by Andrej Bauer on reversing the order of quantifiers
.
I recently also learnt that Coarse structures are a type of "dual" to topologies, in that they capture "global" behaviour versus "local" behvaiour that topologies capture. This viewpoint is explained in:
This answer on MathOverflow to the question "dualizing topology", which writes down the topology axioms using category theory and then dualizes the construction.
In general, the view appears to be held that Coarse Stuctures is the correct way to dualize a topology to study large-scale phenomena. My understanding is that it was used very effectively by Gromov to study hyperbolic groups by considering quasi-isometry.
So, it is natural to ask, "what is the computational equivalent of a coarse structure"? I was hoping the answer would be something like "co-induction" / "making progress", since topology seems to be about "deciding" things. However, I have no clue how to proceed with such a question. I'd love some insight into this.
@YCor, I think your edit changed the meaning of the title. Perhaps the SAT-style analogy notation is unfamiliar? The original title, I think, was not expressing general bafflement as the current one seems to do, but asking "What is to coarse structures as semi-decideability is to topology?" In case I am mistaken, I will leave it to @SiddharthBhat to revert.
@LSpice that was indeed the intention --- However, I was worried the title was too flippant / not well communicated. If the analogy notation is indeed well known, then I'd like to change it back!
Indeed the initial colons were cryptic to me; the OP's new title looks good.
|
2025-03-21T14:48:29.860852
| 2020-02-12T22:15:22 |
352579
|
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|
Stack Exchange
|
"Sub-logarithmic" zero-free regions from Deuring-Heilbronn/Linnik's repulsion theorem
For each $n\in\mathbb{N}$, let:
$\chi_n\pmod{q_n}$ a real non-principal Dirichlet character ($q_1 < q_2 < \cdots$),
$\beta_n$ the largest real zero of $L(s,\chi_n)$,
$\delta_n := (1-\beta_n)\log(q_n)$.
Let $\chi\pmod{q}$ be a Dirichlet character, and consider $s = \sigma + it$ with $|t| < 1$. In p. 206 of Heath-Brown's "Prime twins and Siegel zeros", it is mentioned that the Deuring-Heilbronn phenomenon implies that there is some absolute constant $C>0$ such that, for each $n\in \mathbb{N}$, the region
$$ \bigg\{ s ~\bigg|~ \sigma \geq 1 - \frac{C\log(\delta_n^{-1})}{\log(q)},\ |t| < 1 \bigg\} $$
has no zeros (besides $\beta_1,\ldots,\beta_n$) of $L(s,\chi)$. (At least, that is how I interpreted the assertion "$r_0 \gg L^{-1}\log \eta$" at the mentioned page). Assuming this statement, it follows that:
(Sub-logarithmic zero-free regions (ZFR)) If there exists a sequence of Siegel zeros $\beta_n$ with $\delta_n \to 0$, then all the other zeros $\sigma + i\gamma$ of $L(s,\chi)$ with $|\gamma| < 1$ for Dirichlet characters $\chi\pmod{q}$ satisfy:
$$ \sigma < 1 - \frac{1}{o(\log(q))}. $$
It appears to me that this (or slight variations of this) statement is often used in the literature [the only example I have in mind at the moment is Remark 1 in p. 515 (p. 6 in the link) of Granville & Stark's $ABC$ implies no "Siegel zeros" for $L$-functions of characters of negative discriminant, where it is mentioned that $\delta_n \to 0$ implies $\frac{L'}{L}(1,\chi_n) = (1-\beta_n)^{-1} + o(\log(q_n))$].
However, I am having trouble following the deduction of these "sub-logarithmic ZFRs" from the Deuring-Heilbronn phenomenon alone. Using the Deuring-Heilbronn (Linnik's repulsion theorem) as in Théorème 16, Sec. 6 of Bombieri's "Le grande crible", we get that there are absolute constants $c_1,c_2 >0$ such that, fixing $n\in\mathbb{N}$, it holds:
$$ \sigma < 1 - c_1\log\left(c_2\frac{\delta_n^{-1}}{\log(q_n q)/\log(q_n)}\right) \cdot \frac{1}{\log(q_n q)} $$
(I am just taking $T = q_n q$ in Théorème 16). Assuming we have an infinite sequence $q_n \to +\infty$ with $\delta_n \to 0$, for a given $q\in\mathbb{N}$ we may take $q_k \leq q < q_{k+1}$, so that $\log(q_{k+1} q)/\log(q_{k+1})< 2$, and hence:
$$ \sigma < 1 - c_1\frac{\log(\frac{c_2}{2}\delta_{k+1}^{-1})}{\log(q_{k+1} q)}. $$
It appears, then, that to derive the "sub-logarithmic" ZFRs, it is necessary to have $\log(q_{k+1}) \ll \log(q_k)$ as $k\to \infty$, i.e.: the gaps between the conductors of consecutive exceptional characters need to be polynomially bounded, even if we assume $\delta_n \to 0$.
I do not think my conclusion is correct (e.g., I believe I misinterpreted some aspect of Heath-Brown's paper), but I have not been able to get rid of this condition on the growth of the $q_n$. In short, my question is the following:
Q. Is it really possible to derive "sub-logarithmic" ZFRs from Linnik's repulsion theorem (i.e., without additional growth conditions on the $q_n$)?
Let $\chi$ be a non-principal real Dirichlet character modulo $q$. Let
$$\beta_0=1-\frac{1}{\eta\log q}$$
be a real zero of $L(s,\chi)$ satisfying $\eta\geq 100$ for convenience (Heath-Brown's condition is $\eta\geq 3$). Let $\rho=\beta+i\gamma$ be any zero of $L(s,\chi)$ such that $\rho\neq\beta_0$ and $|\gamma|\leq 1$. We strengthen Heath-Brown's claim $r_0\gg L^{-1}\log\eta$ to (note that $L=\log q)$
$$1-\beta\gg\frac{\log\eta}{\log q}.$$
We shall deduce this from Theorem 2 of Jutila's 1977 paper "On Linnik's constant", which is also Heath-Brown's reference. Let us write $\delta$ for the left hand side. If $\delta>1/60$, then the statement is trivial by $\eta\ll q$. So we shall assume that $\delta\leq 1/60$. Then, Jutila's theorem yields readily (using $D\leq 2q$) that
$$1-\beta_0\geq\frac{1}{10}\cdot\frac{q^{-3\delta}}{\log q}.$$
In other words, $\eta\leq 10 q^{3\delta}$. By the assumption $\eta\geq 100$, this implies that $\eta\leq q^{6\delta}$, or equivalently that
$$\delta\geq\frac{1}{6}\cdot\frac{\log\eta}{\log q}.$$
We have verified Heath-Brown's claim.
Is the $q$ from the exceptional character the same $q$ for the other characters? (This was the exact aspect from Jutila's paper which I did not understand, which was what prompted me to look at Bombieri's "Le Grande Crible")
@Alufat: Jutila's Theorem 2 concerns Dirichlet characters of the same modulus $q$. Similarly, Heath-Brown's Theorem 1 concerns a fixed $q$. Of course this theorem is interesting only when infinitely many $q$'s satisfy the conditions, but $q$ is fixed throughout the proof (and $L$ abbreviates its logarithm). In fact on p.206 of Heath-Brown's paper, the $\rho$'s are zeros of the same $L$-function $L(s,\chi)$, where $\chi$ is as in Theorem 1. They serve to approximate $L'(s,\chi)/L(s,\chi)$, cf. (4.1) and (4.2).
Thank you very much. I believe this was exactly my misconception (re-reading Theorem 1, I saw that Heath-Brown claims uniformity only over a polynomially bounded interval!). The issue of fixing $q$ (which is also present in Tao's blogpost explaining Heath-Brown's proof) confused me because I had the stronger statement (the "sub-logarithmic" ZFR) on the back of my mind while reading. I understand now that the statement I thought "$r_0\ll L^{-1}\log\eta$" meant was actually different; is then the "sub-logarithmic ZFR" (i.e., different $q$s) too much harder than Deuring-Heilbronn?
@Alufat: I believe that your "Sub-logarithmic ZFR" is harder than Jutila's theorem, and I don't know if it has been deduced from $\delta_n\to 0$. Note, however, that $\delta_n\to 0$ implies $\log q_{n+1}/\log q_n\to\infty$, by Corollary 11.9 in Montgomery-Vaughan: Multiplicative number theory I.
Thank you for the reference! (I never got to this chapter of Montgomery-Vaughan). I believe that something like Corollary 11.9 was exactly what was I looking for (I apologize for not being completely clear in the question). I will add the tag 'reference-request' and accept your answer (since it clarified everything I expected). Thanks a lot!
@Alufat: I am glad I could help!
|
2025-03-21T14:48:29.861251
| 2020-02-12T22:40:07 |
352580
|
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"url": "https://mathoverflow.net/questions/352580"
}
|
Stack Exchange
|
Positive Borel measure with empty support on a standard measurable space
Let $X$ be a $T_0$ topological space (not $T_1$), and let $\Sigma_X$ be the Borel $\sigma$-algebra. Assume that $(X,\Sigma_X)$ is a standard measurable space, i.e., measurably isomorphic to the Borel $\sigma$-algebra $(Y,\Sigma_Y)$ of a complete separable metric space $Y$.
Question (amended to honour Dieter Kadelka): Is it true that there does NOT exist a positive measure $\mu$ on $(X,\Sigma_X)$ with empty support, $\mathrm{supp}\mu=\emptyset$?
Note that the measurable isomorphism between $X$ and $Y$ is a priori merely Borel measurable, and can potentially map open sets to sets without interior and vice versa.
Thank you.
Discussion: The question arises from the discrepancy in definitions of the support of a Borel measure on a topological space. I think that the standard definition is this, in which case, unless $X$ is a very good space, you don't have to expect $\mu(X\setminus\mathrm{supp}\mu)=0$. However, in Propositiom 8.6.8 in Dixmier's "$C^*$-algebras", the author defines the support of a measure as the smallest closed subset with negligible complement (the standard analysis definition). Now I wonder if the two definitions coincide in this context. Note that the space in question is only $T_0$ in general, possesses a dense locally compact open subset (Proposition 4.4.5 in the book), and the Borel structure is measurable isomorphic to that of a complete separable metric space (Proposition 4.6.1 in the same book).
Discussion 2: Well, there is more to the spectrum of a separable postliminal $C^*$-algebra - it is locally qusicompact, second countable etc. But the question remains as it is: does the mere Borel isomorphism to a Polish space rule out empty support?
Please clarify the assumptions. What if $X$ is Polish itself? Are you asking for an example of a $T_0$-space with the above property?
Is it better now? I am asking if such a measure can exist, or if it is excluded. It is excluded if $X$ is good enough, but all I have is $T_0$.
The point is that, if the underlying space is not Hausdorff, a non-zero Borel measure can actually have empty support, which is sad. There is an example at the Wiki page referred to above in my question. Now whether or not that can happen in a standard Borel structure is the question. Thank you.
An example of a $T_0$-space which is not $T_1$ is $\mathbb{R}$ with the right order topology $\tau$ (Steen/Seebach: Counterexamples in Topology, 50). The topology $\tau$ is coarser than the usual topology $\tau_1$ on $\mathbb{R}$, and both have the same Borel-$\sigma$-algebra. Since any open neighbourhood of $x \in \mathbb{R}$ w.r.t. $\tau$ is also an open neighbourhood w.r.t $\tau_1$, the support of any $\mu \not= 0$ is not empty (w.r.t. $\tau$). So I think your question should be "Is there a $T_0$ space $X$ and a Borel-measure $\mu$ with the properties in the question".
There exists a counterexample under the Continuum Hypothesis, which implies that the unit interval $[0,1]$ admits a well-order $\preceq$ such that for every $x\in[0,1]$ the initial interval ${\downarrow}x=\{y\in [0,1]:y\preceq x\}$ is at most countable. On $[0,1]$ consider the Hausdorff topology $\tau$ generated by the subbase consisting of the sets $[0,a)$, $(a,1]$ and ${\downarrow}a$ where $a\in [0,1]$. It is easy to see that the Borel $\sigma$-algebra generated by the topology $\tau$ coincides with the Borel $\sigma$-algebra generated by the standard topology on $[0,1]$. We claim that the Lebesgue measure on $[0,1]$ has empty support (in the topology $\tau$). This follows from the fact that each point $x\in [0,1]$ has the open neighborhood ${\downarrow}x$ of Lebesgue measure zero.
|
2025-03-21T14:48:29.861504
| 2020-02-12T23:59:20 |
352583
|
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"url": "https://mathoverflow.net/questions/352583"
}
|
Stack Exchange
|
Hecke algebra relation versus $\operatorname{SL}_2$ trace relation
The quadratic relation in the (type $A$) Hecke algebra is $(T-t)(T+t^{-1}) =0$, which can be rewritten as
$$
T-T^{-1} = t-t^{-1}$$
Suppose $A \in \operatorname{SL}_2(\mathbb{Q})$ with eigenvalues $a,a^{-1}$. Then it isn't hard to show the following identity $$A + A^{-1} = (a+a^{-1}) \mathrm{Id}$$
If $i^2=-1$ then we can "convert" between these two formulas by saying $T=iA$ and $t = ia$.
Is there a conceptual reason for this?
Interesting observation! Unfortunately, this seems to be a coincidence.
The original, algebraically motivated definition of the Hecke algebra gives the quadratic relation as
$$(T-q)(T+1)=0.$$
For $SL_2$, the Hecke algebra has a faithful two dimensional representation. The eigenvalues of $T$ are $q$ and $-1$, and the quadratic relation simply becomes the Cayley--Hamilton theorem.
But any 2x2 matrix $A$ also satisfies its characteristic equation
$$(A-\lambda_1)(A-\lambda_2)=0,$$
where $\lambda_1,\lambda_2$ are the eigenvalues of $A$. If $A$ is nonsingular, then the rescaling
$A\mapsto \sqrt{\lambda_1\lambda_2}A$ transforms this into
$$(A-a)(A-a^{-1})=0,$$
where $a=\sqrt{\frac{\lambda_1}{\lambda_2}}$.
Thus your observation reduces to the fact that $i$ is a "nice" value of $\sqrt{\lambda_1\lambda_2}$ for the Hecke algebra. But in fact the niceness here is rather circular. As mentioned above, the more motivated definition of the Hecke algebra is not symmetric. To reach the modern presentation, we symmetrize the definition by replacing $T\mapsto q^{1/2} T$ to find (with $t=q^{1/2}$)
$$(T-t)(T+t^{-1}) =0. $$
The goal of this rescaling is to make this relation "nice", i.e. transforming nicely under $t\mapsto t^{-1}$. We could have alternatively rescaled $T\mapsto iq^{1/2}$ to get
$$(T-t)(T-t^{-1})=0. $$
These are the only two rescalings if we want the relation to be nice. Thus a priori the only possible values of $\sqrt{\lambda_1\lambda_2}$ for a nice Hecke algebra presentation are $1$ or $i$, explaining your observation. (We pick the $i$ presentation over the $1$ presentation to preserve the behavior of the Hecke algebra as $t\rightarrow 1$.)
Thus the fundamental reason for the similarity between the Hecke algebra presentation and the trace relation in $SL_2$ is that the Hecke algebra is two dimensional. But the dimension of the Hecke algebra is the size of the Weyl group $S_2$. The fact that $|S_2|=2=\mathrm{rank}(SL_2)$ is a coincidence.
|
2025-03-21T14:48:29.861699
| 2020-02-13T00:31:40 |
352588
|
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"authors": [
"Alexey Ustinov",
"Stanley Yao Xiao",
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Stack Exchange
|
Distribution of partial quotients of $\sqrt{d}$ with $X < d \leq 2X$
Recall that for a real number $\alpha$, it has a continued fraction expansion usually written as
$$\displaystyle \alpha = [a_0; a_1, a_2, \cdots].$$
Moreover, $\alpha$ is rational if and only if its continued fraction expansion is finite, and is a quadratic irrational if and only if the continued fraction expansion is eventually periodic. The numbers $a_0; a_1, a_2, \cdots$ are called partial quotients of $\alpha$.
It is known that with respect to Lebesgue measure, almost all real numbers $\alpha$ have the property that the partial quotients of $\alpha$ follow a specific distribution: the frequency that $k$ appears in the sequence $\{a_0; a_1, a_2, \cdots\}$ is $\frac{1}{\log 2} \log \left(\frac{(k+1)^2}{k(k+2)} \right)$.
Consider the interval $I_X = (X, 2X]$ for some (large) positive number $X$. For each integer $n \in I_X$, consider the continued fraction expansion of $\sqrt{n}$, say $\sqrt{n} = [a_0; a_1, a_2, \cdots]$. We know that this sequence is eventually periodic, and in fact $\sqrt{n} = [a_0; \overline{a_1, a_2, \cdots, a_2, a_1, 2a_0}]$ where the bar denotes the periodic part and the string $a_1, a_2, \cdots, a_2, a_1$ denotes a palindrome.
My question is: as $n$ runs over $I_X$, how are the partial quotients of $\sqrt{n}$ distributed? In other words, for each positive integer $k$ what is the frequency of appearance of $k$ as partial quotients of $\sqrt{n}$ as $n$ runs over $I_X$?
Note that the frequency is 0 if $k \gg X^{1/2}$, since it is known that the partial quotients of $\sqrt{n}$ are at most $2 \sqrt{n}$ in size.
It is not an answer, but probably it is the only known information about Gauss-Kuz'min statistics for quadratic irrationals.
For a reduced quadratic irrational $\omega$ (which has a purely periodic representation in the form of a continued fraction) denote by $\rho(\omega)$ the length of $\omega$ which is the length of the corresponding closed geodesic on modular surface $\mathbb{H}/PSL_2(\mathbb Z)$, $\mathbb{H}=\{(x,y):\
y>0\}$ (the projection of the
geodesic joining $\omega$ and $\omega^*$, where
$\omega^{*}$ is the number conjugate to $\omega$). The answer is positive in the case when you average Gauss-Kuz'min statistics over all $\omega$ such that $\rho(\omega)\le X$. More precisely, let $x, y \in [0, 1]$ be real numbers and
$$
r(x,y;N)=\sum_{\substack{\omega\in\mathcal{R}\\
\varepsilon_0(\omega)\leqslant N}}
[\omega\leqslant x,\ -1/\omega^{*}\leqslant y].
$$
Here $\mathcal{R}$ is the set of reduced quadratic irrationals, $\varepsilon_0(\omega)=\frac{1}{2}(x_0+\sqrt{\Delta}y_0)$ is the fundamental solution
of Pell’s equation
$$X^2-\Delta Y^2=4,$$ $\Delta=B^2-4AC$, where $AX^2+BX+C$ is the minimal polynomial of $\omega$; moreover, $[A]$ stands for $1$ if the statement
$A$ is true and for $0$ otherwise. The fundamental solution $\varepsilon_0(\omega)$ is clesely connected to the length: $\rho(\omega)=2\log\varepsilon_0(\omega)$. Then (see Theorem 3 from Spin chains and Arnold's problem on the Gauss-Kuz'min statistics for quadratic irrationals)
$$
r(x,y;N)=\frac{\log(1+xy)}{2\zeta(2)}N^2+{O}(N^{3/2}\log^4N).
$$
I believe there are two differences between your answer and what I am asking about. First, your height function is essentially the size of the fundamental unit of a quadratic order, which is typically much much larger than the discriminant, whereas I am counting by discriminant. Second, you are dealing with all quadratic irrationals, whereas I am focused only on square roots of positive integers (or very roughly speaking, only quadratic forms in the principal class).
@StanleyYaoXiao Yes, but you question looks like unsolved problem, That is why I gave the information which is known for me. I've added this remark to my answer.
|
2025-03-21T14:48:29.861956
| 2020-02-13T01:16:26 |
352589
|
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|
Proportionality constant for the growth rate of $\zeta\left(s\right)$ along the imaginary axis
Going through what it says on page 95 of Titchmarsh's book on the Zeta function, and using his remark about $\mu$ at the bottom of that page, I conclude that there is some constant $C>0$ so that: $$\left|\zeta\left(it\right)\right|\leq C\sqrt{\left|t\right|},\textrm{ }\forall t\in\mathbb{R}$$
In my current work, I need to know an explicit value for this $C$. Graphing the quotient: $$\frac{\left|\zeta\left(2^{t}i\right)\right|}{\sqrt{2^{t}}}$$ gives an apparent global maximum of slightly less than 1.37, occurring at $t≈8.9082$
Is this rigorously justifiable? I.e., is:
$$\left|\zeta\left(it\right)\right|\leq 1.37\sqrt{\left|t\right|},\textrm{ }\forall t\in\mathbb{R}$$
true? More generally, what's the most accurate estimate of this type which is currently known (with explicit constants)?
You need to read the definition of $\mu(0)$ more closely. It is the (greatest) lower bound of the numbers $\xi$ such that $|\zeta(it)|$ is $O(t^\xi)$. Presumably the implicit constant in each Big Oh can depend on $\xi$, and even grow as $\xi$ decreases. Even though $\mu(0)=1/2$, this does not prove $\zeta(it)=O(t^{1/2})$. (The Big Oh may be true, but is does not follow from what's on p.95.)
I was worried about that. Darn.
|
2025-03-21T14:48:29.862080
| 2020-02-13T01:23:36 |
352590
|
{
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"authors": [
"Carlo Beenakker",
"Pritam Bemis",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/150564"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352590"
}
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Stack Exchange
|
Mehta integral and orthogonality
The Mehta integral is the following expression:
$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2}
\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$
To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2}
\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$
It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure
\begin{equation}
\label{eq:ortho}
\frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0.
\end{equation}
Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of
$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)}\sqrt{\int 1 \ d\mu_{n,\gamma}(t)}}$$
where I admit that I use the limit without actually knowing whether it exists.
Case 1:
As one can guess from this thread, if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$
Case 2:
If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.
Case 3:
If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.
My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?
@CarloBeenakker you are right, simplifying the question I accidentally deleted it.
Case 3:
Let me define $t=x\sqrt{2\gamma}$, then it is known from random-matrix theory (see, for example, Forrester's book) that for a fixed $\gamma$ the probability distribution $P(x_1)$ of a single eigenvalue $x_1$ tends in the limit $n\rightarrow\infty$ to the $\gamma$-independent semicircle
$$P(x)=\frac{1}{\pi n}\sqrt{2n-x^2},\;\;|x|\leq\sqrt{2n}.$$
The desired ratio $\nu$ then evaluates to
$$\nu=\frac{\int (2\gamma x^2-1)P(x)\,dx}{\left[\int (2\gamma x^2-1)^2P(x)\,dx\right]^{1/2}}=\frac{\gamma n-1}{\sqrt{2 \gamma n (\gamma n-1)+1}}\rightarrow \frac{1}{\sqrt 2}\;\;\text{for}\;\;n\rightarrow\infty.$$
Case 2:
The case that $n\rightarrow\infty$, $\gamma\rightarrow 0$ at fixed $\gamma n=\alpha>0$ has been studied in The mean spectral measures of random Jacobi matrices related to Gaussian beta ensembles (2014), see also arXiv:1611.09476. The probability distribution $P_\alpha(t)$ is given in this limit by
$$P_\alpha(t)=\frac{e^{-t^2/2}}{\alpha\sqrt{2\pi}}\frac{\Gamma(\alpha)}
{|f(t)|^2},\;\;f(t)=\int_0^\infty x^{\alpha-1}e^{ix t-x^2/2}\,dx.$$
From this the desired $\nu$ can be readily computed,
$$\nu_\alpha=\frac{\int (t^2-1)P_\alpha(t)\,dt}{\left[\int (t^2-1)^2P_\alpha(t)\,dt\right]^{1/2}}=\frac{\alpha}{\sqrt{\alpha (2 \alpha+3)+2}},$$
so for $\alpha=1$ I find $\nu_1=1/\sqrt 7$. The value $\nu=1/\sqrt 2$ of case 3 is reached for $\alpha\gg 1$.
thank you, that's interesting. Do you have any conjectures about case 2 whether the result will be of "order 1" or "tend to zero"?
my "conjecture" would be to substitute $\gamma n=\text{constant}$ in the formula for $\nu$, which would give $\nu=0$ in case 2; incidentally, the plots you show do not take into account the repulsion between the $x_i$'s with $i$ unequal to 1, so I don't think one can draw any conclusion from those.
thank you for your prompt response. This sounds very plausible. I will update the question accordingly.
very interesting, so in Case 3: we do not see $\gamma$ at all in the limit and in this regime $2$, although the value is of order $1$, it vanishes as $\alpha$ tends to zero.
yes, this somehow makes sense if you think of case 2 approaching case 3 when $\alpha\rightarrow\infty$ (hence $\nu\rightarrow 1/\sqrt 2$), and approaching case 1 if $\alpha\rightarrow 0$ (hence $\nu\rightarrow 0$).
|
2025-03-21T14:48:29.862467
| 2020-02-13T03:16:45 |
352595
|
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"authors": [
"Mario Krenn",
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"მამუკა ჯიბლაძე"
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Stack Exchange
|
Parametrization of real-valued SU(N)
I want to construct a $SU(N)$ matrix $V$, with the following property:
All the elements of the first row are given, i.e. $V_{1,j}=a_i$ (with $\sum_i a_i^2=1$)
All matrix elements are real, i.e. $V_{i,j} \in \mathbb{R}$
How can I find a matrix $V$ that satifies the criteria? Specifically, how can I find the matrix elements as a function of $a_i$, i.e. $V_{i,j}(a_i)$?
Special case: SU(2)
$$
V=
\left[ {\begin{array}{cc}
a_1 & a_2 \\
V_{2,1} & V_{2,2} \\
\end{array} } \right]
$$
We easily find $V_{2,1}=-a_2$ and $V_{2,2}=a_1$.
Special case: SU(3)
$$
V=
\left[ {\begin{array}{cc}
a_1 & a_2 & a_3 \\
V_{2,1} & V_{2,2} & V_{2,3} \\
V_{3,1} & V_{3,2} & V_{3,3} \\
\end{array} } \right]
$$
Here already I cannot find any feasible way to represent $V_{i,j}$ as a function of $a_1, a_2, a_3$. I have tried to use the generators of SU(3), the Gell-Mann matrices $\lambda_i$. In particular, $\lambda_2$, $\lambda_5$, $\lambda_7$ are the generators for real-valued SU(3) matrices. However, the resulting equation system involves multiple trigonometric functions for which I cannot solve $V_{i,j}(a_1, a_2, a_3)$.
The matrix $V$ is not unique, I just want any solution.
Generally your functions $V_{i,j}$ will not be continuous in the variables $a_i$. This is because if they were continuous, you would be asking for sections of Stiefel bundles, and these often do not exist. But if you are okay with discontinuous function, yes there are plenty of options. Gram-schmidt is one way to construct such functions.
Thank you for your answer, this is already very interesting. I didn't know that it cannot be done continuously. In my case fortunatly, it is no problem.
Are there any particular cases that are especially important for your work? Or do you need the arbitrary $N$ case? Other than Gram-Schmidt, the holonomy of the tangent bundle also provides a relatively clean solution.
I would be extremly happy with N=3 already, to get an idea how to construct it, and in particular because I could use the N=3 matrix directly in my project.
Since you are requiring all the entries to be real, you are really working within $\mathrm{SO}(n)$, so you can easily see that this is impossible for $n=3$, if you want it to be continuous. The second row would be a nowhere vanishing tangent vector field to the $2$-sphere (the position vector being the first row), and, by a well-known theorem, this is impossible for a continuous tangent vector field. If you don't require continuity, then, of course, it's easy: Do it continuously away from a single point on the $2$-sphere (easy) and then define it however you like at that point. (cont....).
....(cont from above). However, for $n=4$ and $n=8$, you can do this. For example,$$\begin{pmatrix} a_1&a_2&a_3&a_4\-a_2&a_1&a_4&-a_3\-a_3&-a_4&a_1&a_2\-a_4&a_3&-a_2&a_1\end{pmatrix}.$$ The $n=8$ case can be done using the matrix of left multiplication by a unit octonion on the space of octonions, which is a real vector space of dimension 8. Other than $n=1,2,4,8$, it cannot be done continuously, by a famous theorem of Adams.
In addition to the comments I made above about continuous solutions, I thought I'd point out a solution that works for all $n$ with only one point of discontinuity, namely
$$
(a_1\ a_2\ \ldots\ a_n) = (1\ 0\ \ldots\ 0).
$$
Away from this point, one can start with the following formulae:
$$
V_{i,1}= V_{1,i} = a_i\qquad\text{and}\qquad
V_{i,j} = \delta_{ij} - \frac{a_ia_j}{(1-a_1)}\quad\text{when}\ 1<i,j\le n
$$
Note that the resulting matrix is both orthgonal and symmetric. However, it has determinant $-1$, so reversing a single row, say, the last one, will give a matrix in $\mathrm{SO}(n)$.
That's awesome, thank you, i already use it in my codes! How did you find the construction? Did you just use some advanced techniques, or was it obvious for you? (maybe you can add 1-2 sentences, would find it very interesting).
@MarioKrenn: Yes, this is the matrix of the orthogonal, linear transformation that exchanges the vector $a= (a_1\ \ldots\ a_n)$ with the vector $n= (1\ 0\ \cdots\ 0)$ and leaves all vectors perpendicular to them fixed. In other words, it is reflection in the hyperplane orthognal to the difference vector $a-n$. That's why it has determinant $-1$ and why it isn't defined when $a=n$.
Since you state that already the $3\times 3$ case would be very useful to you:
$$
V=\left( \begin{array}{ccc}
a_1 & a_2 & a_3 \newline
-\frac{a_2 }{\sqrt{a_1^2 +a_2^2 } } & \frac{a_1 }{\sqrt{a_1^2 +a_2^2 } } & 0 \newline
-\frac{a_1 a_3 }{\sqrt{a_1^2 +a_2^2 } } & -\frac{a_2 a_3 }{\sqrt{a_1^2 +a_2^2 } } & \sqrt{a_1^2 +a_2^2 }
\end{array} \right)
$$
Not continuous (or even well-defined) at $(a_1,a_2,a_3)=(0,0,\pm1)$.
@RobertBryant - As already pointed out by Ryan Budney, one cannot expect a continuous solution in general, so this behavior is not surprising. Of course, supplementing a choice at $a_1 = a_2 =0$ is trivial.
You are correct, but discontinuities are generally bad for numerical implementations, and your proposed solution is discontinuous at two places rather than the minimal one. Also, as I point out in my remarks to the question itself, there is a very simple continuous solution when $n=1,2,4,8$. Also, there is a simpler solution (for all $n$) that has only one point of discontinuity.
@RobertBryant Thank you - yes, I didn't strive to write down something optimal in terms of continuity. I was mainly reacting to the OP's statement that anything for $N=3$ would be immediately useful and continuity wasn't a concern.
One way to formulate your problem is via Stiefel bundles.
In your case there is the bundle $SU(n) \to S^{2n-1}$ given by taking the first row vector of a special unitary matrix. You are asking for a way to reverse the process, i.e. if you have a unit row vector you want to complete it to not just a Hermitian basis but one that has determinant one.
The idea is to consider complex mirror reflections $M_p$ in the complex hyperplane orthogonal to vectors $p \in S^{2n-1}$. Let $q \in S^{2n-1}$ and let $p \in H_q S^{2n-1}$, this is meant to indicate the hemi-sphere of $S^{2n-1}$ centred at the point $q$. Then the composite
$$M_q \circ M_p$$
should be essentially the hermitian matrix you are looking for. Likely it will only be giving you the first column vector the one you want (depending on how you think about matrices, i.e. perhaps you will need to take the transpose). That vector will be the vector twice as far from p as q is, i.e. take the great circle from p to q, and go twice as far. That's why there is the discontinuity at the point antipodal to p.
Does this sound reasonable? Hmm, on second thought, I'm running a bit on autopilot here. I am not certain if you get every vector in $S^{2n-1}$ via this construction. In the orthogonal group $O_n \to S^{n-1}$ this construction works fine. I'll see if this construction can be fixed for $SU(n)$. But I might need to sleep on it.
There is an additional requirement that everything must be real, so in fact one deals with $SO_n\to S^{n-1}$
|
2025-03-21T14:48:29.862941
| 2020-02-13T03:34:50 |
352596
|
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|
Stack Exchange
|
A question on trig series
Assume $\{a_k\}_{k\ge1}$ is a real sequence such that $u(x) = \sum_{k\ge 1}a_k\sin(kx)$ is a smooth function, and for every $x \in [-\pi, \pi]$
$$\left(\sum_{k\ge 1}\frac{a_k}{k}\sin(kx)\right)\left(\sum_{k\ge1}a_k\cdot k\cos(kx)\right) = \left(\sum_{k\ge1}a_k\sin(kx)\right)\left(\sum_{k\ge1}a_k\cos(kx)\right).$$
What can we say about $\{a_k\}$? Can we show that there exists at most one term of $a_k$ that is nonzero?
Note that if we set $w(x) = -\sum_{k\ge 1}\frac{a_k}{k}\sin(kx)$, we have the relation $u_xw = uw_x$ and $w_x = Hu$. Here $H$ denotes the standard Hilbert transform.
Your sums should be surrounded by parentheses, right? (I initially read them as nested sums, which doesn't make much sense.)
Yes, this says that $fg'=gf'$ (defining $f,g$ as the two series with the sines), so $(f/g)'=0$ and thus also $a_k/k=ca_k$, so $a_k=0$ for $k>1$. (There are probably other ways of doing it too.)
Thanks for your comment, Christian. This is the first intuition to this question, but the problem is that $g$ may have strange zero sets even the function is smooth. The fraction $f/g$ is not always defined. So we cannot use this to prove the argument rigorously.
I think we need to be careful about whether $f/g$ is defined. For example, let $f = x\exp(-\frac{1}{x^2})$. Define the function $g$ as $g = f$when $x\ge 0$, $g = -f$ when $x < 0$. One can check that both $f$ and $g$ are smooth and $f'g = fg'$. But $f/g$ is not a constant in this situation. The problem is that $x = 0$ is a zero point of infinite order. Thus if $g$ has weird zero set, $f/g$ may have strange behaviour.
Do we need $f/g$ to be defined everywhere for Christian's argument to work? Won't an open interval be enough? (Aside: $a_k =0$ for $k\neq 1/c$).
I was aware of this issue, this was just supposed to be a sketch, sorry for not making that clearer. I assumed that since one can also apply the argument to $g/f$, there will be a way around this issue. Michael's idea sounds promising too, especially since $f,g$ will be real analytic for rapidly decaying $a_k$'s.
Yes, you are right, if both $f$ and $g$ are analytic, then everything should be ok. The question is what happens if we only assume smoothness of $f$ and $g$. I think I should modify the question to make it more clear.
This ($f,g$ smooth, but not more) seems an interesting question. I guess (assuming the statement is true) one would have to use that $f,g$ are not just any two functions, but $Hf=g'$, with $H$ denoting the Hilbert transform . I have no clear idea how to proceed though
This is a good observation! I don't know how the Hilbert transform helps this problem, but this is actually where this problem comes from.
If that is where the problem comes from, perhaps it would have been better to transparently give this context in the original post (rather than having $fg'=gf'$ and $Hf=g'$ observed later in the comments)?
By the way, if this problem is solved here in a non-trivial way and later used in a paper, it would be appropriate to cite this mathoverflow post.
Thank you for your suggestion, sharpend! Sure, if any comments or answers here are used in a paper, this post should be cited. I didn't include the original problem since I thought the problem formulated in this way is more interesting and easier to understand. I will edit the post to include this.
@JacobLu Are there updates on this problem? Is still open or you found the solution? As you know Claudio Rea tried to solve it but I do not know if his solution was correct. Claudio died a week ago and thinking to him this problem came again to mind again. Thanks for letting me know.
@GiorgioMetafune I am so sorry to hear the sad news. This is really shocking since I was talking to Claudio by email one month ago. Unfortunately, the question is still open now. The previous answer by Claudio has really good observations, but there were some issues with several technical details. I am still trying to solve it and I will certainly let you know if there are any updates on this problem.
|
2025-03-21T14:48:29.863239
| 2020-02-13T03:36:56 |
352597
|
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|
Stack Exchange
|
Probability of m crossings of 0 before time n of a Gaussian random walk
Let $S_n = \sum_{n=1}^n X_n$ be a Gaussian random walk where $X_n$ are i.i.d random variables with distribution $\mathcal{N} (0,1)$. Could anyone show some hints and reference about how to compute the following probability?
Let $M(n)$ be the number of crossings of $0$ before time $n$, i.e., the largest integer $k$ such that there exists a strictly increasing finite sequence $a_1 < b_1 < a_2 < b_2 < \ldots < a_k < b_k$
of nonnegative integers smaller than or equal to $n$ with the properties $S_{a_i} < 0$ and $S_{b_i} > 0$ for every $i$. Then for any $m <n$, what is
$$
\mathbb{P}\left( M(n) < m \right)?
$$
A good upper bound(e.g. exponentially small) of this probability also provides an answer.
