Recent MathOverflow Questions

Is there an adaptation of the theory of standard forms and Tomita-Takesaki theory to the $\mathbb{Z}_{2}$-graded case?

Math Overflow Recent Questions - Fri, 08/17/2018 - 05:30

Let $A$ be a von Neumann algebra acting on a Hilbert space $H$, and suppose that $\Omega \in H$ is a cyclic and separating vector for $A$. Then in Tomita-Takesaki theory one defines an unbounded operator $S$ to be the closure of the operator \begin{equation} a \triangleright \Omega \mapsto a^{*} \triangleright \Omega. \end{equation} It then turns out that if we consider the polar decomposition \begin{equation} S = J \Delta^{1/2} \end{equation} into an anti-linear unitary involution $J$ and an unbounded positive operator $\Delta^{1/2}$ it follows that the map $A \ni a \mapsto Ja^{*}J \in (A')^{\operatorname{opp}}$ is an isomorphism. (Here the superscript opp stands for opposite algebra.)

Now, let $A$ be any von Neumann algebra, then a standard form for $A$ is a quadruple $(A, H, J, P)$, where $H$ is a Hilbert space in which $A$ acts, $J$ is an antilinear operator in $H$, and $P$ is a self-dual cone, which satisfy the following properties.

  1. $JAJ = A'$,
  2. $JaJ = a^{*}$, if $a \in A \cap A'$,
  3. $Jv = v$ for all $v \in P$,
  4. $aJaJP \subseteq P$ for all $a \in A$.

Theorem: [Haagerup] If $(A,H',J',P')$ is a second standard form for $A$, then there exists a unique unitary $u :H \rightarrow H'$ such that

  1. $uau^{*} = a$ for all $a \in A$,
  2. $J' = uJu^{*}$,
  3. $P' = uP$.

It turns out that if $\Omega \in H$ is cyclic and separating as above, then the quadruple $(A,H,J, P_{\Omega})$, with $P_{\Omega} = \{ aJaJ\Omega \mid a \in A \}$ is a standard form for $A$.

Does there exist an adaptation for the $\mathbb{Z}_{2}$-graded case?

What I mean by this is as follows. Suppose that $A$ is a $\mathbb{Z}_{2}$-graded von Neumann algebra acting on a $\mathbb{Z}_{2}$ graded Hilbert space $H$, which is equipped with a cyclic and separating vector $\Omega \in H$. (Suppose that $\Omega$ is even.) Furthermore, assume that the representation of $A$ on $H$ respects the grading. Define an involution on $A$ by $a^{\sharp} = a^{*}$ if $a$ is even and $a^{\sharp} = -ia^{*}$ if $a$ is odd. We then consider the unbounded operator $S^{\sharp}$ defined to be the closure of the operator \begin{equation} a \triangleright \Omega \mapsto a^{\sharp} \triangleright \Omega. \end{equation} Then consider the polar decomposition $S^{\sharp} = J^{\sharp} \Delta^{1/2\sharp}$.

Is it true that the map $A \ni a \mapsto J^{\sharp} a^{\sharp} J^{\sharp} \in (A^{\backprime})^{\operatorname{opp}}$ is an isomorphism? (Here $A^{\backprime}$ is the graded commutant of $A$.)

Similarly, one can define a graded standard form of $A$, by replacing $A'$ by $A^{\backprime}$ and $*$ by $\sharp$.

Does Haagerup's theorem still hold? If so, what circumstances guarantee that $u$ is even?

(This question was motivated by the analysis done on pages 57 and 58 of these lecture notes on conformal nets.)

What is a true invariant of $G$-crossed braided fusion categories?

Math Overflow Recent Questions - Fri, 08/17/2018 - 03:58

Definition. An invariant of a (spherical) fusion category with extra structure is a number or a set or tuple of numbers preserved under (appropriate) equivalences.

(Spherical) fusion categories have well-known invariants like their global dimension, their rank, the categorical and Frobenius dimensions of their simple objects, or the Frobenius-Schur coefficients. Ribbon fusion categories have some more invariants like the $S$-matrix, the twist eigenvalues, and all the invariants of its symmetric centre. Graded fusion categories have e.g. the size of the group and the dimension as an invariant.

