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This is a follow up to Is the Clifford group perfect (equals its own commutator subgroup)?

Motivation:

Since global phase is unphysical in quantum mechanics we often consider projective representations, where the matrices are only well defined 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.

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1 Answer 1

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No.

For example $ \overline{Cl_3(2)} $ is not superperfect. Its Schur multiplier is cyclic 2 times cyclic 4 $$ C_2 \times C_4 $$

Recall from

Is the Clifford group perfect (equals its own commutator subgroup)?

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 the determinant one subgroup of the single qupit Clifford group $$ S(Cl_1(p)) $$ is always superperfect with Schur multiplier cyclic $ p $, $ C_p $ (with the exception of $ p=2,3 $ for which $ S(Cl_1(p)) $ and $ \overline{Cl_1(p)} $ are not even perfect). For example $$ \overline{Cl_1(5)} \cong PerfectGroup(3000,1) $$ and $$ S(Cl_1(5)) \cong PerfectGroup(15000,3) $$ which is superperfect. 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 $)

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