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most recent 30 from 2017-12-12T17:39:16Z

Why is the catalecticant invariant under coordinate changes?

Fri, 09/15/2017 - 05:49

Let $\mathbf{k}$ be a commutative $\mathbb{Q}$-algebra. (We could play the same game over any commutative ring $\mathbf{k}$, but this would be a bit more technical, so let me avoid it.)

Fix a nonnegative integer $n$. A binary form shall mean a homogeneous polynomial $f=f\left( x,y\right) \in\mathbf{k}\left[ x,y\right] $. Any binary form $f\left( x,y\right) $ of degree $2n$ can be written uniquely in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n}{i}a_{i} x^{2n-i}y^{i}$ for some scalars $a_{0},a_{1},\ldots,a_{2n}\in\mathbf{k}$ (which are simply the coefficients of $f$, rescaled by binomial coefficients). Let $\mathcal{F}_{2n}$ denote the $\mathbf{k}$-module of all binary forms of degree $2n$. The matrix ring $\mathbf{k}^{2\times2}$ acts on the $\mathbf{k} $-module $\mathcal{F}_{2n}$ by the rule \begin{equation} \left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \cdot f=f\left( ax+cy,bx+dy\right) . \end{equation} (In other words, a matrix $\left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \in\mathbf{k}^{2\times2}$ acts on a binary form by substituting $ax+cy$ and $bx+dy$ for the variables $x$ and $y$. This is called a "change of variables", at least if the matrix is invertible.)

The catalecticant $\operatorname*{Cat}f$ of a binary form $f\left( x,y\right) $ of degree $2n$ is a scalar, defined as follows: Write $f\left( x,y\right) $ in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n} {i}a_{i}x^{2n-i}y^{i}$, then set \begin{equation} \operatorname*{Cat}f=\det\left( \begin{array}[c]{cccc} a_{0} & a_{1} & \cdots & a_{n}\\ a_{1} & a_{2} & \cdots & a_{n+1}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n} & a_{n+1} & \cdots & a_{2n} \end{array} \right) =\det\left( \left( a_{i+j-2}\right) _{1\leq i\leq n+1,\ 1\leq j\leq n+1}\right) . \end{equation} A classical result, going back to Sylvester (On the Principles of the Calculus of Forms) in 1852 (who, in turn, ascribes it to Cayley), then says that $\operatorname*{Cat}f$ is a $\operatorname*{GL} \nolimits_{2}\mathbf{k}$-invariant (in an appropriate sense of this word, i.e., $\operatorname*{GL}\nolimits_{2}\mathbf{k}$ transforms $\operatorname*{Cat}$ by multiplication with an appropriate power of the determinant). More precisely, any binary form $f\left( x,y\right) $ of degree $2n$ and every matrix $A\in\mathbf{k}^{2\times2}$ satisfy \begin{equation} \operatorname*{Cat}\left( A\cdot f\right) =\left( \det A\right) ^{n\left( n+1\right) }\cdot\operatorname*{Cat}f. \label{catalecticant1} \tag{1} \end{equation}

Question. Is there a purely algebraic/combinatorial proof of \eqref{catalecticant1}?

These days, \eqref{catalecticant1} is usually seen as a consequence of the fact that if $\mathbf{k}$ is an algebraically closed field of characteristic $0$, then a binary form $f\left( x,y\right) $ of degree $2n$ satisfies $\operatorname*{Cat}f=0$ if and only if $f$ can be written as $\sum_{j=1} ^{n}\left( p_{j}x+q_{j}y\right) ^{2n}$ for some $2n$ elements $p_{1} ,p_{2},\ldots,p_{n},q_{1},q_{2},\ldots,q_{n}\in\mathbf{k}$. This is proven, e.g., in J. M. Selig, Sylvester's Catalecticant. Once this fact is granted, the equality \eqref{catalecticant1} follows using some technical Nullstellensatz arguments (I believe so; the details are somewhat annoying to verify, requiring e.g. to show that $\operatorname*{Cat}f$ is irreducible as a polynomial in the $a_{0},a_{1},\ldots,a_{2n}$ for $n>0$).

