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Aug 1, 2014 at 19:43 comment added jjcale A larger finite subgroup than the Weyl-Heisenberg group is its normalizer, see e.g. arxiv.org/abs/1003.3591v2 .
Jul 19, 2012 at 10:18 comment added Felino Indeed I am expecting that, generically, the Cayley graph of a discrete subgroup will be an expander graph. But maybe this just happens for certain dimensions of the lattice. One way of rephrasing the question would be: take a discrete subgroup of SU(N) at random, Has it property T? Is it expanding? Do we need large N for this to happen?
Jul 17, 2012 at 13:57 comment added Yoav Kallus Why do you expect any peculiarities? One subgroup I know whose order is $O(N^2)$ is the Weyl-Heisenberg group generated by the two maps $|k\rangle\mapsto |k+1\rangle$ and $|k\rangle\mapsto e^{2\pi k/N}|k\rangle$, where $|k\rangle$, $k=0,\ldots, N-1 \pmod{N}$, is a orthonormal basis of $\mathbb C^N$.
Jul 16, 2012 at 13:53 history asked Felino CC BY-SA 3.0