Recent MathOverflow Questions

Lagrangian foliation for a holomorphic symplectic manifold

Math Overflow Recent Questions - Mon, 04/30/2018 - 10:23

I am interested in gathering as many examples as possible for Lagrangian foliations of holomorphically symplectic manifolds $(X, \omega)$, where $X$ is a $2n$-dimensional complex manifold equipped with a closed non-degenerate (2,0)-form $\omega$ on $X$. A trivial example would be the cotangent bundle of a $n$-dimensional complex manifold. Can someone point to a nontrivial example of a Lagrangian foliation for a holomorphically symplectic manifold where one has a "natural" choice of polarisation like in the case of the cotangent bundle? Are there such examples at all? I cannot find any so I wonder. Thanks in advance for your input.

Edit: I changed the word "canonical" with "natural" in order to circumvent misunderstandings.

Left Kan extension along Yoneda of pullback-preserving functor preserving pullbacks

Math Overflow Recent Questions - Mon, 04/30/2018 - 10:07

Let $F\colon C \to D$ be a functor. The Kan Extension of $y_D \circ F$ along $y_C$ yields a functor $F_!: Fun(C^{op},Set) \to Fun(D^{op},Set)$. Here, $y_C$ and $y_D$ denotes the respective Yoneda embeddings. It is well-known that if $C$ and $D$ have finite limits and $F$ preserves them, then so does $F_!$.

Question: Suppose that we only assume that $C$ has pullbacks and that $F$ preserves them (so some pullbacks exist in $D$, too). Is it true that $F_!$ preserves pullbacks as well?

(I'm willing to assume that both $C$ and $D$ have finite products, but in my case, the functor $F$ does certainly not preserve these.)

I went through Lurie's ∞-categorical version of the "well-known" part ( in HTT), but I was not able to verify that the last step works without assuming that $F$ preserves products because this step involves some heavy machinery. I could give a different proof for this last step if $F_!$ preserved monomorphisms, but I believe that this is not the case in general.

Solve z^5 = -1 Give the solution in polar form [on hold]

Math Overflow Recent Questions - Mon, 04/30/2018 - 09:40

Solve z^5 = -1 Give the solution in polar form

I know this involves finding the root then putting it in the form of sin and cos and using this to find the arctan.... but.. I'm completely lost as to how

Could anyone help?

Li-Yorke chaos: the non compact case

Math Overflow Recent Questions - Mon, 04/30/2018 - 09:39

1) Is there any notion of Li-Yorke chaos for non compact (metric) spaces $X$ and non continuous transformation $f:X \rightarrow X$? Could you bring some references?

2) I mean, why are so important the compactness of the space $X$ and continuity of the function $f:X \rightarrow X$ in the Li-Yorke chaos definition?

Just to remember, a pair $x,y \in X$ is called scrabled if $\liminf d(f^{n}(x),f^{n}(y)) = 0$ and $\limsup d(f^{n}(x),f^{n}(y)) > 0$ ($d$ is the metric). A set $S \subset X$ is called scrambled if every pair $x,y \in S$ is scrambled. Finally, a system $(X,d,f)$ is Li-Yorke chaotic if there is a uncountable scrambled set $S \subset X$.

Thanks for your attention

finding values of the limit [on hold]

Math Overflow Recent Questions - Mon, 04/30/2018 - 09:02

Determine the value(s) of a such that a) limx->2 f(x) =1 where f(x) = ax+3, x≤2 and 3-x, x>2 b) limx->x/4 ((a-atan x)/(sin x – cos x)) = 1 c) lim x->0( (sqrt(x^2 +9 ) -3) / ax^2)

Continuity of disintegrations

Math Overflow Recent Questions - Mon, 04/30/2018 - 08:44

Suppose that $\pi:Y\to X$ is a continuous surjection from one compact metric space to another. Given a regular probability measure $\mu$ on $Y$ with pushforward measure $\nu:=\pi^*\mu$, it is known that there is a family $(\mu_x)_{x\in X}$ of probability measures on $Y$ such that $x\mapsto \mu_x(B)$ is Borel measurable for each Borel subset $B$ of $Y$, such that almost every $\mu_x$ lives on the fiber above $x$, and such that $\int_Y f(y)d\mu(y)=\int_{x\in X}\int_{\pi^{-1}(x)}f(y)d\mu_x(y)d\nu(x)$.