I dislike this argument for the technicalities involved, including the passage to an algebraically closed field and the use of the Nullstellensatz (at least for principal ideals). It seems to me that \eqref{catalecticant1} should have a purely algebraic proof, using properties of determinants. Sylvester might have had one, but I find his paper impossible to decipher. Does anyone know such a proof?

On an observation which relates to the exponential sum $\sum_{n=1}^{[\sqrt{t/2\pi}]} n^{-\frac{1}{2}+it}$

Fri, 09/15/2017 - 03:22

This observation is based on the numerical calculation of the exponential sum: $$\sum_{n=1}^{[\sqrt{t/2\pi}]} n^{-\frac{1}{2}+it}$$ It is known that this sum is related to the famous Riemann–Siegel formula.

Now if we denote this sum as $S_1$, and define sum $S_2$ as follows: $$\sum_{n=1}^{[\sqrt[4]{t/2\pi}]} n^{-\frac{1}{2}+it}$$

Numerical calculation shows that there exists a strong similarity between $|S_1|$ and $|S_2|$.

Here is a plot of $|S_1|$ (blue) and $|S_2|$ (green), with $t$ varies in $[10^4-1000,10^4]$

and another with $t$ varies in $[10^6-1000,10^6]$, more calculations have affirmed this phenomenon.

So there goes the question: Does there exist a good mathematical explanation of this phenomenon?

Is there always an G-invariant uniform measure preserving map for countable G?

Thu, 09/14/2017 - 18:00

Suppose $\circ:G \times X \to X$ with $X$ countable. There is a measure preserving dynamical system $([0,1]^X, \lambda^X, G)$ where $\lambda$ is the uniform measure on $[0,1]$ and for $(a_x)_{x \in X} \in [0,1]^X$, $g$ takes $(a_x)_{x \in X}$ to $(a_{\circ(g,x)})_{x \in X}$.

If $X$ has a finite orbit $Y$ then there is an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$. For example

$$(a_x)_{x \in X} \mapsto \sum_{y \in Y} a_y \mod 1.$$

If $X$ has no finite orbits will there still always be an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$?

As a concrete example, is there such an invariant measure preserving map when $X = (\mathbb{Z}, S)$ and $G = $Aut$(\mathbb{Z}, S)$, where $S$ is the successor relation (and there is no other structure on $\mathbb{Z}$)?

Almost certain extinction for a Markov Jump Process

Thu, 09/14/2017 - 09:14

I'm studying a simplification of a biological neuron model with $n$ neurons. We are describing the evolution of the membrane potential of each neuron. Let $(X_t)_{t\geq 0}$ be a Markov Jump Process in $\mathbb{N}_0^n$ with the following generator $Q$:

\begin{equation} \begin{cases} (i)\ \ \ \ q(x,(x)_i^c) = b(x_i) \\ \\ (ii)\ \ \ q(x,x+e_i)=\lambda x_i\ \ \ \ i=1,...,n \\ \\ (iii)\ \ q(x,x-e_i) =\mu x_i \end{cases} \end{equation}

Where $(x)_i^c: = x+c(1,...,1)-(x_i+c)e_i\ \ \ (c\in \mathbb{N})$; $\ \ \mu > \lambda\ $; and $b:\mathbb{R}_{\geqslant 0}\to \mathbb{R}_{\geqslant 0}$ is a non-negative, non-decreasing function with $b(0)=0$.

Looking carefully at $(x)_i^c$, you can see that it represents the discharge of neuron $i$. It goes to zero, whereas the other neurons recieve a fixed potential $c>0$.

The goal is to prove that this process has almost certain extinction. That is, we need to prove that

$$ \mathbb{P}(T_0<+\infty) =1 $$

Where $T_0:=\inf\{t>0:X_t=(0,...,0)\}$

Here are some observations and an idea of what may be a path towards a proof:

For a start, see that if we ignore condition $(i)$, then $(ii)$ and $(iii)$ represent in each coordinate an independent Birth-and-Death process in $\mathbb{N}_0$ with almost certain extinction, so if we ignored $(i)$ then we would have almost certain extinction for $(X_t)$.

