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2025-03-21T14:48:31.079938
| 2020-05-25T13:46:07 |
361323
|
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|
Stack Exchange
|
Matrix inequality : trace of exponential of Hermitian matrix
I want to know whether the following inequality holds or not.
\begin{align}
(\mathrm{Tr}\exp[(A+B)/2])^2\leq(\mathrm{Tr}\exp A)(\mathrm{Tr}\exp B)\tag{1}
\end{align}
where $A, B$ are Hermitian matrices of the dimension $D$.
Note that if $A$ and $B$ commute, we can see (1) holds using the simultaneously diagonalizing basis and Cauchy-Schwarz inequality. The problem is the case where $A$ and $B$ do not commute.
You can prove it using the Golden-Thompson inequality $Tr (e^{A+B}) \leqslant Tr(e^{A} e^{B})$ and then applying the Cauchy-Schwarz inequality.
|
2025-03-21T14:48:31.080016
| 2020-05-25T14:43:22 |
361328
|
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"Noam D. Elkies",
"Steven Stadnicki",
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"paul Monsky"
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|
Stack Exchange
|
Conceptual explanations of the class numbers for the first few $\mathbb{Q}(\sqrt{p})$ with odd conductor
It's known that the class number of $\mathbb{Q}(\sqrt{p})$ is $1$ for all primes $p<229$.
Question: What would it be like for conceptual explanations of $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$ (equivalently, $\mathbb{Q}(\sqrt{p})$ having odd conductor)?
To clarify the question:
A conceptual explanation should treat the first few primes simutaneously, instead of a case-by-case analysis where a case is a single prime. (A case-by-case analysis with a finite number of cases that a priori covers the whole range of primes is allowed, e.g. the cases being p=1, 5 or 9 mod 12.)
For "the first few primes of form $4k+1$", I mean such continuous primes up to a bound, e.g. $5,13,17$ but not $5,17,29$. The argument should be able to cover primes in such a way.
To avoid trivialities, the conceptual explanations should cover at least $5, 13, 17$ and $29$.
An example of conceptual explanation would be like:
By Example 2.9 of Masley's paper Class numbers of real cyclic number fields with small conductor, the class number of such fields are odd.
The Minkowski bound gives $h(\mathbb{Q}(\sqrt{p}))<3$ for $p<36$. Thus we have established $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$: $5,13,17$ and $29$.
This explanation also works for cyclic cubic fields of conductor $7$ and $13$.
Bonus for explanations that are not specialized on real quadratic fields, e.g. the explanation presented above.
Tiny note that there are very few primes of the form p=9 mod 12. :)
We give a uniform approach to $p \leq 61$ by applying
analytic discriminant bounds to the Hilbert class field.
To be sure this is not entirely "conceptual", but then
some computation is needed even to deal with $p < 36$ using Minkowski.
If $p = 4k+1$ is prime then $K = {\bf Q}(\sqrt{p})$ has odd class number $h$,
so either $h=1$ or $h \geq 3$. If $h \geq 3$ then the Hilbert class field
$H_K$ is a totally real field of degree $2h \geq 6$ and discriminant $p^h$.
We can now apply to $H_K$ the known lower bounds on the discriminants of
totally number fields.
The Odlyzko bounds (see Table 4 on page 134 of
his
1990 paper give a lower bound of $7.941$ on the root-discriminant
$|{\rm disc}(F)|^{1/n}$ for any totally real field $F$ of degree $n \geq 6$.
Hence $p > 7.941^2 > 63$. So we have accounted for all $p \leq 61$.
If the zeta function $\zeta_F$ satisfies the Riemann hypothesis,
the lower bound improves to $8.143$. Unfortunately this does not
account for any more primes because $8.143^2 = 66.3+$ and
$65$ is not prime. Since there exists a totally real sextic field
of discriminant $300125 = 5^3 7^4$ (see the
LMFDB entry),
such bounds can never get us past $300125^{1/3} = 66.95+$,
so the primes $p \in [73, 197]$ must be dealt with in some other way.
Some complementary heuristics, too long for a comment.
Again start from the fact that the class number $h$ of $K = {\bf Q}(\sqrt{p})$
is odd if $p$ is a prime of the form $4k+1$. This time we compare with Dirichlet's class number formula, which here gives
$$
L(1,\chi_p) = \frac{2\log \epsilon}{\sqrt p} h,
$$
where the character $\chi_p$ is the Legendre symbol $\chi_p(n) = (n/p)$,
and $\epsilon$ is the fundamental unit of $K$.
We expect that $L(1,\chi_p) \approx 1$, so large $h$ go with small $\epsilon$.
A unit $\epsilon > 1$ in a real quadratic field of discriminant $D$
must be at least as large as $\frac12(m + \sqrt{m^2 \pm 4})$
for some odd integer $m$, with $D = m^2 \pm 4$.
If $D$ is prime then we must use the plus sign
(unless $m=3$, but then the fundamental unit is $(1+\sqrt5)/2$).
$\epsilon > \sqrt{p} - O(1/\sqrt{p})$ and
So, $2\log \epsilon / \sqrt{p} > \log p - O(1/p)$.
Setting $L(1,\chi_p) \approx 1$ and $h=3$ in the class number formula gives
$\sqrt{p} \approx 3 \log p$; the solution $p \approx 289$ is of about
the right size for the minimal example of $h>1$.
We're actually closer here than we deserved to be:
the solution of $L(1,\chi_p) \sqrt{p} = 3 \log p$ is quite sensitive to
the size of $L(1,\chi_p)$, and $L(1,\chi_{229}^{\phantom.}) = 1.075+$
is unusually close to $1$; for example, $L(1,\chi_p) > 2$ for
$p = 193, 241, 313, 337$, while $L(1,\chi_p) < 0.4$ for
$p = 173, 293, 677, 773$.
Most of the early examples of $h > 1$ have small $\epsilon$,
either with $p=m^2+4$ as above or the next-smallest possibility,
$\epsilon = m + \sqrt{p}$ with $p=m^2+1$. Indeed
this
LMFDB list of fields ${\bf Q}(\sqrt p)$ with $p<2000$ and $h>1$
begins with
$$
229 = 15^2 + 4,\
257 = 16^2 + 1,\
401 = 20^2 + 1,\
577 = 24^2 + 1,
$$
$733 = 27^2 + 4$,
and then two exceptions $p=761$ and $p=1009$ and nine further $p$
of which all but $1429, 1489, 1901$ are not of the form $m^2+4$ or $m^2+1$.
Moreover $229$ is the only second-smallest prime of the form $p = m^2 + 4$
that satisfies our analytic bound $p > 63$ --- and the smallest is
$p = 173$, which was our example of an unusually small $L(1,\chi_p)$.
Likewise the next two examples were $293 = 17^2 + 4$ and $677 = 26^2 + 1$,
which are conjectured to be the largest discriminants $p = m^2+4$ and $p = m^2+1$
for which ${\bf Q}(\sqrt p)$ has class number $1$.
Noam: In 2 articles published in Acta Arithmetica in 2003, A. Biro claims to have established the truth of the conjectures (of Yokoi and Chowla) mentioned in your final sentence, using extensive computer calculation. Is his claim generally accepted?
Thank you: I was not aware of this pair of theorems
(and didn't know or remember the attribution of the conjectures
to Yokoi and Chowla). I'll fix this in the next edit.
|
2025-03-21T14:48:31.080813
| 2020-05-25T15:33:22 |
361332
|
{
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"Mohan",
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|
Stack Exchange
|
Existence of a non-degenerate global section is an open property?
Setting:
$X$:projective surface over algebraically closed field $k$.
$T$:scheme over $k$.
$E$: Coherent sheaf on $X \times_k T$ , flat over $T$ and $\forall t \in T$, $E_t$ is rank 2 torsion-free sheaves.
Question
We assume that $\exists t \in T$ s.t. $E_t$ has a non-degenerate global section.
Then, is there $t \in U \subseteq T:$ open s.t. $\forall u \in U$, $E_u$ has a non-degenerate section ?
Originally, I was reading page 3, line 15 of this paper (https://arxiv.org/abs/alg-geom/9312011v1).
Any comment welcome! Thank you.
Edit:$E_t$ has a non degenerate global section means that there exists $s \in H^0(E_t)$ s.t by the $s$ we get a exact sequence
$ 0 \rightarrow \mathscr{O} \rightarrow E_t \rightarrow F \rightarrow 0$
where $F$ is a torsion free sheaf on X
What does non-degenerate global section mean to you?
Sorry.I added the definition of non-degenerate global section.
What if you take the family of $\mathscr L \oplus \mathscr L$ where $\mathscr L$ runs over all degree $0$ line bundles of a curve (or surface if you insist that $X$ is a surface)? Then the trivial line bundle has such a section, but the others don't have any nonzero global sections.
By upper semicontinuity, a much more plausible question would be the opposite: if $E_t$ does not have a [insert adjective] section, then does the same hold in a neighbourhood?
|
2025-03-21T14:48:31.080950
| 2020-05-25T15:38:27 |
361333
|
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"Its_me",
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"Nate Eldredge",
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|
Stack Exchange
|
Maximum of bounded expectations at a certain Borel set?
Assume ${\bf x} \in \mathbb{R}^n$ denotes a real-valued bounded random variable with a distribution measured on the Borel space $(\mathbb{R}^n,\mathcal{B}^n)$. Let $f:\mathbb{R}^n\to\mathbb{R}$ denote a bounded Borel measureable function. Then, the following expectation value for any Borel set $B$
$$
E[f({\bf x}){\bf 1}_{{\bf x}\in B}] = \int_B f({\bf x}) dP
$$
is bounded.
Is there any reason that the $\sup_{B\in\mathcal{B}^n} E[f({\bf x}){\bf 1}_{{\bf x}\in B}]$ occurs at a certain Borel set $B$ or could there just be convergence towards the supremum?
The conditions $f$ bounded and Borel measurable do not imply that the integral of $f$ over any Borel set is finite: just take $n=1$ and the identity function, and $B = \mathbb{R}$. Do you want to assume that in addition that $f$ is integrable over $\mathbb{R}$?
Yes, it's attained. Note that the desired expression can be written as $E[(f 1_B)(\mathbf{x})]$. Then it's clear that we get the maximum by taking $B = \{f \ge 0\}$, so that $f 1_B = f^+$, the positive part of $f$. Indeed, if $A$ is any other Borel set, then $E[(f 1_A)(\mathbf{x})] = E[(f^+ 1_A)(\mathbf{x})] - E[(f^- 1_A)(\mathbf{x})]$. But $f^+ 1_A \le f^+$ and $f^- 1_A \ge 0$.
Previous overly sophisticated argument, please ignore:
Let $\nu(B) = E[f({\bf x}){\bf 1}_{{\bf x}\in B}] = \int_B f({\bf x}) dP$; then $\nu$ is a signed measure on $\mathbb{R}^n$. The Hahn decomposition theorem guarantees that we can partition $\mathbb{R}^n = B_+ \cup B_-$, with $B_+, B_-$ Borel, such that $\nu(A) \ge 0$ for all Borel $A \subset B_+$ and $\nu(A) \le 0$ for all Borel $A \subset B_-$. In particular, if $B$ is any other Borel set, we have $\nu(B) = \nu(B_+) - \nu(B_+ \setminus B) + \nu(B \cap B_-)$. But $\nu(B_+ \setminus B) \ge 0$ by the theorem, and $\nu(B \cap B_-) \le 0$, so $\nu(B) \le \nu(B_+)$. Thus the maximum is attained at $B_+$.
Great, thanks a lot!
Please let me know: Was this question too simple for MathOverflow, therefore the down vote?
I didn't cast that vote, but it looks like someone else has voted to close the question as "not research level", so maybe someone else thinks so.
@Its_me: Oh, it's much simpler than I'm making it. Just take $B = { f \ge 0}$.
Ahh, I see. Thanks.
|
2025-03-21T14:48:31.081136
| 2020-05-25T15:44:01 |
361334
|
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|
Stack Exchange
|
Sampling i.i.d. variables with restrictions
General Problem: Suppose $X_1,\ldots,X_n \sim \mathbb{P}_X^{\otimes n}$ is a finite sequence of i.i.d. (real- or integer-valued) random variables. Suppose $A\subseteq \mathbb{R}^n$ is a set of "admissible configurations".
Are there efficient methods of sampling from the restriction of $\mathbb{P}_X^{\otimes n}$ onto $A$?
More specific context: I understand that above formulation of the problem is hopelessly general, therefore I would like to give more context on my actual setting. In my case, I am considering a model with $\mathbb{P}_X = \mathrm{Poi}(\lambda)$ and where $A$ is given as the solution set of a finite numbers of equations (representing parity constraints on the Poisson variables). More specifically, I am considering the random current model (cf. this survey article by Duminil-Copin).
Therefore a (perhaps more approachable) version of the problem would be as follows: Consider also a finite set of linear functions $F_1,\ldots,F_N \colon \mathbb{R}^n \to \mathbb{R}$ and an "admissible value set" $V_\mathrm{adm}$. Then let $A:= \{\mathbf{x} \in \mathbb{R}^n\mid F_i(\mathbf{x}) \in V_\mathrm{adm}, \, i=1,\ldots,N\}$.
Is it in this context possible to efficiently sample from the restriction of $\mathbb{P}_X^{\otimes n}$ onto $A$?
I would be glad about input either on the general of more specific formulation of the problem. Also hints to general literature on such problems would be very helpful.
EDIT: As requested, I will try to make the random current setting more explicit: One considers a simple graph $G = (\mathcal{V},\mathcal{E})$. For a subset $S\subseteq \mathcal{V}$, we define the "admissible set" $A_S :=\{ \mathbf{n} = \{\mathbf{n}_e\}_{e\in \mathcal{E}}\in \mathbb{N}_0^\mathcal{E}\mid \mathrm{deg}_\mathbf{n}(v) \text{ odd for $v$ in $S$ and even otherwise}\}$, where $\mathrm{deg}_\mathbf{n}(v) = \sum_{vw \in \mathcal{E}}\mathbf{n}_{vw}$.
Can you write down the parity constraints explicitly?
@IosifPinelis : I edited my question to explain the "random current setting" I am interested in.
So, your $n$ is the number of edges?
Exactly, to every edge $e$ there is a number $\mathbf{n}_e$ associated, which follows the law of i.i.d. Poisson processes restricted to $A_S$.
@MattF. :Yes, absolutely. From a more physical point of view, one can think of the numbers as (undirected) current strengths. Then, the vertices of odd degree are "sources/sinks" of the current.
@MattF. : I do not fully understand your comment. What do you mean by "assignment of weight"?
A natural approach would be Markov chain Monte Carlo (MCMC), which involves a Markov chain which has your desired measure as its equilibrium measure. At the very least you need this chain to be irreducible, and then you also want it to converge quickly to equilibrium.
If you are content with a sample that is "close to equilibrium" then you could run the chain for a long time, and take the final state as a sample.
If instead you want do design a way of sampling precisely from equilibrium, it might be a much harder problem, but there are methods such as "coupling from the past" that could have the potential to work.
Here is an example of the sort of chain that might be suitable for your problem. The analysis of the convergence properties will depend on the structure of your graph. To initialise the chain, you need to find some starting configuration which satisfies all the constraints. Then at each step of the chain, choose, from some suitable distribution, a cycle $v_0 \to v_1 \to \dots \to v_{n-1} \to v_0$ in $G$. Now resample the configuration on the edges of that cycle, according to the conditional distribution of those edges with the rest of the configuration held fixed. (In your model, to keep the parity constraints satisfied, you either need to change the parity of each of the values around the cycle, or to keep the parity of each of the values around the cycle unchanged.) If your cycle is small, then this could just be done by rejection sampling (keep sampling independent Poisson variables until they satsify all the constraints) - or there might also be a more tailored method.
This is some sort of "Glauber dynamics".
Alternatively, you can consider a more restricted set of transitions (for example, only steps that change the weight of every edge along the cycle by $\pm 1$ - for your model, such changes preserve the required parity constraints), calculating the probability of a move between two given configurations according to the ratio of the equilibrium probabilities of those configurations. This would be a sort of
Metropolis-Hastings algorithm.
EDITED TO ADD:
I realised there is a considerable simplification. You can sample first a model where the state of any edge is "even" or "odd". For this, you again take product measure, conditioned on satisfying the parity constraints at the vertices; now the product measure is the one that puts probability $p$ on "even" and $1-p$ on "odd", where $p$ is the probability that a Poisson($\lambda$) random variable is even.
Once you have sampled such a configuration (maybe via MCMC) you now assign values to edges, independently. Each "even" edge is sampled from Poisson($\lambda$) conditioned on being even, and each "odd" edge from Poisson($\lambda$) conditioned on being odd.
For the even/odd model, again a natural move for the MCMC would be to flip all the values around a cycle. More generally, if $x$ is a feasible configuration, then another configuration $y$ is feasible iff their sum (or equivalently difference, since everything is mod 2) $z=x-y$ is feasible for the case where all vertices are even. So a general feasible move is to take such a configuration $z$, and flip all the edges where $z$ is odd. The ratio of probabilities of the states before and after the move is easy to calculate (it's just $p/(1-p)^k$ where $k$ is the net change in the number of even vertices), and that's what you need for Metropolis-Hastings.
(I should stress that none of this actually addresses the question of whether the chain converges to equilibrium in a reasonable time. This may
depend on your graph and on the specifics of how you choose the MCMC updates.)
Thank you for your input, James! Using MCMC methods seems like the most natural approach to me as well, the problem being that I do not see how to construct a Markov chain with the correct invariant measure. In particular the "Glauber dynamics" you proposed seems difficult to implement, since the model does not satisfy (as far as I can see) a Domain Markov Property, so the conditional measure is not tractable. (Also, I wouldn't know from which distribution to pick the loops. Perhaps a uniform even subgraph or the Loop-O(1) measure might work ...)
I thought the conditional distribution given the rest of the graph would be easy. Isn't it just independent Poissons, subject to the parity constraints? And once you fix everything outside the cycle, the parity constraints just become constraints on the parity on the values of neighbouring edges in the cycle. There will always be two classes (as I mentioned, if the parity constraints are currently satisfied, then to keep them satisfied you either change the parity of all values around the cycle, or keep the parity of all values around the cycle unchanged). Maybe I misunderstood the model....
For how to choose the loops, the answer must depend on the graph. Suppose you were on a 2-dimensional lattice torus. Then I think a decent option would just be the 4-cycles around 1x1 faces. I think that gives an irreducible chain, because any other cycle can be constructed from a whole bunch of 4-cycles by taking symmetric differences.... but I could be mistaken there....
@PeterWildemann Another way to put it: although you can't calculate the probability $\pi(x)$ that you want, for a given feasible configuration $x$, you can calculate the ratio $\pi(y)/\pi(x)$ where $x$ and $y$ are two feasible configurations - it's just the same as the ratio of the unconditioned probabilities $\mathbb{P}_X^{\otimes n}(y)/\mathbb{P}_X^{\otimes n}(x)$. Knowing these ratios is exactly what you need in order to make Metropolis-Hastings work. (Obviously the game is not over at that point - you still need to deal with the question of convergence somehow....)
@PeterWildemann: I added an edit above
Yes, conditioned on the parity of the edges, the model indeed decouples, although I don't think that this really leads to any major simplification. However, your above comment seems quite helpful. I will spend some time on this and try to come back with my findings.
The idea of first sampling the "parity field" that you proposed actually turned out to lead to a nice solution! According to an article by Lupu and Werner (https://arxiv.org/abs/1511.05524), the odd edges in the random current satisfy the law of the socalled loop o(1) model. This model you can sample via the Markov chain that "flips along elementary loops". Moreover, the random current model is then obtained by an independent Poisson process as you explained.
|
2025-03-21T14:48:31.081756
| 2020-05-25T17:06:37 |
361338
|
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|
Stack Exchange
|
What does play the role of a subobject classifier for quotient objects?
It is known that in the category of sets the dualization of the notion of a subobject classifiers does not work because the only object admitting a morphism into an initial object is the empty set.
But if we look at the idea of a subobject classifier (to index subobjects), then we can see that in the category of sets quotient objects (defined as equivalence classes of epimorphisms that start from a given object) have canonical representatives: those are equivalence relations on the given set.
So, in the category of sets quotient objects of an object $a$ can be classified by the set $E(a)\subseteq \mathcal P(a\times a)$ of all equivalence relations on $a$.
Question. Has the set $E(a)$ of equivalence relations some categorial counterpart, which would index all quotient objects of a given object?
What you've written about sets works just as well in any topos. The equivalence relation $\sim$ associated to an epimorphism $f:a\to b$ can be described as the equalizer of the two maps obtained by composing $f$ with the two projections $a\times a\to a$. In fact, this construction of $\sim$ and the fact that $f$ can be recovered as the coequalizer of the projections restricted to $\sim$ needs much less than a topos; a regular category should suffice. Of course to then talk about $\sim$ being an "element" of $\mathcal P(a\times a)$ requires $\mathcal P$ and thus essentially a topos.
@AndreasBlass It is a bit strange that the existence of a quotient object classifier requires the existence of a subobject classifier. Then the general philosophy whould imply that the existence of a subobject classifier is equivalent to the existence of a (properly defined) quotient object classifier, which is a bit strange.
Check out https://mathoverflow.net/questions/7776/universal-property-for-collection-of-epimorphisms
@StevenGubkin Thank you for the link. Indeed, very close question. I tried to find something relatined on MO before writing this question but without success.
Even in a topos, there's only a partial parallel between subobjects and quotients. For any $a$, one has an "object of subobjects" $\mathcal P(a)$ and an "object of quotients" in the sense of an object of equivalence relations on $a$. One also has a universal subobject, $1\to\Omega$, of which all subobjects (of any $a$) are pullbacks. But there is no universal quotient, of which all quotients are pushouts. This seems (to me) to be an instance of a rather general set-theoretic phenomenon: Quotients are not as nice as subsets. (See next comment.)
Some examples: The subsets of $a$ form a Boolean lattice; the quotients of $a$ form only a modular lattice (or do they have more lattice-theoretic structure?). $\mathbb N$ has only countably many finite subsets but uncountably many finite quotients; as a result, Ramsey's theorem is a lot easier than the dual Ramsey theorem of Carlson & Simpson. If you assume AD instead of AC, all well-orderable subsets of $\mathbb R$ are countable, but $\mathbb R$ has immense well-orderable quotients.
@AndreasBlass This asymmetry concerns only the concrete category of Sets. In other categories there are other phenomena. This is like in cosmology: why the constants of our universe are such-and-such?Because we live in concrete universe with such-and-such contants and also with such category of sets, which is important for us. Very interesting philosophical problems, actually. I did not think that they are related also to mathematics. I thought that mathematics is above all those multiverse speculations. But who knows?
Is it really standard to define "quotient objects" using epimorphisms? for rings it clearly doesn't match the usual notion.
@YCor you could replace 'epimorphism' by any of the more subtle, stronger flavours of epimorphism, all of which are equivalent in $\mathbf{Set}$, but which are different in more general categories. Probably 'regular epi' is a reasonable choice, though I don't know where quotient rings fall on the list, if at all.
@AndreasBlass: Won't you need an effectivity condition in addition to regularity?
@AndrejBauer I think you're right.
One way to write the universal property of this object $E(a)$ is as follows:
a map $x \to E(a)$ is the same as an isomorphism class of epimorphism $x \times a \twoheadrightarrow k$ in $Set/x$, that is a diagram
$$ x \times a \twoheadrightarrow k \to x$$
whose composite is the first projection.
So it does "feel like" a subobject classifier, but it is not the dual notion (the dual of a subobject classifer, would have a universal property specifying what are morphisms out of it, not into it).
For this universal property to make sense in a category $C$, you need pullback of epimorphism to exists, or if you restrict to some specific class of epimorphism, that the class of epimorphism under consideration is stable under pullback.
You can also consider the universal property, equivalent in set, but that might be different in more general category:
a map $x \to E(a)$ is the same an an (isomorphism class of) equivalence relation on $x \times a$ which is included in $\Delta_x \times a \times a$.
Where "equivalence relation" on $z$ means subobject of $z \times z$ satisfying the usual stability properties...
This version of the universal property makes sense more generally in any category with finite limits (pullback of monomorphism is enough actually). The two are equivalent in an exact category.
As pointed out by Andreas Blass in the comment, in an elementary topos, these two universal properties make sense and defines a universal objects.
I have just noticed that in your (two) answers you replace epimorphisms (as isomorphic classes) by two other isomorphism classes. But the question was to avoid using isomorphism classes at all, because in NBG formalization of category theory, it is not allowed to form classes of classes. So, whereas subobject classifiers allow us to define an object indexing all subobjects, I do not see from your answer which object can index quotient objects. Maybe in topoi you can use the subobject classifier to do this job by analogy as it is done in the category of set. But what to do in general case?
@TarasBanakh : I'm affraid I don't understand your point. To me a "subobject classifier" is an object $\Omega$ such that $Hom(X,\Omega)$ identifies (functorially) with the class of isomorphisms class of monomorphisms $A \to X$. The fact that the framework you chose to work in cannot in general consider class of class in general doesn't means that on specific instances you cannot constructs a set isomorphic to it.
These are justs universal property, you can always rephrase them without mentioning these isomorphism class. for example an "epimorphism classifier E(a)" in the first part of my answer is an object coming with an epimorphism $E(a) \times a \twoheadrightarrow k \rightarrow E(a)$ such that for every diagram $X \times a \twoheadrightarrow p \rightarrow X$, there is a unique map $X \to E(a)$ such that the second diagram is obtained as a pull back of diagram above. This is exactly the same thing, but it no longer involves classes in anyway.
Thank you for the explanation. Then this is indeed what I expected: some existing object indexing quotient objects (which are equivalence classes).
|
2025-03-21T14:48:31.082267
| 2020-05-25T17:25:51 |
361343
|
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|
Stack Exchange
|
Is there a known condition for partial sums of a decreasing positive sequence to take all values up to the total sum?
Let $a_0>a_1>\cdots>0$ have the property that, for each positive $a<\sum_{n\in\Bbb N}a_n$ (admitting $\infty$ for the sum), there is $A\subset\Bbb N$ such that $a=\sum_{n\in A}a_n$ . Are there known necessary and sufficient conditions on the $a_n$ (not involving arbitrary partial sums) for this property? To illustrate, $a_n=1/(n+1)$ and $a_n=1/2^n$ possess the required property, but $a_n=1/(2+\varepsilon)^n$ does not for any $\varepsilon>0$.
(This question is adapted from one asked on Mathematics Stack Exchange two months ago which received no answer.)
Even deleting a single term (other than the first term) from the sequence $a_n = 1/2^n$ makes it no longer have your property.
I think it's necessary and sufficient to have $a_n \leq \sum_{m > n}a_m$ for all $n$. Is this the kind of condition you have in mind? I'm not sure what you mean by "not involving arbitrary partial sums".
@WillBrian : Yes, your sum condition doesn't involve indices comprising arbitrary subsets of $\Bbb N$, only terminal segments of $\Bbb N$. So it's not just a rewrite of the originally stated property.
OK, thanks -- I'll see if my idea works, and write an answer if it does :)
I wrote an answer a while ago on Math Stack Exchange that answered exactly this: here. I think it's entirely in line with Will Brian's answer.
@MiloBrandt : Your answer showed the sufficiency of the bound $a_n \leq \sum_{m > n}a_m$, but not the necessity (which wasn't needed to answer the question).
It is necessary and sufficient that
$\lim_{n \rightarrow \infty}a_n = 0$, and
$a_n \leq \sum_{m > n}a_m$ for all $n$.
In other words: the terms go to zero, and no term is bigger than the sum of all the following terms.
Necessity: First, it is necessary that $\lim_{n \rightarrow \infty}a_n = 0$ because you cannot form any sum smaller than $\lim_{n \rightarrow \infty}a_n$.
Suppose $a_n > \sum_{m > n}a_m$ for some $n$. Let $\varepsilon$ be some number with $0 < \varepsilon < a_n - \sum_{m > n}a_m$. Then I claim there is no $A$ such that $\sum_{m \in A}a_m = a_1+a_2+\dots+a_{n-1}+a_n - \varepsilon$. To see this, just consider two cases: (1) if $\{a_1,\dots,a_n\} \subseteq A$, then $\sum_{m \in A}a_m$ is too big, because $\varepsilon > 0$, and (2) if $a_i \notin A$ for some $i \leq n$, then $\sum_{m \in A}a_m$ is too small, because
$\sum_{m \in A}a_m \,\leq\, (a_1+a_2+\dots+a_n) - a_i +\sum_{m > n}a_m \,\leq\, (a_1+a_2+\dots+a_n) -a_n + \sum_{m > n}a_m < (a_1+a_2+\dots+a_n) - \varepsilon.$
Sufficiency: Suppose $a_n \leq \sum_{m > n}a_m$ for all $n$, and let $c$ be any number with $0 \leq c \leq \sum_{m \in \mathbb N}a_m$. Then we can construct the desired $A \subseteq \mathbb N$ recursively, as follows. If it has already been decided for all $m < n$ whether $m \in A$ or not, then put $n \in A$ if and only if $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq c$. (In other words, put $n \in A$ if and only if putting $n \in A$ does not make the sum too big.) Once we have built $A$ according to this rule, it is clear that $\sum_{m \in A}a_m \leq c$, because none of the finite partial sums exceeds $c$.
Now suppose, aiming for a contradiction, that $\sum_{m \in A}a_m < c$, and let $\varepsilon = c - \sum_{m \in A}a_m$. There is some $N$ such that $a_n < \varepsilon$ for all $n \geq N$. For each such $n$, we have $\sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq \sum_{m \in A} a_m = c - \varepsilon$, and hence $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m < c$. By our rule for constructing $A$, this means $n \in A$ for all $n \geq N$. In other words, $A$ is a co-finite subset of $\mathbb N$.
Let $n$ denote the largest member of $\mathbb N \setminus A$. (Note that $\mathbb N \setminus A \neq \emptyset$, because $\sum_{m \in A}a_m < c \leq \sum_{m \in \mathbb N}a_m$.) Then $\left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+a_n \leq \left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+ \sum_{m > n}a_m = \sum_{m \in A}a_m < c$. This is the contradiction we were after, because this tells us that we should have had $n \in A$, although $n$ was supposed to be the largest number not in $A$.
Your condition holds when e.g. $a_n=1+1/(n+1)$, whereas then $a=1$ is not a partial sum.
@IosifPinelis: I was assuming that $\lim_{n \rightarrow \infty} a_n = 0$. I just failed to notice that this wasn't part of the question. I'll edit to clarity. As for the rest of it, it looks like maybe we had pretty much the same idea at the same time!
$\lim_{n\to\infty}a_n=0$ is entailed by the original condition, because $a$ can be arbitrarily close to zero.
@JohnBentin: Yes -- this is the "necessity" argument for that half of my condition. But it should still be included in the answer because, as Iosif Pinelis points out in his comment, the other half of my condition is not enough by itself. It's the conjunction of the two statements that is both necessary and sufficient.
(+1) Thank you for a beautiful answer!
I will add a link to my on answer on [math.se] which contains some references: Surjective Function from a Cantor Set.
In order for all $a\in(0,a_0+a_1+\cdots)$ to be representable as partial sums of of the $a_i$'s, it is necessary that
\begin{equation*}
a_\infty:=\lim_n a_n=0; \tag{1}
\end{equation*}
otherwise, no $a\in(0,a_\infty)$ is a partial sum of the $a_i$'s. So, assume (1).
A sufficient condition is that for all natural $n$
\begin{equation*}
a_{n-1}\le a_n+a_{n+1}+\cdots \tag{2}
\end{equation*}
for all $n$. Indeed, assume (2) holds. Take any $a\in(0,a_0+a_1+\cdots)$. Successively define
\begin{equation*}
S_0:=\{k\ge0\colon a_k\le a\},\quad k_1:=\min S_0
\end{equation*}
and, for $j\ge2$,
\begin{equation*}
S_{j-1}:=\{k>k_{j-1}\colon a_k\le a-s_{j-1}\},\quad k_j:=\min S_{j-1},
\end{equation*}
where
\begin{equation*}
s_j:=a_{k_1}+\cdots+a_{k_j}.
\end{equation*}
If $S_{j-1}=\emptyset$ for some $j=1,2,\dots$, then, by (1), ($j\ge2$ and) $s_{j-1}=a$, so that we are done.
It remains to consider the case when $S_{j-1}\ne\emptyset$ for all $j=1,2,\dots$, so that we have $0\le k_1<k_2<\cdots$. Without loss of generality,
\begin{equation*}
s_j\le a-h
\end{equation*}
for some real $h>0$ and all $j$. By the construction, for each $j$ either
(i) $k_j=k_{j-1}+1$ or
(ii) $a_{k_j-1}>a-s_{j-1}$.
In case (ii), $a_{k_j-1}>a-s_{j-1}=a-s_j+a_{k_j}\ge h+a_{k_j}$. So, if case (ii) holds for infinitely may $j$'s, then, letting $j\to\infty$ and recalling (1), we get $0\ge h+0$, a contradiction.
So, case (i) holds eventually, for all large enough $j$. Then for some natural $m$ and $n$ we have $s_{m-1}+a_n+a_{n+1}+\cdots<a<s_{m-1}+a_{n-1}$, which contradicts (2). $\Box$
Why the downvote? Is there anything wrong with this answer?
(I did not downvote.) Is your first condition flipped? It currently says that it's necessary for the limit to be positive.
@aorq : Oops! Of course, I meant that $a_\infty$ must be $=0$, not $>0$. Thank you for the comment.
|
2025-03-21T14:48:31.082834
| 2020-05-25T17:40:02 |
361344
|
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"Denis T",
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|
Stack Exchange
|
Cohn localization examples
I'm working on my master's thesis, part of which involves an exposition on Cohn localization. (nlab discussion)
In Free ideal rings and localization in general rings, Sec 7.4, Cohn gives a construction for a ring $\Sigma^{-1}R$. Given a set of matrices $\Sigma$ (with a mild closure condition), this ring admits a homomorphism from $R$ which is universal with respect to the property that the image of each matrix in $\Sigma$ is invertible over $\Sigma^{-1}R$.
I understand the construction and its uses in finding conditions for embeddability of domains into skew fields and the existence of universal fields, but I would really like to have some concrete examples of what the construction actually gives.
There are a few trivial examples - if $R$ is commutative, then $\Sigma^{-1}R$ is just the ring of quotients of $R$ with denominator set comprised of the determinants of matrices in $\Sigma$. If $\Sigma$ contains the zero matrix, the Cohn localization is the zero ring.
But neither of these highlights what makes Cohn localization a novel idea or sheds any light on what "matrix inverting homomorphisms" look like away from the commutative case.
Cohn's book also lacks examples. Where else can I see some concrete and informative examples of the ring $\Sigma^{-1}R$?
Let $R$ be an arbitrary ring, and let $M$ be some $n\times n$ matrix over $R$. Universally inverting $M$ consists of first passing to the free product ring $R \ast_{\mathbb{Z}} \mathbb{Z}\langle u_{i,j}, :, 1\leq i,j\leq n\rangle$, and then factoring out the ideal of relations that forces the matrix $(u_{i,j})$ to be the inverse of $M$. Knowing what elements belong to that ideal is, in general, very difficult to determine (without commutativity, even for a single matrix). Cohn's closure conditions are designed to try to simplify those considerations, but generally it is still difficult.
That said, to get an explicit example, take your favorite ring $R$ and your favorite matrix $M$, and first figure out the free product above, and then explicitly work out the ideal of relations.
You might also look up Leavitt path algebras, as some of those algebras are examples of this universal inversion.
@PaceNielsen I'd honestly say that your method with infinte presentation is totally hopeless and provides zero insight on what localization result will be. If you are localizing at matrices invertible in a quotient ring, then there's a method by Vogel that realizes localization as a certain subring in adic completion (under some conditions, of course). There's one example of that approach in Vogel&Farber article "The Cohn localization of the free group ring" and other in "Cohn localization of finite group rings".
@DenisT. It's not totally hopeless in the infinite presentation case, if there is a reduction system in the sense of Bergman. That said, it is much easier in the finite presentation case---but even there the localization is not easy. (I doubt it even has to be finitely presented in general.) When I said "Take your favorite ring" that was a bit tongue-in-cheek, because it might not be his favorite ring anymore, after trying that computation. But it is worth giving it a try, just to see the difficulty.
Cohn localization of group ring of free group on $r$ generators $k[F_r]$ w.r.t. set of matrices which are invertible after augmentation $\epsilon$ is the ring of "noncommutative rational functions", if $k$ is PID. To define what that "rational functions" are, recall that free group ring is embedded into formal series ring $\Gamma_r := k\langle\langle x_1, \dots, x_r \rangle \rangle$ via Magnus homomorphism
$$\mu: k[F_r] \to k\langle\langle x_1, \dots, x_r \rangle \rangle; \, x_i \mapsto 1 + x_i + x_i^2 + \dots.$$
Now consider set of $(\epsilon, Id)$-derivations $\delta_i$ on $\Gamma_r$ which send $x_iw$ to $w$ and $x_jw$ to $0$. They and their composites constitute a ring of operators $D := k[\delta_1, \dots, \delta_r]$ acting on $\Gamma_r$ Now, a rational function is an element $s$ of $\Gamma_r$ for which $Ds$ is a finitely generated $k$-module. One can readily check that the set of rational functions is closed under multiplication; if $r = 1$ that will result in usual rational functions ring.
What makes Cohn localization "novel", as you say? Free group rings do not satisfy Ore condition, so there's no way to construct this ring via usual method of "ring of fractions".
|
2025-03-21T14:48:31.083163
| 2020-05-25T19:20:01 |
361349
|
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|
Stack Exchange
|
Real non-principal 2-blocks for finite groups of Lie type
Is it true that each finite group of Lie type has a non-principal real $2$-block? Here the principal block is the one containing the trivial character and a block is real if it contains the complex conjugates of its characters.
Note: G. O. Michler and W. Willems proved, a long time ago, that each such group has a $p$-block of defect zero, for every prime $p$. However I do not know if they thought about the reality of such blocks.
Have you checked which 2-block contains the Steinberg character (associated the natural characteristic of the Lie type group). This is a rational-valued character (and answers your question for Lie type groups of characteristic 2, though in odd characteristic I'm not sure how to proceeed).
Hi Geoff, for a Lie type group defined in odd characteristic p, the degree of St is a power of p, hence is odd. As St is real valued, it therefore belongs to the principal 2-block. So no joy there.
Srinivasan and Vinroot, in this paper, prove that a character is real-valued iff its label is the 'obvious' form, i.e., (s,chi) where s and its inverse are conjugate, and chi is real-valued. So the question now becomes whether all such characters lie in the principal 2-block. I haven't gone further than this though, although it should be approachable for classical groups, and also for exceptional groups.
Hi David,
as Radha Kessar has pointed out, for $n>1$ and $q$ an odd prime power, the $2$-blocks of $GL(n,q)$ are parametrized by the conjugacy classes of odd-order semisimple elements. Moreover a block is real if and only if the characteristic polynomial of any corresponding semisimple element is self-dual. Using this idea, it is easy to show that, apart from $GL(2,3)$ and $GL(3,3)$, all groups $GL(n,q)$ have a real non-principal $2$-block.
|
2025-03-21T14:48:31.083334
| 2020-05-25T19:32:02 |
361350
|
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|
Stack Exchange
|
Distance of parametrized skew hermitian exponentials
Consider two skew-adjoint matrices $A$ and $A'$, i.e. $A^*=-A$ and $A'^*=-A'$. It is well-known that
$e^{-tA}$ and $e^{-tA'}$ are unitary operators.
I would like to know:
Is it true that $\sup_{t \in \mathbb{R}} \Vert e^{-tA}-e^{-tA'} \Vert = 2(1-\delta_{A,A'})?$
What is $\delta_{A, A'}$?
the kronecker delta that is 1 if $A=A'$ and zero otherwise.
I see, thank you.
So the question is: if two strongly continuous unitary groups are not identical, then must their pointwise distance be 2?
@NikWeaver correct.
Please don't delete your question after it's been given an answer, this is disrespectful.
This is not true in general. E.g., let
$$A:=\left(
\begin{array}{cc}
-i & 0 \\
0 & 0 \\
\end{array}
\right),\quad A':=\frac i2\,
\left(
\begin{array}{cc}
-1 & 1 \\
1 & -1 \\
\end{array}
\right).$$
Then, for real $t$,
$$e^{-tA}=\left(
\begin{array}{cc}
e^{i t} & 0 \\
0 & 1 \\
\end{array}
\right),\quad
e^{-tA'}=\frac12\,\left(
\begin{array}{cc}
1+e^{i t} & 1-e^{i t} \\
1-e^{i t} & 1+e^{i t} \\
\end{array}
\right),$$
and
$$\|e^{-tA}-e^{-tA'}\|=\sqrt{1-\cos t}, $$
so that
$$\sup_{t\in\mathbb R}\|e^{-tA}-e^{-tA'}\|=\sqrt2\ne2(1-\delta_{A,A'}).$$
|
2025-03-21T14:48:31.083445
| 2020-05-25T19:43:22 |
361351
|
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|
Stack Exchange
|
When localization is indecomposable
We know that if $R $ is a domain then any localization of $R $ at any multiplicative subset of $R $ is indecomposable, that is, has no non trivial idempotents. Now let $R $ be a commutative ring with 1 that has no non trivial idempotents. I am looking for an additional condition on $R $ under which any localization $R_r $ to be indecomposable for each non nilpotent element $r\in R $.
Thanks for any help.
This is true if and only if $R$ has a unique minimal prime ideal.
@Mohan, thank you, can you expain a little more. Especialy for the only part.