My question is: suppose in advance that $\mathcal{O}$ is some fixed nonempty open subset of $Y$. Is it true that there is a measure $\mu$ as above for which $\mu(\mathcal{O})>0$ and such that there is a disintegration as in the previous paragraph for which the map $x\mapsto \mu_x(\mathcal{O})$ is continuous?

Minimal Subset Sum Superset

Math Overflow Recent Questions - Mon, 04/30/2018 - 02:20

For a positive integer set $m \in \mathbb{N}_{+}^k$, we define the set of subset sums $$ M = \left\{ \sum_{i \in I} m_i \mid I \subseteq \{1,\dots,k\} \right\}. $$

Given a finite set of positive integers $C \subset \mathbb{N}_+$, what is the minimal $k$ and the corresponding $m$ such that $$ M \supseteq C \textrm{ ? } $$ Let's define $n = \max(C)$. It is easy to see that $k \leq \log_2 ( n +1)$ as $M = \{1,\dots,n\}$ for $m = [1, 2, 4, 8, \dots, 2^{k-1}]$. Also, a solution can be recovered by iterative elimination of $C$. For example: $$\begin{eqnarray}\require{cancel} C = [1, 2, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14] \\ [1, 2, \cancel{3}, 5, 6, 7, 8, 9, 10, 12, 13, 14] \\ [1, 2, 5, \cancel{7}, \cancel{8}, 9, 10, 12, 13, 14] \\ [1, 2, 5, 9, \cancel{10}, \cancel{12}, 13, 14] \\ [1, 2, 5, 9, 13, \cancel{14}], \end{eqnarray}$$ which is viable, but non-optimal compared to $m = [1, 2, 5, 7]$. Is this a known problem and is there an efficient method to find minimal $m$?

Unclear construction in a paper of Ornstein and Weiss

Math Overflow Recent Questions - Mon, 04/30/2018 - 02:17

I originally posted this on math.stack, but no one answered, so im posting here:
I need help understanding the following construction (Taken from the paper "Entropy and isomorphism theorems for actions of amenable groups"):

Begin by taking the direct product of two dyadic odometers. The space for this $\mathbb{Z^2}$-action can be thought of as the collection of all the different ways of organizing $\mathbb{Z^2}$, first into a tiling by
$2$ x $2$ squares, then a tiling by $4$ x $4$ squares, each well tiled by the $2$ x $2$ squares, then a tiling by $8$ x $8$ squares, etc. $\mathbb{Z^2}$ acts on this space of patterns by translations. Next to each such pattern we will associate to each element of $\mathbb{Z^2}$ a random variable taking the values $+-$ with probability $(1/2,1/2)$. These random variables will be independent, subject to the following constraints:

$(1)$ in every $2$ x $2$ square the bottom pair is equal to the top pair;

$(2)$ in every $4$ x $4$ square, the left pair of $2$ x $2$ squares equals the right pair of $2$ x $2$ squares;

$(3)$ in every $2^3$ x $2^3$ square the bottom pair of $2^2$ x $2^2$ squares equals the top pair of $2^2$ x $2^2$ squares;

$(2n)$ in every $2^{2n}$ x $2^{2n}$ square the left half of the square equals the the right half of the square;

$(2n+1)$ in every $2^{2n+1}$ x $2^{2n+1}$ square the bottom half of the square equals the top half of the square.

It is clear that along the $45^∘$ diagonals one sees independent random variables, thus the entropy of $ST$ is at least $\log 2$, while in the vertical and horizontal directions the entropy is killed because we repeat exactly, infinitely many times. Thus $h(S)= h(T)=0$ but $h(ST)≥\log2$.

I don't understand what exactly is the measure space in this construction. Is the underlying set a subset of $\mathrm{odometer}$ x $\mathrm{odometer}$ x $\{+,-\}^{\mathbb{Z^2}}$$?$ What is the measure?
Can someone help me figure it out? it's probably standard.

Here is a picture of the construction, if you prefer.