I expect that the perturbation introduced by $(i)$ does not affect so much: What $(i)$ says is that when a coordinate is too "high", then it increasingly tends to "discharge". The total potential $||X_t||_1$ is increased by $c(n-1)$ and decreased by $x_i(t)$, so it only increases if the potential $x_i(t)$ was below the threshold $c(n-1)$. A discharge of a neuron with such a potential occurs with rate at most $b(c(n-1))$.

The idea I have for a proof is to define some neighborhood around $(0,...,0)$ (say $\mathcal{C}:=\{ ||X||_1 \leqslant k\}$ for some $k>0$) where I can have positive probability of extinction $p$, independently of the starting point. And then I'd like to prove that if you are out of $\mathcal{C}$, then you come back with probability 1.

Finally I would define a Geometric distribution $p$ associated, thinking as the experiment " whether or not $(X_t)$ starting inside $\mathcal{C}$ reaches extinction".

Are my ideas correct?

What about the idea of a proof?

If correct, are there some details I need to be careful with?

Has anything (other than what is in the obituary witten by M. Noether) survived of Paul Gordan's defense of infinitesimals?

Thu, 09/14/2017 - 09:00

Question. Has anything other than what can be guessed from this obituary written by Max Noether survived of the 'defense' of infinitesimals that Paul Gordan gave in his doctoral disputation on March 1, 1862 in Berlin?


  • This is an interesting historical morsel that I accidentally stumbled over and that Mikhail Katz suggested to be made an MO question.

  • It should be noted that the philosophical substance of this 'discovery' is not surprising at all, rather expected and boring in fact (the only interesting result of this question would be a mathematical argument): that Noether says that Gordan thought that infinitesimals are good and that Gordan defended 'implicit solutions better than explicit solutions' only because he had to, is squarely in line with the typology found in the usual 'narrative', i.e.,

Gordan : $\mathsf{explicit}$ , Hilbert : $\mathsf{implicit}$

  • Someone having less reservations against such titles than the writer of these lines might have entitled the question "The King of Invariants and his Defense of the Infinitesimals: a lost Latin swan song?", or something like that.

  • For readers' convenience, here are relevant passages from the obituary. Source is Mathematische Annalen 75, 1914:

My translation:

In Breslau Gordan got the topic of his dissertation; it arose from an article entitled "De linea geodetica", with which in August 1861 Gordan won a competition started by a prize question asked by the faculty. This doctoral dissertation that Gordan defended on March 1, 1862 in Berlin treated the geodetic line on the spheroid of the earth. Of the questions which Gordan defended at this disputation---in a Latin, of course, which was deemed not exactly classical by Kronecker---the following two may be mentioned; the first, because it is in line with Gordan's views both then and later in life; the second, for the opposite reason:

"The method of the Infinitely-Small is, I claim, no less precise than the method of limits."

"It is of greater interest to investigate the implicit properties of a function defined by a differential equation than to investigate in terms of which known functions it can be expressed."

Complex analysis attracted Gordan in autumn 1862 to go to Riemann [Riemann was employed at the university of Göttingen at the time]. He [...]

  • Max Noether writes as if he knew more about Gordan's thoughts on infinitesimals than what he tells readers in the obiturary.

  • Noether insinuates that Kronecker was present at the disputation.

  • I don't know what the regulations for disputations were in the 1860s at University of Berlin. (This is the 'ancestor' of what is by now one of at least three universities in Berlin.) If notes were taken, then they may have survived, though I think this unlikely. (And even if they have survived: anyone who knows what minutes of exams usually look like will not expect there to be much information in them. The best one could hope for is that Gordan later himself wrote carefully about infinitesimals.)

  • One could even hope that this question has (some sort of) mathematical answer; 155 years are not much by historians' standards. Of course, one should not expect there to be much in exam minutes, even conditioning on the, by itself, unlikely event that such were taken and have survived.

