# The adjoint representation and irreducibility of the Clifford group

Let $$P$$ be the Pauli group and $$Cl$$ (the Clifford group) be the normalizer of $$P$$ in the unitary group $$U_d$$ . Consider the representation given by acting the unitary group $$U_d$$ on the Lie algebra $$\mathfrak{su}_d$$ by conjugation (essentially the adjoint representation) $$\text{ad}: U_d \to \text{End}(\mathfrak{su}_d)$$ Since the Clifford group is a subgroup of $$U_d$$ we can restrict $$\text{ad}$$ to get a representation $$\text{ad}:Cl \to \text{End}(\mathfrak{su}_d)$$ Is this representation of the Clifford group an irreducible representation?

• I think you can argue from the fact that the adjoint representation of U is irreducible that the adjoint representation of the Clifford group is also irreducible, since the Clifford group is a 2-design. Commented Mar 20, 2022 at 19:30
• Ya that sounds right but I was curious to see if someone had an easier/different perspective/ or how they would work out the details of the 2-design argument Commented Mar 21, 2022 at 1:54

Let me elaborate. Let $$\tau^{(2)}(U):= U\otimes U$$ be a representation of $$U(d)$$. Then, a subgroup $$G\subset U(d)$$ is a unitary 2-design if and only if $$\tau^{(2)}|_{G}$$ decomposes into the same irreps as $$\tau^{(2)}$$. This is because the 2-design definition reads $$\int_G U^{\otimes 2} (\cdot) (U^{\otimes 2})^\dagger \,dU = \int_{U(d)} U^{\otimes 2} (\cdot) (U^{\otimes 2})^\dagger \,dU.$$ These are both projectors onto the commutant of $$\tau^{(2)}$$ and $$\tau^{(2)}|_{G}$$, respectively. Since the dimension of the commutant is the sum of squared multiplicities of the irreps and $$G$$ is a subgroup, the commutant can only be the same if the irreps of $$\tau^{(2)}$$ are still irreducible under $$G$$.
Note that this implies the same statement for $$(U \mapsto U\otimes\overline U) \simeq (U\mapsto U(\cdot)U^\dagger)$$.
It is easy to see that the latter representation decomposes as $$1 \oplus \mathrm{ad}$$. Clearly, the trivial irrep is irreducible for $$G=\mathrm{Cl}_n$$, so $$\mathrm{Cl}_n$$ is a unitary 2-design if and only if $$\mathrm{ad}|_{\mathrm{Cl}_n}$$ is irreducible.
The usual proof uses that $$\mathrm{Cl}_n$$ acts transitively on traceless Pauli matrices. This already implies the correct dimension of the commutant, see Lemma 2 in [Zhu]. Alternatively, on can evaluate the above integral on the Pauli basis, see e.g. Sec. IV.C in [Gross et al.].