An example for something that is not an invariant is an $F$-matrix. It depends on the choice of basis vectors for trivalent morphism spaces, and that is not preserved by an equivalence. Summing over the appropriate elements of the $F$-matrices yields the Frobenius-Schur coefficients, though, and they are invariants.

A (spherical) $G$-crossed braided fusion category (short: $G\times$-BFC) is a $G$-graded fusion category with a compatible $G$-action and a crossed braiding. This implies e.g. that its trivial degree is a ribbon fusion category. All this gives us access to the invariants I've already mentioned, but I want to know whether a $G\times$-BFC has any new invariants.

Question. Are there any invariants of $G\times$-BFCs that are not invariants of the underlying $G$-graded fusion category, or of the trivial degree? For example, does the crossed braiding contain information beyond the trivial degree, and what information does the $G$-action possess?

Bonus question. Can those new invariants be expressed diagrammatically, i.e. in the graphical calculus of $G\times$-BFCs?

Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$

Math Overflow Recent Questions - Fri, 08/17/2018 - 02:46

The question is to prove: $$ \int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1. $$ Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed.

I also wonder if there is a systematic approach to solve this kind of problem.

Thanks for your help.

Number guessing game with lying oracle

Math Overflow Recent Questions - Thu, 08/16/2018 - 22:19

You are probably already familiar with the usual number guessing game. But for concreteness I restate it.

The usual game

The Oracle chooses a positive integer $n$ between 1 and 1024 (or any power of 2).

At each turn, you make a guess $g$, and the Oracle tells you whether $n \leq g$ or $n > g$. The game ends when you can determine what $n$ is.

It is pretty easy to see that the optimal strategy requires exactly 10 turns.

The modified game

In this game, the game runs almost exactly the same as the usual game. Except that there is a probability $p \in [0,1/2)$ (which is known to you) that the Oracle lies.

More precisely, at each turn, the Oracle flips a (biased) coin and determines, independently of previous actions and other things in the game, whether to lie to you or not. When she lies she gives the exact opposite of the correct answer for the comparison $n \leq g$ or $n > g$.

The game ends when you can determine, to a previously-agreed-upon confidence level, what is the answer $n$. (For argument sake, say that you can say with 95% probability what the value of $n$ is.)

To model this, imagine you starting with 0 knowledge, so that each number between 1 and 1024 is equally likely. At each step you can update the probability distribution using the usual Bayesian updating procedure. The game ends when one of the numbers has probability 95% or higher.

Question

Can we estimate the expected number of guesses before the conclusion of the game, as a function of the lying probability $p$? (... and of the desired confidence level, and the number of bins?)

Obviously, if the confidence level is anything above 50%, and if $p = 0$, the game reduces the usual game.

As $p \to 1/2$, the expected number of guesses tend to infinity (at $p = 1/2$ the Oracle just responds randomly).

So in particular, what are the asymptotics of the number of guesses as $p \to 0$ and as $p \to 1/2$?

Numeric Data

The simulations below using the following naive strategy for guesses:

At each step, based on the current "prior" probability, guess the number $g$ such that the difference $|P(n \leq g) - P(n > g)|$ is minimized.

In the case $p = 0$ this reduces (one can check) to the binary search method. This choice is made to "maximize information gained" from that step. I don't have a proof that this is the optimal strategy.

From numerical simulations, with 1024 numbers, 95% confidence interval, and 3000 trials,

p = 2^-2 avg = 56.806666666666665 p = 2^-3 avg = 23.315 p = 2^-4 avg = 16.112666666666666 p = 2^-5 avg = 13.047666666666666 p = 2^-6 avg = 11.633 p = 2^-7 avg = 10.604333333333333 p = 2^-8 avg = 10.285333333333334 p = 2^-9 avg = 10.156666666666666 p = 2^-10 avg = 10.069 p = 2^-11 avg = 10.029666666666667 p = 2^-12 avg = 10.011333333333333 p = 2^-13 avg = 10.005666666666666 p = 2^-14 avg = 10.0 p = 2^-15 avg = 10.0

which suggests a slope for small $p$ of around 100.

Frequentist versus Bayesian

James Martin brought up a very good point about frequentist versus Bayesian. Let me illustrate that with data from two runs using the strategy described above, with $p = 2^{-8}$.