Update. I now took the time to read the whole of the obituary written by Noether ('op. cit.'). I here summarize all that seems not irrelevant to the present question (to cut a long story short: nothing strictly relevant to the question will follow; the last item will border on the off-topic; Noether is basically saying that Gordan did not use limits simply because he didn't like them, or just didn't care, and avoided them, not because he had a thought-out positive theory about infinitesimals; it seems more of a rather boring avoidance of second-order definitions and a preference for first-order definitions, than any positive theory about infinitesimals, making the chances of any appreciable answer to this question seem even slimmer than they already seem just from thinking about the question; I don't expect there to ever be much of an answer):

  • The most relevant item in this 'update' is the following passage in op. cit., in which Noether describes Gordan as a teacher:

My attempt at a translation:

In his own field [Noether has written in the immediately preceding paragraph that Gordan liked to read German literature] it was not so much the perusal of the work of others than a 'big picture' of the inner connections and an instinctive feeling for the paths and goals of mathematical endeavours which enabled him to distinguish the valuable from the inferior, and small hints were sufficient for this. However, Gordan never did justice to foundational conceptual developments: also, in his lectures he entirely avoided any basic definitions of a conceptual kind, even the definition of 'limit'. [emphasis added] His lectures extended only to mathematics of the common kind, occasionally also to the theory of binary forms; the exercises were preferably taken from algebra. He liked to lecture on the work of Jacobi [Gordan's doctoral advisor], e.g. on functional determinants, yet never about complex analysis, higher geometry or mechanics1; he also did not have anyone give seminar presentations. The lectures [of Gordan], essentially, had an effect rather due to his vivacious way of speaking, and due to a vigour which encouraged further independent study, than due to systematizing and rigor.

  • Noether on the first page of op. cit. names his 'informants', which explains the pretense Noether's of being informed about so much in Gordan's life:


Part of the relevant [i.e.: relevant for the biographical parts of the obituary] data was taken from [Gordan's] dissertation (I of the bibliography at the end). For other bits of information I am indebted to R. Sturm, a fellow student of Gordan at Breslau, C.F. Geiser, one of the opponents at the Berlin disputation in 1862, J. Thomae, a fellow student at Göttingen, and A. Brill, a colleague at Gießen; furthermore, I am indebted to L. Schlesinger for information from files at Gießen.

  • Noether also describes what he thinks is the only non-algebraic work Gordan ever did after his dissertation:2

My translation:

Only once did Gordan ever work on a non-algebraic topic: in 1893 Hilbert had, based on a new idea, given a simple proof of the transcendence of the numbers $e$ and $\pi$; and immediately afterwards Hurwitz had given a modification of the proof, by way of avoiding the integrals used by Hilbert, essentially by using integration-by-parts, and by the use of the simplest case of the intermediate-value-theorem. In an article (54), which by the way only had the aim of giving an exposition of Hurwitz's deduction, 'played backwards', Gordan took one more step in towards an elementary proof, by replacing also the still remaining differential quotients by use of the Taylor series of $e^x$. [emphasis added; by the way: what is nowadays the 'most' elementary proof that $\pi$ and $e$ are transcendental; the only proof I ever studied in detail used integrals, and, as such, limits] Regarding the $r!=h^r$ symbolism, which has been emphasized by various others after Gordan's paper, it must however be said that it does not play any further role in the proof than being a mere abbreviating notation, and that it moreover has simply been lifted from Hurwitz's note. The estimation of the vanishing remainder term, an estimate which is essential for the proof and was left out of the note (54), has been carried out by Gordan in 1900 in an unpublished notebook in Gordan's Nachlass. 3

  • The 'MathSci-Net' and 'Zentralblatt' of the 19th century was the 'Jahrbuch über die Fortschritte der Mathematik'; therein, Heinrich Burkhardt reviewed Gordan's 'trancendent note'; for convenience, since it is not entirely off-topic (it is 'Gordan on the non-algebraic'), I reproduce the review here (it is also something of a self-contained summary of Hilbert's proof, and as such, maybe of interest in and of itself):

The blue link in the yellow lead to a review which is (displayed to me as) empty.

1To my way of thinking, this is all rather expected and unsurprising and in line with the usual reputation of Gordan: all the subjects Noether says Gordan avoided are second-order theories; in particular, in the mechanics of the time variational principles had an important role, and variational principles involve quantifier over sets of functions, i.e., second-order quantifiers; 19th century mechanics/variational principles/principles of least action, etc. to me seem recognizably non-algebraic. Also, one could now very much overinterpret and say something about whether and how much infinitesimals help in complex analysis, and that there, perhaps, limits are more necessary than in real analysis, and Gordan avoided complex analysis because of this. This, however, is evidently an overinterpretation.

