Is the Clifford group perfect?

Question

Let $ {\rm Cl}_n $ be the Clifford group on $n$ qubits. What is the commutator subgroup of $ {\rm Cl}_n $? Respectively, is the Clifford group perfect, that is, does ${\rm Cl}_n$ equal its own commutator subgroup?

My guess is that the Abelianization of $ {\rm Cl}_n $ is an elementary Abelian $ 2 $-group of rank $ 2n $, in other words the vector space $ \mathbb{F}^{2n} $. So I would imagine the commutator subgroup is something like the symplectic group $ Sp_{2n}(2) $ since $ {\rm Cl}_n $ is roughly made of two parts: a symplectic part and a $ 2 $-group part. That's kind of a vague statement but an example of what I mean is that quotienting the Clifford group by the Pauli group gives the symplectic group.

Answer 1

$$ \newcommand{\Sp}{\mathrm{Sp}} \newcommand{\Cl}{\mathrm{Cl}} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} $$ Note that the symplectic group $\Sp_{2n}(2)$ is not a subgroup of $\Cl_n(2)$ (see Is the Clifford group a semidirect product?).

Claim: Suppose $p$ prime and $n\in\mathbb N$ are such that $\Sp_{2n}(p)$ is perfect. Then the projective Clifford group $\overline\Cl_n(p)$ is perfect.

Note that $\Sp_{2n}(p)$ is perfect except for $(n,p)\in\{ (1,2), (1,3), (2,2) \}$.

Proof: The situation is arguably simpler when the local prime dimension $p$ is not $2$. Then the Clifford group is the semidirect product $\Cl_n(p) \simeq \mathcal{P}_n(p) \rtimes \Sp_{2n}(p)$ where $\mathcal{P}_n(p)$ is the generalized Pauli group. This is not the case for $p=2$, but the argumentation still works in this case.

1. We always have that $\mathcal{P}_n(p)$ is a normal subgroup and $\Cl_n(p) / \mathcal{P}_n(p) \simeq \Sp_{2n}(p)$. In terms of cosets, we have for any $U,V\in\Cl_n(p)$: $$ [U\mathcal{P}_n(p), V \mathcal{P}_n(p)] = [U,V] \mathcal{P}_n(p). $$ Note that we can alternatively write the cosets as $C_g$, labelled by elements $g\in\Sp_{2n}(p)$. The isomorphism above then implies $[C_g,C_h]=C_{[g,h]}$.

2. Up to a phase, we can write any element of $\mathcal{P}_n(p)$ as $w(a)$ where $a\in\F_p^{2n}$. Any $U\in C_g$ acts as $U w(a) U^\dagger \propto w(g(a))$. Then: $$ [U,w(a)] = (U^{-1} w(a)^{-1} U) w(a) \propto w(g(a))^{-1} w(a). $$

If $\Sp_{2n}(p)$ is perfect, the commutators $[U,V]$ for $U,V\in\Cl_n(p)$ generate an arbitrary Clifford unitary, up to a Pauli operator (and a phase) by 1), i.e. we can write any $C\in\Cl_n(p)$ as $C = \alpha w(b) [U,V]$ with $\alpha\in Z(\Cl_n(p))$. If $b=0$, we're done, hence let us assume that $b\neq 0$. Since $\Sp_{2n}(p)$ acts transitively on $\F_p^{2n}\setminus 0$, we can use 2) to eliminate at least the Pauli operator $w(b)$ by a commutator of an element in $\Cl_{n}(p)$ and $\mathcal{P}_n(p)$. Concretely, we can choose $a\neq 0,-b$ and then find a $g\in\Sp_{2n}(p)$ such that $g(a) = a+b$. For any $W\in C_g$ we then have $[W,w(a)] \propto w(b)^{-1}$ and hence $C = \alpha' [W,w(a)] [U,V]$. Thus, we have shown that any element in $\Cl_n(p)$ is a product of commutators, up to a global phase.

Remark: I think could hold in a non-projective version if $p\neq 2$, because we can then also try to eliminate the phase $\alpha \in \Z_p$. However, it's a bit more complicated and seems to depend on $p \mod 4$ (as indicated by the determinant of $H$).