Distance function to $\Omega\subset\mathbb{R}^n$ differentiable at $y\notin\Omega$ implies $\exists$ unique closest point

Math Overflow Recent Questions - Mon, 04/30/2018 - 00:40

I am trying to show the following two statements are true:

(1) For any nonempty set $\Omega\subset\mathbb{R}^n$, the set $B$ consisting of points $y\notin\Omega$ where there is not a unique closest point $x\in\partial\Omega$ for each $y$ has $\mathcal{H}^n$-measure 0.

(2) This implies that the number of points $a\in A\subset \mathbb{R}^2$ such that $\mathcal{H}^{1}(\vec{n}(a)\cap B)>0$ is countable, where $\vec{n}(a)$ is the normal to $a\in A$ and $A$ is $C^1$ (i.e. it is locally the image of a $C^1$ function from $\mathbb{R}$ to $\mathbb{R}$).

For (1), I would like help proving only this related fact: if the distance function $f$ to $\Omega$ is differentiable at a point $y\notin\Omega$, then there exists a unique closest point $x\in\partial\Omega$ to $y$.

I thought that I could get at this by saying that the graph of $f$ has a sharp corner wherever there is more than one closest point (and hence is not differentiable there). But, I am told this is not always true.

For (2), I thought that, since we're in $\mathbb{R}^2$, if we have an uncountable number of segments with positive $\mathcal{H}^{1}$ measure, then $B$ would have to have positive $\mathcal{H}^{2}$ measure since $B$ contains these segments. I'm told this is not true either. Please assist.

Semigroups of one-dimensional diffusions

Math Overflow Recent Questions - Sun, 04/29/2018 - 21:49

I have a question about semigroup of one-dimensional diffusions.

Let $X$ be the Ornstein Uhlenbeck process on $\mathbb{R}$. It is known that the $L^2$ semigroup $\{T_t\}$ of $X$ is a compact operator on $L^{2}(\mathbb{R},m)$, where $m$ is the speed measure of $X$. $\{T_t\}$ is extended to strongly continuous contraction semigroup on $L^{p}(\mathbb{R},m)$, $1\le p<\infty$. Moreover, $\{T_t\}$ becomes a compact operator on $L^{p}(\mathbb{R},m)$ for any $1<p<\infty$. $\{T_t\}$ is not a compact operator on $L^{1}(\mathbb{R},m)$.

My question

  • Is there are nontrivial diffusion on $\mathbb{R}$ whose semigroup is compact on $L^{p}(\mathbb{R})$ for any $1 \le p \le \infty$? Consider the following differential operator: \begin{equation*} \frac{d^2}{dx^2}-x^{3}\frac{d}{dx}. \end{equation*} and the diffusion $Y$ associated with the above operator. The semigroups associated with $Y$ is compact on $L^{p}(\mathbb{R},m)$ for any $1 \le p \le \infty$?

Regularity of solutions to $-\Delta u = {\rm div} F$, $F\in L^1$

Math Overflow Recent Questions - Sun, 04/29/2018 - 20:36

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary.

  • What are the regularity results for solutions to $$ -\Delta u= {\rm div }\, F, \qquad F\in L^1(\Omega,\mathbb{R}^n)? $$ Can one find a solution in $W^{1,1}_0(\Omega)$? (I think the answer is no, but I do not know a counterexample). I know that $-\Delta u=f\in L^1$ has a unique solution in $W^{1,p}_0(\Omega)$ for all $1\leq p<n/(n-1)$, (see but now the right hand side is much worse.
  • What are the regularity results for solutions to $$ -\Delta u= {\rm div }\, F, \qquad F\in L^p(\Omega,\mathbb{R}^n), 1<p<\infty? $$ If $1<p<\infty$ we can solve the quation $-\Delta U=F$, $U\in W^{2,p}(\Omega)$. Then $u={\rm div}\, U\in W^{1,p}$ is a solution to $-\Delta u={\rm div}\, F$, but I do not know if we can take $u\in W^{1,p}_0(\Omega)$.

global estimate for biharmonic function

Math Overflow Recent Questions - Sun, 04/29/2018 - 15:32

My question is inspired by the work of Lamm and Rivière : Conservation Laws for Fourth Order Systems in Four Dimensions

Here is the setting of the problem. Let $u\in W^{2,2}(B(0,1),S^n)$, where $B(0,1)$ is the unit ball of $\mathbb{R}^4$ and for a fixed $n\geq 1$. We define the bi-energy $B$ as: $$B(u):=\int_{B(0,1)} \vert \Delta u\vert^2 \, dx $$

(or $B(u):=\int_{B(0,1)} \Vert (\Delta u)^T \Vert^2 \, dx $ , where $(\Delta u)^T$ is the projection of $\Delta u$ on $T_u S^n$).