2 And to me, it seems evident that if there is any hope of a mathematical answer, then this hope comes from the possibility that Gordan might later have written about infinitesimals/used them for something in published printed work. However, unfortunately, I think this is unlikely.

3 It might be an interesting separate historical problem to locate this "unpublished" notebook. Please note, though, that I don't think this relevant for the question: what Noether describes sounds like a rather ordinary estimate of a remainder term, presumably by inequalities; it does not sound like a use that Gordan made of infinitesimals, which don't yield effective estimates anyway, as far as I know. One should be clear that there is no sign that Gordan ever recognizably used infinitesimals, except for his thesis defense.

simplicial structure on a flat fiber bundle

Thu, 09/14/2017 - 08:42

Let $p: E \rightarrow B$ be a flat fiber bundle with fiber $F$ where $E$, $B$, $F$ are nice spaces (say smooth manifolds). Then $E$ has the form of a twisted product

(i) $E \cong \widetilde{B} \times_{\pi_{1}} F$,

where $\widetilde{B}$ is the universal cover of $B$, $\pi_{1}$ is the fundamental group of $B$ and $F$ carries a $\pi_{1}$-action.

Now, how nice can we assume this $\pi_1$-action to be? In particular, is it very restricting to only look at flat fiber bundles of the form

(ii) $E \cong \widetilde{B} \times_{\pi_{1}} |L|$,

where $L$ is a simplicial $\pi_{1}$-complex? What, if we further assume the action on $L$ to be regular?

I can't think of a flat bundle that is not of form (ii). Do you know counterexamples that clarify what (ii) cannot describe? (possibly with relaxed conditions on $E$, $F$, $B$.) And are there theorems that specify exact conditions on a flat bundle to be of form (ii)?

I am particularly interested in the case $B$, $F$ compact, $\pi_{1}$ infinite.

A bounded polynomial having bounded coefficients: several variables

Thu, 09/14/2017 - 01:36

Consider the multivariate case for the question "Approximation theory reference for a bounded polynomial having bounded coefficients" (Approximation theory reference for a bounded polynomial having bounded coefficients)

I tried to find an upper bound for coefficients for each monomial. I tried two methods:

  1. Lagrange interpolation. As in Lemma 4.1 in .

  2. I first only consider the variable $x_1$ and get an upper bound for the coefficients. Now the bound is the upper bound for another polynomial of $x_2,x_3,...,x_n$ and we can repete the above method.

However, using either method above, the upper bound will contain factor $d^n$ or $c^{d*min\{d,n\}}$ where c is a constant, which I think is too large. Is it possible to find a better upper bound?

Is there a non-boolean Eulerian interval of finite groups?

Wed, 09/13/2017 - 17:18

An Eulerian subgroup lattice is boolean (see here), so it is natural to wonder whether it is also true for an interval of finite groups. The smallest non-boolean Eulerian lattice is the following:
It is the face lattice of the square polytope (see here), let's call this lattice $P_4$.

By some tools used in this paper and the computation of Gordon Royle cited here, we can prove that if $P_4=[H,G]$ as a lattice, then $|G:H| \ge 135$ (and so $|G| \ge 270$, because $H \neq 1$). Moreover, we can checked by GAP that $|G| \ge 512$, and if $G$ is simple then $|G| \ge |{\rm PSU}(4,3)|= 3265920$.

The existence of a lattice which is not the lattice of an interval of finite groups is an open problem.
In the following paper (p72): Overgroup lattices in finite groups of Lie type containing a parabolic Michael Aschbacher conjectures (after John Shareshian) that a lattice $L$ such that the poset $\overline{L}:=L \setminus \{\hat{0},\hat{1} \}$ is disconnected with connected components the posets $\overline{B}_{n_1}, \dots , \overline{B}_{n_r}$, with $B_{n_i}$ a boolean lattice of rank $n_i \ge3$ (and $r \ge 2$), is not the lattice of an interval of finite groups. But if all the $n_i$ are equal and odd, then $L$ is a non-boolean Eulerian lattice (here the smallest example).