Certainly not exponential: $P(M(n)=0)\sim 1/\sqrt{n}$
|
2025-03-21T14:48:29.863322
| 2020-02-13T03:37:38 |
352598
|
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|
Stack Exchange
|
Embedding of Verma modules in Kac-Moody Lie algebras
Let $\mathfrak{g}(A)$ be a symmetrizable Kac-Moody Lie algebra over $\mathbb{C}$ and ($\mathfrak{h}$, $\Pi, \Pi^\vee)$ be a realization of the GCM $A$. Assume that $$\mathfrak{g}(A)=\mathfrak{h} \oplus \mathfrak{n}^+ \oplus \mathfrak{n}^-$$
be the standard triangular decomposition. $\Pi = \{\alpha_i, i=1,\cdots,n \}$ is the set of simple roots. Let $\Delta, \Delta^+, \Delta_{re}(\Delta^+_{re}), \Delta_{im}(\Delta^+_{im}) $ be the corresponding root system, the set of positive roots, the set of (positive) real roots, the set of (positive) imaginary roots respectively. Let $W$ be the Weyl group.
For each weight $\lambda \in \mathfrak{h}^*$, the Verma module with respect to $\lambda$ is denoted by $M(\lambda)$, which is defined by the quotient of the universal enveloping algebra $\mathrm{U}(\mathfrak{g})$ and its left ideal generated by $\mathfrak{n}^+$ and $h-\lambda(h)$, for each $h \in \mathfrak{h}^*$.
Let $(\cdot, \cdot)$ be the standard non-degenerated bilinear form of $\mathfrak{h}^*$. Fix $\rho \in \mathfrak{h}^*$, such that $2(\rho, \alpha_i)/(\alpha_i,\alpha_i)=1$, for each $i=1,\cdots,n$. Consider the set $\mathcal{C}$ of non-critical weights of $\mathfrak{g}(A)$ defined as following:
$$\mathcal{C}:=\{\lambda \in \mathfrak{h}^* | 2(\lambda + \rho, \beta) \neq (\beta, \beta), \text{for any imaginary root } \beta \in \Delta_{im}\}.$$
It seems to be true that
$$(*) \quad \mathrm{dim}_\mathbb{C}\mathrm{Hom}_{\mathfrak{g}(A)}(M(w \cdot \lambda), M(\lambda)) \leq 1,$$
for any $\lambda \in \mathcal{C} $ and any $w \in W$ no matter what the orbit $W(\lambda)\cdot \lambda$ contains a dominant weight (antidominant weight) or not, where $W(\lambda)$ is the integral Weyl group of $\lambda$, and $w\cdot \lambda = w(\lambda + \rho)-\rho$.
Questions:
(1) Is the above claim true? If the answer is yes, then how do we show it? If not, can you list any counterexample?
(2) In the critical case, i.e. $\lambda$ lies in $\mathfrak{h}^*-\mathcal{C}$, is $(*)$ true?
Thanks for your time. Please let me know if anything is unclear!
|
2025-03-21T14:48:29.863466
| 2020-02-13T04:04:34 |
352599
|
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|
Stack Exchange
|
Scalar curvature in terms of second fundamental form, reference request
I would like to cite a reference for the following formula for scalar curvature:
If $\Sigma$ is a hypersurface in Euclidean space, then $R=H^2-\lvert A\rvert^2$, where $R$ is the scalar curvature of
$\Sigma$, $A$ is the second fundamental form of $\Sigma$, and
$H=-\mathop{\mathrm{tr}}A$ is the mean curvature of $\Sigma$.
This formula is a special case of a more general formula on Wikipedia where $\Sigma$ is a hypersurface in a manifold rather than Euclidean space. It can be quickly proved from the Gauss equation or by computing both sides in terms of the principal curvatures, but I was unable to find it in the textbooks I checked. Would you happen to recognize this formula and be able to suggest a citation for it?
I would call it a special case of the Gauss-Codazzi equations. It is classical and you can find it on any textbook in Riemanniana Geometry, such as do Carmo, Petersen, Gallot-Hulin-Lafontaine, ...
It's a formula that appears in any paper studying or using the scalar curvature of a hypersurface. Since most textbooks don't discuss specifically results involving the scalar curvature of a hypersurface, they probably don't mention this formula. However, as you say, it's derived easily by simply taking two traces of the Gauss equations. So, if you want to cite it, just say something roughly like that. Or you could google for papers about it and look for one that states the formula explicitly.
|
2025-03-21T14:48:29.863588
| 2020-02-13T05:39:40 |
352602
|
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|
Stack Exchange
|
Nature of Fourier coefficient of a modular form after applying a certain map (trace operator)
Asking this here because of no response at (MathStackExchange).
Let $N|M$, and consider the trace operator $Tr^M_N$ defined on $M_k(\Gamma_0(M))$ - vector space of modular forms of weight $k$ for the congruence subgroup $\Gamma_0(M)$, mapping it to $M_k(\Gamma_0(N))$ (level lowering operator). It is defined as $$(Tr_N^M f)(z) = \sum\limits_{\begin{pmatrix} a &b\\c & d \end{pmatrix}\in \Gamma_0(N) \ \backslash \Gamma_0(M)} (cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right),$$ where the sum varies over representatives. Assume $f = \sum_{n\geq1} a_f(n)q^n$ is a cusp form in $M_k(\Gamma_0(M))$, with $rational$ Fourier coefficients. Is it easy to derive the nature of coefficients of $(Tr_N^M f)(z)$ - i.e., whether they are still rational or not ? Finding out the explicit Fourier expansion is one way, but it seems bit lenghty(and tediuos), hence asking for help.
I think this might be a kinda-related/helpful question in this regard, but I am unsure of it.
Yes, it stays rational. There is a very precise relation on how conjugating coefficients by an element in Aut$(\mathbb{C}/\mathbb{Q})$ interact with a slash operator, e.g. Thm 3.3 of https://arxiv.org/pdf/1807.00391.pdf. Then it becomes just the matter of whether you can choose representatives of $\Gamma_{0}(N)/\Gamma_{0}(M)$ accordingly.
@GTA Thanks for pointing out the reference. I have not come across this one previously. I will look into this and work out the details(and hopefully write an answer ).
The conceptual reason is that the trace map has a purely algebraic definition: in weight 2, this is just the push-forward of differential forms $\Omega^1(X_0(N)) \to \Omega^1(X_0(M))$. Since the map $X_0(N) \to X_0(M)$ is defined over $\mathbb{Q}$, the trace map preserves rationality. (This argument works in weight 2 but can be extended to higher weight.)
@FrançoisBrunault Thanks for an alternate way to see it(I am not much comfortable looking at the trace map in that way, but I will try to see how it can be done in higher weight). Will this work for weight $k+1/2$ modular forms?
I'm not sure about half-integral weight (the metaplectic cover of SL_2 does not give rise to an algebraic curve, if I recall correctly). One trick would be to multiply by $\theta$ and prove something like $\mathrm{Tr}(f \theta) = \mathrm{Tr}(f) \theta$.
@FrançoisBrunault I will try to work it out.
|
2025-03-21T14:48:29.863782
| 2020-02-13T05:54:08 |
352603
|
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|
Stack Exchange
|
Why is modular forms applicable to packing density bounds from linear programming at $n\in\{8,24\}$?
Sphere packing problem in $\mathbb R^n$ asks for the densest arrangement of non-overlapping spheres within $\mathbb R^n$. It is now know that the problem is solved at $n=8$ and $n=24$ using modular forms. I understand some sphere packing and issues going with it but my understanding is most upperbounds come from linear programming and the bounds that are currently proven optimal (including at $n=8$ and $n=24$) already come from linear programming bounds.
How do modular forms become part of the story that provide the lower bound (do they arise naturally from some packing related structure)?
Is there a bigger story that this is just a chapter of that may apply to other upper bounds generated from linear programming? What makes modular forms click for this class of linear programming bounds (perhaps Sphere packing and quantum gravity is of utility?)?
This by Henry Cohn might answer your question: https://mathoverflow.net/questions/235347/understanding-sphere-packing-in-higher-dimensions/235360#235360
@YoavKallus https://arxiv.org/abs/1603.05202 says the properties of the functions involved. However does not say why it has to be a modular form. Perhaps modular forms are not the right objects at $n\not\in{8,24}$ and something else?
@VS. What do you mean by right objects for these n? For most of these n, the LP bound is not expected to be tight. Are you asking whether modular forms are involved in the construction of the smallest LP upper bound?
@YoavKallus No. What I am saying is this. There is an achievability result by construction and an upper bound as Harry Richman says below based on some functions. It is intriguing that at $n\in{8,24}$ there is a modular form candidate for $f$ showing achievability known is the best. At $n=5$ perhaps a different construction needs to be shown optimal. For that I doubt $f$ will be a modular form if similar strategy works.
@VS. Numerical results for n=5 suggest there is no f that gives upper bound matching current best lower bound. For n=8, even before the exact construction was given, the numerical results suggested there was such f. Either LP bounds are not powerful enough in n=5 to show optimality or currently densest known packing is not optimal. More likely the former.
This is a tough question, and I don’t think there’s a definitive answer yet. For some mathematical details, see the following survey articles:
https://arxiv.org/abs/1611.01685
https://arxiv.org/abs/1603.05202
Instead, I’ll focus on the big picture here. Why modular forms? I can see a couple of potential answers:
(1) Why not modular forms? Before Viazovska’s proof, numerical experiments indicated that there were remarkable special functions in 8 and 24 dimensions that would prove the optimality of $E_8$ and the Leech lattice. However, nobody had any idea how to construct them explicitly, or prove existence at all. Modular forms are by far the most important class of special functions related to lattices (in higher dimensions, since arguably trig functions and the exponential function are the most important special functions related to lattices), so lots of people expected that the magic functions for sphere packing should be connected somehow to modular forms. The proof had to wait for Viazovska to come up with a beautiful integral transform, but the fact that it used modular forms wasn’t such a great surprise. I.e., her contribution wasn’t the idea that modular forms should play a role, but rather figuring out how to use them, which was quite subtle and ingenious.
You’re right that nobody has any idea how to use modular forms to optimize the linear programming bound in other dimensions. However, it’s possible that they will continue to play a role. For example, see the example Felipe Gonçalves and I found at the end of Section 2.1 of our paper https://arxiv.org/abs/1712.04438 (which is not a sphere packing bound, but closely related). It really looks like a small perturbation of a function based on modular forms (see https://arxiv.org/abs/1903.05737), and I wouldn’t be surprised if the optimal function has a nice series expansion based in some way on modular forms. From this perspective, the remarkable thing about 8 and 24 dimensions wouldn’t be the appearance of modular forms, but rather the fact that the series collapses to a single term, with a matching sphere packing. However, this is all speculative.
(2) The other perspective is that we have very little understanding of why 8 and 24 dimensions are special in the first place. For example, why shouldn’t sphere packing in 137 dimensions also admit an exact solution via linear programming bounds? It sure doesn’t look like it does, but perhaps we just don’t know the right sphere packing to use, and some currently unknown packing might match the upper bound. That would be very surprising, since our experience is that exceptional phenomena occur in clusters. We would expect to see some sort of remarkable symmetry group, probably a finite simple group, and there aren’t any candidates acting on 137 dimensions. However, this expectation is just based on our limited experience, and mathematics can confound our expectations. So far, nobody has found even a convincing heuristic argument for why there shouldn’t be an exact solution in 137 dimensions, and that’s a major gap in our understanding. The most we can say is that it would have to differ in some important ways from 8 and 24 dimensions, which is far from an explanation of why it can’t happen.
I guess I’d summarize it like this. If you accept that lattices in 8 and 24 dimensions play a special role, then modular forms feel naturally connected. However, we’re missing a deeper explanation of the role of these special dimensions.
Let me add one more specific mathematical comment. The magic functions in 8 and 24 dimensions fit into a general picture of building radial functions that vanish on all but finitely many vector lengths in a lattice, and whose Fourier transforms vanish on all but finitely many vector lengths in the dual lattice. If you can do this in full generality, then Poisson summation lets you solve for the number of lattice vectors of each length. These are the coefficients of a modular form, namely the theta series of $E_8$ or the Leech lattice, so the conclusion is that this family of functions somehow “knows about” the theta series. In other words, you can’t expect to construct the whole family without running into modular forms in some way. This leaves a couple of possibilities: maybe the magic functions for sphere packing are simpler than most functions in this family, and could be constructed without modular forms, or maybe these functions are deeper than modular forms (and require some mysterious special functions not yet known to mathematicians). What we know now is that modular forms suffice, and in a sense are necessary because the magic functions are unique.
Would there be some connection between $3$, $8$ and $24$ or would it be wild to expect one? Since we know the best bounds in all three has it been predicted that some multiplicative behavior might be possible? Just wondering if there could be some underlying arithmetic to these optimality?
The linked paper in post on quantum gravity speculates some other deeper construction than modular forms of which modular forms emerge at $n\in{8,24}$ (I think).
Regarding $3$, $8$ and $24$ it would be nice to know if there is a modular connection to $3$ as well.
Regarding 3, 8, and 24, there are a lot of numerical relationships in this area, some of which seem deeply meaningful and others less so, and it is hard to say. I doubt 3 dimensions plays a significant role in LP or their generalizations (and Hales’s proof uses entirely different techniques), but maybe it will in the future. As for the sphere packing and quantum gravity paper, it’s really interesting and important, and raises as many questions as it answers.
Conformal field theory builds in modular invariance (which amounts to conformal invariance on tori), so it makes sense that modular forms should play an important role there, but then the question is why conformal field theory is connected with sphere packing. So far, it’s unclear whether there’s an even deeper connection, or it’s just that the LP bound for sphere packing is isomorphic to the spinless modular bootstrap.
'Regarding 3, 8, and 24, there are a lot of numerical relationships in this area, some of which seem deeply meaningful and others less so, and it is hard to say.'. 1. Is there an useful reference?
'Conformal field theory builds in modular invariance (which amounts to conformal invariance on tori),'. 2. Is there some theory of 'conformal form' applicable to packing problems on other spaces such as tori? Just wondering.
Unfortunately I don’t know of an overall reference. “Monstrous moonshine” is probably the most dramatic and important case, but there’s a lot of other numerology scattered in the literature. Figuring out what’s going on can be tricky. For example, suppose we’re counting minimal vectors in lattice. For D_4, E_8, and the Leech lattice, we get 24, 240, and 196560, with each number divisible by the previous one. Is there an explanation in terms of group actions? Yes for the first case: the 24 vectors form the units in the Hurwitz integral quaternions, and they act on E_8.
No in the second case: the 240 vectors form the unit integral octonions, but they aren’t associative so the question of whether they act on the Leech lattice doesn’t even make sense (they don’t even act on themselves, let alone other things). Nevertheless, there are octonionic constructions of the Leech lattice that are a little more subtle but explain the divisibility by 240. So sometimes things work the way you’d hope, but sometimes we need more subtle or sophisticated explanations.
For the second question about conformal field theory and packing in other settings, I don’t know of a relationship, but I have no grounds for ruling one out.
Sorry what the connection between $n\in{3,8,24}$ and "Monstrous Moonshine"?
No direct connection; they just both involve trying to figure out what deeper meaning there might be behind numerical relationships.
Could working with Epstein Zeta function directly indicate symmetries of the function necessary to get best packing?
In your opinion is there any other application of modular forms to traditional computational number theory or is the only known use case is in packing problems?
Why is methods from sphere packing have anything to do with uncertainty principle in https://link.springer.com/article/10.1007/s00222-019-00875-4? The problems seem philosophically different? Is there anything that relates both? Perhaps there is a better elucidation of the relation. I can post a new question if needed.
@HenryCohn What is the reading methodology to understand the breakthrough work(s) in Sphere Packing and its applications to potentially (but not yet studied) packing in other spaces coming from information theory such as packing related to zero error information theory?
In my understanding, the connection to modular forms came via a result of Cohn and Elkies in their paper "New upper bounds on sphere packings I," Ann. of Math. 157 (2003) 689-714, also on the arxiv.
Theorem 3.1 (p. 694) states that the density $\rho$ of a sphere packing in $\mathbb R^n$ satisfies
$$ \rho \leq \frac{\pi^{n/2}}{(n/2)!} \frac{f(0)}{2^n \hat f(0)} \tag{$*$}$$
where $f: \mathbb R^n \to \mathbb R$ is
any admissible (i.e. sufficiently decaying) function
such that
the Fourier transform $\hat f$ is admissible,
$f(x) \leq 0$ for $\|x\| \geq 1$,
$\hat f(t) \geq 0$ for all $t$.
However, this result gives an upper bound on packing density rather than a lower bound.
(So I'm not sure whether this answers your question 1.)
In dimensions 8 and 24, the sphere packing problem was later solved by finding a modular-form-inspired function $f$ such that $(*)$ matches the density of the $E_8$ lattice and the Leech lattice, respectively.
Ahh ok. somewhat counterintuitive though. So for other cases still have to go through explicit yet unknown constructions?
Why $f$ should be modular form (it seems it will not be modular forms in $n\not\in{8,24}$ or at least beyond some finite $n$)?
|
2025-03-21T14:48:29.864692
| 2020-02-13T06:01:09 |
352604
|
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Stack Exchange
|
How flexible is the infinite-dimensional torus?
Let $\mathbb T=\mathbb R/\mathbb Z$ be the circle group and $\mathbb T^\omega$ be the infinite-dimensional torus, considered as an abelian compact topological group.
Problem 1. Is it true that for any finite set $F\subset\mathbb T^\omega$ and any neighborhood $U\subseteq \mathbb T^\omega$ of zero there exists an automorphism $\alpha$ of $\mathbb T^\omega$ such that $\alpha(F)\subset U$?
This problem can be reformulated in the language of the special linear groups $SL(n,\mathbb Z)$.
Problem 2. Is it true that for any $n\in\mathbb N$, neighborhood of zero $U$ in $\mathbb R^n$ and vectors $x_1,\dots,x_n$ in $\mathbb R^\omega$ there exists $m>n$ and a matrix $A\in SL(m,\mathbb Z)$ such that $\mathrm{pr}_n\circ A\circ \mathrm{pr}_m(x_i)\in U$ for all $i\in\{1,\dots,n\}$?
Here $\mathrm{pr_k}:\mathbb R^\omega\to\mathbb R^k$, $\mathrm{pr}_k:x\mapsto x{\restriction}k$, is the projection onto the first $k$ coordinates.
Remark 1. For any field $\mathbb F$ and vectors $x_1,\dots,x_n\in\mathbb F^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb F)$ of $\mathbb F^{2n}$ such that $A(\{x_1,\dots,x_n\})\subset\{0\}^n\times\mathbb F^n$.
Remark 2. For any vector $(x,y)\in\mathbb R^2$ and any $\varepsilon>0$ there exists a matrix $A\in SL(2,\mathbb Z)$ such that $(x,y)\cdot A\in (-\varepsilon,\varepsilon)\times\mathbb R$. Such matrix $A$ can be constructed by finding relatively prime integer numbers $p,q$ such that $|xp+yq|<\varepsilon$ and then finding integer numbers $a,b$ such that $pb-qa=1$ (using the extended Euclidean algorithm). Then the matrix $A=\left(\begin{array}&p&a\\q&b\end{array}\right)\in SL(2,\mathbb Z)$ has the required property.
Taking into account Remarks 1 and 2, I would expect that the following stronger form of Problem 2 has an affirmative answer.
Problem 3. Is it true that for any $n\in\mathbb N$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset(-\varepsilon,\varepsilon)^n\times\mathbb R^n$?
Do you have a serious reason to restrict to matrices of determinant 1?
@YCor I far as I know, no. In order to construct a homomorphism $\alpha$ from a matrix $A$ we need that $A$ is invertible, but its determinant can be $-1$.
@AlexRavsky Yes. I just think that restricting to $SL$ can just make some minor uninteresting complications.
This is a draft proof of an affirmative answer to Problem 3.
Proposition. For any $n\in\mathbb N$, $\varepsilon>0$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset[-\varepsilon,\varepsilon]^n\times\mathbb R^n$.
Proof. Let $m=2n$, $x_i=(x_{i1},\dots,x_{im})$ for each $i$ and $X=\|x_{ij}\|$. We shall call a column
of a matrix small, provided all its entries have absolute value at most than $\varepsilon$, and big, otherwise. Let $k$ be the maximal number of small columns in a matrix $XB$, where $B\in SL(m,\mathbb Z)$ and $C$ be an arbitrary matrix in $SL(m,\mathbb Z)$ such that $XC$ has $k$ small columns. It suffices to show that if $k<n$ then there exists a matrix $D\in SL(m,\mathbb Z)$ such that a matrix $XCD$ has $k+1$ small columns. Without loss of generality we can suppose that the big columns $y_1,\dots, y_{l}$ of the matrix $XC$ are from the first to the $l$-th. Pick a positive integer $M$ such that the maximal absolute value of an entry of these columns is at most $M\varepsilon$. Pick an arbitrary positive integer $K>(2lM)^n$ and define a map $f$ from the subset $Q^l$ of points of the set $[-K, K]^l$ with all integer coordinates to $\mathbb R^n$ by putting $f(d)=d_1y_1+\dots + d_ly_l$ for each $d=(d_1,\dots,d_l)\in Q^l$. Since all $|d_i|\le K$ and all entries of columns $y_i$ have absolute value at most $M\varepsilon$, each coordinate of a vector $f(d)$ is at most $lKM\varepsilon$. Therefore the image $f(Q^l)$ can be covered by $(2lKM)^n$ axis-parallel cubes with side $\varepsilon$. Since $|Q^l|=(2K+1)^l>(2K)^{n+1}>(2lKM)^n$, there exist two distinct elements $d’$ and $d’$ of $Q^l$ such that each coordinate of a vector $y=f(d’)-f(d’’)$ has an absolute value at most $\varepsilon$. Put $d=d’-d’’$. Dividing entries of $d$ by their greatest common divisor, if needed, we can assume that the greatest common divisor of entries of $d$ is $1$. It is well-known (see, for instance, this MSE thread]) that there exists a matrix $D’\in SL(l,\mathbb Z)$ whose first column is $d$. Put $D=\begin{pmatrix} D’ & 0\\ 0 & I\end{pmatrix}$, where $I$ is the $(m-l)\times (m-l)$ identity matrix. Then the first column of the matrix $XCD$ is small, whereas its last $k$ columns are the same as in the matrix $XC$. $\square$
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2025-03-21T14:48:29.865008
| 2020-02-13T07:56:18 |
352607
|
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Stack Exchange
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When is every element of a coend of abelian groups contained in one of the summands?
Let $I$ be a small category and let $D : I^{\mathrm{op}} \times I \to \mathsf{Ab}$ be a functor. The coend
$$\int^{i \in I} D(i,i)$$
can be constructed as the direct sum $\bigoplus_{i \in I} D(i,i)$ modulo the equivalence relation generated by
$$D(i,i) \ni D(f,\mathrm{id}_i)(x) \sim D(\mathrm{id}_j,f)(x) \in D(j,j)$$
for $f : i \to j$, $x \in D(j,i)$.
Question. Under which conditions on $I$ and $D$ is every element of $\int^{i \in I} D(i,i)$ the class of some element in some $D(i,i)$? In other words, when $U : \mathsf{Ab} \to \mathsf{Set}$ is the forgetful functor, I ask for a condition that the canonical map
$$\int^{i \in I} U(D(i,i)) \to U\left(\int^{i \in I} D(i,i)\right)$$
is surjective. This is clearly equivalent to the condition that the image of this map is closed under sums.
For example, this is the case when for all $x \in D(i,i)$, $y \in D(j,j)$ there is some span $i \leftarrow k \rightarrow j$ such that $x$ has a preimage in $D(i,k)$ and $y$ has a preimage in $D(j,k)$, since then $x$ and $ y$ are equivalent to elements in $D(k,k)$. This condition is already useful in practice, but obviously it is quite strong. Probably weaker assumptions are also sufficient.
Answers for the following special case are appreaciated as well: Let $A : I^{\mathrm{op}} \to \mathsf{Ab}$, $B : I \to \mathsf{Ab}$ be two functors and let $D(i,j) = A(i) \otimes B(j)$, so that the coend is the tensor product of functors $A \otimes_I B$. In this case, the question is when every element of $A \otimes_I B$ is the class of some element in some $A(i) \otimes B(i)$.
Background. I have to do some computations with some complicated coends, and it won't be sufficient to use the universal property. I really need a better idea of the element structure of the coend. The next step would be to find a description of the kernel of $D(i,i) \to \int^{i \in I} D(i,i)$.
Doesn't the special case of the tensor product (even for $I=1$) show the answer is "almost never"? I mean, it might still be interesting to work out when this happens, but I wouldn't expect to it too in useful cases.
@OmarAntolín-Camarena Actually, for $I=1$ the sufficient condition I mentioned is satisfied. My question is not when every tensor is pure (I corrected that in the post), but when every element belongs to a single tensor product then, or in the general case, to a single $D(i,i)$. I have clarified that in the post as well.
@MartinBrandeburg Oh, right! I was a little careless, I did think it meant the analogue of every tensor being pure. The correct thing has a much better chance of happening. :)
When doing coends of abelian groups, I'm usually thinking of $Ab$-enriched coends, which I think are relevantly different from the $Set$-enriched coends you seem to be using here.
@TimCampion Good point. I need to think about the context again, but there was probably a reason that the coends where normal ($\mathbf{Set}$-enriched) coends. They appeared in the context of certain categories of algebras, which are not $\mathbf{Ab}$-enriched, for instance.
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2025-03-21T14:48:29.865535
| 2020-02-14T23:28:17 |
352741
|
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Stack Exchange
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Deformations that flatten small curvature
I'm trying to show that any 3-dimensional polyhedron with many vertices can be mildly deformed so that its vertices are no longer convexly independent. I suspect it suffices to look at a vertex with low angular defect (guaranteed to exist by Descartes' theorem of total angular defect).
Specifically, I conjecture the following: for all $\epsilon > 0$, there exists $\delta > 0$ such that if $v$ is a vertex with angular defect at most $\delta$, one can construct a map $\rho : \mathbb{R}^3 \to \mathbb{R}^3$ such that:
$\rho$ preserves all Euclidean distances in $\mathbb{R}^3$ up to a multiplicative $(1 + \epsilon)$.
If the neighboring vertices of $v$ are $u_1, \ldots, u_d$, then $\rho(v)$ is a convex combination of $\rho(u_1), \ldots, \rho(u_d)$.
This seems like a statement that should have a simple proof if true, but I'm not sure how to prove it.
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2025-03-21T14:48:29.865646
| 2020-02-15T00:15:21 |
352743
|
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|
Stack Exchange
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Algebraic de Rham cohomology for singular varieties
I would like to know to what extent the naive algebraic de Rham cohomology is a "bad" cohomology theory. If $X$ is smooth then there is a comparison theorem with singular cohomology. If $X$ is singular, then Hartshorne embeds $X$ in a smooth variety, generalises the definition and proves a similar comparison theorem singular cohomology, Poincaré duality and other things (http://www.numdam.org/article/PMIHES_1975__45__5_0.pdf).
In this thread (naive de Rham cohomology fails for singular varieties), we learn that the naive De Rham cohomology is, for sure, not the same as singular cohomology and does not satisfy Poincaré duality.
Despite these negative results, my question is the following.
If $X$ is singular, can we still prove that its naive de Rham cohomology is finitely generated? Can we prove that it is zero above its dimension?
Any known results or references are welcome.
Here is an answer for the naive de Rham cohomology $\mathbb{H}(X, \Omega_X^{\bullet})$ (not the more sophisticated one of the linked article by Hartshorne (which involves chooseing an embedding $X \subset Y$ in a smooth variety $Y$, completion along $X$ etc.)).
We can use the hypercohomology spectral sequence
$$
E_1^{pq} = H^q(X, \Omega_X^p) \implies \mathbb{H}^{p+q}(X, \Omega_X^\bullet)
$$
(a.k.a. the Hodge-to-de Rham spectral sequence). Assuming $X$ is proper all terms on this $E_1$-page are finite dimensional, hence so is the de Rham cohomology. Setting $n = \dim X$, by Grothendieck's vanishing theorem and the fact that the complex $\Omega_X^\bullet$ has length $n$
$$
H^q(X, \Omega_X^p) = 0 \text{ for } p > n \text{ or } q > n
$$
and hence $\mathbb{H}^i(X, \Omega_X^\bullet) = 0$ for $i > 2 \dim X$.
Dear cgodfrey, why is it the case that the complex of Kähler differentials on X has lenghth n? Since X is singular, I cannot see why this holds. Thanks in advance.
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2025-03-21T14:48:29.865803
| 2020-02-15T02:20:33 |
352746
|
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Stack Exchange
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A conjecture (or theorem?) on unit vectors in a Euclidean space
I have heard (if I am not mistaken) that there exists the following conjecture (or theorem?).
Let $u_1,\dots,u_n$ be unit vectors in an $n$-dimensional Euclidean vector space. Then there exists another unit vector $x$ such that
$$(\prod_{i=1}^n |(x,u_i)|)^{1/n}\geq 1/\sqrt{n}.$$
Is it conjecture or theorem? In either case I would be interested to have a reference.
Remark. This post is a continuation of the previous one: Reference to a conjecture on unit vectors in Euclidean space
Yep, this one (to the best of my knowledge) is still open for large $n$ (small $n$ are fine, say, for $n\le 5$, the same idea as I used for the sum works and you can modify it slightly to go up a little bit. It also works for a small perturbation of an orthonormal system, so the counterexample, if it exists, should be rather skewed). Unfortunately, I'm not good with references, so I'll leave that part to someone else.
Not an answer. I tried the log sum inequality applied to the log of the left hand side, but no luck.
Assume the unit vectors $u_i$ are orthogonal and use a change of basis so that $u_i=e_i,$ the vector with the sole nozero entry equal to one in the $i^{th}$ component.
Taking $$x=\frac{1}{\sqrt{n}}(\pm 1,\pm 1,\ldots,\pm 1),$$
achieves your lower bound and perturbing $x$ only increases the LHS of your inequality.
Now consider the $u_i$ below with only two vectors changed and $\theta \in (0,1)$:
$$
\begin{array}{ccccccc}
(& \sqrt{1-\theta^2} & \theta & 0 & \cdots & 0& ),\\
(& \theta &\sqrt{1-\theta^2} & 0 & \cdots & 0& ),\\
(& \sqrt{1-\theta^2} & \theta & 0 & \cdots & 0& ),\\
&&& \ddots &&&\\
(&0 & 0& 0 &\cdots & 1&)\\
\end{array}
$$
and note that once again the left hand side of the inequality increases.
I feel that arguments along this line may prove the inequality.
this was too long to enter as a comment. I'd be interested to see if the downvoter has something constructive to contribute to the question or my comment.
I didn't downvote but I'm confused why you're bringing up whether the downvote can contribute to the question. If they have a reason to downvote your answer, that is enough; they don't need to be able to contribute to the question
For orthogonal vectors the inequality is satisfied by $ x = \frac{y}{||y||}$, with $y = \sum_i u_i$.
|
2025-03-21T14:48:29.866007
| 2020-02-15T02:26:40 |
352749
|
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"Andrés E. Caicedo",
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|
Stack Exchange
|
Which Hilbert's 10th polynomials are known to have solutions?
The Diophantine equation
$$x^3 + y^3 + z^3 = 42$$
was recently solved by
Booker and Sutherland:
Sum of three cubes for 42 finally solved.
Is there a clean partition of the form of those
polynomial equations all of which do have integer solutions,
and those that are known to be undecidable
(following the negative solution of Hilbert's 10th)?
Or—in the absence of a clean partition—can at least
the equations be partitioned into:
$$\{ \textrm{solvable, undecidable, unknown} \}$$
Am I correct that the status
of $x^3 + y^3 + z^3 = c$ is unknown
except for certain values of $c$?
I ask this naively; not my expertise.
Related MO question:
What are the solutions of this Diophantine equation?.
The negative solution of Hilbert's tenth problem means that there is no algorithm that would recognize the solvable polynomial equations. Hence the "clean partition" you are looking for does not exist.
Of course, one can partition the polynomials any way one likes. The partition solvable/non-solvable/undecidable is a legitimate one, except that one "does not know" which polynomials belong to the individual parts.
Thank you. I thought e.g., linear Diophantine equations and Pell's equation fell into the "solvable" class, or at least the "understood" class.
@JosephO'Rourke: Yes, there are well-understood classes of Diophantine equations. However, Matiyasevich's theorem states, in a sense, that we shall never fully understand which equations are solvable/non-solvable/undecidable. Just as we shall never fully understand which statements in number theory are provable/falsifiable/undecidable. We make progress from day to day (this is our job), but we are walking on an infinite path.
"we are walking on an infinite path"---Well phrased!
@MarkSapir: Matiyasevich's theorem also implies that there exists polynomials for which solvability is undecidable within ZFC (assuming ZFC is consistent). Indeed, assume that for each polynomial solvability is decidable within ZFC. Then, given an arbitrary polynomial, you can go through the first order statements in ZFC one by one, and after a while you will arrive at a proof or disproof of solvability (because a proof in ZFC is a first order statement in ZFC). This is a general algorithm for solvability, but Matiyasevich proved that such an algorithm does not exist. Contradiction.
@MarkSapir: Of course you are right that algorithmic undecidability only makes sense for infinite classes of polynomials. The "undecidability within ZFC" is a different kind of notion, but not independent of "algorithmic undecidability" (because ZFC is recursive). This is what I tried to indicate in my remark.
I am not sure this quite answers the question. Yes, for an arbitrary family of polynomials there is not much more to say, but the polynomials in the question are fairly concrete. Can we say something about the decidability of the resulting equations, even if unfeasible?
@AndrésE.Caicedo: The question was about all polynomial equations, not just about $x^3+y^3+z^3=c$. Note also that the OP accepted my answer.
@GHfromMO Sure. But, regardless of what Joseph question was, do you know anything about the specific family mentioned in the problem?
@AndrésE.Caicedo: It is conjectured that every integer not congruent to $\pm 4$ modulo $9$ is a sum of three integral cubes. There are a few positive results, but otherwise the status of these equations is unknown (including their possible independence of ZFC).
|
2025-03-21T14:48:29.866288
| 2020-02-15T02:31:41 |
352750
|
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|
Stack Exchange
|
Definitions of sequential homotopy colimits
Suppose $\mathfrak{M}$ is the category of $S^1$-spectra of simplicial sheaves. I know its sequential homotopy colimits (colimit in $\mathbb{N}$ as usual) coincide with categorial colimits since stable equivalences are preserved under filtered colimits.
But I don't see why they coincide with the homotopy colimits in $\mathrm{Ho}(\mathfrak{M})$, as a triangulated category. That is,
$$\mathrm{Hocolim}(E_i):=\mathrm{Cone}(1\text{-}\mathrm{shift}:\oplus E_i\longrightarrow\oplus E_i)$$.
Does anybody know the reason?
I've worked out.
For any two morphisms $f,g:A\to B$ in $\mathfrak{M}$, we have
$$Cone(f-g)=Cone(((id_A,f),(id_A,g)):A\oplus A\to A\oplus B)$$
by elementary transformations of matrices.
We prove that latter is the homotopy pushout of the diagram
$$A\xleftarrow{id+id}A\vee A\xrightarrow{f+g}B$$
, which is just $Hocoeq(f,g)$. So just assume we have a general pushout diagram
$\require{AMScd}$
\begin{CD}
W @>t>> V\\
@V s V V @VV a V\\
U @>>b> D
\end{CD}
in $\mathfrak{M}$, such that $W$ is cofibrant and $t$ is a cofibration. So $Cone(t)=Cone(b)=D/U=V/W$ by definition and left properness of $\mathfrak{M}$. By Lemma 1.4.3 in 'Neeman Triangulated Category', we have
$$Cone(W\xrightarrow{(s,t)}U\oplus V)=D$$
by possibly modifying $a$. Thus we have proved the claim. Hence we have proved
$$Cone(f-g)=Hocoeq(f,g)$$.
There's probably a more classical way to look at this, but here's one way to think about it. Homotopy colimits in $M$ are the same as $\infty$-categorical colimits in the $\infty$-category $M[W^{-1}]$ obtained from $M$ by localizing at the stable equivalences. In particular, sequential homotopy colimits in $M$ are the same as $\infty$-categorical colimits in $M[W^{-1}]$ indexed by $\omega$ -- the poset of natural numbers.
In 1-category theory, there is a formula which expresses any colimit indexed by a category $J$ as a (reflexive) coequalizer of certain coproducts. Similarly, in $\infty$-category theory, there is a formula which expresses any colimit indexed by a simplicial set $J$ (which need not be a quasicategory) as a geometric realization of certain coproducts -- for quasicategories, you can find this in Chapter 4 of HTT. Schematically, it looks like $\varinjlim F = |[n] \mapsto \amalg_{j \in J_n} F(d_1(\dots(d_n(j)))|$. We can get away with the coproduct being over just the nondegenerate $n$-cells. This doesn't really simplify things when $J = \omega$, since (the nerve of) $\omega$ has nondegenerate simplices of arbitrarily large dimension.
