The Weingarten Calculus

Haar integration as a problem in invariant theory and finite-dimensional linear algebra.

2026-10-01 · 4 min read

This post is inspired by a presentation I gave last year on a topics course in Quantum Learning Theory.

Weingarten calculus is a method for computing polynomial integrals over compact groups. Its central idea is surprisingly simple: averaging a unitary representation over Haar measure produces an orthogonal projection. Once we identify the range of that projection, integration becomes a finite-dimensional linear algebra problem.

We first prove the fundamental theorem, then work through the symmetric group with a toy example, and finally recover the familiar unitary Weingarten formula using Schur–Weyl duality.

The setup

Let $G$ be a compact group with normalized Haar measure $dg$, and let $\mathcal{H}$ be an $N$-dimensional Hilbert space with orthonormal basis

$$ e_1,\ldots,e_N. $$

Suppose

$$ U:G\longrightarrow \mathrm U(\mathcal{H}) $$

is a continuous unitary representation. For $a,b\in[N]$, write

$$ U_{ab}(g)=\langle e_a,U(g)e_b\rangle. $$

Fix $d\in\mathbb N$ and two functions $i,j:[d]\to[N]$. We want to compute integrals of the form

$$ I_{ij}=\int_G\prod_{r=1}^d U_{i(r)j(r)}(g)\,dg. $$

This looks to be quite a hairy integral. The main insight of Weingarten calculus is that it is really a problem about the $G$-invariant subspace of $\mathcal{H}^{\otimes d}$.

For a multi-index $i\in\operatorname{Fun}(d,N)$, define

$$ e_i=e_{i(1)}\otimes\cdots\otimes e_{i(d)}. $$

Then the matrix coefficients of the tensor-power representation satisfy

$$ \begin{aligned} U^{\otimes d}_{ij}(g) &=\langle e_i,U^{\otimes d}(g)e_j\rangle\\ &=\prod_{r=1}^d\langle e_{i(r)},U(g)e_{j(r)}\rangle\\ &=\prod_{r=1}^d U_{i(r)j(r)}(g). \end{aligned} $$

Thus the product in the integrand is just a single matrix coefficient:

$$ I_{ij}=\int_G U^{\otimes d}_{ij}(g)\,dg. $$

Haar averaging is an orthogonal projection

Define the operator

$$ P=\int_G U^{\otimes d}(g)\,dg. $$

The integral is well-defined because $G$ is compact and the representation is continuous. We claim that $P$ is the orthogonal projection onto

$$ (\mathcal{H}^{\otimes d})^G =\{v\in\mathcal{H}^{\otimes d}:U^{\otimes d}(g)v=v\text{ for every }g\in G\}. $$

First, $P$ is idempotent. By Fubini’s theorem and invariance of Haar measure,

$$ \begin{aligned} P^2 &=\int_G\int_G U^{\otimes d}(g)U^{\otimes d}(h)\,dg\,dh\\ &=\int_G\int_G U^{\otimes d}(gh)\,dg\,dh\\ &=\int_G P\,dh\\ &=P. \end{aligned} $$

Second, $P$ is self-adjoint. Since the representation is unitary,

$$ \begin{aligned} P^* &=\int_G U^{\otimes d}(g)^*\,dg\\ &=\int_G U^{\otimes d}(g^{-1})\,dg\\ &=P, \end{aligned} $$

where the final equality follows from invariance of Haar measure under inversion. Therefore $P$ is an orthogonal projection.

Now we find the range. For $h\in G$,

$$ U^{\otimes d}(h)P =\int_G U^{\otimes d}(hg)\,dg =P, $$

so every vector in $\operatorname{im}(P)$ is fixed by $G$. Conversely, if $v$ is $G$-invariant, then

$$ Pv=\int_G U^{\otimes d}(g)v\,dg =\int_G v\,dg =v. $$

Hence

$$ \operatorname{im}(P)=(\mathcal{H}^{\otimes d})^G. $$

This illustrates the conceptual heart of the method: Haar integration projects onto invariant tensors.