In fact it is quite important to notice that

$$\Vert (\Delta u)^T \Vert^2= \Vert \Delta u\Vert^2 -\Vert \nabla u\Vert^4 \; (1)$$

Critical points of this equation are known as biharmonic maps, generalizing harmonic in dimension $4$. The Euler-Lagrange equation is quite complicated see the introduction of [1], but it is mostly like $$\Delta^2 u= \langle \nabla u, \nabla \Delta u\rangle + \vert \nabla^2 u\vert^2 + \vert\nabla u\vert^2 \Delta u +\vert \nabla u\vert^4 $$

In the article they prove an $\epsilon$-regularity result, that is to say if the $W^{2,2}$-norm of $u$ is small enough (in fact in the sphere case $B(u)$ is enough because of the remark above) then you control all the derivatives on $B(0,1/2)$. It is a very powerful result. But I wonder about a global one. Of course you can't expect to control anything more than $W^{2,2}$ on $B(0,1)$ but at least can you prove that $\Delta^2u$ is in $L^1$?

Here is the precise question, does there exist $\epsilon >0$ such that if $\phi\in W^{2,2}$ with $B(\phi)\leq \epsilon$ and $u\in W^{2,2}(B(0,1),S^n)$ is a minimizer of $B$ under the boundary condition $u=\phi,\partial_\nu u=\partial_\nu \phi$ on $\partial B(0,1)$, then $$\Delta^2 u \in L^1$$ or even $$\Vert \Delta u\Vert_1\leq f(\epsilon)$$ with some $f\in C(\mathbb{R}^+)$ with $f(0)=0$.

It would be very surprising to me if not since it is for free for harmonic maps since the equation is $\Delta u =\vert \nabla u\vert^2$, hence the $L^1$-norm of the Laplacian is automatically controlled by the energy.

Is it possible to derive the rules of set theory as transfers from the pure finite set world, and can we extend this further?

Math Overflow Recent Questions - Sat, 04/28/2018 - 03:17

Informally the idea of this question is about whether the rules of set theory can be derived as a transfer of some rules from the hereditarily finite set realm, and whether this transfer principle itself can be coined for notions other than the "finite" notion?

The principle I want to negotiate is: "if $\phi$ is a property that is definable by a formula in the language of set theory that is strictly shorter than the shortest parameter free formula in that language that can define 'finiteness', then if $\phi$ is CLOSED on the the hereditarily finite set world, then it can be generalized over the whole realm of sets"!

The crude informal idea is that if a property that cannot mention finiteness generalizes over the whole hereditarily finite set realm, then it can go beyond it.

To formally capture that, I'll work up in a class theory, so we define "set" as an element of a class, the language of the theory is mono-sorted first order logic with identity and membership, we stipulate axioms of:

  1. Extensionality: as in ZF
  2. Class comprehension schema: $\forall x_1,..,x_n\ \exists x \ [x=\{y|set(y) \wedge \phi(y,x_1,..,x_n)\}] $
  3. The empty class is a set
  4. Singletons: $\forall x [ set(x) \to set(\{x\})]$
  5. Boolean Union: $\forall x,y [set(x) \wedge set(y) \to set (x \cup y)]$

Define: $$ fin(A) \iff \forall K (\exists o (o \in K \wedge \neg \exists m (m \in o)) \wedge \forall x (x \in A \to \exists y (y \in K \wedge x \in y \wedge \forall z (z \in y \to z=x))) \wedge \\ \forall a \forall b (a \in K \wedge b \in K \to \exists c (c \in K \wedge \forall d (d \in c\leftrightarrow d \in a \lor d \in b ))) \\ \to A \in K)$$

In English: $A$ is finite if and only if it is an element of every class $K$ that contains the empty set among its elements, is closed under Boolean union and that has the singletons of all elements of $A$ among its elements.

