Connes' embedding conjecture for uncountable groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:51:59Z http://mathoverflow.net/feeds/question/80220 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80220/connes-embedding-conjecture-for-uncountable-groups Connes' embedding conjecture for uncountable groups Valerio Capraro 2011-11-06T15:56:06Z 2011-11-07T15:19:57Z <p>In this topic, I will use the word <em>uncountable group</em> referring to groups whose cardinality is $\leq|\mathbb R|$.</p> <p><strong>Notation:</strong> $R$ is the hyperfinite $II_1$-factor, $\omega$ is a free ultrafilter on the natural numbers, $R^\omega$ is the tracial ultrapower, $\tau$ is the unique normalized trace on $R^\omega$, $U(R^\omega)$ is the unitary group of $R^\omega$. </p> <p><strong>Definition:</strong> A group $G$ is called hyperlinear if there is a group monomorphism $\theta:G\rightarrow U(R^\omega)$ such that $\tau(\theta(g))=0$ for all $g\neq1$.</p> <blockquote> <p><strong>Question 1:</strong> Does there exist an uncountable non-hyperlinear group?</p> </blockquote> <p>A bit of background: the same question for countable groups is known as Connes' embedding problem for groups and it's still unsolved. When I began my PhD (Nov. 2008), my former advisor Florin Radulescu told L. Paunescu and myself that he would have liked to have a better understanding of the problem in higher cardinality. In particular, he asked us to see whether the free group on uncountable many generators, here denoted by $\mathbb F_c$, and the circle group $S_1$ were hyperlinear. Maybe he expected that one of them was not, but we came up with the general result that every subrgroup of $U(R^\omega)$ is hyperlinear (this is basically proved in <a href="http://arxiv.org/abs/0911.4978" rel="nofollow">http://arxiv.org/abs/0911.4978</a>); in particular $S_1$ is hyperlinear and, with an additional 10-line argument, also $\mathbb F_c$ turns out to be hyperlinear. Moreover, the result is quite general and excludes many possible example a priori, making Question 1, in my modest opinion, non-trivial and interesting. At some point, maybe also talking with someone else (but I don't remember exactly who), I got quite convinced that Question 1 has the same level of difficulty of Connes' problem and that they may be actually equivalent. In this case, I would like to find a formal way to express that. In this view, a positive answer to the following question would not completely solve the problem, but would be a nice starting point.</p> <blockquote> <p><strong>Question 2:</strong> Let $G$ be an uncountable group. Do there always exist countable groups $G_1,G_2,\ldots$ and a free ultrafilter $\omega$ such that $G$ embeds into the algebraic ultraproduct of the $G_i$'s?</p> </blockquote> <p><strong>Update:</strong> as shown my Simon Thomas below, the answer is positive assuming CH. On the other hand, Joel's answer shows that without CH we have some weaker result. For instance, Question 2 has affirmative answer if we allow the sequence $G_i$ to be indexed by a possibly non-countable set $I$. </p> <p>Thanks in advance,</p> <p>Valerio</p> http://mathoverflow.net/questions/80220/connes-embedding-conjecture-for-uncountable-groups/80249#80249 Answer by Simon Thomas for Connes' embedding conjecture for uncountable groups Simon Thomas 2011-11-06T21:56:50Z 2011-11-07T15:19:57Z <p>A partial answer to Alain's question ... Suppose that the Continuum Hypothesis $CH$ holds. Let $G$ be any group of size $2^{\aleph_{0}} = \aleph_{1}$. Then there exists a countable subgroup $H$ of $G$ with the same first order theory $T$. Let $\Gamma = \prod_{\mathcal{U}} H_{n}$ be the ultraproduct such that every $H_{n} = H$. Then $CH$ implies that $\Gamma$ is a saturated model of $T$ and hence $G$ embeds into $\Gamma$.</p> <p>Your question concerning the possibility of embedding groups $G$ with $|G| \leq 2^{\aleph_{0}}$ into ultraproducts $\prod_{\mathcal{U}} H_{n}$ over a countable index set might be interesting in the case when $CH$ fails. There is a similar open problem concerning sofic groups. It is an easily seen that every group is sofic iff $Sym(\mathbb{N})$ embeds in some universal sofic group. However, even if every group is sofic, it is not clear whether or not $Sym(\mathbb{N})$ embeds in a universal sofic group arising from an ultrafilter over $\mathbb{N}$.</p> <p>If you are interested in embeddings using ultraproducts over larger index sets, then you can make use of an ancient result of Malcev, which says that every group $G$ embeds into a suitable ultraproduct of its finitely generated subgroups. Here the index set $I$ is the set of its finitely generated subgroups, which is uncountable if $G$ is uncountable. </p> <p>PS: In Shelah's paper on ultraproducts, he mentions that if Martin's Axiom holds, then there exists an ultrafilter $\mathcal{D}$ over $\mathbb{N}$ such that for every countable structure $M$ for a countable language, the corresponding ultraproduct $\prod_{\mathcal{D}}M$ is saturated. In particular, it is consistent with the failure of $CH$ that every group $G$ with $|G| \leq 2^{\aleph_{0}}$ embeds into an ultraproduct $\prod_{\mathcal{U}} H_{n}$ over a countable index set. </p> http://mathoverflow.net/questions/80220/connes-embedding-conjecture-for-uncountable-groups/80283#80283 Answer by Joel David Hamkins for Connes' embedding conjecture for uncountable groups Joel David Hamkins 2011-11-07T09:22:24Z 2011-11-07T09:48:20Z <p>The general situation, where CH fails, may be informed by the <a href="http://eom.springer.de/k/k110060.htm" rel="nofollow">Keisler-Shelah isomorphism theorem</a>, which asserts that two first-order structures have isomorphic ultrapowers if and only if they have the same first-order theory.</p> <p>In particular, for any infinite group $G$ at all, of any size, we may take a countable elementary subgroup $H$, meaning in particular that they have the same first-order theory, and so there is a nonprincipal ultrafilter $U$ on an index set $I$ such that the ultrapowers $G^I/U\cong H^I/U$ are isomorphic. Since every first-order structure maps elementarily into its ultrapowers, this means in particular that $G$ maps elementarily (and hence monomorphically) into an ultrapower of $H$, a countable group.</p> <p>Thus, this fully answers the version of question 2 in which we allow the ultrafilter to live on a bigger index set: </p> <p><b>Theorem.</b> For every group $G$ there is a countable group $H$ and a free ultrafilter $U$ on a set, such that $G$ embeds into the ultrapower $H^I/U$. </p> <p>If you want to insist that the ultrafilter concentrate on index set $\mathbb{N}$, however, then things become more complicated. If the CH holds, then the Keisler-Shelah theorem shows that any two groups of size at most $2^{\aleph_0}$ and with the same theory have isomorphic ultrapowers by an ultrafilter on $\aleph_0$, and so the desired result is attained. In the non-CH case, however, what we seem to get is that for any cardinal $\lambda$, if $\beta$ is smallest such that $\lambda^\beta\gt\lambda$, then any two groups of size $\beta$ with the same theory have isomorphic utrapowers using an ultrafilter on $\lambda$. Thus, they each map into an ultrapower of the other.</p> <p>The Keisler-Shelah theorem was proved first by Keisler in the case that GCH holds, using saturation ideas as in Simon's answer. The need for the GCH was later removed by Shelah.</p>