However, let $N$ be the same simplicial set which I coincidentally just described answering another question of yours -- its 0-cells are the natural numbers, there is a unique 1-cell from $n$ to $n+1$, and there are no other nondegenerate cells. There is a natural inclusion $N \to \omega$. Using the $\infty$-categorical Quillen's Theorem A (see 3.2 there, or HTT Chapter 4), one easily shows that this inclusion is cofinal (In fact, I think this is explicitly shown somewhere in Chapter 4 of HTT). Therefore, a sequential colimit of a functor $\omega \to \mathcal C$ can be computed by first restricting to get a functor $N \to \mathcal C$ and then evaluating this colimit.
Now when we apply the general formula to our $N$-indexed colimit, since $N$ has no nondegenerate simplices of dimension greater than 1, we only need the first two layers of the simplicial object, so everything simplifies to the coequalizer of two maps $\amalg_{n \in \mathbb N} E_n \rightrightarrows\amalg_{n \in \mathbb N} E_n$, namely the identity and the shift map. In an additive category, this coequalizer can be computed as the (homotopy) cofiber of the difference of these two maps.
Finally, coproducts in $M[W^{-1}]$ are computed as in the homotopy category, and cofibers are computed using the triangulated structure, yielding exactly the formula you described.
If the question is really why the homotopy coequalizer of two maps is the same as the homotopy cofiber of their difference (the last step of the above argument, essentially), then let's go through that. By passing to the opposite category, it will suffice to treat the dual case, and show that a homotopy equalizer of two maps $f,g: X \rightrightarrows Y$ is the same as the fiber of their differences. Homotopy limits, like ordinary limits, are defined representably. That is, a cone in $\mathcal C$ is a limit if and only if it becomes a limit cone after composing with $Hom_{\mathcal C}(C,-): \mathcal C \to Spaces$ for each $C \in \mathcal C$ (where $Hom_{\mathcal C}$ denotes a mappping space). So it will suffice to show this when $\mathcal C = Spaces$, and we assume that $X,Y$ are infinite loop spaces and $f,g$ are infinite loop maps.
Now, a point $(x,\gamma) \in Hoeq(f,g)$ consists of a point $x \in X$ and a path $\gamma$ from $f(x)$ to $g(x)$. A point $(x,\gamma) \in Fib(g-f)$ consists of a point $x \in X$ and a path $\gamma$ from $(g-f)(x)$ to the basepoint 0. A map $Hoeq(f,g) \to Fib(f,g)$ is given by sending $(x,\gamma)$ to $(x,\gamma -f(x))$, where $\gamma-f(x)$ is the path obtained by using the addition on $Y$ to add the constant path at $f(x)$ pointwise to $\gamma$. A map in the other direction is given by sending $(x,\gamma)$ to $(x,\gamma+f(x))$, and these are inverse homotopy equivalenes.
I don't think I need such a heavy context. I find that the question is to prove that the homotopy coequalizer of two morphisms in $\mathfrak{M}$ is just the cone of their difference in $Ho(\mathfrak{M})$.
I’d be curious how you reduce to that statement. It’s quite elementary to show that coequalizers are cofibers of differences.
Maybe I’ll also point out that this infinity categorical argument is exactly the same argument one would make 1-categorically, just lifted to the higher context.
How about the homotopy coequalizer of two identical maps?
Yup, the homotopy coequalizer of two identical maps $X \rightrightarrows Y$ is the cofiber of the zero map $X \xrightarrow 0 Y$, which is none other than $Y \oplus \Sigma X$.
Emm, but we couldn't substract two general morphisms in $\mathfrak{M}$...
But you can subtract them in $M[W^{-1}]$.
First, an elementary manipulation of homotopy colimits
shows that the sequential homotopy colimit
can be replaced by the homotopy coequalizer
of the identity map and the shift map
on the coproduct of all objects.
The latter homotopy coequalizer can be replaced
by the homotopy coequalizer of the zero map
and the difference of the identity and shift map.
Secondly, by definition of a triangulated category
extracted from a stable model category,
the distinguished triangles are defined precisely as the homotopy
coequalizers of the zero map and the given map.
Combined with the fact that homotopy coproducts in a stable
model category coincide with coproducts in its underlying
triangulated category, we obtain the desired result.
The point is that $Ho(\mathfrak{M})$ is an additive category but $\mathfrak{M}$ is NOT, so you could't talk about 'difference of maps in $\mathfrak{M}$'.
@NanjunYang: The difference of maps mentioned in the first paragraph is not induced by any kind of additive structure. Rather, it is induced by the fact that Map(X,Y) is weakly equivalent to Map(X,ΩΣY), which is a group object in spaces, which allows us to compute the difference of two maps.
|
2025-03-21T14:48:29.866878
| 2020-02-15T02:38:13 |
352751
|
{
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}
|
Stack Exchange
|
Viewing a finite group as a group scheme
I've read books which have this statement, without explanation : 'every finite group is an algebraic group'. I'm trying to understand what exactly they mean. The definition I have in my mind of a group scheme $ G $ over a base field $ k $ is a representable functor $$ G : (\text{Sch}/k)^{op} \rightarrow (\text{Grp}) $$ I discussed this with my professor and here's what he explained to me (which I find very beautiful):
Suppose $ G $ is a finite group. Consider the affine scheme $ X_G = \coprod_{g \in G} \text{spec} (k) = \text{spec} ( \prod_{g \in G} k) $ which is the disjoint union of $ |G| $ copies of $ \text{spec} (k) $. This gives the corresponding Yoneda functor and for any $ k $-scheme $ S $, we have $$ X_G(S) = \text{Hom} (S, X_G) = \text{Hom}_{k} \left( \prod_{g \in G} k, \Gamma(S, \mathcal{O}_S) \right) $$ Now, a $ k $-algebra morphism from $ \prod_{g \in G} k $ to $ \Gamma(S, \mathcal{O}_S) $ is determined by knowledge of the images of the idempotents $ e_g \in \prod_{g \in G} k $. So $$ X_G(S) = \{ (s_g)_{g \in G} | s_g \in \Gamma(S, \mathcal{O}_S) , \sum s_g = 1, s_g s_{g'} = 0, (s_g)^2 = s_g \} $$ The group structure is given by $ (s_g) \cdot (t_g) = (u_g) $ where $ u_g = \sum_{hk=g} s_h t_k $. The fact that $ (u_g) $ also satisfies the property of being in $ X_G(S) $ is clear from computation. $ X_G(S) $ indeed becomes a group with identity given by $ (a_g) : a_e = 1 $ and $ a_g = 0 $ when $ g \neq e $ and the inverse of $ (s_g) $ is $ (t_g) $ with $ t_g = s_{g^{-1}} $.
So my (poorly phrased) question: Is this the only way to view a finite group as a group scheme or are there other ways too?
For context, this really came up in a discussion about quotients in algebraic groups. If $ G $ is $ GL_n $ and $ T $ is a maximal torus with normalizer $ N_G(T) $, then $ N_G(T)/T = S_n $, the symmetric group on $ n $ letters.
This is not the only way, but it's the way that's meant unless something else is stated. For example, $\mu_n$ is finite, but it doesn't arise this way. The significance of your construction is that the scheme is constant: i.e., its group of rational points is $G$, regardless of the ground field (whereas, for example, $\mu_3(\mathbb R)$ is trivial while $\mu_3(\mathbb C)$ has order $3$). (I think—this is my naïve understanding, as a mere user of and not expert in group schemes, and hopefully someone will correct me if I've used the words incorrectly.)
I don't think that there is a different natural construction of a group scheme associated to a finite group. You can assume that authors mean this construction unless otherwise stated.
Notice, however, that this construction can be also done in the functorial setup, and a little bit more general and elegant as well:
Let $S$ be any base scheme. If $G$ is any set, define the functor $G(-) : (\mathsf{Sch}/S)^{\mathrm{op}} \to \mathsf{Set}$ by
$$G(X) := \{f : X \to G \text{ locally constant}\}.$$
The action on morphisms is clear. The functor $G(-)$ is representable by the $S$-scheme $\coprod_{s \in G} S$, since a locally constant map $X \to G$ corresponds to a partition $X = \coprod_{g \in G} X_g$ into disjoint open subschemes, which corresponds to an $S$-morphism $X \to \coprod_{g \in G} S$ (since $\mathsf{Sch}$ is extensive).
If $G$ is a group, then the functor $G(-)$ factors over $\mathsf{Grp}$. It follows that $\coprod_{g \in G} S$ carries the structure of a group scheme over $S$. (There is no need to write down the multiplication map etc. We simply get it from the functorial characterization of group schemes.) If $G$ is finite and $S$ is affine, then $\coprod_{g \in G} S$ is affine and of finite type over $S$.
|
2025-03-21T14:48:29.867137
| 2020-02-15T02:50:02 |
352752
|
{
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"Dongyang Chen",
"Nate Eldredge",
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|
Stack Exchange
|
Quotient Banach space whose dual map sends the ball onto a given convex subset
Let $X$ be a Banach space and let $A$ be a closed, convex and balanced subset of $B_{X^{*}}$ (where $B_{X^{*}}$ denotes the closed unit ball of the dual $X^{*}$). Is there a closed subspace $M$ of $X$ such that $Q^{*}_{M}$ maps $B_{(X/M)^{*}}$ onto $A$, where $Q_{M}:X\rightarrow X/M$ is the quotient map?
What happens when $X=\mathbb{R}^n$? Suppose $A$ is any closed convex balanced set with nonempty interior, other than the ball itself. If $M \ne 0$ then $Q_M^$ has rank less than $n$ and so its image cannot cover $A$, and if $M=0$ then $Q_M^$ is the identity map and it maps the ball to itself.
Thanks, Nate. What happens if $X$ is infinite-dimensional?
I was just thinking about that. More generally, the image of $Q_M^*$ will always equal the annihilator of $M$, right? If $M \ne 0$ this is a proper closed subspace. So take any $A$ which is not contained in a proper closed subspace (e.g. any $A$ with nonempty interior) and is not the ball, and I think that is a counterexample.
Indeed, $Q^{}{M}B{(X/M)^{}}$ is equal to the closed unit ball of the annilator of $M$. If we take $A$ with nonempty interior, then $M=0$.
|
2025-03-21T14:48:29.867246
| 2020-02-15T06:06:30 |
352759
|
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"url": "https://mathoverflow.net/questions/352759"
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|
Stack Exchange
|
Convolve a 4D Gaussian function along a plane?
There is a 4D Gaussian function $G(u,s)=G(x|c,\mu,\Sigma )$ where $x=\begin{bmatrix}u\\ s\end{bmatrix}$,$u$ and $s$ is all 2D vector.
Now I want to blur (convolve) it along with $u$ by another 2D Gaussian $g(u|c',\mu',\Sigma')$.
I know that I can write $G(u,s)$ as a 2D Gaussian $g[s](u|c_0,\mu_0,\Sigma_0)$ by fixing $s$ and convolve it by $g(u|c',\mu',\Sigma')$. To do like this means that we calculate a convolution between two 2D Gaussian functions:
$g[s](u|c_0,\mu_0,\Sigma_0)\bigotimes g(u|c',\mu',\Sigma')=g[s](u|c_1,\mu_1,\Sigma_1)=F(u,s)$
But is the convolved result $F(u,s)$ still a 4D Gaussian? If true, what is the new $c$, $\mu$, $\Sigma$?
I am thinking about another method to deal with this question is that we regard the convolution kernel $g(u|c',\mu',\Sigma')$ as a 4D function by extending its covariance matrix
to a 4x4 matrix $\begin{bmatrix}
\Sigma'& O \\
O & O
\end{bmatrix}$. But the question is that the detaminate of this matrix is 0.
|
2025-03-21T14:48:29.867347
| 2020-02-15T07:15:00 |
352760
|
{
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"authors": [
"Dieter Kadelka",
"Manoj Kumar",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/151864"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352760"
}
|
Stack Exchange
|
How to prove space of non-negative Radon measures is complete?
Let $\mathcal{M}^{+}(\mathbb{R}_{+})$ be space of non-negative Radon measures on $\mathbb{R}_{+}$ with bounded total variation and define the metric $\rho$ on $\mathcal{M}^{+} (\mathbb{R}_{+})$ as $$ \rho(\mu,\nu)= \sup \left \{ \int_{\mathbb{R}_{+}} \psi d (\mu - \nu) ~|~ \psi \in C^{1}(\mathbb{R}_{+}), \|\psi \|_{\infty} \le 1 , \|\partial_{x} \psi \|_{\infty} \le 1 \right \} .$$ How to prove $\mathcal{M}^{+}(\mathbb{R}_{+})$ is complete w.r.t. $\rho$. I know that $$ \lim_{n \to \infty} \rho(\mu_{n},\mu) = 0 \iff \mu_{n} \to \mu~ \text{narrowly for}~ n \to \infty.$$
But how above equivalence can help us to prove the completeness ?
Assuming that the above equivalence is valid, all you have to show that $\lim_{m,n \to \infty} \rho(\mu_m,\mu_n) = 0$ implies uniform tightness of the sequence $(\mu_n)$. So let $\epsilon > 0$ be arbitrary. Then there is $n_0 \in \mathbb{N}$ with $\rho(\mu_m,\mu_n) \leq \epsilon$ for $m,n \geq n_0$, in particular $\mu_n(\mathbb{R}_+) - \mu_m(\mathbb{R}_+) < \epsilon$. Since $\mu_{n_0}$ is tight, there is $t_0 \in \mathbb{R}_+$ with $\mu_{n_0}((t_0,\infty)) \leq \epsilon$. Now let $\psi \geq 0$ be any
function with the properties above and with $\psi|{[0,t_0]} \equiv 0$, $\psi|{[t_0+2,\infty]} \equiv 1$ and $\psi$ increasing. Such function exists. Then
$$\mu_n((t_0+2,\infty)) \leq \int \psi~d\mu_n \leq \int \psi~d\mu_{n_0} + \epsilon \leq 2\epsilon.$$
It follows that $(\mu_n)$ is uniformly tight, hence there are limit points $\mu$ with respect to the narrow topology. But then $\mu_{n_k} \to \mu$ weakly for some subsequence $(\mu_{n_k})$, hence $\rho(\mu_{n_k},\mu) \to 0$ by the above equivalence. Since $(\mu_n)$ is Cauchy neceesarily $\mu$ is unique and $\lim_{n \to \infty} \rho(\mu,\mu_n) = 0$.
It is well known that $\mathcal{M}^+$ is Polish. Proving that a concrete metric is complete almost always is done as in my answer.
Thanks for the answer, I just want to know what do you mean by- $\psi$ with the properties above in your answer? Also, will the existence of $\psi$ be shown through the standard argument of mollifiers?
The properties of $\psi$ in your definition of $\rho$, such a $\psi \in C^1$ etc.
Concerning mollifier: You can do it by mollifier and also directly. Note that the interval $[t_0,t_0+2]$ has length $2$.
can you please provide some references, where I can get this kind of proof.
|
2025-03-21T14:48:29.867519
| 2020-02-15T08:12:39 |
352761
|
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"Winnie_XP",
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}
|
Stack Exchange
|
IntersectInP bug of Macaulay2
I am trying to use the intersectInP command in Macaulay2, inside package ReesAlgebra. However, I tried to follow the exact code in the user-guide, but it doesn't run in my Ubuntu app (of win 10). Can anyone explain to me which part I have done wrong ?
Here is the user-guide age (https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.15/share/doc/Macaulay2/ReesAlgebra/html/_intersect__In__P.html)
Here is my code screenshots.
There is no bug. You are trying to follow the documentation of a newer version than the one you are running. You should get a newer version of Macaulay2.
Thanks ! I have been using distinguished commands with 1.10, but I didn't realize that intersectInP is a new package.
|
2025-03-21T14:48:29.867604
| 2020-02-15T08:59:11 |
352762
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352762"
}
|
Stack Exchange
|
For what automorphic representations is Ramanujan-Petersson known?
I had in mind that Ramanujan-Petersson conjecture was essentially unknown in the case of number fields. I however recently heard that
If an automorphic representation on $GL(2)$ is ramified at a finite place, then it is supercuspidal or a twist of a ramified principal series. These two kind of automorphic representations satisfy the Ramanujan-Petersson conjecture.
Is that true or have I misunderstood? I am desperately searching for a clear reference of this fact. More generally, in what cases do we know that the Ramanujan-Petersson conjecture is true?
Denote by $\mathbb{A}$ the adeles over a number field $F$. An cuspidal automorphic representation $\pi$ of $\text{GL}_2(\mathbb{A})$ now factors by the tensor product theorem as $\pi\cong \prod_v\pi_v$, where $\pi_v$ are smooth irreducible unitary representations of $\text{GL}_2(F_v)$.
The Ramanujan-Petersson conjecture now states that the representations $\pi_v$ are tempered for all finite places $v$. (One could include the archimedean places, which then amounts to a generalisation of Selbergs Eigenvalue conjecture.) As far as I know this conjeture is only known for Hilbert modular forms with some mild technical conditions on the weight (see for example Don Blasius: Hilbert Modular Forms and the Ramanujan Conjecture). In general there are only results towards the Ramanujan-Petersson conjecture available. See Blomer, Brumley: On the Ramanujan Conjecture over number fields.
What you might have heard is the following. Say $\pi$ is ramified at a finite place $v$, then $\pi_v$ is either supercuspidal, a (unitary) twist of Steinberg or a unitary principal series representation. It is now true that supercuspidal representations and (unitary) twists of Steinberg are tempered by nature. The same is true for most unitary principal series representations with the following exception. Given a non-tempered unramified representation $\sigma_v$ of $\text{GL}_2(F_v)$. Then a unitary twist $\chi_v\otimes \sigma_v$ is also non-tempered. Also the latter example can theoretically occur in nature as long as the Ramanujan-Petersson conjecture is not known. (Take a cuspidal automorphic representation $\pi$ violating the it at a unramified finite place $v$ and twist it by a suitable Hecke-character.)
To summarise: If $\pi$ ramifies at a finite place $v$, then $\pi_v$ is tempered as long as it is not a twist of an unramified not-tempered representation. One might abuse language and say $\pi$ satisfies the Ramujan-Petersson conjecture at $v$. Howeever this is far from knowing the full conjecture for $\pi$ globally.
From the way you phrase your question, I suspect that you misunderstand something. Being supercuspidal is a local condition. If $\pi$ is an automorphic representation of $\mathrm{GL}_2$ over $\mathbb{Q}$, then $\pi_p$ might be supercuspidal for some ramified primes $p$, and non-supercuspidal for some other ramified primes $p$. The Ramanujan conjecture is a global statement, but it can be phrased as: $\pi_p$ is tempered for any prime $p$.
I will try to summarize for you what is known from local theory. For more details see Schmidt: Some remarks on local newforms for GL(2). Let $\omega$ be the central character of $\pi$, and assume that it is unitary (this is a simple normalization condition). For any prime $p$, there are five possibilities:
$\pi_p$ is induced from two unramified characters of $\mathbb{Q}_p^\times$;
$\pi_p$ is induced from one ramified and one unramified character of $\mathbb{Q}_p^\times$;
$\pi_p$ is induced from two ramified characters of $\mathbb{Q}_p^\times$;
$\pi_p$ is a Steinberg representation of $\mathrm{GL}_2(\mathbb{Q}_p)$ twisted by a character of $\mathbb{Q}_p^\times$;
$\pi_p$ is a supercuspidal representation of $\mathrm{GL}_2(\mathbb{Q}_p)$.
In the first case, $\pi_p$ is unramified, and $\pi_p$ is tempered if and only if the inducing characters of $\mathbb{Q}_p^\times$ are unitary. This is known when $\pi_\infty$ belongs to the discrete series (Deligne's famous theorem), but unknown otherwise.
In the second case, $\omega_p$ is ramified (hence $\pi_p$ is also ramified), and $\pi_p$ is tempered if and only if the inducing unramified character of $\mathbb{Q}_p^\times$ is unitary. Again, this is only known in special cases.
In the remaining three cases, $\pi_p$ is tempered.
|
2025-03-21T14:48:29.867886
| 2020-02-15T09:49:16 |
352769
|
{
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"Michael Engelhardt",
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|
Stack Exchange
|
Bromwich integral transformed to an integral on the real axis
I am new in complex integration and inverse Laplace transforms. I already asked this question on math.se but got no answer.
The author of a textbook claims that the inverse Laplace transform has expression
$$
f(t) = \frac{2\exp(bt)}{\pi}\int_0^\infty\Re\bigl(\hat{f}(b+iu)\bigr)\cos(ut)\,\mathrm{d}u.
$$
He obtains this formula by substituting $s = b+iu$ in the Bromwich integral
$$
f(t) = \frac{1}{2\pi i}\int_{b-i\infty}^{b+i\infty}\exp(st)\hat{f}(s)\,\mathrm{d}s.
$$
However I've numerically checked this formula and it doesn't seem to hold true:
fhat <- function(s) 1/(s+3) # Laplace transform of exp(-3x)
b <- 5
integrand <- function(u, x){
Re(fhat(b+1i*u))*cos(x*u)
}
x <- 2
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)$value
# -0.1124648
exp(-3*x)
# 0.002478752
For $b = -2$ the result is close to the expected value $\exp(-3x)$:
b <- -2
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)$value
# 0.002479138
I understood that $b$ must be choosen anywhere at the right of the singularities of $\hat{f}$ (here $-3$). Am I wrong? Here the result depends on the choice of $b$. Do I misunderstand something, or is there something wrong in this textbook?
Here is the derivation of the formula:
This does seem weird - if I just do the substitution, I get $(\mbox{Re} \hat{f}) \cos ut - (\mbox{Im} \hat{f}) \sin ut $ instead of $2 (\mbox{Re} \hat{f}) \cos ut$ ... maybe some additional assumptions are made about $\hat{f} $?
@MichaelEngelhardt I just edited my post to add the derivation of the formula.
Also here, equation (12).
Ah yes, that was the decisive additional ingredient, that $f$ vanishes for $t<0$ - this makes the terms with $\mbox{Re} \hat{f} $ and $\mbox{Im} \hat{f} $ equal. So, it seems to me that the derivation in the book is fine as long as $\hat{f} $ is sufficiently well-behaved for the integrals to exist. I would rather be sceptical of the numerical evaluation - have you varied the step size? That "subdivisions" variable doesn't seem very large ... and you have oscillatory integrands ...
@MichaelEngelhardt You're right, this is a numerical problem. I've tried with Wolfram and the value indeed does not depend on $b$. Changing subdiv in R has no effect.
Interesting, and a cautionary tale - good to hear you gained further insight.
@MichaelEngelhardt Yes, and this book does not mention the difficulty to numerically evaluate such an integral. That's not serious. I've tried a more powerful numerical integration routine in R and it works.
Too long for a comment:
The following paper give a review of existing inverse Lapplace transform algorithms:
Kristopher L. Kuhlman, "Review of inverse Laplace transform algorithms
for Laplace-space numerical approaches", Numerical Algorithms 63, No. 2, pp. 339-355 (2013), DOI 10.1007/s11075-012-9625-3, MR3057203, Zbl 1269.65134.
and a little bit more specific
Lloyd N. Trefethen, J. André C. Weideman, and Thomas Schmelzer, "Talbot quadratures and rational approximations", BIT Numerical Mathematics, 46(3), pp. 653-670 (2006), DOI 10.1007/s10543-006-0077-9, MR2265580, Zbl 1103.65030.
Of course they do not mention your specific implementation of the Bromwich inversion integral.
Fernando Damian Nieuwveldt has implemented in this code a fine version of the inverse Laplace algorithm (in Python, not in R). Here an excerpt:
Talbot suggested that the Bromwich line be deformed into a contour that begins and ends in the left half plane, i.e., $z \to −\infty$ at both ends.
Due to the exponential factor the integrand decays rapidly on such a contour. In such situations the trapezoidal rule converge extraordinarily rapidly.
For example (in Nieuwveldt's code) here we compute the inverse transform of $F(s) = 1/(s+1)$ at $t = 1$.
If you are interested in more precision you find an algorithm with arbitrary precision in here.
|
2025-03-21T14:48:29.868176
| 2020-02-15T11:41:45 |
352772
|
{
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|
Stack Exchange
|
What is rigidity of Hirzebruch, and Witten genera?
I would like to find some good references (or any insight) that would help me to understand a few articles mentioning rigidity of Hirzeburch genus. One of the consequences of this phenomenon is that whenever you have a Hamiltonian circle action (with isolated fixed points?) on a symplectic manifold $M$, you can calculate the Hirzebruch genus $\chi_y(M)$ from the Betti numbers of $M$. This is
written, for example, in Remark 4.15(i) in https://arxiv.org/abs/1604.00277.
I tried to look into classical papers on rigidity, for example, Bott and Taubes 1989 https://pdfs.semanticscholar.org/e058/c27d3a9d551852d4325739559e069da33598.pdf , but I am not able to see a connection to the article above (due to my very low knowledge in the area). So I wonder if there are some intermediate references, that explain a bit what is rigidity and how it affects the data of fixed points of $S^1$-actions (including the weights) on a manifold.
Also, if you could give some general answer (intuition, reference) for the question in the title, it would be great as well.
Very quickly, the Hirzebruch genus is the generating function for indices $\chi^p$, each of which is the index of the elliptic operator $\bar\partial+\bar\partial^\ast$ on $\bigoplus_{q;\text{even}}\Omega^{p,q}(M)$. Rigidity concerns the $S^1$ acting trivially in some ways on the virtual index (i.e. cancellations in kernel and cokernel of the operator).
Thanks Chris. What you say works for all almost complex manifolds? Do you think there is nice readable reference for this? (if the reference speaks of connection to localisation formulas, even better).
The book "Manifolds and Modular Forms" by Hirzebruch, Berger and Jung may be helpful. See especially section 5.7.
|
2025-03-21T14:48:29.868315
| 2020-02-15T11:42:40 |
352773
|
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|
Stack Exchange
|
Properties of analytic "super-monomials"
Defining as monomials $m(x,n)\,:=\,x^n,\,n\in\mathbb{N}_0$, I denote by an "super-monomial" an analytic function of the form
$$ \overline{m}(x,n,(a))\ :=\ x^n+\sum\limits_{i=1}^\infty \frac{a_{n+i}x^{n+i}}{(n+i)!};\ \frac{d^{n+k}\overline{m}(x,n,(a))}{dx^{n+k}}\not\equiv 0,\, \frac{\left|a_{n+k}\right|}{(n+k)!}\le n+k\ \ \forall k\ge 0$$
Question:
what can be said about the error term of interpolating a function $g(x)$ by a linear combination
$$\sum\limits_{i=0}^{n}\alpha_i\overline{m}(x,i,(a)_i)$$
over the common interval of convergence at $n+1$ equidistant arguments; when compared to polynomial interpoation; will it be generally better or worse?
is it possible to determine optimal sets $\lbrace(a)_i\,|\,0\le i\le n\rbrace$ of coefficient sequences that are optimal for interpolating functions with a given degree of smoothness?
is there already an established name for what I called super-monomials?
|
2025-03-21T14:48:29.868405
| 2020-02-15T14:40:00 |
352781
|
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"reuns"
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|
Stack Exchange
|
Relations between spectral parameters of automorphic representations
Let $\pi$ be an automorphic representation (say, trivial central character) of $GL(2)$. Let $\alpha(p)$ and $\beta(p)$ denote its spectral parameters at the place $p$, that is to say the associated local L-function takes the form
$$L_p(s, \pi) = (1-\alpha(p)p^{-s})^{-1}(1-\beta(p)p^{-s})^{-1}.$$
Using the expression of $L(s, \pi)$ as a Dirichlet series $\sum_n
c_n n^{-s}$ and the Hecke relations, I can deduce the usual relations
$$\alpha(p) + \beta(p) = c_p \qquad \text{and} \qquad \alpha(p)^2 + \beta(p)^2 = c_{p^2} - 1.$$
Is there a relation in a similar fashion for spectral parameters and coefficients in the case of "twisted", like
$$\alpha(p_1)\alpha(p_2) + \beta(p_1)\beta(p_2) \qquad \text{or} \qquad \alpha(p_1)^{v_1}\beta(p_2)^{v_2}+\alpha(p_2)^{v_1}\beta(p_1)^{v_2}$$
for some powers $v_i$, even $1$ or $2$? I have no idea how to get it from L-functions or other arguments (symmetric square or Rankin-Selberg?).
Do you mean $\alpha_1(p)$ instead of $\alpha(p_1)$ ?
If $p_1$ and $p_2$ are distinct primes, then there is no way to write these expressions in terms of polynomials of the form $\sum_{i,j} a_{ij} c_{p_1^i} c_{p_2^j}$ with coefficients $a_{ij} \in \mathbb{C}$. You can only do this if the expressions are symmetric in both $(\alpha(p_1),\beta(p_1)) \mapsto (\beta(p_1),\alpha(p_1))$ and $(\alpha(p_2),\beta(p_2)) \mapsto (\beta(p_2),\alpha(p_2))$
@PeterHumphries This could be already very interesting, if they have such symmetries, how could I get suitable relations as polynomials in the coefficients?
Not sure - it's more of a problem of algebraic geometry. Basically, Hecke eigenvalues are Schur polynomials in $\alpha,\beta$, and Schur polynomials form a basis for the vector space of symmetric polynomials, and so every symmetric polynomial in $\alpha,\beta$ can be written as a linear combination of Hecke eigenvalues.
|
2025-03-21T14:48:29.868539
| 2020-02-15T15:08:08 |
352785
|
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"authors": [
"Arseniy Akopyan",
"Fedor Petrov",
"Gerry Myerson",
"https://mathoverflow.net/users/2158",
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|
Stack Exchange
|
Number of orders of distances between points on a line
Points $a_1, a_2, \dots, a_n$ on a line form a set from $n(n-1)/2$ distances between them. Suppose all that distances are different, numerating them from the shortest to the longest one we obtain some permutation on $n(n-1)/2$ elements.
How many permutations can be obtained by this way?
In other words, you want to count the number of regions in the hyperplane arrangement ${x_i-x_j,x_i+x_j-x_k-x_l}$ (and divide it by $n!$). I doubt that a formula exists.
Can you do it for a few small values of $n$, Arseniy, and then consult the Online Encyclopedia of Integer Sequences?
@GerryMyerson, I will try if I found how to program it.
@FedorPetrov, what about the asymptotic?
immediate upper estimate is like $n^{3n+o(n)}$, is it ok for you?
Is it by any chance http://oeis.org/A004123 ?
Have you had a look at that link, Arseniy?
Yes, thanks for the ref. I do not understand why it could be the same. But I am busy on that week and will look on it later
|
2025-03-21T14:48:29.868648
| 2020-02-15T15:17:16 |
352786
|
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|
Stack Exchange
|
Distribution of signs of automorphic forms
Let's say we have an automorphic form $f$ on $GL(2)$ that is self-dual. In particular, the associated L-function $L(s,f)$ satisfies a functional equation with sign $\varepsilon_F = \pm 1$.
Is it known that the proportion of such automorphic forms with given sign (say $-1$) is exactly $1/2$?
I know many results about distributions of signs for coefficients and eigenvalues of automorphic forms, however when I think of this question I wonder whether it is well-known or difficult?
Paging @Kimball - https://doi.org/10.1016/j.jnt.2018.01.015
For simplicity, let's consider the case of holomorphic modular forms over $\mathbb Q$ of squarefree level and trivial nebentypus. Then one knows from
Iwaniec, Henryk; Luo, Wenzhi; Sarnak, Peter. Low lying zeros of families of $L$-functions. Publications Mathématiques de l'IHÉS, Tome 91 (2000) pp. 55-131.
an asymptotic formula for the dimensions of the subspaces of cusp forms with root number $+1$ and root number $-1$. In particular, the proportion of forms with root number $+1$ tends to $\frac 12$ as you take some combination of the weight and the level to infinity.
As Peter mentions in the comments, I also consider this in my paper
Refined dimensions of cusp forms, and equidistribution and bias of signs. Journal of Number Theory, Vol. 188 (2018), 1-17.
Specifically, I get an exact formula for the dimensions of subspaces with prescribed root number (or prescribed Atkin-Lehner signs), and observe a "strict bias" phenomenon for root number +1: while the proportion of newforms with root number +1 is asymptotically $\frac 12$, in any given space, there are always at least as many newforms with root number +1 as with -1, and it is strictly greater except in a few special situations.
One should similarly be able to prove that the proportion is $\frac 12$ in more general families of automorphic forms, say using a trace formula and simply bounding error terms, though I don't know the most general situation in which this has already been done in the literature. Essentially one needs to know that the trace of the Fricke involution is not large compared to the dimension.
|
2025-03-21T14:48:29.868937
| 2020-02-15T15:46:28 |
352788
|
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"D.S. Lipham",
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|
Stack Exchange
|
The "core" of complete Erdős space
This question is about the Erdős spaces:
$\mathfrak E=\{x\in \ell^2:x_n\in \mathbb Q\text{ for all $n<\omega$}\}$; and
$\mathfrak E_c=\{x\in \ell^2:x_n\in \mathbb P\text{ for all $n<\omega$}\},$
where $\ell^2$ is the Hilbert space, $\mathbb Q$ is the set of rational numbers, and $\mathbb P=\mathbb R\setminus \mathbb Q$.
I would like to know whether the Polish space $\mathfrak E_c$ contains a homeomorphic copy of either $\mathfrak E$ or $\mathbb Q \times \mathfrak E_c$, such that the complement of that copy is zero-dimensional.
Question 1. Let $X\subseteq \mathfrak E_c$ such that $\mathfrak E_c\setminus X$ is zero-dimensional. Can we have $X\simeq \mathfrak E$ or $\mathbb Q \times \mathfrak E_c$?
Note that a separable metrizable space $X$ has a completion with zero-dimensional remainder if and only if $X$ is rim-complete [2, Main Theorem (case $n=0$)].
Question 2. Are $\mathfrak E$ and $\mathbb Q\times \mathfrak E_c$ rim-complete?
Note that $\mathbb Q\times \mathfrak E_c$ is a closed subspace of $\mathfrak E$ by [1, Corollary 9.5], so if $\mathfrak E$ is rim-complete then so is $\mathbb Q \times \mathfrak E_c$.
In the comments Taras Banakh noted that $\mathbb Q \times K$ cannot be rim-complete when $K$ is a compact space of positive dimension.
[1] Dijkstra, Jan J.; van Mill, Jan, Erdős space and homeomorphism groups of manifolds, Mem. Am. Math. Soc. 979, v, 62 p. (2010). ZBL1204.57041.
[2] Aarts, J. M., Completeness degree. A generalization of dimension, Fundam. Math. 63, 27-41 (1968). ZBL0167.51303.
Very good question. I tend to believe (but cannot prove) that you Question should have negative answer.
@TarasBanakh The Lavrentiev approach seems to be more difficult than I previously thought. I think a good sub-question is: If $X$ is a Polish space of positive dimension, and $G$ is a $G_\delta$-subset of $\mathbb R\times X$ containing $\mathbb Q \times X$, then is the dimension of $G\setminus(\mathbb Q \times X)$ necessarily positive?
If $X$ is compact, then the answer to the subquestion is yes.
I have no reference, but the argument is quite simple: If a Polish space $X$ contains a topological copy of $\mathbb Q\times K$ for some compact space $K$, then $X\setminus (\mathbb Q\times K)$ contains a topological copy of $K$. Using the Lavrentiev Theorem we can reduce the problem to the case $X\subset \mathbb I\times K$, where $\mathbb I$ is a compactification of $\mathbb Q$. Then $S=(\mathbb I\times K)\setminus X$ is a $\sigma$-compact subset of $\mathbb I\times K$ and so is its projection $pr(S)$ to $\mathbb I$.
The Baire Theorem implies that $\mathbb I\setminus\mathbb Q$ is not $\sigma$-compact and hence there exists $x\in\mathbb I\setminus\mathbb Q$ such that $x\notin pr(S)$ and hence ${x}\times K\subset X$.
Maybe given any bounded open subset of $\mathfrak E$ we can find a closed copy of $\mathbb Q$ in its boundary?
No, the boundary of any open bounded set in the Erdos space is uncountable.
And the space $\mathbb Q\times \mathfrak E_c$ indeed has a basis consisting of open sets with Polish boundaries.
@TarasBanakh also about $\mathbb Q$, I was only talking about a closed subset of the boundary. For instance, the rectangular open sets in $\mathbb Q\times \mathfrak E_c$ have closed copies of $\mathbb Q$ in their boundaries.
@TarasBanakh I still do not see why $\mathbb Q\times \mathfrak E_c$ is rim-complete. Did you mean rim-$\sigma$-complete or am I missing some small detail?
Ups! You right concerning the boundaries, Rectangular sets indeed have copies of $\mathbb Q$ in their boundaries. But maybe Baire Theorem will help: assuming that some bounded (canonically) open set in $\mathbb Q\times \mathfrak C_c$ has Polish boundary, decompose it into countably many slices and find a " locally isolated" slice and then derive a contradiction with canonical openness, Maybe this way?
@TarasBanakh Yes, or maybe it is possible in $\mathfrak E$ to take a closed copy of $\mathbb Q$ in the bounded open set, and "move" it to a closed (or at least $G_\delta$) copy of $\mathbb Q$ in the boundary.
@TarasBanakh Another idea: Given a bounded open set in $\mathfrak E$, and a $G_\delta$-set $G\subseteq \mathfrak E_c$ containing its boundary, show $G$ contains a point with irrational coordinates.