"I think this is along the shortest way to define "finite set" in the first order language of set theory"

Perhaps the above formula can be shortened further, or perhaps there is another shorter parameter free formulation of "x is a finite set" in the language of set theory, however for the sake of presentation here we'll take this formula to be the shortest formula defining finiteness.

  1. $ \forall x (x \text{ is hereditarily finite } \to set(x))$

Where "x is hereditarily finite" is defined as the transitive closure class of x being finite.

We shall denote the class of all sets by $V$, and the class of all hereditarily fintie sets by $HF$

  1. $HF \in V$

  2. The principle of Transfer from the pure finite world: if $\phi(y,x)$ is a formula shorter than any formula defining finiteness, in which only symbols $``y,x"$ occur free, and those only occur free, then:

$\forall x [x \in HF \to \forall y (\phi(y,x) \to y \in HF)] \to \forall x \in V [ \forall y (\phi(y,x) \to y \in V)]$

Now it is clear that all axioms of Union, Power and Separation over sets are derivable from the above transfer principle, and so $ZC$ is interpretable here. Actaully if we restrict $\phi$ to have no more than three atomic subformulas, we can still interpret the whole of $ZC$. Replacement is not interpretable by this principle. Yet, a minor modification of this principle might succeed in proving replacement over sets, this can be done by changing the closure property to involve only subsets of $HF$, what I call as "proximity closure over HF", so to restate that:

8'.The principle of Transfer from proximity of the pure finite world: if $\phi(y,x)$ is a formula shorter than any formula defining finiteness, in which only symbols $``y,x"$ occur free, and those only occur free, then:

$\forall x [x \in HF \to \forall y \subseteq HF (\phi(y,x) \to y \in HF)] \to \forall x \in V [ \forall y (\phi(y,x) \to y \in V)]$

That replacement is provable can be shown from examining the following formula whose length is shorter than any formula defining finiteness.

$\exists F [\forall m (m \in F \to \exists a,b (a \in A \wedge b \in B \wedge a \in m \wedge b \in m)) \wedge \\ \forall m,n (m \in F \wedge n \in F \wedge \exists k(k \in m \wedge k \in n) \to n=m) \wedge \\ \forall b (b \in B \to \exists m (m \in F \wedge b \in m))]$

Now if $A$ is hereditarily finite and $B$ is a subset of $HF$ that fulfills the above formula, then $B$ is hereditarily finite, this mean that the property defined by the above formula is "proximity closed over the hereditarily finite world"

I don't have any proof of consistency of these principles, but if there is no clear inconsistency of those relative to $ZF$ or $MK$ or some extension of those, then could it be possible to think of extending that principle to properties other than "x is finite"? so we generalize it to some line properties, so for a property $P$ in that line, we stipulate that:

any predicate $Q$ that is closed over the pure $P$ world, would generalize over the whole set world,

or even stronger:

any predicate $Q$ that is proximity closed over the pure $P$ world, would generalize over the whole set world

Of course in both cases $Q$ must be expressible by a formula strictly shorter than the shortest expression defining property $P$, and also we stipulate parallel axioms sufficient to define the property $P$, also axioms asserting the element-hood of all hereditarily $P$ classes, and the existence of a set of all hereditarily $P$ sets. Of course this can only be done for some selected line of properties.

is that possible or it is involved with clear inconsistencies? and what would be the general qualification of such property $P$?

Iterations of proper forcings defined in an inner model

Math Overflow Recent Questions - Fri, 04/27/2018 - 14:17

Are there any examples where a countable support iteration of proper forcing defined in an inner model does crazy things in an outer model? This is vague, so to narrow it down, are there examples of $V\supset M$ and a countable support iteration of proper forcing $\langle P_i, \dot{Q}_j: i\leq \omega, j<\omega\rangle$ defined in $M$ such that for each $j\in \omega$, $P_j$ is proper in $V$, but $P_\omega$ collapses $\omega_1$ in $V$. Of course, when $V$ and $M$ are sufficiently close (for example $V\models M^\kappa\subset M$ for large enough $\kappa$), the iteration will look the same. Since we are taking countable support iteration, I'll impose that (at least) $V\models M^\omega\subset M$.