All these evidences lead to wonder:
Question: Is there a non-boolean Eulerian interval of finite groups?

From an eventual positive answer for the relative version of K.S. Brown's problem and dual (i.e. $\sum_{K \in [H,G]} \mu(K,G)|G:K|$ and $\sum_{K \in [H,G]} \mu(H,K)|K:H|$ are nonzero) and the property here, we can deduce the following extension of (dual) Ore's theorem on boolean intervals:

Conjecture: Let $[H,G]$ be an Eulerian interval of finite groups. Then:

  • $\exists V$ irr. $\mathbb{C}$-rep. of $G$ such that $G_{(V^H)} = H$ (see here).
  • $\exists g \in G$ such that $\langle Hg \rangle = G$ (see here).

This conjecture could help the investigation.

On the divisor function in a summation

Wed, 09/13/2017 - 08:40

I want to compute the limit inferior and superior of the following sum $$f(x):=\sum_{\substack{d \leq x\\P^+(d)\le\sqrt{x}}} \mu(d)\tau(d)\left[\frac{x}{d}\right]$$

Where $μ$ is the Mobius function, $τ$ is the number of positive divisors of $n$, $h(x)=[x]$ is the floor function and $P^{+}(d)$ is the largest prime divisor of $d≥2$, and $P^{+}(1)=0$.

Is there a criterion for compactness in $L^\infty(\Omega)$ with strong topology?

Tue, 09/12/2017 - 21:58

If such criterion exists, since $C(\Omega)$ is closed in $L^\infty(\Omega)$, and if $\Omega$ is bounded and closed, the Ascoli-Arzela theorem has given a sufficient and necessary condition,means this criterion needs more requirements than Ascoli-Arzela theorem. So my questions:

  1. How to generalize the compactness condition in $L^\infty(\Omega)$($\Omega$ is compact in $R^N$)?
  2. If $\Omega$ is not compact, or just $R^N$, what is the compactness criterion?

Thank you!

Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not

Tue, 09/12/2017 - 21:09

The following fact is known:

If there is a measurable cardinal, then there are only countably many constructible reals.

It is also known that if $ZFC$ is consistent, then $ZFC$ + "There exists a (two-valued) measurable cardinal" + $CH$ is also consistent.

If one replaces "There exists a (two-valued) measurable cardinal" with "The cardinality of the continuum is a real-valued measurable cardinal", is

i)$ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" + $CH$ is consistent if $ZFC$ is consistent (it is not, by a result of Banach and Kuratowski)? Also,

ii) Does $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" imply that there are only countably many constructible reals?

Since i) is inconsistent, how is it possible that $ZFC$ + "The cardinality of the continuum is a real-valued measurable" implies $\lnot$$CH$ while $ZFC$+ "There exists a (two-valued) measurable cardinal" is consistent with $CH$ since the two theories are equiconsistent?

Is an Eulerian subgroup lattice boolean?

Tue, 09/12/2017 - 21:05

Let $G$ be a finite group and $\mu$ the Möbius function of the subgroup lattice $\mathcal{L}(G)$.

The reduced Euler characteristic of the order complex of the coset poset $\{ Kg \ | \ K<G, \ g \in G \} $ is $$\chi(G) := -\sum_{H \in \mathcal{L}(G)} \mu(H,G)|G:H|.$$ Gaschütz showed that $\chi(G)$ is nonzero for $G$ solvable and the question whether it is nonzero for any finite group is an open problem motivated by K.S. Brown (see DOI: 10.1016/j.aim.2015.10.018).

Consider a dual version of Brown's problem: the question whether $\hat{\chi}(G)$ is nonzero, with $$\hat{\chi}(G) := -\sum_{H \in \mathcal{L}(G)} \mu(1,H)|H|.$$ We have checked by GAP that $\chi(G)$ and $\hat{\chi}(G)$ are nonzero for $|G| \le 100$.