@Mohan, I can prove that if $R $ has finite many minimal prime ideals, then the property holds. But I cannt prove the converse. Are you sure that the converse is true? Can you give some hints?
I apologize. I am `Noetherian' person and there are only finitely many minimal primes. I should have stressed that.
|
2025-03-21T14:48:31.083547
| 2020-05-25T20:01:12 |
361352
|
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|
Stack Exchange
|
tetrahedral interpolation and integration along a segment
Let's say we have a several tetrahedrons $T_i$ whose faces touch so that each face belong to two tetrahedrons. Each tetrahedron contain a value $V_{i}$.
Given a position $P$ inside the tetrahedron $T_0$, and neighboring tetrahedron are labeled $T_1, T_2, T_3, T_4$.
How to compute the value $V(P)$ such that its value is a linear interpolation between all $V_i$?
Following this, given a direction $\vec{d}$ and the origin $O$ and a scalar $t$ such that $P(t)=O+d*t$, what is the equation giving the interpolated value along this segment $V(t)$, considering only the part where the segment is inside $T_0$?
I tried to use barycentric coordinates, and I think it confused me more than it helped.
What would be a simple explanation for solving such a problem?
what does it mean when you say that a tetrahedron "contains" a value?
Let's say in another way that a value is given to each of them as input data. It can be something like density for instance
If you also have some bounary conditions you could try to find valus for the vertices s.t. for each tetrahedron its value is equal to the average of the values of its vertices.
One common method is to assign values to the vertices of your central tetrahedron $T$,
and then use
Barycentric coordinates to interpolate from the vertices of $T$ to any point $p \in T$.
The link shows how to convert between the coordinates of $p$ to its barycentric
coordinates $\lambda_1,\lambda_2,\lambda_3,\lambda_4$. Then use those $\lambda$'s to form a
weighted version of the values at the corners to $T$ to the value at $p$.
To use this approach, you need values at the vertices of $T$. I assume when you say that
each tetrahedron $T_i$ "contains a value" $v_i$, you mean that $v_i$ is somehow appropriate
throughout $T_i$. Then it makes sense to assign to a vertex $u$ of $T$ the average of the
values $v_i$ for the three tetrahedra incident to $u$, and the
value of your central tetrahedron $T$.
To make this calculation less of a heuristic would require explicit criteria
the interpolation is to achieve.
So:
Compute values for the four corners of $T$.
Compute the barycentric coordinates $\lambda_i$ for $p$. (Requires inverting a $3 \times 3$ matrix.)
Use the $\lambda$'s to weight the vertex values to an appropriate value for $p \in T$.
|
2025-03-21T14:48:31.083739
| 2020-05-25T20:23:47 |
361355
|
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"Nate Eldredge",
"Ryan Hendricks",
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"url": "https://mathoverflow.net/questions/361355"
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|
Stack Exchange
|
Approximation of vectors using self-adjoint operators
Let $T$ be an unbounded self-adjoint operator.
Does there exist, for any $\varphi$ normalized in the Hilbert space, a constant $k(\varphi)>0$ and a sequence of normalized $(\varphi_n)$ such that $$ \lim_{n \rightarrow \infty} \Vert \varphi-\varphi_n \Vert=0 $$
and $\Vert T \varphi_n \Vert \le k(\varphi).$
Somehow this looks strange, if you think of $\varphi \notin D(T)$ as then $$\Vert T\varphi \Vert"="\infty$$
all of a sudden, on the other hand, maybe you only have to choose $k$ in a suitable way.
Sorry, why can't we have $k(\varphi)=|T\varphi|$, and $\varphi_n=(1-\frac{1}{2^n})\varphi$?
@fierydemon: $\varphi$ is not assumed to be in the domain of $T$.
No, this only holds when $\varphi \in D(T)$.
It's enough to assume that $T$ is closed and densely defined, so that $T^{**} =T$. Let $\psi \in D(T^*)$ be arbitrary. If the hypothesis holds then we have
$$ |\langle \varphi, T^* \psi \rangle| = \lim_{n \to \infty} |\langle \varphi_n, T^* \psi \rangle| = \lim_{n \to \infty} |\langle T \varphi_n, \psi \rangle| \le k(\varphi) \|\psi\|$$
which is exactly the definition of $\varphi \in D(T^{**}) = D(T)$. One can also see that in fact $T \varphi_n \to T \varphi$ weakly.
Note the argument still goes through even if we only assume that $\varphi_n \to \varphi$ weakly instead of strongly.
|
2025-03-21T14:48:31.083858
| 2020-05-25T20:55:50 |
361359
|
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"Asaf Karagila",
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"url": "https://mathoverflow.net/questions/361359"
}
|
Stack Exchange
|
Constructing a section of an equivalence relation compatible with the intersection
Let $X$ be a (countably) infinite set and define an equivalence relation $\sim$ on the power set $P(X)$ of $X$ by defining two subsets $A$ and $B$ of $X$ to be equivalent if they differ by at most finitely many elements (i.e., $A \sim B$ if the symmetric difference $A \Delta B$ is finite).
Let $[-]\colon P(X) \to P(X)/_{\sim}$ denote the quotient map onto the equivalence classes. Does there exist a section $s \colon P(X)/_{\sim} \to P(X)$ of it with the following three properties?
$s([\emptyset]) = \emptyset$.
$s([X]) = X$.
For every $A$ and $B$ in $P(X)$ we have $s([A]) \cap s([B]) = s([A \cap B])$.
I tried around constructing such a section by using the Lemma of Zorn, but I couldn't figure out how to make the crucial step in the argument work (enlarging the domain of such a section, but which is defined only on a subset of $P(X)/_{\sim}$, by at least one more missing element from $P(X)/_{\sim}$).
This would fit better on [math.se].
No such section exists. The main point in the proof is that there exist uncountably many (in fact continuum many) infinite subsets of $\mathbb N$ such that the intersection of any two of them is finite. (Such sets are called almost disjoint.) In order to preserve $\cap$ and $\emptyset$, a section would have to send the equivalence classes of these almost disjoint sets to genuinely disjoint subsets of $\mathbb N$, and there aren't enough of those.
To produce the claimed almost disjoint sets, it suffices to produce them as subsets of some other countably infinite set rather than of $\mathbb N$. Take the countably infinite set $\mathbb Q$ and associate to each real number $r$ a sequence $A_r$ of distinct rationals converging to $r$. Distinct $r$'s give almost disjoint $A_r$'s.
|
2025-03-21T14:48:31.083993
| 2020-05-25T21:20:43 |
361361
|
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"url": "https://mathoverflow.net/questions/361361"
}
|
Stack Exchange
|
The Jacobson radical as a bimodule
Let $A$ be a finite dimensional algebra with Jacobson radical $J$.
Question 1: In case $A$ is a Nakayama algebra with a linear quiver corresponding to a Dyck path $D$ (via its Auslander-Reiten quiver), then it seems that the number of indecomposable summands of $J$ as a bimodule is equal to the number of touch points of $D$ with the x-axis excluding the origin. (see the statistic http://www.findstat.org/StatisticsDatabase/St000011 )
Is this true and is there a simple explanation for this?
In case this is true, $J$ as a bimodule would be indecomposable if and only if the corresponding Dyck path $D$ is indecomposable.
The question has a positive answer for all Nakayama algebras with at most 6 simple modules or equivalently all Dyck paths from (0,0) to (10,0).
A representation-theoretic interpretation for this number is the number of simple non-projective direct summands of $J$ as a one-sided module.
Question 2: Is there a simple criterion when the Jacobson radical $J$ of a general algebra $A$ as a bimodule is indecomposable? Can the number of indecomposable summands of $J$ be counted by a simple formula?
|
2025-03-21T14:48:31.084091
| 2020-05-26T00:10:54 |
361369
|
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|
Stack Exchange
|
Uniform integrability contradicts convergence to $L^2$ subspace
The following question was asked at https://mathoverflow.net/questions/361367/uniform-integrability-contradicts-convergence-to-l2-subspace :
Let $V$ be a finite-dimensional subspace of $L^2(\mathbb{R})$.
Assume that $f_n$ is a sequence of square-integrable functions with
$\Vert f_n \Vert_{L^2}=1$ that satisfies two properties:
1.) $d(f_n,V) \rightarrow 0$ that is the distance to $V$ vanishes in the limit
2.) There exists a uniform (in $n$) constant $k$ and a strictly positive function $g$ such that the following uniform integrability
condition holds $$\int_{\mathbb{R}} g(x) \vert f_n(x) \vert^2 \ dx \le k.$$
I want to show that if for all $v \neq 0$ in $V$ we have
$$\int_{\mathbb{R}} g(x) \vert v(x) \vert^2 \ dx=\infty$$ then such a
sequence $f_n$ cannot exist.
The intuition is that the $f_n$ are more and more supported in $V$
where every element has infinite integral against $g$, so the uniform
integrability condition cannot hold.
EDIT: If we knew for example that $f_n$ would not just converge to $V$
but to a fixed element $f$ in $V$, then it would follow that for a
subsequence of the $f_n$ we would have $f_n \rightarrow f$ almost
everywhere and thus get a fast contradiction using Fatou's lemma.
The question was then deleted by the OP while I was typing the answer. I thought the question might still be of some interest and will give an answer to it below.
Let $g_n:=P_V f_n$, where $P_V$ is the orthoprojector onto $V$. Then $\|g_n\|_2\le1$, $g_n\in V$, and $V$ is finite dimensional. So, passing to a subsequence, without loss of generality we may assume that $g_n\to v$ for some $v\in V$. Also, $\|f_n-g_n\|_2=d(f_n,V)\to0$. So, $f_n\to v$ in $L^2$ and hence $\|v\|_2=1$, so that $v\ne0$ and hence
$$\int g|v|^2=\infty. \tag{1}$$
Also, in view of the condition $f_n\to v$ in $L^2$, passing again to a subsequence, without loss of generality we may assume that $f_n\to v$ almost everywhere. Now the Fatou lemma and condition (1) imply
$$\int g|f_n|^2\to\infty,$$
as desired.
|
2025-03-21T14:48:31.084257
| 2020-05-26T00:39:02 |
361372
|
{
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"Dmitry Vaintrob",
"Robert Bryant",
"Vít Tuček",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Curvature of nonsymmetric metric tensors?
Consider a smooth manifold $M$ of arbitrary dimension. We have notions of psuedo-Riemannian or Riemannian metrics on a manifold, and they differ in the slightest way of being positive-definite or not. However, what happens if we drop positive-definite AND symmetry? For example, if we had a nondegenerate bilinear form $g_p: T_p M \times T_p M \to \mathbb{R}$ that varied smoothly between points. Has this been explored in depth? It appears to me at surface level that one could still concoct connections, curvature, and possibly a notion of parallel transport in this flavor of 'smooth geometry'. A motivation for me to ask is as follows. Suppose $R$ is an $S$-algebra where $\Omega_{R/S}$ is reflexive and the canonical isomorphism $\phi: \Omega_{R/S} \to \Theta_{R/S}$ is an isomorphism of $R$-modules (i.e. nonsingular varieties). There exists a canonical map $\Omega_{R/S} \times \Theta_{R/S} \to R$, which is $R$-bilinear and nondegenerate, and is given by $\langle \omega, V \rangle = l(\omega)$ where $l:\Omega_{R/S} \to k$ such that $l \circ d_{R/S} = V$. This induces a morphism $$\Theta_{R/S} \times \Theta_{R/S} \xrightarrow{\phi^{-1} \times 1}\Omega_{R/S} \times \Theta_{R/S} \to R.$$ Natural questions that arise are is this composition $R$-bilinear nondegenerate, and when is it symmetric? Which symmetric $R$-bilinear forms factor through $\phi^{-1} \times 1$? When we work with a manifold and have a metric tensor that is bilinear and nondegenerate, just how interesting is this flavor of curvature (whatever it is supposed to mean)?
There is no functorial way to define a connection on $TX$ starting from a non-symmetric nondegenerate form. As an extreme opposite case to a metric you can consider a symplectic manifold, which is equipped with a nondegenerate form. It is not the case that a symplectic manifold has a canonically definable connection on its tangent bundle: for example, if this were the case then every manifold would carry a connection on $TX \oplus T^*X$ which is invariant under autodiffeomorphisms, and this is not the case e.g. for $\mathbb{R}$.
Consider a bilinear form $b \in \mathcal{C}^\infty (T^*M\otimes T^*M, \mathbb{R})$ and an affine connection $\nabla \colon \mathcal{\Gamma}^\infty(TM) \to \mathcal{\Gamma}^\infty(T^*M\otimes TM)$ whose parallel transport preserves $b$. That can be expressed by the condition $b(\nabla_X Y, Z) + b(Y, \nabla_X Z) = 0$ for all $X, Y, Z \in \Gamma(TM).$
When $b$ is symmetric nondegenerate tensor (i.e. a pseudo-Riemannian metric), then this $\nabla$ is caleld metric connection. It always exists, but it is not unique! To get a unique connection one has to impose also that $\nabla$ has zero torsion tensor.
When $b$ is antisymmetric nondegenerate tensor (i.e. a presymplectic form), then $\nabla$ is known as symplectic connection but this time the uniqueness is not saved by torsion-freeness.
The way to handle these problems in general is to consider one torsion free $b$-connection $\nabla$ and study its modification $\nabla + A$ where $A \in \Gamma(\mathrm{End}(TM)).$ Such modified connection preserves $b$ if and only if $b(A(X)Y, Z) + b(Y, A(X)Z) = 0.$ Torsion-freeness is equivalent to $A(X)Y - A(Y)X = 0.$ From this it is obvious that the affine space of torsion-free $b$-connections is governed by the representation theory of Lie algebra of the isotropy group of $b$. Sometimes you get just the trivial representation, sometimes you get bigger space.
The punchline here is that connection preserving some tensorial objects are generally not unique. You can either add some additional data to fix them, or you can try to construct invariants which do not depend on the possible choices.
A few remarks:
First, in a sense, (special cases of) this (are) is very commonly studied. Because a bilinear differential form $g$ as you have defined it can naturally be written as a sum $g = \sigma + \alpha$ where $\sigma$ is symmetric and $\alpha$ is skew-symmetric, you are, equivalently, asking about the geometry of the pair $(\sigma,\alpha)$. The most famous example is that of Kähler geometry, i.e., where $\sigma$ is positive definite and $\alpha$ is nondegenerate and parallel with respect to the Levi-Civita connection of $\sigma$, but there are many variations on this (pseudo-Kähler, Hermitian, unitary...) that essentially have the same nature. In general, when $\sigma$ is non-degenerate, you know that there is the canonical Levi-Civita connection of $\sigma$, but, of course, depending on the assumptions you make about $\alpha$, there could be other `canonical' connections.
Second, it is not clear how you ought to define 'non-degenerate' for a general $g$, because there are different notions, and it might well be that what you want to call 'non-degenerate' depends on the applications you have in mind. For example, if $M=\mathbb{R}^2$, and $g = \mathrm{d}x\otimes\mathrm{d}y$, then $g$ is 'degenerate' in the naïve sense (since it has tensor rank $1$ instead of $2$), but both $\sigma = \tfrac12(\mathrm{d}x\otimes\mathrm{d}y+\mathrm{d}y\otimes\mathrm{d}x)$ and $\alpha=\tfrac12(\mathrm{d}x\otimes\mathrm{d}y-\mathrm{d}y\otimes\mathrm{d}x)$ are non-degenerate in the usual senses for symmetric and anti-symmetric quadratic differential forms. In particular, since $\sigma$ has a 'functorial' connection (being non-degenerate), $g$ does also. You'd want to count such a $g$ as 'non-degenerate', no?
Finally, as Dmitry has already remarked, you do need some hypotheses beyond simple algebraic 'non-degeneracy' in order to have a 'functorial' (i.e., invariant under all diffeomorphisms)construction of a connection. Such considerations will enter into any discussion of the right hypotheses, as Vit's discussion above makes clear. I'll just add that sometimes it's important to take higher order derivatives of $g$ into consideration in order to define a class of structures for which a 'functorial' connection exists. For example, as Dmitry pointed out, a closed non-degenerate $2$-form on $M$ does not determine a 'functorial' affine connection on $M$. However, if you drop the assumption 'closed' and replace it with an appropriate higher order non-degeneracy condition, there sometimes is an associated functorial connection. For example, there is an open set $\mathcal{F}$ of germs of $2$-forms on $4$-manifolds that is perserved by (local) diffeomorphisms that does have the property that any $2$-form $\alpha$ on $M^4$ whose germ at every point belongs to $\mathcal{F}$ possesses a functorial torsion-free connection $\nabla^\alpha$, but the definition of $\mathcal{F}$ (and of $\nabla^\alpha$) depends on higher derivatives of $\alpha$ than just first derivatives. (In particular, $\mathrm{d}\alpha$ will be nowhere vanishing if $\alpha$ has its germs belonging to $\mathcal{F}$ at every point.)
Remark: I was asked about how the above $\mathcal{F}$ is defined and how it works. Here's a quick sketch of the result of applying Cartan's method of equivalence to this question:
Start with a non-degenerate $2$-form $\alpha$ on a $4$-manifold $M$. First, there exists a unique $1$-form $\beta$ on $M$ such that $\mathrm{d}\alpha = \beta\wedge\alpha$. Let $\gamma = \mathrm{d}\beta$. Then $0 = \mathrm{d}(\beta\wedge\alpha) = \gamma\wedge\alpha$. Second, there exists a uniqe function $F$ on $M$ such that $\gamma^2 = F\,\alpha^2$, and the first condition defining $\mathcal{F}$ is that $F$ should be nowhere vanishing. For simplicity, I'm going to continue the analysis under the assumption that $F<0$ (there is a similar branch when $F>0$, but I'll leave that to the interested reader). Set $F = -f^2$ where $f>0$. Then, using $\gamma\wedge\alpha= 0$ and $\gamma^2 = -f^2\,\alpha^2$, we see that $(\gamma\pm f\alpha)^2 = 0$, but $(\gamma+f\alpha)\wedge(\gamma-f\alpha) = -2f^2\,\alpha^2\not=0$, so, setting $\gamma\pm f\,\alpha = \pm 2f\,\alpha_{\pm}$, we have
$$
\alpha = \alpha_+ + \alpha_-\quad\text{and}\quad
\gamma = f\,(\alpha_+ - \alpha_-),
$$
where $\alpha_\pm$ are a pair of (nonvanishing) decomposable $2$-forms whose wedge product is nonvanishing.
Finally, there are unique decompositions
$$
\beta = \beta_+ + \beta_- \quad\text{and}\quad \mathrm{d}f = \phi_+ + \phi_-
$$
where $\beta_\pm \wedge\alpha_{\pm} = \phi_\pm \wedge\alpha_{\pm} = 0$, and so there will be unique functions $g_\pm$ such that
$$
\beta_\pm\wedge\phi_\pm = g_\pm\,\alpha_\pm\,.
$$
The final 'open' conditions on $\alpha$ needed to define $\mathcal{F}$ are that $g_+$ and $g_-$ be nonvanishing.
In this case, the $1$-forms $\beta_+$, $\beta_-$, $\phi_+$, and $\phi_-$ define a coframing on $M$ that is functorially associated to $\alpha$. Once one has such a 'canonical' coframing, it is easy to define a connection, in fact, a large family of connections, on $M^4$ (for example, one of these connections (with torsion) will make the given coframing parallel) including some that are torsion-free.
Concerning nondegeneracy: is not just an isomorphism between the tangent and the cotangent bundle the most natural and uniquely determined notion of a nondegenerate "nonsymmetric"?
@მამუკაჯიბლაძე: Ah, but the question is which map are you going to ask to be an isomorphism? If you have a bilinear form $g:V\times V\to\mathbb{R}$ that is neither symmetric nor anti-symmetric, there are two possibilities: $\alpha(v)(w) = g(v,w)$ or $\beta(v)(w) = g(w,v)$. When $g$ is neither symmetric nor anti-symmetric, one of $\alpha:V\to V^$ or $\beta:V\to V^$ could be an isomorphism without the other being an isomorphism. Which one are you going to use?
Ooooh. Life is tough :) Thank you for the explanation. Indeed $V\to W^$ and $W\to V^$ are safely distinguishable, but when $V=W$, there is a problem, yes.
Still, I believe one of $\alpha$, $\beta$ is an isomorphism iff the other one is. At least if the canonical map $V\to V^{**}$ is an isomorphism - which is, I believe, always the case with finite rank vector bundles, no?
@მამუკაჯიბლაძე: Oh, yes. You are right, for a bilinear tensor, the left rank is the same as the right rank. I was remembering that this wasn't true for general trilinear tensors, and forgot that it is true for bilinear tensors. I guess what I should have written is that you could have taken $\alpha(v)(w) = g(v,w)$ or $\beta(v,w) = g(v,w)+g(w,v)$ (or some other combination, such as $\gamma(v,w) = g(v,w)-g(w,v)$, and one of these could be an isomorphism $V\to V^*$ and the others not. Thanks for pointing out my error. I'll fix my answer above.
@RobertBryant Could you please add some references to that higher order symplectic example?
@VítTuček: Well, I'm not sure of the best place to see a reference to Cartan's equivalence problem applied to 'generic' non-degenerate $2$-forms on $4$-manifolds, but this is a well-known exercise in applying Cartan's method of equivalence. I'll add a paragraph at the end, that sketches how this goes.
|
2025-03-21T14:48:31.085014
| 2020-05-26T02:45:55 |
361376
|
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|
Stack Exchange
|
Rates of convergence to Tracy-Widom?
$\renewcommand{\!}{\mathbf}
\renewcommand{\Ai}{\operatorname{Ai}}$
One can define the Tracy-Widom distribution as the Fredholm determinant $F_2(t)=\det(\mathbf I-\mathbf A)$ where
$$\mathbf A(x, y)=\begin{cases}
\frac{\Ai(x) \Ai'(y)-\Ai'(x) \Ai(y)}{x-y} & \text {if } x \neq y \\
\Ai'(x)^{2}-x \Ai(x)^{2} & \text {if } x=y
\end{cases} \text{ for } \Ai(x) = \frac 1{\pi} \int_0^\infty \cos\Big(\tfrac 13 t^3 + xt\Big) \, d t$$
It's well known that the largest eigenvalue of an $n\times n$ GUE matrix (appropriately scaled) converges in distribution to $F_2$. Also, by the celebrated Baik-Deft-Johansson theorem, the length of the longest increasing subsequence of a random permutation $\in S_n$ (appropriately scaled) also converges in distribution to $F_2$. I'm sure there are many other examples of situations in which $F_2$ appears as the limiting distribution.
However, I have been trying to find references about how fast the things that converge to $F_2$ converge to $F_2$. My searches have turned up this paper: https://arxiv.org/pdf/0803.3408.pdf and this paper: https://arxiv.org/pdf/1901.05235.pdf, both of which seem to answer related questions but not exactly what I'm looking for.
Does anyone know the rates at which the largest eigenvalue of a GUE matrix and/or the longest increasing subsequence of a random permutation converge to $F_2$? If not exactly, are there references that have done numerical computations and put forward a conjecture about such rates?
I think it is an open question to design a Stein's method framework when the limiting distribution is a Tracy-Widom one.
Accuracy of the Tracy-Widom limits for the extreme eigenvalues in white Wishart matrices studies the rate of convergence for Wishart matrices $WW^T$, with $W$ an $n\times p$ matrix with i.i.d. Gaussian matrix elements. The error is ${\cal O}[\min(n,p)]^{-2/3}$.
Here is an error plot, showing for two percentiles of the exact distribution the relative error of the Wishart distribution. The errors are smaller further out in the tail of the distribution.
|
2025-03-21T14:48:31.085171
| 2020-05-26T03:08:14 |
361377
|
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"Jens Reinhold",
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|
Stack Exchange
|
Alexander duality for Homology sphere which is the Geometric realization of a finite simplicial complex
The Alexander duality Theorem is usually stated for a triangulable pair $(\mathbb S^n, Y)$ where $Y$ is a subset of the standard sphere $\mathbb S^n$. My question is: Does the duality also hold if we rather replace $\mathbb S^n$ by a compact orientable Homology sphere (without boundary) (https://en.m.wikipedia.org/wiki/Homology_sphere) ? I'm mainly interested in the cases $n=2$ and $3$. I'm willing to assume that the Homology sphere is the Geometric realization of a finite abstract simplicial complex (https://en.m.wikipedia.org/wiki/Abstract_simplicial_complex)
Thanks
This seems to be true, cf. Massey, "A generalization of the Alexander duality theorem" Indiana Univ. Math. J. 30 (1981).
You have to use Poincaré-Lefschetz duality : Let $M$ be a compact orientable $n$-manifold, $Y\subset M$ be a closed subset then we have an isomorphism
$$\check{\mathrm{H}}^p(M,Y)\cong H_{n-p}(M-Y)$$
induced by the cap product with the fundamental class of $M$ (the left hand side is Cech cohomology).
You also have
$$\check{\mathrm{H}}^p(Y)\cong H_{n-p}(M,M-Y).$$
In fact these isomorphisms are compatible with the long exact sequences of the pairs $(M,Y)$ and $(M,M-Y)$.
In your case, if $M$ is a triangulated manifold and $Y$ is a subpolyhedron of $M$, cech cohomology groups are nothing but singular cohomology groups.
You can have a look at Bredon's book "Topology and Geometry" (chapter VI, section 8).
|
2025-03-21T14:48:31.085288
| 2020-05-26T06:15:57 |
361379
|
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|
Stack Exchange
|
Number of points in a lattice and an oblong box
I have a very simple question in geometry of numbers. (It is a slight modification of Counting points on the intersection of a box and a lattice .) There's a bound I can easily prove, and it's good enough for my purposes, but it may be embarrassingly suboptimal.
Let $S = (N_1,2 N_1] \times \dotsb \times (N_n,2 N_n]$, where $N_i\geq M\geq 1$. Define the lattice $L$ as the preimage of
$r_1 \mathbb{Z} \times \dotsb r_n \mathbb{Z}$ under an affine linear map $\vec{v} \mapsto A \vec{v} + \vec{b}$, where $r_i\geq M$ are integers and $A=\{a_{i,j}\}$ is a non-singular $n$-by-$n$ matrix such that $a_{i,j}\in \mathbb{Z}$, $|a_{i,j}|\leq C$. How do you bound the number of points $|S\cap L|$ in $S\cap L$?
It is simple to show (chopping $A S$ into hypercubes of side $M$) that
$$|S\cap L| \leq (4 C n)^n \prod_{i=1}^n \frac{N_i}{M}.$$
How much better can one do? Can one replace $(4 n)^n$ by $2^n n!$, say? Or (much more ambitiously) $\prod_{i=1}^n N_i/M$ by $\prod_{i=1}^n N_i/r_i$?
(It would be interesting, for starters, to
combine the argument above with Davenport's Lemma (as in Counting number of points on a lattice in a hypercube), but doing so in such a way as to obtain a real improvement doesn't seem obvious.)
It does seem to me that one can do better in some circumstances by using Davenport's Lemma after all. We recall (see Davenport's On a principle of Lipschitz): for $B$ convex,
$$ \bigl(|B\cap\mathbb{Z}^n|-\mathrm{vol}(B)\bigr) \leq \sum_{m=0}^{n-1} V_m,$$
where $V_m$ is the sum of the $\mathrm{vol}(\pi(B))$ under all projections $\pi$ obtained by setting $n-m$ coordinates to $0$.
Let $B = R (A S + \vec{b})$, where $R(x_1,\dotsc,x_n) = (x_1/r_1,\dotsc,x_n/r_n)$.
Then $|S\cap L| = |B\cap \mathbb{Z}^n|$, and $$\mathrm{vol}(B) = \frac{\det(A)\cdot \mathrm{vol}(S)}{\prod_{i=1}^n r_i}\leq C^n n! \prod_{i=1}^n \frac{N_i}{r_i}.$$
Now, the ($m$-dimensional) volume of $\pi(B)$ for $\pi$ a projection obtained by setting $n-m$ coordinates to $0$ is at most $2^{m-n}$ times the sum of the $\mathrm{vol}(\pi(R P))$ over all $m$-dimensional sides $P$ of the parallelepiped $AS$. It is clear that $\mathrm{vol}(\pi(R P)) =\mathrm{vol}(\pi(P))/\prod_{i\in I} r_i$, where $I$ is the set of coordinates that $\pi$ does not set equal to $0$. We know that $\mathrm{vol}(\pi(P)) \leq \mathrm{vol}(P)$ because, in general, (orthogonal) projections do not increase volumes. We also know that
$\mathrm{vol}(P) \leq \prod_{i\in I'} C n N_i$, where $I'$ is the set of indices that vary as we traverse the side $P$. (Here $C n N_i$ is an upper bound on the
length of the image under $A$ of the side $(N_i,2 N_i]$.) Hence
$$\textrm{vol}(\pi(R P)) \leq (C n)^m \frac{\prod_{i\in I'} N_i}{\prod_{i\in I} r_i}.$$
We are summing over $\binom{n}{m}$ projections and $\binom{n}{m}$ sides. Thus, the contribution of a given $m$ to the right side of Davenport's Lemma is at most
$$2^{-n} (2 C n)^m \binom{n}{m}^2 \cdot \frac{\prod_{i=1}^m N_i}{\prod_{i\in I} r_i},$$
where we assume that $N_1,N_2,\dotsc$ are sorted in decreasing order, and $I$
is the set of $m$ indices corresponding to the $m$ smallest values of $r_i$. We sum over all $m\leq n$, and, obtain a total of
$$\begin{aligned}
2^{-n} \sum_{m=0}^{n-1} (2 C n)^m \binom{n}{m}^2 \cdot
\max_{|I|=|I'|=m} \frac{\prod_{i\in I'} N_i}{\prod_{i\in I} r_i}&\leq
\frac{(C n)^{n-1}}{2} \binom{2n}{n} \cdot
\max_{|I|=|I'|<n} \frac{\prod_{i\in I'} N_i}{\prod_{i\in I} r_i}\\
\lesssim
\frac{(2 C n)^{n-1}}{\sqrt{2 \pi}} \cdot
\max_{|I|=|I'|<n} \frac{\prod_{i\in I'} N_i}{\prod_{i\in I} r_i}.
\end{aligned} $$
We conclude that $$|S\cap L| \leq C^n n! \prod_{i=1}^n \frac{N_i}{r_i} + O\left( (2 C n)^{n-1} \max_{|I|=|I'|<n} \frac{\prod_{i\in I'} N_i}{\prod_{i\in I} r_i}\right).$$
That doesn't improve on the original bound in full generality, however (except in so far as it replaces $(4 C n)^n$ by $C^n n! + O((2 C n)^{n-1})$.
|
2025-03-21T14:48:31.085522
| 2020-05-26T06:25:05 |
361381
|
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|
Stack Exchange
|
Multiplication formula in Grassmannian cluster categories
Grassmannian cluster categories are studied in A categorification of Grassmannian cluster algebras and Cluster categories from Grassmannians and root combinatorics. The category $CM(B_{k,n})$ of Cohen-Macaulay modules over certain algebra $B_{k,n}$ is an additive categorification of the Grassmannian cluster algebra $\mathbb{C}[Gr(k,n)]$. A cluster character $M \mapsto \psi_M$ on the Grassmannian cluster category $CM(B_{k,n})$ is defined in Definition 9.1 A categorification of Grassmannian cluster algebras.
I think that the following multiplication formula is true:
For any Auslander-Reiten sequence $\tau M \to N \to M$ in the Auslander-Reiten sequence of $CM(B_{k,n})$,
$$
\psi_{\tau M} \psi_M = \psi_N + \prod_i \psi_{P_i}, \quad (1)
$$
where $P_i$ are some projective modules in $CM(B_{k,n})$ (they corresponds to frozen variables in $\mathbb{C}[Gr(k,n)]$).
Is the formula (1) already known in literature? Thank you very much.
|
2025-03-21T14:48:31.085607
| 2020-05-26T07:22:58 |
361386
|
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|
Stack Exchange
|
How to prove this combinatorial identity?
If $n \in \mathbb N \setminus \{0\}$ and $x,y,z \in \mathbb R$ such that $x+y+z=n-1$, show that
$$\dfrac{(-4)^n}{\binom{2x}{n}}\sum_{r+s=n,r,s\in Z}\dfrac{\binom{y}{r}\binom{y-a}{r}\binom{z}{s}\binom{z+a}{s}}{\binom{2y}{r}\binom{2z}{s}}
=\sum_{j\ge 0}\binom{n}{2j}\dfrac{\binom{-\frac{1}{2}}{j}\binom{a-\frac{1}{2}}{j}\binom{-a-\frac{1}{2}}{j}}{\binom{x-\frac{1}{2}}{j}\binom{y-\frac{1}{2}}{j}\binom{z-\frac{1}{2}}{j}}$$
for every $a \in \mathbb R$.
What is the name of the book?
This problem is from china book 走向IMO 2017 page 127 https://detail.tmall.com/item.htm?id=570447384104
Check out the book $A=B$ by Zeilberger, Wilf, and Petkovsek.
It’s all about how to prove essentially any combinatorial identity like this that you can think of.
The moral of the story is you type it into your computer, hit enter, and let it verify it for you and produce a certificate. These algorithms are all definitely available as packages for Maple and Sage. It might even be part of the standard Maple library.
Added:
"Science is what we understand well enough to explain to a computer. Art is everything else we do. During the past several years an important part of mathematics has been transformed from an Art to a Science: No longer do we need to get a brilliant insight in order to evaluate sums of binomial coefficients, and many similar formulas that arise frequently in practice; we can now follow a mechanical procedure and discover the answers quite systematically. I fell in love with these procedures as soon as I learned them, because they worked for me immediately. Not only did they dispose of sums that I had wrestled with long and hard in the past, they also knocked off two new problems that I was working on at the time I first tried them. The success rate was astonishing. ..." -- Donald Knuth (taken from the Foreward of $A=B$)
The full text for $A=B$ is available here free of charge, and the authors wanted to make sure anyone who wanted it could have it without paying.
(It really is magic, by the way. Give it a try!)
I can think of identities with binomial coefficients as exponents. Can the book deal with those? Gerhard "Tongue Is Not In Cheek" Paseman, 2020.06.19.
Oof. I think not (though I'm not an expert). By the way... @GerhardPaseman, I was recently talking with another mathematician about this site, and he remarked "Oh, I love how Gerhard Paseman always signs his stuff with 'middle names' relevant to the subject!" (For the record, me too!)
In these troubled times, such appreciation is very important to me. (OK. It's important to me at other times too. ) Thank you very much for it. Gerhard "This Signature Is For You" Paseman, 2020.06.19.
|
2025-03-21T14:48:31.085811
| 2020-05-26T07:26:45 |
361387
|
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|
Stack Exchange
|
On a particular proof of "if the sharp of every real exists and every club contains a club constructible from a real, then $\delta^1_2 = \omega_2$"
I am referring to the proof of (4) implies (1) in Theorem 3.16 of Woodin's The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal. His proof leverages on the fact that if the sharp of every real exists, then $\delta^1_2 = u_2$, where $u_2$ is the second uniform indiscernible (the least ordinal above $\omega_1$ which is an $x$-indiscernible for every real $x$).
This is roughly how it goes:
Fix any ordinal $\alpha$ strictly between $\omega_1$ and $\omega_2$, and a well-ordering $<_{\alpha}$ of $\omega_1$ of ordertype $\alpha$.
Find a club $C$ of $\omega_1$ such that for every $\gamma \in C$, the rank of $<_{\alpha}$ restricted to $\gamma$ is less than the least ordinal in C greater than $\gamma$.
By hypothesis, there is a $D \subset C$, $D$ a club of $\omega_1$, such that $D$ is constructible from a real $z$. We can assume $D$ is definable from $z$ and $\omega_1$ in $L[z]$ since $z^\sharp$ exists.
By reflecting the definition of $D$ downwards, we have that for every $\gamma \in D$, $rank(<_{\alpha} \restriction \gamma) < rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$, where $\mathcal{M}(z^{\sharp}, \gamma + 1)$ stands for (quoting Woodin) "the $\gamma$[$+1$] model of $z^{\sharp}$". I am not exactly sure what that quote means, so I assumed $rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$ is the least $z$-indiscernible above $\gamma$, which works for the arguments hitherto. Did I make a mistake in my interpretation here?
Now, following the previous inequality of ranks, he immediately concluded that $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$, which finishes the proof.
It is at the last step (the final bullet point) that I got lost. How does $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$ follow from the previous steps?
EDIT: Alternatively, I would appreciate it if anyone can point me to another proof of the statement "if the sharp of every real exists and every club contains a club constructible from a real, then $\delta^1_2 = \omega_2$"
I think what's involved in the last step is indiscernibility. Since the Silver indiscernibles w.r.t. $z$ form a club, you have the next-to-last bullet point for some indiscernibles $\gamma$, and you can replace such a $\gamma$ with $\omega_1$ in any statement that $L[z]$ understands. This replacement will turn $<\alpha\upharpoonright\gamma$ into $<\alpha$.
But we do not know if $<_{\alpha}$ is constructible from any real, so the set of inequalities in the penultimate bullet point may not be expressible in L[z]. Or am I mistaken here?
Let $\kappa$ be a sufficiently large regular cardinal. Take a countable elementary substructure $H$ of $H(\kappa)$ containing $z$ and $<_\alpha$ with $\omega_1\cap H\in D$. Let $\pi : M\to H(\kappa)$ be the inverse of the transitive collapse of $H$. To see the fifth bullet point, it suffices to show that in $M$, $\pi^{-1}(<_\alpha)$ has rank less than $(\text{rk}(\mathcal M(z^\#,\omega_1^M+1)))^M$. But this follows from the fourth bullet point since $\omega_1^M = \omega_1\cap H \in D$, $\pi^{-1}({<}_\alpha) = {<}_\alpha \restriction \omega_1^M$, and $M$ correctly computes $\text{rk}(\mathcal M(z^\#,\omega_1^M+1))$ since $z^\#$ is in $M$.
Thanks for the clear explanation!
|
2025-03-21T14:48:31.086049
| 2020-05-26T08:50:41 |
361390
|
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|
Stack Exchange
|
What is known about the duals of cyclic polytopes?
What is known about the duals of cyclic polytopes, in particular, their facets (or equivalently, the vertex-figures of cyclic polytopes)?
In even dimensions, all facets of the dual are combinatorially equivalent. Are these facets themselves duals of cyclic polytopes?
I think this cannot be true in odd dimensions $\ge 5$. For example, a 5-dimensional cyclic polytope is 2-neighborly, so its vertex figures are only 1-neighborly (is this true?), but 4-dimensional cyclic polytopes are 2-neighborly as well. So when does it happen that the facets are again duals of cyclic polytopes, and when are they all combinatorially equivalent?
In general, is there some kind of classification of the combinatorial types of these facets?
Many properties of the vertex figures of cyclic polytopes can be obtained from Gale evenness condition.
Let $P=C(n,d)$ be a cyclic $d$-polytope, and let $v_1<\cdots<v_n$ be its vertices ordered according to the moment curve. The following follows from Gale evenness condition.
In even dimensions, the vertex figure of $P$ at every vertex is a cyclic $(d-1)$-polytope.
In odd dimensions, every facet contains $v_1$ or $v_n$.
In odd dimensions, for $d\ge 5$ and $n\ge d+2$, the vertex figures of $P$ at $v_1$ and $v_n$ are cyclic $(d-1)$-polytopes, but the vertex figure at some other vertex is not.
The proofs of 1) and 2) follow straight from the condition.
For the proof of 3) use counting. First, for every $d\ge 4$ count in two different ways the vertex-facet incidences in $P$ in the case that each vertex-figure is a cyclic polytope, and obtain the following:
$$df_{d-1}(C(n,d))=nf_{d-2}(C(n-1,d-1))$$
Second, for odd $d\ge 5$, establish the following $$f_{d-1}(C(n,d))=2f_{d-2}(C(n-1,d-1))-f_{d-3}(C(n-2,d-2))$$
Comparing the two previous expressions gives 3).
Duality gives the results you are after.
Regards,
Guillermo
Thank you very much for your answer! I am not too familiar with applying Gale's criterion. Can you say a word about how it helps you to deduce something about the vertex figures, since it primarily makes statements about the facets of a cyclic polytope?
The facets of P containing $v_i$ are in one-to-one correspondence with the facets of $P/v_i$. That is, a set $X'$ of $d-1$ vertices corresponds to a $(d-2)$-face of $P/v_i$ if and only if the set $X:=X'\cup {v_i}$ is a facet of $P$.
|
2025-03-21T14:48:31.086498
| 2020-05-26T09:26:39 |
361391
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Size issues (small/large categories) when defining stacks in the Algebraic/differentiable/topological setting
Angelo Vistoli in the notes Notes on Grothendieck topologies, fibered categories and descent theory starts the section of category theory with the following note:
We will not distinguish between small and large categories. More
generally, we will ignore any set-theoretic difficulties. These can be
overcome with standard arguments using universes.
Question : Which of the notions introduced in Angelo Vistoli's notes assumes that the category is small? In particular their application to Algebraic/differentiable/topological stacks?
For example, Behrang Noohi puts the following extra condition in his notes on topological stacks:
Throughout the paper, all topological spaces are assumed to be compactly generated.
This could be because, the category $\text{Top}$ of all topological spaces is not a small category.
Are there any places one has to be careful to not allow large categories?
Some references to support this question :
nlab says "In technical terms, a site is a small category equipped with a coverage or Grothendieck topology". It also says (Remark $2.3$ at same page) "Often a site is required to be a small category. But also large sites play a role."
David Metzler in Topological and Smooth Stacks defines (page $2$) a site as a small category equipped with Grothendieck topology. It further says "We will want to discuss, for example, “the category of stacks on the category of all topological spaces,” but strictly speaking this does not exist, since the category of topological spaces does not have a set of objects, but rather a proper class. To avoid this problem we will consider throughout some fixed category $\mathbb{T}$ of topological spaces which has a set of objects, or at least, is equivalent to such a category".
