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

Let $$Cl_n$$ be the Clifford group on $$n$$ qubits. What is the commutator subgroup of $$Cl_n$$?

It is definitely not all of $$Cl_n$$ since $$Cl_n$$ is not perfect. My guess is that the abelianization of $$Cl_n$$ 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 $$Cl_n$$ 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.

$$\newcommand{\Sp}{\mathrm{Sp}} \newcommand{\Cl}{\mathrm{Cl}} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$$ 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 $$H$$).

• Wow I'm actually shocked that the clifford group is perfect that was actually what I was hoping for but it seemed way too good to be true. I'm changing the title of the question to celebrate Apr 26 at 16:42
• Ok so I'm trying to think about what special properties of the clifford group you really used in this case. It seems like you showed that if $H$ is a perfect group and $(\pi,V)$ is a representation of $H$ such that the action of $H$ on the nonzero elements of $V$ is transitive then the semidirect product $V \rtimes_\pi H$ is perfect. Apr 27 at 12:01
• However for the $p=2$ (qubit) case you do even better proving something along the lines of if $1 \to V \to G \to H \to 1$ is a group extension with $H$ perfect and $V$ abelian and if moreover the natural action of $H$ on $V$ is transitive on the nontrivial elements of $V$ then $G$ must be perfect. (the action of $H$ on $V$ works by conjugation in $G$ by a preimage of $h$ the reason it is well defined is because $V$ is abelian and normal, this is a general fact about extension by abelian groups see the remark below equation 17.28 in section 17.4 of Dummit and Foote ). Apr 27 at 12:14
• Now I'm wondering about perfect central extensions/ schur multiplier. Is it possible that $\overline{Cl_n}$ is its own covering group / universal perfect central extension (in other words schur multiplier is 0)? It seems a bit crazy especially since the center is trivial (quantumcomputing.stackexchange.com/a/24008/19675) but occasionally this sort of thing happens. where a group has trivial center and is its own universal cover for example this is true in the Lie theory sense for $E _8, F_4$and $G_2$. I'll probably make a new question but just curious. Apr 27 at 12:44
• @IanGershonTeixeira About the Schur multipliers: I don't know much about this, but apparently, there are some results due to Stein and Steinberg which show $H_2(\mathrm{Sp}_{2n}(p), \mathbb Z) = 0$ if $p$ is prime (and $n\geq 3$). Maybe one can again exploit this result? Apr 28 at 7:35