Let $G$ be a finite group such that $\mathcal{L}(G)$ is an Eulerian lattice, then $\mu(1,H) = \mu(1,G) \mu(H,G)$ for any $H \in \mathcal{L}(G)$ (see this post). Then $\hat{\chi}(G) = - \mu(1,G) \varphi(G)$ with $$\varphi(G) = \sum_{H \in \mathcal{L}(G)} \mu(H,G)|H|.$$ But by Crosscut Theorem and inclucion-exclusion principle $\varphi(G) = | \{g \in G \ | \ \langle g \rangle = G \} |.$ So if $\hat{\chi}(G)$ is nonzero as suggested by the dual Brown's problem, then so is $\varphi(G)$, which means that $G$ is cyclic and $\mathcal{L}(G)$ distributive; but it is assumed Eulerian, so it is boolean. Conclusion, the existence of a non-boolean Eulerian subgroup lattice would give a negative answer to the dual Brown's problem.

Question: Is an Eulerian subgroup lattice boolean?

Warning: By googling "Eulerian subgroup lattice" you find:
J.P. Bohanon and L. Reid, Families of finite groups with Eulerian subgroup lattices, in progress.
It seems that this works deals with Eulerian graphs, not with Eulerian posets.

The "strong" measure number

Tue, 09/12/2017 - 21:03

Beyond measure zero we have yet another measure-y notion of smallness: strong measure zero. A set $S\subseteq\mathbb{R}$ is strong measure zero if, for any $f:\mathbb{N}\rightarrow\mathbb{R}_{>0}$, there is a sequence $U_i$ of open sets with

  • the diameter of $U_i$ is $<f(i)$, and

  • $S\subseteq\bigcup_{i\in\mathbb{N}} U_i$.

(Various intermediate notions of nullness are also considered.) At least for me, it wasn't immediately obvious that there are measure zero sets which aren't strongly measure zero, but in fact the Cantor set is provably not strong measure zero.

Strong measure zero-ness raises a natural cardinal characteristic-style question:

  • What is the smallest size $\mathfrak{s}_-$ of a non-strongly measure zero set? What is the supremum $\mathfrak{s}_+$ of the cardinalities of the strong measure zero sets?

Clearly every countable set is strong measure zero, so $\mathfrak{s}_-$ is uncountable. The converse, Borel's conjecture, is consistent with ZFC (this was proved by Laver) and implies $\mathfrak{s}_-=\omega_1$ and $\mathfrak{s}_+=\omega$ (this reveals that $\mathfrak{s}_+$ is a bit weird; still, it seems a natural thing to consider, even if it's not a "kosher" CCC). Meanwhile, Sierpinski showed that CH implies that $\mathfrak{s}_+=\mathfrak{s}_-=2^{\aleph_0}$.

My question is:

What is known about $\mathfrak{s}_-$ and $\mathfrak{s}_+$? In particular, where does $\mathfrak{s}_-$ sit amongst other cardinal characteristics (say, those in Cichon's diagram)?

measure spaces and sigma algebra

Tue, 09/12/2017 - 20:36

Is the function P(· | Ψ) a probability measure?

What makes the transfinite derived series terminate at identity?

Tue, 09/12/2017 - 20:22

Let $G$ be a group and $G^{(n)}$ be the $n$th derived series of $G$. Suppose $G^{(n)}/G^{(n+1)}$ is infinite for $n\geq 0$. Then the derived series of $G$ doesn't stabilize at finite $n$. Does there exist a limit ordinal $\omega$ such that $G^{(\omega)}$ is trivial?

Corlette-Simpson correspondence over nodal curves

Tue, 09/12/2017 - 20:07

As we know that over a smooth projective variety $X/\mathbb{C}$, there is a correspondence between semi-simple flat bundles and polystable Higgs bundles with vanishing Chern classes, through differential geometric objects-Harmonic bundles.

On Open varieties $X-D$, with $D$ normal crossing, we also have a modified correspondence given by Simpson and T. Mochizuki: The category of polystable regular filtered flat bundles with trivial Chern classes is natural equivalent to the category of polystables regular filtered Higgs bundles with trivial Chern classes, via the category of Tame Harmonic bundles over the open variety $X-D$. The tameness condition of Harmonic bundles gives a good estimate for the upper bounded of curvatures associated to two metric connections (Which is called acceptable), which make it possible to define a functor from category of Tame harmonic bundles to the category of regular filtered Higgs bundles and the category of regular filtered flat bundle, resp.