So, it "looks like", even though one can define a site over a large category, and then a stack over a site (which was defined on a large category), one often restricts (for computational purposes or personal interests) to a small categories and stacks on them. Is this what it is or am I misunderstanding something here?
I am not asking how does one modify definitions of Angelo Vistoli's notes so that there is no issue when dealing with large categories...
If I am not confusing something, "using universes" in this context just means the following: if you encounter a functor that should be representable but is not, you enlarge your category by adjoining a representing object for your functor. From this point of view, whether you distinguish small/large or not, there still remains a distinction: a functor is either representable or not.
@მამუკაჯიბლაძე Hi, thanks for the response. I am not aware of what you have mentioned. I would like to read more if you can point me to some reference regarding making something a representable functor by adding an object representing it.. Idea is clear but some more details might clear somethings for me...
Well when you use subcanonical Grothendieck topologies, you basically do just that - make some non-representable functors representable. So any text on topos theory might be viewed relevant for that. I will come back if I figure out something more to the point.
@მამუკაჯიბლაძე sue, thank you :)
I don't think assuming that topological space are compactly generated has anything to do with a size problem. The category of compactly generated topological space is not small either. It probably has to do with wanting to consider exponential of topological space (maping space).
@SimonHenry oh. Sorry. I did not think if the category of compactly generated topological spaces is small or not.. I have done a rough reading of the paper by Behrang Noohi, could not find where exactly they use this restriction.. I will do another reading and see if they have done something similar to what you have mentioned. Thanks for the tip..
I wonder why there is no tag for “topological stacks”..
I was misled by such statements early in my career - universes do not allow you to ignore the difference between small and large categories. Rather, universes allow us to replace the distinction between set and proper class with the distinction between a set belonging to a fixed universe $\mathcal{U}$ and a set (still not a class) outside that universe. This means we can still apply all set-theoretic operations to "large" sets, because we never use proper classes. The definition of complete and cocomplete category must be adapted to use diagrams from $\mathcal{U}$.
@RobertFurber I don’t have much experience with universes.. It would be useful if you can make your comment as an answer with some more details.
can you make your comment as an answer adding some details which you think might be relevant. @SimonHenry Reading some places revealed that this condition on Topological spaces has something to do with Cartesian closedness
Yes. The category of topological space is not cartesian closed but the category of compactly generated is. This has nothing to do with size problem, but with properties of the compact-open topology. I'm affraid I don't really have more details to add to my comment, I think David Robert's answer pretty much cover everything.
@SimonHenry Ok. Thank you :)
Are there any places one has to be careful to not allow large categories?
No. For the purposes of forming the 2-category of algebraic/topological/differentiable stacks, or more generally, some kind of presentable stacks over a large category there are no size issues. Naively, the 2-category of stacks on $S$ is carved out from the presheaf category $[S^{op},\mathbf{Cat}]$ (or $[S^{op},\mathbf{Gpd}]$), which does present size issues for $S$ not essentially small. However, the 2-category of presentable stacks (of groupoids, say, which is the case you are looking at) is equivalent to the bicategory of internal groupoids and anafunctors (and transformations). This can be defined elementarily from the 2-category of internal groupoids, functors and natural transformations. Given a quite weak size condition on the site structure—that is, the size of generating sets of covering families—this bicategory is even locally essentially small. The only case 'in the wild' that I know of that fails this weak condition is the fpqc topology on categories of schemes, and algebraic geometers are a bit wary of that: see tag 0BBK. They are happy to say a single presheaf (of sets, modules, groupoids) is a stack for the fpqc topology, but generally talk about sheaves/stacks for the fppf topology at the finest: see the definition in tag 026O.
Added For a large site not satisfying the condition WISC, the sheafification or stackification functors might not exist. This problem, however, does not impact considering presentable stacks, only when one is wanting to think about arbitrary stacks. For an example of how bad this can get, Waterhouse's paper
Basically bounded functors and flat sheaves, Pacific Journal of Mathematics 57 (1975), no. 2, 597–610 (Project Euclid)
gives an example of a presheaf on the fpqc site that does not admit any sheafification. The following quote from the Stacks Project is relevant:
The fpqc topology cannot be treated in the same way as the fppf topology. Namely, suppose that R is a nonzero ring. We will see in Lemma 34.9.14 that there does not exist a set $A$ of fpqc-coverings of $Spec(R)$ such that every fpqc-covering can be refined by an element of $A$. If $R=k$ is a field, then the reason for this unboundedness is that there does not exist a field extension of $k$ such that every field extension of $k$ is contained in it.
If you ignore set theoretic difficulties, then you run into presheaves which do not have a sheafification, see [Theorem 5.5, Waterhouse-fpqc-sheafification]. A mildly interesting option is to consider only those faithfully flat ring extensions $R\to R'$ where the cardinality of $R'$ is suitably bounded. (And if you consider all schemes in a fixed universe as in SGA4 then you are bounding the cardinality by a strongly inaccessible cardinal.) However, it is not so clear what happens if you change the cardinal to a bigger one. (Tag 022A)
I should add that differentiable stacks can be defined using a site with countably many objects, namely the Euclidean spaces $\mathbb{R}^n$, so all this discussion is not needed for that case.
A side question.. David Metzler in his paper https://arxiv.org/abs/math/0306176 assumes a site to be defined over a small category.. nlab says similar thing "Remark 2.3. Often a site is required to be a small category. But also large sites play a role" at https://ncatlab.org/nlab/show/site.. It looks like even defining Grothendieck topology on a large category seems to be avoided by some people.. Any specific reason for that typo of choice?
What I have understood as of now is that, even if the base category $\mathcal{S}$ is a large category, there are no size issues if I restrict my attention to the $2$-category of representable stacks over $\mathcal{S}$.. It is because this $2$-category of representable stacks over $\mathcal{S}$ is "equivalent" to the "bicategory of internal groupoids and anafunctors"..
I'm not sure what your first question is asking ("typo"?). Defining the structure of a site on a large category has no real issues, the problem is that there might not be a sheafification/stackification functor (a real size issue), even though general individual sheaves/stacks make sense. For your second question, I don't mean the 2-category of representable stacks, since they turn out to be sheaves of sets, but the 2-category of presentable stacks—this is a general term for algebraic/topological/differentiable/etc stacks over a general site. We don't need stackification for this.
I mean to say "Any specific reason for that type of choice?" For your second explanation, I think I understand what you mean.. There may not exists sheafification functor $\text{Presheaves}/\mathcal{S}\rightarrow \text{Sheaves}/\mathcal{S}$ when $\mathcal{S}$ is a large category.. I would like to read more about the precise difficulty in sheafification if you can point me to some reference.. I request you to add those lines in your previous comment to your answer.. nlab page definitio of sheafifcation starts with "Let $(C,J)$ be a site in the sense of: small category equipped with a coverage"
Waterhouse's paper Basically bounded functors and flat sheaves in Pacific J. Math. (https://projecteuclid.org/euclid.pjm/1102906018) gives an example of a presheaf on the fpqc site that does not admit a sheafification.
Thanks for the reference.. it is a short paper,I think I will be able to learn something from it..
added link for WISC. Your answer also had a link but it was not "clickable". I do not know why it is like that...
OK, thanks. Must have been a markdown formatting mistake on my part.
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2025-03-21T14:48:31.087284
| 2020-05-26T09:47:58 |
361395
|
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|
Stack Exchange
|
Number of solutions to linear diophantine equations, with natural coefficients in a box
Let c, k, d $ \in \mathbb{N} $, let a, x $ \in \mathbb{N}^k $ suppose for all i $ \leq $ k, $ x_i \leq d $, $ a_i \in \mathcal{O}(d2^i) $ and $ \sum{a_ix_i} = c $ my question is for the value of c that maximizes the number of solutions to this equation, what is a tight bound for the number of solutions to this equation for said c. I know that in the related problem of the number of solutions for a given c, when the $x_i's$ are unrestrained as c -> $ \infty $ the number of solutions is roughly $ \mathcal{O}(c^k) $.
However, my intuition tells me that constraining the $x_i$'s should reduce the maximal number of solutions to roughly $ \mathcal{O}(2^k) $ for any c, since the bounds for the $ x_i's $ guarantees zero solutions after a certain point, and the bound the $a_i's$ spreads the solution over a relatively wide range.
I know this question has some relation to work done by Richard Stanley, and I'm wondering if its actually a special case of the work of him, or someone else.
I'm kind of an abstract math noob so please ask clarifying questions if part of my question seems confusing. Also please go easy on me if this is actually a really easy problem, or related to a well known result.
Let $k,a_i$ be fixed, and $f(c)$ denotes the number of solution to $\sum_{i=1}^k a_i x_i=c$. Then by inclusion-exclusion the number of solutions bounded by $x_i\leq d$ equals
$$\sum_{A\subseteq{a_1,\dots,a_k}} (-1)^{|A|} f\big(c-(d+1)\cdot\sum_{a\in A} a\big).$$
Can you explain please. It seems like this is using inclusion exclusion on d, but it seems like it uses f(c) in the answer since that is the term in the sum for the empty set.
Yes, $A$ can be the empty subset, which corresponds to term $f(c)$.
Oh I see thats the number of solutions to the unconstrained problem, that makes sense. Thank you so much!
Do you happen to know the asymptotics of this function in terms of a_i's, d, k? Not the f function I know that one, but the inclusion exclusion sum you gave.
There is a lot known on this sort of thing. You’re asking for the “atom probabilities” for a sort of Rademacher sum (see also “anticoncentration”). It has to do with the additive structure of the coefficients $a_i$. It gets into something called “inverse Littlewood-Offord theory” with folks like Terry Tao and Van Vu.
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2025-03-21T14:48:31.087491
| 2020-05-26T10:47:34 |
361399
|
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|
Stack Exchange
|
Is the 2-dimensional Gauss-Bonnet theorem applicable in higher dimensions?
This is a cross-post of this MSE post that users commented that it is appropriate for MO.
I want to know
Question: Is the 2-dimensional Gauss-Bonnet theorem applicable (any topological or geometrical obstruction) in higher dimensions?
My idea is that one can consider 2-dimensional embedded submanifolds of $(M^n,g)$ and then applying Gauss-Bonnet theorem to all of such submanifolds then collecting these information together somehow and finding a topological or geometrical property (like fundamental group, Homology groups, etc.). Is that possible at all?
See Wikipedia on the Chern–Gauss–Bonnet theorem, which is an even-dimensional generalization of Gauss-Bonnet.
I know about Chern-Gauss-Bonnet Theorem.I want to use 2−dimensional Gauss-Bonnet to deduce a higher dimensional result.
Isn't the Chern-Gauss-Bonnet Theorem a higher dimensional result???
I am not aware of any proof of Chern--Gauss--Bonnet, or any other theorem, that arises by simultaneous gluing of all surfaces inside the manifold. I think that nothing is known about that question.
I don't know why this question has been down-voted? What kind of details needed to clarify the problem and prevent from closing? Is this a vague question?
I think it is down-voted for being too vague. In the present form, it is not answerable. As for how many details you need to add, I do not know, since I am not sure what the question really is.
The valid @ThiKu's answer shows that the question is not vague for him and who up-voted his answer. My question is: Can one get a nice result (topological or geometrical) by knowing curvature and topology of all 2-dim sub-manifolds of $(M^n,g)$?
I think the wording of the question is unclear. You're asking whether the 2-dimensional Gauss-Bonnet theorem is ever used to prove a theorem about higher dimensional manifolds.
When you say "higher dimensions" what do you mean? i.e. higher than what?
For arbitrary dimensions higher than 2.
There are some applications of Gauß-Bonnet in foliations and laminations of 3-manifolds.
The first result that comes to mind is Candel’s Uniformization Theorem, which gives necessary and sufficient conditions for a lamination to admit a leafwise hyperbolic metric. The proof uses Gauß-Bonnet in a nontrivial way. (A shorter exposition is in Chapter 7 of Calegari: “Foliatuons and the Geometry of 3-manifolds”.)
|
2025-03-21T14:48:31.087702
| 2020-05-26T11:37:51 |
361400
|
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|
Stack Exchange
|
Can a covering space of the $p$-adic disc split over the circle?
Let $D = {\rm Sp}\, \mathbb{C}_p\langle x\rangle$ be the affinoid unit disc over $\mathbb{C}_p$.
Is there an example of a connected finite etale cover of $D$ whose restriction to the "unit circle" ${\rm Sp}\, \mathbb{C}_p\langle x, x^{-1}\rangle \subseteq D$ is disconnected?
Just to point out something obvious, but maybe could be useful to someone trying to construct a counter example: I'm pretty sure the cover needs to not come from the special fiber of a normal integral model. In particular, any of the strange covers of $\mathbb{A}^1_{\overline{F}_p}$ won't help.
Does the proof of 7.5 here https://pdfs.semanticscholar.org/4f28/729a874c3864afe9fba31764d7f4e37443ce.pdf show that the answer is no?
@AriyanJavanpeykar Does the statement itself not imply the claim since the Gauss point is contained in the unit circle and the usual fact about fundamental groups :"surjective iff remains connected upon pullback"?
@AriyanJavanpeykar thank you, this answers the question, at least if "becomes disconnected" is replaced with "splits completely". To record the argument here: let $D' = {\rm Sp}, \mathbb{C}_p\langle x^{-1}\rangle$ so that ${D, D'}$ is an admissible covering of $\mathbb{P}^1$ and $D\cap D'$ is the unit circle. Given a finite etale cover which splits completely over $D\cap D'$, we can extend it to a finite etale cover of $\mathbb{P}^1$ by pasting in disjoint copies of $D'$. As $\mathbb{P}^1$ is (algebraically) simply connected, the extended cover is trivial, and hence so is the original one.
$\newcommand{\Sp}{\mathrm{Sp}\,}\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathrm{m}}$There seem to exist such examples and this does not contradict de Jong's argument because his proof only shows that the homomorphism $G:=\pi_1(\Sp C\langle x^{\pm 1}\rangle)\to H:=\pi_1(\Sp C\langle x\rangle)$ satisfies the following property: if an action of $H$ on a finite set becomes trivial when restricted to $G$ then it is trivial. Equivalently, the normal subgroup generated by the image of $G$ is equal to the whole of $H$.
Take $$A= C\langle x,y\rangle/(y^{p+1}-xy-p^{1/2})$$ The discriminant of this polynomial in $y$ is $-(p+1)^{p+1}p^{p/2}-p^px^{p+1}$ up to a unit and is invertible in $C\langle x\rangle$. Therefore, $C\langle x\rangle \to A$ is a finite etale extension.
To check that $\Sp A$ is connected it is enough to show that the polynomial $y^{p+1}-xy+p^{1/2}$ is irreducible in $C\langle x\rangle[y]$. If $y^{p+1}-xy-p^{1/2}=f_1(y)\cdot f_2(y)$ is a factorization into a product of monic polynomials then both $f_1,f_2$ have to lie in $\cO_C\langle x\rangle[y]$ so their reductions modulo the maximal ideal $\fm_C\subset\cO_C$ provide a factorization of $(y^{p}-x)y$. Hence, we may and will assume that $\deg f_1=1$ and $y^{p+1}-xy-p^{1/2}$ has a root $f(x)\in \cO_C\langle x\rangle$. A root has to have $f(0)^{p+1}=p^{1/2}$ and, taking the derivative of the equation $f(x)^{p+1}-xf(x)+p^{1/2}=0$ we also get $(p+1)f'(x)f(x)^p-f(x)-xf'(x)=0$ so $f'(0)=\frac{1}{p+1}f(0)^{1-p}\not\in \cO_C$ which is a contradiction. Therefore, $\Sp A\to D$ is a connected finite etale cover.
On the other hand, this cover splits over $C\langle x,x^{-1}\rangle $: we can find a root of $y^{p+1}-xy-p^{1/2}=0$ by the Hensel's lemma arguing by induction on $n$: suppose that $y_n\in \cO_C/p^{n/2}[x^{\pm 1}]$ is a root of this equation modulo $p^{n/2}$ that reduces to $0$ mod $p^{1/2}$. To lift it over $p^{(n+1)/2}$ it is enough to be able to supply, for any given $a\in\cO_C/p^{1/2}[x^{\pm 1}]$, an element $z$ such that $(p+1)y_n^p\cdot z-x\cdot z=a$ in $\cO_C/{p^{1/2}}[x^{\pm 1}]$. This is possible because $y_n^p-x$ is a sum of a nilpotent and invertible element, hence is invertible.
This example is motivated by the behavior of the $p$-torsion in a family of elliptic curves: given such family $\mathcal{E}\to S$ over a $p$-adic formal scheme $S$ over $\cO_K$ let $S_0(p)\to S_K$ be the etale covering of the generic fiber parametrizing $1$-dimensional $\mathbb{F}_p$ subspaces in $\mathcal{E}_K[p]$. When restricted to the generic fiber of the ordinary locus $S_{ord}$ this covering admits a section by the theory of canonical subgroup but it need not have one over the whole of $S_K$. In other words the restriction of the representation $\pi_1(S_K)\to GL_2(\mathbb{F}_p)$ to $\pi_1(S_{ord})$ lands inside a Borel subgroup.
Using Katz-Mazur-Drinfeld integral model the etale cover $S_0(p)\to S_K$ extends to a flat cover of $S$ itself and the special fiber splits into two irreducible components exactly as in the example above.
|
2025-03-21T14:48:31.088023
| 2020-05-26T11:58:46 |
361401
|
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|
Stack Exchange
|
How to combine global standard deviation given several sample statistics?
Is there any approximation formula (best guess?) to calculate global std given multiple set statistics (size, mean, std)?
I have an aggregated statistics from several sets.
set, size, mean, std, median
1, 40, 12, 4, 11
2, 12, 5, 3, 4.5
3, 28, 8, 13, 8
It is easy to calculate global mean as
$$\mu = \frac{\sum{n_i \cdot \mu_i}}{\sum{n_i}}$$
but unfortunately this won't work for std and median.
You can find the individual variances and so means of second moments easily enough, then the joint mean of second moments in the way you suggest and then find the overall variance and standard deviation
See e.g. pooled variance, pooled standard deviation.
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2025-03-21T14:48:31.088112
| 2020-05-26T15:43:03 |
361408
|
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|
Stack Exchange
|
Computing $\pi_2$ of the complement of a 2-knot or spaces with aspherical splittings?
As far as I know, it is an open question if the complement of a ribbon disk $D^2 \subset B^4$ is aspherical. In reading "Some remarks on a problem of J.H.C Whitehead" by Howie, it is noted that there has been at least one published incorrect proof of this result, namely, in this paper.
I was looking at "The homotopy groups of knots. I. How to compute the algebraic 2-type." by S. J. Lomonaco, Jr. and I noticed that in this paper the author seems to be using the fact that ribbon disk complements are aspherical in his calculations (the problem seems to start at Theorem 7.4).
Any 2-knot $K \subset S^4$ is the union of two ribbon disks, namely if we think of $S^4 = B_1 \cup B_2$ as the union of two 4-balls there are ribbon disks $D_1,D_2 \subset B_1,B_2$ with $K$ isotopic to $D_1 \cup D_2$ - namely, with respect to the standard height function restricted to $K$, isotope $K$ so that there are an equal number of minima and maxima, minima come before maxima, and rearrange the 1-handles so that all of the fusion bands come before all of the fission bands. If we assume that ribbon disk complements are aspherical, then the complement of $K$ is the union of two aspherical spaces along an aspherical space (since classical knot complements are aspherical).
The primary work in the aforementioned paper by Lomonaco goes towards proving Theorem 7.1, which gives a method for computing the homology of the universal cover of a space that can be decomposed as the union of two connected aspherical spaces along another connected aspherical space. Additionally, in this theorem, the author shows how to compute the homology of the cover as a $\mathbb{Z}\pi_1$-module and the $k$-invariant. This theorem is then applied to the above "bridge position" of a 2-knot in the previous paragraph as well as to Heegaard splittings of 3-manifolds.
I have no real reason to suspect that Theorem 7.1 is incorrect (although I have not checked through the homological algebra), but I am coming here for a little reassurance. Is this the case? Also, are there any generalizations of Theorem 7.1 for more complicated aspherical splittings (say with three connected aspherical components all with pairwise connected aspherical intersection and with a connected aspherical triple intersection)? I imagine the answer here would be some sort of spectral sequence.
Additionally, am I correct in saying that the calculations in say Figure 5, 6, 7, and 9 are incorrect? Or maybe they are just not correctly justified (relying on the unproven conjecture above) but still true? If you can show that the results to the calculations are incorrect then there is either another error in the paper that I am not aware of, or you've got a disproof of the above mentioned conjecture (and hence also the Whitehead asphericity conjecture). Therefore, I am imagining that the results are true but a different technique is needed to justify them.
My original answer (now deleted) did not address the question posed, so a portion is now a comment.
It is not the case that all 2-knots have aspherical complements. A paper of E. Dyer and A. T. Vasquez, Can. J. Math. 25 (1973), 1132–1136 shows that if a higher dimensional knots has aspherical complement then the knot group is $\mathbb{Z}$! In dimension 2 this implies topologically unknotted; Dyer-Vasquez show that asphericity implies unknotted in higher dimensions as well. For example, any nontrivial fibered 2-knot complement has nonzero $\pi_2$.
Yes, the Lomonaco paper has a gap (at least), due to its reliance on the Whitehead conjecture for ribbon disk complements, which is still open. You should see "On 2-Dimensional Homotopy Invariants of Complements of Knotted Surfaces" by J.F. Martins instead; he shows how to compute the algebraic 2-type of the complement of a knotted surface in S^4 without this assumption.
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2025-03-21T14:48:31.088425
| 2020-05-26T16:04:46 |
361410
|
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|
Stack Exchange
|
Automorphisms of $\mathbb{C}[x, y, z]$ over $\mathbb C[x]$
What are the automorphisms of $\mathbb{C}[x, y, z]$ fixing $\mathbb{C}[x]$? I.e. those automorphisms $\phi:\mathbb{C}[x, y, z]\to\mathbb{C}[x, y, z]$ s.t. $\phi(x) = x$. I am interested in a complete description of this group, similar to the description of $\operatorname{Aut}_{\mathbb{C}}(\mathbb{C}[x, y])$, which is known.
It embeds in $\mathrm{Aut}_{K}(K[x,y])$ for $K=\mathbf{C}(x)$. If the latter group has a known amalgam decomposition, it at least yields an action of the smaller group on a tree, which ought to yield another description, possibly as amalgam as well.
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2025-03-21T14:48:31.088504
| 2020-05-26T16:08:13 |
361411
|
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|
Stack Exchange
|
Distribution of the direction of Gaussian random variable
Let $X$ be a complex normal random variable. (Or, equivalently, a 2D real normal.) Is it possible to say anything useful about the distribution of the phase of $X$? Is it possible to do estimation on it?
What about the multivariate case? That is, I have a multivariate complex normal, and would like to understand the multivariate distribution of the phases of the individual components.
Does anything become easier if the real and imaginary parts are uncorrelated? (In the multivariate case, uncorrelated for each component.)
This problem arises immediately in MIMO radar direction estimation. See, for example, MUSIC
In the 2D case, you can write $X=AZ$, where $A$ is a $2\times2$ nonsingular real matrix, $Z:=[Z_1,Z_2]^T$, and the $Z_j$'s are iid standard normal. You want to find
$$p:=P(n_1\cdot X>0,\;n_2\cdot X>0),$$
where $n_1$ and $n_2$ are unit vectors in $\mathbb R^2$ and $\cdot$ is the dot product.
By the rotational symmetry of the distribution of $Z$, you have
$$p=P(m_1\cdot Z>0,\;m_2\cdot Z>0)=\frac1{2\pi}\arccos\frac{m_1\cdot m_2}{|m_1|\,|m_2|},$$
where $m_j:=A^T n_j$ and $|\cdot|$ is the Euclidean norm on $\mathbb R^2$.
When the dimension is $>2$, the problem similarly reduces to finding the probability that a standard normal random vector is in a polyhedral cone. This is a difficult problem, admitting a certain recursive solution, which can be resolved more or less explicitly for dimensions $\le4$. See e.g. Plackett and references there, notably to Schläfli.
I don't necessarily want to limit myself to the case where $\mathbb{E}(X) = 0$, and this method seems limited to that case, correct?
@ElenaYudovina : Yes, this approach works only for $EX=0$. However, the problem is hardly just with this approach. Even for the 2D case, Mathematica cannot find a closed-form expression for general $EX\ne0$ -- I think it does not exist. You can try to use the dimension reduction as in the linked paper by Plackett, but I think an ordinary integral which cannot be taken in closed form will then still remain. Alas, not many integrals can be taken in closed form!
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2025-03-21T14:48:31.088678
| 2020-05-26T16:29:29 |
361415
|
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|
Stack Exchange
|
How do solutions to this quadratic congruence distribute as the number of factors grows?
This is a question that arose during a conversation with a colleague regarding Landau's fourth problem, which asks whether there are infinitely many primes of the form $n^2+1$. The Conjectured asymptotic for
$$\sum_{n\leq X}\Lambda(n^2+1)$$
is
$$-\sum_{n\leq x^2+1}\frac{\mu(n)\rho(n)\log n}{n}\sim \frac{CX}{\log X}$$
where $\rho(n)$ counts the number of solutions to the congruence $a^2+1\equiv 0 \pmod n$ and $C$ is specific arithmetic constant. Although Landau's fourth problem is considered to be out of reach, the question I want to ask seems to be a natural prerequisite. I have not proved this implication yet, but it seems to be a nice problem anyway.
Question
Let $n\in\mathbb{N}$ be squarefree and such that $p|n\Rightarrow p=4m+1$ for some $m\in\mathbb{N}$ so $-1$ is a square $\pmod n$ with $2^{\omega(n)}$ square roots $\pmod n$. Do these roots equidistribute $\pmod n$ as $\omega(n)\rightarrow\infty$?
Weyl's criterion asserts that equidistribution is equivalent to having
$$\sum_{a^2+1\equiv 0\pmod n}e\left(\frac{ak}{n}\right)=o\left(2^{\omega(n)}\right)$$
as $\omega(n)\rightarrow\infty$ for each $0<k<n$. By the chinese remainder theorem, the above statement equates to having
$$\prod_{p|n}\cos\left(\frac{2\pi k \overline{(n/p)}\rho_p}{p}\right)=o(1)$$
where $\rho^2_p+1\equiv 0\pmod p$ and $\overline{x}$ denotes the multiplicative inverse of $x \pmod p$.
Since $0$ and $p/2$ cannot be roots $\pmod p$, the limits of such trigonometric products will be zero if unboundedly many of the arguments do not converge to $0$ or $\pi$ as $\omega(n)\rightarrow\infty$, which amounts to having unboundedly many of the sequences
$$\frac{n}{p}\pmod p\hspace{1cm}p|n$$
diverge. Is it possible that all but finitely many of them could converge?
Yes, this is a well-known. In fact one can prove the following:
Let $\{\alpha_n\}$ be an arbitrary complex sequence with $|\alpha_n| \leq 1$ for all $n \geq 1$ and finite $\ell^2$-norm. Then for any positive numbers $D, N$ one has
$$\displaystyle \sum_{d \leq D} \sum_{v^2 + 1 \equiv 0 \pmod{d}} \left \lvert \sum_{n \leq N} \alpha_n e \left(\frac{vn}{d} \right)\right \rvert^2 \ll (D + N) \lVert \alpha \rVert_2^2,$$
where $\alpha = \{\alpha_n\}$ and $\lVert \cdot \rVert_2$ is the $\ell^2$-norm.
The proof goes as follows. It clearly suffices to consider dyadic intervals of $D$, so we may assume that $D < d \leq 2D$ say. We then consider the equidistribution of the numbers $v/d$, with $v^2 + 1 \equiv 0 \pmod{d}, 0 \leq v < d$. For each such $v$ there exist positive integers $r,s$ such that $r^2 + s^2 = d$ and $r \equiv vs \pmod{d}$. This implies that we have an equation of the form
$$\displaystyle r - vs = \ell d,$$
which we rearrange as
$$\displaystyle \frac{r}{sd} - \frac{\ell}{s} = \frac{v}{d}.$$
By exploiting symmetry we can always assume $|s| \geq |r|$, and that $s > 0$. Since $r^2 + s^2 = d$ this implies
$$\displaystyle \frac{|r|}{sd} = \frac{|r|}{s(r^2 + s^2)} \leq \frac{|r|}{s(2|r|s)} \leq \frac{1}{2s^2},$$
with equality only when $r = s = 1$. Hence if we have two pairs $(v_1, d_1), (v_2, d_2)$ say we get the equality
$$\displaystyle \frac{v_1}{d_1} - \frac{v_2}{d_2} = \frac{r_1}{s_1 d_1} - \frac{r_2}{s_2 d_2} - \frac{\ell_1}{s_1} + \frac{\ell_2}{s_2},$$
which implies that the mod 1 distance between $v_1/d_1, v_2/d_2$ is $\gg 1/(s_1 s_2) \gg 1/D$ (this is because $s_i \gg d_i^{1/2}$, and $d_1, d_2 \in [D,2D)$). The large sieve inequality then gives the conclusion above.
With (a lot) more work one can even show that the fractions $v/p$ equi-distribute, where we take prime moduli only. See the following paper by Duke, Friedlander, and Iwaniec.
This is very helpful regarding equidistribution on average, thank you. Yet I don’t see how these kind of results imply that one has equidistribution for sequences of $d$s with $\omega(d)$ tending to infinity as in the question. Or is this something I am missing?
|
2025-03-21T14:48:31.089055
| 2020-05-26T16:45:12 |
361416
|
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|
Stack Exchange
|
Real-world example of a Banach *-algebra with a nonzero *-radical
Is there a real-world example of a Banach *-algebra with a nonzero *-radical (intersection of kernels of all *-representations)? Textbooks give examples of finite-dimensional algebras with degenerate involutions.
How do you define "real-world"?
I guess I'll take what I can get, with the bare minimum being "not specifically constructed for the purpose of providing an example", but it would be nice if there was some application. My best hope for a fully "natural" example is an algebra constructed from a group that only has a zero *-radical when the group has some approximation property.
Hi Cam, is the Volterra algebra with conjugation of functions as the star-operation good enough? :)
In similar vein, I vaguely recall that the disc algebra with involution given by "take complex conjugates of all the Taylor coefficients" has bad properties as a star-algebra, this crops up somewhere in Palmer vol 2 but unfortunately I don't have a copy at hand
Hey Yemon! The Volterra algebra is actually a pretty great example. I'm surprised it's not in Palmer volume 2 along with the others. The disc algebra example is 9.7.25 in Palmer, but it has a faithful *-representation. I'm curious whether there's a infinite-dimensional non-commutative example that isn't somehow based on a finite-dimensional or commutative example. Maybe a semigroup algebra of some kind?
|
2025-03-21T14:48:31.089180
| 2020-05-26T16:49:58 |
361417
|
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"url": "https://mathoverflow.net/questions/361417"
}
|
Stack Exchange
|
Sections of non-reduced schemes
Let $X$ be an affine, irreducible (complex), generically reduced, scheme containing an embedded point, say at $x \in X$. Suppose further dimension of $X$ is strictly positive (can assume to be one dimensional). Assume that the associated reduced scheme $X_{\mbox{red}}$ is non-singular. Note that, there is a natural closed immersion $X_{\mbox{red}} \to X$. Then,
1) Are there examples of $X$ as above such that there exists a morphism $X \to X_{\mbox{red}}$ such that the composition $$ X_{\mbox{red}} \to X \to X_{\mbox{red}}$$ is the identity? I know of such examples in zero dimension. I am interested to see if such a phenomenon can happen in the setup of embedded points in schemes of positive dimension.
2) If such a map $X \to X_{\mbox{red}}$ exists, then does the structure sheaf of $X$ decompose (as an $\mathcal{O}_{X_{\mbox{red}}}$-module) as $\mathcal{O}_{X_{\mbox{red}}} \oplus I\mathcal{O}_{X_{\mbox{red}}}$ for some nilpotent ideal $I$?
There's plenty of examples. Let $k = \mathbf{C}$. Take $R = k[x_1,...,x_n]$, an artinian local $k$-algebra $A$ with residue field $k$, and consider the fibre product $S = R \times_k A$, where the map $A \to k$ is the obvious one and $R \to k$ sets all $x_i = 0$. Geometrically, you've tacked on a copy of $\mathrm{Spec}(A)$ at the origin in $\mathbf{A}^n_k$. So the projection $S \to R$ is a nilpotent thickening with kernel $\mathfrak{m}A$ (and there's a section). For (2), in what category do you want the splitting? There's such a splitting after pushing down to $X{red}$ for trivial reasons...
@Anonymous Is the scheme in your example really generically reduced? It seems it is generically non-reduced.
@Jana fibre product, not tensor product. If $A = k[\varepsilon]/(\varepsilon^2)$, then $S \cong R[y]/(x_1y, \ldots, x_ny, y^2)$, which has an embedded point at the origin and the reduction is $S[y]/(y) = R$. Anonymous's construction is just a generalisation of that example.
@R.vanDobbendeBruyn Thanks. I was thinking of tensor product.
|
2025-03-21T14:48:31.089343
| 2020-05-26T17:13:14 |
361420
|
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"Francois Ziegler",
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"url": "https://mathoverflow.net/questions/361420"
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|
Stack Exchange
|
Alternate proof of uniqueness of integral curves to vector fields
Let $V$ be a continuous vector field on an open set $U \subset \mathbb{R}^n$ and let $p_0 \in U$. There are many ways to construct local integral curves of $V$ through $p_0$, i.e. differentiable maps $\gamma\colon (-\epsilon,\epsilon) \rightarrow U$ with $\gamma(0) = p_0$ whose derivatives equal $V$. This is another way of stating the existence of solutions to a system of first order ODE's given an initial value.
If $V$ is Lipshitz, then such integral curves are unique. Is there a direct way of seeing this, perhaps with a stronger smoothness assumption (e.g. $V$ being $C^{\infty}$)? The usual proof gives both existence and uniqueness by recasting this as an integral equation and showing that the solution to the integral equation is a fixed point for a contracting map on some appropriate space of functions. This has always seemed a little magical to me, in contrast to some of the more geometric arguments for existence of integral curves (but in contrast to existence, I don't know any other proofs of uniqueness).
There is Cauchy’s original method of “majorants”.
Osgood's criterion is known to be a necessary (and sufficient) condition for uniqueness, since there are non-Osgood vector-fields for which uniqueness does not hold (e.g. $V(x)=|x|^{\alpha}$ for any $\alpha\in(0,1)$ in dimension 1). Of course existence requires basically nothing, except continuity, by the Cauchy-Peano theorem.
|
2025-03-21T14:48:31.089481
| 2020-05-26T17:14:46 |
361421
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361421"
}
|
Stack Exchange
|
Radon transform range theorem and radial functions
(UPDATED for rapid decay considerations + new question)
In dimension 2, the Radon transform range theorem states that a rapidly decaying (Schwartz) function $g(t,\theta)$ can be represented as a Radon transform of some function $f(x,y)$ (i.e. $g=R[f]$) if and only if, for all integers $n\geq0$
$$P_n(\theta) := \int\limits_{-\infty}^{\infty} t^n g(t,\theta) dt$$ is a homogeneous polynomial of degree $n$ in $\cos\theta$ and $\sin\theta$. This is often referred to as the moment or Cavalieri conditions. See e.g. Helgason's book p.5, Lemma 2.2 (2011 ed.) for the property that $P_n$ must be a homogeneous polynomial of degree $n$.
Question 1: If $f$ is a radial function then its Radon transform $g=R[f]$ is known to be independent from $\theta$. Therefore all moments $P_n(\theta)$ are also independent from $\theta$, in apparent violation of the property that $P_n$ is a homogeneous polynomial in $\cos\theta, \sin\theta$ of degree $n$ when $n\geq1$. What am I missing?
For example, consider $f(x,y) = e^{-x^2-y^2} / \sqrt\pi$. Then $g(t,\theta) = e^{-t^2}$ which is independent from $\theta$ as expected of radial functions. The first moment is 0 which is NOT a homogeneous polynomial of degree 1, the second moment is $\sqrt\pi/2$ which is NOT a homogeneous polynomial of degree 2, and so forth.
Question 2: What happens when the above integral does not converge? Usually this happens when there is no solution to the Radon transform inverse problem, but consider $g(t,\theta) = (1-e^{-1/t^2})/|t|$ which is independent from $\theta$. After calculations, the inversion formula for radial function gives $$f(x,y) = f_0\!\left(\sqrt{x^2+y^2}\right), \qquad f_0(r) = \frac2{\pi r^3}\mathfrak D\!\left(\frac 1r\right)$$ where $\mathfrak D(x) := e^{-x^2}\int_0^x e^{t^2} dt$ is Dawson's function. So there exists $f$ such that $g=R[f]$, and yet
$$
P_n(\theta) = \int\limits_{-\infty}^{\infty} t^n \frac{1-e^{-1/t^2}}{|t|} dt
$$
does not converge for $n\geq 2$.
My partial answer to Q2: This specific example is not a Schwartz function. Any references to range theorems for non-Schwartz functions appreciated. I found "A Range Theorem for the Radon Transform" (Madych and Solmon, 1988), other suggestions very appreciated.
Thanks!
p.
To answer Q1: There are trig identities at play. First, 0 is usually accepted under the definition of "homogeneous polynomial" (i.e., it's a polynomial whose coefficients are all zero) so there is no contradiction there. But for the case of the second moment being constant, we have the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, so actually a constant is representable by a homogeneous trig polynomial of degree 2 polynomial, so again, no contradiction.
Edit: for Q2, The moment conditions can indeed be violated when the function is not Schwartz. A good reference for this is the paper:
Solmon, D. C. (1987). Asymptotic formulas for the dual Radon transform and applications. Mathematische Zeitschrift, 195(3), 321-343.
The main theorem of this paper shows that the inverse Radon transform maps any even Schwartz function $\phi$ over $\mathbb{S}^{d-1}\times \mathbb{R}$ to a $C^\infty$-smooth function on $\mathbb{R}^d$ that decays like $O(\|x\|^{-d})$ as $\|x\|\rightarrow \infty$, i.e., absolutely integrable along hyperplanes, but not necessarily absolutely integrable over all of $\mathbb{R}^d$. Here $\phi$ does not have to satisfy any of the moment conditions, even though the defining integrals are convergent since $\phi$ is Schwartz.
Thanks for your answer. I realized this later. However: the degree of the zero polynomial is $-\infty$, not 1. In addition, if Pn is considered as a "spherical" homogeneous polynomial subject to trigonometric identity simplifications, how would you define its degree?
I don't think there is a uniquely defined degree in this case. But I don't think that matters in defining the moment conditions. Simply take $H_n$ to be the space of all functions realizable as a homogeneous polynomial of degree $n$ evaluated on the sphere. Then the moment conditions say that $P_n$ must belong to $H_n$.
To be more clear, the proof of the moment conditions amounts to the identity:
$\int Rf(w,t) t^n dt = \int_{\mathbb{R}^d} f(x) \langle w,x\rangle^n dx$
for all $w = (w_1,...,w_n) \in \mathbb{S}^{d-1}$. The right-hand side may be formally expanded into a homogeneous polynomial of degree $n$ in the variable $w$, and this is what is meant by a "degree n" spherical polynomial.
Then I think it would be less ambiguous to say that $P_n$ is a polynomial of degree <= n
That's not true, though. For example, $H_1$ as I've defined it does not contain constant functions, since there is no linear combination of $\sin(\theta)$ and $\cos(\theta)$ that gives a constant. Likewise,$H_2$ does not contain $\sin(\theta)$ or $\cos(\theta)$, since there is no linear combination of $\sin^2(\theta)$, $\cos(\theta)\sin(\theta)$, $\cos^2(\theta)$ to give you those functions, etc.
Ok I think your $H_n$ can probably be characterized as all homogeneous polynomials of degree $n-2k, k\in{0, 1, \cdot, [n/2]$ where $[\cdot]$ denotes the floor function. Or am I missing something else?
I think that's essentially correct. But, actually, there is a cleaner way to characterize the $H_n$ using spherical harmonics: $H_n$ is the space of harmonic polynomials up to degree $n$, where a harmonic polynomial is any polynomial whose Laplacian vanishes. See Theorem 1.1.3. of Bai and Xu's "Approximation Theory
and Harmonic Analysis
on Spheres and Balls": https://link.springer.com/content/pdf/10.1007/978-1-4614-6660-4.pdf
|
2025-03-21T14:48:31.089837
| 2020-05-26T19:45:30 |
361425
|
{
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"Alexandre Eremenko",
"David Roberts",
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|
Stack Exchange
|
Determining the most popular MSC codes of new papers
How can I tell what are the most popular MSC codes of new papers (using arXiv and/or publications in MachSciNet indexed journals)? I would like to compare general areas (e.g. number theory vs dynamical systems) as well as specific codes like 11Y05, with date ranges like 1 year, 5 year, 10 year.
This question was answered in Quantitatively speaking, which subject area in mathematics is currently the most research active? about 10 years ago, and I think an update is needed.
Welcome to MathOverflow! May I suggest to change the title in order to reflect the content of the question more precisely? It seems debatable whether the number of publications is a good measure of "popularity" of a field. (And since MathOverflow is probably not the right place to discuss what is and what is not a good measure of popularity, I suggest to avoid this discussion by rephrasing the title.)
You can use "Citations" and then "Top 10" buttons in Mathscinet to estimate this. It gives you most cited papers in every MSN subject. Just compare how many times these top cited papers are cited. Compare between the subjects.