The fundamental theorem

Choose any basis $a_1,\ldots,a_m$ of $(\mathcal{H}^{\otimes d})^G$, and define the matrix

$$ \mathbf A=\big[\langle e_i,a_x\rangle\big]_{ i\in\operatorname{Fun}(d,N),\,x\in[m]}. $$

The columns of $\mathbf A$ are the invariant basis vectors written in the tensor-product basis. Since these columns are linearly independent, their Gram matrix

$$ \mathbf G=\mathbf A^*\mathbf A $$

is invertible. The orthogonal projection onto their span is

$$ P=\mathbf A(\mathbf A^*\mathbf A)^{-1}\mathbf A^*. $$

The Weingarten matrix is defined by

$$ \mathbf W=(\mathbf A^*\mathbf A)^{-1}. $$

Since $I_{ij}=P_{ij}$, taking matrix entries gives the fundamental formula

$$ \boxed{ I_{ij}=\sum_{x,y=1}^m \mathbf A_{ix}\mathbf W_{xy}\mathbf A^*_{yj} }. $$

So every Weingarten calculation has the same recipe:

Linearize the integrand using a tensor-power representation.
Find a basis for the invariant subspace.
Compute the Gram matrix of that basis.
Invert the Gram matrix and read off the desired projection entry.

A toy example: the symmetric group

Let $G=S_N$ act on $\mathcal{H}=\mathbb C^N$ by permuting the standard basis:

$$ U(\sigma)e_x=e_{\sigma(x)}. $$

Haar integration is now normalized counting measure, so

$$ I_{ij} =\frac{1}{N!}\sum_{\sigma\in S_N} \prod_{r=1}^d U_{i(r)j(r)}(\sigma). $$

The action of $S_N$ on a function $i:[d]\to[N]$ can rename the values of $i$, but it cannot change which arguments share the same value. This motivates the equivalence relation

$$ r\sim_i s\quad\Longleftrightarrow\quad i(r)=i(s). $$

The equivalence classes form a set partition of $[d]$, which we call $\operatorname{type}(i)$. For example, $(1,1,2)$ and $(4,4,7)$ have the same type, while $(1,2,1)$ and $(1,2,3)$ do not.

For a set partition $\lambda$ of $[d]$ with at most $N$ blocks, define the orbit sum

$$ a_\lambda =\sum_{\substack{i\in\operatorname{Fun}(d,N)\\ \operatorname{type}(i)=\lambda}}e_i. $$

These orbit sums form a basis for $(\mathcal{H}^{\otimes d})^{S_N}$. Distinct orbit sums have disjoint support, so they are orthogonal. If $\lambda$ has $k$ blocks, then assigning distinct labels from $[N]$ to those blocks gives

$$ \langle a_\lambda,a_\lambda\rangle =(N)_k =N(N-1)\cdots(N-k+1). $$

Thus the Gram matrix and its inverse are diagonal:

$$ \mathbf G_{\lambda\mu} =\delta_{\lambda\mu}(N)_{|\lambda|}, \qquad \mathbf W_{\lambda\mu} =\frac{\delta_{\lambda\mu}}{(N)_{|\lambda|}}. $$

Moreover,

$$ \mathbf A_{i\lambda} =\langle e_i,a_\lambda\rangle =\mathbf 1\{\operatorname{type}(i)=\lambda\}. $$

Substitution into the fundamental theorem yields

$$ \boxed{ I_{ij} =\frac{\mathbf 1\{\operatorname{type}(i)=\operatorname{type}(j)\}} {(N)_{|\operatorname{type}(i)|}} }. $$

This answer also has a direct probabilistic interpretation, which I’m always looking for. A permutation can carry $j$ to $i$ exactly when $i$ and $j$ have the same collision pattern. If their common type has $k$ blocks, a uniformly random permutation realizes the required matching with probability $1/(N)_k$.

