Is the Clifford group superperfect?

Question

This is a follow up to this question. Since global phases are unphysical in quantum mechanics we often consider projective representations, where the matrices are defined only modulo a global $ e^{i\theta} $, instead of true linear representations. It turns out that the projective representations of $ G $ correspond exactly to the linear representations of the universal cover which is a central extension of the original group. For example $ SO_3 $ has universal cover $ SU_2 $ and projective representations of $ SO_3 $ correspond to half-integer spin in quantum mechanics.

• This story for semisimple Lie groups has an analogue in the theory of perfect finite groups. For a perfect group $ G $ there is a universal central extension, sometimes called the universal cover, with the property that the projective representations of $ G $ are in exact correspondence with the linear representations of the universal cover.
• In the theory of semisimple Lie groups, a group which is its own universal cover is called simply connected. This is equivalent to the fundamental group of $ G $ being trivial. For a semisimple Lie group the fundamental group is always a finite Abelian group.
• In the theory of perfect groups a group which is its own universal cover is called superperfect, this is equivalent to the Schur multiplier being trivial. For a perfect finite group the Schur multiplier is always a finite Abelian group.

The Clifford group $ \overline{Cl}_n$ (the automorphism group of the Pauli group $ P_n $ ) is a perfect group which is important in quantum computing. I want to know if $ \overline{Cl}_n$ is superperfect or if the there exists some nontrivial perfect central extensions (i.e. the Schur multiplier is nontrivial). If the Schur multiplier is nontrivial I would certainly be curious which finite Abelian group it is.

Answer 1

The answer is no; consider, for example, $ \overline{Cl_3(2)} $ which is not superperfect. Its Schur multiplier is cyclic 2 times cyclic 4 $ C_2 \times C_4 $. Recall that every $ \overline{Cl_n(p)} $ is perfect with the exception of $$ \overline{Cl_1(2)},\overline{Cl_1(3)},\overline{Cl_2(2)}. $$ $ \overline{Cl_1(2)} \cong S_4 $ is solvable as is the Hessian group $ \overline{Cl_1(3)} $. $ \overline{Cl_2(2)} $ has a perfect subgroup of index 2 (its commutator subgroup).

In general it seems that the Schur multiplier of $ \overline{Cl_n(p)} $ is size $ p^n $ (Schur multiplier is always a finite Abelian group, so there are a limited number of Abelian groups of size $ p^n $).

It seems that (excluding $ p=2,3 $ which are solvable) $ \overline{Cl_n(p)} $ is perfect with Schur multiplier cyclic $ p $, $ C_p $, and moreover the universal cover of $ \overline{Cl_n(p)} $ for $ p \neq 2,3 $ is the determinant one subgroup of the standard single qupit Clifford group $ S(Cl_1(p)) $, which is superperfect. For example $$ \overline{Cl_1(5)} \cong {\rm PerfectGroup}(3000,1) $$ is perfect and has Schur multiplier $ C_5 $ and $$ S(Cl_1(5)) \cong {\rm PerfectGroup}(15000,3) $$ is superperfect (i.e. perfect and trivial Schur multiplier). These are GAP IDs.

My guess is that only for single qupit cliffords is the determinant 1 subgroup ever perfect or super perfect. For example $ S(Cl_3(2)) $ is not perfect. My guess also is that the Schur multiplier of $ \overline{Cl_n(p)} $ is never cyclic for multi-qubit Cliffords ( meaning $ n \geq 2 $)