For example, in the subject 65 ten top papers have more than 1000 citations, while in the subject 31 a typical "top ten" paper is cited only about 200 times.
I edited the title in accordance with Jochen Glueck's suggestion.
"People with subscription access can use "Citations" and then "Top 10" buttons in Mathscinet to estimate this." FTFY
|
2025-03-21T14:48:31.089989
| 2020-05-26T20:34:02 |
361430
|
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|
Stack Exchange
|
when a set of roots extend to a system of simple roots
Given a set of roots in a root system, assume that the pairing of each two roots in this set is not positive. Then clearly the set gives a closed root subsystem. My question is, when this set extends to a system of simple roots in the original root system. Of course, it is not always true, for example taking closed subsystem of long roots in B_2. Is it true for any simply-laced case. We are particularly interested in the exceptional type E_6.
By the way, it's unclear what you mean by "the set gives a closed root subsystem."
Let $\alpha_1,\ldots,\alpha_6$ be the simple roots of $E_6$, and suppose that $\alpha_3$ is the root corresponding to the trivalent node of the Dynkin diagram. Let $\theta$ be the highest (positive) root. Then $\{\alpha_1,\alpha_2,\alpha_4,\alpha_5,\alpha_6,-\theta\}$ is a system of simple roots for a root system of type $A_2 \times A_2 \times A_2$, so it clearly does not extend to a system of simple roots for $E_6$.
This example is coming from Borel-de Siebenthal theory, which basically says that the maximal rank sub-root systems of a root system are given by taking the extended Dynkin diagram of the Dynkin diagram, and deleting some node. The "affine node" of the extended Dynkin diagram corresponds to $-\theta$ the negative highest root.
(By the way, if you insist that the system of simple roots give an irreducible root system, then the answer changes in this case of $E_6$, because the only connected Dynkin diagrams we can get from the extended Dynkin diagram by deleting a node are of type $E_6$.)
EDIT: Let me mention a positive result, since you seem interested in that as well. Let $\Phi$ be a root system in a vector space $V$. Let $V'\subseteq V$ be a subspace spanned by a subset of roots. Set $\Phi' := \Phi\cap V'$, a sub-root system of $\Phi$. Let $S'$ be a system of simple roots for $\Phi'$. Then $S'$ can be extended to $S$, a system of simple roots for $\Phi$. This is for instance Bourbaki, "Lie groups and Lie algebras," Chapter VI, Section 1.7, Proposition 24.
Thanks. If we already know the set of roots generates a root system whose Dynkin diagram is a subdiagram of E_6, can we conclude that this is a Levi subsystem of E_6?
@JJH: What does "Levi subsystem" mean?
It exactly means, the set of roots can be extended to a system of simple roots.
Hmm, I'm not sure. My guess would be not necessarily. For instance, take $S={\alpha_1,\alpha_2,\alpha_4,\alpha_5,-\theta}$, where $\alpha_6$ corresponds to the node adjacent to the affine node in the extended Dynkin diagram. Then $S$ is a system of simple roots for a Dynkin diagram of type $A_2\times A_2\times A_1$, which is a sub-diagram of $E_6$, but I don't see an obvious way to extend $S$ to a system of simple roots for all of $E_6$ (OTOH, I don't see why this is impossible either).
@JJH: see my edit for a positive result in this direction.
Thanks a lot for the reference.
|
2025-03-21T14:48:31.090189
| 2020-05-26T20:47:36 |
361432
|
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|
Stack Exchange
|
Asymptotic density of sums of consecutive primes
Call a positive integer respectable if it is a sum of consecutive prime numbers. For example, every prime numbers is respectable. So are $3+5=8$, $2+3+5=10$, $5+7=12$, $3+5+7=15$, $2+3+5+7=17$, $7+11=18$, but $3+11=14$ is not. So are $13+17+19=49$ and $23+29=52$.
Define the respectable counting function $F$ by defining $F(X)$ to be the number of respectable numbers less than or equal to $X$.
Can you find an asymptotic formula for $F(X)$ in terms of elementary functions? For example, is there a real number $r$ such that $F(X) \sim rX/\ln{X}$? Of course, $r \ge 1$, by the prime number theorem.
These are tabulated at http://oeis.org/A034707. The complementary sequence, numbers not a sum of consecutive primes, is http://oeis.org/A050940 where it says, "this sequence is infinite with lower density at least $1 - \log 2 = 0.306\dots$" with the reference Leo Moser, Notes on number theory. III. On the sum of consecutive primes, Canad. Math. Bull. 6 (1963), pp. 159-161. The link to Moser's paper doesn't work.
The lower bound will be $$(\sum_{k=0}^{x_1} \pi(\frac{x}{k})) - \frac{x_1(x_1-1)}{2}$$, where $x_1$ is such that $\pi(\frac{x}{x_1})<x_1$ but $\pi(\frac{x}{x_1-1})>x_1$
We should add $\frac{1}{2}$ with the lower bound.
The Moser paper is available at https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/on-the-sum-of-consecutive-primes/99CB1E53FB57B2D2084D58D2383B1905
Please use a high-level tag like "nt.number-theory". I added this tag now. Regarding high-level tags, see https://meta.mathoverflow.net/q/1075/
This is only a partial answer, but probably too long to be a comment. One can count, for fixed $k$, the number of sums of $k$ consecutive primes.
Since we are concerned with consecutive primes, each sum is indexed by the smallest summand. For a prime $p$ let us write $p = p^{(1)}$, and $p^{(2)}, \cdots, p^{(k)}$ to be the next $k-1$ primes in order. We are thus considering the counting function
$$\displaystyle \pi^{(k)}(X) = \# \{p : p^{(1)} + \cdots + p^{(k)} \leq X \}.$$
Note that for fixed $k$ there is no problem with double counting, since the sums are strictly increasing.
We now use the fact that we can control the gaps between consecutive primes. Indeed, one can show that for sufficiently large $X$ the interval $[X, X + X^{0.525})$ contains a prime (this is a theorem due to Baker, Harman, and Pintz). Therefore, it follows that
$$\displaystyle p^{(1)} + \cdots + p^{(k)} = kp^{(1)} + O \left(p^{0.525} \right).$$
Hence, we see that
$$\displaystyle \pi^{(k)}(X) = \pi(X/k) + E(X),$$
where $E(X)$ counts the number of primes $p$ such that $kp \leq X$ but $p^{(1)} + \cdots + p^{(k)} > X$. This only happens for $p \in [X/k - O(X^{0.525}), X/k)$ and as long as $k$ is fixed, we can estimate the number of such primes by counting primes in short intervals, getting that there are $O(X^{0.525}/\log X)$ such primes. This error is far smaller than the error inherent in $\pi(X/k)$ when pulling out the main term $(X/k)/\log(X/k)$, so we can ignore it. Hence
$$\displaystyle \pi^{(k)}(X) \sim_k \frac{X}{k \log X}.$$
The arguments above give quite a bit of flexibility in terms of allowing $k$ to tend to infinity with $X$, but the exact range of uniformity probably takes a bit of work to obtain. The issue of repeat representations however will likely be a thornier issue.
|
2025-03-21T14:48:31.090448
| 2020-05-26T21:02:43 |
361434
|
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|
Stack Exchange
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Translation of Marsden-Weinstein-Meyer into classical mechanics language
The Marsden-Weinstein-Meyer theorem is expressed in a too general way to be understood by a mean square physicist, as me. However, if we limit the scope to a Hamiltonian mechanics, it should be possible to express it in the Hamiltonian language, at least to some extent.
So my question is what the M.W.M. theorem says on the following problem. I have a Hamiltonian $H$, and I know that it is in involution with another function $J$, i.e. $\{H,J\}=0$.
1) What are the additional hypothesis that must be met by $H$ and $J$, so that the M.W.M. theorem can be applied?
2) Under those hypothesis, what is the thesis?
But: since the problem is expressed in simple and practical terms, also the answer must be absolutely practical. Derivatives, Poisson parenthesis, solutions of differential equations are allowed; everything that can be followed by a physicists is allowed; but no Lie group, no symplectomorphism, no coadjoint action is allowed in the answer. Where this is not possible, examples should be given.
Literature references are also welcome!
Edit: I already asked related questions, which are however different from this one. Here, I'm asking what are exactly the hypothesis and the thesis of MWM limited to a specific situation. Not a general explanation of reductions. In particular, I would like to know: what are the hypothesis on $J$, if the result is globally valid, and what is the relation with Poincarè reduction.
...physicists don't understand Lie groups?
Well, the meaning of my question is different. Since I'm asking about a given J, which generates a Hamiltonian flow, I want that the answer refers to that flow, and not to a general mathematical concept which somehow has to do with it. Mathematicians are great when it is a matter of generalizing, but they have difficulties in dealing with practical cases. I can show you a thread on MO where no one was able to answer a very simple question related to MWM, although they were likely able to discuss MDM in a very general way. That's why I'm now asking.
In cases when your Lie group is 1-dimensional and simple connected, i.e. the real number line, i.e. when there is precisely one function $J$ as the moment map, i.e. the cases you want to know about, then MWM is essentially Poincare reduction: locally change variables to get $J=p_n$, and then $H$ turns out not to depend on $q_n$, and $p_n$ is constant along flow of $H$, so on level sets of $p_n$, $H$ reduces to one fewer variable. However, the MWM story is not purely local. MWM requires, even here, global hypotheses, and gives a global conclusion. The function $J$ perhaps cannot be globally made into $p_n$, since Darboux coordinates are only local. However, if the flow lines of $J$ on a level set of $J$ can be parameterized by a smooth manifold, we can make a global statement as below. I don't know a reference for this.
A regular point $x_0$ of a function $y=f(x)$ is a point at which at least one partial derivative $\partial f/\partial x_i$ is not zero. A regular value $y_0$ of a function $y=f(x)$ is a point so that every point $x_0$ at which $f(x_0)$ is equal to $y_0$ is a regular point. By a theorem of Sard, almost every value of a smooth function is a regular value.
Take a Hamiltonian function $H$ on a symplectic manifold $X$, i.e. a Hamiltonian system with Hamiltonian $H$. If we pick a regular value $J_0$ of $J$, then the level set $X_{J_0}\subset X$, i.e. the set of points where $J=J_0$, is a submanifold of $X$ invariant under the Hamiltonian flow of $J$. Suppose that the set of flow lines are parameterized by a smooth manifold $Y$ of dimension one less than the dimension of the level set. Let $\varphi\colon X_{J_0} \to Y$ be the map taking each point $x\in X_{J_0}$ to the flow line of $J$ through that point $x$. Then there is a function $h$ on $Y$, so that $H(x)=h(\varphi(x))$ for any point $x\in X_{J_0}$. (We say that $H$ descends down to $Y$, and write $h$ as $H$.) This $h$ is a Hamiltonian of a Hamiltonian system on that smooth manifold $Y$, for a natural symplectic structure.
Simplest example: if $J=p_n$ in global Darboux coordinates, i.e. $H$ is independent of $q_n$, then we can use coordinates $q_1,\dots,q_{n-1},p_1,\dots,p_{n-1}$ for that quotient manifold. In that case, we can write $H(p_1,\dots,p_n,q_1,\dots,q_n)$ as a function $h(p_1,\dots,p_{n-1},q_1,\dots,q_{n-1})$.
A symplectic structure on $Y$ means that there is some way to set up a Hamiltonian system on $Y$, but a precise definition requires a familiarity with differential forms or some other mathematical structure which I can't perhaps give you. The point is then that $\varphi\colon X_{J_0} \to Y$ takes Hamiltonian paths of $H$ on $X$ (which, when they start on $X_{J_0}$, always stay on $X_{J_0}$) to Hamiltonian paths of $h$ on $Y$, for the associated Hamiltonian system with $h$ as Hamiltonian function.
It is true, as Michael says, that this essentially says that ignorable things are ignorable, i.e. that if you don't use $q_i$, you can skip using $p_i$ too. If you would like an easy way to see that, note that near any regular point of a function $J$, there are Darboux coordinates in which $J=p_n$. So that reduce the local picture to the study of Hamiltonians $H$ for which $\{p_n,H\}=0$, and this you can work out trivially by hand.
Let me give an example, as the result is perhaps still not clear. If $H(p_1,q_1,p_2,q_2)=p_1^2+q_1^2+p_2^2$, and $J(p_1,q_1,p_2,q_2)=p_2$, then $h(p_1,q_1)=p_1^2+q_1^2+J_0^2$.
The first example where we cannot use global Darboux coordinates to work this out is the harmonic oscillator $J=(1/2)(p_1^2+q_1^2+\dots+p_n^2+q_n^2)$ where $X=\mathbb{R}^{2n}$ with usual Darboux coordinates. The regular values $J_0$ of $J$ are any nonzero values. The level sets $X_{J_0}$ of $J$ are spheres: $J=J_0$ is a sphere of radius $\sqrt{2J_0}$. The quotient space $Y$ of flow lines is a complex projective space: if we take $z_j=p_j+\sqrt{-1}q_j$, then there is a unique flow line on $X_{J_0}$ for each complex line spanned by a vector $z=(z_1,\dots,z_n)$. The symplectic structure on the complex projective space is the famous Fubini--Study symplectic structure. See Arnold, Mathematical Methods of Classical Mechanics, p. 24, for this example with $n=2$, and Appendix 3 for the general construction of this symplectic structure.
Could you please extend a bit? "Regular value" is a value that is not - say - a maximum or minimum? "Descends" means that you take the values of H on the manifold? "For a natural symplectic structure" means that there is a canonical transformation such that this happen? Yes, we physicists are simple minded people!
Oh - seems like what it's saying is that one can ignore an ignorable coordinate (using its conserved conjugate momentum as an input parameter) ... but maybe I'm being too simple-minded ...
There is still something unclear. It seems that Y parametrizes the flow lines only on a given level set (in the example in Darboux coordinates, the level set p_n=P has dimension n-1, and the flow lines on it are parametrized with n-2 dimensions). Then, phi should be a function from the level set to Y, not X -> Y. Could you please explain?
@DorianoBrogioli: correct, I will fix that.
Naively, I would describe this procedure saying that we make a canonical transformation such that one of the Darboux coordinates is equal to J; this is the case in your example, or e.g. when we exploit the conservation of angular momentum in three-body problem. Instead you say that it it is not so easy and it requires more complex mathematical structures. Are you sure that there is at least one case of H and J in which the MWM holds, but the result cannot be obtained with a properly chosen canonical transformation?
It is not a canonical transformation, because the number of variables in $h$ is smaller than in $H$. Canonical transformations preserve dimension. But as I said, for the problem as you posed it, without Lie groups and with only one real valued function $J$, the result of MWM is just the obvious result above, when written in the appropriate Darboux coordinates.
Phi is not a canonical transformation, but the whole procedure can be described in the following way. You make a canonical transformation which brings J into p_n. Then, you consider H as a function of p_1, p_n-1 , q_1, .. q_n-1 ; you treat p_n as a fixed parameter and neglect q_n. This is the procedure called "Poincarè reduction" in Arnold's books. Here, I'm (kindly) challenging you to provide an example of H and J that can be treated with MWM but not with the Poincarè reduction.
In cases when your Lie group is 1-dimensional and simple connected, i.e. the real number line, i.e. when there is precisely one function $J$ as the moment map, i.e. the cases you want to know about, then MWM is Poincare reduction.
Wonderful. Can you please add this statement in your answer? I would be grateful if you can also provide a reference where this is stated explicitly. I never found it. Btw, most mathematicians are not able to give this simple answer, so, yes, the question and the answer are not too trivial for MO.
So MWM is only locally valid? Well, this is surprising! For this reason. In Arnold's books, it is proven that Poincarè's reduction can be done locally for any conserved quantity J. Here, instead, you are requiring in the hypothesis that J has a property: "the flow lines on X_J0 can be parametrized by a smooth manifold Y of dimension one less than the dimension of the level set." This is not true for any J (I have an example). So it seems that MWM hypothesis is more restrictive, while the thesis is the same. Is this true?
I have tried to add some detail to clarify the local versus global aspect. The conclusion is not just local; the manifold $Y$ of flow lines has a global symplectic structure, and the map $\varphi$ matches up the symplectic flow of $H$.
Fine. So there are cases in which J cannot be brought into p_n by a global canonical transformation, but still MWM gives a global reduction. This reduction is expressed in terms of symplectic structure (I guess, by giving a suitable bilinear form) but not in terms of canonical transforms. I'm still kindly challenging you to give an example of such a J.
I’m not sure if this is what you are looking for, but you can think of this as fixing the values of conserved quantities. You have a function on phase space which commutes with the
Hamiltonian using the Poisson bracket. Therefore, the value of this function is conserved. Pick a value, and you can describe the equations of motion using one fewer variables.
As a simple example, take the cotangent bundle to $S^1 \times \mathbb{R}$ with the usual Hamiltonian. Then, $J$ is the momentum along the circle, which is conserved. Fix that to the value $j$. Then the motion of a particle can be described solely using the phase space $T^*\mathbb{R}$ with the Hamiltonian $\frac{p^2 + j^2}{2m}$. This is arises from the full phase space as the space $J^{-1}(j)/U(1)$, where the quotient arises because we have fully described the motion in the circle direction. This is precisely the quotient space in the theorem.
This is called "Poincarè reduction" in the Arnold's books. It is demonstrated that it can be done locally. I was wondering if MWM can say something more on the topic. So, is MWM globally valid? And what are exactly the hypothesis on the conserved value?
Yes, it's global. The hypotheses are mostly to ensure that the inverse image is a nice space that is acted on nicely by the residual symmetries.
The hypothesis you mention is on the Hamiltonian flow of J? @Ben McKay says that "the set of flow lines are parameterized by a smooth manifold Y of dimension one less than the dimension of the level set." Is this the hypothesis? Is so, please add it to your answer.
The condition is that the value of the conserved variable is a regular value, which means that the inverse image is a manifold. You also have to assume that the symmetry (sticking to the single conserved quantity case) acts nicely on that manifold so the quotient is also a manifold.
"the symmetry acts nicely on that manifold so the quotient is also a manifold" ... For example, it "acts nicely" for J=p_n, but it "does not act nicely" if J is a non-integrable Hamiltonian. Is this correct? Can you try to express the condition in terms of simpler things like integrability?
The theorem is stated in terms of group actions, ie, your conserved quantity is related to a continuous symmetry. That seems pretty simply to me. Probably the condition is just that the Hamiltonian vector field associate with J is complete, but I might be forgetting something.
Yes, it sounds simple also to me. But I'm asking what it implies in terms of classical mechanics. The class of the J such that their Hamiltonian flow has the required property of "acting nicely" is clearly characterized. But now I'm asking: is it a class already known in classical mechanics, e.g. is it the class of integrable or superintegrable Hamiltonians, or can it be characterized by looking at the Lyapunov exponents, or has chaotic trajectories? Or it is completely unrelated? It is curious that the mathematicians find this question so strange and that there is no literature!
It sounds like you want to ask the question: is there an intrinsic characterization of those conserved quantities that give rise to a complete Hamiltonian vector field. Maybe ask a new question to make that clear?
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2025-03-21T14:48:31.091406
| 2020-05-26T22:14:06 |
361442
|
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Stack Exchange
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history of geometric mechanics
I was thinking about the foundations of geometric mechanics and its precursors.
I wondered who was the first to realized the equivalence between Riemannian geometry and Lagrangian mechanics. In particular:
Solutions (trajectories) of Lagrange equations are geodesics of Levi-Civita connection.
The moment of inertia tensor can be viewed as a Riemannian metric tensor.
I would be grateful for any answers and relevant references to this topic.
I’m voting to close this question because it's better suited to hsm.stackexchange than here.
Looks like a perfectly reasonable question to me, offering someone the opportunity to explain some interesting mathematics. Why is there no "vote against" on the closure link?
(Anonymity? Man up.) There are knowledgble users of MO who refuse to use HSM, so better to post both places with a cross-ref.
Darboux was perhaps the first to argue fully explicitly that “the general problem of mechanics is nothing but the generalisation to an arbitrary number of variables of the problem of the study of geodesics” (Leçons sur la théorie générale des surfaces, Paris, 1889, p. 500). A very useful article on the (pre)history of this is Lützen, Interactions between mechanics and differential geometry in the 19th century.
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2025-03-21T14:48:31.091540
| 2020-05-26T23:06:47 |
361445
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Stack Exchange
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Example of primitive substitution with two rationally independent eigenvalues?
I am looking for an example of a primitive substitution $\sigma$, not Pisot, such that the associated subshift $X_\sigma$ has two irrational and rationally independent eigenvalues. Equivalently, a substitution $\sigma$ such that $X_\sigma$ has a minimal irrational rotation of the 2-dimensional torus as a factor.
There are well-known criteria for deciding whether a given complex number is an eigenvalue of $X_\sigma$, and a lot of examples of substitutive systems in the literature. But I couldn't build nor find an example of a substitution not Pisot with this kind of eiganvalues.
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2025-03-21T14:48:31.091609
| 2020-05-26T23:14:40 |
361446
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361446"
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Stack Exchange
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Upper bound $\tau_C := \int_{\|x\| \le 1}(vol(C \cap (x + C))/vol(C))dx$ for a convex body $C \subseteq \mathbb R^n$, by reducing to a ball
Let $C$ be a convex body in $\mathbb R^n$, i.e a bounded convex subset of $\mathbb R^n$ which has nonempty interior, and which is (A) open, or (B) closed (I'm not sure one makes more sense; choose the one which works best for you). Consider the number $\tau_C \in [0, 1]$ defined by
$$
\tau_C := \frac{1}{\omega_n}\int_{\|x\| \le 1}\frac{vol(C \cap (x + C))}{vol(C)}dx = \frac{1}{\omega_n}\int_{\|x\| \le 1}\mu_C(x + C)dx,
$$
where $\omega_n$ is the volume of the unit ball in $\mathbb R^n$ and $\mu_C$ is the uniform measure of $C$. In general, estimating $\tau_C$ would be challenging, since $C$ can be very arbitrary. An exception is when $C$ is euclidean ball. The goal is then to obtain good upper-bounds on $\tau_C$.
The baseline approach. Let $C'$ be the Steiner symmetrization of $C$ w.r.t a coordinate axis. In this post (claim 4; also see comments under the question), it was shown that $\tau_C \le \tau_{C'}$. On the other hand, it is well-known that $C$ can be symmetrized into a ball $B$ of same volume as $C$, via a process $C \rightarrow C' \rightarrow \ldots \rightarrow C^{(k)} \rightarrow \ldots \rightarrow B$ (which is convergent in Hausdorff distance), it follows that $\tau_C \le \tau_B$.
Question
Is there an alternative way to establish that $\tau_C \le \tau_B$ (where $B$ is a ball of same volume as $C$) without using Steiner symmetrization, but perhaps, using another mapping / process. Can this be done using, for example the Brenier map between $C$ and $B$ ?
N.B.: Breniers map because, after all, such maps are used to prove the minimality of the, say, the surface of a ball (via Brunn-Minkowski), and so maybe they could also be of help here.
Is there perhaps a way to directly upper bound $\tau_C$ as a function of its volume ?
Thanks in advance for your help!
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2025-03-21T14:48:31.091760
| 2020-05-26T23:46:13 |
361449
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Stack Exchange
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Do unitarily equivalent normal operators always have same cyclic vectors?
Let two normal operators A and B be unitarily equivalent, i.e. A = U*BU where U is unitary operator.
It follows that A and B have or do not have cyclic vectors at the same time, but does it follow that these vectors are the same?
It seems not, but I cannot create a counterexample.
Won't the cyclic vectors of $U^{-1}BU$ be just $U^{-1}v$ for all cyclic vectors $v$ of $B$?
So to create a counterexample, let $u$ and $v$ be unit vectors such that $u$ is a cyclic vector for $A$ while $v$ is not, let $U$ be a unitary so that $U u = v$, and let $B = U A U^*$.
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2025-03-21T14:48:31.091832
| 2020-05-27T03:21:28 |
361454
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"Brendan McKay",
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Stack Exchange
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Minimum mean over all random variables subject to logarithm constraint
Does the following problem have a solution?
$$
\min_X \mathbb{E} X
\quad\text{subject to}\quad
\mathbb{E} \log X = C.
$$
Here, the minimization is with respect to all integrable random variables $X$ and $C$ is some constant. Alternatively, instead of minimizing over random variables, one may equivalently view this as optimizing over the space of probability measures.
Put $X=e^Y$. By Jensen's inequality, $E e^Y\ge e^{EY}=e^C$. Since concentrating $X$ at $e^C$ achieves that bound, it is the best.
Assume $X>0$ a.s. (so the constraint can be satisfied) and write $Y=\log X$.
By Jensen's inequality (https://en.wikipedia.org/wiki/Jensen%27s_inequality),
$\mathbb{E} X \ge e^{\mathbb{E} Y}=e^C$, so the minimum is attained for $X$ such that $\mathbb{P}(X=e^C)=1$.
I beat you by 3 minutes, but I hope we didn't just do someone's homework.
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2025-03-21T14:48:31.091925
| 2020-05-27T03:42:40 |
361455
|
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"Pietro Majer",
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Stack Exchange
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Integral convergence implies pointwise
This is a simplified version of a question I am looking at, but embarrassingly I can't do this one anyway.
Let's assume $f(x)$ is a decreasing positive function, $f(0)$ is infinite and, moreover, $f(x)$ is comparable to $1/x$ near $x=0$. We look only at positive $x$'s.
Question. Assume we know that the limit
$$
\lim_{a\to 0^+}\int\limits_{a}^{2a} f(t)dt
$$
exists as a positive finite number. Does it imply that
$$
\lim_{n\to\infty} f(2^{-n})\cdot 2^{-n}
$$
exists?
Note: it definitely does not imply that the limit of $xf(x)$ exists.
If you are ok with $f$ being not continuous and not strictly decreasing, here's a simple enough counter-example. (With a bit more work you can make it continuous and strict, but it distracts from the point.)
Let $f(x)$ be the step-function defined by
$$ f(x) = \begin{cases}
2^{k} & x \in ( 2^{-k}, 2^{1-k}) \\
2^{k} & x = 2^{-k}, \quad k \text{ is odd} \\
2^{k+1} & x = 2^{-k}, \quad k\text{ is even}
\end{cases} $$
You have that the integral
$$ \int_a^{2a} f(t) ~dt = 1 $$
for any $a$.
You have that $2^{-k} f(2^{-k})$ alternating between 1 and 2 and does not converge.
Incidentally, the "comparability to $1/x$" in your question is superfluous. Since you assumed decreasing, you have that
$$ a f(2a) \leq \int_a^{2a} f(x) ~dx \leq a f(a) $$
which implies that
$$ \liminf xf(x) \geq \lim \int_a^{2a} f(x) ~dx $$
$$ \limsup xf(x) \leq 2 \lim\int_a^{2a} f(x)~ dx$$
so comparability is automatic, and also the existence of a sequence $x_n \to 0$ such that $x_n f(x_n)$ converges. (Of course $(2^{-n})$ need not be such a sequence.)
if one wants a strictly decreasing example, adding 1/x to your f(x) will do it
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2025-03-21T14:48:31.092059
| 2020-05-27T05:51:51 |
361459
|
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|
Stack Exchange
|
Base change of normalization map and scheme-theoretic surjectivity
Let $C$ be an affine, integral curve and $f: \widetilde{C} \to C$ be its normalization. Let $g:D \to C$ be a finite, affine, surjective morphism (note $D$ need not be reduced, but can assume generically reduced). Denote by $D'$ the base change of $\widetilde{C}$ by the morphism $g$ and $g': D' \to D$ the resulting morphism. Is the morphism $g'$ scheme-theoretically surjective i.e., the induced ring homomorphism
$\mathcal{O}_D \to \mathcal{O}_{D'}$ is injective?
No. Let $A \to B$ be a ring map. Let $M$ be a finite $A$-module such that $M \to M \otimes_A B$ is not injective. Then with $A' = A \oplus M$ where $M$ is an ideal of square zero and $B' = A' \otimes_A B$ the base change, we see that $A' \to B'$ is not injective.
Apply this with $A = k[t^2, t^3]$ where $k$ is a field, $B = k[t]$ is the normalization of $A$ and $M = k[t^2, t^3]/I$ where $I$ is the ideal generated over $A$ by $t^5$. Then $t^6 \not \in I$ but $M \otimes_A B = B/J$ where $J$ is the ideal in $B$ generated by $t^5$ and so $t^6 \in J$. Hence $M \to M \otimes_A B$ is not injective.
Sorry, forgot to add the condition that the morphism $g$ is surjective. I have edited the question.
The morphism $\text{Spec}(A \oplus M) \to \text{Spec}(A)$ is surjective as it has a section. This is my last comment on this question.
I am sorry, I do not think your example works. Ofcourse $M \to M \otimes_A B$ is not injective. But, this does not imply $A' \to B'$ is not injective. In particular, in your example you claim that the morphism $k[t] \to (k[t^2, t^3] \oplus k[t^2, t^3]/(t^5)) \otimes k[t]$ is not injective, where the tensor product is over $k[t^2, t^3]$. But, note that, $t^a$ for any $a$ maps to $(1 \oplus 1) \otimes t^a=(t^a \oplus t^a) \otimes 1$ which is never zero. I am assuming that the ring homomorphism from $k[t^2, t^3] \to k[t^2, t^3] \otimes k[t^2, t^3]/(t^5)$ sends $f$ to $(f,f \mod t^5)$.
@Jana you consider the map $A \to B'$ instead of the map $A' \to B'$.
@JoshuaMundinger Sorry, I wrote it in a confusing manner, but you can use the arguments to check that the map from $A'$ to $B'$ is injective. The important point to note is that as $g$ is assumed to be surjective, the most natural map from $A$ to $A \oplus M$ is the one that takes $f$ to $(f,f \mod t^5)$ (I think this is the map that Johan considers). But, for such a map, the induced map from $A'$ to $B'$ is injective. Probably, you can come up with a different injective map from $A$ to $A \oplus M$, under which $A' \to B'$ is not injective, but I could not think of such a map.
@Jana if you mean the map $A \to A'$ over which we take the tensor product, this is $a \mapsto (a,0) \in A \oplus M$.
@JoshuaMundinger The map you give is not a ring homomorphism. Remember, $1 \in A$ must go to $(1,1)$ in $A \oplus M$.
@Jana there is no $1 \in M$. The identity of $A \oplus M$ is $(1,0)$. I suggest writing down the ring operation in $A \oplus M$ and checking that this map is a homomorphism.
@JoshuaMundinger $M$ is the quotient of a polynomial ring, so of course contains $1$.
@Jana The construction given in the question was for an arbitrary module $M$, which has no $1$. The quotient of a polynomial ring is thought of as a module, not a ring. Thus, the right map to take is $a \mapsto (a,0)$.
@JoshuaMundinger You want $A'$ to be a ring (in the notation of the question, this should be the coordinate ring of $D$). For an arbitrary module $M$ there is no product structure and $A'$ will not be a ring.
@Jana There is a product structure: $(a,m)(a',m') = (aa', am' + a'm)$. That is the structure in the question.
@JoshuaMundinger Ok, I understand the ring structure. But I still cannot reconstruct the example of Johan using the maps you suggested. Perhaps you could write your arguments as an answer along with a concrete example. The ring structure that you suggested was far from my first guess, so I think it would be a good idea to write these details as an answer to complement the answer of Johan
|
2025-03-21T14:48:31.092607
| 2020-05-27T05:56:15 |
361460
|
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|
Stack Exchange
|
Image of extension ideal classes homomorphism in ideal class group under Artin map in class field theory
Let $K/P$ be a finite extension of number fields and $\epsilon_{K/P}:[\mathfrak{a}] \in Cl(P) \rightarrow [\mathfrak{a}.\mathcal{O}_K]\in Cl(K)$ be the ideal class transfer homomorphism. It's well known that there exists an isomorphism $\gamma:Cl(K) \rightarrow Gal(H_K/K)$ by $[\mathfrak{b}] \mapsto \left( \frac{H_K/K}{\mathfrak{b}}\right)$ (the Artin symbol at $\mathfrak{b}$), for the Hilbert class field $H_K$ of K. Is there any result (s) about the combination $\gamma \circ \epsilon_{K/P}$?
This is known as the "Verlagerung" map (German for "transfer"). It doesn't have a particularly nice description purely on the Galois side. If you google for "class field theory" and "Verlagerung" you'll find plenty of references.
Many thanks for your response.Using the "transfer map", I want to see what happens in $P \subseteq F \subseteq K$ a tower of finite extensions of number fields. Roughly speaking, what is the image of "combination" of transfer maps?
Harvey Cohn has written repeatedly about transfer maps in class field theory in his books on algebraic number theory and the construction of class fields.
|
2025-03-21T14:48:31.092722
| 2020-05-27T05:56:45 |
361461
|
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"url": "https://mathoverflow.net/questions/361461"
}
|
Stack Exchange
|
What is known about the irrationality of ratios and products of logarithms of integers?
Let $a,b,c,d$ be positive natural numbers such that $\{a,b\} \neq \{c,d\}$ and such that none are perfect powers. Is it true that
$$\frac{\log a \log b}{\log c \log d} \notin \mathbb{Q} ?$$
The theorems of transcendental number theory, such as the Lindemann-Weierstrass theorem, say a fair amount about the transcendence and irrationality of linear combinations of log's of natural numbers, but it isn't clear to me what one can say about products/quotients of logs of natural numbers. What, if anything, is known here?
Even less ambitiously, can one show that there is no non-trivial solutions to expressions such as
$$ \frac{a b \log c \log d} {cd\log a \log b } =1$$
in the natural numbers?
A minor MathJAX note: I think you should write that out as
$$\frac{\log a \log b}{\log c \log d} \not\in \mathbb{Q},$$
@MarkLewko?
Isn't it already hard to say anything about $\log2\log3$?
There is a new ($218$ pages long though) preprint by Calegari, Dimitrov, Tang in which the authors claim (among many other things) that $\log (1+1/m)\log (1+1/n)$ is irrational as long as $|\frac{m}{n}-1|<10^{-6}, m,n$ integers not $-1,0$; see https://arxiv.org/abs/2408.15403 Theorem $C$ page $4$
|
2025-03-21T14:48:31.092841
| 2020-05-27T06:06:40 |
361462
|
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|
Stack Exchange
|
Cobordism monopole Floer homology
From the famous book: Monopole and three manifold, Kronheimer and Mrowka(https://www.maths.ed.ac.uk/~v1ranick/papers/kronmrowka.pdf). It is known that:
Let $Y$ be a closed oriented $3$ manifold, choosing a spinc structure $\mathfrak s$ and metric $g$ and a generic perturbation $p$, one can construct the monopole Floer homology groups:
$$\check{HM}_*(Y,\mathfrak s, g,p),~\hat{HM}_*(Y,\mathfrak s, g,p),~\overline{HM}_*(Y,\mathfrak s, g,p).$$
The groups are graded over a set $\mathbb J_s$ admitting a $\mathbb Z$ action.( details are given in Section 20-22). We define the negative completions(Definition 23.1.3 of the book) by
$$\check{HM}_\bullet(Y,\mathfrak s, g,p),~\hat{HM}_\bullet(Y,\mathfrak s, g,p),~\overline{HM}_\bullet(Y,\mathfrak s, g,p).$$
If we want to consider all spinc
structures at the same time, we need to consider the completed monopole Floer homology
$$\check{HM}_\bullet(M,F;\mathbb F)=\bigoplus_\mathfrak s\check{HM}_\bullet(M,F,\mathfrak;\mathbb F).$$
To show that these homology groups are independent of the metric and the perturbation, the authors gave a property: a cobordism between 3-manifolds gives rise to homomorphisms between
their monopole Floer homologies(see Section 23-26). They construct a homomorphism from $\check{HM}_\bullet(Y,g_1,p_1)$ to $\check{HM}_\bullet(Y',g_2,p_2)$, where there is a cobordism from $Y$ to $Y'$.
Q I do not understand the two points below:
Why the authors use the negative completion, where we need it?
If we just want to show that the monopole Floer homology $\check{HM}_*(Y,\mathfrak s)$ is independent of the metric and perturbation, can we just using the trivial cobordism $[0,1]\times Y$ to show a homomorphism $\check{HM}_*(Y,\mathfrak s,g_1,p_1) \to \check{HM}_*(Y,\mathfrak s, g_2,p_2)$? The homomorphism is given by counting the number of solutions of the zero-dim moduli space $M([a_1],W^*,[b_2])$, where $W^*=(-\infty,0]\times Y\cup I\times Y\cup[1,\infty)\times Y$, and $[a_1]$ and $[b_2]$ are the critical points of $(Y,\mathfrak s,g_1,p_1)$ and $(Y,\mathfrak s,g_2,p_2)$ respectively. I think the arguments of Section23-25 also work before taking the negative completion .
PS Let $G_*$ be an abelian group graded by the set $\mathbb J$ equipped with a $\mathbb Z$-action. Let $O_a(a\in A)$ be the set of free $\mathbb Z$-orbits in $\mathbb J$ and fix an element $j_a\in O_a$ for each $a$. Consider the subgroups
$$G_*[n]=\bigoplus_a\bigoplus_{m\geq n} G_{j_a-m},$$
which form a decreasing filtration of $G_*$. We define the negative completion of $G_*$ as the topological group $G_\bullet\supset G_*$ obtained by completing with respect to this filtration.
This is explained in the remark below Theorem 23.1.5. However, if the restriction map $\text{Spin}^c(W) \to \text{Spin}^c(Y) \times \text{Spin}^c(Y')$ has finite fibers, as is the case for $W = Y \times I$, then there is a cobordism map at the level of the uncompleted grading. So you get invariance of the uncompleted groups, but not full functoriality. (The topological energy, which you need to bound to get compactness, is bounded by the expected dimension of the resulting moduli space and a quantity depending on the underlying spin^c structure.)
The first bullet is definitely explained in the book! Surely around where it was introduced, it has to do with summing over all spin-c structures. We need to pass to the completion because the 4-manifold can have infinitely many spin-c structures that would need to be used.
The second bullet, yes. In general we should not expect results concerning completions of a graded group to also hold for the uncompleted group. But here we consider the trivial cobordism, and spin-c structures on $[0,1]\times Y$ are the same as spin-c structure on $Y$, so no completion is needed in this situation.
|
2025-03-21T14:48:31.093085
| 2020-05-27T06:49:50 |
361463
|
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|
Stack Exchange
|
Equivariant resolution of singularities with equivariant centres
From what I understand, given a complex projective variety X inside a compact complex manifold Y, according to Hironaka, there is a sequence of $r$ blowups $Y_i$ of Y along complex submanifolds (centres) $C_i$ such that the proper transform $\tilde{X}$ of $X$ in $Y_r$ is smooth. By functoriality, if $G$ is say, a compact connect Lie group, acting on $Y$ such that $X$ is invariant, then there is a lift of this action to $Y_r$ such that $\tilde{X}$ is also invariant.
My questions are as follows :
1. Does the above work even if $X$ is reducible ?
2. Can the centres $C_i$ be chosen to be invariant under $G$ ?
I do not believe the original proof was stated as functorial in the way that you write. However, it is now known that there is a resolution algorithm that is "functorial" for smooth morphisms, e.g., for actions of a (necessarily smooth) complex Lie group. This should be in the book by Koll'ar on resolution of singularities. Cheers Vamsi -- I hope that this helps!
|
2025-03-21T14:48:31.093185
| 2020-05-27T07:08:29 |
361464
|
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|
Stack Exchange
|
Rank of a sum with free products
Let $G$ be a finitely generated group. Does there exist a constant $\kappa$ depending only on the rank of $G$ such that, if $G \simeq F_1 \oplus \cdots \oplus F_n$, then at most $\kappa$ factors are non-trivial free products?
Motivation. In order to optimize some of the results from my preprint, I would to like to say that a finitely generated group can be decomposed as a graph product over a finite graph with factors which do not split non-trivially as graph products. This cannot be done as stated because there exist finitely generated groups isomorphic to their own square, so I introduced the concept of graphically irreducible groups: a group $G$ is graphically irreducible if, for every graph $\Gamma$ and every collection of groups $\mathcal{G}$ indexed by $V(\Gamma)$ such that $G$ is isomorphic to the graph product $\Gamma \mathcal{G}$, the graph $\Gamma$ must be complete.
Expected result: A finitely generated group decomposes as a graph product over a finite graph with graphically irreducible factors.
The idea is to argue by induction over the rank. If $G$ is not graphically irreducible, then it is isomorphic to a graph product $\Gamma \mathcal{G}$ where $\Gamma$ is finite and not complete. The graph $\Gamma$ can be decomposed as a join $\Gamma_0 \ast \Gamma_1 \ast \cdots \ast \Gamma_n$ where $\Gamma_0$ is complete and where each graph among $\Gamma_1, \ldots, \Gamma_n$ contains at least two vertices and is not a join. It is not difficult to see that the factors indexed by the vertices in $\Gamma_1 \cup \cdots \cup \Gamma_n$ have smaller ranks (compared to $G$). But we have no information on the subgroup $\langle \Gamma_0 \rangle$ generated by the factors indexed by $V(\Gamma_0)$. However, a positive answer to the question mentioned above would imply that we can take $n$ maximum in the previous decomposition, and the expected result is proved. Indeed, for every $1 \leq i \leq n$, the graph $\Gamma_i$ is not complete, so the subgroup $\langle \Gamma_i \rangle$ surjects onto a non-trivial free product $F_i$. Consequently
$$G \simeq \Gamma \mathcal{G} \simeq \langle \Gamma_0 \rangle \oplus \langle \Gamma_1 \rangle \oplus \cdots \oplus \langle \Gamma_n \rangle$$
surjects onto $F_1 \oplus \cdots \oplus F_n$, which implies that this sum has rank $\leq \mathrm{rank}(G)$.