But you consider $\mathfrak E$ as embedded into $\mathfrak E_c$ (by an ``irrational shift"). Right?
I have an idea how to prove what you need: just apply Kuratowski-Ulam Theorem on meager sections (https://en.wikipedia.org/wiki/Kuratowski%E2%80%93Ulam_theorem)!
Unfortunately, at the moment I should prepare slides for tomorrow's presentation, so I cannot write the solution immediately. Please wait just a bit.
Unfortunately, my approach using Kuratowski-Ulam does not work. But nonetheless I believe (without proof yet) that $\mathbb Q\times\mathfrak E_c$ cannot be rim-complete.
@TarasBanakh Well thanks for trying. It seems to be a very tricky problem, and it may even be one of those cases where our intuition is wrong.
@TarasBanakh These questions are so simple but I am still struggling to make heads or tails of them. Very frustrating.
Would you define -- pretty please -- "rim-complete"?
@WlodAA It means there is a basis of open sets with completely metrizable (absolute $G_\delta$) boundaries. Don't know if any other terms exist for this.
Thank you. (These definitions get sophisticated).
It seems that your question reduces (at least in one direction) to the problem: let $\gamma:C\to[0,1]$ be an upper semicontinuous function on the Cantor set $C$ whose graph $\Gamma$ is a complete Erdos space. Is there a countable subset $Q\subset [0,1]$ without isolated points such that $(Q\times C)\cap\Gamma$ is Polish?
@TarasBanakh What do you mean by $(Q\times C)\cap \Gamma$? Also if $\Gamma$ is homeomorphic to $\mathfrak E_c$ then the domain of $\gamma$ has to be first category. The domain cannot be compact, but there is a $\sigma$-compact subset $S\subseteq C$ such that $\mathfrak E_c$ is the graph of a USC function $\gamma:S\to[0,1]$.
Maybe you just mean $\Gamma$ is the positive graph in $C\times [0,1]$, and you want to know if $(C\times Q)\cap \Gamma$ is Polish?
@D.S.Lipham I had in mind the complete Erdos space, modeled as the set of end-points of the Lelek fan. The Lelek fan with removed origin can be seen as the hypograph of an upper semicontinuous function on the Cantor set, and the end-points of the Lelek fan then lie on the graph $\Gamma$ of this upper semi-continuous function. Because of that, the complete Erdos space can be identified with the graph of this upper semicontinuous function. Am I right with this description?
Essentially yes. $\gamma$ is sometimes called a Lelek function. $\Gamma\setminus (C\times {0})$ is complete Erdos space. The complement of the zero-set of $\gamma$ is $\sigma$-compact and first category.
@TarasBanakh Do you think it is true that if $X$ is a zero-dimensional subset of the graph of a USC function $\varphi:C\to [0,\infty)$, then $\mathbb R\times(0,\infty)\setminus X$ is path-connected? This would imply negative answers to all questions above (and it is weaker than my recent conjecture in: https://mathoverflow.net/questions/361671/zero-dimensional-functions-in-the-plane).
|
2025-03-21T14:48:29.869378
| 2020-02-15T15:49:40 |
352789
|
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|
Stack Exchange
|
Area of a surface confined by a sphere
Let $S$ be a hypersurface enclosed inside the unit sphere in $R^n$. We may assume that every ray $\{t x: t \geq 0 \}$ intersects $S$ at most once.
Under what extra condition is ${\rm Area}(S) \leq {\rm Area}(S^{n-1})$ ?
(I am mostly interested in the 2-dimensional case.)
Thanks.
Should be able to use Radon transforms. Try the paper of Alvarez Paiva and Fernandez.
I think you mean the unit ball, don't you?
If you regard the hypersurface $S$ as a radial graph over a domain $D\subseteq S^{n-1}$, i.e, $$ S = { r(p),p\ |\ p\in D,}$$ for some function $r:D\to[0,1]$, then the standard formula for the $(n{-}1)$-content of $S$ is $$A_{n-1}(S) = \int_D r^{n-2}(r^2+|\nabla r|^2)^{1/2} ,\mathrm{d}V_{n-1},$$ so it's a question of what you want to assume about the function $r$.
Thanks very much. It seems like a useful tool. I am trying to find a reference for this formula. (For instance, do you mean $D$ is somehow centered at the origin? I also wonder how to interpret the gradient of $r$ exactly, as it is defined on $D$.) BTW, i refined the question in: https://mathoverflow.net/questions/352836/area-of-a-surface-confined-by-a-sphere-ii
$D$ is a domain (i.e., open set) in $S^{n-1}$. (I imagine that you want $D$ to be all of $S^{n-1}$, but you didn't say that.) I'm not sure what you mean by 'centered at the origin'. If $D$ is a hemisphere in $S^{n-1}$, is it 'centered at the origin'? Gradient has its usual meaning for a function defined on a hypersurface: Extend the function to be constant in the normal direction and compute the gradient as usual in $n$-space. I saw the other question (and, really, you should have just modified or added to this one, they are so close). What do you mean by 'elliptic'?
I see now. Thanks for the clarification. By a elliptic point I mean the two principle curvatures are of the same sign, or that the Gauss curvature is positive. (I am assuming that the dimension n is 2.)
|
2025-03-21T14:48:29.869539
| 2020-02-15T16:24:05 |
352791
|
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"Laurent Moret-Bailly",
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|
Stack Exchange
|
Reference request - conjugacy classes over local fields
Is there a nice reference for reductive groups over local fields, which for example contains discussion of things such as: Given a semisimple element in $G(F)$, its $G(F)$-conjugacy class is closed in the analytic topology of $G(F)$, and related things?
I don't know such a reference. However, I would suggest to ask a reference for the following assertion: the $G(\overline F)$-conjugacy class of any semisimple element in $G(\overline F)$ is Zariski-closed. The tags in such a question should include invariant-theory.
Now it remains to show that if $F$ is a local field and $X$ is a homogeneous space of $G$, then the orbits of $G(F)$ in $X(F)$ are closed in the analytical topology. Proof: the number of orbits is finite (see Serre, Galois Cohomology), and they are open, hence the complement of each orbit is open, hence each orbit is closed.
@MikhailBorovoi Dear Michael, the first fact I knew, but your next comment helped me understand better how to think about this rational situation. Thank you!
@MikhailBorovoi Do you know if something goes wrong if $F$ has char. p? I think of $G(F)$ etc. as analytic manifolds, so don't have a feeling of what can go wrong in char. p
Serre (GC, III.4.4, Thm. 5) assumes that $F$ is perfect when proving that there are finitely many $G(F)$-orbits in $X(F)$.
In char 0, since $X$ is homogeneous, for any $x\in X(F)$ the map $$\phi_x\colon G\to X, \quad g\mapsto g\cdot x$$ is smooth, and hence the map on $F$-points $G(F)\to X(F)$ is open by the implicit function theorem, and hence the $G(F)$-orbit of $x$ is open.
However, I think that all will work also in char p, if the (scheme-theoretical) stabilizer of $x$ in $G$ is a smooth connected reductive group.
@MikhailBorovoi OK, thank you, I will think about that.
More generally, if $F$ is a henselian valued field, $X$ a homogeneous space under an algebraic $F$-group $G$, and the stabilizers are smooth, the $G(F)$-orbits are open and closed in $X(F)$. This follows from Proposition 3.4.1 in this paper: note that if $x\in X(F)$ the corresponding orbit map $G\to X$ is a $G_x$-torsor over $X$.
@LaurentMoret-Bailly Thank you!
|
2025-03-21T14:48:29.869720
| 2020-02-15T16:44:40 |
352792
|
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|
Stack Exchange
|
Independent conditions imposed by points in different planes
Let $H_1$ and $H_2$ be two planes in $\mathbb{P}^3$.Let $P$ be a set of $9$ points such that no three lie on a line. Suppose $H_1$ contains 4 of them and $H_2$ contains remaining $5$ points. Is it true that $P$ imposes independent conditions on quadrics ?
No, because the points lie on the quadric $Q = H_1 \cap H_2$.
Thanks for the answer. But if $Q= H_1\cap H_2$ is the only quadric , then it imposes 9 independent conditions. Thus it does not give any contradiction.
Sorry, I misunderstood the question. Indeed, if you take 5 and 4 points in general position on the planes they impose independent conditions on quadrics.
Thank you very much for the reply. Could you please give a sketch of the proof or at least a reference? If no three points lie in a line, then i hope they are in general position in plane.
Let $C'' \subset H_2$ be the unique conic on $H_2$ containing the 5 points and let
$$
\{C'_t\}_{t \in \mathbb{P}^1} \subset H_1
$$
be the pencil of conics containing the 4 points. Let
$$
L = H_1 \cap H_2.
$$
The points impose independent conditions on quadrics if and only if there is no $t$ such that
$$
C'_t \cap L = C'' \cap L.\tag{*}
$$
Indeed, any quadric passing through these points (except for $H_1 \cap H_2$) intersects the planes along conics $C'_t \cup C''$ for some $t$, and it intersects the line $L$ along a 2-point scheme (jr contains $L$), so $(*)$ follows.
The converse follows from the exact sequences
$$
0 \to \mathcal{O}_{\mathbb{P}^3} \to \mathcal{O}_{\mathbb{P}^3}(2) \to \mathcal{O}_{H_1 \cup H_2}(2) \to 0
$$
and
$$
0 \to \mathcal{O}_{H_1 \cup H_2}(2) \to \mathcal{O}_{H_1}(2) \oplus \mathcal{O}_{H_2}(2) \to \mathcal{O}_{L}(2) \to 0.
$$
Thank you very much for the answer. I have a silly question: why there is no $t$ such that (*) hols ?
Assume for simplicity that the pencil has no conics containing $L$. Then $t \mapsto C'_t \cap L$ defines a map $\mathbb{P}^1 \to S^2L = \mathbb{P}^2$ (its image is, in fact, a line). On the other hand $C'' \cap L$ defines a point of $S^2L$. So, the extra generality assumption that you need is that the above line in $S^2L$ does not contain the above point.
Now i have understood. Thank you very much.
Can we impose that generality assumption from the generality assumption on points in the plain ? sorry for bothering you again
Sorry I got the answer. Probably we can move the point in the 2nd plane or in 1st plane to get that generality hypothesis. thanks.
|
2025-03-21T14:48:29.869914
| 2020-02-15T17:04:07 |
352796
|
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"authors": [
"Gael Meigniez",
"Moishe Kohan",
"https://mathoverflow.net/users/105095",
"https://mathoverflow.net/users/39654"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352796"
}
|
Stack Exchange
|
Foliation of $\mathbb R^n$ by connected compact manifolds
Does there exist a smooth nontrivial fiber bundle $p: F \hookrightarrow \mathbb R^n \to B$ such that $F$ and $B$ are connected manifolds with $F$ compact? "Nontrivial" here means the fiber $F$ is not a point.
The title says "foliation", the body of the question says "fibration": Which one do you actually mean? The question about foliations was asked earlier (one and half ears ago) at MSEhttps://math.stackexchange.com/questions/2842525/folitaion-of-mathbb-rn-by-compact-leaves
There does not, even if you don’t require the fiber and base to be manifolds (or even connected, just that $F$ is not a single point). See
Borel, Armand; Serre, Jean-Pierre,
Impossibilité de fibrer un espace euclidien par des fibres compactes,
C. R. Acad. Sci. Paris 230 (1950), 2258–2260.
On the other hand, if you only mean "foliation" as in your title, and not "fibration", then there is Vogt's foliation of R^3 by circles! (But it is not C^1, only differentiable).
Vogt, Elmar, "A foliation of R3 and other punctured 3-manifolds by circles",
Publications Mathématiques de l'IHÉS, Tome 69 (1989), p. 215-232
http://www.numdam.org/item/PMIHES_1989__69__215_0/
And it appears that the existence of a $C^1$-foliation by circles is an open problem.
Thank you Moishe, I should have added that Vogt's foliation is not C^1. It's corrected.
|
2025-03-21T14:48:29.870036
| 2020-02-15T17:14:35 |
352797
|
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"AlexArvanitakis",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352797"
}
|
Stack Exchange
|
Differential operators and vector fields
Let $M$ be a smooth manifold. It is well known that there is a bijective correspondance between vector fields on $M$ and differential operators of order 1. My question is: if we take a differential operator of order larger than $1$, what is the "corresponding object" ?
You can form tensors of any desired rank.
perhaps you want jet bundles?
The "well known" bit isn't quite correct, I think because of a slight misunderstanding of the "order" of a differential operator: order is a filtration, not a grading. The idea still makes some sense, however, and the correction points the way to a correct answer: instead of looking for something corresponding to the operator itself, look for something in the associated graded algebra. The relevant concept is the principal symbol.
|
2025-03-21T14:48:29.870131
| 2020-02-15T17:30:38 |
352799
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/68463",
"sharpe"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/352799"
}
|
Stack Exchange
|
References for Neumann eigenfunctions
I am looking for references on eigenfunctions with Neumann boundary condition.
In an article, the author wrote in introduction that when a domain is planar polygon, the second eigenfunctions on it have critical points at vertices. But, I could not find the source. I think this is probably a well-known fact. But I am very interested in the proof.
It may be a famous fact for experts, but I would appreciate it if you could tell me.
This is quite straightforward once we know that the gradient is well-defined and continuous. Indeed: by the Neumann boundary condition, the gradient of an eigenfunction at a boundary point is parallel to the boundary. Therefore, at a vertex, the gradient is parallel to two different directions, and hence it is necessarily zero.
@MateuszKwaśnicki Thank you very much for your comments. I am convinced. I didn't think of it until you said. The comments of Carlo Beenakker is in a different direction, right?
I think so, it deals with non-vertex critical points. Still, it is an excellent answer!
@MateuszKwaśnicki I think so too!
From this recent paper I would conclude the statement is false: the second Neumann eigenfunction of an acute triangle has one non-vertex critical point.
This was a Polymath problem.
Thank you for your comment. Do you mean that the second Neumann eigenfunction on an acute triangle does not have critical points at vertices?
that is not how I understand the cited paper: the statement is that there is exactly one critical point that is not at a vertex, I don't think critical points at vertices are excluded.
|
2025-03-21T14:48:29.870272
| 2020-02-15T17:47:16 |
352801
|
{
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"authors": [
"Bugs Bunny",
"Jeremy Rickard",
"YCor",
"abx",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/352801"
}
|
Stack Exchange
|
Faithfully flat modules over a group algebra
Suppose we have the following data:
1) A group ring $\mathbb{Z}[G]$, where $G$ is a torsion free group.
2) $M_{\bullet}$ a bounded (above and below) chain complex of $\mathbb{Z}[G]$-modules such that each $M_{i}$ is a finitely generated free $\mathbb{Z}[G]$-module.
My question is the following:
If the homology of $M_{\bullet}\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ is trivial ie
$H_{n}(M_{\bullet}\otimes_{\mathbb{Z}[G]}\mathbb{Z})=0$ for all $n\in \mathbb{Z}$ does it imply that $$H_{n}(M_{\bullet})=0$$ for all $n\in \mathbb{Z}$
I am not sure if this is a research-level question, btw.
Why "faithfully flat" in the title? it doesn't reappear in the question.
No way. Let $G={\mathbb Z}$ so that ${\mathbb Z}[G]={\mathbb Z}[x,x^{-1}]$. Then use the complex
$$\ldots \rightarrow 0 \rightarrow 0 \rightarrow {\mathbb Z}[x,x^{-1}] \xrightarrow{1-x+x^2} {\mathbb Z}[x,x^{-1}] \rightarrow 0 \rightarrow 0 \rightarrow \ldots$$
The complex $M_{\bullet}\otimes _{\mathbb{Z}[G]}\mathbb{Z}$ doesn't seem to be exact.
Point taken. It is just wrong polynomial. We need $f(1)=1$, not $f(1)=0$.
Let $G$ be infinite cyclic, generated by $x$.
Let $M_\bullet$ be a free resolution of the $\mathbb{Z}[G]$-module $U=\mathbb{Z}/3\mathbb{Z}$ with $x$ acting by multiplication by $-1$. For example, take $M_\bullet$ to be
$$\dots\to0\to\mathbb{Z}[G]\stackrel{\pmatrix{-3\\x+1}}{\longrightarrow}\mathbb{Z}[G]\oplus\mathbb{Z}[G]\stackrel{\pmatrix{x+1&3}}{\longrightarrow}\mathbb{Z}[G]\to0\to\dots$$
Then $M_\bullet$ is not acyclic, but the homology of $M_\bullet\otimes_{\mathbb{Z}[G]}\mathbb{Z}$ is $\text{Tor}^{\mathbb{Z}[G]}_\bullet(U,\mathbb{Z})$, which is zero, since tensoring $U$ with the projective resolution
$$\dots\to0\to\mathbb{Z}[G]\stackrel{x-1}{\longrightarrow}\mathbb{Z}[G]\to0\to\dots$$
of $\mathbb{Z}$ gives the complex
$$\dots\to0\to\mathbb{Z}/3\mathbb{Z}\stackrel{-2}{\longrightarrow}\mathbb{Z}/3\mathbb{Z}\to0\to\dots,$$
which is acyclic.
As Bugs Bunny's corrected answer shows, this is unnecessarily complicated.
|
2025-03-21T14:48:29.870420
| 2020-02-15T17:52:07 |
352802
|
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|
Stack Exchange
|
Is the set of images of an open subset of full-rank matrices an open subset of the Grassmannian?
$\DeclareMathOperator\Gr{Gr}$ Let $\Gr(k,n)$ be the set of $k$-dimensional subspaces in affine space $\mathbb{A}^n$ over an algebraically closed field. If $U\subseteq (\mathbb{A}^n)^{\times k}$ is an open subset of $n \times k$ matrices of rank $k$, is $\{\text{Im}(u): u \in U\} \subseteq \Gr(k,n)$ open?
Here is my convoluted attempt at a solution. Via the Plücker embedding, one can view $\Gr(k,n)$ as the set of totally decomposable elements of $\bigwedge^k_n$, which can be shown to form a projective variety.
$\Gr(k,n)$ can also be viewed as the set of $n \times k$ matrices of rank $k$, quotiented by the (equivalence relation induced by) the right action of $\text{GL}(k)$. Since $\text{GL}(k)$ is a reductive algebraic group, I believe this quotient is again an affine variety with induced topology given by the quotient topology. Is this correct? If so, then I don't think it's hard to prove that these two varieties corresponding to $\Gr(k,n)$ are isomorphic, which would seem to imply my desired result.
Related question.
At least one thing seems wrong in what you said: the Grassmannian is a projective variety, not an affine variety.
@SamHopkins Yep, its projective. I meant that I am taking Gr(k,n) to be the projective variety of k-planes in affine n-space, as opposed to projective n-space (although the distinction doesn’t really matter)
I was referring to this bit: "Since GL(k) is a reductive algebraic group, I believe this quotient is again an affine variety with induced topology given by the quotient topology."
Oh, thanks for catching that. As a set, Gr(k,n) can be viewed as the set of rank k $n\times k$ matrices, quotiented by the right action of GL(k). Can we view this set as a projective variety? Does the induced quotient topology agree with the topology of the Plücker embedding?
The map $U \to \mathbb{G}(k,n)$ that you are interested in is flat, and a flat finitely presented map is well known to be open.
Flatness can be checked in several ways: for example, the map is generically flat, by the generic flatness theorem. But it is also $\mathrm{GL}_n$-equivariant, and the action of $\mathrm{GL}_n$ on $U \to \mathbb{G}(k,n)$ is transitive. Alternatively, it follows from the fact that the fibers are equidimensional, $\mathbb{G}(k,n)$ is regular and $U$ is also regular, hence Cohen--Macaulay.
|
2025-03-21T14:48:29.870607
| 2020-02-15T18:10:53 |
352803
|
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|
Stack Exchange
|
What do you call a map of spaces which is weakly left orthogonal to all $n$-connected maps?
$\let\op=\operatorname$In $\op{Set}$, we have an $(\op{Epi},\op{Mono})$ orthogonal factorization system. Strikingly, if we reverse the roles, we get the no-less-important $(\op{Mono},\op{Epi})$ weak factorization system.
In the $\infty$-category $\op{Spaces}$ of spaces, the most direct analog of the $(\op{Epi},\op{Mono})$ orthogonal factorization system on $\op{Set}$ is the $(\text{Effective Epi}, \op{Mono})$ orthogonal factorization system, but this is just the $(-1)$th in a whole tower: for each $n \in \mathbb Z_{\geq -2}$, we have an $(\text{$n$-connected}, \text{$n$-truncated})$ factorization system [1].
It seems that, just as in the analogous case in $\op{Set}$, one can take the left half of each these orthogonal factorization systems, and view it as the right half of a weak factorization system $(\mathcal L_n, \text{$n$-connected})$ [2]. To see this, one shows that the $n$-connected maps are precisely the maps which are weakly right orthogonal to the maps $\{S^k \to 1 \mid -1 \leq k \leq n\}$, and applies the small object argument to obtain factorizations.
In $\op{Set}$, we have the cute fact that the resulting weak factorization system $(\op{Mono},\op{Epi})$ is just the original orthogonal factorization system $(\op{Epi},\op{Mono})$ with the left and right classes swapped. This is not the case in $\op{Spaces}$, even when $n=-1$: a map $A \xrightarrow i B$ of spaces is weakly left orthogonal to the effective epimorphisms if and only if it is a coproduct inclusion $A \to A \amalg S$ where $S$ is discrete; this is more restrictive than being a monomorphism [3]. I don't know how to characterize the left class $\mathcal L_n$ for $n\geq 0$ as cleanly. In fact, unlike the case in $\op{Set}$, I don't think we have either containment $\mathcal L_n \subseteq \text{$n$-truncated}$ or $\text{$n$-truncated} \subseteq \mathcal L_n$ in general. This leads to my
Questions: Let $n \in \mathbb Z_{\geq -2}$.
Is there a good characterization of the class of maps $\mathcal L_n$, i.e., the maps of spaces which are weakly left orthogonal to the $n$-connected maps?
What would be a good name for the maps of $\mathcal L_n$?
Note that by the small object argument, the maps of $\mathcal L_n$ are precisely the retracts of transfinite composites of cobase-changes of coproducts of the maps $\{S^k \to 1 \mid -1 \leq k \leq n\}$. So in some sense this is a quite explicit class of maps. By "characterization" I suppose I mean something which can be "checked directly" without having to find all the data of a construction of this form.
[1] Here a map is said to be $n$-truncated or $n$-connected if its fibers are all so. This convention is off by one from the most classical convention.
[2] Some care should be taken with the definition of a weak factorization system $\infty$-categorically: say that a morphism $A \xrightarrow i B$ is weakly orthogonal to a morphism $X \xrightarrow p Y$ if the map $\op{Hom}(B,X) \to \op{Hom}(B,Y) \times_{\op{Hom}(A,Y)} \op{Hom}(A,X)$ is an effective epimorphism. Spelled out, this says that if we have a commutative square—i.e., morphisms $A \xrightarrow u X$, $B \xrightarrow v Y$ along with a homotopy $\gamma: pu \sim vi$, then there exists a lift, i.e., $B \xrightarrow w X$ and homotopies $\alpha: wi \sim u$, $\beta: pw \sim v$ and (here's the only subtle part) a homotopy of homotopies from the composite $\beta \ast \alpha$ to $\gamma$. Then a weak factorization system is, as usual, a pair of classes of morphisms $(\mathcal L, \mathcal R)$ which are complements to each other with respect to weak orthogonality, such that every morphism admits a factorization as a morphism in $\mathcal L$ followed by a morphism in $\mathcal R$.
[3] Recall that a monomorphism of spaces is a coproduct inclusion $A \to A \amalg S$ where $S$ may be an arbitrary space.
Good grief. $\mathcal L_n$ is just the class of maps which are retracts of relative cell complexes of dimension $\leq n+1$. I'd probably call these "retracts of $n+1$-skeletal maps" or something.
Is this your answer to your own question? I have trouble parsing the meaning of "good grief" here ….
@LSpice. Yeah, I think I answered my own question... Maybe I should delete it, but I spent so much time typing it out... And now you've invested time in prettying it up, too! (Thanks, btw) Actually, I suppose it's still an interesting question how to characterize these maps, but $\mathcal L_n$ is so familiar that if a good characterization were known it would have been known in the '50's, and I don't think such a thing is known... Although maybe in the simply-connected case you can say something about dimension vs. homology dimension or something...
To clarify, "Good grief" is my sheepish exclamation when I realized that I found a ridiculous roundabout way to ask for a characterization of the retracts of relative $n+1$-dimensional complexes without recognizing that that's what I was asking!
Yes, there is a characterization in terms of (co)homology in most cases. Of course for a map in this class the relative homology and cohomology vanishes in degrees above $n+1$. Conversely, if relative $H^j$ vanishes for all $j\ge n+1$, for all coefficient systems on the codomain, then the map is in that class. This holds for all $n>2$, at least.
@TomGoodwillie Thanks! Just to be clear, you're saying that if $H^{\geq 5}(Y,X; \mathcal L) = 0$ for all coefficient systems $\mathcal L$ on $Y$, then $X \to Y$ is (homotopy equivalent to) a retract of a relative $\leq 4$-dimensional cell complex, but you're not sure whether $\forall \mathcal L, H^{\geq 4}(Y,X;\mathcal L) = 0$ implies that $X \to Y$ is (homotopy equivalent to) a retract of a relative $\leq 3$-dimensional cell complex, right? I'm curious now how the dimensionality affects things.
@Tim Campion I seem to have gotten confused by the notation. What I meant was better by two dimensions. Suppose that $(Y,X)$ is cohomologically $d$-dimensional in the sense that $H^{>d}$ vanishes for all coefficient systems. Let's show that it is homotopically $d$-dimensional in the sense that it is a retract of a relative cell complex with no cells in dimension $>d$. We need $d\ge 2$. Any map of spaces $X\to Y$ can be factored $X\to Z\to Y$ with $i:X\to Z$ a relative cell complex with no cells in dimension $>d$, and with $p:Z\to Y$ a $d$-connected map (homotopy fibers are $(d-1)$-connected).
Now use obstruction theory to make a section of $p$ that restricts to the given $i$ in $X$. (Think of a Postnikov tower for $p$ and lift up the tower using the vanishing of cohomology with coefficients in the homotopy groups of the fiber of $p$.) This works because the fibers are simply connected ($p$ is a $2$-connected map).
This question was answered in the comments $\mathcal L_n$ comprises those maps which are retracts of relative $\leq n+1$-dimensional relative cell complexes.
Tom Goodwillie explains in the comments a cohomological characterization of $\mathcal L_n$ for sufficiently large $n$.
|
2025-03-21T14:48:29.871175
| 2020-02-15T18:22:28 |
352806
|
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|
Stack Exchange
|
The exact constant in a bound on ratios of Gamma functions
The answer to another question (Upper bound of the fraction of Gamma functions) gave an asymptotic upper bound for an expression with Gamma functions:
$$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\!\leq \,C\,\frac{a+b}a, \forall a,b\geq\frac12$$
What is the best possible value for the constant $C$ in that statement?
The optimal $C$ is $\mathrm{e}$.
Proof:
We have
$$\ln C \ge \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
-\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
Let
$$F(a, b) := \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
-\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
We have
$$\frac{\partial F}{\partial b}
= - \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a} \ge 0 \tag{1}$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
The proof of (1) is given at the end.
Fixed $a\ge 1/2$, we have
$$G(a) := \lim_{b\to \infty} F(a, b) =
\ln a
+ \frac{
-\ln a - \ln\Gamma(a)}{a}
$$
where we have used
$$\lim_{b\to \infty} -\ln(a + b) + \frac{\ln \Gamma(a + b) - \ln\Gamma(b)}{a} = 0.$$
(Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
We have
$$
a^2 G'(a) = a - 1 - a\psi(a) + \ln a + \ln \Gamma(a) \ge 0.\tag{2}
$$
The proof of (2) is given at the end.
We have
$$\lim_{a\to \infty} G(a) = 1.$$
(Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
Thus, the optimal $C$ is $\mathrm{e}$.
Proof of (1):
Using Theorem 5 in [1]: for all $u > 0$,
$$\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2},$$
we have
\begin{align*}
&- \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a}\\
\ge{}&- \frac{1}{a+b} + \frac{1}{a} \left(\ln (a+b) - \frac{1}{2(a+b)} - \frac{1}{12(a+b)^2}\right)\\
&\qquad
- \frac{1}{a}\left(\ln b - \frac{1}{2b} - \frac{1}{12(b+1/14)^2}\right)\\
={}& \frac{1}{a}\ln(1 + a/b) - \frac{1}{a+b} - \frac{1}{2a(a+b)} - \frac{1}{12a(a+b)^2} + \frac{1}{2ab} + \frac{1}{12a(b+1/14)^2}\\
\ge{}& \frac{1}{a}\left(\ln(1 + a/b) - \frac{a/b}{1 + a/b}\right)
+ \frac{1}{2ab} - \frac{1}{2a(a+b)} + \frac{1}{12a(b+1/14)^2} - \frac{1}{12a(a+b)^2}\\
\ge{}&0
\end{align*}
where we use $\ln(1+x) \ge \frac{x}{1+x}$ for all $x \ge 0$.
We are done.
$\phantom{2}$
Proof of (2):
Using
$\Gamma(x) \ge \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}$ and
$\psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2}$ for all $u > 0$ (Theorem 5 in [1]),
we have
\begin{align*}
&a - 1 - a\psi(a) + \ln a + \ln \Gamma(a)\\
\ge{}& a - 1 - a \left(\ln a - \frac{1}{2a} - \frac{1}{12(a+1/14)^2}\right) + \ln a + \frac12\ln(2\pi) + (a-1/2)\ln a - a\\
={}& \frac12\ln(2\pi a) - \frac12 + \frac{a}{12(a+1/14)^2}\\
\ge{}& 0.
\end{align*}
We are done.
Reference
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.
Too long for a comment. You know certainly that the so-called Beta function $B$ ( a classical special function) is defined by
$$
B(a,b)=\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}=\int_0^1 t^{a-1}(1-t)^{b-1} dt
\quad\text{ for $\Re a>0, \Re b> 0$.}
$$
You are thus looking for a lowerbound for the $B$ function. When $a, b$ are real-valued and large the Beta function is equivalent to
$$
\sqrt{2π}\frac{a^{a-\frac 12}b^{b-\frac 12}}{(a+b)^{a+b-\frac 12}}.
$$
You can also fix $b$ and consider $a$ large to get the equivalent
$$
\Gamma (b) a^{-b}.
$$
Both asymptotics are obtained from Stirling's formula.
Taking $f(a,b)=\frac{a+b}{a}$ as in the other question, I conjecture that the optimal $C$ is $e$.
One bit of evidence is that the limit of the ratio for $a=cb$ and $b\to\infty$ is $(c+1)^{1/c}$ which converges to $e$ as $c\to\infty$. Also the limit for $a=b^{1/2}$ as $b\to\infty$ is exactly $e$ and the ratio seems to be increasing.
|
2025-03-21T14:48:29.871662
| 2020-02-15T19:11:26 |
352808
|
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"url": "https://mathoverflow.net/questions/352808"
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|
Stack Exchange
|
Magnitude of ADR algebras
Let $A$ be a connected quiver algebra with $n$ simple modules and Jacobson radical $J$ and Loewy length $n+1$ (that is $J^{n+1}=0$ and $n$ is minimal with this property).
The ADR-algebra $B_A$ of $A$ is defined as $B_A:=End_A(A \oplus A/J \oplus A/J^2 \oplus ... \oplus A/J^n)$.
Recall that the magnitude (see http://www.tac.mta.ca/tac/volumes/31/3/31-03.pdf ) of an algebra of finite global dimension and Cartan matrix $C$ is defined as the sum of all entries of the inverse of $C$.
Question 1: Is the magnitude of $B_A$ equal to $n$?
In the special case when $A$ is a commutative Frobenius algebra,
we have that the Cartan matrix $C$ of $B_A$ has entries $C_{k,t}=\dim(Hom_A(A/J^k,A/J^t))$, which should be equal to $dim(J^{max(0,t-k)}/J^t)$ in case I made no mistake, so it has an elementary interpretation.
Question 2: Is the magnitude of $B_A$ equal to one in case $A$ is a commutative Frobenius algebra?
The questions are tested for many algebras with the computer, but of course there are too many algebras to have really good evidence.
(Question 2 is of course a special case of question 1, but here the Cartan matrix has a more elementary form to calculate explicitly).
For example let $A=K[x,y]/(x^2,y^2)$, then the Cartan matrix of $B_A$ is given by $C$=[ [ 1, 2, 1 ], [ 1, 3, 3 ], [ 1, 3, 4 ] ].
The inverse matrix is
[ [ 3, -5, 3 ], [ -1, 3, -2 ], [ 0, -1, 1 ] ], where the sum of all entries is equal to 1.
|
2025-03-21T14:48:29.871780
| 2020-02-15T19:31:56 |
352809
|
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|
Stack Exchange
|
Naturality up to (inner) automorphism?
Let $C$ be a small poset category. I'm coming across a category where objects are (certain) functors valued in groups $F\colon C\to\operatorname{Grp}$, and morphisms between two objects $F$ and $G$
are almost natural transformations $\eta\colon F \Rightarrow G$.
The difference is that if $f\colon e \to v$ is an arrow of $C$, we are allowed an inner automorphism $\Phi_e\colon Gv \to Gv$ such that the following diagram commutes
$$\require{AMScd}\begin{CD} Fe @>\eta_e>> Ge \\
@VVFfV @VV\Phi_eGfV \\ Fv @>\eta_v>> Gv.\end{CD}$$
I'm curious if this "naturality up to automorphism" condition occurs elsewhere in the wild and if you know of a name for it?
Assuming the inner automorphisms you get are assumed to satisfy some further coherency conditions, your morphisms should amount to pseudonatural transformations. This requires thinking of the category of groups as a 2-category.
To clarify: groups are seen as a $2$-category, where arrows are group morphisms, and two cells from $f$ to $h$ are element $g$ such that $g f g^{-1}= h$ (or the other way arround). It is equivalent to the $2$-category of connected groupoids.
Oh this is very neat! I think the coherency conditions end up being vacuous in my setting ($C$ is a small category without loops coming from a graph), and I learned a new word!
|
2025-03-21T14:48:29.871898
| 2020-02-15T21:10:20 |
352815
|
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|
Stack Exchange
|
Schur polynomials with zeros in an infinite geometric progression
Let $a_1,...,a_n\in \overline{\mathbb{Q}}^\times$ be algebraic numbers, such that $\frac{a_i}{a_j}$ is not a root of unity for $i\neq j$. Furthermore, let $m\in\mathbb{N}$, $m>1$.
Question: Is there a homogeneous polynomial $P\in \overline{\mathbb{Q}}[X_1,...,X_n]$ in $n$ variables of some degree $k$, such that $P(a_1^{m^\ell},…,a_n^{m^\ell})=0$ (edit: where $P$ is a certian Schur polynomial) for all $\ell\in\mathbb{N}_0$?
Edit: Originally, I asked myself this question where is a certain Schur polynomial. Since there are indeed examples of such polynomials in general (as suggested by user44191 in the comments below), I should add this assumption to my question.
Choose $\alpha_3 = \frac{\alpha_2^2}{\alpha_1}$; then $\alpha_1 \alpha_3 = \alpha_2^2$, so $\alpha_1^a \alpha_3^a = \alpha_2^{2a} = \left(\alpha_2^a\right)^2$ for any $a$, including $a = m^\ell$ for any $m, \ell$. So $P = X_1 X_3 - X_2^2$ should answer your question, if I'm not mistaken.
Would you like to tell us what a Schur polynomial is?
I'm not sure if it is necessary to fully define Schur polynomials. They can be defined by a semistandard Young tableux (you can find this in the Wikipedia entry of Schur polynomials, right at the beginning of the Properties section). What seems to be relevant is that these polynomials are symmetric, though.
Probably not. I don't see an easy proof in general. Similar questions have been asked in Arithmetic Dynamics (look for dynamical analogues of Mordell-Lang). Clearly, the answer is no for $n=2$. For $n=3$, if you had such a situation then you would get infinitely many points on a fixed number field on the curves of the form $P(x_1^{m^k},x_2^{m^k},x_3^{m^k})=0$ (for any $k$) and these curves will have all components of genus at least two if $k$ is large enough, so Mordell-Faltings prevents this, so the answer is no for $n=3$ also. (BTW, I assume $m>1$ and I guess you do too).
Edit: The counterexample of user44191 is correct. What goes wrong with my proof? $x^ny^n = z^{2n}$ factors as $\prod_{\zeta^n =1} (xy -\zeta z^2)$. So my assertion that you get curves of higher genus is wrong. It's only wrong if $P$ is a binomial, though. In this case $P=0$ is a translate of a subgroup of the torus $\mathbb{G}_m^2$.
I can construct counterexamples with non-binomials, e.g. by adding binomials together (e.g. $x_1^n y_1^n - z_1^{2n} + x_2^n y_2^n - z_2^{2n}$); you still should get the torus subgroup, but it happens in more cases than binomials.
@user44191 Yes, but with more variables. I am only claiming stuff with $n=3$.