Another test question:

Suppose $V\models M^\omega \subset M$. Is it true that countable support iteration of Cohen forcing defined in $M$ is equivalent to the one defined in $V$?

A question on Hilbert polynomial and flat morphisms

Math Overflow Recent Questions - Fri, 04/27/2018 - 13:09

Note: this question was previously asked at math.stackexchange (under the same title) to no avail.

I have been recently reading Dr. Kaledin's notes on algebraic geometry. There is a statement in lecture 16 about which I feel confused.

Оказывается, что для пучков на проективном пространстве, полином Гильберта это единственный существенно дискретный инвариант: как только он зафиксирован, можно построить такое плоское семейство над конечномерной нётеровой базой $Y$, что любой пучок с данным $P(\mathcal{F},l)$ появляется в нем как слой, причем только один раз.

It can be translated as "It turns out that for sheaves on projective space Hilbert polynomial is the only essentially discrete invariant: once it is fixed, it is possible to construct a flat family over a finite-dimensional Noetherian base Y such that any sheaf with given $P(\mathcal{F}, l)$ appears in it as fiber exactly once".

I struggle to translate this remark into a precise mathematical statement. In particular, I don't understand what does 'exactly once' mean; if we have two points $y_1, y_2 \in Y$, how can we compare sheaves on $f^{-1}(y_1)$ and $f^{-1}(y_2)$? Can someone provide me the precise statement that Kaledin probably had in mind?

Graph-theoretic Algorithm for Path with Minimum Average Edge Length

Math Overflow Recent Questions - Fri, 04/27/2018 - 12:34

I am looking for a graph-theoretic algorithm, that determines among all simple paths $\mathcal{P}_{ab}$ that connect vertex $a$ with vertex $b$ the one, that has minimal average edge-length, i.e. $$\ p_{ab}\in\mathcal{P}_{ab}: \frac{\ell(p_{ab})}{card(p_{ab})}\ \le\ \frac{\ell(q_{ab})}{card(q_{ab})}\ \forall q_{ab}\in\mathcal{P}_{ab}$$

By "graph-theoretic" I mean algorithms in the vein of those that are likely to be found in publications about algorithmic graph-theory; on the contrary, mathematical programming is not what I am looking for.

Schubert variety containment and strong versus weak Bruhat order

Math Overflow Recent Questions - Fri, 04/27/2018 - 12:09

First consider Schubert cells in the full flag variety $Fl_n$. Then for a permutation $w\in S_n$, the poset of Schubert varieties contained in the Schubert variety corresponding to $w$ is isomorphic to the interval $[e,w]$ in the strong Bruhat order.

Now consider Schubert cells in the Grassmannian $Gr(k,n)$. Then for a Grassmannian permutation $w\in S_n$, the poset of Schubert varieties contained in the Schubert variety corresponding to $w$ is isomorphic to the dual $[e,w]^*$ of the interval $[e,w]$ in the weak Bruhat order.

Is there some intuitive explanation for this duality between strong and weak Bruhat orders?

Closure in the strong dual topology

Math Overflow Recent Questions - Fri, 04/27/2018 - 11:29

Originally asked on MSE.

Let $E$ be a metrizable locally convex topological vector space and let $E^{*}$ be its dual space endowed with the strong topology = topology of uniform convergence on (closed convex balanced) bounded subsets of $E$. Let $F\subset E^{*}$.

Is it true that $\overline{F}=F_1$, where $F_1$ is the set of all linear functionals on $E$, whose restrictions on every (closed convex balanced) bounded subset of $E$ are continuous in the weak $\sigma(E,F)$ topology?

It seems that I can adapt the proof of Grothendieck's completion theorem to prove this. Indeed, every linear functional continuous on a closed set $B$ in $E$ can be uniformly approximated on $B$ by an element of $E^{*}$ (I guess this is Grothendieck's lemma). Hence, we only need to show that $F_1\subset E^{*}$. But this follows from the fact, that since $E$ is metrizable, $E^{*}$ is complete, and so it is a closed subset of $E'$ (the algebraic dual), andowed with the uniformity of uniform convergence on bounded sets.

If this is correct, I hope this result is contained in some textbook on locally convex spaces, and so a reference would be highly appreciated.


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