No.
Indeed, write $C_p\ast C_p=\langle u_p,v_p\mid u_p^p=v_p^p=1\rangle$.
Let $J$ be any finite set of primes, and $G_J=\prod_{p\in J}C_p\ast C_p$. Then $G$ has generating rank two, regardless of $J$: indeed, it is generated by $u_J=\prod_{p\in J}u_p$ and $v_J=\prod_{p\in J}v_p$. While $G_J$ is a direct product of $|J|$ groups, each of which is a nontrivial free product.
Nice example, thank you.
|
2025-03-21T14:48:31.093484
| 2020-05-27T07:09:21 |
361465
|
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|
Stack Exchange
|
Undecidable statements of second-order arithmetic
What are examples of statements of second order arithmetic (SOA) that are undecided by that theory? How do they relate to the existence of large cardinals?
For example, you can ask whether there exists a real coding a model of ZFC.
What kind of second-order arithmetic? Arithmetic with full second-order semantics is categorical and therefore complete. Axiomatic second-order arithmetic such as $Z_2$ is, despite the name, an ordinary first-order theory in a two-sorted language, so you can take e.g. $\mathrm{Con}_{Z_2}$ by Gödel’s theorem.
The first thing I thought of was the continuum hypothesis, but I'm not sure if it can be stated in SOA.
@none there is such a statement. $\aleph_1$ (or rather $\aleph_1+1$) is naturally a quotient of $2^\omega$ (or rather $2^{\omega\times\omega}$) under the definable relation of isomorphic wellorderings. So an injection from $2^\omega$ into $\aleph_1$ makes sense. However, we shouldn't hope for a definable injection of this kind but we may posit one by expanding the language using a new symbol.
@FrançoisG.Dorais No, there is no such statement. CH is a third-order $\Sigma^2_1$ statement that is not ZFC-provably equivalent to any $\Pi^2_1$ statement, and a fortiori to any second-order statement. If you expand the language with an extra symbol it is no longer the language of arithmetic.
@EmilJeřábek Yes, you are correct. I misused the word "statement" in its more general sense. Expanding the language is necessary as you explain. (Of course, any $\Sigma^2_1$ statement can be made second-order by adding a symbol for the witness.)
If you are talking about $\mathsf{Z}_2$, then my favorite example of an independent sentence is Borel determinacy. Although, there are plenty of other examples of independent sentences coming from descriptive set theory.
@FedorPakhomov Even $\Delta^0_4$ determinacy is beyond $\mathsf{Z}_2$.
|
2025-03-21T14:48:31.093665
| 2020-05-27T07:37:19 |
361467
|
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|
Stack Exchange
|
Partial sums of $\sum_0^\infty z^n$
Let $z$ be a complex number with $|z|<1$. For every subset $A\subset\mathbb N$, the series $\sum_{m\in A}z^m$ is convergent. Denote $S(A)\in\mathbb{C}$ its sum and $\Sigma_z$ the set of all numbers $S(A)$. Remark that the cardinal of $\Sigma_z$ is (likely) that of ${\cal P}({\mathbb N})$, the continuum.
Is it possible that $\Sigma_z$ be a neighbourhood of the origin ?
Clarification question: do you mean $\Sigma = {\sum_{m\in A} z^m : |z|<1}$ for fixed $A$? (So $\Sigma$ depends on $A$ and $S(A) = \sum_{m\in A} z^m$ depends on $z$? Or is it the other way around?)
@FrançoisG.Dorais: for fixed, non-trivial $A$ the function $z\mapsto \sum_{m\in A}z^m$ is a non-constant holomorphic function, thus open.
Aha! I think I had it the wrong way around: $z$ is fixed and $\Sigma = { S(A) : A \subseteq \mathbb{N}}$. My confusion.
The question is: Does there exist $|z|<1$ such that $\Sigma_z={\sum\limits_{n\in A} z^n: A\subseteq \mathbb N_0}$ is a neighbourhood of $0$.
I believe that for $z=re^{2\pi i\alpha}$ two necessary conditions are $r\ge 1/2$ and $\alpha\notin \mathbb Q$.
@JochenWengenroth Why is $\alpha\notin\mathbb Q$ necessary when $r<1$?
To record what I just understood from @JochenWengenroth's comment: $r \geq 1/2$ is necessary because $\Sigma = z\Sigma \cup (1 + z\Sigma)$ and therefore $\mu(\Sigma) \leq 2r\mu(\Sigma)$ where $\mu$ is Lebesgue measure. ($\Sigma$ is evidently compact and therefore of finite measure.)
I agree with @EmilJeřábek. Wouldn't $z = i/\sqrt{2}$ work knowing that every integer has a negabinary representation? https://mathworld.wolfram.com/Negabinary.html
You are right, @EmilJeřábek. That was too quick, and in view of Francois Dorais' answer indeed wrong.
@JochenWengenroth your intuition might still be correct in some way: perhaps $1 > r > 1/2$ is sufficient when $\alpha$ is irrational?
I think every $z$ with $|z|$ large enough (I mean, larger than some explicit number $<1$) works, by a Hausdorff dimension argument? at least showing $\Sigma_z$ has nonempty interior. A closely related problem has been quite studied, see https://lamington.wordpress.com/2014/12/21/roots-schottky-semigroups-and-bandts-conjecture/#more-2372. Btw $\Sigma_z$ is either a Cantor (automatic for $|z|<1/2$) or is compact connected, so is of cardinal $2^{\aleph_0}$ in all cases.
@YCor Not exactly, because $z$ must not be real !
@DenisSerre of course $z$ is not real in the linked blog.
@FrançoisG.Dorais $\mu(z\Sigma) = r^2 \mu(\Sigma)$, right? So that gives $r \geq 1/\sqrt 2$.
@BartMichels Yes, absolutely! Thanks for correcting my tiredness! (I need to go to bed now.)
The number $z = i/\sqrt2$ seems to work!
Given $x \in [-2/3,4/3]$ we can find a "negabinary" expansion
$$x = \sum_{k=0}^\infty (-1)^k\frac{b_k}{2^k},$$ where each $b_k \in \{0,1\}$.
Similarly, given $y \in [-2/3\sqrt2,4/3\sqrt2]$ we can find
$$y = \frac{1}{\sqrt2}\sum_{j=0}^\infty (-1)^j\frac{c_j}{2^j},$$
where each $c_j \in \{0,1\}$.
Therefore,
$x + iy = \sum_{n \in A} z^n$ where
$$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$
As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $|z| \geq 1/\sqrt2$ is necessary. By definition, $$\Sigma_z = z\Sigma_z\cup(1+z\Sigma_z).$$
If $\mu$ denotes Lebesgue measure, then $\mu(z\Sigma_z) = |z|^2\mu(\Sigma_z)$ thus $\mu(\Sigma_z) \leq 2|z|^2\mu(\Sigma_z)$. Since $\Sigma_z$ is compact, it has finite measure and thus if $\mu(\Sigma_z)>0$ then we must have $|z|^2 \geq 1/2$.
It remains open whether $|z|\geq1/\sqrt2$ and $z \notin \mathbb{R}$ is sufficient for $\Sigma_z$ to contain $0$ in its interior.
I'm a bit tired so the interval bounds might be a bit off... Please check my arithmetic!
Denis: thank you for correcting my accidental abuse of the letter $i$. For what its worth, I think the case $z= i/\sqrt2$ is anecdotal and doesn't fully answer your question.
François, it answers my question. Of course, it raises the more general question of which z has this property ?. By the way, how do you prove that $\Sigma_z$ is measurable ?
Denis, $\Sigma_z$ is compact since it is a continuous image of $2^{\mathbb N}$. Therefore it is measurable and has finite measure.
See Davis, Chandler; Knuth, Donald E. Number representations and dragon curves-I. J. Recreational Math. 3 (1970), no. 2, 66–81. and part II in J. Recreational Math. 3 (1970), no. 3, 133–149.
|
2025-03-21T14:48:31.094000
| 2020-05-27T08:21:23 |
361469
|
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|
Stack Exchange
|
Veblen function with uncountable ordinals & beyond
Disclaimer: I am not a professional mathematician.
Background: I have been researching large countable ordinals for awhile & I think the Veblen function is particularly eloquent. My understanding is that $\Gamma_0$, the small Veblen ordinal & the large Veblen ordinal are all significantly smaller than the first uncountable ordinal $\omega_1$. Having some extra time during quarantine, I had an idea to extend the Veblen function to the domain of uncountable ordinals & created the following notation. I would like to know how far this notation reaches & if anything similar already exists.
Note: For the sake of brevity I have omitted numerous steps from the hand written derivation of this notation.
Consider $\phi_0'(\alpha)=\omega_\alpha$ such that:
$$\phi_0'(0)=\omega_0=\omega$$
$$\phi_0'(1)=\omega_1$$
Nesting these functions results in:
$$\phi_0'(\phi_0'(0))=\omega_\omega$$
$$\phi_0'(\phi_0'(\phi_0'(0)))=\omega_{\omega_\omega}$$
Next, consider the supremum of the previous nestings:
$$\phi_1'(0)=\sup\{\omega, \omega_\omega, \omega_{\omega_\omega},...\}$$
$\phi_1'(0)$ is then the first fixed point of $\phi_0'(\alpha)$ which correlates to $\phi_1(0)=\varepsilon_0$ being the first fixed point of $\phi_0(\alpha)=\omega^\alpha$ in the original Veblen function.
Continuing as in the original case, we eventually hit the limit of our single variable function. At this point ($\Gamma_0$ in the original), we turn to the multivariable function:
$$\phi_{1,0}'(0)=\phi'(1,0,0)=\sup\{\phi_1'(0),\phi_{\phi_1'(0)}'(0),\phi_{\phi_{\phi_1'(0)}'(0)}'(0),...\}$$
Again, like in the original case with the small Veblen ordinal, we eventually get stuck. At this point we move to the version of the Veblen function with a transfinite number of variables.
$$\phi'(1@\omega)=\sup\{\phi'(1,0),\phi'(1,0,0),\phi'(1,0,0,0)\}$$
Eventually this notation reaches as cap as well. In the orginal case, this is called the large Veblen ordinal & is the cap of the original Veblen function. In the expansion, we simply iterate our 'jump' operator:
$$\phi_0''(0)=\sup\{\phi'(1@0),\phi'(1@\omega),\phi'(1@\varepsilon_0),...\}$$
We can keep going by iterating the base function such that:
$$\Phi_0(0)=\sup\{\phi_{0}'(0), \phi_0''(0), \phi_0'''(0),...\}$$
Given the general form $\alpha_\gamma^\beta(\delta)$ we are essentially:
maxing out $\delta \leadsto$ incrementing $\gamma$
maxing out single variable $\gamma \leadsto$ multivariable $\gamma$
maxing out multivariable $\gamma \leadsto$ incrementing $\beta$
maxing out $\beta \leadsto$ incrementing $\alpha$
Repeating the process a couple more times results in:
$$\sup\{\Phi_0(0),\Phi_0'(0),\Phi_0''(0),...\}=\psi_0(0)$$
$$\sup\{\psi_0(0),\psi_0'(0),\psi_0''(0),...\}=\Psi_0(0)$$
Looping repeatedly reminded me of the original Veblen function process & so I created the following function:
$$\Xi(\alpha, \beta, \gamma, \delta)=\alpha_\gamma^\beta(\delta)$$
Such that:
$$\Xi(0,0,0,0)=\phi_0(0)=1$$
$$\Xi(0,0,0,1)=\phi_0(1)=\omega$$
$$\Xi(0,0,1,0)=\phi_1(0)=\varepsilon_0$$
$$\Xi(0,1,0,0)=\phi_0'(0)=\omega$$
$$\Xi(0,1,0,1)=\phi_0'(1)=\omega_1$$
$$\Xi(1,0,0,0)=\Phi_0(0)$$
$$\Xi(2,0,0,0)=\psi_0(0)$$
$$\Xi(3,0,0,0)=\Psi_0(0)$$
If you made it this far, thank you for taking the time. To reiterate, how far does this notation reach & does anything like this already exist?
What is your question?
@Wojowu Is there any problems with this notation & how far does this reach?
I'm afraid MathOverflow is not the place to ask about general thoughts on some idea of yours (see this meta discussion), though if you have some specific questions it will be more on-topic. As to how far it reaches - this is hard to answer, because there aren't any common notations which reach cardinals this large, bar any variations of the one you have just presented. No notation of this kind can reach an inaccessible, nearly by definition. High level indication is that ZFC proves such notations have limits, but it can't prove inaccessibles exist.
@Wojowu Based on your feedback & this meta question I have updated my question. [Aside: I'm a big fan of your Higher order set theory]
@meowzz I am not sure whether I correctly understood all the definition that you proposed. Nevertheless the idea of making the analogue of Veblen functions by starting from $x\longmapsto \aleph_x$ instead of $x\longmapsto \omega^x$ seems to be pretty straightforward. Although, I am not familiar with any particular works that studied this, I personally have considered an analogue of Veblen functions based on $x\longmapsto \beth_x$ when I was studying superintuitonistic propositional logics that require frames of large cardinality for Kripke completeness (this never got to a publication).
@meowzz Nevertheless, if I understood correctly what you are doing here, this cardinals definitely have been considered before. Namely, if you look at stronger ordinal notation systems based on collapsing functions, in the same time they provide a notation system for (relatively) large cardinals. In particular consider the notation system that is used in analysis of $\Delta^1_2\text{-}\mathsf{CA}_0+\mathsf{BI}$ and $\mathsf{KPi}$ that collapses first inaccessible cardinal. In particular it has notations for your analogues of $\Gamma_0$, small Veblen ordinal, and large Veblen ordinal.
I haven't looked at the post in enough detail, but my impression upon a cursory look is that the following should be a safe upper limit. In OTM model add an extra tape or two basically serving as an "oracle" to calculate $\omega_i$ given $i$ number of $1$'s on the "advice tape" [shouldn't be too difficult to give equally powerful additional commands in other models]. I suppose this should be a safe upper-limit to cover the kind of cases mentioned in OP? Or am I missing something here?
You might find 4.33 (p. 152) of Levy's Basic Set Theory of interest. Let $A^{(0)}$ be the class of alephs, let $A^{(1)}$ be the class of ordinals $\gamma$ such that $\aleph_{\gamma}=\gamma,$ let $A^{(2)}$ be the class of ordinals $\gamma$ such that $\gamma$ is the $\gamma$-th ordinal $\alpha$ such that $\aleph_{\alpha}=\alpha,$ let $A^{(3)}$ be the class of ordinals $\gamma$ such that $\gamma$ is the $\gamma$-th ordinal in the enumeration of $A^{(2)},\ldots$ (process continues transfinitely so that $A^{(\beta)}$ is defined for all ordinals $\beta).$
Now take the diagonal intersection of the $A^{(\beta)}$'s to obtain the class of ordinals $\gamma$ such that $\gamma$ is the $\gamma$-th ordinal $\alpha$ such that $\alpha \in A^{(\beta)}$ for all $\beta < \alpha,$ and now repeat the process with the class of ordinals in the diagonal intersection, and continue for another diagonal intersection, and another and another into the transfinite, then look at ordinals $\gamma$ such that $\gamma$ doesn't show up as a least (nonzero?) ordinal in any of these classes until you've done $\gamma$ many diagonal intersections, $\ldots$
A useful google search is Mahlo + cardinal + hyperinaccessible.
@FedorPakhomov I would be interested in reading what you wrote, if at all possible. I haven't seen much about ordinal collapsing functions with inaccessibles (I normally see them with uncountables) - but it seems interesting & I will look into it more. If you have any resources you would reccomend, I'm open to suggestions.
@SSequence I have looked at ITTMs a bit & have been wanting to look into OTMs. If you have any good reccomendations for resources I'd appreciate it.
@DaveLRenfro That portion of Levy's work does indeed look interesting! Also, I have made that Google search numerous times. Still having a bit of a hard time understanding even the smallest of the large cardinals..
having a bit of a hard time understanding even the smallest of the large cardinals --- I've pretty much given up on getting much of an understanding of how large they are (and they're a bit outside my main areas of expertise anyway), and I now mainly view them as larger than I could ever hope to reach by the kinds of vaguely explicit or the least bit constructive (in a very loose sense) modes of travel that I know about. It's as if I only know how to walk (proceed step-by-step) and I want to reach some of the distant quasars; another mode of travel is needed.
@DaveLRenfro Well put! I would upvote your comment more times if I could.
@meowzz I am not sure what would be the best source for this. But you could check the paper "Ordinal notations based on a hierarchy of inaccessible cardinals" by Wolfram Pohlers.
What I believe is the slowest "mode of travel" faster than Veblin hierarchy methods was developed (I think) by Heinz Bachmann in early 1950s. The best introductory survey I know of (in English) is Normal functions and constructive ordinal notations by Larry William Miller. See also, at least for the references, A survey on ordinal notations around the Bachmann-Howard ordinal by Wilfried Buchholz (2016).
It's a bit long for a comment, but I'll make several points.
These are not uncommon ordinals.
I've seen them used in Rathjen's ordinal collapsing function involving Mahlo cardinals, which he denotes $\Phi$. As the comments point out, they appear in various places.
This is not at all how the multivariable Veblen function behaves (before the edit).
Your $\phi_{1,0}'(0)$ is simply $\phi_{\phi_1'(0)}'(0)$. It would be akin to saying that $\Gamma_0=\phi(\phi(1,0),0)$, which is not at all true.
To explain how the multivariable Veblen function works, I recommend seeing it as recursively closing over itself on lexicographically smaller arguments. In short, left-most arguments are more significant than right-most arguments. That is, we have things like $(1,0,0)>_L(\omega,0)>_L(3,0)>_L(2,\omega)>_L(1,0)$. From this, one can see that $\Gamma_0=\phi(1,0,0)$ is greater than $\phi(\alpha,\beta)$ for any $\alpha,\beta<\Gamma_0$. This can be shown to be equivalent to
$$\phi(1,0,0)=\sup\{\phi(1,0),\phi(\phi(1,0),0),\phi(\phi(\phi(1,0),0),0),\dots\}$$
but makes more sense when considering transfinitely many arguments.
As far as I can tell, it's significantly smaller than the usual Veblen function modified with $\phi(\alpha)=\omega_\alpha$.
The Veblen function is already optimal, as far as this kind of recursion goes. Thus, the fact that your functions have significantly less arguments than the general Veblen function will make it much smaller. A quick look and I'd say only 5 or 6 arguments of the Veblen function would be needed to outperform your functions.
Thanks for the feedback! I had a feeling I was not annotating $\phi_{1,0}'(0)$ correctly. I updated the equation, however am still unsure if this is the proper fundamental sequence. Wrt "the usual Veblen function modified w/ $\phi(\alpha)=\omega_\alpha$" - do you know of any literature that does this? I've only seen the idea mentioned a few times online & haven't actually seen anyone work through it. (Aside: did you create the $\omega_\star$ notation here or does it come from somewhere?)
I would recommend steering away from fundamental sequences. These ordinals are not even countable, so even if they aren't regular, they may not have countable cofinality. Personally, I find it a lot simpler to work with the concept of closure: "what operations can apply to smaller ordinals that will not result in ordinals larger than $\phi_{1,0}'(0)$?" As I mentioned, I've seen it used in Rathjen's papers. I can try to pull this up later. Per the aside, it was made up on a whim.
@SimplyBeautifulArt Rathjen's notation system for $\mathsf{KPM}$ definitely covers this cardinals. But as I pointed out in my comment to this question, this cardinals would be already present if we collapse below the first inaccessible.
@FedorPakhomov That is likely true, but I don't remember off the top of my head if Rathjen had used $\Phi$ there.
I just want to note that I already understood what you put about the multivariable Veblen function. I just am not as sure about the notation of the version with transfinitely many variables. After looking a bit more I found something that put it as $\phi(1@\omega)$ which seems logical to me. I may modify this post soon to integrate that notation. My idea was that $\phi_0''(0)$ would be the supremum of the transfinitely many variables version. Also, while $\phi_\alpha'(\beta)$ might have similar strength (more, less or equal) to the normal function modified w/ $\omega_1$, what about $\Xi$?
@meowzz The Veblen function on transfinitely arguments works the same way as I have already described. With your fixes though, if I am understanding them correctly, $\phi_0''(0)$ would simply be $\phi(\omega@\omega)$. It is not clear what you mean to have after that, but everything is probably not past the Veblen function on $\omega+k$ arguments for maybe $k=2$ or $k=3$.
How would you annotate the large Veblen ordinal?
@meowzz With what? And I think we're kind of going a bit far off topic. If you want to, we could enter a chat room.
Chat would be great.
@meowzz We can chat in here if you'd like.
I am hoping an expert would answer this question so as to shed light on deeper or more profound points. As such, this is a basic answer covering some easy to understand points. This is based upon number of things I thought about years ago (it seems that some of those observations can be used in this question).
So let's start with your question "how far does this notation reach". I don't know what would be the answer to the question. It seems that to be able to answer though one would have to frame the question much more precisely (and I am not certain what that framing would be). Meanwhile the specific constructions you are posting (and far beyond that) are easily understood thinking in terms of generalized notion of being able to do complex calculations on ordinals.
For example, let's talk about something specific. In the beginning of your post you mention a way of starting with the function $x \mapsto \omega_x$ and how to arrive at an ordinal that is analogous to $\Gamma_0$. This analogy can be made precise using infinite programs that are sufficiently powerful. How so? Assume that a function $f:\mathrm{Ord} \rightarrow \mathrm{Ord}$ is "given" to the program. Exactly the same program that takes one to $\Gamma_0$ (given $f(x)=\omega^x$) will take one to "analogue of $\Gamma_0$" that you mention in your question. The only difference is that the function $f$ "given" to the program now is $f(x)=\omega_x$.
Now the same observations apply to bigger ordinals. I haven't studied the original Veblen paper so I am not 100% sure if the correspondences that I mention below are exact or not (so please correct if they aren't).
One way to think about SVO is in terms of a function $F:(\omega_1)^\omega \rightarrow \omega_1$. For example, writing $\omega_1=w$, we will have $\mathrm{SVO}=\mathrm{sup}\{\,F(w^i) \,\, | \,\, 1 \leq i<\omega\}$. This is analogous to thinking $\Gamma_0$ in terms of $F:(\omega_1)^2 \rightarrow \omega_1$. So, we will have $\Gamma_0$ as the first fixed point of the ordinal function $x \mapsto F(\omega_1+\omega_1 \cdot x)$. Quite informally, I use the term "storage-functions" for these functions $F$. The $\omega_1$ isn't quite relevant in the sense that we just need an ordinal "big enough" ($\omega_{CK}$ would be sufficient in the above two cases). But anyway, that's besides the point. The point here is that when a function $x \rightarrow \omega^x$ alongside with a command of form $u:=\omega_1$ is given to us, then there is a specific infinite program which can compute the storage function (in input-output sense).
Is this relevant to your question? Yes. The same program that gives us SVO when given the function $x \mapsto \omega^x$ will take us to the "analogue of SVO" in the question (using the function $x \mapsto \omega_x$). But the issue of "storage function" seems to become trickier in this "analogue case".
EDIT: I am not suggesting to gloss over several important aspects such as equivalence of different definitions. If we are being fully detailed, I will admit the paragraphs above are quite insufficient. END
Finally, very briefly, towards the end you mention "extension" of transfinite variable. In the case of original hierarchy these kind of basic extensions would be handled by extending the domain of the "storage function" by a very modest amount. For example, from $F:(\omega_1)^{\omega_1} \rightarrow \omega_1$ to $F:(\omega_1)^{\omega_1} \cdot \omega \rightarrow \omega_1$ etc. Similarly observations made earlier in this post about the "same" program taking us to the "analogue" of corresponding ordinal would apply (when given $x \mapsto \omega_x$ instead of $x \mapsto \omega^x$).
EDIT2: To OP (as a precaution): Please note that just writing $F:(\omega_1)^{\omega_1} \rightarrow \omega_1$ (or anything of that sort) doesn't mean that the underlying function has been fully well-defined and neither I meant to imply that. In the given specific cases, precise definition can either be descriptive or based upon a (infinite) program which computes the function (given an extra command of form $u:=\omega_1$). Showing that the given def. satisfy certain desirable/required properties is bound to be more work. END
How time consuming it would be to write the detail of storage functions? For $(\omega_1)^2 \rightarrow \omega_1$ (starting with $x \mapsto \omega^x$) taking us to $\Gamma_0$ it should be fairly simple (though still a bit long to post all of it here). And then it gets lengthier, as it gets more complicated.
Nice approach! Do you have any good resources for infinite programs?
I do not know. A lot of them are quite technical. The usage of infinite programs here is quite basic. I am merely suggesting they can preserve the details of operations (in the same natural number) of using $x \mapsto \omega^x$ to go to $\Gamma_0$ and using $x \mapsto \omega_x$ to go to "analgoue of $\Gamma_0$" which you mention in the question.
The use of $F$ and $\omega_1$ as you describe is more or less equivalent to Chris Bird's ordinal collapsing function, and the analog of $\omega_1$ for the modified Veblen function would be an inaccessible cardinal or something of similar nature.
@SimplyBeautifulArt "and the analog of $\omega_1$ for the modified Veblen function would be an inaccessible cardinal or something of similar nature" ...... Interesting observation. Yeah I wasn't sure (since I don't know anything about bigger cardinals etc.) about what would be the large-enough "container value" (analogous to $\omega_1$) when we are considering "storage function" in an extended manner such as in OP. That's why I left this point. Though I am reasonably certain that for the cases such as in OP we should be able to take some kind of supremum (continued)
of halt-values (of suitable programs) as an analogue of $\omega_1$ (for "container functions"). But I haven't really thought about it carefully enough. Regarding collapsing functions, I don't know anything about them (except that they are based on closure) ..... but anyway, that's besides the point. Anyway, I hope this comment of mine wasn't too confusing.
An inaccessible cardinal $I$ is regular and a fixed-point of $x\mapsto\omega_x$, which ensures that $F:I^\omega\to I$ is definable on all inputs (i.e. we never go beyond $I$).
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2025-03-21T14:48:31.095193
| 2020-05-27T09:03:39 |
361472
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Stack Exchange
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Deformations of the Riemann zeta function
Consider the Dirichlet series (for fixed $0 < a \leq 1$):
$$\zeta_a(s) = \sum_{n\geq 1}\frac{a^n}{n^s}$$
which reduces to the Riemann zeta function for $a=1$. What is known about this function, in terms of functional equations or relations to other standard Dirichlet series? If we try to do the standard trick with Mellin transforms to relate this to a Jacobi theta function, it works and we get that:
$$\pi^{-s/2}\Gamma(s/2)\zeta_a(s) = \int_0^\infty\frac{\nu(z,iy)-1}{2}y^{s/2}\frac{dy}{y}$$
for $\nu(z,\tau)$ the Jacobi theta function and $\exp(2\pi iz) = a$:
$$\nu(z,\tau) = \sum_{n\in \mathbb Z}\exp(\pi in^2\tau + 2\pi inz).$$
Is there some nice functional equation with non zero $z$ for $\nu(z,\tau)$?
The motivation for this post comes from this blog post of Matt baker where he shows that for $a$ integral, $\sum_{d|n}\mu(n/d)a^d \equiv 0 \pmod n$ as a generalization of Fermat and Euler's theorem.
This suggests that the arithmetic function $n \to a^n$ has interesting arithmetical properties and perhaps the corresponding Dirichlet function would be interesting. Unfortunately, $\zeta_a(s)$ is divergent for $a > 1$, the regime we actually care about!
So perhaps, we should instead look for p-adic analogues. More precisely, for $0 < a < 1$ a rational number such that $a-1$ is a p-adic unit. I believe I can show in this case that for $n$ a negative integer, $\zeta_a(n)$ takes rational values that are moreover p-adically integral and that there is a p-adic analytic function that interpolates these values.
Have these functions been studied before?
This is a version of the Lerch zeta function, see https://en.wikipedia.org/wiki/Lerch_zeta_function
In the space of (degree $1$..) Dirichlet series with functional equation we can take linear combinations so it is continuous, but when we add the Euler product requirement it becomes a discrete space. Also the RH has a formulation which holds for linear combination of L-functions, replacing the convergence/asymptotic of the Euler product by the convergence/asymptotic of $\sum_{n,\gcd(n,k!)=1} a_nn^{-s}$ as $k\to \infty$ (so it doesn't correspond to the zeros of $\sum_n a_nn^{-s}$ but to the zeros of each L-function in the linear combination)
This is the Polylogarithm, valid for arbitrary complex order $s$ and for all complex arguments $a$ with $|a| < 1$; it can be extended to $|a| \ge 1$ by the process of analytic continuation. It is related to the Lerch zeta function. The functional equation is originally due Jonquiere in 1889. The fullest investigation of the functional equation I can find is in this talk by Lagarias. There is also this monograph by Laurincikas.
The polylogarithm is $\zeta_a(s)$ as a function of $a$, for fixed $s\in\mathbb{N}$.
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2025-03-21T14:48:31.095537
| 2020-05-27T09:23:12 |
361473
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Stack Exchange
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Trace inequality normal derivative
For $v(\Omega) \in W^1_2$ and $\Omega \in C^1$ we have a trace inequality:
$$\Vert v \Vert _{L_2(\partial \Omega)} \leq C_\Omega \Vert v \Vert _{W_2^1},$$
which can be found in many places in the literature.
Is it then true that for a normal derivative we have:
$$\left \Vert\frac{\partial v}{\partial n}\right \Vert _{L_2(\partial \Omega)} \leq C_\Omega \Vert v \Vert_{W_2^2}?$$
I am not able to find a reference for that.
Just apply the first inequality to $\nabla v$.
Perhaps there is a typo, but as Lorenzo pointed out the answer seems pretty trivial at the moment. So I feel that this is not quite a research-level question?
Thank you for the answers! So we have:
$$||\frac{\partial v}{\partial n}||{L_2(\partial \Omega)} = ||\nabla v||{L_2(\partial \Omega)} \leq C_\Omega || \nabla v ||{W_2^1(\Omega)} \leq C\Omega ||v||_{W_2^2(\Omega)} ?$$
First equality should be $\leq$: $\partial v / \partial n$ is pointwise $\leq |\nabla v|$.
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2025-03-21T14:48:31.095649
| 2020-05-27T09:26:46 |
361474
|
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"Carlo Beenakker",
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"MUH",
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Stack Exchange
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Basic research problems references
I have been looking for research problems in pure mathematics that I can try to solve for publishing papers. I am quite aware that it takes a lot of time and effort to get to a level where I can do some "serious" research work. Meanwhile, I was thinking that, may be, I can try to solve problems which are at the level of 1st year graduate level (or even Master's level). By "level", I mean that a 1st year grad student should know all the terms used in the paper.
I understand that maintaining a list of unsolved/open problems is an extremely tough task and probably the best method is to look for journals/papers where authors include open/unsolved/extension of the problems.
I would appreciate if you can include here:
(i) List of problems (I have also collected a list of unsolved/open problems from the internet which I would be happy to share here in one of the answers).
(ii) Journals which concerns articles at the "level" mentioned above (For e.g. I hope that this list will exclude annals of math, Inventiones Mathematicae or any top journals which mostly include high level research papers. The American Mathematical Monthly is an excellent source.)
Remarks:
1) I admit that it seems like a very vague question but I am also confident of the fact that many other fellow students are looking for such a source. If the community feels that it should be closed, I will also second it.
2) One of the great sources for finding research problems are conferences (or proceedings of ICM ) where the leading researchers include open problems, but those problems, in my view, seem to be untouchable for most of the 1st year graduate students.
3) From one point view, there is a related question about letters in mathematics. This is definitely one of the answer I would have expected for this question.
I will be glad to provide any details if needed. Kindly let me know if I should post this question somewhere else.
I can hardly imagine how this might work without an advisor to guide you towards meaningful and doable projects; lists of open problems, almost by definition, are problems which experts have thought about and were not able to solve, so these do not seem suitable for someone entering a field.
@CarloBeenakker On one hand, I do believe that getting in touch with an advisor is a good way, but I also feel we should have some idea about what we want to do. Sometimes, it seems very unprofessional to approach any researcher without anything in your mind. You would be entirely at the mercy of the researcher. While, if you have something to talk about, he/she may be interested in, at least, listening to your proposal. (PS I don't mean only open problems but expositions and other means for publications as well)
This sounds a little bit like “Where can I find money?”
It is not only about open/unsolved problems but also about the journals which include papers that grad students can check.
Dear MUH, Contemporary mathematics is so advanced that there is no way one can do a quality research without years of study and a good advisor who would guide you through the literature. There are open problems with elementary and easy to understand statements, but the reason why they are still open is because they are too difficult. Are you a graduate student? You can reach me for a private conversation by an email. You can find it online.
G. Eric Moorhouse's list of open problems, mainly focusing on finite geometry.
Douglas B. West's list of open problems in Graph theory.
The open problem garden.
Mohammed Ghomi has a beautiful list of open problems on curves and surfaces, but perhaps most of them are out of reach.
Wow! I like this list a lot.
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2025-03-21T14:48:31.095937
| 2020-05-27T09:27:04 |
361475
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Stack Exchange
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A couple of points in a proof about of $\infty$-toposes
I wanted to have a better understanding of the geometric interpretation of $\infty$-toposes, and in particular learn something about étale morphisms, but I got stuck trying to unravel two points in the proof of HTT <IP_ADDRESS>.
At the end of the proof, it is claimed that the $\infty$-category of Cartesian sections of $\mathcal{Z}'' \times_S (N(\Delta)^{op} \times \{ 1 \}) \to N(\Delta)^{op}$ is even isomorphic to $\mathcal{X}^{/U_{\bullet}}$, and I really don't see how. After writing things as explicitly as I can, it seems to me that objects in the former are functors $N(\Delta)^{op} \times N(\Delta)^{op} \times \Delta^1 \times \Delta^1 \to \mathcal{Z}$ + conditions, whereas objects in the latter are functors $N(\Delta)^{op} \times \Delta^1 \to \mathcal{Z}$ + conditions, and none of these conditions collapses any of those factors. What am I missing? Is there a formal way to obtain an isomorphism there?
Just below this, at the beginning of Remark <IP_ADDRESS>, it is claimed that we can identify $Shv_{\hat{\mathcal{S}}} (\mathcal{X})$ with $S^{-1} \mathcal{\hat{P}(C)}$ by using <IP_ADDRESS> and <IP_ADDRESS>. What these directly lead to is that $Shv_{\mathcal{\hat{S}}}(\mathcal{X})$ can be identified with the full subcategory of $Fun^R (\mathcal{P(C)}^{op}, \mathcal{\hat{S}})$ spanned by the functors sending the morphisms in $S$ to equivalences in $\mathcal{\hat{S}}$. How should I conclude from there?
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2025-03-21T14:48:31.096079
| 2020-05-27T10:04:03 |
361477
|
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|
Stack Exchange
|
Smallest subgroup of unitary group, containing diagonal matrices and a fixed unitary matrix is the whole group
Suppose that $U_n(\mathbb{C})$ is the group of unitary matrices of dimension $n$ over complex numbers. Fix a unitary matrix $A \in U_n(\mathbb{C})$ and consider the smallest closed subgroup $K \subseteq U_n(\mathbb{C}) $, that contains all diagonal matrices (maximal torus group) and also $A$. It seems that apart from some exceptions for $A$ (like when $A$ is a diagonal matrix), $K = U_n(\mathbb{C}) $. Do you have any idea, if this is true, how to prove it and how to derive the exception cases?
Particularly, I'm interested in the case where $A$ is a Circulant matrix, i.e. it has the following form:
$$A=F^{-1}\cdot L\cdot F,$$
where $L$ is a diagonal matrix and $F$ is the DFT matrix. I'm not sure if this restriction simplifies the problem or not.
P.S. I have asked the question here in math.stackexchange group, but I guess, the question should be more relevant to this group.
For large families of familiar counterexamples to the first paragraph, take $A$ in the normalizer of the diagonal torus $T$, or in the centralizer of any subtorus $S\subset T$.
Here is a suggestion: If you want $A$ and the diagonal matrices to generate the full unitary matrix, then, in particular, the only matrices which commute with both $A$ and all the diagonal matrices should be scalar matrices, using Schur's Lemma. Furthermore, it is easy to see that the only matrices which commute with all diagonal matrices are themselves diagonal. Hence a necessary condition for $A$, together with diagonal matrices to generate the full unitary group is that $A$ commutes with no non-scalar diagonal matrix.
It is true that $A$ has this property if $A$ is the permutation matrix associated to the $n$-cycle $(12 \ldots n)$ (or any other $n$-cycle would do). However, in this case $A$, together with diagonals, does not generate the full unitary group, but generates a group of "monomial matrices" (matrices with one non-zero entry in each row and column).
Thanks. It can be verified that this necessary condition is satisfied for Circulant matrices iff it has at least one non-zero non-diagonal element.
|
2025-03-21T14:48:31.096252
| 2020-05-27T10:33:39 |
361480
|
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|
Stack Exchange
|
Cyclic vector of holomorphic vector bundle with flat connection over compact Riemann surface
I originally posted the question on math.stackexhange, but there doesn't seem to be an answer. I apalogize in advance for cross posting.
Let $E\rightarrow X$ be a holomorphic vector bundle over a compact Riemann surface with a holomorphic connection $\nabla:E\rightarrow E\otimes K$, where $K$ is the canonical bundle of $X$. Since the holomorphic connection is necessarily flat, its sheaf of local holomorphic sections $\mathcal{E}$ defines a (holonomic) $D$-module. Every holonomic D-module is locally cyclic, i.e. for any point $z_0$ there exists a neighborhood $U$ s.th. $\mathcal{E}(U)$ has a cyclic generator as a $D$-module (see e.g. Proposition 3.1.5. in Björk: Analytic $D$-Modules and Applications). Suppose we are given a coordinate $z$ on $U$ and identify $D(U)\cong D_1$, with $D_1=\mathbb{C}\left\lbrace z \right\rbrace \left\langle \partial_z \right\rangle$ (differential operators with coefficients in convergent power-series). So locally it holds $\mathcal{E}(U)\cong D_1/ I$, where $I$ is the ideal of differential operators annihilating the cyclic generator. This ideal is in general generated by two elements $P,Q$, with $P$ an operator of smallest possible degree in $I$ and furthermore $I/D_1P$ is of torsion type, i.e. for any $D\in I$ it holds $z^nD\in D_1P$ for some $n$ (Proposition 5.1.4 and Remark 5.1.5 in Björk: Analytic $D$-Modules and Applications). This implies for the dual $D$-module $\hom_{D_1}(D_1/I,\mathbb{C}\left\lbrace z\right\rbrace)=\left\lbrace f\in \mathbb{C}\left\lbrace z\right\rbrace \, \middle|\, Pf=Qf=0\right\rbrace=\left\lbrace f\in \mathbb{C}\left\lbrace z\right\rbrace \, \middle|\, Pf=0\right\rbrace$.
So far so good. Now on $U$ the holomorphic connection reads $\nabla|_U=\partial+A$ with $A$ some matrix of holomorphic functions.The dual bundle naturally comes with a holomorphic connection, too, which in local coordinates takes the form $\partial-A^T$. The whole discussion above shows that locally flat sections ($(\partial-A^T)Y=0$) are in one to one correspondence with solutions of $Pf=0$.
On the other hand there is Deligne's lemma of a cyclic vector. One way to formulate it, is to say that locally on a coordinate neighborhood $U$, for a vector bundle with holomorphic connection there exists $G\in \mathrm{GL}(n,\mathcal{O}(U))$, s.th.
\begin{equation} \partial_z G-G A^T=\tilde{A}G \end{equation} with $\tilde{A}$ in companion form. Here $\partial-A^T$ is the local form of the holomorphic connection. But in general the non unit entries $a_i$ in $\tilde{A}$ are only meromorphic and $G$ might not be invertible as a holomorphic matrix.
It is clear that a system of linear differential equations $\partial_z Y=A^T Y$ with $A^T$ in companion form corresponds to a single $n$-th order scalar differential equation $Qf=0$. So from Deligne's cyclic vector lemma I get an $n$-th order scalar differential equation, but the corresponding differential operator might not be in $D_1$, but in $\mathbb{C}\left\lbrace z\right\rbrace [z^{-1}]\left\langle \partial_z\right\rangle$.
Q: Is there any relation between the differential operator I get from the discussion in the first paragraph applied to the dual bundle and the differential equation I get from Deligne's cyclic vector lemma?
I guess they are the same, maybe after imposing further constraints on the open $U$. It might very well be that the relation is obvious and just shows my lack of understanding.
I think I found the relation. The paper A Simple Algorithm for Cyclic Vectors by N. Katz seems crucial. The cyclic vector lemma is mostly stated for differential fields (see e.g. section 2 in Galois Theory of Linear Differential Equations), but theorem 1 in the paper by Katz actually doesn't need a field. To be more precise, let $R$ be a commutative, local ring with a derivation $\partial:R\rightarrow R$, an element $z\in R$ s.th. $\partial(z)=1$ and $(V,D,\mathbf{e})$ a freely, finitely generated $R$-module, where $D$ is an additive mapping $D:V\rightarrow V$ satisfying the usual Leibniz rule $D(rv)=\partial(r)v+rD(v)$ and $\mathbf{e}=(e_0,\dots e_{n-1})$ is an $R$-basis. Then under the assumption that $R$ is an $\mathbb{Z}\left[\frac{1}{(n-1)!}\right]$-algebra and $z$ is in the unique maximal ideal of $R$, there is a cyclic vector $c(\mathbf{e},z)$, i.e. $(c,Dc,\dots , D^{n-1}c)$ is an $R$-basis in $V$.
