This a lemma used in (Scott 2006) when discussing complex projective t-designs.

Let $\pi(x)\equiv|x\rangle\!\langle x|$ be the projection onto some pure state (represented as an element of the complex projective space), denote with $\mu$ the uniform measure in the complex projective space, and let $\Pi_{\rm sym}^{(t)}$ be the projector onto the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$. Then $$\int_{{\Bbb CP}^{d-1}}d\mu(x)\pi(x)^{\otimes t}=\binom{d+t-1}{t}^{-1} \Pi_{\rm sym}^{(t)}.\tag1$$ The author mentions that can be proved as follows:

Use Schur's lemma. The LHS of the equation is invariant under all unitaries $U^{\otimes t}$ which act irreducibly on the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$.

While explanation makes sense, I'm looking to see some of the underlying details be written down more explicitly. In particular, we seem to be considering the group action of $\mathbf U(d)$ on $\mathbb{CP}^{d-1}$. I guess this is the action $(U,[v])\mapsto [Uv]$ for $U\in\mathbf U(d)$ and $v\in \mathbb{C}^d$, $[v]\in\mathbb{CP}^{d-1}$. This action is the one used to define the measure $\mu$, which is characterised by its being invariant under this action.

But then, we consider the action of $\mathbf U(d)$ on the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$, and using Schur's lemma with this action. I'm assuming the version of Schur's lemma we're using is the fact that, given a complex finite-dimensional irreducible representation $\rho:G\to\operatorname{Lin}(\mathbb{C}^n)$, if a linear operator $f:\mathbb{C}^n\to \mathbb{C}^n$ is such that $[f,\rho(g)]=0$ for all $g\in G$, then $f=\lambda I_n$. I know that one way to use this lemma is to build explicitly elements that commute with the irrep by taking averages, and using this observation we get results of the form: $$\sum_{g\in G} \rho(g)A \rho(g)^\dagger = \frac{|G| \operatorname{Tr}(A)}{n} I_n,$$ for any linear operator $A:\mathbb{C}^n\to\mathbb{C}^n$.

This is clearly very close to (1), especially observing the fact that $\binom{d+t-1}{t}$ is the dimension of the totally symmetric subspace we're using, and $\Pi_{\rm sym}^{(t)}$ is essentially the identity, when we're restricting the domain on said space. Still, the LHS of (1) is averaging over the elements of another set over which $\mathbf U(d)$ acts, rather then over the group elements themselves, which is what is used in Schur's lemma, so what gives?


2 Answers 2


An element $[v]\in\mathbb{CP}^{d-1}$ can be represented by $\pi(v) = |v\rangle\langle v|$, which is a linear operator on $\mathbb{C}^d$. A unitary action $(U,[v])\mapsto [Uv]$ on ${\Bbb CP}^{d-1}$ then corresponds to the map $|v\rangle\langle v| \mapsto U|v\rangle\langle v|U^\dagger$ on the space of linear operators.

The expression $A = \int_{{\Bbb CP}^{d-1}}d\mu(x)\pi(x)^{\otimes t}$ is a linear operator on $(\mathbb{C}^d)^{\otimes t}$. An action $U$ on ${\Bbb CP}^{d-1}$ then corresponds to $$ A \mapsto U^{\otimes t} \cdot A \cdot (U^\dagger)^{\otimes t} = \int_{{\Bbb CP}^{d-1}}d\mu(x)\pi(U(x))^{\otimes t}. $$

Since $\mu$ is the Haar measure we have that $A = U^{\otimes t} \cdot A \cdot (U^\dagger)^{\otimes t}$ for any unitary $U$. It's also easy to see that the support of $A$ is in the symmetric subspace. We then apply Schur lemma to the operator $A$ and the representation $U^{\otimes t}$ of the unitary group acting on the symmetric subspace.


Thinking about this a bit more: the point is that we're not using the common way to build linear operators commuting with the representation by taking averages over the group elements. That is, we're not using objects of the form $\sum_g \rho(g)A\rho(g)^\dagger$ for some $A$. We're stil lusing the more general result about the fact that if $f$ commutes with group elements then $f=\lambda I$, but we build such an operator in a different way.

More precisely, we take the action of $G\equiv \mathbf U(d)$ on the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$. Explicitly, this is $(U,v)\mapsto U^{\otimes t}v$ for all $U\in \mathbf U(d)$ and $v\in\mathbb{C}^{dt}$ totally symmetric vectors. This is an action because if $v$ is symmetric then $U^{\otimes t}v$ is as well.