Starting game with 1024 bins. Running game now with secret answer: 857 Guessing 512 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 768 with probability 0.003875850705270716 Guessing 895 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999999 My best guess is that the secret answer is 768 with probability 0.007721187532297775 Guessing 831 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 832 with probability 0.015441436261097105 Guessing 863 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999986 My best guess is that the secret answer is 832 with probability 0.030880965812145108 Guessing 847 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000013 My best guess is that the secret answer is 848 with probability 0.0617618727298829 Guessing 855 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999987 My best guess is that the secret answer is 856 with probability 0.12256258895055303 Guessing 859 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999982 My best guess is that the secret answer is 856 with probability 0.24704724796624977 Guessing 857 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999997 My best guess is that the secret answer is 856 with probability 0.4903092482458648 Guessing 856 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999998 My best guess is that the secret answer is 857 with probability 0.9881889766208873 I did it! And it only took me 10 tries; the Oracle lied 0 times.

Note that in binary, the number 857 is 1101011001 with more or less even distributions of zeros and ones. For 513 whose distribution is more extreme:

Starting game with 1024 bins. Running game now with secret answer: 513 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000033 My best guess is that the secret answer is 513 with probability 0.003875733349545551 Guessing 639 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999948 My best guess is that the secret answer is 513 with probability 0.007721187532297778 Guessing 575 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999996 My best guess is that the secret answer is 513 with probability 0.015323132493622977 Guessing 543 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.030180182919776057 Guessing 527 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.05857858737666496 Guessing 519 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.999999999999996 My best guess is that the secret answer is 513 with probability 0.1106260320552282 Guessing 515 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.19905831824030176 Guessing 513 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999991 My best guess is that the secret answer is 513 with probability 0.3315926624554764 Guessing 385 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.6614346507956576 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999979 My best guess is that the secret answer is 513 with probability 0.9902152355556972 I did it! And it only took me 11 tries; the Oracle lied 0 times.

Incidentally, this also explains (partly) the drop between $2^{-13}$ and $2^{-14}$ seen in the numerical results above. When $p = 2^{-13}$, when $n = 513$, after 10 truthful responses the confidence that 513 is the correct answer is "only" 94%. But the 95% threshold is breached when $p = 2^{-14}$.

Are there examples of conjectures supported by heuristic arguments that have been finally disproved?

Math Overflow Recent Questions - Thu, 08/16/2018 - 10:53

The idea for this comes from the twin prime conjecture, where the heuristic evidence seems just so overwhelming, especially in the light of Zhang's famous result from 2014 about Bounded gaps between primes and its subsequent improvements.

Are there examples of conjectures with some more or less "good" heuristic arguments, but where those arguments have finally not been "strong" enough?

I don't mean heuristic in the sense that some conjecture holds for small numbers, e.g. the fact that $li\ x-\pi (x) $ was thought to be always positive, before Littlewood showed that not only it eventually changes sign, but does so infinitely often later on. So my question is different from the question about eventual counterexamples.

Likewise, the fact that the first $10^{15}$ or so zeros of the zeta function "obey" RH does tell us something, but not a whole lot compared with infinity. So again this is not what I mean by heuristic.

Question on a proof of density of periodic orbits

Math Overflow Recent Questions - Thu, 08/16/2018 - 08:53

In page 215 and 216 of the book "Introduction to the Modern Theory of Dynamical Systems" by Anatole Katok, Boris Hasselblatt, there is a theorem stated as following:

Theorem: Let $\Gamma$ be a discrete group of fixed-point-free isometries of $\mathbb{D}$ such that $M:=\Gamma \setminus \mathbb{D}$ is compact. Then the periodic orbits of the geodesic flow on $SM$ are dense in $SM$.

