My question:

Is the center of $ \overline{\text{Cl}_n} $ trivial?

Recall that the algebra generated by the Pauli group is the full matrix algebra. So any matrix that commutes with the Pauli group must be in the center of the full matrix algebra, which is the diagonal copy of $ \mathbb{C}^* $. Moreover, any unitary matrix which commutes with the Pauli group must be in the diagonal copy of $ U(1) $. So for example the center of the unitary group is exactly the diagonal copy of $ U(1) $.

The Clifford group is often defined as a subgroup $ \text{Cl}_n $ of the unitary group. In particular as the normalizer of the Pauli group. Any normalizer contains the center. So $ \text{Cl}_n $ contains a full $ U(1) $ subgroup. In particular, the Clifford group is infinite as a subgroup of the unitary group. Moreover, since $ \text{Cl}_n $ contains the Pauli group the center is at most the diagonal copy of $ U_1 $. And thus the center is exactly the diagonal copy of $ U(1) $.

Another way to define the Clifford group is as automorphisms of the Pauli group. Conjugation by any element of $ \text{Cl}_n $ is an automorphism of the Pauli group. This conjugation action induces a homomorphism $$ \text{adj}: \text{Cl}_n \to Aut(\text{P}_n) $$
the kernel of this map is exactly the matrices that commute with the Pauli group. But as stated above, that is just the diagonal copy of $ \mathbb{C}^* $ intersected with the Clifford group: in other words the diagonal copy of $ U(1) $.

Let $ \overline{\text{Cl}_n} $ be the subgroup of the projective unitary group which is the image of the Clifford group under this map $$ \overline{\text{Cl}_n}=\text{Cl}_n/U(1) $$ This group is also sometimes called the Clifford group.

I am curious about this second version of the Clifford group, which is quite different in some ways. For example, it is a subgroup of the automorphism group of a finite group (the Pauli group), and so is itself finite. A formula for the exact number of elements is given in the answer to this question.

I know that $ \overline{\text{Cl}_n} = \text{Cl}_n/Z(\text{Cl}_n) $ and sometimes modding out by the center gives a quotient group with trivial center but also sometimes it doesn't. For example the center of the Pauli group is exactly the multiples of the identity and modding out by that does not give a group with trivial center. In fact the quotient group is actually abelian! On the other hand modding out a unitary group by its center does give a quotient group with trivial center (the projective unitary group). So it can go both ways. That leads me to ask:

Is the center of $ \overline{\text{Cl}_n} $ trivial?

UPDATE: Here is the hint from the answer worked out. The answer shows that if $ [U] $ is in the center of $ \overline{\text{Cl}_n} $ then $$ U=a_QQ $$ for some $ Q \in \mathcal{P}_n $ and $ a_Q \in \mathbb{C} $ (indeed $ a_Q \in U(1) $ since $ U,Q $ are both unitary). In other words, if $ [U] \in \overline{\text{Cl}_n} $ is central then we must have $$ [U] = [Q] $$ for some $ Q \in \mathcal{P}_n $.

For any $ R \in \mathcal{P}_n $ we have $ R^2=1 $ so $$ \mu_R:=\frac{I+iR}{\sqrt{2}} $$
has inverse $$ \mu_R^{-1}=\frac{I-iR}{\sqrt{2}} $$
which in particular shows that $ \mu_R $ is unitary (recall every $ R \in \mathcal{P}_n $ is self-adjoint). Then conjugating any $ P \in G_n $ by $ \mu_R $ we have two cases. Since $ P,R \in G_n $ they either commute or anticommute. If they commute then $$ \mu_R P \mu_R^{-1} = P $$ if they anticommute then $$ \mu_R P \mu_R^{-1} =\frac{I+iR}{\sqrt{2}} P \frac{I-iR}{\sqrt{2}}= \Big( \frac{1+iR}{\sqrt{2}} \Big)^2P=iRP $$ either way the result is in $ G_n $ so we can conclude that $ \mu_R \in \text{Cl}_n $. Now suppose for the sake of contradiction that the $ Q $ found earlier is not the identity, $ I $. Then we can choose some $ R \in \mathcal{P}_n $ that anticommutes with $ Q $ and thus $$ [\mu_R][U][\mu_R^{-1}]=[\mu_R][Q][\mu_R^{-1}]=[iRQ]=[R][Q] \neq [Q]=[U] $$ which contradicts the centrality of $ [U] $. So we can conclude that $ Q=I $ and $$ [U]=[Q]=[I] $$ Since $ [U] $ was an arbitrary element of the center this proves that $ \overline{\text{Cl}_n} $ has a trivial center, as desired.


1 Answer 1


Yes, the center of $\overline{\text{Cl}_n}$ is trivial.


The proof consists of two parts. First, we establish that if $[U]\in Z(\overline{\text{Cl}_n})$ is an element of the center then $U$ is a Pauli operator. This is accomplished by expanding $U$ in the Pauli basis and using conjugation to flip the signs of coefficients, effectively killing all but one of them. Second, we remark that the only suitable Pauli operator is the identity. The second part is easy and is left as an exercise for the reader.


By definition of the quotient group, the elements of $\overline{\text{Cl}_n}$ are cosets of $U(1)$. Thus, every element of $\overline{\text{Cl}_n}$ can be written as $[U]=\{e^{i\theta}U|\theta\in[0,2\pi)\}$ for some $U\in\text{Cl}_n$. Moreover, $[U][V]=[UV]$ and $[U]^{-1}=[U^{-1}]$ for any $U,V\in\text{Cl}_n$ and $[U]=[V]$ if and only if there exists $\theta\in[0,2\pi)$ such that $U=e^{i\theta}V$.