Since locally on a trivializing open $U$ with coordinate $z$, the sheaf of sections $\mathcal{E}(U)$ is a free $\mathbb{C}\left\lbrace z \right\rbrace$ module and $\left(\mathbb{C}\left\lbrace z \right\rbrace,\partial_z,z\right)$ is a local ring in which $(n-1)!$ is invertible one can apply the theorem to this situation.
Moreover the path trough the dual bundle seems unnecessary from this. Since $(c,\nabla_{\frac{\partial}{\partial z}}c,\dots, \nabla^{(n-1)}_{\frac{\partial}{\partial z}}c$) is a local frame, its corresponding matrix $C=\left(c\,\nabla_{\frac{\partial}{\partial z}}c\, \cdots\,\nabla^{(n-1)}_{\frac{\partial}{\partial z}}c\right)$ is invertible as a holomorphic matrix. Applying the gauge transformation $C^{-1}\left(\partial_z+A\right)C$ to the local form of the connection gives a connection in companion form and hence an $n$-th order scalar differential equation.
On the other hand, from the $D$-module point of view, it is clear that there is an equation \begin{equation} \nabla^{(n)}_{\frac{\partial}{\partial z}}c=\sum_{i=0}^{n-1}f_i(z)\nabla^{(i)}_{\frac{\partial}{\partial z}}c\end{equation}, thus $P\equiv \partial_z^n-\sum_{i=0}^{n-1}f_i \partial_z^i$ is in the ideal $I$ of $D_1$, killing the cyclic generator. In addition I think $I=D_1P$, since the leading coefficient in $P$ is just $1$ and $\mathcal{E}(U)$ is locally free $\mathbb{C}\left\lbrace z\right\rbrace $-module. Furthermore, $P$ is the differential operator one gets from the companion form of the connection.
|
2025-03-21T14:48:31.096599
| 2020-05-27T10:39:48 |
361482
|
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|
Stack Exchange
|
BGG Category $\mathcal{O}$ is not closed under extension
What is the reason for the BBG category $\mathcal{O}$ failing to be closed under extensions i.e which of the 3 axioms of $\mathcal{O}$ does not hold under taking extensions?
Is there a prototype of a counterexample?
You can usually extend two modules from $\mathcal{O}$ by a module which is not semisimple for the Cartan subalgebra (i.e. fails to be a weight module). See Exercise 3.1. in [J. E. Humphreys, Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$. Graduate Studies in Mathematics, 94].
Thank you very much!
|
2025-03-21T14:48:31.096667
| 2020-05-27T11:44:08 |
361487
|
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|
Stack Exchange
|
Effect of collapsing two vertices of distance $2$
Motivating example. Consider the graph $G=(V,E)$ with $V = \{0,1,2,3\}$ and $E = \big\{\{i,i+1\}: i\in \{0,1,2\}\big\}$. We have $\chi(G) = 2$, but if we collapse $0$ and $3$, we get the complete graph on $3$ points (having chromatic number $3$). Note that $0$ and $3$ have distance $3$.
Problem statement. If $G = (V,E)$ is a simple, undirected graph and $v\in V$, let $N(v) = \{w\in V: \{v,w\}\in E\}$. We say $x\neq y \in V$ have distance 2 if $\{x,y\}\notin E$ but $N(x)\cap N(y) \neq \varnothing$. If $x\neq y \in V$ and $\{x,y\}\notin E$ we let define the graph $G/\{x,y\}$ by $V(G/\{x,y\}) = V \setminus \{x\}$ and $$E(G/\{x,y\}) = \{e \in E: y \notin e\}\cup\big\{\{x,z\}: z\in N(y)\big\}.$$
What is an example of a finite connected graph $G=(V,E)$ and $x\neq y\in V$ such that $x,y$ have distance 2 and $\chi(G) < \chi(G/\{x,y\})$?
Add to your motivating example a fifth vertex called $4$ and make it adjacent to all of $0,1,2,3$. This graph is $3$-colorable (by the $2$-coloring of your example and a third color for vertex $4$) but if you collapse $0$ with $3$ you get the complete graph on $4$ vertices, which needs four colors. And the distance from $0$ to $3$ is $2$, via vertex $4$.
|
2025-03-21T14:48:31.096793
| 2020-05-27T11:51:16 |
361488
|
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"Alexandre Eremenko",
"Christian Remling",
"J W",
"J. M. isn't a mathematician",
"Jochen Glueck",
"Michael Renardy",
"Piyush Grover",
"Robert Israel",
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"https://mathoverflow.net/users/33741",
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"https://mathoverflow.net/users/7934",
"leo monsaingeon"
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|
Stack Exchange
|
What do we learn from the Wronskian in the theory of linear ODEs?
For a real interval $I$ and a continuous function $A: I \to \mathbb{R}^{d\times d}$, let $(x_1, \dots, x_d)$ denote a basis of the solution space of the non-autonomous ODE
$$
\dot x(t) = A(t) x(t) \quad \text{for} \quad t \in I.
$$
The mapping
$$
\varphi: I \ni t \mapsto \det(x_1(t), \dots, x_d(t)) \in \mathbb{R}
$$
is usually called the Wronskian of the basis $(x_1,\dots,x_d)$, and it seems to be an obligatory topic in every ODE course or book that I've seen.
So in an ODE course that I am currently teaching, I'm facing the following problem:
(1) Despite its prevalence in courses and textbooks, I've rarely (not to say never) encountered any situation where the Wronskian of an ODE is used in a way that sheds non-trivial insight onto the problem at hand - in particular not in any of the books where I've read about it. (Of course, I have also searched on the internet for it, but without any success.)
(2) I feel quite uneasy to teach a concept which I am unable to motivate properly.
(3) I'd feel even more uneasy to just omit it from the course, since chances are that my not knowing of an application of the Wronskian is just due to my ignorance.
Well, what I did is to merely mention the Wronskian in a remark - but of course (and fortunately) I did not get away with it, because quite soon a student asked what the Wronskian is good for.
So this is the
Question: What is the Wronskian (in the context of linear ODEs) good for?
Remarks.
One can show that $\varphi$ satisfies the differential equation
$$
\dot \varphi(t) = \operatorname{tr}(A(t)) \varphi(t),
$$
and since this a one-dimensional equation we have the solution formula
$$
(*) \qquad \dot \varphi(t) = e^{\int_{t_0}^t \operatorname{tr}(A(s)) \, ds} \varphi(t_0)
$$
for it (for any fixed time $t_0$ and all $t \in I$). This is nice - but still I can't see how to explain to my students that it is useful.
I've often seen discussions to the end that $(*)$ implies that "the Wronskian is non-zero at a time $t_0$ if and only if it is non-zero at every time $t$" - but I find this somewhat straw man-ish: the fact that $(x_1(t), \dots, x_d(t))$ is linearly independent at one time $t_0$ if and only if it is linearly independent at every time $t$ is an immediate consequence of the uniqueness theorem for ODEs, without any reference to the Wronskian.
One can give a geometrical interpretation of $(*)$: For instance, if all the matrices $A(t)$ have trace $0$, and it follows that the (non-autonomous) flow associated with our differential equation is volume preserving. However, I'm not convinced that this serves as a sufficient motivation to give the mapping $t \mapsto \det(x_1(t), \dots, x_d(t))$ its own name and to discuss it in some detail.
Maybe a word on the notion "good for" that occurs in the question: I'm pretty comfortable with studying and teaching mathematical objects just in order to better understand them, or for the sake of their intrinsic beauty. However, whenever we do so, this usually happens within a certain theoretical context - i.e., we build a theory, introduce terminology, and this terminology somehow contributes to the development (or to our understanding) of the theory.
Some my question could be rephrased as:
"I'm looking either (i) for applications of the Wronskian of ODEs to concrete problems (within or without mathematics) or (ii) for ways in which the concept 'Wronskian' facilitates our understanding of the theory of ODEs (or of any other theory)."
The term 'Wronskian' also seems to be used with a more general meaning (see for instance this Wikipedia entry). However, I am specifically interested in the Wronskian for the solutions of a linear ODE.
The notion of "Evans functions" comes to my mind. Apart from that I have always felt that making this whole Wronskian thing such a big deal in a regular ODE course was mostly due to long habis dying hard, more than real usefulness of the concept. Of course history is important, and perharps there are actually good reasons for studying Wronskians, but I've never really encountered them in any deeply necessary way in my research as applied mathematician/analyst/PDE. So +1 from me, I'm looking forward to enlightning answers!
The geometric interpretation is quite handy via the liouville's theorem, not just when volumes are preserved (stablilty implies shrinking volumes etc.... ). The geometric picture of how volumes evolve under ODE flow is enlightening, and I believe most ODE students don't really see that picture unless taught using e.g., Arnold's ODE book. Wrronskian is part of that discussion, but not the whole thing ofcourse. –
If $\textrm{tr}: A=0$, then the Wronskian is a preserved quantity by (*), which is certainly nice to have. In my line of work, it is frequently used in this way (mainly for equations with $d=2$). On the other hand, I don't think there's anything outrageous about not mentioning the Wronskian in an introductory ODE course.
In undergraduate texts, Wronskians are usually introduced in the context of second order equations, where all you have to do to prove linear independence of two solutions is show that their ratio is not constant. Unfortunately it seems that this is usually not pointed out to the students.
You might be interested to know that we have a Mathematics Educators Stack Exchange site, although it does not preclude one from asking educational questions on MO or Mathematics Stack Exchange.
@JW: Thank you for your comment. I was thinking about posting the question there, but I eventually chose MathOverflow since the ODE course I mentioned was mainly the trigger for asking, but I've also been wondering about this question in its own right for quite some time.
Here is a typical use in an undergraduate textbook: to prove that for distinct $\lambda_j$ the exponentials $e^{\lambda_jt}$ are linearly independent. It has some applications on the more advanced level, but you were asking about undergraduate textbooks. Also notice: uniqueness theorem, even for linear ODE is rarely proved in undergraduate textbooks, at least in the USA. So for linear equations with constant coefficients, the notion of Wronskian permits you to find $n$ linearly independent solutions without an appeal to the unproved uniqueness theorem. Same applies to the proof that cosines with distinct frequencies are linearly independent.
Another application. How to write a linear differential equation of order $n$ satisfied
by $n$ given functions $f_1,\ldots,f_n$? Here is how:
$$\left|\begin{array}{cccc}w&f_1&\ldots&f_n\\ w'&f_1^\prime&\ldots&f_n^\prime\\
\ldots&\ldots&\ldots&\ldots\\
w^{(n)}&f_1^{(n)}&\ldots&f_n^{(n)}\end{array}\right|=0.$$
Expanding with respect to the first column, we obtain that the Wronskian
$W=W(f_1,\ldots,f_n)$
is the coefficient at $w^{(n)}$, in particular, if all $f_j$ are analytic then the singular points of the equation are the zeros of $W$.
The significance of the Wronskian is not limited to differential equations.
Consider a finite-dimensional vector space $V$ consisting of functions.
(For example, polynomials of degree at most $n$). Suppose we have a basis
$f_1,\ldots,f_n$. How to expand a function $f\in V$ in this basis? Write
$$f=c_1f_1+\ldots+c_nf_n,$$
differentiate $n-1$ times and solve the linear system with respect to $c_j$.
The determinant of this system is the Wronskian. This was the original goal of Heine-Wronski when he invented it.
For less elementary applications, type "Wronski map" in the cell "Anywhere"
or in the cell "Title" in Mathscinet search.
See my answer here for a simple inductive proof that the exponentials are linearly independent, without any appeal to Wronskians.
+1 Thank you very much your answer! Although I tend to agree with @RobertIsrael that his "direct" proof is a bit more natural than a proof that uses the Wronskian, I find this answer very useful anyway: a few such facts which can also be proved with appeal to the Wronskian provide at least some context for the Wronskian. (And context is probably one of the most important things in order to motivate a mathematical concept.)
Concerning the uniqueness theorem: Indeed, the point that the Wronskian allows us to deal with linear ODEs without appeal to the uniqueness theorem also briefly occurred to me. But I think this is mostly a point about the academic system: in universities in Germany (where I teach) the existence and uniqueness theorem is typically taught (and proved) in most introductory ODE courses. (I'd guess it's similar in many other European countries, but I should probably be careful not to extrapolate too much.)
@Jochen Glueck: sorry I read another proof in the link given by Robert Israel, the first one.
I really like all the answers, so I couldn't make up my mind which one to accept - hence, I finally accepted this answer since it was the first one that was posted.
Quite an important use of the Wronskian arises in the spectral analysis of the Hill operator
$$\frac{d^2}{dx^2}+q(x)$$
when $q$ is periodic. This is the search of Floquet exponents.
+1 Very interesting answer, thanks a lot! (Embarrasingly, Floquet theory has not been on my radar at all.)
Curiously, I don't recall Arscott ever discussing the Wronskian in his treatise. (I have still upvoted this answer.)
The crucial quantity for the spectral analysis of this operator really is the trace of the transfer matrix, not its determinant.
This is in the same spirit as Piyush Grover's comment. The determinant $\det(x_1(t),\ldots,x_n(t))$ definitely deserves a name (not only in the context of linear ODE's). In such a lecture students could (and ,in my opinion, should) learn the meaning of the divergence of a vector field $F$. Having learned Picard-Lindelöf they are ready to understand the flow $\phi(t,x)$ as he solution of the initial value Problem $\phi'(t,x)=F(\phi(t,x))$, $\phi'(0,x)=x$, and for a small cube
$x+[0,r]^n$ you can throw the edges into the flow to get after a short time almost the parallelepiped with edges $\phi(t,x+re_j)-\phi(t,x)$ whose (oriented) volume compared to the volume of the cube is $$v(t,r)=\det[\phi(t,x+re_1)-\phi(t,x),\ldots,\phi(t,r+e_n)-\phi(t,x)]/r^n$$ If you take the derivative $\partial_t$ at $0$ and the limit $r\to 0$ you get the divergence of the vector field (no problem to take time dependent vector fields).
The desire to make this more precise also motivates the theorems about differentiability of the solutions of initial value problems with respect to the initial values. Then you can throw quite arbitrary small sets into the flow and compare the evolved (oriented) volume with the original one by calculating them with the $n$-dimensional substitution rule.
Thanks a lot for your answer! (+1) I like this geometrical point of view very much - so you nearly convinced me to include this kind of reasoning the next time when I teach such a course (for the current semester, the opportunity is most likely gone since it would have required me to organise the contents of the course in a somewhat different way than how I have actually done it).
|
2025-03-21T14:48:31.097659
| 2020-05-27T11:57:23 |
361489
|
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"zeraoulia rafik"
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|
Stack Exchange
|
Primality test for specific class of $N=8kp^n-1$
My following question is related to my question here
Can you provide a proof or a counterexample for the following claim :
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ . Let $N=8kp^n-1$ such that $k>0$ , $3 \not\mid k$ , $p$ is a prime number, $p \neq 3$ , $n > 2$ and $8k<p^n$ . Let $S_i=P_p(S_{i-1})$ with $S_0=P_{2kp^2}(4)$ , then:
$$N \text{ is a prime iff } S_{n-2} \equiv 0\pmod{N}$$
You can run this test here.
EDIT
I have verified this claim for $k \in [1,500]$ with $p \leq 139$ and $n \in [3,50]$ .
I think you should refer this question to that your new one.
P_m(x) = D_m(x,1) (the m-th Dickson polynomial of the first kind with parameter 1), so P_a(P_b) = P_{ab}, so S_0 = D_{(N+1)/4}(4,1) mod N. You might want to check out the literature on Dickson pseudoprimes.
This is a partial answer.
This answer proves that if $N$ is a prime, then $S_{n-2}\equiv 0\pmod N$.
Proof :
It can be proven by induction that
$$S_i=(2-\sqrt 3)^{2kp^{i+2}}+(2+\sqrt 3)^{2kp^{i+2}}\tag1$$
Using $(1)$ and $2\pm\sqrt 3=\bigg(\frac{\sqrt{6}\pm\sqrt 2}{2}\bigg)^2$, we get
$$\begin{align}&2^{N+1}S_{n-2}^2-2^{N+2}
\\\\&=(\sqrt 6-\sqrt 2)(\sqrt 6-\sqrt 2)^{N}+(\sqrt 6+\sqrt 2)(\sqrt 6+\sqrt 2)^{N}
\\\\&=\sqrt 6\bigg((\sqrt 6+\sqrt 2)^{N}+(\sqrt 6-\sqrt 2)^{N}\bigg)
\\&\qquad\qquad +\sqrt 2\bigg((\sqrt 6+\sqrt 2)^{N}-(\sqrt 6-\sqrt 2)^{N}\bigg)
\\\\&=\sqrt 6\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i+(-\sqrt 2)^i\bigg)
\\&\qquad\qquad +\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i-(-\sqrt 2)^i\bigg)
\\\\&=\sum_{j=0}^{(N-1)/2}\binom N{2j}6^{(N+1-2j)/2}\cdot 2^{j+1}+\sum_{j=1}^{(N+1)/2}\binom N{2j-1}6^{(N-2j+1)/2}\cdot 2^{1+j}
\\\\&\equiv 6^{(N+1)/2}\cdot 2+2^{(N+3)/2}\pmod N
\\\\&\equiv 12\cdot 2^{(N-1)/2}\cdot 3^{(N-1)/2}+4\cdot 2^{(N-1)/2}\pmod N
\\\\&\equiv 12\cdot (-1)^{(N^2-1)/8}\cdot \frac{(-1)^{(N-1)/2}}{\bigg(\frac N3\bigg)}+4\cdot (-1)^{(N^2-1)/8}\pmod N
\\\\&\equiv 12\cdot 1\cdot \frac{-1}{1}+4\cdot 1\pmod N
\\\\&\equiv -8\pmod N\end{align}$$
So, we get
$$2^{N+1}S_{n-2}^2-2^{N+2}\equiv -8\pmod N$$
It follows from $2^{N-1}\equiv 1\pmod N$ that
$$S_{n-2}\equiv 0\pmod N$$
|
2025-03-21T14:48:31.097801
| 2020-05-27T12:44:21 |
361491
|
{
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"authors": [
"Nishant Chandgotia",
"https://mathoverflow.net/users/13774"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361491"
}
|
Stack Exchange
|
Size of a family of sets of $k$-separated functions over $\{0,1,\ldots,n-1\}$
For $n\ge 1$ we write $[n]$ to denote the set $\{0,1,\ldots,n-1\}$. Let $2^{[n]}$ be the set of all functions from $[n]$ to $\{0,1\}$. Let $\mathcal{F}$ and $\mathcal{G}$ be two nonempty subsets of $2^{[n]}$. Fix $1\le k\le n$. Take $0\le j \le n-k$. We say that $f,g\in 2^{[n]}$ disagree over $[j,j+k)$ if the restrictions of $f$ and $g$ to $\{j,j+1,\ldots,j+k-1\}$ are different, that is, $f|_{\{j,j+1,\ldots,j+k-1\}}\neq g|_{\{j,j+1,\ldots,j+k-1\}} $. We say that $f\in 2^{[n]}$ disagree with a nonempty $\mathcal{G}\subset 2^{[n]}$ over $[j,j+k)$ if for every $g\in\mathcal{G}$ we have that $f$ and $g$ disagree over $[j,j+k)$. That is, for some fixed $j$ the function $f$ disagress over $[j,j+k)$ with all functions in $\mathcal{G}$.
We say that $\mathcal{F}$ and $\mathcal{G}$ are $k$-separated if there exists $0\le j \le n-k$ such that
either there exists $f\in\mathcal{F}$ which disagrees with $\mathcal{G}\subset 2^{[n]}$ over $[j,j+k)$ or there exists $g\in\mathcal{G}$ which disagrees with $\mathcal{F}\subset 2^{[n]}$ over $[j,j+k)$.
Let $\mathscr{T}_{n,k}$ be the largest possible family of pairwise $k$-separated nonempty subsets of $2^{[n]}$, that is, $\mathscr{T}_{n,k}$ is a family of maximal cardinality among all families $\mathscr{F}$ such that if $\mathcal{F},\mathcal{G}\in \mathscr{F}$ and $\mathcal{F}\neq \mathcal{G}$ then $\mathcal{F}$ and $\mathcal{G}$ are $k$-separated.
Keeping $k$ fixed I am looking for an asymptotic behaviour of the cardinality, $|\mathscr{T}_{n,k}|$ of $\mathscr{T}_{n,k}$ as $n\to\infty$. In particular, I hope that
$$
\frac{|\mathscr{T}_{n,k}|}{2^{2^n}}\to 1\text{ as }n\to\infty.
$$
1: Hopes. Let me begin by taking away your hope - that is, disproving the conjectured asymptotic, $|{\mathscr{T}_{n,k}}|/2^{2^n} \to 1$.
I will say that a family $\mathcal{F}$ is $k$-rich if for each $j \in [n]$ and $w \in \{0,1\}^k$, there is $f \in \mathcal{F}$ with $f|_{[j,j+k)} = w$. Accordingly, I will say that a pair $(j,w) \in [n] \times \{0,1\}^k$ "bad" for $\mathcal{F}$ if $f|_{[j,j+k)} \neq w$ for all $f \in \mathcal{F}$, so that $\mathcal{F}$ is $k$-rich if it has no "bad" pair $(j,w)$.
Claim. No two $k$-rich families are $k$-separated.
Proof. For any sequence $g \in \{0,1\}^n$, any $j \in [n]$ and any $k$-rich family $\mathcal{F}$, there exists $f \in \mathcal{F}$ such that $g|_{[j,j+k)} = f_{[j,k+j)}$. Hence, $g$ does not disagree with $\mathcal{F}$ over $[j,j+k)$. The rest follows by unwinding definitions.
Claim. The number of families that are not $k$-rich is $o(2^{2^n})$.
Proof. Recall taht for each family $\mathcal{F}$ that is not $k$-rich, there exist a "bad" pair $j \in [n]$, $w \in \{0,1\}^k$. Given $j \in [n]$, $w \in \{0,1\}^k$, the number of $f \in \{0,1\}^{[n]}$ with $f|_{[j,j+k)} \neq w$ is $(1-2^{-k})2^n$. Hence, the number of families $\mathcal{F}$ for which the pair $(j,w)$ is bad is $2^{(1-2^{-k})2^n}$. By the union bound, the number of families that are not $k$-richis at most $2^{2^n} \times {2^k n}/{2^{2^{n-k}}}$. If $k$ is kept constant and $n \to \infty$, then ${2^k n}/{2^{2^{n-k}}} \to 0$.
2: Asymptotics. We will show that $\lim_{n \to \infty} \log |\mathscr{T}_{n,k}|/n = 2^{k(1+o(1))}$, where the $o(1)$ term tends to $0$ as $k \to \infty$.
For a given family $\mathcal{F}$, let $B(\mathcal{F})$ denote the set of ``bad'' pairs $j,w$:
$$B(\mathcal{F}) = \{ (j,w) \ : \ f_{[j,j+k)} \neq w \quad \forall f \in \mathcal{F}\} \subset [n] \times \{0,1\}^k.$$
Claim: Two families $\mathcal{F}, \mathcal{G}$ are $k$-separated if and only if $B(\mathcal{F}) \neq B(\mathcal{G})$.
Proof: Suppose that $f \in \mathcal{F}$ and $j$ is such that $f$ disagrees with $\mathcal{G}$ on $[j,j+k)$. Then $(j,f|_{[j,j+k)}) \in B(\mathcal{G}) \setminus B(\mathcal{F})$. Conversely, if $(j,w) \in B(\mathcal{G}) \setminus B(\mathcal{F})$ then there exists $f \in \mathcal{F}$ such that $w=f|_{[j,j+k)}$, and since $(j,w) \in B(\mathcal{G})$, $f$ disagrees with $\mathcal{G}$ on $[j,j+k)$.
Since each two sets in $\mathscr{T}_{n,k}$ are $k$-separated, each of them corresponds to a different subset of $[n] \times \{0,1\}^k$, and so $$|\mathscr{T}_{n,k}| \leq 2^{n2^k}.$$
In the opposite direction, let $\mathscr{B}$ denote the family of all sets $B \subset [n] \times \{0,1\}^k$ such that $k \mid j$ and $\sum_{i} w_i \equiv 1 \bmod{2}$ for all $(j,w) \in B$.
Claim: For each $B \in \mathscr{B}$ there exists a family $\mathcal{F}_B$ such that $B(\mathcal{F}_B) = B$.
Proof: Let $\mathcal{F}$ consis of all sequences $f \in \{0,1\}^n$ such that $f|_{[j,j+k)} \neq w$ for each $(j,w) \in B$. Clearly, $B \subset B(\mathcal{F})$ so it remains to show that $B(\mathcal{F})$ contains no other pair. Let $(i,u) \in [n] \times \{0,1\}^k \setminus B$ be any such tentative pair. We need to construct $f \in \mathcal{F}$ with $f_{[i,i+k)} = u$. On each interval $[j,j+k)$ with $k \mid j$ and $[i,i+k) \cap [j,j+k) = \emptyset$, put $f|_{[j,j+k)} = 0^k$. If $i < j < i+k$, $k \mid j$, then set $f|_{[i,j)} = u_{[0,j-i)}$, $f_j = \sum_{t=<j-i} u_t \bmod{1}$ and $f|_{(j,j+k)} = 0^{j-i-1}$. If $j<i<j+k$, $k \mid j$, define $f|_{[j,j+k)}$ in the analogous manner. This construction guarantees that $\sum_{t=0}^{k-1} f_{j+t} \equiv 0 \bmod{2}$ for each $j$ with $k \mid j$, so $f|_{[j,j+k)} \neq w$ for each $(j,w) \in B$.
Since the set $\mathscr{T} = \{ \mathcal{F}_B \ : \ B \in \mathscr{B}\}$ is $k$-separated, we have the bound:
$$
|\mathscr{T}_{n,k}| \geq |\mathscr{B}| \simeq 2^{n 2^{k-1}/k}.
$$
At the end of the day, we have the nearly matching lower and upper bounds:
$$ 2^{k} \geq \lim_{n \to \infty} \log |\mathscr{T}_{n,k}|/n \geq 2^{k-1}/k = 2^{k - o(k)}.$$
Great answer Jakub!
|
2025-03-21T14:48:31.098390
| 2020-05-27T12:48:45 |
361492
|
{
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"Taras Banakh",
"Tobias Fritz",
"https://mathoverflow.net/users/27013",
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|
Stack Exchange
|
Categories admitting singleton-classifiers and characterization of the category $\mathbf{Set}$
Trying to characterize categories equivalent to the category of sets, I have discovered (for myself) that instead of requiring that the coprojection morphism $\mathsf{true}:1\to \Omega=1\sqcup 1$ is a subobject classifier, it suffices to require that this morphism is an singleton classifier, which means that for every morphism $x:1\to X$ there exists a morphism $\chi_x:X\to \Omega$ such that for any morphism $y:1\to X$ the equality $\chi_x\circ y=\mathsf{true}$ is equivalent to $x=y$.
Question. Is the notion of a singleton classifier essentially weaker than that of subobject classifier? Has it been already considered in the literature and if yes, under which terminology?
Using element classifiers I can prove the following characterizations in von Neumann-Bernays-Godel axiomatic system:
Theorem 1. A category $\mathcal C$ is equivalent to the category of sets if and only if $\mathcal C$ has the following properties:
1) $\mathcal C$ is locally small;
2) $\mathcal C$ is balanced (mono+epi = iso);
3) $\mathcal C$ has a terminal object $\mathtt 1$;
4) $\mathtt 1$ is a $\mathcal C$-generator;
5) $\mathcal C$ has equalizers;
6) $\mathcal C$ has arbitrary coproducts;
7) $|\mathsf{Mor}(\mathtt 1,\mathtt 1\sqcup \mathtt 1)|=2$;
8) the morphism $\mathsf{true}:\mathtt 1\to \mathtt 1\sqcup\mathtt 1$ is an singleton classifier in $\mathcal C$.
$\phantom{m}$
Theorem 2. A category $\mathcal C$ is isomorphic to the category of sets if and only if $\mathcal C$ has the following properties:
$(1)-(8)$ from Theorem 1;
(9) $\mathcal C$ has a unique initial object;
(10) for any noninitial $\mathcal C$-object $x$ the class of $\mathcal C$-objects that are isomorphic to $x$ is a proper class.
I have a feeling that these characterizations are known. If yes, to whom should they be attributed?
Added in Edit. I have found something quite close to the above characterizations in nLab.
I think that Axiom 5) is redundant. For if $0$ and $1$ were isomorphic, then $1 \sqcup 1$ would again be initial and hence terminal, forcing $|\mathrm{Mor}(1,1\sqcup 1)| = 1$. Even better, there is no morphism $1 \to 0$ at all, because if there was one then it would already have to be an isomorphism.
This is quite interesting. Is there a direct way to see that such a category has products, perhaps using only a subset of the axioms?
@TobiasFritz I do not know, The proof just establishes that each object is isomorphic to the coproduct of its elements. And this gives an equivalence of categories. Products appear a posteriory due to the Axiom of (Global) Choice in the category of sets.
@TobiasFritz Thank you for your comment concerning $0\ne 1$. Then the conditions related to 0 can be removed at all, since the existence of $0$ follows from the existence of coproducts. I will allow myself to make the corresponding changes in the question.
I don't think this has been considered. Mainly I've never seen it, but also there are specific feature of this notions that makes it unlikely to be a relevant category theoretic notion independently of your other conditions:
It is not really a universal property, in the sense that it does not characterize what are morphisms to $\Omega$ as not all morphisms to $\Omega$ classify a singleton.
Your uniqueness conditions only involves the behavior of the map $X \to \Omega$ on elements of $X$. This is of course very natural in your situation as all map in the category of sets are determined by their value on elements (and this is implied by your axiom that $1$ is a generator) but this is a very weird condition in category where $1$ is not assumed to be a generator. For example, I don't think a subobject classifier will be an element classifier in general.
Regarding an example where this is different from a sub-object classifier:
If I have a model $M$ of IZF or CZF, then in the category $S$ of sets of $M$, $2=1 \coprod 1$ classifies only complemented subobjects. (In a model of IZF there would be an actual subobject classifier $\Omega$, with $2 \subset \Omega$, but not necessarily in a model of CZF).
If I restrict to the full subcategory $D \subset S$ of objects that are decidable (i.e. the set $X$ such that the diagonal inclusion $X \to X \times X$ is complemented) It is a classical category theoretic fact that $D$ is stable under finite limits (because $S$ being an extensive category decidable will be stable under finite product, and a sub-object of a decidable object is decidable).
Now, all singleton in $D$ are complemented as a singleton $a:1 \to X$ can be written as pullback of $1 \times X \to X \times X$ along $X \to X\times X$, so $2= 1 \coprod 1$ will indeed be an 'element classifier'. Explicitely, as $X$ is decidable, there is a map $\delta:X \times X \to 2$ that classifies the diagonal, and given $x: 1 \to X$, $\{x\}$ is classified by $\delta(x, \_ )$.
However it does not classify all subobjects (unless the law of excluded middle holds in $M$ of course) for example as soon as there are a non-complemented subset of $\mathbb{N}$ in $M$, then as $\mathbb{N}$ is always a decidable object that will give you a subobject not classified by $2$.
Note: I initially tried to use a sheaf model to get a more explicit category, but I got into trouble due to the fact that in a sheaf model $1$ is almost never a generator. And as I mentioned at the begining, your definition is a bit unnatural if we do not assume that $1$ is a generator. I realized I do not know how to construct well-pointed toposes not satisfying LEM, other than going through the kind of filter-quotient constructions that produces models of IZF from sheaves model...
Note 2: What do you think of the name "singleton classifier" instead of "element classifier" ?
|
2025-03-21T14:48:31.098782
| 2020-05-27T13:17:25 |
361493
|
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"Arno Fehm",
"Lucien",
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"https://mathoverflow.net/users/50351",
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"tomasz"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361493"
}
|
Stack Exchange
|
Saturated differentially closed field
What means "saturated" in "saturated differentially closed field" ?
Presumably saturated in the sense of model theory? See https://en.wikipedia.org/wiki/Saturated_model
Yes, but still? The model theoretic material discussed in this wikipedia page is quite abstract. I'm looking for a more concrete explanation, specific to the case of a diff closed field.
Maybe you should say more precisely what you are interested in. Do you need to prove saturation? Do you want to apply it? What is the characteristic of your differentially closed field?
By stability, there are saturated dcf of characteristic zero of each uncountable cardinality, and saturated DCF of each positive characteristic of each cardinality $\kappa$ with $\kappa^{\aleph_0}=\kappa$. Since DCF eliminates quantfiers, a $\kappa$-saturated model (for uncountable $\kappa$) is just one which all families of fewer than $\kappa$ constructible sets with the finite intersection property have nonempty intersections. A saturated model is one which is saturated in its own cardinality.
|
2025-03-21T14:48:31.098885
| 2020-05-27T13:21:23 |
361495
|
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"Iguana",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361495"
}
|
Stack Exchange
|
Integral over an exponential sum with squares
How should I estimate the following integral
$$I = \int_0^1 \left( \sum_{n=0}^{p-1} e(n^2t) \right)^2 dt $$
where $p$ is a prime?
Here is the method I followed:
\begin{align*}
I & = \int_0^1 \left( 1+ \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\
& = 1 + 2 \sum_{n=1}^{p-1} \left( \int_0^1 e(n^2t) dt \right) + \int_0^1 \left( \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\
& = 1 + 2 \sum_{n=1}^{p-1} \left( \frac{e(n^2)-1}{2\pi i n^2} \right) + \int_0^1 \left( \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\
& = 1+O((p-1)^{1+\epsilon})
\end{align*}
where the third term is estimated using Hua's lemma.
This estimate is not good enough for my purpose. Is it possible to get a better error term than the one here?
The integral is equal to exactly one - just expand out the square and use orthogonality to get
$$ I = \sum_{0\leq n,m<p} \int_0^1 e((n^2+m^2)t) \mathrm{d} t = \sum_{0\leq n,m<p}1_{n^2+m^2=0}=1.$$
Ah, of course. Thanks! Don't know why I overcomplicated it.
A humble question: What did you mean with orthogonality? I do not see how to get $ \int_0^1 \exp ((n^2 + m^2)t) dt = 1_{n^2 + m^2 = 0}$.
Here $e(x)$ is shorthand for $e^{2\pi ix}$, so by orthogonality I mean the fact that $\int_0^1 e^{2\pi i tn}\mathrm{d}t$ is $1$ if $n=0$ and $0$ if $n$ is any other integer.
|
2025-03-21T14:48:31.098998
| 2020-05-27T14:41:32 |
361504
|
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"authors": [
"Tomasz Kania",
"dreamwave",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361504"
}
|
Stack Exchange
|
Monotone series of projections converging to 1 in von Neumann algebra
The following statement is being used a lot in the literature, and I wonder how to prove it.
Let $M$ be an infinite-dimensional von Neumann algebra (with unit element), show that there is an increasing series of projections {$p_i$}$_{i=1}^\infty$ such that $\forall n : p_n \neq 1$, and $p_n\stackrel{n\to\infty}\longrightarrow 1$.
My thoughts so far:
I know that in an infinite-dimensional von Neumann algebra there is an infinite set of pairwise orthogonal projections. so taking their partial sums would be an increasing series of projections, however I'm not sure if they converge to 1.
Thanks in advance!
Take a unitary element $u$ so that $\sigma(u)=\mathbb T$, the unit circle. Then ${1,u}^{\prime\prime}\subset M$ is isomorphic to $L_\infty(\mathbb T)$. Can you do it in $L_\infty(\mathbb T)$? (Simply take an essentially increasing family of sets of positive measure that cover $\mathbb T$ and their indicators are the sought projections.)
@TomaszKania - Thanks for your comment, could you explain why does {$1,u$}$''$ isomorphic to $L_\infty(\mathbb{T})$?
That's the spectral theorem for normal elements; see Borel functional calculus in any operator algebra textbook.
|
2025-03-21T14:48:31.099105
| 2020-05-27T15:10:13 |
361507
|
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"authors": [
"Andrei Smolensky",
"Nathan Reading",
"YCor",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361507"
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|
Stack Exchange
|
Regarding $F_4$ and $G_2$ Lie algebras, do there exist $F_n$ or $G_n$ families of Lie algebras?
For example, $E_6$ exceptional Lie algebra is part of the $E_n$ series of Lie algebras (Kac-Moody algebras). Are $F_4$ or $G_2$ maybe also parts of some $F_n$ or $G_n$ series of Lie algebras or are they isolated cases?
I'd guess $F_n$ has two natural extrapolations $o-o\stackrel{4}-o-o-\dots-o$, with the two possible orientations for the 4-arrow (irrelevant for the Coxeter version), and similarly $G_2$ has two natural extrapolations $o\stackrel{6}-o-o-\dots-o$ with either orientations for the 6-arrow.
The families suggested by @YCor have the nice property that they contain the affine Lie algebras related to finite types $F_4$ and $G_2$. (In Kac's notation, these are $F_4^{(1)}$, $E_6^{(2)}$, $G_2^{(1)}$, and $D_4^{(3)}$.) Of course, the $E_n$ series does not contain the affine $E_6^{(1)}$ or $E_7^{(1)}$ Lie algebras, so maybe that's not a necessary feature.
Another way to look at the extrapolations suggested by @YCor is by means of diagram folding. Namely, $G_2$ is an order 3 folding of $D_4$, while $F_4$ is an order 2 folding of $E_6$. This can be continued to the 2- and 3-foldings of the diagrams of types $Y_{n,n,n}$ and $Y_{n,n,1}$ respectively. Other options include $n$-folding of the diagram with $n$ nodes connected by simple edges to a single vertex (then one gets $C_2$, $G_2$, $A_{2}^{(2)}$, $H(1,5)$, etc.) But since there is no description of a single KM-algebra but by generators and relations, this is all just speculations.
|
2025-03-21T14:48:31.099241
| 2020-05-27T15:22:50 |
361508
|
{
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"Gabe K",
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|
Stack Exchange
|
Reference on noncommutative PDE
I would like to ask if there is reference on semi-linear parabolic PDE (or more generally any kinds of PDE) with non-commutative unknown variable. For example, assume $u$ is a matrix-valued function (regularity assumption / integrability assumption to be specified), consider PDE (as an example): $\partial_t u - \Delta u + F(u) = 0$, where $F(u)$ is a nonlinear function of $u$ (if linear then this would not be much different from commutative case). For example, drawing analogy with PDE coming from harmonic map, we can take $F(u) = \langle\nabla u, \nabla u\rangle = \sum_i \nabla_i u\nabla_i u$ where the product is matrix product. What can we say about PDE of this kind?
In the matrix case, this would be a system of semi-linear parabolic PDE. What are some general things we know about such systems of semi-linear PDEs? For example, is there a general maximum principle for such matrix valued case?
Furthermore, what do we know if we restrict u from general matrices to classical matrix Lie groups (such as $\mathrm{O}(n)$, $\mathrm{U}(n)$, $\mathrm{Sp}(n)$)?
On the other direction, what do we know if we consider more general non-commutative variable than $u$? Say $u$ is taken from a tracial von Neumann algebra. What can we say in that case?
Thanks a lot for the help.
I don't know about a reference for general theory, but here is a paper that studies one such equation (continuum (nonlinear advection)-type model for rotation matrices in 3D): Degond, Pierre, Amic Frouvelle, and Sara Merino-Aceituno. "A new flocking model through body attitude coordination." Mathematical Models and Methods in Applied Sciences 27.06 (2017): 1005-1049.
And here's another one (It studies coupled system, one variable is a vector and another is a symmetric tensor):Chen, Gui-Qiang, et al. "Global existence and regularity of solutions for active liquid crystals." Journal of Differential Equations 263.1 (2017): 202-239.
There are a lot of work done on PDEs with vector-valued unknowns (or systems). Generally vectored valued functions don't even admit a multiplication (they do not form an algebra) and therefore is more general than the "noncommutative" case, so if something holds for the vector valued cases it should also hold for your case. Maximum principles for weakly coupled parabolic systems are known, see e.g. Dow (1975) "STRONG MAXIMUM PRINCIPLES FOR WEAKLY COUPLED SYSTEMS OF QUASILINEAR PARABOLIC INEQUALITIES" (caps due to copy-pasting from article).
One special case of an equation similar to your example is Ricci flow. This has been studied in depth and Hamilton established a tensor maximum principle that plays a central role in the analysis. I wonder if some of that theory might be useful for your purposes.
My coauthor Dmitry Vorotnikov and Yann Brenier have been working on optimal transport of matrices, see https://arxiv.org/abs/1808.05064
|
2025-03-21T14:48:31.099435
| 2020-05-27T16:45:51 |
361511
|
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|
Stack Exchange
|
2-d geometric Brownian motion hitting time distribution
I am trying to solve following problem: Given two independent geometric Brownian motions
$\frac{d x_t}{x_t}=\mu_x dt + \sigma_x dw_t^x$
and
$\frac{d y_t}{y_t}=\mu_y dt + \sigma_y dW_t^y$
and some initial point $(x_0, y_0)$ to the right of the line $y=\phi_0 + \phi_1 x$
($y_0 < \phi_0 + \phi_1 x_0)$
I want to know what the distribution of the time it takes to hit the line is.
I think I know how to solve it if they are Brownian motions: I apply Bayes' rule to compute the probability of hitting the line after and before the hitting time, and that helps. But I need the symmetry of the Brownian motion at some point which I don't have in the GBM.
Any help on how to get started is very helpful.
|
2025-03-21T14:48:31.099509
| 2020-05-27T17:33:15 |
361513
|
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"Neil Strickland",
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|
Stack Exchange
|
A certain semi-simplicial space
I would like to understand whether the following construction has been studied. Let me model the pointed sphere as $S^n = I^n / \partial I^n$, and let $\Omega^n (-) := map((I^n, \partial I^n), (-, *))$.