Thanks for the comments. Originally, I asked myself this question where $P$ is a certain Schur polynomial. But I was wondering, if this can be negated even in general.
|
2025-03-21T14:48:29.872094
| 2020-02-15T22:40:37 |
352820
|
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|
Stack Exchange
|
Area of a elliptic surface confined by a sphere
Let $S$ be a surface enclosed inside the unit sphere in $R^3$. If every point of S is elliptic, then must $\operatorname{Area}(S)≤\operatorname{Area}(S^2)$?
Just thought I'd mention that the term elliptic surface has a very specific meaning in algebraic geometry, and it's not the one you're using! An elliptic surface is a surface (2-dimensional algebraic variety) $S$ that admits an algebraic map $f:S\to C$ to a curve such that all but finitely many of the fibers are elliptic curves. (Usually it's assumed that there is a section, in which case the set of sections is a group.)
I do not think so. Just imagine that you peel a large orange whose surface area is much bigger than that of a unit sphere. Then you "spiral" the peel to make it arbitrarily small and place it inside a unit sphere. Imagine a surface that looks more or less as this one:
It has positive curvature so very point is elliptic.
Here you remove a small cylinder around the vertical axis so there is no problems with the curvature.
Since you can have as many "turns" as you want, its surface area can be arbitrarily large while the surface occupies a small region in space.
Thanks. Originally I had a ray condition on S in mind. Your example shows I cannot dispense with it. I reformulated the question in
https://mathoverflow.net/questions/352836/area-of-a-surface-confined-by-a-sphere-ii
|
2025-03-21T14:48:29.872219
| 2020-02-15T22:59:08 |
352821
|
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|
Stack Exchange
|
Generalization of approximate tangent spaces to subsets of arbitrary manifolds?
I'm anything but an expert in Geometric Measure Theory, so please forgive me if I'm asking a trivial question.
Let $(M^n, g)$ be a smooth Riemannian manifold, $d \in \mathbb{N}$ and $A ⊂ M$ be a subset which is measurable w.r.t. the Hausdorff-$d$ measure $^d$ coming from the metric $g$. Embed $(M, g)$ isometrically into some $ℝ^N$. For $^d$-measurable subsets of $ℝ^N$ (like $A$) there exists the notion of approximate tangent spaces.
So suppose $A ⊂ M ⊂ ℝ^N$ has an approximate tangent space $T_x A$ at $x ∈ A$. Are the following statements true?
$T_x A$ is a subspace of $T_x M$.
$T_x A$ is independent of the isometric embedding.
(Assuming 2 is true), $T_x A$ is independent of the choice of $g$. (This requires that $^d$-measurability of $A$ is independent of the choice of $g$ which should be the case, unless I'm mistaken.)
Put differently, the question I'm asking is: In how far is the approximate tangent space of $A ⊂ M$ at $x$ an intrinsic concept, depending only on the structure of $A$ as a subset of $M$, and $M$ as a differentiable manifold? The main reason I'm asking is that I frequently need to switch between metrics (and thus isometric embeddings) and don't want to worry about approximate tangent spaces changing when I do that.
We say that $x\in A$ is a density point if
$$
\lim_{r\to 0} \frac{\mathcal{H}^d(A\cap B(x,r))}{\omega_d r^d}=1.
$$
It is well known that $\mathcal{H}^d$-almost all points of $A\subset M$ are density points.
At the density points we simply have that $T_xA=T_xM$. The density point is independent of a choice of a metric or an isometric embedding.
For an overview of basic results in geometric measure theory I would recommend
F. Morgan, Geometric Measure Theory: A Beginner's Guide.
Thanks for you answer, Piotr! I think I should have chosen less confusing variable names for the dimensions because judging from your statement $T_x A = T_x M$ I think you assumed $\mathop{dim} M = d$. In any case, this shouldn't matter too much. If $A$ is $d$-rectifiable, it it is contained in countably many $d$-dimensional manifolds $N_i$ up to a set of $^d$-measure zero, and then for $^d$-a.e. $x \in A$ the approximate tangent space $T_x A$ agrees with the tangent space of one of the $N_i$ and this is obviously an invariant notion and independent of any embedding. Thanks again!
|
2025-03-21T14:48:29.872396
| 2020-02-16T00:14:10 |
352826
|
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|
Stack Exchange
|
Positively curved metric with uniformly positive scalar curvature
Can we find a complete noncompact Riemannian manifold $(M^n,g)$ with bounded geometry satisfying the following conditions?
the curvature operator $Rm>0$;
the scalar curvature $R \ge 1$.
Notice that any such manifold must be diffeomorphic to $\mathbb R^n$.
Yes, this is possible. Note that a strictly convex hypersuface in $\mathbb R^{n+1}$ has positive $Rm$.
To get an example, consider the following graph $H\subset \mathbb R^{n+1}$ over the open unit $n$-disk in $\mathbb R^n$:
$$H:=\left(x_{n+1}=\frac{1}{1-\sum_{i=1}^n{x_i^2}}\right).$$
Clearly, $H$ is convex. Moreover, the scalar curvature of $H$ tends to the scalar curvature of the unit $n-1$-sphere as $x_{n+1}$ tends to infinity. In particular the scalar curvature is greater than a certain positive $c>0$ on $H$. So if one scales down this hypersurface by a constant (i.e takes the hypersurface $\varepsilon\cdot H\subset \mathbb R^{n+1}$), we get $R(\varepsilon\cdot H)>1$. The example works if $n\ge 3$.
|
2025-03-21T14:48:29.872499
| 2020-02-16T00:28:51 |
352828
|
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|
Stack Exchange
|
A question related to supercuspidal representations of $\operatorname{GL}_2$ over local fields
I was learning about the representation of $\operatorname{GL}_2$ over local fields and came to know something like: if the residual characteristic of the local field is an odd prime, then every supercuspidal representation is a dihedral supercuspidal representation. I know what is a dihedral supercuspidal representation (I read it in Bump's Automorphic forms book).
But I could not find the previous statement when residual characteristic is odd. Please suggest some good references for this.
From the first two lines it is pretty clear that if the residual characteristic of the local field is even, then there are non-dihedral supercuspidal representations. I also want to know about them. Please suggest some references for this too.
What is a dihedral supercuspidal representation? Where are you learning about supercuspidals of $\operatorname{GL}_2$? (The second question just to know what might be useful further references.)
Let $E/F$ be a quadratic extension of non-Archimedean locan fields and let $\zeta$ be a quasicharacter of $E^$ that does not factor through the norm map $N: E^ \to F^*$. Using this yoy will be able to construct irrducible admissible representation of $GL(2,F) $_+ and induce that to $GL(2,F)$. The induced representation is irreducible and supercuspidal which is called digedral supercuspidal representation. For more details, you can look page no 541 from Bump's automorphic form and representation book.
When the residue characteristic is odd, all supercuspidal representations of GL(2,F) arise from admissible pairs, and via the LLC they correspond two dimensional irreducible continuous representations of Weil group of F.
So I think the word dihedral here should mean that the image of the corresponding Galois representation (in PGL(2, C)) is dihedral.
For details you may check Bushnell—Henniart’ book on GL_2.
Hi, are you talking about the book The Local Langlands Conjecture for GL(2)??
@Kiddo yes, that is the book mentioned in my comment above.
I take your word that 'dihedral' is Bump's terminology (I don't have the book to hand), but I think that the modern terminology for such supercuspidals is 'tame'. I don't know the first place where it is proven that all supercuspidals of $\operatorname{GL}_2$ are tame in odd residual characteristic; certainly it's proven in Bushnell and Henniart - The local Langlands conjecture for $\operatorname{GL}(2)$ (MSN), as @Peng mentions, but that's not the earliest place—although it is probably the best modern reference if this exact fact is what you want to learn. The result you want in the odd-residual-characteristic case is Theorem 20.2; and the non-dihedral (I think usually called exceptional) case is discussed in §45 ff.
The idea of constructing supercuspidals by induction goes back to Mautner in the '60s, although I don't know a reference (Mautner - Spherical functions over $\mathfrak p$-adic fields (MSN) is related, but seems to be about the spherical case); and I think an exhaustion result was first proven in Gel'fand and Graev - Irreducible unitary representations of the group of second-order unimodular matrices with elements in a locally compact field (MSN) and by Shalika - Representations of the two by two unimodular group over local fields (MSN), but for $\operatorname{SL}_2$ (which is harder). (I for one am grateful that authors are no longer afraid of using math mode in titles.) I believe that the current state of the art in such results is Kim - Supercuspidal representations: An exhaustion theorem (MSN), where one should also mention the work of Fintzen, specifically Fintzen - Types for tame $p$-adic groups, which improves on Kim's hypotheses. I think one of the first references for what you call the non-dihedral case is Kutzko - The exceptional representations of $\operatorname{Gl}_2$ (MSN).
|
2025-03-21T14:48:29.872743
| 2020-02-16T00:48:46 |
352829
|
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|
Stack Exchange
|
A fiber bundle of the Euclidean space over an orbifold
Consider a fiber bundle $p: F\hookrightarrow
E \to B$, where $E$ and $F$ are smooth manifolds and $B$ is a smooth orbifold. More precisely, each point $b \in B$ has an orbifold chart $U=\tilde U/\Gamma$ and a finite group $\Lambda$ acting freely on $\tilde U \times F$ such that $p^{-1}(U)=(\tilde U \times F)/\Lambda$ and the projection map $\pi_1:\tilde U \times F \to \tilde U$ is $(\Lambda, \Gamma)$-equivariant. Hence the following diagram is commutative:
$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$
$$
\begin{array}{c}
& \tilde U \times F & \ra{} & p^{-1}(U) & \\
& \da{\pi_1} & & \da{p} & \\
& \tilde U & \ra{} & U &
\end{array}
$$
If $E$ is diffeomorphic to $\mathbb R^n$ and $F$ is compact, can we prove that the fiber bundle must be trivial? That is, $F$ is a single point.
|
2025-03-21T14:48:29.872833
| 2020-02-16T00:59:03 |
352830
|
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|
Stack Exchange
|
Characterizing the left / right classes of (weak) factorization systems in locally presentable categories
Let $\mathcal M \subseteq Mor(\mathcal C)$ be a class of morphisms in a locally presentable category.
It's well-known that $\mathcal M$ is the left half of an accessible orthogonal factorization system iff $\mathcal M$ is accessible (as a full subcategory of the morphism category $\mathcal C^{[1]}$), and is closed under colimits in $\mathcal C^{[1]}$, cobase-change, composition, and isomorphisms.
Analogously (though perhaps this is less well-known), $\mathcal M$ is the right class of an accessible orthogonal factorization system iff it is accessible and accessibly-embedded in $\mathcal C^{[1]}$, and closed under limits in $\mathcal C^{[1]}$, base-change, composition, and isomorphisms. The proof of course is not dual -- one observes that under these conditions, $\mathcal M$ is accessibly-reflective in $\mathcal C^{[1]}$, shows that one leg of each unit map must be an isomorphism, so that the reflector provides factorizations, and then verifies a few things.
In the case of weak factorization systems (wfs), the situation can't be quite so simple. For one thing, not all accessible wfs on a locally presentable category are cofibrantly-generated, so any "small generation" argument is going to be more delicate.
More to the point, the left class of a wfs can be accessible without the wfs being accessible, and conversely (at least under anti-large-cardinal hypotheses) the left class of an accessible wfs need not be accessible. So even though one might guess that closure under coproducts, cobase-change, isomorphisms, composition, transfinite composition, and retracts should nearly characterize left classes of accessible wfs on locally presentable categories, it's not clear what kind of "accessibility" hypothesis to add to get a characterization.
Nevertheless, there might be more hope for characterizing the right classes of wfs on locally presentable categories. In particular, the following guess seems reasonable:
Question:
Let $\mathcal M \subseteq Mor(\mathcal C)$ be a class of morphisms in a locally presentable category. Suppose that $\mathcal M$ is accessible and accessibly embedded in $\mathcal C^{[1]}$, and closed under products, base change, isomorphisms, composition, co-transfinite composition, and retracts. Does it follow that $\mathcal M$ is the right class of an accessible weak factorization system on $\mathcal C$? If not, is there a characterization along similar lines?
Presumably the proof of any characterization will proceed by using some form of Garner's small object argument, but beyond that it's unclear to me.
Actually, there's a stronger condition than closure under dual transfinite composition which might be needed: say that a morphism in $\mathcal M \subseteq \mathcal C^{[1]}$ (corresponding to a commutative square in $\mathcal C$ with two opposite legs lying in $\mathcal M$) is $\mathcal M$-cartesian if the comparison map into the pullback lies in $\mathcal M$. If $\mathcal M$ is the right class of a wfs, I believe it's the case that the wide subcategory of $\mathcal M$ whose morphisms are the $\mathcal M$-cartesian squares is closed in $\mathcal C^{[1]}$ under cofiltered limits. We might have to include this condition in our putative characterization.
I'm very interested by this question. I might be something though: is it clear that right class of accessible weak factorization are accessible and accessibly embedded ? They are full images of accessible functors, but as far as I can tell this is their only accessibility property that I know of...
But maybe full accessible image of accessible categories might a good accessibility condition...
Some care is needed, but I feel like full accessible image might really be an interesting condition for left classes. If you have a left class that is a full image of a $\lambda$-accessible functor $F: V \rightarrow C^\rightarrow$ it sounds tempting to consider the algebraic weak factorization system generated by $V_\lambda$... I would have to think more about it, but it sounds like a promising strategy... Unless you already know this does not work ?
@SimonHenry I'm not sure. Maybe you're right!
Both the left and the right class of an accessible wfs are full images of an accessible functor. Assuming the existence of a proper class of almost strongly compact cardinals, they are preaccessible. Consider a wfs $(Emb,Top)$ in $\bf{Pos}$. Here $Emb$ is accessible and, if this wfs is accessible, complete lattices (= $Emb$-injectives) are a full image of an accessible functor. Assuming the existence of a proper class of almost strongly compact cardinals, complete lattice are accessibly embedded in posets, which is not true. Thus, under this assumption, $(Emb,Pos)$ is not accessible.
|
2025-03-21T14:48:29.873211
| 2020-02-16T02:37:30 |
352831
|
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|
Stack Exchange
|
When is a holomorphic map from a complex surface to a Riemann surface is a holomorphic family of Riemann surfaces?
A holomorphic family of Riemann surfaces of type $(g, n)$ is a triple $(M, \pi,B)$ defined as follows:
$\bullet$ M is a $2$-dimensional complex manifold (topologically, a $4$-manifold);
$\bullet$ $B$ is a Riemann surface;
$\bullet$ $\pi: M \to B$ is a holomorphic map;
$\bullet$ for all $t\in B$, the fiber $S_t =\pi^{-1}(t)$ is a Riemann surface of genus $g$ with $n$
punctures;
$\bullet$ the complex structure on $S_t$ depends holomorphically on the parameter $t$.
I wonder when is a holomorphic map $\pi:M\to B$ from a complex surface to a Riemann surface is a holomorphic family of Riemann surfaces of type $(g, n)$? In particular, why is a Kodaira fibration a holomorphic family of Riemann surfaces? (It's not clear to me that $S_t$ is holomorphic on the parameter $t$.)
A Kodaira fibration is a compact complex surface X endowed with a holomorphic submersion onto a Riemann surface $\pi: X\to\Sigma$ which has connected fibers and is not isotrivial.
What is your precise definition of "S_t depends holomorphically on the parameter $t$"? For me, it means that the fibration define a holomorphic curve in the moduli space, and this is more or less immediate by (semi)-universality of the latter (well, up to some technical point necessary to deal with curves with automorphisms).
In fact, recall that the total space of a Kodaira fibration is always algebraic (because it is of general type).
@FrancescoPolizzi I think it means we have a holomorphic map from $B$ to the Teichmuller space of the fiber $F$. I can’t see why that is naturally holomorphic. And what’s the semi-universality?
Google "moduli space" and look for "fine moduli space". The moduli space of curves is not fine, but you can deal with this by suitable technical constructions (using rigidifications or passing to stacks). This essentially gives to you the map $B \to \mathcal{M}_g$, that turns out to be algebraic (since the family is so). Then you can lift it to the Teichmuller space as an holomorphic map, using the fact that $\mathcal{T}_g$ is the orbifold universal cover of $\mathcal{M}_g$.
There are some technical issues, but morally things go in this way.
@FrancescoPolizzi for the lifting to the Teichmuller space, do you use the lifting criterion from the algebraic topology? But lifting criterion says we have this lift iff the image of the $\pi_1(B)$ in $\pi_1(M_g)$ is a subgroup of $\pi_1(T_g)$, however $\pi_1(T_g)$ is trivial while $\pi_1(B)$ is not, how do you see the image of $\pi_1(B)$ in $\pi_1(M_g)$ is trivial then?
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2025-03-21T14:48:29.873388
| 2020-02-16T03:16:03 |
352834
|
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|
Stack Exchange
|
Simultaneous Hahn-Banach theorem
Let $C(\mathbb{T})$ be the Banach algebra of continuous functions on the unit circle. Let $n \in \mathbb{N}$ and let $P_n(\mathbb{T})$ be the subspace of trigonometric polynomials of degree at most $n$.
Let $N \in \mathbb{N}$ and let $\ell_1,\ldots,\ell_N$ be positive linear functionals on $P_n(\mathbb{T})$. Define \begin{equation} A_{j,k} = \sup_{0 < p \in P_n(\mathbb{T})} \frac{\ell_j(p)}{\ell_k(p)}\end{equation}
Do there exist positive linear extensions $\widetilde{\ell}_1,\ldots,\widetilde{\ell}_N$ of $\ell_1,\ldots,\ell_N$ to all of $C(\mathbb{T})$ such that $\widetilde{\ell}_j(f) \leq A_{j,k} \widetilde{\ell}_k(f)$ for all nonnegative $f \in C(\mathbb{T})$ and all $j,k \in \{1,\ldots,N\}$?
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2025-03-21T14:48:29.873458
| 2020-02-16T03:26:20 |
352836
|
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|
Stack Exchange
|
Area of a surface confined by a sphere II
[A followup on two related posts:
Area of a surface confined by a sphere
Area of a elliptic surface confined by a sphere
. Thanks to all the inputs so far.]
Let $S$ be a surface enclosed inside the unit sphere in $R^3$. If
every point of $S$ is elliptic and
there is a point $p$ inside the unit sphere so that every half-ray emanating from $p$ intersects $S$ at most once,
then must it be the case that $\operatorname{Area}(S)\le \operatorname{Area}(S^2)$?
I do not think it is a good idea to ask almost identical question that many times.
Are your assumptions equivalent with S being a subset of the boundary of a convex body? If so, then the inequality follows e. g. by monotonicity of mixed volumes.
I think the answer is negative. (See below.) Thanks for your suggestion, that is indeed what I think I need.
The two conditions in the question do not imply that the convex hull of S and p has S as part of its boundary. Moreover, I believe that the two conditions do not imply the desired conclusion. Here is a counterexample to the 1-D version of the question; it seems obvious how to extend it to the 2-D case.
The outer part of the red curve is almost as long as the circle, it makes a sharp turn to add a lot of lengths but without violating either condition.
In the 2-D case, make a surface with an outer part that is close to the confining sphere, then at a 'slit' the surface makes a sharp turn and adds a flat plate to its outer part.
p.s. I apologize for posting many similar questions.
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2025-03-21T14:48:29.873585
| 2020-02-16T08:44:22 |
352842
|
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|
Stack Exchange
|
Nascent formal group law
$\DeclareMathOperator\FGL{FGL}$The formal group law (cf. Wikipedia, Ex. 1.6 of nLab, Hazewinkel) derived from an analytic function or formal series $f(x) = x + a_2 x^2 + a_3 x^3 + ...$ and its formal compositional inverse, perhaps derived from Lagrange inversion, $f^{(-1)}(x)$ is defined by
$$\FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)]$$.
I'm interested in earlier investigations of this formula and the closely related formula
$$F(x,t)=\exp[t \cdot g(x)d/dx]x = f[f^{-1}(x)+t],$$
where $g(x) = [df^{(-1)}/dx]^{-1}.$
Both occasionally pop up in MO (see Q1, Q2, O3).
(Edit 9/19/22: In "Coefficient rings of formal group laws", Buchstaber and Ustinov give a nice presentation of some history of FGLs and their applications in algebraic geometry, algebraic topology, and mathematical physics.)
So far the earliest presentations I know of are
by Abel for the $\FGL(x,y)$ in 1826 (cf. "From Abel's heritage: Transcendental objects in algebraic geometry and their algebraization" by F. Catanese, p. 6, Thm. 2.1, and "The Work of Niels Henrik Abel" by Houzel, p. 24, Eqn. 5.)
by Abel for $F(x,t)$ in 1826 (cf. Abel equation, 1826).
by Charles Graves for $F(x,t)$ in 1853 in "A generalization of the symbolic statement of Taylor's theorem" in the Proceedings of the Royal Irish Academy, Vol. 5, (1850-1853), p. 285-287 (cf. p. 13 in The Theory of Linear Operators ... , Principia Press, 1936, by Harold T. Davis and MO-Q4).
Aware of any earlier presentations than Abel's or Graves?
Hazewinkel link is to "On formal groups. The functional equation lemma and some of its applications."
Inselberg in "Superpositions of nonlinear operators. I. Strong superpositions and linearizability" on p. 502 states, "Abel in 1826 was the first to show that if $F(x, y)$ is an abelian group operation on the reals and satisfies certain other conditions, then there exist a one-to-one function, $f$, of one variable such that
$F(x, y) = f^{ -1}( f (x) + f(y))$."
N. H. Abel, Recherche de fonctions de deux quantites variable independantes
x et y, telles que $f(x, y)$, qui ont la propriM que $f(z,f(x, y))$ est une fonction symttrique de $z, x$ et $y$, J. Reine Angew. Math. 1 (1826), 11-15 [Oeuvres completes, I, pp. 61-65, Christiania 1881].
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2025-03-21T14:48:29.873763
| 2020-02-16T08:47:53 |
352843
|
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|
Stack Exchange
|
Promonoidal categories as $S$-algebras
A promonoidal category is a pseudomonoid in the monoidal bicategory $\bf Prof$, where the monoidal structure is given (on objects) by the product of categories.
I would like to show that promonoidal categories are, as a consequence, algebras for a pseudomonad $S: {\bf Prof}\to{\bf Prof}$.
My ultimate goal is to understand what is a co/lax/pseudo/morphism of algebras in this case, i.e. a "promonoidal profunctor" $p : B°\times A\to Set$ that "preserves the promonoidal structure".
I expect these two conditions to be equivalent:
$p$ is a pseudo-algebra morphism
$p$ induces a strong monoidal functor $\hat p : [A°,Set]\to [B°,Set]$, where presheaf categories are endowed with the Day convolution product.
If the promonoidal structure on $A$ or $B$ is representable, this is in turn equivalent to
The mates $p^\lhd : B° \to [A,Set]$ or $p^\rhd : A \to [B°,Set]$ are strong monoidal, if the codomain has the convolution monoidal product.
Suitably laxified, 2. and 3. define co/lax algebra morphisms for the same $\bar S$.
Using a general theorem about lifting monad to Kleisli categories of other monads, I can show that there is a lifting $\bar S : {\bf Prof} \to {\bf Prof}$ of the monad $S$ on Cat whose algebras are monoidal categories; but a $\bar S$-algebra seems to be just a multicategory, and not of the particular kind that corresponds to promonoidal categories.
What am I missing?
I might have managed to get the extra structure from the definition of $\bar S$.
|
2025-03-21T14:48:29.873889
| 2020-02-16T09:27:06 |
352844
|
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|
Stack Exchange
|
Pulling back a functor, it becomes monadic
$\require{AMScd}$I am in the following situation: the diagram
$$
\begin{CD}
\cal M @>r>> [{\cal B},Set] \\
@VuVV @VVf^*V \\
\cal D @>>N_g> [{\cal A}^\text{op},Set]
\end{CD}
$$
is a (strict) pullback in $\bf Cat$; moreover, $f : \cal A^\text{op}\to B$ is bijective on objects (and $f^*$ is the "inverse image" functor given by precomposition). In turn, $g$ is the "nerve" induced by a functor $g : \cal A \to D$. $N_g$ is fully faithful because $g$ happens to be dense, and so $r$ is f.f. as well. If needed, $\cal D$ is complete and cocomplete.
This gives $\cal M$ a very explicit description: it is a reflective subcategory of $[{\cal B},Set]$ made by the functors $F : {\cal B}\to Set$ such that $F(fA)={\cal D}(gA,D)$ for some $D\in\cal D$.
Is all this sufficient to imply that $u$ is monadic? If not, what additional assumptions are needed?
If $A$ and $B$ are small and $D$ is locally presentable then $u$ is a monadic right adjoint. More generally, what is non-trivial is the construction of a left adjoint of $u$. As soon as $u$ has a left adjoint, $u$ is monadic.
The proof is relatively simple: the forgetful functor $f^*$ is monadic, so it satisfies all the conditions of Beck monadicity criterion, and all the conditions of the Beck criterion except the existence of an adjoint are automatically satisfied for its pullback ($f^*$ is an isofibration, so the square is both a strict and pseudo-pullback).
Now when all the category involved are locally presentable as both $f^*$ and $N_g$ are accessible right adjoint functors, the square is also a pullback in the category of locally presentable categories and accessible right adjoint functor (which is known to have all limits, and these limits are preserved by the forgetful functor to set by an old results, which appears in Bird's phd thesis, though I think it was known before)
For more details, you can also have a look to John Bourke and Richard Garner Monads and theories this is how they construct their functor from "pre-Theory to Monads" (which is left adjoint to the Kleisli category functor)
Perfect answer! Thanks. I hoped there was an explicit description for the left adjoint in terms of a universal property, but I can get away with the mere existence...
If you try to give an explicit construction of the left adjoint of $u$, using for example a coend formula similar to the one for the left adjoint of $f^*$, it does not quite work on the first try, the object you get is only partially equipped with the appropriate structure, so you need to "iterate" in an appropriate when, possibly a transfinite number of time... And you need some kind of assumption (local presentability being the nicest one) to ensure convergence.
My initial idea was to prove that $Ran_u1$ exists and it is preserved by $u$; too naive?
|
2025-03-21T14:48:29.874100
| 2020-02-16T09:34:27 |
352845
|
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|
Stack Exchange
|
Kurtz randomness and supermartingales with infinite *limit*
Suppose you replace the usual success conditions for a supermartingale (lim sup is infinite) with the requirement that the actual limit is infinite, e.g. a supermartingale $B$ succeeds on $X \in 2^\omega$ just if
$$\lim_{n \to \infty} B(X\restriction_n) = \infty$$
I'm 90% sure that you don't get a 'valid' randomness notion because any sufficiently generic real should qualify. However, if I had to guess I'd think this notion turns out to be equivalent to Kurtz 'randomness' (avoids all $\Pi^0_1$ null sets). Since I'm guess this is a known result and my brain is feeling super foggy I thought I would ask
Assuming you mean a c.e. (super)martingale, this is just Martin-Löf randomness. What needs to be shown is that, for every non-random real, there is a martingale which limit-succeeds on the real. You can see this from the proof of the equivalence of the martingale definition and the Kolmogorov complexity definition:
Let $U$ be a universal prefix-free machine. For $\sigma \in 2^{<\omega}$, define $M_\sigma$ to be the (computable) martingale that begins with capital 1 and bets it all on $\sigma$, and bets evenly afterwards. So $M(\tau) = 2^{|\sigma|}$ for any $\tau$ extending $\sigma$. Define $M = \sum_{U(\rho)=\sigma} 2^{-|\rho|}M_\sigma$. This is a c.e. martingale (with starting capital $\Omega$). For any non-random $X$, for any $d$, there is an $n$ with $K(X\upharpoonright n) < n - d$. So there is some $\rho$ with $|\rho| < n - d$ and $U(\rho) = X\upharpoonright n$. So $2^{-|\rho|} M_{X\upharpoonright n}$ is a summand in the definition of $M$, and $2^{-|\rho|}M_{X\upharpoonright n}(\tau) > 2^d$ for any $\tau$ extending $X\upharpoonright n$. So $\lim_{n \to \infty} M(X\upharpoonright n) = \infty$.
Note that this construction shows why your generic intuition is wrong: for every $d$, there is a neighborhood of $X$ on which the martingale never again dips below $2^d$. So generics can't get out of it.
i kept thinking this was the case but assumed I was missing something when Nies listed the equivalent conditions and they all mention infinitely often or order functions which confused me. But maybe this one was just less useful
|
2025-03-21T14:48:29.874257
| 2020-02-16T09:41:21 |
352846
|
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|
Stack Exchange
|
On Kellner's result and the Erdos-Moser equation
Let $m, k$ be positive integers. Consider the Erdos-Moser expression $S_{k}(m) = 1^k + 2^k + ... + (m-1)^k$. By a result of Kellner, we know that if $m | S_{k}(m)$, then $m|B_k$, where $B_r$ denotes the $r-$th Bernoulli number.
Question: Since $B_k$ is not an integer for any even $k$, does Kellner's result entail that $k$ must be odd if $m | S_{k}(m)$ ?
PS: If the answer to the above question is "yes", it would follow that if $S_{k}(m)=am^k$ for some integer $a$, then $k=1=a$ (the generalised Erdos-Moser conjecture).
Here is a reference to Kellner's result: https://www.google.com/url?sa=t&source=web&rct=j&url=https://pure.mpg.de/rest/items/item_3122423/component/file_3122424/content&ved=2ahUKEwigtZns2dXnAhUJ8hQKHXvFBsk4ChAWMAN6BAgCEAE&usg=AOvVaw2raZVEqq9tXOsT1cH4P5S8
|
2025-03-21T14:48:29.874336
| 2020-02-16T10:22:04 |
352851
|
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"shubhankar"
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|
Stack Exchange
|
Modern context for hypercohomology spectra
In Thomason's paper Algebraic K-theory and étale cohomology, (Ann. ENS 1985, Numdam link) Thomason develops an elaborate theory of hypercohomology spectra, $\mathbb{H}(X,\mathcal{F})$ for presheafs of spectra $\mathcal{F}$ on the étale site of the scheme $X$.
As per my current understanding these hypercohomology spectra come equipped with a 'descent' spectral sequence, which in the particular case of algebraic $K$-Theory, considered as a presheaf of spectra on the étale site of $X$, starts from the étale cohomology (with Tate Twisted $\mathbf{Z}_l$ coefficients) of $X$ and converges to the localization of $K(X)$ with respect to mod $l$ complex $K$-theory. (We assume $l$ is invertible in $X$.)
My question is if there exists a more modern reformulation of this theory.
I am quite certain that since the paper was written in the 80's most of the techniques must have become standard and the results should follow from the homotopy theory of spectra valued presheafs on a site (in a more modern language).
I would be particularly interested in a reformulation of this work in the language of $\infty$-categories.
The modern version, in its optimal form, is found in the work of Clausen and Mathew arxiv:1905.06611
@Denis-CharlesCisinski Thank you! This seems to be exactly what I wanted.
|
2025-03-21T14:48:29.874464
| 2020-02-16T10:33:29 |
352852
|
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"Denis Serre",
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"sharpe"
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|
Stack Exchange
|
Regularity of Neumann eigenfunctions at vertices of polygons
Given a bounded polygonal domain $D$ in $\mathbb{R}^2$, the Neumann eigenfunctions have continuous version on $\overline{D}$. The eigenfunctions also have critical points at vertices of $D$ (I have been taught the simple proof
by Prof. Mateusz Kwaśnicki at this page).
However, the modulus of continuity of the eigenfunctions at vertices seem not obvious. For example, when $D$ is an equilateral triangle and $\phi$ is a Neumann eigenfunction on $D$, can we describe the modulus of continuity of $\phi$ at a vertex $p$ ? It holds that $|(\nabla \phi)(p)|=0$. Can we expect that $|\phi(x)-\phi(p)|=o(|x-p|^\alpha)$ as $x \to p$ for some $\alpha>1$.
If so, what is the optimal exponent?
Grisvard's book is a standard reference for this.
@MichaelRenardy Thank you very much for your comment. Is the title of the book "Elliptic Problems in Nonsmooth Domains"? I'll check it.
To complete Michael's comment, the order of the singularity at a corner is explicit and depends only upon the angle of this corner. Its magnitude can be expressed as an integral, but this is not so much explicit.
@DenisSerre Thank you very much for your comment. For example, can you tell me the order for Neumann eigenfunctions on equilateral triangle?
I suspect that the order is greater than or equal to $2$.
Let me develop a little my comment. Consider a corner of the domain, with tip at the origin and one edge horizontal. Denote $\alpha$ its angle of aperture. The leading singularity of a solution of $\Delta u=f$ with a smooth $f$, and Neumann condition, is $A\Re(z^\kappa)$. With $z=e^{i\theta}$, its normal derivative is $r^{-1}\partial_\theta$. Whence the condition $\sin(\kappa\alpha)=0$, where we must take the smallest root $\kappa$, in absence of better information:
$$\kappa=\frac\pi\alpha.$$
For $\alpha=\frac\pi3$, this gives $\kappa=3$, which is not so bad as the eigenfunction is $C^{3-\epsilon}$. But the larger $\alpha$, the less regular are the eigenfunctions.
Edit. When $f=\lambda u$, which is in $H^1(\Omega)$, it has enough "smoothness" that the above applies.
Thank you for your kind reply. I have a question on it. In the proof, you assume that $f$ is smooth, right? In your definition, "smooth" is $C^1(\overline{D})$? I am also considering the regularity of eigenfunctions of the Neumann Laplacian and the Neumann semigroup. Hence, it is inconvenient to assume that f is smooth in advance.
|
2025-03-21T14:48:29.874665
| 2020-02-16T10:38:29 |
352853
|
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|
Stack Exchange
|
The "Chaos Game" as a particular series of i.i.d. random variables
Fix a parameter $\alpha\in(0,1)$ and take an i.i.d. sequence $X_0,X_1,\ldots$ of $\mathbb{R}^n$ valued random variables. Construct the limiting random variable
$X_\infty = (1-\alpha)\sum_{k=0}^\infty \alpha^k X_k.$
Is any general result known about this kind of limit? What if the $X_i$ follow a well known distribution like uniform/Rademacher?
I was motivated by this sum after running into: https://en.wikipedia.org/wiki/Chaos_game. For example if the $X_i$ are uniformly distributed on the 3 vertices of a triangle and $\alpha = 1/2$ the limiting distribution is supported on the associated Sierpinski Triangle. The fact that finite support distributions can give fractal shapes from this construction leads me to believe this is a non-trivial question.
My apologies if this ends up being an exercise in some well known textbook on probability theory. If it is, I'd appreciate a reference for that textbook.
Edit: I was able to locate http://u.math.biu.ac.il/~solomyb/RESEARCH/Bernotes.pdf
$\newcommand\al{\alpha}$Let us drop the factor $1-\al$, by considering
$$Y:=X_\infty/(1-\al)=\sum_{k=0}^\infty\al^k X_k.$$
By Kolmogorov's three-series theorem, this series will converge almost surely (a.s.) unless at least one of the tails of the distribution of $X_0$ is too heavy.
Assume that the series indeed converges a.s.
Then, obviously,
$$Y\overset D=X+\al Y,$$
where $\overset D=$ denotes the equality in distribution and $X$ is an independent copy of the $X_k$'s. So, we have the functional equation for $F_Y$:
$$F_Y(y)=\int_{-\infty}^\infty F_Y((y-x)/\al)\,dF_X(x)$$
for real $y$, where $F_Z$ denotes the cdf of $Z$. Equivalently, we have the functional equation for $f_Y$:
$$f_Y(t)=f_Y(\al t)\,f_X(t)$$
for real $t$, where $f_Z$ denotes the characteristic function of $Z$. Of course, we can also write
$$f_Y(t)=\prod_{k=0}^\infty f_X(\alpha^k t)$$
for real $t$.
In the particular case when $X$ is Rademacher, the distribution of $Y$ is the well-studied Bernoulli convolution.
In the particular case when $X$ is $U(0,1)$ and $\alpha=1/2$, $F_Y$ is the well-studied Fabius function.
Before asking my question I'd not gone much further beyond getting that characteristic function expression. I had not known of the Bernoulli convolution, looks like this has been considered before.
For now I'll content myself to generating Python fractals with this. Consider the question answered. Thank you.
I was able to locate http://u.math.biu.ac.il/~solomyb/RESEARCH/Bernotes.pdf if anyone else is new to the subject and wants further reading.
|
2025-03-21T14:48:29.874834
| 2020-02-16T10:48:35 |
352854
|
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"Peter Gerdes",
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|
Stack Exchange
|
Non-relativized, Computable and Schnor randomness w.r.t a measure
Riemann and Slaman have some great work classifying what reals are 1-random with respect to a measure $\mu$ relative to $\mu$. In that paper they cite Levin and Kautz (but not to refs I can find) for the claim that if $X$ is 1-random with respect to a computable measure $\mu$ then $X$ is of 1-random degree. They explain that the argument works by showing that $\mu$ is the image of $\lambda$ (Lebesgue measure) under a computable bijection $f$ and noting that $f^{-1}(X)$ will be 1-random if $X$ is 1-random with respect to $\mu$.
