Recent MathOverflow Questions

Minimum Simple Burger-Mozes Type Group

Math Overflow Recent Questions - Fri, 03/01/2019 - 07:21

Burger and Mozes constructed (Burger and Mozes - Lattices in products of trees) infinite, finitely presented, torsion-free simple groups which split as amalgams of two finitely generated free groups over a finite index subgroup embedded by specific embeddings obtained through studying actions on product of regular trees. It is an interesting problem to make this amalgam as small as possible. For example, $F_{217} ∗_{F_{75601}} F_{217}$ is the smallest example constructed by Burger and Mozes; this was later improved to $F_9 \ast_{F_{81}} F_9$ by Rattaggi, and very recently to $F_5 \ast_{F_{25}} F_5$ by Caprace-Radu. These examples are crucially based on non-abelian finite simple groups, and in particular, the fact that $A_6$ is a non-abelian finite simple group such that the stabilizer of one letter in the action on $\{1, \dots, 6\}$ is isomorphic to another non-abelian finite simple group, namely $A_5$. For more details on this, see Rattaggi's excellent compendium.

My question, then, concerns the following: is this the smallest (with respect to some ordering on the ranks of the factors and the amalgamated subgroup) possible example of such a Burger-Mozes amalgam? It is clearly (?) not doable using only the original techniques of Burger and Mozes, as $A_5$ is of course the smallest non-abelian finite simple group.

I am also interested in any extensions of the original arguments to more general settings, e.g. by extending their normal subgroup theorem to a broader class of lattices.

Note that the ordering chosen on the factors and the amalgamated subgroup may have some relevance. Furthermore, some amalgams are isomorphic; the original Burger-Mozes example, for example, satisfies $F_{217} ∗_{F_{75601}} F_{217} \cong F_{349} ∗_{F_{75865}} F_{349}$.

Bertrand's Paradox (probabilities) [on hold]

Math Overflow Recent Questions - Fri, 03/01/2019 - 04:55

Bertrand's paradox in probability has been declared insoluble but the difference between 1/2 ad 1/4 has a very simple reason: 1/2 results from selecting random points on a Polar system and 1/4 in a Cartesian system. A deeper reason is that both systems act different at differential levels. Are these statements correct?

Details are at YouTube’s “Bertrand´s Paradox.(probabilities) Solution. Big mistery good by. 7/7”.

Compact operators on Banach spaces and their spectra

Math Overflow Recent Questions - Fri, 03/01/2019 - 02:10

I have a question about compact operators on Banach spaces.

Let $B$ be a real Banach space and $L$ a closed linear operator on $B$. We assume that $L$ generates a contraction semigroup $\{T_t\}_{t>0}$ on $B$ .

If $B$ is a Hilbert space and $L$ is self-adjoint, the following assertions are equivalent:

(1) The spectrum of $L$ is discrete (the essential spectrum $\sigma_{ess}(L)=\emptyset$).

(2) $T_t$ is compact for any $t>0$.

(3) $T_t$ is compact for some $t>0$.

(4) $R_{\lambda}:=(\lambda-L)^{-1}$ is compact for any $\lambda \in \rho(L)$.

(5) $R_{\lambda}$ is compact for some $\lambda \in \rho(L)$.

Here, $\rho(L)$ is the resolvent set of $L$.

Even if $B$ is not a Hilbert space, (2)$\Rightarrow$(4), (4)$\Leftrightarrow$(5), (5)$\Rightarrow$(1).

My question

In what follows, we further assume that $\{T_t\}_{t>0}$ is strongly continuous and $B$ is a $L^1$ space on a measure space.

Does (1)$\Rightarrow$(5) hold? or

Under what conditions, does (1)$\Rightarrow$(5) hold?

By the way, I am particularly interested in situations where $\{T_t\}_{t>0}$ is generated by a symmetric Markov process on a locally compact metric measure space $(X,\mu)$. In this case, for each $1\le p <\infty$, $\{T_t\}_{t>0}$ is extended to a strongly continuous contraction semigroup $\{T_t^p\}_{t>0}$ on $L^{p}(X,\mu)$ and it holds that $T_t^p f=T_tf$ for any $t>0$ and $f \in L^{1}(X,\mu) \cap L^{p}(X,\mu)$.

Higher $\infty$-categories

Math Overflow Recent Questions - Thu, 02/28/2019 - 15:16

Is there a reason we consider $\infty$-categories to be the $\omega^{th}$ step in the 2-internalization inside Cat (or enrichment over Cat if you prefer)* process made invertible above some finite ordinal, and don't continue on to higher steps in the recursion? Is there nothing to be gained, or is the $\omega^{th}$ step already mysterious enough that going further is foolhardy?

For example, it seems (very naively) that something like a $(\omega_1,\omega)$-category or higher categories defined up to large cardinals that become invertible at smaller large cardinals might be interesting, or in a $\neg CH$ universe we could ask about $(\omega_1,\mathfrak{c})$-categories and the like. My apologies if this question is trivial, but I couldn't find a discussion/explanation in the literature.

*This is an incorrect characterization of how to arrive at a 'fully weak' $\infty$-category (thanks Mike for catching the error), and it appears as though it's an open question wether we can give an algebraic definition of a fully weak $\infty$-category. For details on how to correctly iterate internalization to arrive at a correct definition for weak $n$-categories for all $n$, see this excellent paper by Simona Paoli.

Limitations of the splitting construction and SFT

Math Overflow Recent Questions - Thu, 02/28/2019 - 09:39

I am trying to understand the so-called symplectic field theory (SFT) machinery used in symplectic topology. As I understand it, one of the applications of SFT (or rather, of the splitting construction) is to the questions of Lagrangian isotopy, as exemplified by these papers. Splitting construction allows one to move, in a Hamiltonian fashion, Lagrangian submanifolds to a complement of certain divisor; in the complement of the divisor, the problem typically becomes much easier.

My question is: what are some limitations of SFT or of the splitting construction? Is there an example of a symplectic topological problem where one might naively expect them to be useful but they turn out not to be?


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