Q: I think it should be nature to ask: is there a definition of "tame Harmonic bundles" over nodal curve $X$? The naive thoughts are: (1) It should come from harmonic bundles on normalization $\eta: \tilde{X}\rightarrow X$ with gluing along preimage of nodes. Or (2) It should come from tame harmonic bundles on $\tilde{X}-\eta^{-1}(S)$, where $S$ is set of nodes on $X$, with compatible conditions along the two punctures corresponding to the same node.

Any references are appreciated.

More on divisibility

Tue, 09/12/2017 - 19:08

This is a fuzzier follow-up to this question. Again, we construct the graph whose vertices are integers from $1$ to $n,$ and two vertices are connected whenever one of the corresponding integers divides the other, and then we lay the graph out radially, so the vertex corresponding to $1$ is at the center of the circle, and the others are clockwise around the circumference This is what we get (for $n=180$). There are obviously patterns, but how do we explain them?

Can you consistently add axioms about the Busy Beaver function to ZF?

Tue, 09/12/2017 - 18:54

Consider a Turing Machine with N states which checks all theorems of ZF and halts upon finding a contradiction. If ZF were consistent and could prove the value of Busy Beaver (N), then it would be able to prove its own consistency, which Godel proved impossible; so either ZF is inconsistent or the value of BB(N) is independent of ZF.

But what if we add to ZF an extra axiom K, which specifies the exact (true) value for BB(k), for some large k? If ZF is consistent (edit: and sound), then ZFK is consistent (else a contradiction in ZFK would be a proof in ZF of ~K). Now assume that there is a Turing Machine with k states or fewer which checks all theorems of ZFK. If it halts, then ZFK is inconsistent, so ZF is inconsistent or unsound. If it doesn't halt after BB(k) steps, then it has proved the consistency of ZFK, which is impossible by Godel.

It seems like I've shown that either ZF is inconsistent or unsound or that there is no such k-state Turing Machine which proves the theorems of ZFK. But it seems obvious that for sufficiently large K, there are such Turing Machines, since all they need to do is symbolic manipulation of finite axioms. What's going on?

Linear and numerical equivalence

Tue, 09/12/2017 - 18:30

What are the weakest hypothesis one can put on a smooth projective complex variety $X$ in such a way that two effective divisors $D_1,D_1$ on $X$ are numerically equivalent if and only if $D_1$ and $D_2$ are linearly equivalent? Do the varieties with this property have a name in the literature?

I know this is implied by $h^1(X,\mathcal{O}_X)=0$.

packing with special sets in high dimensional Euclidean space

Tue, 09/12/2017 - 18:21

Let $\lambda$ be Lebesgue measure on $[0,1]$. For $\mathbf{x}=(x_1,x_2,..,x_k)\in[0,1]^k$, define $$A(\mathbf{x}):=\{(y_1,\dots,y_k)\in [0,1]^k: \text{there exist intervals }I_1,\dots,I_k \text{ in }[0,1]$$ $$\text{ such that } x_i,y_i\in I_i \text{ and } \lambda(\cup_iI_i)\leq\frac12\}$$

My question: Is it true that the maximum number $n$ such that there exists $\mathbf{x}_1,..,\mathbf{x}_n\in[0,1]^k$ that satisfy $A(\mathbf{x}_1),..,A(\mathbf{x}_n)$ are mutually exclusive has at most exponentially growth speed in $k$?

That is if we denote $$n^*=\max\{n\in\mathbb{N}:\exists \text{ $\mathbf{x}_1,..,\mathbf{x}_n\in[0,1]^k$ that $A(\mathbf{x}_1),..,A(\mathbf{x}_n)$ are mutually exclusive }\} ,$$ can we prove $$n^*\leq C^k$$for some constant $C>1$?

First I tried a simple volume argument. I guessed for any $\mathbf{x}\in[0,1]^k$ we have $\text{vol}(A(\mathbf{x}))\geq c^k$ for some constant $c<1$. But it turns out not true. See the answer here lower bound volume of a set.

Any thoughts and comments are welcome!

Thanks for your help!