I have two questions related to this and since I've pretty much avoided randomness since grad school I figured they might have obvious answers to those working in the area.
Does this claim generalize to computable (computable martingales) or Schnor (tests with uniformly computable measures) even in the presence of atoms? I'd be happy just with a reference to the proof so I can check without having to recreate it. If I was only considering continuous measures I'd probably have taken this for granted but without reading/reconstructing the proof it's not obvious to me this can be done in the presence of atoms.
If we drop the restriction to computable measures but don't relativize the notion of randomness to a representation of $\mu$ (i.e. we demand our $\mu$ ML tests be made of actual c.e. sets not sets c.e. relative to $\mu$) is there a degree $\mathbf{d}$ such that every real $X$ is 1-random with respect to a $\mathbf{d}$ computable measure $\mu$? The result in Riemann and Slaman tells us that we can take $\mu$ to be computable in $X''$ but they are demanding that $X$ pass all $\mu$-c.e. ML tests. Can we make $\mathbf{d} \leq_T 0'$?
That's the main thrust of my question and if this is published somewhere I'd love to know. It might also be nifty to know what's the best we can do (any high degree? a low degree?). However, I have some vague memory that if you don't relativize you get a pathological notion so there might be a really trivial answer here.
P.S. I tagged this with martingales as that is how one defines computable randomness but I'm not sure if the tag should be used (even if technically accurate) to the notion as used in algorithmic randomness and couldn't figure out where to ask/discuss.
It's theorem 6.12.9 in Downey & Hirschfeldt. The answer should be yes, with the caveat that it won't hold if $X$ is an atom of $\mu$.
Levin proved the existence of a $\mu$ such that every real passes all $\mu$-c.e. ML tests, which is stronger than what you're requiring. For what you're asking, I believe we can use c.e. permitting to construct such a $\mu$ below every non-computable c.e. degree. The goal is to ensure that there are no c.e. $\mu$-ML tests. Choose some $2^{-d}$ of measure to attack the $e$th test. Assign it somewhere with some large use, and wait for the $(d+1)$-level of the $e$th test to enumerate some neighborhood. When it does, wait for permission to shift that $2^{-d}$ of measure to this neighborhood (thus proving that this is not an ML-test). Meanwhile, begin again with a larger $d$ in the usual c.e. permitting fashion. In fact, I believe this could be made to work with hyperimmune permitting.
Thanks so much! I appreciate it. As I said it's been awhile since I touched randomness.
Ohh and I see your answer to 2 explains my vague memory that something goes very very wrong.
|
2025-03-21T14:48:29.875187
| 2020-02-16T11:02:42 |
352855
|
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"authors": [
"Ali",
"Chee",
"https://mathoverflow.net/users/14390",
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|
Stack Exchange
|
Conjugate points and Jacobi matrices
Let $(M,g)$ be a smooth compact Riemannian manifold of dimension $n \geq 3$ and let $\gamma:[-2,2]\to M$ be a geodesic that does not contain any conjugate points on $[-2,2]$.
I have two questions, as follows.
(i) Is it possible to construct transversal (to $\gamma$) Jacobi fields $J_1, \ldots, J_{n-1}$ such that the determinant of the $(n-1)\times (n-1)$ matrix $X(t)$ with columns $J_1(t),\ldots,J_{n-1}(t)$ on the smaller interval $[-1,1]$ only vanishes at the point $t=0$?
(ii) Is it possible to construct transversal (to $\gamma$) Jacobi fields $J_1, \ldots, J_{n-1}$ such that the determinant of the $(n-1)\times (n-1)$ matrix $X(t)$ with columns $J_1(t),\ldots,J_{n-1}(t)$ on the smaller interval $[-1,1]$ only vanishes at the point $t=0$, and additionally rank of $X(0)$ is $n-3$?
hi Ali, just a comment: (a) look at the 2nd part of the proof of Proportion 2.5.8 and the 2nd paragraph of the proof of Theorem 2.1.12 in 2nd edition of Klingenberg's book "Riemannian geometry". These give sufficient conditions (for conjugate points) on the determinant of the matrix formed by Jacobi fields normal to a geodesic; (b) look at the proof of Lemma 5.4 of Sakai's "Riemannian geometry". This shows how to use the tangent map of the exponential map to construct such Jacobi fields (a classic technique) and how the determinant comes into play; (c) modify the skills in (b) if needed
Many thanks Chee.
|
2025-03-21T14:48:29.875299
| 2020-02-16T11:52:07 |
352856
|
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|
Stack Exchange
|
Set of points such that the distance is achieved by at least two geodesics is closed
If $(M,g)$ is a complete Riemannian manifold and let $p\in M$. Define a function $$d^2:M\to \mathbb{R},~q\mapsto \mathrm{dist}^2(p,q).$$ Consider the set $$A=\text{Set of all points $q\in M$ such that the distance is achieved by at least two geodesics. }$$ I need to prove that the set $A$ is closed.
I tried to go by the sequence approach (as $M$ has a metric). Let $\{q_n\}$ be the points in $A$ and $q_n\to q$. I need to prove that $q\in A$. Now let $\gamma_n$ and $\eta_n$ be geodesics such that $$d(p,q_n)=l(\gamma_n)=l(\eta_n).$$ Now is it possible (or a similar idea) to prove that limit of the two geodesics converge to the geodesic and as the length is a continuous function $$l(\gamma_n)\to l(\gamma),~~\text{and}~~ l(\eta_n)\to l(\eta),$$ which will prove that $q\in A$ and we are done.
Please tell me if this approach is valid or if any other approach I have to go through.
Where were you told that this subset is closed?
I guessed that the set where the geodesics are unique is open but that was somewhat hard to prove that is why tried to go through the closedness.
This subset is not closed in general. See for example Figure 13 here: https://en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid. The set of limit points of your set may consist of points, conjugate to $p$, where there is only one minimizing geodesics. Your argument is incorrect as the two limit geodesics may be the same.
@Raziel, Sorry, I didn't get you. Isn't the set $A$ is the cut-locus, which is closed?
No, the cut locus in Riemannian geometry can be defined as the closure of your set. But $A$ is not closed in general. Again, see the example of the ellipsoid.
@Raziel, Then https://math.stackexchange.com/questions/3547834/the-set-a-is-open is also false, as it is the complement of the set $A$.
That's correct.
|
2025-03-21T14:48:29.875455
| 2020-02-16T12:57:30 |
352859
|
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|
Stack Exchange
|
Finding a real-valued function not differentiable in any $B_{r}(\mathbf{x})$
Is there a function $f:O\to \mathbb{R}$, $O$ is an open subset in $\mathbb{R}^2$, which satisfies both $(1)$ and $(2)$ ?
$(1)$ All of its second partial derivatives are defined on $O$ and continuous at $(x_0,y_0)\in O$ ;
$(2)$ It is not differentiable in any neighborhood of $(x_0,y_0).$
Obviously, $(1)$ involves the differentiability of $f$ at $(x_0,y_0)$.
The original post come from here. In order to express my purpose clearly, I simplify $n$-variables to two-variables.
The answer is no. Indeed, if your condition (1) holds, then
all of the second-order partial derivatives of $f$ are bounded on an open neighborhood $U$ of $(x_0,y_0)$. So, by Lemma 1 below, both of the first-order partial derivatives of $f$ are Lipschitz and hence continuous on $U$, and therefore your condition (2) cannot hold.
Lemma 1. If $|g''_{xx}|+|g''_{yy}|+|g''_{xy}|+|g''_{yx}|\le M$ on $U$, then
$|f'_x(x,y)-f'_x(a,b)|\le M|x-a|+M|y-b|$ and $|f'_y(x,y)-f'_y(a,b)|\le M|x-a|+M|y-b|$ for all $(x,y)$ and $(a,b)$ in $U$.
Proof. For any $(x,y)$ and $(a,b)$ in $U$,
$$|f'_x(x,y)-f'_x(a,b)|\le|f'_x(x,y)-f'_x(a,y)|+|f'_x(a,y)-f'_x(a,b)| \\
\le M|x-a|+M|y-b|.$$
Similarly,
$$|f'_y(x,y)-f'_y(a,b)|\le M|x-a|+M|y-b|.\quad\Box$$
|
2025-03-21T14:48:29.875556
| 2020-02-16T13:58:04 |
352863
|
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"Pietro Majer",
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|
Stack Exchange
|
Proof of the du Bois-Reymond lemma "by approximation"
I'm curious about the following argument in Morrey ("Multiple integrals in the calculus of variations", Lemma 2.3.1). Suppose $f\in L^1[0,1]$ and $$\int_0^1 fg\,dx=0$$ for every test function $g\in C^\infty_c[0,1]$. Then, "by approximations," the same is true if $g$ is merely bounded and measurable, in which case we may take $g=\mathrm{sign}(f)f,$ which implies $f=0$ almost everywhere.
How exactly is one supposed to approximate? Certainly not in the $L^\infty$ norm because the closure of $C^\infty_c[0,1]$ in this norm is not $L^\infty$ itself. Furthermore, $\mathrm{sign}(f)f$ need not be in $L^\infty$, so how does this argument work?
How exactly is one supposed to approximate?: Indeed, not in the $L^\infty$
norm, but in the weak* topology of $L^\infty$.
One way to run this argument is to note that you can approximate a bounded measurable function by $C^\infty_c$ functions, almost everywhere and boundedly. That is, given a bounded measurable $h$, you can find a uniformly bounded sequence $g_n \in C^\infty_c([0,1])$ with $g_n \to h$ almost everywhere.
There are many ways to see this. For example, look up your favorite proof that $C^\infty_c([0,1])$ is dense in $L^1([0,1])$. So you can find a sequence $g_n \to h$ in $L^1$. Passing to a subsequence, you can get it to converge almost everywhere. Truncating smoothly at a level higher than $\sup |h|$, you can get the $g_n$ to be uniformly bounded.
So then by the dominated convergence theorem, under the given assumptions, you can conclude that $\int f h = 0$ for all bounded measurable $h$. Now take $h = \operatorname{sign} f$, so that $fh=|f|$.
I suspect the reference to $\operatorname{sign}(f) f$ is a typo.
I will prove a slightly more general results.
Theorem. If $f\in L^1_{\rm loc}(\Omega)$, $\Omega\subset\mathbb{R}^n$, and
$$
\int_\Omega f\phi=0
\quad
\text{for all $\phi\in C_c^\infty(\Omega)$},
$$
then $f=0$ a.e.
Proof.
Suppose that $f\neq 0$. We can assume $f$ is positive on a set of
positive measure (otherwise we replace $f$ by $−f$). Then there is a compact
set $K ⊂ \Omega$, $|K| > 0$ and $ε > 0$ such that $f ≥ ε$ on $K$.
Let $G_i$ be a sequence of open sets such that $K ⊂ G_i ⊂⊂ Ω$, $|G_i \setminus K| → 0$
as $i → ∞$. Take $ϕ_i ∈ C^∞_0(G_i)$ with $0 ≤ ϕ_i ≤ 1$, $ϕ_i|_K ≡ 1$. Then
$$
0 = \int_Ω
fϕ_i ≥ ε|K| − \int_{G_i\setminus K} |f| → ε|K| ,
$$
as $i → ∞$, which is a contradiction. The proof is complete.
|
2025-03-21T14:48:29.875744
| 2020-02-16T14:24:15 |
352865
|
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|
Stack Exchange
|
Sine-Gordon transformation and functional integrals
In the past months, I've been trying to understand the so-called Sine-Gordon transformation, so I've posted some questions here about this topic. I also did an extensive research about this subject, so I came to some conclusions. Still, I have some questions that I'd like to share with you. First, I'll draw the general picture and some of my conclusions.
We consider a function $V: \mathbb{R}^{n}\times \mathbb{R}^{n}$ which is continuously differentiable, satisfies $\sup_{x,y \in \mathbb{R}^{n}}|V(x,y)| \le K$ and
$$ \langle f,Vg \rangle := \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}f(x)V(x,y)g(y)ddy \ge 0 \tag{1} $$
for every $f,g \in L^{2}(\mathbb{R}^{n})$. If we define $B: \mathcal{S}(\mathbb{R}^{n})\times \mathcal{S}(\mathbb{R}^{n})$ to be $B(f,g) \equiv \langle f, Vg\rangle$, the associate quadratic form $f \mapsto B(f,f)$ is non-negative, so that, by Minlos' Theorem, there exists some (Gaussian) measure $\mu_{V}$ on $\mathcal{S}'(\mathbb{R}^{n})$ such that
$$ W(f) := e^{-\frac{1}{2}B(f,f)} = \int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)e^{iT(f)}$$
Because $\mathcal{S}(\mathbb{R}^{n})\subset \mathcal{S}'(\mathbb{R}^{n})$, each $f \in \mathcal{S}(\mathbb{R}^{n})$ induces a distribution in $\mathcal{S}'(\mathbb{R}^{n})$. Thus, if we fix $\epsilon_{1},...,\epsilon_{N}\in \mathbb{R}$ and $x_{1},...,x_{N}\in \mathbb{R}^{n}$, we can choose sequences $\{f_{l}^{(j)}\}_{l\in \mathbb{N}}$ such that $f_{l}^{(j)} \to \epsilon_{j}\delta_{x_{j}}$, for each $j=1,...,N$. We can prove that:
$$\lim_{l\to \infty}\int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)\prod_{j=1}^{N}:e^{iT(f_{l}^{(j)})}:_{V} = e^{-\sum_{1\le i< j \le N}\epsilon_{i}\epsilon_{j}V(x_{i},x_{j})}$$
where $:e^{iT(f)}:_{V} := e^{iT(f)}e^{\frac{1}{2}B(f,f)}$. Let's introduce the notation:
\begin{eqnarray}
lim_{l\to \infty}\int_{\mathcal{S}'(\mathbb{R}^{n})}d\mu_{V}(T)\prod_{j=1}^{N}:e^{iT(f_{l}^{(j)})}:_{V} \equiv \bigg{\langle}\prod_{j=1}^{N}:e^{i\epsilon_{j}T(x_{j})}:_V\bigg{\rangle}_{V} \tag{2} \label{2}
\end{eqnarray}
The right hand side of (\ref{2}) does not make sense, once $T$ cannot be evaluated pointwise. However, this is just a notation. Now, the partition function of a system in the grand-canonical ensemble is given by:
\begin{eqnarray}
\Xi_{\Lambda}(\beta, z) = 1+\sum_{N=1}^{\infty}\frac{z^{N}}{N!2^{N}}\sum_{\substack{\epsilon_{j}=\pm 1 \\ j=1,...,N}}\int_{\Lambda^{N}}dx_{1}\cdots dx_{N}e^{-\beta \sum_{1\le i<j\le N}\epsilon_{i}\epsilon_{j}V(x_{i},x_{j})} \tag{3}\label{3}
\end{eqnarray}
Thus, we can rewrite (\ref{3}) using the notation in (\ref{2}):
\begin{eqnarray}
\Xi_{\Lambda}(\beta,z) = 1+\sum_{N=1}^{\infty}\frac{z^{N}}{N!2^{N}}\sum_{\substack{\epsilon_{j}=\pm 1 \\ j=1,...,N}}\int_{\Lambda^{N}}dx_{1}\cdots dx_{N}\bigg{\langle}\prod_{j=1}^{N}:e^{i\sqrt{\beta}\epsilon_{j}T(x_{j})}:_V\bigg{\rangle}_{V} \tag{4}\label{4}
\end{eqnarray}
This motivates another notation simplification. We interpret the integrand in (\ref{4}) as a $N$ iterated integrals, so that we write:
\begin{eqnarray}
\frac{1}{2^{N}}\sum_{\substack{\epsilon_{j}=\pm 1 \\ j=1,...,N}}\int_{\Lambda^{N}}dx_{1}\cdots dx_{N}\bigg{\langle}\prod_{j=1}^{N}:e^{i\sqrt{\beta}\epsilon_{j}T(x_{j})}:_V\bigg{\rangle}_{V} \equiv \bigg{(}\frac{1}{2}\bigg{\langle}\sum_{\epsilon = \pm 1}\int_{\Lambda} dx :e^{i\sqrt{\beta}\epsilon T(x)}:_{V}\bigg{\rangle}_{V}\bigg{)}^{N} \equiv \bigg{(}\bigg{\langle}\int_{\Lambda}:\cos \sqrt{\beta}T(x):_{V}dx\bigg{\rangle}_{V}\bigg{)}^{N} :=\langle C_{\Lambda,\beta}\rangle_{V}^{N} \tag{5}\label{5}
\end{eqnarray}
Finally, we have:
\begin{eqnarray}
\Xi_{\Lambda}(\beta,z) = \sum_{N=0}^{\infty}\frac{z^{N}}{N!}\langle C_{\Lambda,\beta}\rangle_{V}^{N} \equiv \langle\exp(z C_{\Lambda, \beta})\rangle_{V} \tag{6}\label{6}
\end{eqnarray}
Relation (\ref{6}) is called the Sine-Gordon transformation. Now, this being said, I'd like to raise some questions.
(1) If my reasoning is correct, the Sine-Gordon representation is formal, in the sense that it does not represent a proper Gaussian integral; instead, it is just a matter of notation. If this is the case, I'm ok with that, but I don't get the point here. If (\ref{6}) is just a matter of notation, why is it useful? If I draw any conclusion having (\ref{6}) as a starting point, why should it actually hold if all this is formal written? I know Gaussian integrals are useful tools but this is not a proper Gaussian integral, right?
(2) Is it possible to give precise meaning to (\ref{6})? Is there any construction where $\Xi_{\Lambda}(\beta,z)$ is an actual Gaussian measure and, in case there is, how to proceed? (It's not unusual to come across some paper where the Sine-Gordon transformation is treated as a real mathematically meaningful representation, so I wonder if this I'm just misreading it or if there is actually have a meaningful version of it).
(3) In practice, notation (\ref{2}) is useful because it allows us to perform some formal operations such as interchanging integrals and products as in (\ref{5}) and (\ref{6}). Can some of these operations be properly justified? In other words, to what extent does (\ref{6}) holds only as a formal series?
Remark: For completeness, I've based this post mainly on Fröhlich's work and Fröhlich and Park's work. Other good references are Brydges and Federbush's work and Dimock's work.
is this question really different from https://mathoverflow.net/q/350040/11260 ?
Yes. Let me elaborate. In the first question I tried to understand the mathematical meaning of (2). This is because I thought there might be such meaning. At the moment, I think the Sine-Gordon transform is meant to be formal, as many papers state. But this raises other questions such as (1) what's the point of such a formal definiton, (2) is this the only construction?
Of Course, there are some intersections between these two questions. The answer I got in the prévios question was great, but the authors seem to understand these topics in a different way (none of them, for instance, make explicit use of formal series) so I'm trying to understand what am I missing and how should I read all these properly.
A possible approach to make sense of the formula which is not just formal: take your random $T$, convolve it with a rescaled mollifier and see if the result converges as a random variable on $\Omega=S'$ with values in some Besov or other function space made of true functions inside $S'$. If $V$ is sufficiently regular, you may be able to show in this way the measure $d\mu_V$ is supported on a space of function instead of all of $S'$. See https://arxiv.org/abs/1502.07335 and https://arxiv.org/abs/1506.05740
Thanks for your comments and references. Let me ask you something: in general, the sine-gordon is used as I stated in only formal way? I hardly never see different approaches.
|
2025-03-21T14:48:29.876125
| 2020-02-20T01:44:39 |
353124
|
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|
Stack Exchange
|
Is the Wasserstein kernel positive definite?
Define a point cloud $X=\{x_i\}_{1\leq i\leq n}$, for $x_i\in\mathbb R^d$. Define the Wasserstein kernel as $$W(X,Y)=\max_{T}\frac{1}{n}\sum_{kl}T_{kl}\langle x_k,y_l \rangle$$
where $T$ is any doubly stochastic matrix.
Consider point clouds $X_1,...X_m$ each of size $n$ and their Gram matrix $(W(X_i,X_j))_{ij}$. Is it positive semi-definite?
What we know: There always exists a permutation matrix $T$ maximizing the above.
It is not positive semi-definite.
Take $m=4, n=2, d=2$.
I define $u_i = (\lfloor i / 2 \rfloor, i \% 2)$ for $i=0\dots3$.
I take $X_1 = \{ u_0, u_1\}, X_2 = \{u_0, u_2\}, X_3 = \{u_0, u_3\}, X_4 = \{u_1, u_2\}$
$W(X_i, X_j) = 0$ means that all vectors in the two sets are orthogonal. It can only happen for $\{i, j\} = \{1, 2\}$.
$W(X_i, X_j) = 1$ implies that either $i=j=3$ or $i=j=4$.
Hence, $$2 G = \begin{bmatrix}
1 & 0 & 1 & 1\\
0 & 1 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{bmatrix}$$
One can easily show that $|G| = -1/16$ (see on wolframalpha).
Hence, it must have a negative eigenvalue, which implies the kernel is not positive semi-definite.
|
2025-03-21T14:48:29.876226
| 2020-02-20T02:25:58 |
353126
|
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|
Stack Exchange
|
Metrization of quotient spaces defined by sequences of continuous functions
Let $K$ be a compact Hausdorff space and let $C(K)$ be the space of all scalar-valued continuous functions on $K$. Let $(f_{n})_{n}$ be a sequence in $C(K)$ satisfying $\sup\limits_{n}\sup\limits_{t\in K}|f_{n}(t)|<\infty$. We define an equivalence relation $R$ on $K$ by $$t_{1}Rt_{2}\Leftrightarrow f_{n}(t_{1})=f_{n}(t_{2}), \forall n.$$
Let $K_{1}:=K/R$ be the quotient space with the quotient topology $\tau$. That is, $\tau=\{V\subseteq K_{1}:Q^{-1}(V)$ is open in $K\}$, where $Q:K\rightarrow K_{1}$ is the quotient mapping. It is easy to see that $(K_{1},\tau)$ is compact. My question is the following:
Question. Is $(K_{1},\tau)$ metrizable?
Yes, it is metrizable. Assume the scalar field is real; if it is complex, replace the functions $f_n$ with their real and imaginary parts. WLOG each $f_n$ maps $K$ into $[0,1]$. Amalgamate the $f_n$ into a single function $f: K \to [0,1]^{\omega}$. Then $K_1$ is homeomorphic to $f(K)$ with topology induced from $[0,1]^{\omega}$. Metrizability of $K_1$ now follows from metrizability of the cube.
|
2025-03-21T14:48:29.876317
| 2020-02-20T04:25:13 |
353129
|
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|
Stack Exchange
|
Eigenvector of Hadamard matrix functions
Let $X\in\mathbb{R}^{n\times n}$ with SVD $X=UDV^T$. Are there known results regarding the eigenvectors of $Y=X^{\odot g}$? I am mainly interested in simple functions such as $g(z)=z^2$, i.e. $Y_{ij}=X^2_{ij}$.
I have tried finding it in Topics in Matrix Analysis without any luck so far.
Any help will be greatly appreciated!
|
2025-03-21T14:48:29.876363
| 2020-02-20T06:18:41 |
353133
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353133"
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|
Stack Exchange
|
A question about a non-linear ODE
Given the equation $x''(t)+x'(t)+x^3(t)=0$ with $x(0)=1$ and $x'(0)=0$ how to prove that
$ \mathop {\lim }\limits_{t \to + \infty } x\left( t \right) = 0 $.
For every $t>0$ it is $x(t)>0$.
The first question is easy. With the second question I'm stuck, at least analiticaly.This problem is from G. Gallavotti "The elements of Mechanics"
|
2025-03-21T14:48:29.876427
| 2020-02-20T07:23:35 |
353135
|
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"Dima Pasechnik",
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|
Stack Exchange
|
What is known about the non-existence of strongly regular graphs srg(n,k,0,2)?
Only few strongly regular graphs with parameters $\lambda=0$ (triangle-free) and $\mu=2$ (any two non-adjacent vertices
have exactly two common neighbors) are known, see the wikipedia page: the 4-cycle, the Clebsch graph and the Sims-Gewirtz graph.
I am looking for any information about the potential existence of more such graphs. For which values of $n$ and $k$ are they known not to exist?
a keyword here is "rectagraph" (trinagle-free graph with every pair of vertices at distance 2 having exactly 2 common neighbours).
Thanks, I did not know this name. So basically srg(n,k,0,2) graphs are rectagraphs of diameter 2.
Yes. A canonical reference is https://doi.org/10.1016/S0304-0208(08)73275-4
(can't get through the paywall though :- ( )
Example 1 in A.Neumaier paper says in partcular that the vertex degree in this case must be $k=t^2+1$, for $t$ not divisible by 4. As well, the number of vertices is $v=1+k+\binom{k}{2}$. The examples you list correspond to $t=2,3$. The next possible parameter set corresponds to $t=5$, so you have $v=352$, $k=26$. A.Brouwer's database lists this tuple of parameters as feasible, but no examples known. Similarly for $t=6,7$ you have feasible sets of parameters $v=704,1276$, resp. $k=37,50$, but no examples known.
To see that $k=t^2+1$, note that the 2nd eigenvalue of the adjacency matrix is
$$
r:=\frac{1}{2}\left[(\lambda-\mu)+\sqrt{(\lambda-\mu)^2 + 4(k-\mu)}\right]=-1+\sqrt{k-1},\quad \text{i.e. $t^2:=(r+1)^2=k-1.$}
$$
Similarly, the 3rd eigenvalue is $s:=-1-\sqrt{k-1}$, and one can compute their multiplicites, see e.g. Brouwer-van Lint, p.87
to rule out the case $t$ divisible by 4.
Namely, the multiplicity of $r$ is given by
$$
-\frac{k(s+1)(k-s)}{(k+rs)(r-s)}=\frac{k\sqrt{k-1}(k+1+\sqrt{k-1})}{4\sqrt{k-1}}=\frac{(t^2+1)(t^2+2+t)}{4},
$$
which cannot be an integer if $4|t$.
Great, thanks! Actually the condition k=s^2+1 is referenced from this paper (but I don't see now where to derive this condition from): https://www.sciencedirect.com/science/article/pii/0012365X75900576 Perhaps you could edit your answer to refer to that.
see my edit - as well, I replaced $s$ with $t$ to avoid notation clash.
Great, thank you!
|
2025-03-21T14:48:29.876581
| 2020-02-20T10:11:46 |
353142
|
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|
Stack Exchange
|
Joint typicality of sequences
I know that for two i.i.d distributions $P$ and $Q$ the probability that $Q$ will produce a length $n$ sequence that is $\epsilon$-typical according to $P$ is bounded by
$$Q(T_{P,\epsilon})\leq e^{-nD(P||Q)-n|\mathcal{X}|\epsilon}$$
where $D(P||Q)$ is the KL-dvergence and $T_{P,\epsilon}$ is the (strong) $\epsilon$-typical set of the distribution $P$, i.e. sequences with empirical distribution that is $\epsilon$-close to $P$.
Will this bound hold for $Q$ that is not i.i.d? For example, a Markov source?
That is, will I have
$$Q^n(T_{P^n,\epsilon})\leq e^{-D(P^n||Q^n)-n|\mathcal{X}|\epsilon}$$
where $Q^n$ is the distribution on sequences of length $n$?
Essentially, yes, although it is ambiguous the way you stated it (so it is not clear to me what $T_{P^n}$ really means). The general statement is
that you have a LDP for the empirical process $n^{-1}\sum \delta_{\theta^i X}$ where
$X$ is the infinite sequence and $\theta$ is the (left shift), in the product topology
(ie, you can look at any types of $k$ successive letters, with $k$ fixed but arbitrary) and the rate function is the specific entropy $$\lim_{m\to\infty} \frac1m D(P_m|Q_m)$$ where $P_m$ is the restriction to $m$ letters of $P$ (assuming $P$ is stationary, otherwise the rate is infinity). If $Q$ is Markov and you are interested in types of length $k$ then you can restrict in the above formula $m$ to be equal $k+1$. For this and related statements look at ``LDP for empirical processes" in large deviations books, or look at the original Donsker-Varadhan papers.
|
2025-03-21T14:48:29.877179
| 2020-02-20T12:27:20 |
353154
|
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"Lennart Meier",
"Noah Snyder",
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"https://mathoverflow.net/users/40804",
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|
Stack Exchange
|
Conceptual proof of classification of surfaces?
Every compact surface is diffeomorphic to $S^2$, $\underbrace{T^2\#\ldots \#T^2}_n$, or $\underbrace{RP^2\#\ldots \#RP^2}_n$ for some $n\ge 1$.
Is there a conceptual proof of this classification theorem?
Here, the term "conceptual" a little bit up for interpretation...
One may take it to mean: that doesn't rely on unintuitive facts.
For example, a proof that doesn't need $T^2\#RP^2\cong RP^2\#RP^2\#RP^2$ as an ingredient, but which has the property that this diffeomorphism comes out as a consequence would definitely count as conceptual.
The Zip proof is probably what you are looking for.
Take a look at this explanation of the Zip proof: https://www.maths.ed.ac.uk/~v1ranick/papers/francisweeks.pdf
You might find this helpful: Gallier, Jean, and Dianna Xu. A guide to the classification theorem for compact surfaces. Springer Science & Business Media, 2013. PDF download.
@Ben McKay: Lemma2 of the ZIP proof is exactly what I wanted to avoid...
Zeeman's proof is detailed in this note by Andy Putman. It includes a reference to Hatcher's proof of triangulation via the torus trick.
I'm not sure if this is André's motivation, but one reason why one might want to better understand conceptually where 3 cross caps = crosscap + handle comes from, is that this relation makes the classification of non-oriented 2d TQFTs extremely weird. In particular, this doesn't look like a "homotopy fixed point" condition, in contrast to the intuition for fully extended TFTs.
I don't agree that your stated diffeomorphism is non-conceptual. Connect-sum with T^2 means: "Do surgery on an oriented embedding of D^2 x S^0 in a single chart", i.e, delete those and attach a handle. Connect-sum with K means: "Do surgery on an embedding of D^2 x S^0 in a single chart which is oriented on one factor and unoriented on the other." We use the single chart to make sense of oriented, here. These are the same because you can move one of your discs around the orientation-reversing loop in RP^2 to make an oriented embedding unoriented. I suppose this leaves why K = RP^2 # RP^2.
I guess the most conceptual proof is the one using Morse theory:
Take a Morse function on the (closed, orientable) surface S. If it has no saddle points, then (using the gradient flow) $S\cong S^2$. Assume by induction that a surface with k-1 saddle points is $S^2$ with finitely many handles added. For the inductive step, consider a Morse function with k saddle points. A neighborhood of a saddle point is a pair of pants, with the Morse function constant on boundary curves. Consider the (possibly disconnected) surface $S^\prime$ obtained by cutting off the pair of pants and gluing in three disks. The Morse function extends (constant on the disks). By induction, each component of $S^\prime$ is $S^2$ with finitely many handles added, so the same is true for $S$.
There are other nonclassical proofs.
It is actually not hard to prove by purely topological arguments that for a (closed, orientable) surface S with $\chi(S)<2$ there is another surface S‘ with $\chi(S^\prime)=\chi(S)+2$. Thus the proof of classification boils down to show that a (closed, orientable) surface with $\chi(S)=2$ must be $S\cong S^2$. This can be seen from the Riemann-Roch theorem, which for $\chi(S)=2$ implies existence of a meromorphic function with only one pole of order 1, thus a biholomorphic map to $P^1C$. (For this argument you need that every orientable surface is complex; this follows from the homotopy equivalence $GL^+(2,R)\sim GL(1,C)$ and the obvious vanishing of the Nijenhuis tensor.)
Yet another approach would be to use uniformization and thus reduce the classification of surfaces to classification of discrete, torsion-free subgroups of PSL(2,R). (Uniformization has, besides the classical proof via the Dirichlet principle, by now more conceptual proofs via circle packing or via Ricci flow.)
Finally, if you believe that every surface can be triangulated, then the classification of surfaces is proven by a not so difficult combinatorial inductive argument. The triangulability of surfaces is of course not so easy, its proof in Moise uses the Schoenflies theorem.
By purely what arguments? (also, if you could give a reference or a sketch of those arguments I would love to see them)
Sorry, purely topological arguments. You just get S‘ from S by cutting along a homological nontrivial curve without self-intersections. The existence of a homologically nontrivial curve follows from the Euler characteristic. If that curve has self-intersections, you decompose it into closed curves without self-intersections. At least one of this curves is homologically nontrivial. Cutting along a homologically nontrivial curve yields a connected surface S‘. Thus S is the connected sum of S‘ with a torus.
@ThiKu: Does ^2#^2≅^2#^2#^2 follow from the type of arguments that you describe?
The third argument via uniformisation works also in this case: both surfaces are hyperbolic, hence aspherical, hence determined by their (isomorphic) fundamental groups. The other arguments use some kind of induction and thus probably require to know the homeomorphism beforehand. In any case, this homeomorphism has an elementary, hands-on, topological proof, so using other arguments for it would be shooting with canons at sparrows, in my opinion.
Once you have a smooth structure (needed for Morse theory, hyperbolization, uniformization, etc), finding a triangulation is trivial.
I know that every smooth manifold has a triangulation, but trivial?
I guess it depends on what you consider trivial, but it's pretty simple (as must be the case since it is true with no further hypotheses, in every dimension!). If I recall correctly, there is a nice discussion of one way to do it in Thurston's book
The proof of Zeeman described in this note is by a substantial margin the easiest and most conceptual proof I know. To simplify the exposition I restrict to orientable surfaces in the note, but it is trivial to also do the non-orientable case (and see the edit below for one description of how to arrange this to avoid using the fact that three cross caps is a handle plus a cross cap).
What I particularly like about it is that it proves not only that the usual classification is the complete list of surfaces, but also at the same time that the Euler characteristic is a complete invariant of orientable surfaces. Indeed, the proof is by (descending) induction on the Euler characteristic, with the base case the 2d Poincare conjecture: all compact connected surfaces have Euler characteristic <=2, with equality iff the surface is a sphere (and you can see the sphere very clearly in the proof, which directly produces from equality a decomposition of the surface into two discs meeting along their boundary).
As evidence of how simple this proof is, on many occasions I have explained it at a chalk board (with full details) in about 10 minutes.
EDIT: I've thought a little bit more about whether or not you can avoid having to prove that the connect sum of $3$ projective planes is isomorphic to the connect sum of a torus and a projective plane. Here is one way to arrange the proof that avoids directly proving this.
As in the proof in my notes, you prove the theorem by downward induction on the Euler characteristic. More precisely, what you prove by induction is the following:
Every connected surface has Euler characteristic less than or equal to $2$.
If a surface is orientable, then its Euler characteristic is even and if it equals $2-2g$, then the surface is a connect sum of $g$ tori.
If a surface is not orientable and if its Euler characteristic is $2-g$, then the surface is a connect sum of $g$ projective planes.
Notice that this implies as a consequence that the connect sum of $3$ projective planes is isomorphic to the connect sum of a torus and a projective plane (which will not be used directly in its proof).
Anyway, you follow the proof in my notes to deal with the base case (which combines $1$ above and the fact that the sphere is the only connected surface of Euler characteristic $2$). You then follow my notes in the inductive case to the point where you find the nonseparating simple closed curve $\gamma$. There are then several cases:
a. If your surface is orientable, then $\gamma$ is a $2$-sided curve, so you can cut and cap off to increase the Euler characteristic by $2$, and be done by induction.
b. If your surface is nonorientable and $\gamma$ is a $1$-sided curve such that cutting along $\gamma$ gives a nonorientable surface, then you can do just like in a (but cutting along $\gamma$ and capping only increases the Euler characteristic by $1$).
b. If your surface is nonorientable and $\gamma$ is a $1$-sided curve such that cutting along $\gamma$ gives an orientable surface, then you can cut cap and induct and deduce that your surface is isomorphic to $\Sigma_g \# \mathbb{P}^1$ for some $g \geq 1$. Of course, this is not what you want; however, if you draw the picture you can easily find a simple closed $1$-sided curve $\gamma'$ on $\Sigma_g \# \mathbb{P}^1$ such that cutting along $\gamma'$ gives a nonorientable surface. This is the curve you should have been using all along! Replace $\gamma$ with $\gamma'$, and go back to case b.
d. Finally, if your surface is nonorientable and $\gamma$ is a $2$-sided curve, then the complement of $\gamma$ must be nonorientable, so cutting and capping we find that the surface must be isomorphic to $\Sigma_1 \# (\#_{k} \mathbb{P}^1)$ for some $k$. Just like in case c, we can find a $\gamma'$ that is $1$-sided such that the complement is nonorientable, and then go back and use that one in case b.
Could you say a little bit more about the non-orientable case? There's gotta be something a little bit tricky because Euler char 0 doesn't imply you can split off a torus without assuming orientability, but Euler char -1 does imply you can split off a torus despite non-orientability. So you need three different lemmas in general? Or two lemmas plus a separate proof that three cross caps = handle + cross-cap.
You modify the proof as follows. Without assuming orientability, then there are two cases. (1) The neighborhood of the loop gamma is a cylinder (which is the case described in Andy's note) or (2) it is a Mobius strip. In case (2) you cut it out and glue in a single disk. This still gives a surface with larger Euler characteristic (increased by 1 this time), so is still covered by induction.