Let $X$ be a based space. Write $X_p := S^{p+1} \wedge \Omega^{p+1} X$. There are maps
$$d_i : S^{p+1} \wedge \Omega^{p+1} X \longrightarrow S^{p} \wedge \Omega^{p} X$$
given by the formula $d_i([t_0, ..., t_p], f) = ([t_0, ..., \hat{t_i}, .., t_p], f\vert_{I^i \times \{t_i\} \times I^{p-i}})$. These form the face maps for a semi-simplicial space $X_\bullet$. Furthermore, evaluation yields an aumentation $\epsilon: X_\bullet \to X$. On geometric realisation we therefore obtain a map
$$|X_\bullet| \to X.$$
Is this construction familiar to anyone?
I don't think the ordering is playing a role here: $S^p\wedge \Omega^pX$ is contravariantly functorial for arbitrary injections of the sets ${1,\dotsc,p}$. I've not seen this variant but it is dual to the fact that $\Omega^p(S^p\wedge X)$ is covariantly functorial for the same maps. This comes up a lot (implicitly or explicitly) when setting up the model category of symmetric spectra.
|
2025-03-21T14:48:31.099728
| 2020-05-27T17:45:16 |
361515
|
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|
Stack Exchange
|
Quantifier swap in Banach space theory
The uniform boundedness principle and its corollaries from a logical point of view are statements of when one can swap quantifiers in Banach spaces. Take for instance the principle of condensation of singularities; if $V$ is a Banach space and $\{\phi_{ij} \}_{i, j \in \mathbb N} \subseteq V^*$ is a family of continuous linear functionals, then
\begin{equation}
(\forall i \in \mathbb N)(\exists f_i \in V)(\sup_{j \in \mathbb N} |\phi_{ij} (f_i)| = \infty) \iff (\exists f \in V)(\forall i \in \mathbb N)(\sup_{j \in \mathbb N} |\phi_{ij} (f)| = \infty).
\end{equation}
The forward is the non-trivial direction, and one of the standard proofs of the like use Baire category, where after one plays around with quantifiers, unions and intersections, we can show that in fact the set of $f \in V$ satisfying the property on the right is in fact generic (in a category sense) in $V$.
I was wondering if there's some kind of deeper interplay of logic, topology and functional analysis going on here, perhaps a general principle for the types of statements about Banach spaces in which we can swap quantifiers $(\forall \exists \iff \exists \forall)$. If so, perhaps this gives a logical heuristic for why the uniform boundedness principle holds; equivalence of weak boundedness and strong boundedness by itself is a surprising result. My intuition says something of the sort should exists, but I haven't been able to dig anything up on the subject.
Swapping quantifiers depends a lot on the statement that comes after the quantifiers, and not just on the Banach space. Even in a finite set, $\forall x \exists y ; A(x,y)$ may or may not be equivalent to $\exists y \forall x ; A(x,y)$, depending on $A$.
Is there a logical interpretation of other quantifier swaps in analysis, e.g. any of the more basic compactness arguments?
See my answer to "Techniques for reversing the order of quantifiers". While you're reading it I'll try to see if your example is also about compactness.
Should the last $f_i$ in the displayed formula be $f$?
It might be worthwhile to note that the very definition of a Banach space - i.e., completeness - can be phrased in terms of a quantifier swap: A metric space $(M,d)$ is complete if and only if, for every sequence $(x_n)$ in $M$, the following two assertions are equivalent: $\exists x \in M ; \forall \varepsilon > 0 ; \exists n_0 ; \forall n \ge n_0: ; d(x_n,x) < \varepsilon$, and $\forall \varepsilon > 0 ; \exists x \in M ; \exists n_0 ; \forall n \ge n_0: ; d(x_n,x) < \varepsilon$. (Since the second assertion is equivalent to $(x_n)$ being a Cauchy sequence.)
@AndrejBauer yes, thanks for catching the typo
|
2025-03-21T14:48:31.099937
| 2020-05-27T17:49:25 |
361517
|
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|
Stack Exchange
|
Schrodinger operator with matrix potential
This question is inspired by attempts to extend in some way Shen's 1999 "On fundamental solutions of generalized Schrödinger operators" to Schrödinger operators $- \Delta + V $ with some matrix potential $V : \mathbb{R}^d \rightarrow \mathbb{M}^{n \times n}$ that is symmetric and positive definite a.e. on $\mathbb{R}^d$.
Suppose we have some matrix function $M : \mathbb{R}^d \rightarrow \mathbb{M}^{n \times n}$ that is symmetric and positive definite a.e. on $\mathbb{R}^d$ (intuitively, some sort of "Agmon function" on $\mathbb{R}^d$ related to our matrix potential). If we are lucky and have $n = d$ then $M$ canonically induces a metric on $\mathbb{R}^d$. Namely, just define \begin{equation} d(x, y) = \inf_\gamma \int_0^1 \langle M(\gamma(t)) \gamma '(t), \gamma'(t) \rangle_{ \mathbb{R}^d} ^\frac12 \, dt \tag{1}\label{Metric}\end{equation} where the infimum is over all absolutely continuous curves $\gamma : [0, 1] \rightarrow \mathbb{R}^d$ where $\gamma(0) = x$ and $\gamma(1) = y$.
If $n \neq d$ then there seems no natural way to do this while "preserving" the matrix structure of $M$. That is, I don't want to look at something like $$d(x, y) = \inf_\gamma \int_0^1 \|M(\gamma(t))\| |\gamma '(t)|_{\mathbb{R}^d} \, dt. $$
Surely $M$ can induce a bundle metric on some rank $n$ vector bundle, but it seems impossible to do this canonically in a way that reduces to \eqref{Metric} when $n = d$.
Are there any papers in the Schrödinger operator literature that touch on this? I can't imagine an issue like this hasn't come up somewhere before.
Is the Schrodinger operator only for motivational purposes? It looks to me as though your question is just about geometry, and I don't see why you tagged elliptic-pde, and I don't see why you are looking for papers in the Schrodinger operator literature. Should the metric have something to do with the Schrodinger operator in some way?
Really it's just for motivational purposes (and is also my general reason for asking, as that's what I'm researching right now). Also I find it hard to imagine this situation has't appeared in the theory of Schrodinger operators. I tagged it elliptic-pde because you can ask the same question for more general elliptic operators with a matrix potential and again it's hard to imagine such an issue has never came up in that context (at least to a non-expert like myself. I mostly do classical harmonic analysis).
Though, maybe this also has appeared naturally in other contexts too...
But what is this metric supposed to tell you about the original Schrodinger operator? Why does it make sense to start from a matrix potential and ask about this metric?
To answer your other question, I'm essentially assuming apriori that a "useful" matrix valued Agmon function can be defined (analogous to how it is in Zhongwei Shen's 1999 "On fundamental solutions of generalized Schrodinger operators"), but this is naturally n times n valued, and thus doesn't naturally define an Agmon metric on $\mathbb{R}^d$... Here, it depends on the potential. What exactly this "Agmon function" is, is still work in progress.
I want upper and lower exponential decay estimates for the fundamental solutions of such matrix potential Schrodinger operators. Usually this is in terms of the Agmon metric.
Please edit your post to include the above ^^ two comments. They provide important context.
What does "a.e." mean in the first sentence? "almost everywhere" doesn't seem to match. I also don't understand "defined on $\mathbb{R}^d$ acting on $\mathbb{R}^n$ valued functions defined on $\mathbb{R}^d$", is something repeated by error?
That said, why do you think that for the matrix schrodinger operator the "Agmon function" would end up being an $n\times n$ matrix? Whatever definition you guess it should reduce to the standard form when $n = 1$, no?
@YCor: Concerning the last part of your comment: I think the sentence is to be understood as "a matrix-valued function $M: \mathbb{R}^d \to \mathbb{R}^{n \times n}$ that acts on vector-valued functions $u: \mathbb{R}^d \to \mathbb{R}^n$ by means of $u \mapsto Mu$.
@JoshuaIsralowitz: I would be interested in what kind of lower decay estimates you hope for: the lower estimates in Shen's 1999 paper imply, in particular, positivity of the fundamental solution - which is not true for Schrödinger systems that are coupled by matrix-valued potentials (unless the potential satisfies very specific properties).
@WillieWong Edit made. And to answer your question, it's not clear what exactly would constitute an appropriate "Agmon function", but thinking about this for a while, I think any kind of scalar function would throw away too much of the matrix structure of $V$. This is just a hunch.
@JochenGlueck We (Blair Davey and I) are imposing some kind of matrix "reverse Holder" condition, or if that's not enough, an additional "matrix A${}_\infty$ condition (something akin to that in Dall'Ara's 2015 paper "Discreteness of the spectrum of Schrodinger operators with non-negative matrix-valued potentials"). What exactly a lower bound would look like, if even possible, is unclear at the moment.
@JoshuaIsralowitz: I see, thanks a lot for your reply!
@JochenGlueck Sure! So far, our efforts have been devoted to proving an appropriate "matrix Poincare inequality" akin to Shen's in his 1999 paper. This has been more or less successful (and is very much akin to the "matrix weighted Poincare inequality" in my IUMJ paper "Matrix weighted Poincare inequalities..." with Kabe Moen). Now we are seeing what's possible with this "Poincare inequality".
What is an "Agmon function", for those not in the know?
@Igor Khavkine, this probably isn't the standard interpretation of an "Agmon function" (and I'm using the word "a" instead of "the" intentionally), but the way I interpret it is (and this is important for our project):
To prove upper/lower decay estimates in Shen's paper, a key step is what's called a Fefferman-Phong inequality, which practically speaking is a global weighted Poincare inequality (specifically for weights, aka scalar positive potentials, in certain reverse Holder classes, which is just for technical reasons).
What Shen then does, is apply this local Poincare inequality on Euclidean balls B(x, r(x)) where the radius r(x) is judiciously picked, multiply both sides by a "local to global correction factor" (i.e. an Agmon function), and integrate both sides with respect to x, to get the Fefferman-Phong inequality. Any "correction factor" satisfying certain properties works.
Again this is might not be standard thinking, but this is how I interpret the situation.
|
2025-03-21T14:48:31.100350
| 2020-05-27T18:16:25 |
361518
|
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|
Stack Exchange
|
Is every finitely-presentable group a finite colimit of copies of $F_2$?
Let $F_n$ be the free group on $n$ generators. Of course, every finitely-presentable group $G$ is a finite colimit of copies of $F_n$, where $n$ is allowed to vary. But is $G$ a finite colimit of copies of $F_2$?
Of course, because $F_{2n}$ is a finite coproduct of copies of $F_2$, we have that any finitely-presentable group $G$ is a finite colimit of finite colimits of copies of $F_2$ -- a "2-fold" finite colimit of copies of $F_2$. But I'm curious about the 1-fold case.
To make the question a bit more concrete, let's unwind what it means to be a finite colimit of copies of $F_2$:
Let $G$ be a group. Then $G$ is a finite colimit of copies of copies of $F_2$ if and only if $G$ admits a presentation of the following description:
There are $2n$ generators coming in pairs $x_1,y_1, \dots x_n, y_n$;
There is a finite set of generating relations, each of the form $w(x_i,y_i)=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i \leq j \leq n$.
So for example, $x_1y_1^2x_1^{-1} = y_2^{-1}x_2$ is a permissible generating relation (with $i=1,j=2$) but $x_1 x_2 = x_3$ is not a permissible generating relation because only 2 different subscripts are allowed to appear in a permissible generating relation. So my question is:
Question: Let $G$ be a finitely-presented group.
Is $G$ a finite colimit of copies of $F_2$?
Equivalently, does $G$ admit a presentation of the above form?
Edit:
The form of the presentation can be constrained even further, to look like this:
There is a finite set of generating relations, coming in pairs each of the form $x_i = w(x_j,y_j)$, $y_i=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i, j \leq n$.
Other variations are possible too; I'm not sure what the most convenient description to work with might be.
Here is how to present your illegal relation $x_1 x_2 x_3 = 1$ in a legal way. That group has $2n$-presentation $\langle x_1, y_1, x_2, y_2, x_3, y_3 \mid x_1 x_2 x_3, y_1, y_2, y_3 \rangle$. To make $x_1 x_2 x_3$ into a legal relation, introduce $x_4,y_4, x_5, y_5$, and relations $x_1 = x_4, x_2 = y_4$, $x_5 = x_4 y_4$, $x_3 = y_3$; then the legal presentation is $$\langle x_1, \cdots, y_5 \mid x_1 = x_4, x_2 = y_4, x_3 = y_3, y_1 = y_2 = y_3 = 1, x_4 y_4 = x_5\rangle.$$ This seems to lead to the algorithm in Pace Nielsen's answer.
@MikeMiller Thanks for pointing that out! (And yes, that was exactly what motivated the solution I produced.)
I believe the answer is yes. Assume, by way of contradiction, that some finitely presented group cannot be so expressed. Then we can choose such a group $G$ where for any generating set of the form $x_1,y_1, x_2,y_2,\ldots, x_n,y_n$ the number of non-permissible relations needed to define $G$ (together with some finite number of permissible relations) is minimized; say those non-permissible relations are $w_1=1, w_2=1,\ldots, w_m=1$. Write $w_1=z_1z_2\cdots z_p$ where each $z_j\in \{x_1^{\pm 1},y_1^{\pm 1},\ldots, x_n^{\pm 1},y_n^{\pm 1}\}$, where we may also assume that $p$ has been minimized. Note that since $w_1=1$ is not permissible, we must have $p\geq 3$.
Add new generators $x_{n+1},y_{n+1},x_{n+2},y_{n+2}$. The relations $x_{n+1}=z_{p}$, $y_{n+1}=z_{p-1}$, $x_{n+2}y_{n+2}=1$ are permissible. The relation $x_{n+2}y_{n+1}x_{n+1}=1$ is also permissible (since it is equivalent to $x_{n+2}=x_{n+1}^{-1}y_{n+1}^{-1}$). Asserting these relations still gives us the same group (since our new relations merely tell us how to write the new generators in terms of the old ones). The relation $w_1=1$ is equivalent to $z_1z_2\cdots z_{p-2}y_{n+2}=1$, but it is now shorter, a contradiction.
Ah, you solved my notational problem (this may be standard). I incorporated it as well to make my answer more readable.
Thanks! Incidentally, I think this algorithm generalizes to show that for any variety (in the sense of universal algebra) with all operations of arity $\leq n$, any finitely-presented algebra is a finite colimit of copies of the free algebra on $n$ generators. I wonder if this is known and written somewhere...
Er -- perhaps the generalization I suggested is correct, but if it is, I now think it requires at least a bit more work.
@TimCampion Your generalization sounds like the kind of question that George Bergman is an expert at. Perhaps he would know the answer.
If only you did not unecessarily say "proof by contradiction", you would have an algorithm in your hands for converting any presentation to a permissible one.
@AndrejBauer Given that Mike Miller, Tim Campion, and yourself all recognized that there was an algorithm present, I'll count this solution as a success. :-)
With a slight tweak, the same algorithm works for either the variety of groups or the variety of monoids. That is, we add $x_{n+1}, y_{n+1}, x_{n+2}, y_{n+2}$ along with the relations $x_{n+1} = z_p, y_{n+1} = z_{p-1}$, but tweak the last two relations to read $x_{n+2} = 1$ and $y_{n+2} = y_{n+1}x_{n+1}$; we still have $w_1 = z_1\cdots z_{p-2} y_{n+2}$. I wonder if it can also be done for semigroups....
@TimCampion Semigroups are even easier. Suppose you have a relation $x_1 x_2 \cdots x_m = y_1 y_2 \cdots y_n$. Create new elements $z_1,z_2,\ldots, z_{m-1}$ and $w_1,w_2,\ldots, w_{n-1}$ and add the relations $z_1=x_1x_2$, $z_2=z_1x_3$, $z_3=z_2x_4$, and so forth up to $z_{m-1}=z_{m-2}x_m$. Do the something similar for the $w$'s. This defines the $z$ and $w$ variables in terms of previous generators. Finally, asserting $z_{m-1}=w_{n-1}$ gives the original relation.
I believe this simplifies the previous argument (since the relations defining inverses are already good, and defining $1$ is also doable in terms of good relations).
Nice! While I'm thinking about this, maybe it's worth pointing out that the algorithm one gets is only effective if there is an effective way to compute, given a group word, an equivalent group word of minimal length. I'm not sure if this is possible for groups, and I certainly presume it isn't possible for more general varieties.
@TimCampion The word problem for groups is not decidable. In particular, one cannot tell whether a group word is trivial (i.e., equal to $1$). But, given any finite presentation, the algorithm can still easily be made effective. Rather than working by contradiction, one instead just reduces word lengths (ala what Andrej Bauer said earlier).
Ah -- of course you're totally right.
@TimCampion I just noticed that your solution for general varieties is essentially what I did in the semigroup case, but I was an hour late! (Just add parentheses, so the left side becomes $(\cdots((x_1x_2)x_3)\cdots x_n)$. Now follow your algorithm.)
Here's a pretty direct way to do this. Choose any presentation by generators $x_1,\ldots,x_n$ and relations $r_1,\ldots,r_m$; say $r_i = z_{i,1} \cdots z_{i,k}$ for $z_{i,1},\ldots,z_{i,k} \in \{x_1^{\pm 1}, \ldots, x_n^{\pm 1}\}$. Firstly, we may assume all $r_i$ have length $k = 3$: the new variables
$$x_{i,j} = z_{i,1} \cdots z_{i,j}$$
for $0 \leq j \leq k$ are subject only to the relations
\begin{align*}
x_{i,0} = e = x_{i,k}, & & & & & & x_{i,j} = x_{i,j-1}z_{i,j} & & (1 \leq j \leq k).
\end{align*}
If $k < 3$, we can eliminate the variable $z_{i,1}$ at the expense of replacing all $z_{i,1}^{\pm 1}$ by $z_{i,2}^{\mp 1}$ (if $k = 2$) or $e$ (if $k = 1$) in the other relations, so we may assume all $r_i$ have length exactly $3$. Then introduce new variables $x_{n+1},\ldots,x_{n+m}$ as well as variables $y_1,\ldots,y_{n+m}$, subject to the relations
\begin{align*}
x_{n+i} &= z_{i,1},& & 1 \leq i \leq m,\\
y_i &= e & & 1 \leq i \leq n,\\
y_{n+i} &= z_{i,2}, & & 1 \leq i \leq m,\\
x_{n+i}y_{n+i} &= z_{i,3}^{-1}, & & 1 \leq i \leq m,\\
\end{align*}
This gives a presentation of the desired form. $\square$
Thanks! It's interesting to see this from a more top-down perspective.
Ok, I think the above answers have pointed the way to a proof of the obvious generalization:
Theorem: Consider a variety $V$ (in the sense of universal algebra) generated by operations of arity bounded by some $N \in \mathbb N$, and let $\mathcal C$ be the category of $V$-algebras and homomorphisms. Let $F$ be the free algebra on $N$ generators. Then every finitely-presented $A \in \mathcal C$ is a finite colimit of copies of $F$.
Proof: By using dummy variables, we may assume that every basic operation in $V$ is of arity exactly $N$. As described in the case of groups in the Question, we are looking for a presentation of $A$ by generators $(x_{11}, \dots, x_{1N},\dots, x_{n1}, \dots, x_{nN})$ modulo "permissible" relations $w(x_{i 1}, \dots x_{i N}) = v(x_{j 1},\dots, x_{j N})$, were $w, v$ are (possibly composite) operations in the variety $V$.
As in Pace Nielsen's answer, it will suffice to show that if $w(x_{11},\dots, x_{nN}) = v(x_{11},\dots,x_{nN})$ is a non-permissible relation, we can, after adding more variables $(x_{n+1,1},\dots, x_{n+1,N})$, replace it with relations using only shorter words (in the sense that the total number of basic operations of which each word is composed is smaller -- the base case is a word $x_{ij}$ composed of no operations; note that a relation $x_{ij} = x_{kl}$ is permissible) and a relation of the form $f(x_{n+1,1},\dots, x_{n+1,N}) = v(x_{11},\dots, x_{nN})$.
To this end, we may write $w(x_{11},\dots, x_{nN}) = f(w_1(x_{11},\dots,x_{nN}), \dots, w_N(x_{11},\dots,x_{nN}))$ where $f$ is a basic operation and the $w_i$'s are shorter words. We impose the relations $x_{n+1,i} = w_i(x_{11},\dots,x_{nN})$ (which use only shorter words) along with the relation $f(x_{n+1,1},\dots, x_{n+1,N}) = v(x_{11},\dots, x_{nN})$, as desired.
I think the argument generalizes, with basically just notational changes, to the case of a many-sorted variety with finitely many sorts (such as the 2-sorted variety of (ring, module) pairs -- showing that any such pair which is finitely presentable is a finite colimit of copies of $\mathbb Z[x,y]$ acting on itself).
|
2025-03-21T14:48:31.100915
| 2020-05-27T19:24:03 |
361520
|
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|
Stack Exchange
|
What is the state-of-the-art for solving polynomials systems over fields that are not algebraically closed?
I am not working in the field of algorithmic algebraic geometry - yet, for my current work, I need some results from it.
More specifically, what is the state-of-the-art when it comes to solving (whatever "solving" means in this case) system of polynomials of fields that are not algebraically closed, whose ideal has dimension $>0$?
Could you recommend a survey paper that summarizes what has been achieved so far?
For the case of $0$-dimensional ideals, there seems to exist many heavily cited papers, like "Solving Zero-dimensional Algebraic Systems" by D. Lazard, which seem mostly to be concerned with finding ways of to display the system of polynomials in a nice way (e.g. triangularly). Are these articles already superseded, or does it make sense to read them?
Edit: In particular, I'm interested in the field $\mathbb{R}$, since most of my example will come from here (but $\mathbb{Q}$ might be also useful; and perhaps even the ring $\mathbb{Z}$; I don't yet know where the results I will get for $\mathbb{R}$ will take me).
Also worth making more precise: In the case of positive dimension of the ideal, I'm interested in methods that tell me, if I project to whole, infinite solution space down to a single variable and I'm interested in, in what set this variable lies. More formally, if $V(f_1,\ldots,f_s)\subseteq F^n$ is my solution variety, with $f_i \in F[x_1,\ldots,x_n]$, and I'm interested in some specific variable, say $n_0$, what methods are there that describe $\mathop{\rm proj}_{n_0}(V(f_1,\ldots,f_s))$?
This question is way too broad to give any kind of meaningful answer. If you can be more specific about how this question is relevant to your work, that should help to give the question more focus.
In particular, "solving" might be interpreted in various communities as either finding a single solution or finding all solutions. This gets murkier for positive-dimensional solution sets: how should one represent "all solutions"?
@RP_ hm... I don't think it would be that helpful to describe how solving polynomial systems arise in my work, since there are no restrictions on the structure of polynomials systems, so I'm still left with the "how can I solve general polynomials systems over the reals". more specifically, I'm investigating certain classes of neural nets and some metric properties of them, that lead to the questions whether some polynomial systems have solutions.
That is not what you asked: you asked about "fields that are not algebraically closed." Now it turns out you are only interested in the reals. Better edit your question.
@RP_ Well, I'm not exclusively interested in the reals, but it does consist in the most important case - see my edit.
@tim yes, that is indeed problem in positive dimension. I'm looking for any kind of parametrization of solutions that help my isolate specific variables, as mentioned in my edit.
To give you an idea of the difficulty, deciding whether a plane curve $P(x,y)=0$ with $P \in \mathbb{Z}[x,y]$ has a real point or not, is a non-trivial task in general, you need the notion of cylindrical decomposition from real algebraic geometry. Yet this existence problem and many more from algebraic geometry can be solved and have even been implemented -- see the softwares Qepcad and Redlog. In particular, it's possible to determine whether an arbitrary system of polynomial equations in $\mathbb{Z}[x_1,\ldots,x_n]$ has a solution in $\mathbb{R}^n$ or not.
Do the polynomials of your system have exact coefficients (i.e., algebraic numbers) or are they approximations?
You can also compute projections using the softwares mentioned above. But you need exact coefficients (e.g. integers). And the output will be a boolean combination of polynomial inequalities in 1 variable.
These problems are hard. If there isn't any structure to exploit, you will run out luck quickly.
@user43263 In general it's better to know things. It sounds like you already know the "dimension" is greater than zero. Do you mean dimension over the reals? I would interpret "structure" pretty broadly. Methods like CAD, Groebner Bases, resultants, and homotopy continuation have different capabilities and may outperform each other in various regimes. One should not expect a "best method" due to computational complexity, as alluded to in my previous remark. Your opportunities for success are slim for very large problems -- but you might only need a local method like Newton iteration anyway.
The equation over the quaternions $z^2 = q$ for arbitrary fixed $q \in \mathbb H$ is in a sense unsolvable. See here: https://mathoverflow.net/questions/342041/is-it-possible-to-constructively-prove-that-every-quaternion-has-a-square-root
@FrançoisBrunault That sounds actually very interesting to me, thank you for pointing that out. In case you are interested, as I learned through this question that the theory is more involved, I wrote a follow-up question, that details my problem more precisely: https://mathoverflow.net/questions/361642/existance-of-solutions-of-polynomials-systems-and-their-rough-shape-over-m
@tim I think I might have some hope to tackle this - perhaps asking for the state-of-the art was too much. I outlined as best as I could the structure my problem has in a follow-up question (to which I linked in the comment above).
For the reals, I particularly like the book by Sturmfels mentioned by Alexandre Eremenko. For the rational numbers, you can hardly do better than Bjorn Poonen's book Rational Points on Varieties, which is available for browsing via his homepage.
For dimension $1$ specifically, Poonen also has a set of lecture notes on rational points on curves, although I always have trouble finding it. Moreover he has several expository articles (listed as such on his page) dealing with rational points on curves.
Restricted to the case of the field of rational numbers and dimension $1$ alone, this is a huge question. Restricting only to the field of rational numbers makes it even huger. Dropping any restrictions on the field entirely makes it well-nigh impossible to answer in full geberality...
It might be good to highlight the difference that, for the reals, the existence of an algorithm has been known for a long time, since work (I think) of Tarski, and it's "just" a question of improving the speed to be practical. For the rationals, the problem is completely open even in some of the simplest cases and one can easily organize multiple large conferences on non-overlapping aspects of the problem.
@WillSawin Yes absolutely, I kind of assumed that had already been mentioned somewhere...
Thanks, I wasn't aware of Poonen's book. After doing some reading I realized, given how comprehensive the theory is, that it is absolutely necessary to outline my problem in as much detail as possible. I have written a follow-uip question in that regard (https://mathoverflow.net/questions/361642/existance-of-solutions-of-polynomials-systems-and-their-rough-shape-over-m ) , since giving all that detail it seemed outside the scope of this question.
@WillSawin Just for fun, could you link one paper (perhaps a survey paper, if there is one) that describes some unknown things in case of the rationals?
@user43263 Here is a list of open problems concerning K3 surfaces, which are sometimes advertised as being the frontier of rational points research in dimension $2$: https://sites.math.washington.edu/~bviray/openproblems.pdf Although this is an arguable point, since there are plenty of questions left about rational surfaces and generally about surfaces of Kodaira dimension $-1$.
For the real field:
MR2830310
Sottile, Frank
Real solutions to equations from geometry.
University Lecture Series, 57. American Mathematical Society, Providence, RI, 2011.
MR2275625 Mikhalkin, Grigory Tropical geometry and its applications. International Congress of Mathematicians. Vol. II, 827–852, Eur. Math. Soc., Zürich, 2006.
MR1108621
Khovanskiĭ, A. G.
Fewnomials.
American Mathematical Society, Providence, RI, 1991.
MR1659509 Bochnak, Jacek; Coste, Michel; Roy, Marie-Françoise Real algebraic geometry. Springer-Verlag, Berlin, 1998.
For other fields:
MR2247966
Vakil, Ravi
Schubert induction.
Ann. of Math. (2) 164 (2006), no. 2, 489–512.
Also:
MR1925796
Sturmfels, Bernd
Solving systems of polynomial equations.
CBMS Regional Conference Series in Mathematics, 97. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 2002.
where the real field is also discussed.
Thank you very much for all these references, in particular Sturmfels' book seems to contain useful theorems for me.
|
2025-03-21T14:48:31.101497
| 2020-05-27T20:32:53 |
361524
|
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|
Stack Exchange
|
Semisimplicity for tensor products of representations of finite groups
Let $G$ be a group and $k$ a field of characteristic $p>0$. Let $$\rho_i: G\to GL(V_i),~ i=1,2$$ be two finite-dimensional semisimple $k$-representations of $G$, with $\dim(V_1)+\dim(V_2)<p+2.$ Then a 1994 theorem of Serre tells us that $\rho_1\otimes\rho_2$ is semisimple.
I was wondering -- is there is an easier proof in the case that $G$ is finite?
Specializing Serre's proof to the case of finite groups seems not to yield any real simplification; one has to apply the so-called saturation technique to replace the subgroup of $G$ generated by elements of order a power of $p$ by a linear-algebraic group.
I suspect the answer is "no" -- it seems to me that the general case reduces to the case of finite groups (by a spreading out and specialization argument), so it's hard to believe this case could be substantially easier -- but I figured it was worth asking. I'd also be interested in a proof with worse bounds, e.g. with $p+2$ replaced by any increasing function of $p$.
I don't think this is obvious in the finite case, and I don't think that the finite case was known in that generality before Serre's proof appeared. If a simpler proof were found for the finite case, I am sure there would be significant interest in it.
That's my instinct as well -- as I remark in the question, I don't think it's too hard to reduce the general case to the case of finite groups, so by the principle of "conservation of work," the case of finite groups should be hard. Still, it's a bit surprising to me that such an innocent statement should be so difficult.
@DanielLitt "Still, it's a bit surprising to me that such an innocent statement should be so difficult." - Possibly a definition for groups.
I'm not a specialist in modular representation theory, and, probably because of that, there is no statement about modular representation theory (other than the ones that are exactly as in complex representation theory) that seems innocent to me.
@LSpice: to be fair the analogous statement to Serre’s result in characteristic zero (that tensor products of semisimple representations of arbitrary groups are semisimple) is non-trivial, and relies on the theory of algebraic groups (it’s due to Chevalley). This claim about finite groups in characteristic p implies that result about arbitrary groups, though!
What makes tensor product special? Where does this fail for arbitrary representation?
@VítTuček: If you take $G=\mathbb{Z}/p\mathbb{Z}$ then the $2$-dimensional $\mathbb{F}_p$-representation sending $x$ to the upper triangular matrix with $1$'s on the diagonal and $x$ in the upper-right-hand corner is not semisimple.
There is a result of D. S. Passmann and D. Quinn in "Burnside's theorem for Hopf algebras", Corollary 8, which says the following:
If $A$ is a finite-dimensional Hopf algebra, then the set of semisimple
$A$-modules is closed under tensor product if and only if the Jacobson
radical $J(A)$ is a Hopf ideal of $A$.
If you require all semisimple $A$-modules to be closed under tensor products, then the question reduces to something about the Jacobson radical.
In a paper of M. Lorenz, "Representations of Finite-Dimensional Hopf Algebras", he makes the following comments:
Remarks and Examples. 1) If all simple $H$-modules are
1-dimensional (equivalently, $H/J\simeq k^r$ as $k$-algebras for some $r$), then all
tensor products of simple $H$-modules are 1-dimensional as well, and
hence condition (2) of the lemma is clearly satisfied. Thus $J$ is a
Hopf ideal in this case. ......
2) If $H=kG$ is a finite group algebra, then $J$ is a
Hopf ideal precisely if $G$ has a normal Sylow $p$-subgroup [M].
[M] R. K. Molner, "Tensor products and semisimple modular representations of finite groups and restricted Lie algebras", Rocky Mountain J. Math. 111981 , 581-591.
This seems to address a different question. Serre's result is about tensoring two low dimensional semisimple representations to get another semisimple representation. It doesn't require that all semisimple representations are closed under tensoring.
As I said in my post, "If you require all semisimple -modules to be closed under tensor products, then the question reduces to something about the Jacobson radical." I was just thinking, if you have more, then certainly you would have less.
|
2025-03-21T14:48:31.101927
| 2020-05-27T21:20:38 |
361528
|
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|
Stack Exchange
|
On full rank submatrices of a construction
Take two matrices $T_1$ and $T_2$ in $\mathbb Z^{n\times n}$ with entries uniformly in $[-b,b]\cap\mathbb Z$ at some $b>0$. The matrices will be of rank $n$ each with probability at least $1-\frac1{\Omega(n^2)}$. Form a matrix $T_3$ of size $n^2\times n$ with rows of $T_3$ being difference of rows of $T_1$ and $T_2$. Since $T_1$ and $T_2$ are random we will have all rows of $T$ distinct with probability $1-o(1)$.
For every row $i\in\{1,\dots,n^2\}$ in $T_3$ form an $n\times n$ matrix $M_i=T_3[i]'\times T_3[i]$ where $'$ is transpose and $T_3[i]$ is $i$th row of $T_3$ and consider the matrix $N_i=(M_i+M_i')-Diag(M_i)$ where $Diag(M_i)$ is diagonal matrix consisting of diagonal entries of $M_i$. Form a matrix $T$ of size $n^2\times\Big(n+\frac{(n-1)n}2\Big)$ with $i$th row of $T$ being the diagonal and top right entries of $N_i$.
What is the probability that the rank of
$T$ is $\Big(n+\frac{(n-1)n}2\Big)$ for a given $b$?
a random $\Big(n+\frac{(n-1)n}2\Big)\times\Big(n+\frac{(n-1)n}2\Big)$ submatrix is $\Big(n+\frac{(n-1)n}2\Big)$ for a given $b$?
The probability that your matrix $T_1$ is singular is in $e^{-O(n^2)}$ (for a fixed $b$).
e^n^2 is technically Omega(n^2) and so I didn't bother since latter suffices for purpose of problem..
To clarify... $M_i$ is symmetric (yes?). And to say it with fewer matrices, you start with random vectors $u_i$ and $v_j$, then for each pair, you cook up matrices $N_{i,j}$ and say $S$ is the collection of these $n^2$ matrices. You want to know the chance that $S$ spans the space of symmetric matrices. And you also care about the chance that a random subset (of correct size) is a basis. Right?
I figured you don't get any with this concoction.
|
2025-03-21T14:48:31.102077
| 2020-05-28T00:17:12 |
361534
|
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|
Stack Exchange
|
$ \int_{E}^{*}{\psi (t) d\mu(t)}=\int_{E}{\phi (t) d\mu(t)} $
Let $(T, \mathcal{A}, \mu)$ be an arbitrary measure space.
The outer integral over $(T, \mathcal{A}, \mu)$ of a (possibly nonmeasurable) function $\psi: T\to (-\infty, +\infty]$ is defined by:
$$
\int_{T}^{*}{\psi (t) d\mu(t)}:=\inf\{\int_{T}{\phi (t) d\mu(t)}~|~\phi:T\to \mathbb{R}\text{ integrable, },\psi\leq \phi\}
$$
Problem:
For a function $\psi:T\to \mathbb{R}$ such that $\int_{T}^{*}{\psi (t) d\mu(t)}<\infty$. Can we say that there exists an integrable function $\phi:T\to\mathbb{R}$ such that:
$$
\int_{T}^{*}{\psi (t) d\mu(t)}=\int_{T}{\phi (t) d\mu(t)}
$$
"Weak Convergence and Empirical Processes" by van der Vaart and Wellner has a chapter on outer integrals, in which you can find this and many related results.
Yes. By definition of the inf, we can find a sequence of integrable functions $\phi_n$ such that $\psi \le \phi_n$ for every $n$, and $\int \phi_n \to \int^* \psi$. Set $\phi = \liminf \phi_n$, which is again integrable and satisfies $\psi \le \phi$. As such, $\int^* \psi \le \int \phi$. On the other hand, by Fatou's lemma, $\int \phi \le \liminf \int \phi_n = \int^* \psi.$
|
2025-03-21T14:48:31.102173
| 2020-05-28T01:10:47 |
361538
|
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|
Stack Exchange
|
Asymptotic behavior of maximum of bessel function
Let $J_n$ be the Bessel function of the first kind. Let $J_n^{(\max)} = \max_{x>0} J_n(x)$. What is known about the asymptotic behavior of $J_n^{(\max)}$ at large $n$? Specifically, I am looking for a lower bound. It is ok if the result only holds for integer and half-integer $n$.
(It is potentially helpful to note that $J_n^{(\max)} = J_n(j'_n)$ where $j'_n$ is the smallest positive zero of $J'_n$, i.e. the global maximum of $J_n$ occurs at the "first" local maximum.)
The answer is given in Formula 9.5.25 of Abramowitz and Stegun, available here:
https://pdfs.semanticscholar.org/1a2a/68cac86cb55abb9eb1858b3b58c4a1b16434.pdf
|
2025-03-21T14:48:31.102248
| 2020-05-28T02:49:39 |
361540
|
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|
Stack Exchange
|
Anti-symmetric operators for the Dirac or Majorana spinors
In a Zoom lecture given by a mathematical physics professor, if I recalled correctly, he explained that the in 1+1 dimensional spacetime (or 2 dimensions in short), the "action" of fermions (spinors) has an anti-symmetric Dirac operator.
Say, if the $\psi$ is a Dirac spinor, he wrote down an action
$$
\int d^2x \sqrt{g} \bar{\psi} (i \gamma^\mu D_\mu) \psi
$$
and (I think) he claims that the operator $(i \gamma^\mu D_\mu)$ is an anti-symmetric matrix.
Say, if the $\chi$ is a Majorana spinor, he wrote down an action
$$
\int d^2x \sqrt{g} {\chi} (i \gamma^\mu D_\mu) \chi
$$
and (I think) he claims that the operator $(i \gamma^\mu D_\mu)$ is also an anti-symmetric matrix.
Is this true that the anti-symmetric matrix has something to do with these fermions (spinors)? or fermion statistics? Why?
p.s. Maybe the first case the $D$ operator is complex, and the second case that the $D$ operator is real (?).
This seems difficult to parse. Anti-symmetric in what indices? The usual requirement is that $\gamma^{0} $ be Hermitean and the $\gamma^{i} $ anti-Hermitean. One can find Hermitean matrices that are symmetric, and ones that are antisymmetric ... perhaps a definite representation of the Dirac matrices is intended? Certainly, $iD_{\mu } $ is Hermitean. And all this shouldn't depend crucially on this being in 1+1 dimensions.
In terms of the Hamiltonian $H$ the antisymmetry follows simply: Majorana fields $\Psi(x,t)$ are real, and they satisfy a real wave equation
$$\partial\Psi/\partial t =-iH\Psi,$$
where $iH$ is real, hence $H$ is imaginary. Since $H$ must also be Hermitian, it means that $H^T=-H$ (antisymmetric).
More explicitly, $H=\gamma_{\rm M}^\mu \partial_\mu$, with $\gamma_{\rm M}^\mu$ the Dirac matrices in the Majorana representation, for which $\gamma_{\rm M}$ is a purely imaginary $4\times 4$ matrix. The antisymmetry of $H$ becomes manifest if we discretize the derivative operator so that $H$ becomes a matrix and $H_{nm}=-H_{mn}$.
The plot thickens ... now, $H$ is constructed from the spatial components only - furthermore, since $H$ is related to $i\gamma_{M}^{i} D_i $ through a factor $\gamma_{M}^{0} $, which is antisymmetric, this would actually imply that $i\gamma_{M}^{i} D_i $ is symmetric, not antisymmetric. Could it be that the spoken word in a Zoom lecture, remembered from memory, is not quite as reliable as the written word?
I think the answer is very nice and seems perfectly fine -- I will accept it in a week if no other better answers. (Voted up)
thank you again - accepted
Sorry, may I confirm: What do we mean by the matrix is real or imaginary?
(1) when we say matrix $P$ is Hermitian, it mens $P^\dagger =P$,. when we say matrix $P$ is anti Hermitian, it mens $P^\dagger = -P$,.
(2) when we say matrix $P$ is symmetric, it mens $P^T =P$,. when we say matrix $P$ is anti symmetric, it mens $P^T =-P$.
(3) What do we mean by the matrix $P$ is real or imaginary? to be mathematical precise?
(PS. Hermitian matrix has real eigenvalues.)
a matrix is real if the matrix elements are $\in\mathbb{R}$; a Hermitian matrix $P$ is real if $P=P^T$.
The answer seems to be in general and independent of space time dimensions, yes or no? thanks!
yes it is as you say.
|
2025-03-21T14:48:31.102495
| 2020-05-28T03:15:58 |
361544
|
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|
Stack Exchange
|
Reference request: Variational techniques for complex "iterated" Lagrangians
I am interested in solving variational problems of the form
$$
\min_u \int \Big\{L(x,y,u(x,y)) + \phi\Big(\int J(z,y,u(z,y))\,dz\Big)\Big\} p(x,y)\,dx\,dy.
$$
for some known, smooth functions $L,J,\phi,p$, and the minimization is with respect to functions $u:\mathbb{R}^m\times\mathbb{R}^n\to\mathbb{R}$. We can assume e.g. $u$ is square integrable or even bounded; this part isn't crucial. Note that the derivatives of $u$ are not involved, which helps simplify things somewhat.
For example, a particular form that has come up is the following:
$$
\min_u \int \Big\{g(x) + \Big(\int h(z) u(z,y)\,dz\Big)^r\Big\} p(x,y)\, dx\,dy
$$
There is some structure here (e.g. the double integral involving two Lagrangian functions) that seems ripe for exploitation. But the second, inner integral seems to break the standard setup of the calculus of variations (what is the derivative of $\int J(z,y,u(z,y))\,dz$ with respect to $u$?), but maybe I am missing something.
