12
$\begingroup$

A compact topological group $G$ is called Van der Waerden if each homomorphism $h:G\to K$ to a compact topological group is continuous. By a classical result of Van der Waerden (1933) the groups $SO(2n+1,\mathbb R)$ are Van der Waerden, i.e., admit no discontinuous homomorphisms to compact topological groups.

On the other hand, Kallman proved that for every $n\in\mathbb N$ the Lie group $GL(n,\mathbb R)$ admits an injective homomorphism to the symmetric group $Sym(\mathbb N)$.

Problem. Does each infinite compact topological group $K$ admit a discontinuous homomorphism $h:K\to P$ to a Polish group $P$?

The answer to this problem is affirmative in three cases:

1) $K$ is solvable. In this case $K$ admits a homomorphism into an infinite abelian-by-finite group in which it is easy to construct a normal subgroup of countable index and the quotient homomorphism by this subgroup will be a discontinuous homomorphism to a countable discrete (and hence Polish) group;

2) $K$ is non-metrizable. In this case $K$ is not Wan der Waerden (see page 412 in the paper of Comfort) and hence $K$ admits a discontinuous homomorphism to some compact Lie group;

3) $K$ is not zero-dimensional. In this case it is possible to construct a continuous homomorphism $h:K\to O(n)\subset SL(n,\mathbb R)$ for some $n$ whose image $h(K)$ is not finite and being a Lie group is not zero-dimensional. Then the Kallman injective homomorphism $\varphi:k(K)\to Sym(\mathbb N)$ is necessarily discontinuous and so is the homomorphism $\varphi\circ h:K\to Sym(\mathbb N)$.

Those partial answers reduce the problem to the following its partial case.

Problem 0. Let $K$ be an infinite compact metrizable zero-dimensional topological group. Does $K$ admit a dicontinuous homomorphism to a Polish group?

Remark. If a compact topological group $K$ has no discontinuous homomorphism to a finite group, then for every $n\in\mathbb N$ it has only finitely many subgroups of index $n$. This can be shown using the limits by ultrafilters (see p.9 of https://arxiv.org/pdf/1108.5130.pdf). An example of a compact group with finitely many subgroups of a given finite index is the product $\prod_{n\ge 5}A_n$.

Question. Let $(k_n)_{n=5}^\infty$ be a sequence of natural numbers. Does the compact group $\prod_{n\ge 5}A_n^{k_n}$ admit a discontinuous homomorphism to a Polish group?

As was pointed out by @YCor, the Question is a partial case of a more general Conjecture 6.4 of Thomas and Zapletal.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I added the tag profinite-groups, since it's the same as being compact zero-dimensional. $\endgroup$
    – YCor
    Commented May 6, 2017 at 8:43
  • 1
    $\begingroup$ Automatic continuity (of homomorphisms into Polish groups) for infinite products of finite simple groups is discussed in this paper by Thomas-Zapletal (math.univ-lyon1.fr/~confluentes/CM/04/0402/Thomas.pdf). They say that no such group is known to satisfy automatic continuity, but that indeed products of growing alternating groups are plausible examples. $\endgroup$
    – YCor
    Commented May 6, 2017 at 8:51
  • 1
    $\begingroup$ For a compact group, an obvious obstruction to Automatic Continuity is the existence of a subgroup of infinite countable index. If $(S_n)$ is a sequence of non-abelian groups, it's a result of Simon Thomas (J. Group Theory 1999) that $\prod S_n$ has no non-closed subgroup of countable index iff the rank of $S_n$ tends to $\infty$ (rank tends to $\infty$ is in the classical sense, described for instance here: mathoverflow.net/a/163690/14094). $\endgroup$
    – YCor
    Commented May 6, 2017 at 9:05

0

Reset to default

You must log in to answer this question.