Elements of the center are Pauli operators

Suppose that $[U]\in Z(\overline{\text{Cl}_n})$. Then for any $[V]\in\overline{\text{Cl}_n}$, we have

$$ \begin{align} [U]&=[V][U][V]^{-1}\tag1\\ [U]&=[VUV^{-1}]\tag2\\ U&=e^{i\theta}VUV^{-1}.\tag3\\ \end{align} $$

In other words, every representative $U$ of an element $[U]$ of the center of $\overline{\text{Cl}_n}$ commutes with all Clifford operators up to global phase.

Now, $\mathcal{P}_n=\{\bigotimes_{i=1}^n P_i|P_i\in\{I,X,Y,Z\}\}$ is a basis, so we can write

$$ U=\sum_{P\in \mathcal{P}_n}a_P P\tag4 $$

for some $a_P\in\mathbb{C}$. We will show that $U\in G_n$ where $G_n$ is the Pauli group on $n$ qubits. First, note that $U\ne 0$, so not all coefficients in $(4)$ are zero. If $a_I\ne 0$ then set $Q:=I$. Otherwise, set $Q$ to some other element of $\mathcal{P}_n$ such that $a_Q\ne 0$. We will show that all other coefficients $a_P$ are zero. To that end, choose any $R\in\mathcal{P}_n-\{Q\}$. If $R=I$ then $a_R=0$. Otherwise, we can rewrite $(4)$ as

$$ U=a_Q Q + a_R R + \sum_{P\in \mathcal{P}_n-\{Q,R\}}a_P P.\tag5 $$

Now, since $R\ne I$ and $R\ne Q$, we can find$^1$ $S\in\mathcal{P}_n$ that commutes with $Q$ and anticommutes with $R$. Then

$$ SUS^{-1}=a_Q Q - a_R R + \sum_{P\in \mathcal{P}_n-\{Q,R\}}b_P P.\tag6 $$

Substituting $V:=S$ in $(3)$, we have

$$ \begin{align} U&=e^{i\theta}SUS^{-1}\\ a_Q Q + a_R R + \dots&=e^{i\theta}a_Q Q - e^{i\theta}a_R R + \dots \end{align}\tag7 $$

where $\dots$ stand in for irrelevant sums over $\mathcal{P}_n-\{Q,R\}$. Since $\mathcal{P}_n$ is a basis, we have $\theta=0$ and $a_R=-a_R$. Therefore, $a_R=0$. But $R\in\mathcal{P}_n-\{Q\}$ was arbitrary, so $a_Q$ is the sole non-zero coefficient in $(4)$ and thus $U\in G_n$.

Elements of the center are not non-identity Pauli operators

Finally, for any non-identity Pauli operator $P\in\mathcal{P}_n$ we can find$^2$ a Clifford operator $V\in\text{Cl}_n$ such that $PV\ne e^{i\theta}VP$ for all $\theta\in[0,2\pi)$. Therefore, $U=I$ and consequently $Z(\overline{\text{Cl}_n})=\{[I]\}$ is trivial.

$^1$ Hint: Suppose $Q$ and $R$ differ at position $k$. Consider three cases...
$^2$ Promised exercise for the reader. Hint: If $Q\in\mathcal{P}_n$ then $\frac{I\pm iQ}{\sqrt2}\in\text{Cl}_n$.

  • 1
    $\begingroup$ Thanks so much for the lovely answer! Just pointing out there is a tiny typo in your second footnote. I think you want $ Q \in \mathcal{P}_n $. A general $ Q \in G_n $ might square to minus the identity, in which case $ \frac{I \pm iQ}{2} $ is not even invertible let alone unitary or in the Clifford group! $\endgroup$ Feb 9 at 16:23
  • $\begingroup$ Good catch! Fixed. Thank you for a great question! Also, good work on the exercise for the reader! :-) $\endgroup$ Feb 9 at 17:20
  • $\begingroup$ This is a bit late, but "an element of the center commutes with all Clifford operators up to global phase" means that the representative lies in a 1-dimensional subrepresentation of $U\mapsto U ( \cdot ) U^\dagger$. However, there's only the trivial irrep and this one is spanned by the identity, so the center is trivial. $\endgroup$ Mar 21 at 16:21
  • $\begingroup$ @MarkusHeinrich It seems to me that you are considering the rep $ ad: Cl_n \to End(\mathfrak{u}_n) $ and then passing to the faithful representation $ ad: \overline{Cl_n} \to End(\mathfrak{u}_n) $. And thinking about how $ ad $ decomposes into irreps. To finish your argument it seems that you either need to know that every 1d irrep of $ \overline{Cl_n} $ is trivial (so $ \overline{Cl_n} $ has trivial abelianization i.e. is a perfect group) or you would need to know something specific about the adjoint representation, for example that $ \mathfrak{su}_n $ is an irrep of $ \overline{Cl_n} $. $\endgroup$ Mar 23 at 20:05
  • 1
    $\begingroup$ @IanGershonTeixeira Yes, you need to know something about the rep theory of $Cl_n$. Note that the representation is fixed, so you only need to know how $U\mapsto U(\dot)U^\dagger$ decomposes (nothing about its 1d irreps). It is not hard to see that the representation decomposes as $1 \oplus \mathrm{ad}$ where $\mathrm{ad}$ is the "adjoint rep" on traceless matrices (i.e. $\mathfrak{su}_n$). There multiple ways show that the latter is an irrep, e.g. using the proof that $Cl_n$ is a unitary 2-design. Importantly, this is completely independent of the characterization of its centre. $\endgroup$ Mar 24 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.