In the proof, they stated a fact that:

given $\epsilon >0$ there exists $\delta>0$ such that when $p \in \mathbb{D}$ is in a $\delta$-neighborhood of $\partial \mathbb{D}$ then any two geodesics through $p$ of Euclidean length greater than $\epsilon$ have a mutual angle of at most $\pi/4$

After that, they used this fact to say that: "most geodesics through $z$ are entirely contained in $U$", with figure:

I could not understand this argument. So I hope everyone will help me! I am new in this subject. This is the full proof:

   

Understanding a group of transformations of the plane $\mathbb{Z} \times \mathbb{Z}$

Math Overflow Recent Questions - Thu, 08/16/2018 - 07:54

I'm doing some research in visualizing arithmetic sets (resp. properties, resp. sequences of integers). I try to create patterns (in which I hope to observe some symmetries) by injectively mapping $\mathbb{N}$ on $\mathbb{Z}\times \mathbb{Z}$ in different ways and highlighting the numbers that have a given property (belong to a given set, are in a given sequence).

You can find a little tool with which I'm playing around here.

What I came up with has nothing to do with the initial approach (i.e. with arithmetic sets) but with transformations of $\mathbb{Z}\times \mathbb{Z}$. It's possibly interesting, but I'm missing conceptual tools to understand and explain it.

What I observed was that different spirals (i.e. regular bijective mappings $m: \mathbb{N} \rightarrow \mathbb{Z}\times \mathbb{Z}$ with $m(0) = (0,0)$) seem to be related by some kind of "quasi-rotations", combined with some scaling, translation and "disturbance" in the middle of the spiral. These transformations can be watched "in action" and indeed "feel" like rotations, sometimes accompanied by some characteristic intermediate "shrinking" (have a look).

I'd like to put these findings into mathematical terms and understand how they relate to each other and to the parameters of the spirals.

Let a spiral $S^i_j$ be parametrized by the width and height of its rectangular "eye", i.e. pairs of integers $(i,j) \in \mathbb{Z}\times \mathbb{Z}$. The "eye" of a spiral is a rectangle of integer side length with one corner in the origin, around which the spiral winds clockwise. This is the eye of the spiral $S^5_2$:

Each transition $T^k_l$ between spirals can be described by a pair of numbers $(k,l) \in \mathbb{Z}\times \mathbb{Z}$ such that

$$T^k_l(S^i_j) = S^{i+k}_{j+l}$$

Obviously, the transitions $T^k_l$ form a group. A transition seems to be related to

  • an angle of rotation $\frac{n\ \pi}{2}$ (see $S^i_j \leftrightarrow S^{i\pm 1}_j$, $S^i_j \leftrightarrow S^i_{j\pm 1}$) (depending on what and how?)

  • the relative size of a "disturbed" zone in the middle of the affected spiral (depending on what and how?)

  • some scaling and translation of "patterns" outside the disturbed zone that remain "intact" otherwise (depending on what and how?)

  • possibly an isolated translation of the upper-right half plane (see $T^{-2}_1$ which sends $S^3_0$ to $S^1_1$)

[What's possibly not characteristic for a given transformation – but possibly depends on the spiral being transformed – is the intermediate "shrinking" factor which becomes visible only when watching the transformation "in action" (see $T^0_1$ sending $S^1_0$ to $S^1_1$, or $T^2_0$ sending $S^1_1$ to $S^3_1$, which both are not accompanied by a rotation).

Compare this scenario (as a vague analogy) with the physical scenario of "Big Collapse followed by Big Bang"]

My question is:

How can I treat these ephemeral quasi-rotations in the context of "hard" group theory? Where in group theory is a place for transformations of the form "rotation + X" with a hard to grasp X?

Can we realize a graph as the skeleton of a polytope that has the same symmetries?

Math Overflow Recent Questions - Thu, 08/16/2018 - 05:18

Given a graph $G$, a realization of $G$ as a polytope is a convex polytope $P\subseteq \Bbb R^n$ with $G$ as its 1-skeleton.

A realization $P\subseteq \Bbb R^n$ is said to realize the symmetries of $G$, if for each graph-automorphism $\phi\in\mathrm{Aut}(G)$ there is an isometry of $\Bbb R^n$ inducing the same automorphism on the edges and vertices of $P$.

Question: Let $G$ be the 1-skeleton of a polytope. Is there a realization of $G$ that realizes all its symmetries?

Note that if we know $G$ to be the skeleton of an $n$-polytope, then the symmetric realization does not have to be of the same dimension. In fact, this is not always possible. The complete graph $K_n,n\ge 6$ is realized as a neighborly 4-polytope, but its only symmetric realization is the simplex in $n-1\ge5$ dimensions. However, this is the only example I know of.