Chris beat me to it! If you want to avoid proving that three cross caps is a handle and a cross cap, you have to prove that if the surface is non-orientable, then you can find a one-sided simple closed curve that also has the property that cutting along it does not give an orientable surface. This requires a small extra argument (since you can't just pick any cycle in the dual graph). Probably it is easier and more natural to just go ahead and prove that 3 cross caps = handle + cross-cap (which I don't think of as being a very hard fact).
I just edited the answer with an approach that avoids proving the three cross caps is a handle and a cross cap (basically by using the inductive hypothesis a little better).
@AndyPutman You say that Zeeman's proof proves that Euler characteristic is a complete invariant of the (oriented) surface. I would say that it doesn't, due to the issue I describe in my answer. Do you disagree about this?
@LennartMeier: I disagree in the sense that I think that foundational manifold issues (like the well-definedness of the connect sum, for which I like the treatment in Kosinski’s book on manifolds) are separate issues from the classification, and should be dealt with before you even think about stating it. Mixing all these things together makes the proof unnecessarily complicated and obscures its main idea.
(let me put it a different way: the well-definedness of the connect sum really has nothing to do with surfaces per se, and in the classification I think it is best to isolate things that are special about surfaces from things that are really general facts about manifolds. Perhaps when I teach it to undergraduates I might give a proof of the well-definedness of connect sums that uses things that are special about surfaces, but I don't think this is the right way for experienced topologists to think about these issues.)
@AndyPutman I agree that this is the correct perspective if you are an experienced topologist and care mostly about the differentiable case. My slight annoyance stems from the fact that neither Zeeman's document nor the Zip proof are aimed at experienced topologists and also don't work in the differentiable setting -- they give the appearance of a proof accessible to undergraduates while in reality I am not aware of a simple proof of the classification of triangulated surfaces.
@LennartMeier: So I have to admit that I've never carefully read either the Zip proof or Zeeman's document. I first learned the classification as an undergraduate from Massey's book, which I thought of as kind of ugly and unintuitive. I then learned Zeeman's proof in conversations with Benson Farb (who did not know the origin of it), and I obsessively search the literature until I found its source. The document I wrote was written not for teaching undergraduates, but for experienced mathematicians. Even though many people think of it as an undergraduate topic (and thus beneath them),
[continued] I personally think that I should know the "best" proof of every result I routinely use, not matter how elementary. And given how many papers I have written about the mapping class group of a surface, the classification of surfaces is one of my most-used tools.
ps: I also think that topological=smooth=PL in dimension 2 is an important result, separate from the classification (and that Hatcher's proof using the torus trick is the best way to do this, though it is not well-suited for undergraduates).
This is more an extended comment than an answer to the question. The first thing to note is that there are different strenghts of the classification theorem for surfaces. Of course, there are the differentiable, triangulated and topological setting. But even if we choose such a setting, there are two statements one has to prove (at least in one approach):
Every closed surface is isomorphic to a sphere with handles or cross-caps attached.
The isomorphism type only depends on the number of handles and cross-caps attached.
Both the Zip-proof and the Zeeman proof refered to above in the comments only prove the first part and not the second part. The second part is essentially equivalent to the well-definedness of the connected sum. Especially in the topological setting, this is a subtle point, requiring even in dimension 2 a kind of Schönflies theorem (and being a very difficult theorem in dimension 4). Out of frustration about this state of affairs I wrote up a variant of Zeeman's proof (based on a treatment by Thomassen), but including the well-definedness of attaching handles/cross-caps. (It took me a long time to thus actually understand the argument of Zeeman.)
This is not really an answer to the original question, both because this proof is in the triangulated setting and does not avoid the isomorphism of the original question. But I really want to stress the point that well-definedness of attaching handles or connected sum is something one has to prove and not just hide by abuse of notation.
With the Zip proof the problems you are worrying about do not actually arise. To begin, fix a set of standard model surfaces $M(a,b,c,d)$ with $a$ handles (tubes), $b$ cross-handles, $c$ cross-caps, and $d$ boundary circles. One just has to show that if one takes a collection of disjoint model surfaces and identifies two oriented boundary arcs, then the result is again homeomorphic to a collection of disjoint model surfaces. (continued)
Once one specifies the two oriented arcs to be identified, the choice of an orientation-preserving homeomorphism identifying them doesn't matter since each orientation-preserving homeomorphism of one of the arcs extends to a homeomorphism of the model surface (before identification) using a collar neighborhood of the boundary. (continued)
After all the pairs of boundary arcs have been identified one has a model surface $M(a,b,c,d)$ with $d=0$, so all that remains to show is that if $b$ or $c$ is nonzero then all handles and cross-handles can be replaced by cross-caps, which isn't hard for the standard models. The operation of forming a connected sum plays no role in this proof, nor does the induction step involve attaching handles, cross-handles, or cross-caps in the interior of a surface.
Using a little bit of real algebraic geometry, there is a conceptual proof at least in the critical case $\chi=-1$, i.e. the case you're talking about explicitly. Indeed, let $S$ be a compact connected smooth surface without boundary with $\chi(S)=-1$. Choose a conformal structure on $S$. Since $\chi$ is odd, $S$ is nonorientable. Let $\tilde S$ be its orientation cover. Then $\tilde S$ becomes a complex algebraic curve endowed with an antiholomorphic involution $\sigma$. Since $\chi(\tilde S)=2\chi(S)=-2$, the curve $\tilde S$ is of genus $2$. In particular, $\tilde S$ is hyperelliptic. Let $$\pi\colon \tilde S\rightarrow \mathbf P^1(\mathbf C)$$ be the hyperelliptic covering. Since the hyperelliptic covering is canonical, the image curve $\mathbf P^1(\mathbf C)$ acquires a real structure, i.e., an antiholomorphic involution $\tau$. In general, there are $2$ possibilities: either $\tau$ has fixed points, or it doesn't.
Let us prove that $\tau$ does have fixed points. If it hadn't any, the ramified double covering $\pi$ would induce a ramified double covering of the quotients $$\bar\pi\colon\tilde S/\sigma=S\rightarrow \mathbf P^1(\mathbf C)/\tau=\mathbf P^2(\mathbf R).$$ Both quotients are compact connected smooth surfaces without boundary as both $\sigma$ and $\tau$ act without fixed points. Their Euler characteristics are equal to $-1$ and $1$, respectively. It follows that $\bar\pi$ is ramfied over exactly $2\times 1-(-1)=3$ points, which is absurd. Therefore, $\tau$ does have fixed points. Hence $\tau$ is conjugate to the usual real structure, and we may assume that $\tau$ is equal to the usual real structure on $\mathbf P^1(\mathbf C)$.
It follows that the curve $\tilde S$ is isomorphic to a curve $C$ of the form $y^2=-p(x)$, for some real polynomial $p$ of degree $6$ taking only strictly positive values on $\mathbf R$ and having simple roots in $\mathbf C$, the isomorphism being an isomorphism of real algebraic curves, i.e., respecting the antiholomorphic involutions on $\tilde S$ and $C$.
Now, the fact that the set of all such polynomials $p$ is connected implies that all smooth surfaces with $\chi=-1$ are diffeomorphic to each other!
The argument works generally for any smooth surface $S$ of odd Euler characteristic, after arguing that one can always deform the conformal structure to a hyperelliptic one, which I believe can be proven without using the classification of smooth surfaces. For nonorientable surfaces of even Euler characteristic, one cannot exclude $\tau$ to act fixed point freely, and I do not see anything conceptual in this case.
|
2025-03-21T14:48:29.878437
| 2020-02-20T12:30:18 |
353155
|
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|
Stack Exchange
|
A decision problem from sheaf set theory?
Let $V^{X}$ be a sheaf model of ZF set theory, where $X$ is a topological space as it is defined in [1].
Let $T(y_1,\ldots,y_n)$ be an $B(T)$-free algebra as it is defined in [2], where $B(T)$ is the equational class of all finite heyting algebras with unary opertors from the finite partial order set $T$.
Since $B(T)$ is first-order definable then $B(T)$ is in $V^{X}$.
We have a decision problem,
$$"V^{X}\Vdash_{u} T[v_1,\ldots,v_n] "\label{*}\tag{*}$$
Where $\Vdash_{u}$ is the satisfaction relation of the sheaf model $V^{X}$ and $u$ is an open subset of $X$, which is a truth witness of $T(y_1,\ldots,y_n)$,$y_1,\ldots, y_n$ are variables and $v_1,\ldots,v_n$ are sets of codified sentences from the $V^{X}$-language .
I think \eqref{*} is a decision problem for two reasons,
First, $B(T)$ is equivalent to an algebraic variety generated by the algebra $\{0,1\}(T)$, ([2], Theorem 10, p. 8).
Second, any finite free algebra of $B(T)$ is a Boolean product of $\{0,1\}(T)$-subalgebras,([2] Corollary 16, p. 11).
My first question is this: are these hypotheses sufficient to find a function
$$
f:\{0,1\}^\ast\to \{\mathrm{yes,no}\}
$$ to make \eqref{*} a decision problem?
My second question is this: since the Heyting algebras are models for intuitionistic logic, could we say that \eqref{*} is equivalent to the decision problem of intuitionistic logic, which is pspace-complete?.
References
[1]. Benavides, John, "Sheaf logic, Quantum set theory and the interpretation of Quantum Mechanics", arxiv 1111.2704 (2011).
[2]. Sanjuan, Eric, "Heyting algebras with Boolean operators for rough sets and information retrieval applications", Discrete Applied Mathematics 156, No. 6, 967-983 (2008), MR2395615, ZBL1134.06007.
Note:- $T(y_1,\ldots,y_n)$ and $B(T)$ are defined in the first-order language of $V^{X}$. $T$ in $T(y_1,\ldots,y_n)$ is different from $T$ in $B(T)$. $T(y_1,\ldots,y_n)$ is a first-order set theory sentence about the $B(T)$-free algebra on $T$.
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2025-03-21T14:48:29.878589
| 2020-02-20T12:48:33 |
353157
|
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|
Stack Exchange
|
Associativity and truncatedness of a $\mathbb N$-filtered object
$\require{AMScd}$Consider the monoid monad: $S : Set \to Set$ sends a set $X$ to the free monoid on $X$:
$$
SX := \coprod_{n\in\mathbb N} X^n
$$ the empty sequence $()\in X^0$ is the identity element for the concatenation operation $X^n\times X^m \to X^{n+m}$.
Now, what is an algebra for this monad? It is a set $M$, with a map $a : SM\to M$ such that axioms hold. But then such $a$ corresponds to a family of maps
$$
(a_n : X^n \to X \mid n \in\mathbb N )
$$ prescribing an $n$-ary operation on $X$ for each $n\in\mathbb N$.
However, now, all $a_n$ for $n\ge 3$ are determined by $a_0,a_1,a_2$ and their composite, because the associativity axiom for $a$ now implies that $a_n$ has a unique reduction/rewriting as a composition
$$
(w_1\times\cdots\times w_n)\circ (w_1\times\cdots\times w_{n-1}) \circ \cdots\circ (w_1\times w_2)\circ a_2
$$
For example, $a_3 : X\times X \times X \to X$ must be either $(a_2\times 1)\circ a_2$ or $(1\times a_2)\circ a_2$; the two are equal, by the very statement of associativity. And $a_4$ is
$$
(1\times a_3)\circ a_3 = (1 \times ((a_2\times 1)\circ a_2))\circ ((a_2\times 1)\circ a_2)
$$
or
$$
(a_3\times 1)\circ a_3 = (((1\times a_2)\circ a_2)\times 1)\circ ((1\times a_2)\circ a_2)
$$ These are equal, again by associativity.
So,
From the very shape of the monad, we have been able to deduce that an algebra structure is completely determined by $a_{\le 2}$.
This is an interesting and distinctive property of semigroups/monoids over other kinds of nonassociative structures: if we represent the composition of a tuple under a binary operation as a tree (whose root is the full composition of the tuple), of course different parenthesizations give different results. This is reflected in the shape of the monad of magmas (whose algebras are sets with a closed binary operation $m : X\times X \to X$),
$$
\begin{cases}
(MX)_0 := X\\
(MX)_n := \sum_{p+q=n} (MX)_p \times (MX)_q \\
MX := \sum_{n\in\mathbb N} (MX)_n
\end{cases}
$$ There is no $n$ that determines $MX$ completely; something that follows from the definition of magma, of course, but also from the very shape of the monad that has magmas as algebras.
Let me try to formalise better what it means "determine completely" a structure: let again $S$ be the monoid monad, and let us consider the $\mathbb N$-indexed set
$$
\begin{CD}
\coprod_{n\in\mathbb N} X^n \\
@VpVV\\
\mathbb N
\end{CD}
$$ the fibers $p^{-1}n$ for $n\ge 3$ now are "uniquely determined" by $p^{-1}\{0,1,2\}$ in the sense that each element in $u\in p^{-1}n$ is a unique push-forward $p_*\bar u$ of an element $\bar u\in p^{-1}\{0,1,2\}$. This is the "truncatedness" condition we were looking for.
All this consists of very elementary observations, I am clearly rediscovering the wheel; and yet I have never seen it worded out explicitly in terms of a truncatedness condition.
I am clearly looking in the wrong place though, because the more I think about it, then, the more I feel inclined to link it to Mac Lane's coherence theorem for monoidal categories ("all diagrams commute" if you give me unitor and associator), and to another kind of important internal monoids we all work with: categories (once you specify binary compositions in a 2-coskeletal quasicategory, i.e. a category, you specify it everywhere).
So, questions:
What's going on here? What is the property of $S$ that $M$ does not have, and that makes it possible to determine an algebra structure on just some $a_n$? Maybe my above argument can be formalised. If so, how?
What is the relation between this statement, and other "truncatedness" conditions on monoid-like structures?
I don't see why your two examples should be different in the way you describe. The action map $M(X) \to X$ of a magma is also completely determined by the restruction $M_2(X) \to X$.
Yeah, you're right, it is not exactly what I wanted to say. Any clue on the final questions?
What do you mean by "the shape of the monad"? Does this refer to the underlying endofunctor?
|
2025-03-21T14:48:29.878878
| 2020-02-20T13:54:13 |
353163
|
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|
Stack Exchange
|
Generating all graphs of order 4 with the help of Collatz
Given a set of positive integers, its common divisor graph ( CD-graph) is the graph whose vertices are the integers, two of which are joined by an edge if (and only if) they have a common divisor greater than 1.
Given the same set of integers, its nth Collatz iterate is the set resulting of applying the Collatz recipe (multiply by 3, and add 1 if odd; divide by 2 if even) to each of its elements.
Does a set of four integers exist such that the CD-graph of itself and that of its first ten iterates are precisely the 11 graphs on 4 vertices?
Notice that the set {7,10,13} does the trick for graphs on 3 vertices: all four graphs are generated.
Do all 11 graphs occur as CD-graphs (regardless of whether or not in this particular way)? When the Collatz map is not injective, do you regard it as collapsing the number of vertices in the CD-graph, or do you allow multiple vertices with the same label (then presumably connected)?
@LSpice: All graphs are the CD graph of some set. No collapsing.
So {5, 32} goes to a 2-vertex graph, not a 1-vertex graph?
@LSpice: That is correct, a 2-vertex graph.
Have you done any computer searching? It seems like it would be straightforward to test all ~4 million quartets of distinct numbers less than a hundred, for instance...
@Steven Stadnicki: My collegue Freddy Barrera has done some computer search which so far has yielded at most 9 distinct graphs.
|
2025-03-21T14:48:29.879022
| 2020-02-20T14:08:04 |
353164
|
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|
Stack Exchange
|
functors reflecting "isomorphism relations"?
Consider a functor $F:C\to D$ between two categories $C$ and $D$. Suppose $F$ satisfies the following property: for any $a, b\in C$, $F(a)\cong F(b)\iff a\cong b$.
Of course, $a\cong b\Rightarrow F(a)\cong F(b)$, so it is the other direction tricky.
The question is then: is there a name for such functors? Have they been studied?
I know that there are conservative functors, which reflects isomorphisms. However, the property of $F$ mentioned above only reflects "isomorphism relations." This question was asked here before, but I don't see a satisfactory answer.
Thanks for the coming help!
Just to highlight the difference between conservative functor and the notion you're talking about, if we let $\text{Set}{n}$ be the full subcategory of $\text{Set}$ whose objects have exactly $n$ elements, then every functor $\text{Set}{n}\to D$ for any category $D$ has the property you're describing, but will usually not be conservative. (And of course there's nothing special about sets here--the same holds for any $C$ and full subcategory on a given isomorphism class.)
Some authors name this "isomorphism reflecting", for example in Noncommutative rings and their applications (p. 153) and Models, Modules and Abelian Groups: In Memory of A. L. S. Corner (p.480). But it is kind of dangerous to use this since some (maybe even more!) authors also mean "conservative" by "isomorphism reflecting", for example in Basic concepts of enriched category theory (p.8) and Homology and homotopy in semi-abelian categories (p.9), also the nlab.
I don't know if there is an English term for the isomorphism relation, in German it's just "Isomorphie" which could be translated to "isomorphy". Then the property should be called "isomorphy reflecting".
Of course, you could always just say "injective on isomorphism classes".
At several places in the book "Category Theory in context", E. Riehl calls this "$F$ creates isomorphisms" (in analogy to the more common term "$F$ creates (co)limits").
I am not sure if this analogy fits well. $F$ creates isomorphisms should mean that if $F(f)$ is inverse to $g$, then there is a unique morphism $f'$ with $g=F(f')$ such that $f$ is inverse to $f'$, right? In other words, $F$ is conservative.
@MartinBrandenburg I guess it depends on the definition of creating limits, there seems to be some debate around that as well: https://mathoverflow.net/questions/103065/what-is-the-correct-definition-of-creation-of-limits If you use Def. 1 in the above link, then I agree with what you are saying. If you use Def. 2, then it fits better this version (then the statement in both cases is "if it exists in the target, then it exists in the domain, and $F$ preserves it").
However, it seems to be the case that either way, "creating isomorphisms" should also include reflecting them, so there is (should be) some difference between this notion and "injectivity on isom. classes."
One paper about this is Elliott's Towards a theory of classification, where such functors are called classification functors, and there is a nice collection of examples.
(Though note that a category theorist would perhaps approach some of the material in that paper a bit differently, e.g. by considering inner automorphisms as 2-morphisms. And Elliott's claim that every category is concrete is wrong for size reasons.)
|
2025-03-21T14:48:29.879405
| 2020-02-20T14:53:52 |
353168
|
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|
Stack Exchange
|
History- calculating convolution by tabular method
I often see a trick for calculating convolution of discrete data by a so-called Tabular method. There are a lot of Youtube videos and many Indian textbooks on Signal Processing [Books].1
Basically, if we have two series P = [1 3 5 3 1] and Q is [ 0 0 0 0.5 0.8 1 1 0 0 0]. One can make a table and multiply elements and sum up the diagonals as illustrated below. I color coded the diagonals. This process results in correct number of elements 10+5-1=14 elements.
Does anyone know who came up with this short cut? It seems like a nice approach for calculating convolution, correlation, cross correlation of discrete data.
Thanks.
This reference claims to have invented the tabular method as a "novel method":
A novel method for calculating the convolution sum of two finite length sequences, J.W. Pierre (1996).
Three variations of the tabular method are discussed in The use of spreadsheets to calculate the convolution sum of two finite sequences (2004), citing a 1990 text book.
Thank you for the first reference on 1990. It is very nice. The 2004 reference shows the same approaches as I did in Excel with color coding but he cites reference 11, which is another South Asian author's textbook. I am sure, this approach must be older than 1990s. Also it seems this tabular of approach is quite popular in Indian engineering schools (as deduced from Youtube videos).
indeed, added the link to the text book.
|
2025-03-21T14:48:29.879552
| 2020-02-20T15:10:34 |
353169
|
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|
Stack Exchange
|
Full automorphism group of a Bruhat-Tits building
If we start with a semisimple algebraic group $G$ defined over a non-archimedean local field and want to understand the relationship of this group with the full type-preserving automorphism group of the associated Bruhat-Tits building, is there any reference in the literature which would help me get a better handle on that? Specifically I'm interested in any sufficient conditions for when it's an open subgroup, if that does indeed ever occur.
I guess you mean the Euclidean building, on which $G_k$ acts properly cocompactly. In finite characteristic it's never an open subgroup. The reason is that the automorphism group $K$ of the underlying topological group sits in the isometry group, and has finite intersection with the action of the (group of points of the) algebraic group. In characteristic zero if $G$ has no rank $1$ factor it's an open subgroup, I think it follows from results of Tits (which also provide a full description in finite char.).
By the way in finite characteristic, it's an open question whether there exists $G$ (centerfree, with no rank 1 factor) such that some lattice in $\mathrm{Isom}(X_G)$ doesn't meet $G$ in a lattice, $X_G$ being the Euclidean building.
|
2025-03-21T14:48:29.879670
| 2020-02-20T15:27:18 |
353171
|
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|
Stack Exchange
|
Analogues over finite fields of certain integers defined multiplicatively in $\mathbb Z$
For any irreducible polynomial $f$ of degree $d$ in finite field $\mathbb F$ we have that $$x^{|\mathbb F|^d}-x\equiv0\bmod f$$ and thus the polynomial $x^{|\mathbb F|^d}-x$ is the analog of factorial in finite fields. Are there analogues of binomial coefficients which can be represented nicely? Would it be reasonable to seek analogues which can be represented nicely of denominator of Bernoulli numbers $B_{2n}$ which are $\prod_{(p-1)|2n}p$?
Maybe a good place to look at is "Basic structures of function field arithmetic" by David Goss.
|
2025-03-21T14:48:29.879752
| 2020-02-20T16:53:28 |
353177
|
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|
Stack Exchange
|
Geometry of algebraic curve determined by point counts over all number fields?
Let $C$ be a smooth (geometrically irreducible) projective curve of genus $g>1$ over a number field $K$. The Mordell conjecture (first proved by Faltings) says that for any finite field extension $L/K$ (say, inside a fixed algebraic closure $\overline{K}$) there is a function $\{$(isomorphism classes of) finite extensions $L/K$ inside $\overline{K}\} \rightarrow\ \{$non-negative integers$\}$ defined by $L \mapsto \#C(L).$ Does this function determine (the isomorphism class of) $C/\overline{K}$, or otherwise any geometric properties of $C$ such as the genus?
The answer is no for a silly reason, the curves $y^2 = x^5 \pm ix +1$ have the same number of points over any extension of $\mathbb{Q}(i)$. But once one identifies Galois conjugate curves, the question comes back.
Right, thanks @FelipeVoloch, I had $K=\mathbb{Q}$ in mind when I wrote the question.
Just a remark: the answer is no for affine curves: consider $C-{P}$ for a projective curve $C$ without automorphisms and 2 different $P \in C(K)$.
You can try to recover the gonality of the curve by looking at how the number of points behaves over extensions of $K$ of a fixed degree. For instance, lots of quadratic points should imply that the curve is hyperelliptic.
@damiano That's a nice plausible idea. Here are a couple more notions: (1) higher genus should mean maximum number of points in all extensions of discriminant $\leq d$ grows more slowly with $d$ (conjecturally the maximum over all extensions of bounded degree is finite); (2) extra automorphisms implies more points in generic extensions with smaller Galois groups.
A variant: Suppose we start with an elliptic curve E over, say Q. Then the function that associates to a number field K the rank of the group E(K) makes sense. Does this function determine the elliptic curve?
@Asvin rank $E(K)$ is $K-$isogeny invariant
@david that's right! So maybe the isogeny class is determined?
small remark: the result of Caporaso Harris Mazur + Pacelli gives that if Lang conjecture is true then #C(L) is bounded only in terms of the genus of C and the degree $[L:\mathbb{Q}]$ so this should impose some sort of restriction on the informations you can recover? (note that there are unconditional results in this direction using Chabauty method by Stoll, Katz-Rabinoff-Zureick-Brown...)
@FelipeVoloch I don't think this is true - given a field extension of $\mathbb Q(i)$, applying the nontrivial automorphism of $\mathbb Q(i)$ may produce a different extension.
@Asvin Given an elliptic curve $E$ defined by a cubic equation $y^2 = f_3(x)$ for a cubic polynomial $f_3$, the average rank of $E'$ over $\mathbb Q( \sqrt{ f_3(x)})$ should equal the rank of $E'$ over $\mathbb Q$ plus $1/2$ unless $E'$ is isogenous to $E$, and $3/2$ if $E'$ is isogenous to $E$.This follows from the $L$-function equidistribution conjecture of Sarnak, Shin, and Templier (at least the naive-height-based version). So this should indeed determine the isogeny class.
@WillSawin Yes, I think you are right.
Heuristically, we can compute the genus. This follows essentially the srategy of damiano and David Lampert in the comments.
First note that it is possible to compute from this function the number $\#'C(L)$ of points over $L$ with field of definition exactly $L$. (Here the field of definition is the smallest field containing $K$ that the point is defined over.) This follows from inclusion exclusion.
Now observe that if we let $$ r_d = \lim \sup_{X\to \infty} \frac{\log \sum_{L/K, |\Delta_{L/K}|<X} \#'C(L) }{\log X} $$
and let $d$ be minimal with $r_d>0$, I think it is reasonable to expect that $d$ is the gonality of $C$ and that $r_d = \frac{1}{ g+ d-1}$, so that the genus of $C$ is $\frac{1}{r_d} +1-d$.
Why do I think this? Consider first a degree $d$ map $\pi: C \to \mathbb P^1$. Associated to a point of $x\in \mathbb P^1(K)$, of which there are $\approx Y^2$ of height $<Y$, we obtain a degree $d$ divisor in $C$. By Hilbert's irreducibility theorem, for almost all such points, the Galois action on this divisor is transitive, so this divisor defines a point of $C$ with field of definition a degree $d$ field extension $L$ of $K$.
Associated to $\pi : C \to \mathbb P^1$ is a branch divisor counting the ramification of $\pi$ over each point. By Riemann-Hurwitz, it has degree $2g-2+ 2d$. The ramification points of $L$ are at most the intersection points of $x$ with the branch divisor in the scheme $\mathbb P^1_{\mathcal O_K}$, and the ramification index is at most the intersection number, with equality if the intersection is transverse.
Thus the discriminant of $L$ over $K$ is at most the arithmetic intersection number, which is simply a degree $2g-2+2d$ polynomial evaluated at $x$, and thus is $O( Y^{2g-2+2d})$. It follows that the number of points arising this way with discriminant less than $X$ is at least $\approx X^{ \frac{1}{ g-1+d}}$.
Furthermore, for points which are generic in the sense that their intersection with the branch divisor is transverse, and which are not too close to the branch divisor in the archimedean sense, this is sharp, and so it's reasonable to expect that the number of points arising this way with discriminant less than $X$ is $\approx X^{ \frac{1}{ g-1+d}}$. This is the first heuristic.
Furthermore, for points arising from covers of an elliptic curve, because the number of rational points on an elliptic curve with height $<Y$ is $O( Y^{\epsilon})$, it's reasonable to expect the number of points with discriminant $<X$ is $O(X^{\epsilon})$. This is the second heuristic.
Combining these, and using Falting's theorem, we see that all the points of degree less than the gonality are explained by finitely many points and elliptic curves and thus $r_d=0$ for $d$ less than the gonality, and for $d$ equal to the gonality we have finitely many covering maps to $\mathbb P^1$, all of which have the same number $X^{ \frac{1}{g+d-1}}$ of points, and so $r_d = \frac{1}{g+d-1}
It might be possible to extend this idea further to calculate the isogeny class of the Jacobian of $C$ by using the average rank of simple abelian varieties when pulled back to fields $L$ of degree $d$, weighted by $\#'C(L)$, but this would be even more heuristic.
Thanks @WillSawin for this interesting answer which I'll try to digest.
|
2025-03-21T14:48:29.880243
| 2020-02-20T16:55:11 |
353178
|
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|
Stack Exchange
|
What types are to mathematical proofs as types à la Martin-Löf are to constructive proofs, and what's wrong with them?
The question is motivated by this surprising sentence from Freek Wiedijk's The QED Manifesto Revisited.
I agree that the QED-like systems that exist today are not good enough
to start developing a library as is described in the QED manifesto.
Surprising to me, that is, who had not heard of the QED project before.
(Edited after A. Bauer's answer) OK, time I mopped up.
I was 1st introduced to the Curry-Howard correspondence back in school days long, loooooong ago; I understood it as s/th along the lines of "Any computer program is a constructive proof that inhabitation of its input type entails inhabitation of its output type"; my very next question was, "OK then, if instead of any program we restrict ourselves to programs that model mathematical proofs in classical logic, what would be the proper type system(s) for them?". I guessed it had to be existential types, guarded subtypes (meaning types restricted by any formula; so that you would write e. g. "let x: {r : Real | !(r : Rational)}" if all you mean is to express that x is irrational), or such like stuff; then I let the matter rest and turned to more pressing ones, to wit, getting a degree.
If I had pressed it further at the time &, instead of the title question, asked "What types are to classical logic as simple types are to constructive logic?", maybe I'd have been answered "There are no such types: there is no need to extend or otherwise modify simple types for classical logic, since it is encompassed in constructive logic. Go get some understanding of type theory, and learn how to do with them. "
If I had asked again at the time the QED manifesto was issued, a plausible answer might have turned to "They're called refinement types. What's wrong with them is, they're a pointless gadget: the problem they solved was not a foundational one, and introducing them did not present type theory with any new challenge. So all the papers they deserve have been written already. "
Returning to it now, it seems to me the natural way to attack the QED challenge would be: 1. to set forth a serious answer to my old question, 2. design the corresponding language, 3. build the proper language tools, including, in the end, a full-fledged proof assistant (probably posing as a type checker), 4. look what happens. In a programming language with refinement types, the C-H correspondence now works as: subprograms returning (refinements of) void model conjectures, their bodies model proof outlines, assertions in the bodies model steps in the proof, Skolem functions arise naturally if handlers for failed assertions are allowed. Then, verified theorems in predicate logic are those subprograms in which the type checker has been able to reduce all assertions to tautologies.
What hopefully would happen if a suitable extension to such a language had become popular is, useful tools could start developing, such as doclets to turn fragments of source code to academic papers; people might get used to writing in it just to convert formulas from .pdf to .ps, or to save themselves the hassle of adjusting to the style guides of every other journal they submit to; later, we might dream of replacing arXiv with a global GitHub repository of their works. This could take place long before any proof assistant worth mentioning existed for the language.
Given Wiedijk's paper though, the real story must have been quite different. It reads like people trying to endow dependent types with refinement soon hit some snag and, instead of developing proof assistants (Edit: read formal languages and tools) for classical logic, they went happily to rewrite classical math for constructive logic.
(Edit) With predictable results: mathematicians have been accepting a certain style of proof outline for some time now and non-mathematicians are used to it, so if it is too arcane work to make it understandable by a automated checker, they will rather dispose of the proof checker than of the proof.
Turn away for a minute from what dependent types mean, and look how they would be used in a language purporting to be strongly typed. The user is student X with a major in control systems, a hobby in Java programming, currently at grips with an intro to topological groups. So far, he has managed to translate "pick any dyadic integer " as "let x: Integer(2)" for proof-checking freeware he downloaded a minute ago, then "consider first the case of a rational integer" to "let x1:Integer(2)*Rational"; now attacking "pick some irrational number x and consider the fractional parts of the p.x's, p ranging over Z: obviously, blah blah blah". Not understanding a word of type theory, he first looks for a way to say "let x: Real - Rational", finds none, looks harder, still finds none, then let go & turns back to more pressing matters, to wit, control systems.
Of course, if student X had access to such a language as I sketched, he would have given up long before he got any proof automated. No harm done, though: what he wants is to understand TG's, not that a computer program understand it in his stead, nor to understand type theory. All he needs is to keep his fingers busy while reading, and I advocate giving him a productive way to do just that.
My read of Wiedijk's paper is, if you bar student X from modelling things the way he understands them, he will turn to what he considers more pressing matters, regardless of how pressing your need for computer-checkable proofs, and how deep your understanding of what the modelling language guarantees. It explains how we still lack a free-access, widely-known, low-quality, multi-audience, computer-interpreted codebase of mathematical theories the way we have a free-access, widely-known, low-quality, multi-audience, human-readable textbase of common lore.
Edit Just to clarify: the codebase of mathematical theories in question would still be as far away from being computer-checked, as is WP from auto-generating ontologies for the subject matter of its articles; light-yrs away. The (hopeful) progress lies in the rest of the road being mostly for metamathematicians & software engineers to tread, with the part of "ordinary" mathematicians perceived as non-problematic.
Hence my question: what type systems are the natural ones to express assertions in classical logic? Or maybe I should have titled this post "What types are to dependent types, as refinement types à la Freeman-Pfenning are to simple types?". Anyway, what have they that makes their theory so unsavory?
I'm confused as to why you think restricting the 'types' under consideration will make the program corresponds to classical proof. In this correspondence type corresponds to propositions, and proposition are the same in classical and constructive mathematics.
This question combines what seems like a basic misunderstanding of how predicate logic relates to type theory (as per Simon Henry’s comment), a disinterest in learning further about it (“turned to more pressing matters”), and some quite unnecessary inflammatory tone (the suggestion that typical mathematicians “loathe constructive logic with all their guts”). That’s not exactly a good approach for engaging with people who work in the area.
@PeterLeFanuLumsdaine it is quite possible I misunderstand what type theory is: I come from computer science, not math. You're quite welcome to enlighten me, and it seems to me I'll make it easier for you, if I state right from the start what my (mis)understanding is. You are quite right that turning to more pressing matters shows disinterest in learning further: the pressing matter was to get a degree in mining, & that was 30yrs ago. It seems to me though, that self-teaching in type theory does witness some interest in it.
@PeterLeFanuLumsdaine My prose about J. R. Mathgeek was not in earnest, and I am surprised, & sorry, that it could set anyone aflame; I'll edit the post.
Joyal proved that a cartesian-closed category that has double-negation elimination is a preorder, i.e. there is at most one morphism between any two objects. Todd Trimble mentioned this here: https://mathoverflow.net/a/43285/61785
@Simon Henry I hope the latest edits clarify that I do not consider restricting "the types under consideration". I consider extending a type system with a mechanism to express subtyping relationships in addition to relationships of type dependence. In this extension, premisses correspond to the subtypes of program arguments, and conclusions to the returned subtype, typically a refinement of the unit type.
I still don't understand were you are going. There is no difference between "classical assertion" and "constructive assertion", the only difference is between "classical proof" and "constructive proof". So the types are the same. The difference is between what you accepte as a proof that something is a terms. (the terms as proof break down in classical logic). Also constructive logic can perfectly emulate constructive logic: a classical proof of P is the the same as a constructive proof of LEM $\Rightarrow$ P or of the double negation translation of $P$
Constructing proof assistant for classical logic has never been a problem: take a constructive proof assistant and add the law of excluded middle as an axiom or assumption. This is really not the reason why proof assistant are not universally used today.
@SimonHenry Oooookay... I think my basic misunderstanding was, what the sentence "build the corresponding assistant" would mean to s/o other than myself. Try "build the corresponding formalization platform" instead, if it helps. I agree the kind of logic that underlies the proof assistant is not a problem.
@SimonHenry It's all one to me whether I write a proof assistant from scratch, or translators to interface with an existing one: in any case I picture myself building a proof assistant, after s/o else built the theory behind it. As A. Bauer commented further below, type theory is not the culprit.
A good starting point to learn about type theories for classical logic is the $\lambda\mu$-calculus introduced in 1992 by Parigot in λμ-Calculus: An algorithmic interpretation of classical natural deduction, which extend the $\lambda$-calculus to give a computational interpretation of classical natural deduction.
For the next step, I would recommend reading the research summary on Hugo Herbelin's home page. Chasing some of the references listed there, you will learn how incorporating control operators into type theory yields a computational understanding of excluded middle.
For a modern take on the role of excluded middle in Martin-Löf type theory I recommend reading at least Section 3.4 of the HoTT book. The important bit to take away from there is the fact that excluded middle can coexist with the rest of homotopy type theory quite naturally.
You speak of proving correctness of programs, possibly using classical logic. For that you can use a system based on refinement types, such as F*. Also note that proving correctness of programs is an activity unlike formalization of traditional mathematics, so do not expect a single tool to work equally well for both of them.
In my opinion the presence or lack of excluded middle in a proof assistant has very little to do with it being useful for formalization. Of course, if you want to formalize classical mathematics then you need excluded middle (but much less frequently than most mathematicians expect). Most modern proof assistants let you postulate excluded middle quite directly and use it to your heart's content, so that cannot be the show stopper. One of the most popular proof assistants is Isabelle/HOL, which has excluded middle built in, but that does not make it significantly more successful in comparison with other formalization tools.
It is a bit difficult to understand what precisely you are getting at in your long discussion, but I would venture to say that you've misidentified negation and excluded middle as the culprit. I am not even sure whether you think that Martin-Löf type theory has no notion of negation – of course it does! In fact, the problems faced by your student X has nothing whatsoever to do with type theory. Student X would face essentially the same obstacles if they used any other formal system, such as first-order logic and set theory, or higher-order classical logic. Your student X is quite naive to think that formalization of mathematics is just a simple exercise in translation from English prose. It may be true that usability is decided by the users, but the users' abilities play a role as well.