We now want to build a linear operator on the symmetric subspace. To do this, we consider the projector $\pi(x)\equiv |x\rangle\!\langle x|\in\operatorname{Lin}(\mathbb{C}^d)$ for some $x\in\mathbb{CP}^{d-1}$ (actually, we should probably more precisely say that $\pi(x)$ is the linear operator on $\mathbb{C}^d$ corresponding to $x\in\mathbb{CP}^{d-1}$, built as $\pi(x)\equiv |y\rangle\!\langle y|/\|y\|^2$ for any representative element $\mathbb{C}^d\ni y\in x$). How do these operators interact with the unitary group? We have, $U\in\mathbf U(d)$, $$U\pi(x) U^\dagger = \pi(Ux),$$ or more precisely, $U\pi(x)U^\dagger = \pi(\tilde Ux)$, where $\tilde U$ is the projective unitary operator corresponding to $U$ through the canonical injection $\mathbf U(d)\to\mathbf{PU}(d)$.

Now, because $\mathbf U(d)$ acts irreducibly (and in particular, transitively) on $\mathbb{CP}^{d-1}$, via the abovementioned action, the integrating $\pi(Ux)$ over all (projective) unitaries $U$ is the same as integrating $\pi(x)$ over all $x\in\mathbb{CP}^{d-1}$. Thus, the operator $\int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(x)$ commutes with $U$, and by Shur's lemma we must have $$\int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(x) = \lambda I,$$ with $I$ the identity in $\mathbb{C}^d$. Taking the trace, and using the normalisation conditions on the LHS, we see that $\lambda d=1$ and thus $\lambda=1/d$.

The reasoning in the above paragraph remains unchanged when considering the action in the symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$. We now consider the operator $\int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(x)^{\otimes t}$, which interacts with symmetric unitaries of the form $U^{\otimes t}$ as follows: $$ U^{\otimes t} \left(\int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(x)^{\otimes t}\right) (U^{\otimes t})^\dagger = \int_{\mathbb{CP}^{d-1}} d\mu(x)\, (U\pi(x)U^\dagger)^{\otimes t} \\ = \int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(Ux)^{\otimes t} = \int_{\mathbb{CP}^{d-1}} d\mu(x)\, \pi(x)^{\otimes t}. $$ In other words, the operator commutes with all matrices of the form $U^{\otimes t}$, for any $U\in\mathbf U(d)$. Even though $\pi(x)^{\otimes t}$ are defined as linear operators in $(\mathbb{C}^d)^{\otimes t}\simeq \mathbb{C}^{dt}$, we can see that $\pi(x)^{\otimes t}v$ is totally symmetric whenever $v\in\mathbb{C}^{dt}$ is. This means that we can consider the restricted action of $\pi(x)^{\otimes t}$ on the totally symmetric subspace of $(\mathbb{C}^d)^{\otimes t}$. This is useful because we can now use the irreducible representation of $\mathbf U(d)$ in this space, and reach the conclusion from Schur's lemma.

  • $\begingroup$ The action of $\mathbf U(d)$ on $\mathbb{CP}^{d-1}$ is surely irreducible. But we need the irreduciblity of $U^{\otimes t}$ on the symmetric subspace. It doesn't follow automatically. $\endgroup$
    – Danylo Y
    Oct 9, 2022 at 15:05
  • $\begingroup$ @DanyloY mh, good point. Do you know a good way to see that? In $(\mathbb{C}^2)^2$ we have basis elements $|00\rangle,|11\rangle$, and $|01\rangle+|10\rangle$. The first two are clearly sent to one another via some $U\otimes U$, hence they must be in the same invariant subspace. Question is then whether you can get out of $\alpha|00\rangle+\beta|11\rangle$, the answer to which is clearly yes, as eg $H\otimes H|00\rangle=|+,+\rangle=\frac12(|00\rangle+|11\rangle+|01\rangle+|10\rangle)$. I don't know if this simple argument generalises easily though $\endgroup$
    – glS
    Oct 9, 2022 at 15:24
  • $\begingroup$ The simplest proof is probably in theorem 5 in arxiv.org/abs/1308.6595 $\endgroup$
    – Danylo Y
    Oct 9, 2022 at 15:51

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