One way to construct a counter-example would be to find an $n$-polytope with a skeleton $G$ that is not more than $n$-connected (e.g. $n$-regular), but where $\mathrm{Aut}(G)$ is no subgroup of the point group $O(n)$. Its symmetric realization must be in dimension $\ge n+1$, but it cannot be because of Balinski's theorem.

Hopf dual of the Hopf dual

Math Overflow Recent Questions - Wed, 08/15/2018 - 09:12

Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^∗$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^∗$ has a natural Hopf algebra structure.

Question: Is the Hopf dual of the Hopf dual of $A$ isomorphic to $A$. It is not obvious for me that it is. If not, then do we know in which cases it is true?

On the set of indices of irreducible depth 3 subfactors

Math Overflow Recent Questions - Wed, 08/15/2018 - 06:50

Let $I_n$ be the set of indices of (finite index) irreducible depth $n$ subfactors. Then $I_2 = \mathbb{Z}_{>0}$.

Question 1: Is it true that $I_3$ has no accumulation point?

If so:
Question 2: Is it true that $|I_3 \cap [x-1,x]|\le x$ (for $x$ large enough)?

If so, let $\epsilon_n$ be the infimum of the length of the gaps in $I_n$ (obviously, $\epsilon_2 = 1$), i.e. $$\epsilon_n := \inf \{ |a-b| \text{ such that } a,b \in I_n \text{ and } a \neq b \}. $$

Question 3: Is it true that $\epsilon_3>0$?

Guess: Q1 yes, Q2 yes and Q3 no. Same questions at depth $n > 3$?

Partitioning the vertices of a graph into induced trees

Math Overflow Recent Questions - Wed, 08/15/2018 - 02:10

I am looking for previous work regarding graphs whose vertices can be assigned colours (not necessarily a proper colouring) in such a way that each colour class induces a tree.

In particular I am most interested in results for planar graphs.

I have searched quite hard and there is a fairly extensive body of work on vertex-arboricity and linear vertex arboricity which relate to colouring a graph so that each colour class is a forest or a disjoint union of paths. (Rather than using the phrase "colouring the vertices", the vertex-arboricity etc is often defined as "partitioning the vertices" but of course they are the same thing.)

There are numerous extensions of the vertex-arboricity concept where the colour classes have to be nearly equal size, or when other restrictions are placed on the monochromatic subgraphs. In fact, put any of the words {equitable, list, star, critical, fractional} in front of "vertex-arboricity" and there is a paper on it.

The other relevant thing that came up was the phrase Yutsis graphs which refer to graphs that can be partitioned into two induced trees. However there are perhaps two or three papers on these, and they focussed primarily on cubic graphs, and did not help me.

So my question is:

Is there an accepted graph-theoretic term and/or associated literature related to the concept of partitioning the vertices of a graph into induced trees?

I am primarily concerned with exact structural results (things like characterisations of families of graphs that can be partitioned into very few trees) that I may be able to extend and/or adapt for my purposes.

I am not really interested in complexity results (e.g. NP-completeness) or anything to do with infinite graphs; there are numerous vertex-arboricity papers on complexity.

Totally geodesic submanifold of codimension 1

Math Overflow Recent Questions - Wed, 08/15/2018 - 00:40

This question is inspired by question in reference.

Question : If $M$ is a simply connected closed Riemannian manifold of nonnegative sectional curvature, then there is a totally geodesic submanifold $S$ of codimension 1

Def : ${\rm conv}\ X$ is a smallest closed convex set containing $X$. And a subset $X$ is convex if for $x,\ y\in X$, any shortest geodesic between them is in $X$.

Construction of $S$ : Let $X_1=B(p,\delta)$ to be a closed ball, which is convex. If $p_1$ is not in $X_1$, then let $X_2={\rm conv}\ X_1\bigcup\{p_1\}$ s.t. $d_H(X_1,X_2) >0$ is small and ${\rm vol}\ X_2-{\rm vol}\ X_1 >0$ is small. Hence if we repeat this process, then $X_n$ goes to $ X_\infty$.

