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The answer is yes for separable Hilbert spaces. If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\ …

Using the $\Delta^2- \epsilon \Delta$ approach, Tim Netzer and I have verified Kazhdan's property (T) for ${\rm SL}(3,\mathbb Z)$. For the standard generators $e_{ij}$ ($i\neq j$) we can show a spectr …

Let $G$ be a group and set $[x,y]:= x^{-1}y^{-1}xy$ as usual and consider it as a binary operation. Question: Is there a description of the identities that the operation $[.,.]$ satisfies for all …

It is an old result of Schützenberger that in a free group, a basic commutator cannot be a proper power. A look at the original reference M.-P. Schützenberger, Sur l'équation $a^{2+n} = b^{2+m}c^{2+p …

Consider $G=\mathbb Z$ and $\chi = \delta_{-1} + \delta_1 \in \mathbb C[\mathbb Z]$. Then $\chi=\chi^*$. One can check that $$(\chi^{2n})_k= \binom{2n}{n+k}.$$ Hence, with your definition (and the rem …

One more answer: The atomless Boolean algebra is just the algebra of clopen subsets of $\{0,1\}^{\mathbb N}$, now pick any discrete embedding $F_2 \subset {\rm Sym}(\mathbb N)$ - for example letting $ …

If $n:=[G:H]$, then $\mathbb C[G] \subset M_n \mathbb C[H]$, where $g \in G$ maps to a permutation matrix decorated with elements from $H$ and the embedding depends essentially only on a choice of a t …

The action of a group $G$ on a set $X$ is called oligomorphic if the diagonal action on $X^n$ has finitely many orbits for each $n$. Question: Is there an infinite (maybe even finitely generated) …

Let $n \in \mathbb N$ and $\{\sigma,\tau\} \subset {\rm Sym}(n)$ be a generating set. Question: What is the maximal possible girth (if one varies $\sigma, \tau$) of the associated Cayley graph? …

If $G$ is amenable, then every weakly mixing representation $\pi$ has almost invariant finite-dimensional subspaces. This means that $\pi \otimes \bar \pi$ has almost invariant vectors. Results like …

The following theorem was proved in [Helmut Wielandt. Eine Verallgemeinerung der invarianten Untergruppen. Mathematische Zeitschrift 45 (1939): 209-244.] a long time ago: Theorem: (Wielandt) Ther …

There is no continuous surjection from $\hat Z[[t]]$ to $(\mathbb Z/3\mathbb Z)[\mathbb Z/2\mathbb Z] = \mathbb Z/3\mathbb Z \oplus \mathbb Z/3\mathbb Z$.

Let $$ \cdots \to \Gamma_n \to \Gamma_{n-1} \to \cdots \to \Gamma_0$$ be an inverse system countable groups and let's assume (for this post) that all homomorphisms in such an inverse system are surjec …

Question: Does the unitary group $U(\ell^2 \mathbb N)$, equipped with the strong operator topology, contain a countable dense subgroup which is amenable as a discrete group? I would be also inter …

For abelian groups, you need to find a number $k$ with the property that at least one entry of each $a \in A$ is not divisible by $k$. In the simplest case, when $n=1$ and $A=\{m\}$ a prime of size ro …