Thanks for the refs! I agree that my discussion need not be made any clearer by its sheer length; I think you mostly got it right, and if I can compress it to a comment on your answer I'll remove it from my post.
Experience shows that (1) it is possible to formalize proofs written in English, (2) ordinary mathematicians do not know enough formal logic to do it without some amount of training (this is a matter of fact, not a criticism), and (3) large-scale formalization is a software engineering problem that mathematicians have not met before in their work. By the way, your $x : \lbrace z : A \mid \phi(z)\rbrace$ is refinement types that I mentioned above (or dependent sums if you prefer dependent type theory).
your 3 counts summarize my point more clearly than I could do myself. I'll only add to (2): ordinary mathematicians and non-mathematicians, and (4): none of the above is sufficient ground to discourage them before they even tried their hand at it, produced s/th useful for metamathematicians & software engineers to start with, and only then, discovered they lack the training to fully formalize their output al by themselves.
@FrançoisJurain: of course, I am not arguing against formalization. In fact, I work in the area and am trying to promote it. But I also try to have a critical position on what I am doing.
I understand Wiedijk as also critical on what you are doing, whereas I do not deem myself competent as a critic. The way I read his paper, in spite of your best efforts to promote formalization, discouragement is just part of the current landscape. I suggest it comes from the strangeness (to the non-type-theorist eye) of the available languages, not from arguments.
Type theory is not the culprit, as set-theory based provers experience the same problems (or worse). It's a combination of two things: software of "academic quality" coupled with early stages of the technology of theorem provers.
|
2025-03-21T14:48:29.881331
| 2020-02-20T18:13:02 |
353184
|
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|
Stack Exchange
|
The "higher topology" of countable Scott sets
Fix some computable bijection $b$ between $\omega$ and $2^{<\omega}$. For $r\in 2^\omega$, let $$[r]=\{f\in 2^\omega: \forall\sigma\prec f(b^{-1}(\sigma)\in r)\}$$ be the closed subset of Cantor space coded by $r$. For $M\models PA$ nonstandard, let $\mathcal{S}(M)$ be the standard system of $M$ thought of as a topological space (namely, as a subspace of Cantor space).
Say that a closed set pattern on a topological space $(X,\tau)$ is an assignment $c$ of $\tau$-closed sets to points in $X$.
EDIT: while it doesn't impact this question or the followup question, it seems natural in retrospect to also add the condition that the relation "$x\in c(y)$" be closed in the product topology; that is, the pattern itself should also be closed.
Every nonstandard $M\models PA$ has a corresponding closed set pattern on $\mathcal{S}(M)$ given by $$c_M: r\mapsto [r]\cap SS(M).$$ If $M$ is countable the space $\mathcal{S}(M)$ is homeomorphic to the rationals, so any interesting behavior is concentrated on $c_M$.
I would like to understand how closed set patterns of the form $c_M$ behave, and the following seems a good starting point. Say that closed set patterns $c_1,c_2$ on $\mathcal{X},\mathcal{Y}$ respectively are equivalent (and write $c_1\sim c_2$) if they differ by a homeomorphism - that is, if there is an $H:\mathcal{X}\cong\mathcal{Y}$ satisfying $$x\in c_1(y)\leftrightarrow H(x)\in c_2(H(y))$$ for all $x,y\in\mathcal{X}$. My question is:
Are there countable nonstandard $M,N\models PA$ such that $c_M\not\sim c_N$?
The knee-jerk approach to a positive answer would be a back-and-forth argument, but since the assignment of closed sets to reals isn't continuous in any good sense that doesn't seem to work here. On the other hand, I don't even see how to start approaching a negative answer.
Note that any countable Scott set is the standard system of a model of PA so I could have talked about those instead; my choice of language reflects my motivation ($c_M$ can be thought of as a "type-2 standard system" of $M$ in a sense), as does my choice in tags. The "reverse-math" tag is rather dubious, though, and I have no real objection to it being deleted if that connection is considered too weak.
I'm also happy to alter the coding $r\mapsto [r]$, but that does seem to be the most natural choice at the moment.
I have an idea that would seem to indicate that the answer is yes, but there's some computability facts you're going to have to fill in for me. The idea is to build a Scott set that is closed under Cantor-Bendixson derivatives, i.e. for any $r$ for every countable $\alpha$, there is an $s$ such that $[s]$ is the $\alpha$th Cantor-Bendixson derivative of $[r]$. This is only really countably many additional sets, so you can easily build a Scott set with this property.
So the point is that for any $\Pi^0_1$ class $F$ in this Scott set, the Scott set computes every point in $F$ with ordinal CB rank. I want to say that a dumb Scott set, such as one in which everything is low, does not have this property, and I also want to say that this property is visible in corresponding closed set patterns (because of the easy observation that the topological closure of $c_M(r)$ in $2^{\omega}$ is $[r]$).
Actually that last part is only easy if you're requiring that the homeomorphism is the restriction of an ambient homeomorphism, but it might still work without that requirement.
@JamesHanson That computability-theoretic fact is certainly true - the degrees of ranked points of lightface $\Pi^0_1$ classes are arbitrarily high in the hyperarithmetic hierarchy. The topological aspect is less clear to me, and I'd rather not bring in the ambient Cantor space (although it certainly wouldn't be uninteresting to do so).
I'm thinking that in a dumb Scott set there should be a relatively low bound on the Cantor-Bendixson rank of $c_M(r)$, as computed in $SS(M)$, but the smart Scott set will have sets with much larger Cantor-Bendixson ranks. If we want to be more certain that this will work we can throw a copy of a very large countable ordinal into the starting set when we build the smart Scott set. (Or more specifically, a code for a closed set with a very large CB rank.)
Actually really that should be the starting point. There's a bound on the CB ranks of $c_M(r)$ (computed in $SS(M)$) for $r$ in some given Scott set and then we can build another Scott set containing much larger CB ranks in that way.
Given any topological space $X$ and subset $F\subseteq X$, define the Cantor-Bendixson sequence of $F$ in $X$ as:
$F^{(0)} = F$
$F^{(\alpha +1)} = F^{(\alpha)} \setminus \{x \in F^{(\alpha)} : x \text{ is isolated in }F^{(\alpha)}\}$
$F^{(\beta)} = \bigcap_{\alpha < \beta} F^{(\alpha)}$, $\beta$ a limit ordinal.
Now we'll define the CB-rank of $F$, written $CB(F)$, to be the least ordinal $\alpha$ such that $F^{(\alpha)} = F^{(\alpha +1)}$ (I think this is a slightly non-standard definition). Note that this doesn't actually depend on the ambient space $X$. The typical argument gives us that for second countable $F$, $CB(F) < \omega_1$ (specifically, take a countable base for the topology on $F$, each set in this base can only be removed at most once in the sequence), and crucially $CB(F)$ only depends on the topological properties of $F$.
Fix a non-standard model $M$ of $PA$. Now, since $\mathcal{S}(M)$ is countable, we have that $\gamma = \sup _{r \in \mathcal{S}(M)}CB(c_M(r))$ is also a countable ordinal.
Now fix a countable closed subset $F \subseteq 2^{\omega}$ with $CB(F) > \gamma$ (such a set always exists, since $\alpha$ is countable). Pick a real $r$ such that $[r] = F$ and now take a countable model $N$ of $PA$ such that $r \in \mathcal{S}(N)$ and $F \subseteq \mathcal{S}(N)$. This is always possible by your comment that every Scott set is the standard system of some countable model of $PA$. (EDIT: But also just compactness and the downward Löwenheim–Skolem theorem, since we don't really care about the particular Scott set in question.)
So now clearly we have $c_N(r) = F$, so $CB(c_N(r)) = CB(F) >CB(c_M(s))$ for every $s \in \mathcal{S}(M)$, and thus we have $c_M\not\sim c_N$.
Here's another way to apply Cantor-Bendixson derivatives (following James Hanson): only some $M$s have the property that $ran(c_M)$ is closed under (single) Cantor-Bendixson derivatives. Specifically, let $T$ be the downward closure of the set of strings of the form $0^n1^k0^s$ such that $s=0$ or $\Phi_n(n)$ has halted by stage $k$ - so that the non-isolated paths of $T$ (besides the all-$0$s path) are those of the $0^n1^\infty$-form for $n$ in the halting problem. Any tree $S$ with $[S]=CB([T])$ would enumerate the complement of the halting problem: $n$ is not in the halting problem iff the part of $S$ above $0^n1$ eventually dies out. In particular, if $\mathcal{S}(M)$ doesn't contain the halting problem then $ran(c_M)$ won't contain $[CB(T)]\cap \mathcal{S}(M)$.
We can also use prunings. For $r\in \mathcal{S}(M)$, let $B_r=\{s\in\mathcal{S}(M): c_M(s)\supseteq c_M(r)\}$. Then we have that $B_r\in ran(c_M)$ for all $r$ iff in $\mathcal{S}(M)$ every tree has a pruning (= subtree with no dead ends and the same paths), which is of course equivalent to being arithmetically closed.
The right-to-left direction is essentially immediate: if $P$ is pruned then $[T]\not\supseteq [P]$ iff for some $\sigma\in P$ we have $\sigma\not\in T$, which is an open condition. In the left-to-right direction, note that a code for $B_r$ lets us enumerate the extendible nodes of the tree coded by $r$ ($\sigma$ is extendible in the tree coded by $r$ iff the real coding the tree of strings $\not\succcurlyeq\sigma$ is not in $B_r$), and the non-extendible nodes of the tree are a priori (relatively) computably enumerable.
Two final remarks:
Note that when we shift attention to the $\omega$-models of $WKL_0$ given by the standard systems, the two arguments above are pointing at $ACA_0$: for $M\models PA$ nonstandard, $ran(c_M)$ is closed under (single) Cantor-Bendixson derivatives iff $B_r\in ran(c_M)$ for all $r\in \mathcal{S}(M)$ iff $\mathcal{S}(M)$ is arithmetically closed. I've followed up on this line of thought here.
All these arguments so far leave open the problem of whether we can have $c_M\not\sim c_N$ for "finer" reasons. Specifically, for $c_1,c_2$ closed set patterns on $\mathcal{X},\mathcal{Y}$, write $c_1\approx c_2$ if there is some $H:\mathcal{X}\cong\mathcal{Y}$ such that $ran(c_2)=\{H[A]: A\in ran(c_1)\}$; then we can ask whether there countable nonstandard $M,N\models PA$ with $c_M\not \sim c_N$ but $c_M\approx c_N$, and I don't see how to attack this at the moment.
|
2025-03-21T14:48:29.882065
| 2020-02-20T18:47:32 |
353186
|
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|
Stack Exchange
|
Rigidity for the category of comodules over a Hopf algebra
On this page
https://ncatlab.org/nlab/show/rigid+monoidal+category
there is a discussion of rigidity (left-right duality) for the catagory of
modules over a Hopf algebra. What happens if we look at
the category of comodules over a Hopf algebra? Do we still
get a rigid category? If so, does there exist any general relationship between the left and right duals?
For modules for a Hopf algebra you need to be careful about finite dimensionality. Even for the trivial Hopf algebra (whose modules are vector spaces) only the subcategory of finite dimensional modules is rigid. I think the situation for comodules is cleaner than for modules because for comodules finite dimensionality can be rephrased in terms of finite generation. So I think you get that the category of finitely generated (presented?) comodules is rigid despite not getting that result for modules.
The category of the finite dimensional comodules of a hopf algebra over a field, is a rigid, monoidal category. (just as the category of the finite dimensional modules).
If we take a fin dim, right comodule $X$, of the hopf algebra $H$ then its dual vector space $X^*$, equipped with the unique comodule structure satisfying $e_V(\phi_{(0)}\otimes v)\phi_{(1)}=e_V(\phi\otimes v_{(0)})S(v_{(1)})$ provides the corresponding left dual object, i.e. the right $H$-comodule $X^*$ which is the left dual of the right $H$-comodule $X$. (I am not sure however for the last part of your question: if there exists some general relationship between left and right dual objects).
In fact, more can be shown. In On hopf algebras and rigid monoidal categories, Isr. J. of Math. v.72, p.252–256, (1990), Ulbrich has shown that:
An ordinary bialgebra $H$, over a field $k$, is a hopf algebra if and only if its category of finite dimensional comodules is rigid.
There, the author essentially uses the dual object to "reconstruct" an antipode for the original bialgebra.
|
2025-03-21T14:48:29.882257
| 2020-02-20T21:48:38 |
353193
|
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|
Stack Exchange
|
Dual of a projective module
Let $R$ be a noncommutative ring with unit, let $P$ be a projective left $R$-module, and denote $^{\vee}\!P := \,_R\mathrm{Hom}(P,R)$. One often sees it written that projectivity implies an isomorphism
$$
\mathrm{ev}:\,^{\vee}\!P \otimes_R P \to R, ~~ \phi \otimes p \mapsto \phi(p).
$$
But I don't see that this is well defined! Consider
$$
(\phi.r) \otimes p \mapsto \phi.r(p) = (\phi(p))r
$$
where we have used the definition of the right $R$-module structure of $^{\vee}\!P$. Compare this with
$$
\phi \otimes (r.p) \mapsto \phi(r.p) = r \phi(p),
$$
where we have used the fact that $\phi$ is a left $R$-module map. Now since $R$ is not commutative, these two values are not guaranteed to be equal, and the evaluation map is not guaranteed to be well-defined. What is wrong here?
Sorry for my misunderstanding earlier. Your question, and the answers below, seem to be connected to Exercise 2.20(4), in T.Y. Lam's "Lectures on Modules and Rings". When $P$ is a f.g. projective left $R$-module, then $P^{\ast}\otimes_R P\cong {\rm End}(_RP)$.
Did you mean to say "morphism" instead of "isomorphism"?
You are right. There is no such a map as the one you are trying to describe.
Here is a map that actually exists. Let $R$ and $S$ be two noncommutative rings with units, and let $P$ be an $R$-$S$-bimodule. Consider the $S$-$R$-bimodule $Q={}_R\mathrm{Hom}(P,R)$. Then the evaluation is an $R$-$R$-bimodule map
$$
\mathrm{ev}\colon P\otimes_S Q\to R, \qquad (p\otimes\phi)\mapsto \phi(p),
$$
with the tensor product taken over the ring $S$.
In particular, if you do not have an $R$-$S$-bimodule but only a left $R$-module $P$, you can take $S=\mathbb Z$. Then you get an $R$-$R$-bimodule map
$$
\mathrm{ev}\colon P\otimes_{\mathbb Z}Q \to R,
$$
with the tensor product taken over the ring integers.
Notice that the $S$-$R$-bimodule $Q$ is (generally speaking) very different from the $S$-$R$-bimodule $Q'=\mathrm{Hom}_S(P,S)$. For the bimodule $Q'$, the evaluation is an $S$-$S$-bimodule map
$$
\mathrm{ev}\colon Q'\otimes_R P\to S, \qquad (\psi\otimes p)\mapsto \psi(p).
$$
Thanks for the answer. But it seems that $Q \otimes_R P$ is well defined as a $\mathbb{Z}$-module. So is the issue here that $Q = ,_R\mathrm{Hom}$ is really a "right dual" for $P$ and not a "left dual"? (Here I mean right and left dual in rough analogy with the monoidal category sense.)
I've added a paragraph at the end of the answer which is relevant to your question in the comment. Yes, this is the difference between the left and right duals.
I think that clears it up - thanks!
Doc, you ain't write no evaluation map. If $R$ is commutative, you write the trace map. If $R$ is noncom, god knows what you write. The evaluation map, that is an isomorphism for a finitely generative projective generator and a homomorphism of $R$-$R$-bimodules, in general, is
$$
p \otimes \phi \mapsto \phi (p), \ P \otimes_{End_RP} P^{\vee} \rightarrow R \, .
$$
Use it with care.
In terms of Leonid's answer, you have a canonical $S:=End_RP$, lying around.
Taking $S$ to be the endomorphism ring of $P$ over $R$ is a good idea, but still the evaluation map is not an isomorphism for every finitely generated projective left $R$-module $P$ (not even in the commutative case). Take $P=0$ to see that the evaluation map does not need to be an isomorphism for a finitely generated projective. I guess the evaluation (with $S={}_R\mathrm{Hom}(P,P)^{\mathrm op}$) is an isomorphism when $P$ is a finitely generated projective generator of $R{-}Mod$.
Right! I will add this condition
What is a "finitely generative projective generator "?
A generator, which is finitely generated and projective. See https://en.wikipedia.org/wiki/Generator_(category_theory)
A projective left $R$-module $P$ is a generator of $R{-}Mod$ if and only if for every nonzero left $R$-module $M$ there exists a nonzero $R$-module morphism $P\to M$. Equivalently, for every left $R$-module $M$ there exists a surjective left $R$-module morphism onto $M$ from a direct sum of sufficiently many copies of $P$. The general definition of what it means for an object of a category to be a generator is slightly more complicated; see e.g. the link above.
|
2025-03-21T14:48:29.882566
| 2020-02-20T21:48:59 |
353194
|
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"Dieter Kadelka",
"Jacob Lu",
"Mateusz Kwaśnicki",
"Nate Eldredge",
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|
Stack Exchange
|
Construct a random time such that the strong Markov property of Brownian motion fails
Let $\{B_t, \mathcal{F}_t; t\ge 0\}$ be a standard, one-dimensional Brownian motion. Can we construct a random time $S$ such that $P[0\le S < \infty] = 1$ and $W_t = B_{S+t} - B_S$ is not a Brownian motion?
Of course we can! For example, set $S$ to be the last time $t$ such that $B_t = -t$. Then $W_t > -t$ for every $t$, and hence $W_t$ is not a Brownian motion.
Or, perhaps even simpler: let $S$ be the first time $t$ such that $B_{t+1} = B_t$. Then $W_1 = 0$ with probability one.
I don't understand. Is not $P[S < \infty] = 0$?
@DieterKadelka: You mean in Mateusz's first example, where $S$ is the last time that $B_t = -t$? No, this is finite a.s., using for instance the strong law of large numbers.
Thank you, guys! Your comments are really helpful!
|
2025-03-21T14:48:29.882659
| 2020-02-20T22:02:25 |
353195
|
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"Richard Stanley",
"Sam Hopkins",
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"https://mathoverflow.net/users/2807"
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"url": "https://mathoverflow.net/questions/353195"
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|
Stack Exchange
|
Is the order complex of open Bruhat intervals polytopal?
Let $P$ be the Bruhat order of a Coxeter group, and let $s<t$ in
$P$. The set $\Delta(s,t)$ of all chains of the open interval $(s,t)$
(called the order complex of $(s,t)$) is a simplicial complex. By
the work of Björner and Wachs, the geometric realization of
$\Delta(s,t)$ is homeomorphic to a sphere. What is known about when
$\Delta(s,t)$ is polytopal, i.e., is the boundary complex of a
simplicial (convex) polytope?
Related question: Bjorner-Wachs show that [s,t] is the face lattice of a regular cellular ball. Is that cell complex ever polytopal?
@SamHopkins: there is some information at https://arxiv.org/pdf/2001.05011.pdf. In particular, $[s,t]$ is polytopal if and only if it is a lattice if and only if it is a boolean algebra.
|
2025-03-21T14:48:29.882753
| 2020-02-20T23:17:32 |
353199
|
{
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"Alexandre Eremenko",
"Lira",
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|
Stack Exchange
|
Analytic function on $\mathbb{C}$
Is there is a holomorphic function $g:\mathbb{C}\to \mathbb{C}$ so that $$\frac{|g'(z)|}{1+|g(z)|^2}=\frac{c}{1+|z|^2}$$ for some $c>1$ and all $z\in \mathbb{C}$?
There is no such function. Indeed, $h(z):=g(1/z)$ satisfies the same functional equation
$$\frac{|h'(z)|}{1+|h(z)|^2}=\frac{c}{1+|z|^2}.$$
In particular, $|h'(z)|>c/2$ for $|z|<1$, hence $h'(z)$ does not have an essential singularity at $z=0$. So $g(z)$ is a polynomial, and then letting $z\to\infty$ in the original functional equation shows that $g(z)=az+b$ with $|a|=1/c$. Then, in the original functional equation, the maximum of the left hand side equals $1/c$, while the maximum of the right hand side equals $c$. So $1/c=c$, which contradicts $c>1$.
No, there are no such functions with $c>1$. Your condition means that they expand the spherical metric. If you look at the area of the whole sphere, you obtain a contradiction.
More precisely, looking at the area of the sphere and the area of its image
we conclude that the area of the image is finite, therefore your function is rational (and $c$ is its degree). But a rational function of degree $>1$
must have a critical point the derivative is 0 at this point, which is in contradiction with your assumption.
If you take $w=z^2$, you get formula $$\int_{\mathbb{C}}|w'(z)|^2/(1+|w(z)|^2)^2 d x dy = 2\pi > \int_{\mathbb{C}}1/(1+|z|^2)^2 d x dy=\pi$$, so why this is a contradiction?
Because $(dw/dz)(0)=0$ in your example, while the assumption excludes this.
|
2025-03-21T14:48:29.882893
| 2020-02-20T23:53:04 |
353201
|
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"url": "https://mathoverflow.net/questions/353201"
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|
Stack Exchange
|
Dimension of $\ell$-adic Eilenberg-Maclane space
I'm currently studying the $\ell$-adic cohomology functor, i.e. the functor $$F:X \rightarrow H^i_{ét}(X,\mathbb{Q}_{\ell}).$$
In some sense, this is a representable functor, i.e. there exists an $\ell$-adic Eilenberg-Maclane space (see Representability of Weil Cohomology Theories in Stable Motivic Homotopy Theory). I would like to know the dimension of this space. The normal way to compute the dimension of a moduli space is to study deformations. However, since the étale site of a scheme is determined by the underlying reduced structure, we see that
$$F(X[\epsilon])=H^i_{ét}(X[\epsilon],\mathbb{Q}_{\ell})\cong H^i_{ét}(X,\mathbb{Q}_{\ell})=F(X).$$
However, confusingly, this implies that this analogue of the Eilenberg-Maclane space is zero dimensional, which seems very wrong to me. My guess would be that square-zero extension do not compute the dimension of a motivic spectra. Therefore it seems natural to ask how we can compute this dimension.
I see this is one of your first question on MO -- welcome! The topic of the question is certainly interesting but I think you need to put a little more effort in future questions into being clear (first and foremost with yourself) about the nature of the objects you're asking about. You don't need to understand the details of all the definitions in order to ask a question, but you shouldn't conflate a topological space and a vector space, for example, and if you have a fuzzy understanding of the distinction, ask a separate question about that first.
In particular there does not exist an $\ell$-adic Eilenberg MacLane space. Rather, there exists an ($\ell$-adic etale) Eilenberg-MacLane motive, which is an object of a different category from spaces. If you want to think of it as a "space-like" object, i.e., get an analog of the algebraic topology functor "Connective Spectra" --> "Topological spaces" which takes a spectrum to the underlying infinite loop spaces while replacing "Connective Spectra" with "Motivic spectra" (with some connectivity condition), you can absolutely not replace "spaces" with "algebraic varieties" (so, you do not expect motivic cohomology theories to be representable by algebraic varieties: think for example of the motive $B\mathbb{G}_m$, which admits nontrivial maps from a curve which are trivial outside a point). The weakest modification you could hope to make is to replace "spaces" with "algebraic varieties up to $\mathbb{A}^1$ homotopy equivalences", which already destroys the notion of dimension (since $*\cong\mathbb{A}^1$). Alternatively, you could replace spaces by "$\infty$-stacks" (essentially, simplicial spaces up to a notion of homotopy equivalence), but again the notion of dimension would get lost.
Moreover, even in topology, the Eilenberg-MacLane spaces can be extremely infinite-dimensional. As you perhaps know, the second Eilenberg-MacLane space of the integers, $K(\mathbb{Z}, 2),$ is $\mathbb{C}P^\infty,$ which has infinitely many nonzero cohomology groups. Things only get worse when you replace $\mathbb{Z}$ by $\mathbb{Q}_\ell$ and pass to motives.
This is not to say that you can't assign a dimension to this motive. For example dualizable objects in symmetric monoidal categories have a notion of dimension, and in the symmetric monoidal category of etale $\ell$-adic motives, the object you are interested in is the unit object and has dimension one.
|
2025-03-21T14:48:29.883420
| 2020-02-21T01:49:59 |
353210
|
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"Alexey Ustinov",
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|
Stack Exchange
|
Estimating certain short Kloosterman sums
Recall that for the classical Kloosterman sum
$$ K(a,b,p^t):= \sum_{x \in (\mathbb{Z}/ p^t \mathbb{Z})^* } \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$
where $\psi(x)=e^{2\pi ix}$, $a,b,t$ are natural numbers and $p$ is a fixed prime number. We have the Weil bound, i.e.
$$ \vert K(a,b,p^t) \rvert \leq (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$
Now if we consider the following short Kloosterman sum
$$ K'(a,b,p^t):= \sum_{x \in A} \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$
where $A:=\left \{ x \rvert x \equiv 1 \mod p^m,\;x \in (\mathbb{Z}/ p^t \mathbb{Z})^* \right \}.$ Here m is a fixed positive integer and we can also assume that $t>>m$. My question is that can we also achieve an analogy of Weil bound for the above short Kloosterman sum $K'(a,b,p^t)$? Hopefully, I expect the following bound
$$ \vert K'(a,b,p^t) \rvert \leq A_m \cdot (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$
Here $A_m$ is a positive constant only depend on the choice of $p$ and $m$.
Since we assume that $t>>m$, then the number of elements in the finite set $A$ is greater than $\sqrt{p^t}$. Then from the “short Kloosterman sums" in Wikipedia, we may achieve an analogy of Weil bound for the above short Kloosterman sum. However, I cannot find any references. So any ideas or references for the bound of above short Kloosterman sum are welcome.
If $x=1+p^my$ ($1\le y\le p^{t-m}$) then
$$\psi \left(\frac{ax+bx^{-1}}{p^t} \right)=\psi \left(\frac{a(1+p^my)+b(1+p^my)^{-1}}{p^t} \right)=\psi \left(\frac{a+b}{p^t} \right)\psi \left(\frac{f(y)}{p^{t-m}} \right),$$
where $f(y)=ay+b(-y+y^2p^m-y^3p^{2m}+\cdots)$. The sum
$$\sum_{y=1}^{p^{n}} \psi \left(\frac{f(y)}{p^{n}} \right)$$
can be calculated explicitly, because $f(y)$ has a nice form (almost all coefficients are divisible by $p$), see Lemma 2.1 in Generalized Twisted Kloosterman Sum Over ℤ[i] by S. Varbanets. The main idea is to take $y=y_0+y_1p^{n-1}$, where $1\le y_0\le p^{n-1}$, $1\le y_1\le p.$
The sum becomes linear over $y_1$. Apply this idea twice.
For $n\ge 1$
$$f(y)\equiv f(y_0)+f'(y_0)y_1p^{n-1}\equiv f(y_0)+(a-b)y_1p^{n-1}\pmod{ p^n}.$$ So
$$S_n(a,b)=\sum_{y=1}^{p^n}e_{p^n}(f(y))=\sum_{y_0=1}^{p^{n-1}}\sum_{y_1=1}^{p}e_{p^n}(f(y_0)+(a-b)y_1p^{n-1})=p\delta_p(a-b)\sum_{y_0=1}^{p^{n-1}}e_{p^n}(f(y_0)),$$
where $e_N(x)=e^{2\pi ix/N}$ and $$\delta_q(x)=\begin{cases}
1,& \text{ if }q\mid x;\\
0,& \text{ if }q\nmid x.
\end{cases}$$
This sum does not vanish if $a\equiv b\pmod{p }$. We may also assume that $a\equiv b\not \equiv0\pmod{p }$ because otherwise original sum can be simplified: for $a=pa_1$, $b=pb_1$
$$\sum_{x \in (\mathbb{Z}/ p^n \mathbb{Z})^*}e_{p^n}(ax+bx^{-1})=p\sum_{x \in (\mathbb{Z}/ p^{n-1} \mathbb{Z})^*}e_{p^{n-1}}(a_1x+b_1x^{-1}).$$ Let $a=b+p^\alpha a_1$, $\alpha\ge 1$, $(a_1,p)=1$. Then
$f(y)=a_1p^\alpha y+b(y^2p^m-y^3p^{2m}+\ldots)$, and
$$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-1}}(f(y)p^{-1}).$$
If $\alpha\ge m$ then
$$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-m}}(g(y))=p^m \sum_{y=1}^{p^{n-m}}e_{p^{n-m}}(g(y)),$$
where $$g(y)=a_1p^{\alpha-m}y+b(y^2-y^3p^{m}+\ldots),$$
and one can apply Lemma 2.1 from the cited article.
If $\alpha<m$
then
$$S_n(a,b)=p^\alpha \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y)),$$
where $$g(y)=a_1y+b(y^2p^{m-\alpha}-y^3p^{2m-\alpha}+\ldots).$$
Again $y=y_0+y_1p^{n-\alpha-1}$, $1\le y_0\le p^{n-\alpha-1}$, $1\le y_1\le p$
$$g(y)\equiv g(y_0)+g'(y_0)y_1p^{n-\alpha-1}\equiv g(y_0)+a_1y_1p^{n-\alpha-1}\pmod{ p^{n-\alpha}}.$$
$$S_n(a,b)= \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y))= \sum_{y_0=1}^{p^{n-\alpha-1}}\sum_{y_1=1}^{p}e_{p^{n-\alpha}}(g(y_0)+a_1y_1p^{n-\alpha-1})=0,$$
because sum over $y_1$ vanishes.
Thanks a lot! Is there any similar result for $p$ equals 2?
I think maybe we cannot apply Lemma 2.1 in your reference. Since Lemma 2.1 requires that the coefficient of $y^2$ should be relatively prime to $p$. However, the polynomial in your calculation does not satisfy this condition.@Alexey Ustinov
@JACK You may repeat the same calculations for $p=2$. Lemma 2.1 not always applied directly. I've added more details in my answer.
Thanks! @Alexey Ustinov I think for $p$ equals 2 case, we may not repeat the same calculations since Lemma 2.1 cannot apply to this case. So maybe we need some modifications.
@JACK For $p=2$ take $y=y_0+y_1p^{n-\alpha-2}$. It works for $m>1$. For $m=1$ take $y=y_0+y_1p^{n-\alpha-3}$.
Dear Alexey, for $p=2$ case, you still need to consider whether $\alpha \geq m$ or not. In $\alpha \geq m$ case, you need to apply Lemma 2.1 in your reference. However. I do not think that your Lemma 2.1 has a $p=2$ version. We need to do some modification. @Alexey Ustinov
|
2025-03-21T14:48:29.883665
| 2020-02-21T02:57:50 |
353214
|
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"Dmitry Vaintrob",
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"url": "https://mathoverflow.net/questions/353214"
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|
Stack Exchange
|
Finiteness results in the category of schemes up to $\mathbb{A}^1$-homotopy
In algebraic geometry, we know that there exist geometrical conditions on a scheme $X/k$ for having finitely many rational points when $k$ is a number field. Namely for curves there is the Mordell conjecture, and in higher dimensions, the Lang conjecture states that if $X(\mathbb{C})^{an}$ is hyperbolic, then $|X(k)|$ is finite. I would be interested to generalize this train of thought to the category of schemes up to $\mathbb{A}^1$-homotopy. Concretely, my question is whether or not there exists a (conjectural) geometrical condition on $[X]$, an $\mathbb{A}^1$-homotopy class, such that the set of morphism $\text{Hom}_{\mathbb{A}^1\text{-hom}}([\text{Spec } k],[X])$ in the category of schemes up to $\mathbb{A}^1$-homotopy is finite.
A naive translation of Faltings Theorem of course doesn't work, as for instance $\mathbb{A}^1_{\mathbb{Q}}$ has infintely many rational points, but $\text{Spec }\mathbb{Q}$ does not.
This mapping space is (I think) the same as the space of rationally connected components. In particular naive Mordell still works, since no two different points of a curve of genus $\ge 1$ are rationally connected!
|
2025-03-21T14:48:29.883784
| 2020-02-21T03:39:26 |
353215
|
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"authors": [
"Anonymous",
"Todd Eisworth",
"Tomasz Kania",
"Wlod AA",
"YCor",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353215"
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|
Stack Exchange
|
Separability of subspaces of homogeneous topological spaces
Let $\ X\ $ be a homogenous separable topological space (i.e. for every $\ x\ y\in X\ $ there exists a homeomorphism $\ f:X\to X\ $ such that $\ f(x)=y,\ $ and there is a countable dense subset of $\ X).\ $
Question Is every subspace of $\ X\ $ separable?
The question allows several variations by considering different stronger kinds of homogeneity. The other direction would be to consider different separability properties like Hausdorff, normal, etc. (No, forget the metric case :) ).
I would conjecture that in the simple case of homogeneity there should be an example of $\ X,\ $ and of its subspace $\ Y\ $ which is not separable while $\ X\ $ is.
This fails even for topological groups. For example, the separable topological group $\mathbb{Z}^{\mathfrak{c}}$ contains a subgroup of uncountable cellularity (hence certainly not separable!). This is a result of my colleague Vladimir Uspenskij in the paper below:
Uspenskij, V. V., On the Suslin number of subgroups of products of countable groups, Acta Univ. Carol., Math. Phys. 36, No. 2, 85-87 (1995). ZBL0854.20064.
Be warned that I am not an expert on topological groups, even though I have heard many talks on them!
The homogeneous compact space ${0, 1}^{\frak c}$ is separable but contains the non-separable $\Sigma$-product with any base point.
That the subgroup $\mathbb{Z}^{(\mathfrak{c})}$ of finitely supported elements is not separable is immediate.
Todd, thank you for your answer. (I've upvoted it, and now I wait patiently or less patiently for more answers). Only after seeing your answer I recalled the Szpilrajn-Marczewski's theorem about the separability of the product of continuum of separable spaces. A generalization of this theorem for general cardinals (and arbitrary "separability") must be known, I am sure (it's also in my old UW script on topology for 2nd-year students).
Any compact homogeneous hereditarily separable space has size at most $\mathfrak{c}$ (this is a result of Ismail). Thus any compact homogeneous separable space of bigger size provides a counterexample (e.g. $2^\mathfrak{c}$ as in Anonymous' comment).
There are Banach spaces $X$ for which the dual space $X^*$ is separable in the weak* topology yet the unit ball $B_{X^*}$ of $X^*$ is not. A notable example is $X=J\!L_2$, the Johnson–Lindenstrauss space.
W.B. Johnson, and J. Lindenstrauss, Some remarks on weakly compactly generated Banach spaces, Israel J. Math. 17 (1974), 219–230. Correction ibid 32
No. 4 (1979), 382–383.
Tomek, is the said unit ball not separable under the topology induced by the weak* topology of $X^*$?
I trust you. :)
@WlodAA, yes, the ball is not weak*-separable as $JL_2$ does not embed into $\ell_\infty$.
|
2025-03-21T14:48:29.884110
| 2020-02-21T04:04:15 |
353217
|
{
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"authors": [
"Dongyang Chen",
"https://mathoverflow.net/users/41619"
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"url": "https://mathoverflow.net/questions/353217"
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|
Stack Exchange
|
A subspace generated by closed $G_{\delta}$ sets in $K$ between $C(K)$ and $C(K)^{**}$
Let $K$ be a compact Hausdorff space. We let $Z$ be the closed subspace generated by $\{\chi_{F}:F$ closed $G_{\delta}$ sets in $K\}$ in $C(K)^{**}$. My question is the following:
Question 1. Is $C(K)\subseteq Z\subseteq C(K)^{**}$?
Since each $f\in C(K)$ is Baire measurable, each such $f$ is the unifrom limit of a sequence of simple Baire measurable functions. But I want a stronger result:
Question 2. For $f\in C(K)$ and every $\epsilon>0$, does there exist $g=\sum\limits_{k=1}^{n}a_{k}\chi_{F_{k}}$ (where each $F_{k}$ closed $G_{\delta}$ sets in $K)$ such that $|f(t)-g(t)|<\epsilon$ for all $t\in K$?
Question 2 has a positive answer. For any $f \in C(K)$, let $a = \inf_{t\in K} f(t)$ and let $b = \sup_{t\in K} f(t)$. For any $n \in \mathbb{N}$ with $n > 0$, we can set $$ g_n = a + \sum_{k=1}^n \frac{b-a}{n} \chi_{F_k},$$ where $ F_k = \{t \in K : f(t) \geq a + \frac{k}{n}(b-a)\} $, which is a closed $G_\delta$ set ($F_k = \bigcap_{\ell \in \mathbb{N}}\{t \in K : f(t) > a + \frac{k}{n}(b-a) - \frac{1}{\ell + 1}\}$). Since the constant $a$ can be thought of as $a\chi_K$, $g_n$ is of the required form. By construction, $|f(t) - g(t)| \leq \frac{b-a}{n}$ for every $t\in K$, which we can clearly make arbitrarily small.
So Question 1 also holds true. Thanks, James.
|
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