If $p_\infty$ is not in $X_\infty$, and $d_H(X_\infty,{\rm conv}\ X_\infty\bigcup\{p_\infty\})$ is large, then $\partial X_\infty$ is a desired one.

If not, we let $X_1=X_\infty$ and repeat these process. That is, ${\rm vol}\ X_\infty$ is supremum of volumes of convex sets containing $p$.

Example : If we starts this process at a point of negative sectional curvature in torus of revolution, then $X_\infty$ is closure of set of all points of negative sectional curvature.

Additional question : We have alvarezpaiva's answer. But I want to know in dimension $3$. Or I want to know why my argument fails in $\mathbb{C}P^2$.

Reference : Convex sets in Alexandrov spaces

Maximal surfaces in pseudo-Riemannian manifolds

Math Overflow Recent Questions - Tue, 08/14/2018 - 10:40

Do you know a reference (with proof) for the second variation of the area of a space-like surface with vanishing mean curvature embedded in a pseudo-Riemannian manifold?

Is it known that if the ambient manifold is negatively curved (for instance $\mathbb{H}^{p,q}$), then these surfaces always maximize the area, among all surfaces with the same boundary? I know a similar result holds in codimension 1.

Relation between left projections

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:21

Let $A$ be a Baer *-ring. Let $x$ be in $A$, $L(x)$ is the left projection of $x$ that is the smallest projection with $L(x)x=x$.

Q. Let $p,q$ are projections in $A$ with $p\leq q$. I feel both of the following points are true but cannot prove them.

1- $L(pe)\leq L(qe)$.

2- $L(ap)\leq L(aq)$.

Can sufficiently symmetric polytopes be uniquely reconstructed from their 1-skeleton?

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:15

General convex polytopes can not be uniquely reconstructed from their 1-skeleton, as explained here. Not even the dimension is known from the skeleton, as e.g. the complete graph $K_n,n\ge 5$ is the 1-skeleton of neighborly polytopes that exist in dimensions $\ge 4$.

But how well works the reconstruction from the 1-skeleton if we restrict to sufficiently symmetric polytopes. Obviously, vertex-transitivity is not enough as seen from the existence of vertex-transitive neighborly polytopes. But what if we add edge-transitivity, uniformity, arc-transitivity or some requirements on the edge-lenghts? E.g. only simplices are vertex- and edge-transitive polytopes with $K_n$ as their 1-skeleton.

Question: Given a "symmetric" polytope $P$ (replace "symmetric" with the sufficiently strong symmetry requirements of your choice). Can there be a different "symmetric" polytope with the same 1-skeleton as $P$?

I asked a similar question on Math.SE, without much success.

Proof of an inequality

Math Overflow Recent Questions - Tue, 08/14/2018 - 09:01

I can't prove one of the inequality in a paper:

$\sum_{w=1}^v w^{\lambda}=O(v^{\lambda+1})$ for $\lambda \neq -1$.

When $\lambda>0$ I can use the AM-QM inequality to prove it, but when it comes to $\lambda<0$, I have no idea.

Automorphy of families of motives

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:55

I have a couple of elementary questions regarding automorphy of Galois representations arising from geometric families.

Suppose we have an algebraic family of varieties over a number field, and Galois representations arising from $l$-adic cohomology of one of the varieties in the family are known to be automorphic. What can one say about automorphy of (Galois representations attached to $l$-adic cohomology of) other varieties in the family?

Alternatively, and more specifically, on the moduli space of algebraic curves (of a fixed genus) over $\mathbb{Q}$, take a small/formal neighborhood of a curve which is automorphic in the above sense. What can be said about automorphy of curves in the neighborhood?

One can also ask a related global question: suppose one somehow knows that the universal curve over the moduli space of algebraic curves of genus $g$ is automorphic. Would it follow that all algebraic curves of genus $g$ are automorphic?

Partition theorems for located words

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:51

In this paper Bergelson, Blass, and Hindman prove the following

Theorem 1.2 Let $W(\Sigma; v)$ be colored with finitely may colors and let $\bar s$ be an infinite sequence from $W(\Sigma; v)$. Then $\bar s$ has a variable extraction $\bar t$ such that the set of extracted words of $\bar t$ is monochromatic.