I am looking for any references covering these types of problems, or related ones. I would also be interested in (even more complicated) formulations involving derivatives of $u$, but the simpler version seems difficult enough!
EDIT: It seems that the most salient aspects of this problem can be exposed by considering the simpler form
$$
\min_u \int \phi\Big(\int J(z,y,u(z,y))\,dz\Big)\,dy.
$$
From this angle, it is clear that $\phi$ is the complication (e.g. if $\phi$ were not involved, this would be a standard calculus of variations problem).
It will.be better for reader comprehension if you replace the argument x in J with some other dummy variable.
The keyword you should search for is 'nonlocal variational calclus'
Nonlocal problems are a pretty broad topic, so if you have any more specific suggestions I'd love to take a look!
When $\phi$ is the $r$-th power, with $r\in\mathbb{N}$, you could rewrite it as a regular variational problem by "splitting" $z$ into $r$ different variables $(\int F(z)dz)^r =\int F(z_1)\cdots F(z_r)dz_1 \cdots dz_r$.
At least if everything is sufficiently smooth,
$$
\frac{\delta }{\delta u(s,t)} \int dy\ \phi \left( \int dz\ J(z,y,u(z,y)) \right) = \phi^{\prime } \left( \int dz\ J(z,t,u(z,t)) \right) \frac{\partial J}{\partial u} (s,t,u(s,t))
$$
|
2025-03-21T14:48:31.102664
| 2020-05-28T03:18:50 |
361545
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361545"
}
|
Stack Exchange
|
Orthogonal projections of permutation matrices are symmetry group of regular simplex?
Each permutaion $\pi\in S_n$ corresponds to a permutation matrix $P_\pi$ with $(0,1)$ entries. To be more specific, $P_\pi = [e_{\pi(1)},\ldots,e_{\pi(n)}]$, where $e_1,\ldots,e_n$ are standard basis of $\mathbb{R}^n$.
Note that $P_\pi \mathbb{1}=\mathbb{1}$, where $\mathbb{1}$ is a vector with all $1$ elements. Then we can block diagonalize $P_\pi$ for $\pi\in S_n$ simultaneously. See this problem about block diagonalization. There is an orthogonal change of basis $V$ such that
$$V^t P_\pi V = \begin{bmatrix}
\hat{P_\pi} & 0\\
0 & 1
\end{bmatrix}
$$
for all $\pi\in S_n$.
I'm reading a paper, which says that $\{\hat{P}_\pi,\pi\in S_n\}$ can be interpreted as the symmetry group of a regular simplex in $\mathbb{R}^{n-1}$. I need more details about why it is true. Thanks for reading! Any comments are appreciated.
|
2025-03-21T14:48:31.102749
| 2020-05-28T03:22:58 |
361546
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361546"
}
|
Stack Exchange
|
How to proceed in this Boundary value problem where Eigen values are calculated numerically?
While solving a boundary value problem (background provided in the Context section) I reach the following variable separated two equations ($F(x)$ and $G(y)$)
\begin{eqnarray}
\lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F &=& 0, \tag 1\\
V \lambda_c G''' - 2 V \lambda_c \beta_c G'' + \left( (\lambda_c \beta_c - 1) V \beta_c + \mu \right) G' + V \beta_c^2 G &=& 0,\tag 2
\end{eqnarray}
with some separation constant $\mu \in \mathbb{R}$.
with boundary conditions as:
For G: $G'(0)=0, G(0)=0$ and $\frac{G''(1)}{G'(1)}=\beta_c$
For F: $F'(0)=0$ and $\frac{F''(1)}{F'(1)}=\beta_h$
The non-homogeneous condition in $F$ is: $\beta_h e^{-\beta_c y}G'(y)F(0)=1$
Since the boundary conditions for $(2)$ are all homogeneous, I have calculated the eigenvalues $\mu_i$ for $(2)$ numerically. For some realistic parameters like βc = 0.921, λc = 1.775*10^-4, V=1, these eigen values are
0.834041, 0.845661, 0.864286, 0.888675, 0.916951, 0.94696, 0.977271, 1.0079, 1.03972, 1.07361, 1.11015,...
Now since these eigenvalues are numeric in nature, I cannot figure out how to move forward with this problem.
I know from standard PDE problems that these eigen values should be utilized to build the $G$ solution and then should be employed in the $F$ non-homogeneous condition to determine the constants. But how am I supposed to even use orthogonality here ? I would really appreciate if someone could give a step-wise way forward for such problems where the EVs are numeric.
Context
I had the following system of PDEs
$$\frac{\partial \theta_h}{\partial x}+\beta_h (\theta_h-\theta_w) = 0 \tag A$$
$$\frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c-\theta_w) = 0 \tag B$$
$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V\frac{\partial^2 \theta_w}{\partial y^2}-\frac{\partial \theta_h}{\partial x} - V\frac{\partial \theta_c}{\partial y} = 0 \tag C$$
with the boundary conditions ($\beta_h, \beta_c, V, \lambda_h, \lambda_c$ are constants)
$$\theta_w(0,y)=1, \theta_w(x,0)=0$$
$$\frac{\partial \theta_w(1,y)}{\partial x}=\frac{\partial \theta_w(x,1)}{\partial y}=0$$
$$\theta_h(0,y)=1, \theta_c(x,0)=0$$
Using the transformation $\theta_{h1}(x,y):=\theta_h (x,y)-1$ (so that $\theta_w(0,y)=0$ which is needed to get an additional homogeneous condition on $F$) where
\begin{eqnarray}
\theta_{h1}(x,y) &=& \beta_h e^{-\beta_h x} \int e^{\beta_h x} (\theta_w(x,y)-1) \, \mathrm{d}x,\\
\theta_c(x,y) &=& \beta_c e^{-\beta_c y} \int e^{\beta_c y} \theta_w(x,y) \, \mathrm{d}y.
\end{eqnarray}
which are substituted in $(C)$ to get:
$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} +( -\beta_h - V \beta_c )\theta_w +\beta_h^2 e^{-\beta_h x} \int e^{\beta_h x} \theta_w(x,y) \mathrm{d}x + \beta_c^2 e^{-\beta_c y}\int e^{\beta_c y} \theta_w(x,y)\mathrm{d}y = 0 \tag D$$
Using the ansatz $\theta_w(x,y) = e^{-\beta_h x} f(x) e^{-\beta_c y} g(y)$ on $(D)$ we obtain two linear third-order ODEs ($1,2$) with constant coefficients for $F(x) := \int f(x) \, \mathrm{d}x$ and $G(y) := \int g(y) \, \mathrm{d}y$ given in the original question
|
2025-03-21T14:48:31.102935
| 2020-05-28T07:26:22 |
361554
|
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"authors": [
"Arno Fehm",
"https://mathoverflow.net/users/50351"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361554"
}
|
Stack Exchange
|
Under what conditions is the compositum of two rational cubic Kummer extensions a rational function field?
Let $k$ be an algebraically closed field of characteristic $p > 3$ and $F = k(x)$ be the rational function field in the variable $x$. Consider two Kummer extensions $F_1 = F(\sqrt[3]{g_1})$, $F_2 = F(\sqrt[3]{g_2})$ of degree $3$, where $g_1, g_2 \in F^*$. Suppose that $F_1, F_2$ are also rational, that is $F_1 = k(x_1)$, $F_2 = k(x_2)$ for some $x_1 \in F_1$, $x_2 \in F_2$.
Under what conditions is the compositum $F_1F_2 = k(\sqrt[3]{g_1}, \sqrt[3]{g_2})$ a rational function field?
Should that not be a simple application of Riemann-Hurwitz? It only depends on the number of distinct ramification points, which one can read off from $g_1$ and $g_2$.
Essentially never. By Hurwitz formula, the covering corresponding to $F_i/F$ (for $i=1,2$) has 2 branch points $a_i,b_i\in \mathbb{P}^1$. The covering corresponding to $F_1F_2/F$ has degree 9. If $(a_1,b_1)\cap (a_2,b_2)=\varnothing$, it has 4 branch points in $\mathbb{P}^1$, with 3 double ramification points above each of these; this makes genus 4. If $(a_1,b_1)\cap (a_2,b_2)$ consists of one point, we get genus 1. Finally if $(a_1,b_1)= (a_2,b_2)$, we have $F_1=F_2=F_1F_2$ — it is rational but this is of course a trivial case.
Incidentally, note that your functions $g_1$ and $g_2$ have only one zero and one pole, so they are homographies.
|
2025-03-21T14:48:31.103044
| 2020-05-28T08:16:41 |
361556
|
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"authors": [
"Pierre PC",
"Ville Salo",
"YCor",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/14094"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361556"
}
|
Stack Exchange
|
Is there a prefix-continuous bijection between finite words and eventually zero words?
Let
$$ X = \{x \in \{0,1\}^{\omega} \;|\; \exists m: \forall i \geq m: x_i = 0\} $$
(one-way infinite eventually zero words). Let $\{0,1\}^*$ denote the finite (not necessarily nonempty) words over $\{0,1\}$, and write $\{0,1\}^{\leq k} = \{w \in \{0,1\}^* \;|\; |w| \leq k\}$ where $|w|$ denotes length.
Is there a bijection $\phi : X \to \{0,1\}^*$ such that
$$ \exists n \in \mathbb{N}: \forall a \in \{0,1\}: \forall x \in \{0,1\}^{\mathbb{N}}: \exists b, c \in \{0,1\}^{\leq n}: \exists y \in \{0,1\}^*: \phi(x) = b \cdot y \wedge \phi(a \cdot x) = c \cdot y $$
holds, where $\cdot$ is concatenation?
This is a kind of coarse uniformity / bornologousness assumption: $\phi$ needs to be bornologous between the two sets, seen as metric spaces with the path metric of the graph structure where $x$ and $y$ are adjacent if $y = ax$ or $x = ay$ for some $a \in \{0,1\}$. This seems vaguely familiar to me but I don't know from where, and I'm not seeing how to construct $\phi$. The straightforward idea of cutting out the zero tail doesn't work because it's not surjective, and I run into trouble trying to fix that. But I also didn't manage to prove impossibility because there's a lot of freedom.
The question arises in some (leisurely) research, so asking here instead of math.SE even if it might be safer to start there with this one. Geometric group theory tag because this is related to Thompson's $V$, even if I didn't elaborate and I doubt it's useful (every countable group acts freely on $\{0,1\}^*$).
The condition can be stated as $\phi$ being bornologous between the two sets, seen as metric spaces with the path metric of the graph structure where $x$ and $y$ are adjacent if $y = ax$ or $x = ay$ for some $a \in {0,1}$.
The condition "$\exists n$, $\forall x$ $\exists c,d,e$, $\exists y$, $\phi(x)=cy$, $\phi(0x)=dy$, $\phi(1x)=ey$" is a priori a bit stronger (as in your condition you require one $y$ for each pair in the triple ${\emptyset,0,1}$), but is there really a reason you don't ask in this simpler way?
To clarify, do we agree they are equivalent? If not, I wonder if I've made a mistake and will have to take another look at what I've written.
I wrote it the way that looked most symmetric to me, the point is a small prefix change should become a small prefix change in the image, for some modulus of bornologuity.
It might be equivalent... it's just that quantifying over pairs of a 3-element set is puzzling unless there's a serious reason to do so. It looks like there's one $y=y_{\emptyset,0}$ working for $\emptyset,0$, one $y_{\emptyset,1}$, and one $y_{0,1}$.
Is that better?
You passed from quantifying over all pairs ${\emptyset,0}$, ${\emptyset,1}$, ${0,1}$, to only the two pairs ${\emptyset,0}$, ${\emptyset,1}$.
Yes. Somehow I felt the first version was more symmetric and thus clearer, but I guess I agree with you now. Thanks!
I believe the following works, but I might be missing something.
If $x$ has at least two 1s, then $\phi(x)$ is the sequence cut just before the last 1:
$$\phi(0011101000\cdots) = 001110 $$
If $x$ has at most one 1, then $\phi(x)$ is the sequence cut before the one, with an additional zero:
$$ \phi(000\cdots) = \varnothing,\qquad \phi(001000\cdots) = 000 $$
The inverse bijection is $\psi(\varnothing)=000\cdots$, $\psi(y)=y\cdot1000\cdots$ when $y$ has at least one 1, and $\psi(y\cdot0)=y\cdot 1$ for $y$ consisting only of zeros (possibly $y=\varnothing$).
Then $\phi(a\cdot x)=a\cdot\phi(x)$ if $x$ has at least two 1s, and the case of $000\cdots$ is unimportant, up to taking a large $n$ (although $n=1$ works so far). Now if $x$ consists only of zeros,
$$\phi(0\cdot y\cdot 1000\cdots)=0 \cdot y\cdot 0=00\cdot y\qquad\text{ and }\qquad\phi(1\cdot y\cdot 1000\cdots)=1\cdot y$$
while $\phi(y\cdot1000\cdots)=y\cdot0=0\cdot y$.
So the hypotheses are satisfied with $n=1$.
Seems to work, nice! However, full disclosure, this is a sort of toy model of the actual problem I'm interested in, though I only realize that now that I see your solution. I'll perhaps make another question later with stricter requirements. (I can also think of a possibly more natural and general version of the present question, I'll see if it makes a good MO question too.)
@VilleSalo Very well, I'm glad I could help you understand your problem better.
|
2025-03-21T14:48:31.103328
| 2020-05-28T09:56:56 |
361559
|
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"LSpice",
"Mikael de la Salle",
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"leo monsaingeon"
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"url": "https://mathoverflow.net/questions/361559"
}
|
Stack Exchange
|
A square root inequality for symmetric matrices?
In this post all my matrices will be $\mathbb R^{N\times N}$ symmetric positive semi-definite (psd), but I am also interested in the Hermitian case. In particular the square root $A^{\frac 12}$ of a psd matrix $A$ is defined unambiguously via the spectral theorem.
Also, I use the conventional Frobenius scalar product and norm
$$
\newcommand\abs[1]{\lvert#1\rvert}\newcommand\pair[2]{\langle#1, #2\rangle}\pair A B:=\operatorname{Tr}(A^tB),
\qquad
\abs A^2:=\pair A A.
$$
Question: is the following inequality true
$$
\abs{A^{\frac 12}-B^{\frac 12}}^2\leq C_N \abs{A-B}\quad ???
$$
for all psd matrices $A,B$ and a positive constant $C_N$ depending on the dimension only.
For non-negative scalar number (i-e $N=1$) this amounts to asking whether $\abs{\sqrt a-\sqrt b}^2\leq C\abs{a-b}$, which of course is true due to $\abs{\sqrt a-\sqrt b}^2=\abs{\sqrt a-\sqrt b}\times \abs{\sqrt a-\sqrt b}\leq \abs{\sqrt a-\sqrt b} \times \abs{\sqrt a+\sqrt b}=\abs{a-b}$.
$\DeclareMathOperator\diag{diag}$If $A$ and $B$ commute then by simultaneous diagonalisation we can assume that $A=\diag(a_i)$ and $B=\diag(b_i)$, hence from the scalar case
$$
\abs{A^\frac 12-B^\frac 12}^2
=\sum\limits_{i=1}^N \abs{\sqrt a_i-\sqrt b_i}^2
\leq \sum\limits_{i=1}^N \abs{a_i-b_i}
\leq \sqrt N \left(\sum\limits_{i=1}^N \abs{a_i-b_i}^2\right)^\frac 12=\sqrt N \abs{A-B}.
$$
Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrassingly not even sure that the statement holds true and I cannot even get started. Since I am pretty sure that this is either blatantly false, or otherwise well-known and referenced, I would like to avoid wasting more time reinventing the wheel than I already have.
Suvrit's answer to Subadditivity of the square root for matrices and answer to Ratio sum comparison on operators seem to be related but do not quite get me where I want (unless I missed something?)
Context: this question arises for technical purposes in a problem I'm currently working on, related to the Bures distance between psd matrices, defined as
$$
d(A,B)=\min\limits_U \abs{A^\frac 12-B^\frac 12U}
$$
(the infimum runs over unitary matrices $UU^t=\operatorname{Id}$).
Is there a reason you call your matrices $U$ unitary rather than orthogonal? (They surely are unitary, but I am used to this choice of terminology only when one does not want to insist a priori on the matrices being real, as you do.)
Not really, I came across the question while reading some papers from Quantum-based optimal transportation, and I guess in this mathematical physics context the terminology "unitary" is predominant? But maybe not, and perhaps I was only unconsciously influenced by the paper I was reading at the time. I confess I was not even aware of the subtle real/complex difference in vocabulary that you pointed out, actually I htough unitary/orthogonal was rather a question of finite/ininite dimension, matrices vs. operators. Guess I was wrong!
The classical operator generalization of the scalar inequality $|\sqrt{a}-\sqrt{b}|^2 \leq |a-b|$ is the Powers-Størmer inequality, which involves two different norms : the trace norm $\|X\|_1 = \operatorname{Tr}|X|$ and the Froebenius norm $\|X\|_2 = (\operatorname{Tr}(X^* X))^{\frac 1 2}$, where $|X| = (X^* X)^{\frac 1 2}$ is the usual absolute value of matrices. It says that for all positive matrices $A,B$ (or operators on a Hilbert space),
$$ \|\sqrt{A} - \sqrt{B}\|_2^2 \leq \|A-B\|_1.$$
It implies a positive answer to your question with $C_N = \sqrt{N}$, because by Hoelder's inequality, $\|A-B\|_1 \leq \sqrt{N} \|A-B\|_2$. The constant is optimal (take $A=\operatorname{Id},B=0$).
The Powers-Størmer surprisingly does not have a wikipedia page yet, but it probably appears in most textbooks on operator algebras or matrix analysis. The original reference is R. T. Powers, E. Størmer, Free states of canonical anticommutation relations, Commun. Math. Phys. 16, 1-33 (1970).
Great, merci beaucoup Mikael! This $L^1$ version is even better than what I needed: I really was trying to control by $L^1$, but for some resaon I was convinced that $L^2$ should be used as an intermediate step. I guess I was wrong. Thank you again.
Content que ça puisse t'être utile !
Oh yes, merci encore, ça me tire une belle épine du pied! By the way, since you appear to be the expert here, and mostly out of curiosity: Have you ever heard of the Bures distance in the infinite-dimensional/operator-theoretical framework? It seems to be quite popular in information theory these days, and apparently plays a role in quantum mechanics, so I was just wondering... (Although I must admit I didn't particularly search the literature for this)
@leomonsaingeon : I do not remember having heard of the Bures distance before your question (but I am no expert in Quantum information theory). But its definition makes perfect sense for infinite dimensional spaces (and more generally for any tracial von Neumann algebra), and I guess that most properties known for finite matrices remain true in this setting.
|
2025-03-21T14:48:31.104103
| 2020-05-28T10:55:21 |
361560
|
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"Alexi",
"Emil Jeřábek",
"Federico Poloni",
"Zach Teitler",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361560"
}
|
Stack Exchange
|
Solving multilinear equations
Let $N=\{1,2,\ldots,n\}$. Suppose we are given $n$ equations, with each equation taking the form $\sum_{A\subseteq N}\left(c_A \prod_{i\in A}x_i \right) = 0$, where each $c_A$ is a real number constant. (So each equation contains at most $2^n$ terms.) An example for $n=3$ is:
$$2x_1x_2x_3 - 4x_1x_2+5x_3+2=0$$
$$7x_1x_3 - 6x_2-4=0$$
$$-x_1x_2x_3 + x_1 - 2x_2 +9 = 0$$
We want to find a solution $(x_i)_{i\in A}$ such that $0\leq x_i\leq 1$ for all $i$ (assuming we know such a solution exists).
Is there an algorithm that solves this in time bounded in $n$?
This is as hard as solving general polynomial equations.
When you say "solve", are you asking for a numerical solution, or what?
@ZachTeitler Can we possibly solve this exactly (like we can do for linear equations), or is there some reason why this is impossible?
@Alexi Exactly, no. With the trick in Vladimir Dotsenko's answer you can easily write a degree-5 polynomial equation in this form, by adding extra variables for $x^2, x^3, x^4$. Evil Abel foils our plans again.
In numerical practice, I would use something like Newton's method or homotopy continuation.
Multilinear equations are hardly easier than general equations. For instance, the multilinear equations
$$
\begin{cases}
x_0-x_1=0,\\
x_0x_1-x_2=0,\\
x_0x_2-x_3=0,\\
\ldots\\
x_0x_{n-1}-x_n=0
\end{cases}
$$
simply tell you that $x_k=x_0^k$ for all $k=1,\ldots,n$. Using this, it is very easy to replace any system of polynomial equations by a system of multilinear ones, so I assume that the most standard method (Gröbner bases) would be the main tool to use.
|
2025-03-21T14:48:31.104246
| 2020-05-28T11:56:14 |
361562
|
{
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"Alessandro Andretta",
"Emil Jeřábek",
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|
Stack Exchange
|
An analogue to Robinson's theorem for Kalmar-elementary functions
Julia Robinson proved that the family of all unary computable total functions is the smallest class containing $S$, $\mathrm{Exc}$, and closed under composition, addition and inversion of surjective functions. (Here $\mathrm{Exc}(x)=x-\lfloor\sqrt x \rfloor^2$, and the inversion of a surjection $f \colon \mathbb{N} \to \mathbb{N}$ is $f^{-1}(m) ={}$the least $n$ such that $f(n)=m$.)
Raphael Robinson proved a similar result for the family of unary primitive recursive functions: it is the smallest class containing $S$, $\mathrm{Exc}$, and closed under composition, addition and iterations. These results are proved in Monk's book Mathematical Logic.
Is there a similar result for the family of unary elementary functions?
The question is too vague. What is a “similar result”? In particular, elementary functions are defined as the smallest class that contains some basic functions and is closed under composition, bounded summation, and bounded products. Why is that not enough for you?
I am not familiar with the proof of the results that you mention. However, giving this two results, I think that I have a natural conjecture of what an analogue for elementary functions could be. Namely my conjecture would be that the class of all unary elementary functions is the least class containing, $S,\mathrm{Exc},\mathsf{exp}$ and closed under compositions, additions, and bounded inversions. Here $\mathsf{exp}(x)=2^x$ and $g$-bounded inversion of $f$ is $x\longmapsto \min({g(x)}\cup{y\mid f(y)=x})$.
Let me explain what I mean by "similar result". The theorems I mentioned provide a definition of the collections of unary computable/primitive recursive functions. I was wondering if anyone is aware of a similar characterization for the unary elementary functions. The suggested conjecture is indeed natural.
|
2025-03-21T14:48:31.104386
| 2020-05-28T12:15:20 |
361564
|
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}
|
Stack Exchange
|
Can we extend a function from the diagonal matrices to an orthogonally-invariant function on $\text{GL}_n$?
This is a cross-post.
Let $g:(0,\infty)^n \to [0,\infty)$ be a symmetric function -i.e. $g(\sigma_1,\dots,\sigma_n)$ does not depend on the order of the $\sigma_i$, with $g(1,\dots,1)=0$.
We identify $(0,\infty)^n$ with the space $\text{PDiag}$ of real positive-definite diagonal $n \times n$ matrices.
Question: Does there exist a function $f:\text{PDiag} \times \text{GL}_n^+ \to\mathbb R^{\ge 0}$
satisfying
$$ \,\,\,\,f(\tilde \Sigma,U\Sigma V)= f(\Sigma,U^{-1}\tilde \Sigma V^{-1}) \, \, \text{ for every } \, U,V \in \text{SO}_n \, \, \text{and } \Sigma, \tilde \Sigma \in \text{PDiag},
\text{ and }\,\, \, \, f(\Sigma,\text{Id})=g(\Sigma)$$
Also, If we assume that $g$ is continuous, can we create a continuous extension $f$?
(Here $\text{GL}_n^+ $ refers to the group of real invertible $n \times n$ matrices with positive determinant).
If that matters, I am ready to replace $\text{SO}_n $ with $\text{O}_n $ and assume "bi-$\text{O}_n$-invariance" in the first requirement instead of "bi-$\text{SO}_n$-invariance".
It seems to me that only trivial functions can satisfy your invariance property. Or perhaps you made a typo wrt $\Sigma$ and $\widetilde{\Sigma}.$
Hi, I am not sure why you think that. My "inspiration" comes from bi-SO(n)-invariant distances; You can take $f$ to be the restriction of such a distance on $GL \times GL$, and $g(\Sigma)$ to be the distance of $\Sigma$ from the identity matrix. Or am I wrong somewhere?
If $g$ factors through $\det$ then $f(a,b)$ can be $g(a)+g(b)$, $g(a)g(b)$, $|g(a)-g(b)|$.... I'm not sure this is the only $g$ for which you can find $f$ but proving it looks like a dive into linear algebra because you need to understand eigenvalues of $U\Sigma V$ and $U^{-1}\tilde \Sigma V^{-1}$ for all $U, V$. You could use some CAS to see if there is any algebraic relation in low dimensions ($2 \le n \le 5$).
|
2025-03-21T14:48:31.104544
| 2020-05-28T12:40:59 |
361567
|
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}
|
Stack Exchange
|
A binomial coefficient identity involving two parameters
In a recent calculation I obtain a result involving the following expression depending on two integers $n,m\geq 0$:
$$S(n,m):=\frac{(n+m+1)!}{n!m!}\sum_{l=0}^{n+m}\frac{1}{n+m-l+1}\sum_{\substack{j+k=l\\ 0\leq j\leq n\\0\leq k \leq m}}(-1)^{j}{n \choose j}{m \choose k}.$$
Numerical evidence suggests that one has
$$
S(n,m)=\begin{cases}
\sum_{k=0}^{m/2}{n+m+2 \choose 2k}, \quad & m\in \{0,2,4,\ldots\},\\
-\sum_{k=0}^{(m+1)/2-1}{n+m+2 \choose 2k+1}, & m\in \{1,3,5,\ldots\}.\end{cases}
$$
Besides being much simpler, this formula has the great advantage that it makes obvious that $S(n,m)$ is an integer and non-zero.
How would I go about proving the claimed identity?
I guess it must be quite simple compared to many other binomial coefficient identities discussed here on MO. My hope is that experts in this business immediately see an "obvious" simplification step that I am not aware of.
The $L^+/L^-$, $\ell^+/\ell^-$ notation is very hard to parse (at least for me). I think it would be clearer to just use several different variables.
@SamHopkins: Thank you, I just renamed the variables.
Taking into account that $\binom pq=0$ for nonnegative integers $p$ and $q$ such that $q>p$, write
$$S(n,m)=\frac{(n+m+1)!}{n!m!}T(n,m),$$
where
\begin{align*}
T(n,m)&:=\sum_{l\ge0}\frac{1}{n+m-l+1}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\
&=\sum_{l\ge0}\int_0^1 dx\,x^{n+m-l}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\
&=\int_0^1 dx\,x^{n+m}\sum_{l\ge0}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk x^{-j} x^{-k} \\
&=\int_0^1 dx\,x^{n+m}\sum_{j\ge0}\binom nj(-x^{-1})^j\;\sum_{k\ge0}\binom mk x^{-k} \\
&=\int_0^1 dx\,x^{n+m}(1-x^{-1})^n(1+x^{-1})^m\ \\
&=\int_0^1 dx\,(x-1)^n(1+x)^m\ \\
&=\int_0^1 dx\,(x-1)^n\,\sum_{k=0}^m\binom mk x^k\ \\
&=(-1)^n\sum_{k=0}^m\binom mk\int_0^1 dx\,(1-x)^n x^k\ \\
&=(-1)^n\sum_{k=0}^m\frac{m!}{k!(m-k)!}\frac{k!n!}{(k+n+1)!}\ \\
&=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{k=0}^m\binom{m+n+1}{m-k}\ \\
&=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{j=0}^m\binom{m+n+1}j.
\end{align*}
So,
\begin{equation}
S(n,m)=(-1)^n\sum_{j=0}^m\binom{m+n+1}j,
\end{equation}
which is simpler than your desired expression (which seems to differ from the latter expression for $S(n,m)$ only by the sign).
Thank you very much! I was blocked by the complication that one has to take into account the conditions $j\leq n$ and $k\leq m$ in the sum... of course it makes life easier to get rid of them as you did.
@BK : I am glad this helped.
|
2025-03-21T14:48:31.104713
| 2020-05-28T12:43:23 |
361568
|
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"David White",
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"url": "https://mathoverflow.net/questions/361568"
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|
Stack Exchange
|
The correct homotopically relevant notion of ideals of dg-algebras (or $\mathbb E_1$-rings)
I'm trying to figure out what an ideal of a, say, dg-algebra (or, if you prefer, $\mathbb E_1$-ring) $R$ is in a homotopically relevant fashion, but I can't actually figure it out. I can assume that $R$ is concentrated in cohomologically nonpositive degrees (or homologically nonnegative degrees). I have stumbled upon a few possibilities:
There is a notion of monomorphism in an $\infty$-category; hence, I would consider the derived category $\mathsf D(R)$ of $R$-dg-modules as an $\infty$-category and say that $I \to R$ is an "ideal" if it is a monomorphism according to that notion. This notion is also used in Spectral algebraic geometry (Remark C.2.3.4. page 1965) in the framework of Grothendieck prestable $\infty$-categories.
On the other hand, I find other sources such as this and this. From the first one, I quote:
There are other things that are weird for commutative ring spectra. Quite often, we end up working with ideals in the graded commutative ring of homotopy groups, but as we saw above, this is not a suitable notion of ideal.There is a notion of an ideal in the context of (commutative) ring spectra [53] due to Jeff Smith, but still several algebraic constructions do not have an analogue in spectra.
Given that, I'm pretty confused. Perhaps the notion of monomorphism (1) is fine, but in the case of commutative ring spectra it does not work really well, hence the issues I found (2)? I've tried to skim through some literature on derived algebraic geometry, but still I couldn't find any satisfying answer...
Homotopy fibre of an $H_0$-surjective morphism of algebras is probably the best you'll manage, and that only works if the algebras are connective. In the derived setting, ideals have much more structure than modules - see for instance Ciocan-Fontanine and Kapranov's papers on derived Hilb and derived Quot.
I characteristic zero, I have suggested a definition of "ideal" in my https://arxiv.org/abs/1608.08598. Namely, I suggest that the definition of "idea" should be "homotopy fibre of a homomorphism", and the paper gives a universal description of the algebraic structure carried by such "ideals" for homomorphisms of any (di)operad. I should warn that that paper needs some rewriting, and my interests have shifted a bit, so I'm not sure when I will do it. I also have not done a good enough job comparing my proposals with the rest of the literature.
In (2), you linked to Mark Hovey's paper on Smith ideals, and mentioned "the commutative framework." But Hovey explicitly writes "we have not dealt with the commutative situation at all," so I don't know what you mean. However, if you do want a theory of commutative Smith ideals, you can find this in my first paper. Also, if you want a theory of Smith $O$-algebras, for an operad $O$ (e.g., $O = E_n$), then you can find this in a paper of mine with Donald Yau.
One of the crucial aspects of the story is that the algebraic structure on the morphism $f: I\to R$, viewed as an object in the arrow category, matches the algebraic structure on the cofiber of $f$ (at least, in stable settings where taking the cofiber makes sense). This is proven in section 4 of Hovey's paper, and in Theorem 4.4.1 of my paper with Donald Yau (with lots of examples occupying the rest of the paper). I'd like to write more, but have to run off to a Zoom meeting now for the rest of the day. Hopefully this observation gets you started. It's an important justification for the approach of both (1) and (2).
This is even more confusing, namely, here: https://www.math.uni-hamburg.de/home/richter/commutativeringspectra.pdf I quote "There is a notion of an ideal in the context of (commutative) ring spectra [53] due to Jeff Smith, but still several algebraic constructions donot have an analogue in spectra." That's why I thought it was a notion meant for the "commutative framework". Also, I don't see how the approaches of (1) and (2) relate.
Probably Birgit had in mind that I'd already sorted out the commutative case, and I was Hovey's student (graduating in 2014). So, she might have just kind of given him credit there for the part of my first semester that was about commutative smith ideals. Not a big deal. Sorry if that caused confusion.
My previous comment should have said "the part of my first paper" - sorry, this is what I get for typing on mathoverflow during a Zoom meeting about the fall semester.
At least in commutative situations, I would argue that a good notion is simply an ideal in $H^0(R)$.
For example, the theory of local cohomology works just as well as it does for commutative rings, as long as you do it with respect to ideals in $H^0(R)$.
Similarly, you can take "derived quotients" with respect to a finite sequence of elements in $H^0(R)$, by taking the Koszul complex with respect to such a sequence.
See for example my very recent paper from last week:
"Koszul complexes over Cohen-Macaulay rings"
https://arxiv.org/abs/2005.10764
Welcome to MathOverflow! I tend to disagree with your answer for the case of spectra. The philosophy of Jeff Smith and Mark Hovey, and the applications to K-theory Smith envisioned, really need some kind of object at the point-set level, not just in the homotopy category. That said, Proposition 1.7 in Hovey's paper unwinds the definition of a Smith ideal into the language of R-bimodules, and might help bridge between your answer and mine.
@DavidWhite, thanks! my answer was about DG-algebras. I have no experience with spectra.
|
2025-03-21T14:48:31.105078
| 2020-05-28T13:20:21 |
361570
|
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|
Stack Exchange
|
Physics applications of quantum logic
Are there any examples of quantum logic being applied to solve actual physical questions, in particular to predict the physical properties (spectrum etc.) of some quantum-mechanical system? (Note that I'm not considering applications in quantum computing here.)
EDIT: To clarify what I mean, consider a simple system like the harmonic oscillator. Using real variables $q_t$ and $p_t$ indexed by real-valued time $t$, and starting from energy conservation in a form like $\forall t\forall s p_s\cdot p_s+\omega\cdot\omega\cdot q_s\cdot q_s=p_t\cdot p_t+\omega\cdot\omega\cdot q_t\cdot q_t$ and from the relation $\dot{q}=p$ expressed using a form "$\forall\epsilon\ldots<\epsilon$", I have no doubt that most of the properties of the classical system could be laboriously derived using the axioms for the real numbers and classical logic. (Of course if I'm wrong and there is some obstacle that means that solving even such a simple dynamical system cannot be logically formalized, that would be an interesting answer, too!) Has anyone ever tried to derive the properties of the corresponding quantum system using some appropriate form of quantum logic?
What do you mean by "quantum logic"? If you just mean the rules of quantum mechanics, you better use those if you want to correctly treat any quantum mechanical system.
quantum computing is not a part of physics? it exploits the laws of physics, doesn't it?
@CarloBeenakker: So does classical computing ... but you're right, I could have phrased that better.
@MichaelEngelhardt: I always took "quantum logic" to be a well-defined term for propositional logics built on non-distributive lattices.
Yes, and no. Until you try to solve some of the problems and find out for yourself, you don't know the answer. :P
@MattF., fair point, but quantum logic takes its inspiration from quantum physics, so some special applicability would not be a particular surprise.
FYI: https://en.wikipedia.org/wiki/Quantum_logic
|
2025-03-21T14:48:31.105230
| 2020-05-28T13:39:53 |
361572
|
{
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"Julien Marché",
"Moishe Kohan",
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"url": "https://mathoverflow.net/questions/361572"
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|
Stack Exchange
|
Find a 4-manifold bounding a 3-torus with any abelian representation in SL_2
Fix $x,y,z\in \mathbb{C}^*$ and let $M=S^1\times S^1\times S^1$ with $\rho:\pi_1(M)\to \operatorname{SL}_2(\mathbb{C})$ mapping the three generators to diagonal matrices with entries $(x,x^{-1})$, $(y,y^{-1})$, $(z,z^{-1})$ respectively.
The question is can you construct a (smooth) 4-manifold $W$ bounding $M$ with a representation $\tilde{\rho}:\pi_1(W)\to \operatorname{SL}_2(\mathbb{C})$ extending $\rho$?
Comments:
1) Such a 4-manifold exists by a non-constructive argument based on the fact that the map $H_3(\mathbb{C}^*,\mathbb{Z})\to H_3(\operatorname{SL}_2(\mathbb{C}),\mathbb{Z})$ induced by the inclusion of diagonal matrices in $\operatorname{SL}_2(\mathbb{C})$ is almost 0.
2) I have a homological argument saying that the extension $\tilde{\rho}$ cannot be abelian.
3) By standard arguments, the same manifold should work for $(x,y,z)$ in a Zariski open subset of $\mathbb{C}^*\times\mathbb{C}^*\times \mathbb{C}^*$.
4) The general motivation is to illustrate topologically the low-dimensional homology of linear groups...
While you surely know this: A similar result fails in one dimension lower, see here.
Thanks for your comment. Although I indeed knew it (I wrote a paper on the topic with L. Liechti see https://arxiv.org/abs/1903.11418), I was not aware of the discussion in MO.
|
2025-03-21T14:48:31.105343
| 2020-05-28T13:54:19 |
361573
|
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"Gerhard Paseman",
"Joseph O'Rourke",
"alesia",
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"https://mathoverflow.net/users/6094"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/361573"
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|
Stack Exchange
|
Hanging a cube with string
This is a variation on a (much) earlier MO question, Hanging a ball with string.
Here instead the task is to arrange a net of string to hang
a unit cube. Assume:
The string is inelastic.
There is no friction between the string and the surface.
There is an attachment point for attaching a vertical hanging string.
The cube cannot "slip out" under small jostlings.
The net has minimal total length among all such nets.
The aspect that I do not know how to formalize is slipping-out.
Intuitively I want it to be robust under nudging.
Formally there should be a requirement that the cube live in some
positive radius ball in a configuration space such that within that ball,
the cube remains confined by the net.
String length: $2+5\sqrt{2} \approx 9.07$
(excluding yellow piece).
I offer two nets as examples to improve. In the first above, the blue rectangle corners are at edge midpoints, and
the four red paths to the attachment point are geodesics.
This net seems to satisfy the no-slipping-out condition,
and I have verified this is optimal among nets that slide
the blue rectangle corners in concert up or down those cube edges.
But many aspects are unclear. In particular:
Q1. Remove one red path, leaving the other three. Can the cube now
slip out? Note the blue paths are not geodesics, in that they do not
unfold straight around a corner.
There are six $\sqrt{5}$-geodesics between opposite cube corners $v_1$ and $v_7$. Choose four, perhaps those shown below; note that $4 \sqrt{5} \approx 8.94$
is slightly shorter than the first net above.
Q2. Does it avoid slipping-out? If so, can the net be reduced to three geodesics
and still avoid slipping-out?
String length: $4 \sqrt{5} \approx 8.94$
(excluding yellow piece).
Q3. What is a more formal definition of the no-slipping-out condition that captures the intuition, and allows settling some
of the questions above?
Of course the main question is: What is the minimum length net?
Is there an example showing that the condition is not equivalent to saying that the graph is a locally unique length minimizer?
@alesia: Nice question! I don't know; I'll think about it.
(Or more accurately, any locally unique minimizer satisfies the condition, and the best solution is a locally unique minimizer. Perhaps.)
Not minimal, but a cuboctahedron (diamond on each cube face) gives an Euler graph that is realized with a single string loop. Don't know how stable it is. Gerhard "Doesn't Know A out Self Stability" Paseman, 2020.05.28.
|
2025-03-21T14:48:31.105552
| 2020-05-28T14:41:21 |
361580
|
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|
Stack Exchange
|
Some sort of twisted group homomorphism
Start with a group $G$ that acts on a set $X$, and a second group $H$. We want to consider functions $\varphi: G \times X \to H$ such that $\varphi(g g', x) = \varphi(g, g'x) \varphi(g', x)$ for all $g$, $g'$, and $x$. Note that if $X$ is singleton (or if $G$ acts trivially), then $\varphi$ is essentially just an ordinary group homomorphism. Is there a name for a function like this?
If I'm correct it's widely called "cocycle", maybe have a look for instance at https://arxiv.org/abs/math/0303236 (Fisher-Margulis). The keyword "cocycle rigidity" often points to this meaning of cocycle.
Specifically, to fit in the usual set-up for these things, I'd call it a co-cycle for $G$ with values in $\operatorname{Func}(X, H)$, where the latter is viewed as a $G$-set in the obvious way. (More generally, whenever you find yourself saying "twisted …", cohomology is probably a good place to look to describe your objects.)
If H = Aut(Y) then your phi can also be described as an action G -> Aut(X x Y) which is compatible with the action G -> Aut(X) relative to the projection X x Y -> X.
Am I wrong to assume that this use of the word "cocycle" in a different from the sense of group cohomology cocycle? Such a cocycle can be described as a group homomorphism G -> G |x M (a section in fact), which looks a little similar to G x X -> H but not very.
(I'm asking because I've also heard the word "twisted homomorphism" used in the context of group cohomology cocycles - specifically here: https://www.kurims.kyoto-u.ac.jp/~motizuki/The%20Etale%20Theta%20Function%20and%20its%20Frobenioid-theoretic%20Manifestations.pdf .)
|
2025-03-21T14:48:31.105683
| 2020-05-28T15:22:32 |
361581
|
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"Alex M.",
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|
Stack Exchange
|
Wave front set and differential operator
A simple result of microlocal analysis would be, if P is a differential operator then $WF(Pu) \subseteq WF(u)$, where WF denotes the wave front set of the distribution.
Can anyone give me an example where the inclusion is strict.
Thanks in advance.
Take $P = 0$ and $u$ anything with a non-empty wavefront set?
If you don't like $P =0$: work on $\mathbb{R}^2$ and let $P = \partial_y$ and $u = H(x)$ where $H$ denotes the Heaviside function.
@hsm: If you have equality, then $P$ is hypoelliptic, by definition. In order to have strict inclusion, then, it is enough to choose $P$ not hypoelliptic; the example that immediately comes to mind is the wave operator $P = \partial_t^2 - \partial_x^2$.
|
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