The unitary group

Now take $G=\mathrm U(N)$ with its defining representation on $\mathcal{H}=\mathbb C^N$. The center of $\mathrm U(N)$ acts on $\mathcal{H}^{\otimes d}$ by a nontrivial scalar when $d\geq1$, so this tensor power has no invariant vectors. To obtain nonzero moments, we pair entries of $U$ with entries of its complex conjugate and study integrals of the form

$$ \int_{\mathrm U(N)} \prod_{r=1}^d U_{i_rj_r}(U) \prod_{r=1}^d \overline{U_{i'_rj'_r}(U)}\,dU. $$

These products arise as matrix coefficients of the conjugation representation on $\operatorname{End}(\mathcal{H}^{\otimes d})$. Its invariant subspace is the commutant

$$ \operatorname{End}_{\mathrm U(N)}(\mathcal{H}^{\otimes d}) =\{T:T U^{\otimes d}=U^{\otimes d}T \text{ for every }U\in\mathrm U(N)\}. $$

Schur–Weyl duality identifies this commutant with the span of the permutation operators

$$ A_\sigma(v_1\otimes\cdots\otimes v_d) =v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(d)}, \qquad \sigma\in S_d. $$

The Hilbert–Schmidt Gram matrix of these operators has a particularly nice form:

$$ \begin{aligned} \langle A_\rho,A_\sigma\rangle &=\operatorname{Tr}(A_\rho^*A_\sigma)\\ &=\operatorname{Tr}(A_{\rho^{-1}\sigma})\\ &=N^{\#\operatorname{cycles}(\rho^{-1}\sigma)}. \end{aligned} $$

Therefore

$$ \mathbf G =\left[N^{\#\operatorname{cycles}(\rho^{-1}\sigma)} \right]_{\rho,\sigma\in S_d}. $$

When $N\geq d$, the permutation operators are linearly independent and the unitary Weingarten matrix is simply $\mathbf G^{-1}$. When $N

Because $\mathbf G_{\rho\sigma}$ depends only on $\rho^{-1}\sigma$, its inverse does as well. We write

$$ \mathbf W_{\rho\sigma} =\operatorname{Wg}_N(\rho^{-1}\sigma). $$

In terms of irreducible characters of $S_d$,

$$ \boxed{ \operatorname{Wg}_N(\pi) =\frac{1}{d!} \sum_{\substack{\lambda\vdash d\\\ell(\lambda)\leq N}} \frac{f^\lambda\chi^\lambda(\pi)} {\displaystyle\prod_{\square\in\lambda}(N+c(\square))} }. $$

Here $f^\lambda$ is the dimension of the irreducible $S_d$-module indexed by $\lambda$, $\chi^\lambda$ is its character, and $c(\square)$ is the content of a box in the Young diagram of $\lambda$.

Finally, for multi-indices $\mathbf i,\mathbf j,\mathbf i',\mathbf j'\in[N]^d$, the unitary moment formula is

$$ \boxed{ \begin{aligned} &\int_{\mathrm U(N)} \prod_{r=1}^d U_{i_rj_r}(U) \prod_{r=1}^d \overline{U_{i'_rj'_r}(U)}\,dU\\ &\qquad= \sum_{\rho,\sigma\in S_d} \left(\prod_{r=1}^d\delta_{i_r,i'_{\rho(r)}}\right) \left(\prod_{r=1}^d\delta_{j_r,j'_{\sigma(r)}}\right) \operatorname{Wg}_N(\rho^{-1}\sigma). \end{aligned} }. $$

Two small unitary moments

For $d=1$, the Gram matrix is $[N]$, so $\operatorname{Wg}_N(e)=1/N$. We recover the familiar identity

$$ \int_{\mathrm U(N)}U_{ij}\overline{U_{i'j'}}\,dU =\frac{1}{N}\delta_{ii'}\delta_{jj'}. $$

For $d=2$ and $N\geq2$, the two Weingarten values are

$$ \operatorname{Wg}_N(e)=\frac{1}{N^2-1}, \qquad \operatorname{Wg}_N((12))=-\frac{1}{N(N^2-1)}. $$

Every fourth-order moment of Haar unitary entries is obtained by combining these two numbers with the Kronecker-delta matchings in the boxed formula above.

The big picture

Weingarten calculus separates a difficult looking integral into two tasks:

Invariant theory: determine the tensors or operators fixed by the group action.
Linear algebra: compute and invert their Gram matrix.

For $S_N$, the invariant basis is indexed by set partitions and the Gram matrix is diagonal. For $\mathrm U(N)$, Schur–Weyl duality supplies permutation operators, and the inverse Gram matrix is governed by the character theory of $S_d$. In both cases, the integral itself is just an entry of an orthogonal projection.