I would like to know if there is an easy counter example to the claim that the components of $\bar t$ are obtained concatenating those of $\bar s$ in increasing order. (As stated, the theorem also allows substitution of the variable $v$ with elements of $\Sigma$ as long as all components of $\bar s$ contain at least a variable.)

Notation $W(\Sigma; v)$ is the set of words in the alphabeth $\Sigma\cup\{v\}$ that contain at least one occurrence of the variable $v$.

Noether-Lefschetz for Kummer-quartic surfaces with curves of bounded degree

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:16

The Noether-Lefschetz theorem states, that a very general quartic surface $S$ in $\mathbb{P}^3$ has $Pic(S)\cong\mathbb{Z}$ generated by $\mathcal{O}_{\mathbb{P}^3}(1)_{|S}$, so all curves on $S$ are complete intersections. (We work over $\mathbb{C}$).

Do we have similar results for non-very general quartic surfaces, if we restrict to curves with bounded degree?

$\textbf{Question:}$ Given natural numbers $k,d\in \mathbb{N}$ with $2\leq k\leq 4$. Is it possible to find a Kummer-quartic surface $S$ in $\mathbb{P}^3$ with $\rho(S)=k$ such that all curves $C$ in $S$ with $deg(C)<d$ are complete intersections? Here Kummer means $S$ has exactly the maximal number of 16 nodes.

If yes: can we somehow describe all of theses surfaces in the moduli space of Kummer-quartic surfaces in $\mathbb{P}^3$?

For example: Can we find a Kummer-quartic surface $S$ in $\mathbb{P}^3$ with $\rho(S)=2$ such that every curve $C$ in $S$ with $deg(C)<16$ is a complete intersection?

About a pattern on prime numbers

Math Overflow Recent Questions - Tue, 08/14/2018 - 08:15

I have read in https://www.sciencealert.com/mathematicians-discover-a-strange-pattern-hiding-in-prime-numbers that says: "But this doesn't explain the magnitude of the bias the team found, or why primes ending in 3 seemed to like being followed by primes ending in 9 more than 1 or 7. Even when they expanded their sample and examined the first few trillion prime numbers, the mathematicians found that - even though the bias gradually falls more in line with randomness - it still persists."

I did a numerical analysis for the pairs (n3, n7), (n3, n9) where the numbers n3, n7, n9 which are integers ending at 3, 7, 9 respectively. These numbers (n3, n7, n9) may be (A) composites, multiples of 3, (B) composites, but not multiple of 3 or (C) prime numbers p3, p7, p9. We are looking for pairs of prime numbers (1) p3 with next p7 or (2) p7 with previous p3, (3) p3 with next p9 or (4) p9 with previous p3.

We take into account, that the prime numbers (random or not) are ranked in ascending order. ie after p1 series has p3 after p7 and then p9. Αfter 1 cycle again 1.

We will now see how a pair fails to be (p3, p7) or (p3, p9). This occurs in 4 cases.

(1) ((A) -> n3, p7) (#), ((A) -> n3, p9) (##)
(2) ((B) -> n3, p7), ((B) -> n3, p9)
(3) (p3, (A) -> n7), (p3, (A) -> n9)
(4) (p3, (B) -> n7), (p3, (B) -> n9)

(5) (p3, p7), (p3, p9)

We knows for big N have about the same amount of p3's, p7's, p9's in primePi(N).

We see (#) that the primes p7 can have losses in all four categories (1)(#), (2), (3), (4). But the p9's (##) only in 3 categories -> (2), (3), (4) Because the (1)(##) never happens, the reason is that p9 = ((A) -> n3) + 6 is multiple of 3, ie never is prime number. So p9's not consume their prime numbers in this category. For this p9's have more opportunities (since they are limited to 3 categories) than p7's (which spread into four) to reach and pair with p3's, see (5). That's why we find more pairs (p3, p9) than (p3, p7). See example in comments. But this phenomenon also seems to be very intense in a small sample of prime numbers, because they are more dense. In huge samples all categories normalized and approach at 25%. So, I will make a conjecture, that when N -> infinity, then (p3, p1), (p3, p3next), (p3, p7), (p3, p9) have a probability of 25%. As we would expect from random numbers.
Question: Am I right or not?

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