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.


1 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 $H$).

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    $\begingroup$ 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 $\endgroup$ Commented Apr 26, 2022 at 16:42
  • $\begingroup$ 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. $\endgroup$ Commented Apr 27, 2022 at 12:01
  • $\begingroup$ 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 ). $\endgroup$ Commented Apr 27, 2022 at 12:14
  • $\begingroup$ 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. $\endgroup$ Commented Apr 27, 2022 at 12:44
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    $\begingroup$ @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? $\endgroup$ Commented Apr 28, 2022 at 7:35

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