4d. Weingarten formula

[math] \newcommand{\mathds}{\mathbb}[/math]

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Our aim now is to go beyond what we have, with results regarding the truncated characters. Let us start with a general formula coming from Peter-Weyl, namely:

Theorem

The Haar integration over a closed subgroup [math]G\subset_vU_N[/math] is given on the dense subalgebra of smooth functions by the Weingarten type formula

[[math]] \int_Gg_{i_1j_1}^{e_1}\ldots g_{i_kj_k}^{e_k}\,dg=\sum_{\pi,\nu\in D(k)}\delta_\pi(i)\delta_\sigma(j)W_k(\pi,\nu) [[/math]]

valid for any colored integer [math]k=e_1\ldots e_k[/math] and any multi-indices [math]i,j[/math], where [math]D(k)[/math] is a linear basis of [math]Fix(v^{\otimes k})[/math], the associated generalized Kronecker symbols are given by

[[math]] \delta_\pi(i)= \lt \pi,e_{i_1}\otimes\ldots\otimes e_{i_k} \gt [[/math]]

and [math]W_k=G_k^{-1}[/math] is the inverse of the Gram matrix, [math]G_k(\pi,\nu)= \lt \pi,\nu \gt [/math].

Show Proof

This is something very standard, coming from the fact that the above integrals form altogether the orthogonal projection [math]P^k[/math] onto the following space:

[[math]] Fix(v^{\otimes k})=span(D(k)) [[/math]]

Consider now the following linear map, with [math]D(k)=\{\xi_k\}[/math] being as in the statement:

[[math]] E(x)=\sum_{\pi\in D(k)} \lt x,\xi_\pi \gt \xi_\pi [[/math]]

By a standard linear algebra computation, it follows that we have [math]P=WE[/math], where [math]W[/math] is the inverse of the restriction of [math]E[/math] to the following space:

[[math]] K=span\left(T_\pi\Big|\pi\in D(k)\right) [[/math]]

But this restriction is the linear map given by the matrix [math]G_k[/math], and so [math]W[/math] is the linear map given by the inverse matrix [math]W_k=G_k^{-1}[/math], and this gives the result.

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In the easy case, we have the following more concrete result:

Theorem

For an easy group [math]G\subset U_N[/math], coming from a category of partitions [math]D=(D(k,l))[/math], we have the Weingarten formula

[[math]] \int_Gg_{i_1j_1}^{e_1}\ldots g_{i_kj_k}^{e_k}\,dg=\sum_{\pi,\nu\in D(k)}\delta_\pi(i)\delta_\nu(j)W_{kN}(\pi,\nu) [[/math]]

for any [math]k=e_1\ldots e_k[/math] and any [math]i,j[/math], where [math]D(k)=D(\emptyset,k)[/math], [math]\delta[/math] are usual Kronecker type symbols, checking whether the indices match, and [math]W_{kN}=G_{kN}^{-1}[/math], with

[[math]] G_{kN}(\pi,\nu)=N^{|\pi\vee\nu|} [[/math]]

where [math]|.|[/math] is the number of blocks.

Show Proof

We use the abstract Weingarten formula, from Theorem 4.40. Indeed, the Kronecker type symbols there are then the usual ones, as shown by:

[[math]] \begin{eqnarray*} \delta_{\xi_\pi}(i) &=& \lt \xi_\pi,e_{i_1}\otimes\ldots\otimes e_{i_k} \gt \\ &=&\left \lt \sum_j\delta_\pi(j_1,\ldots,j_k)e_{j_1}\otimes\ldots\otimes e_{j_k},e_{i_1}\otimes\ldots\otimes e_{i_k}\right \gt \\ &=&\delta_\pi(i_1,\ldots,i_k) \end{eqnarray*} [[/math]]

The Gram matrix being as well the correct one, we obtain the result.

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Let us go back now to the general easy groups [math]G\subset U_N[/math], with the idea in mind of computing the laws of truncated characters. First, we have the following formula:

Proposition

The moments of truncated characters are given by the formula

[[math]] \int_G(g_{11}+\ldots +g_{ss})^kdg=Tr(W_{kN}G_{ks}) [[/math]]

where [math]G_{kN}[/math] and [math]W_{kN}=G_{kN}^{-1}[/math] are the associated Gram and Weingarten matrices.

Show Proof

We have indeed the following computation:

[[math]] \begin{eqnarray*} \int_G(g_{11}+\ldots +g_{ss})^kdg &=&\sum_{i_1=1}^{s}\ldots\sum_{i_k=1}^s\int_Gg_{i_1i_1}\ldots g_{i_ki_k}\,dg\\ &=&\sum_{\pi,\nu\in D(k)}W_{kN}(\pi,\nu)\sum_{i_1=1}^{s}\ldots\sum_{i_k=1}^s\delta_\pi(i)\delta_\nu(i)\\ &=&\sum_{\pi,\nu\in D(k)}W_{kN}(\pi,\nu)G_{ks}(\nu,\pi)\\ &=&Tr(W_{kN}G_{ks}) \end{eqnarray*} [[/math]]

Thus, we have reached to the formula in the statement.

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In order to process now the above formula, and reach to concrete results, we must impose on our group a uniformity condition. Let us start with:

Proposition

For an easy group [math]G=(G_N)[/math], coming from a category of partitions [math]D\subset P[/math], the following conditions are equivalent:

• [math]G_{N-1}=G_N\cap U_{N-1}[/math], via the embedding [math]U_{N-1}\subset U_N[/math] given by [math]u\to diag(u,1)[/math].
• [math]G_{N-1}=G_N\cap U_{N-1}[/math], via the [math]N[/math] possible diagonal embeddings [math]U_{N-1}\subset U_N[/math].
• [math]D[/math] is stable under the operation which consists in removing blocks.

If these conditions are satisfied, we say that [math]G=(G_N)[/math] is uniform.

Show Proof

The equivalence [math](1)\iff(2)[/math] comes from the inclusion [math]S_N\subset G_N[/math], which makes everything [math]S_N[/math]-invariant. Regarding [math](1)\iff(3)[/math], given a subgroup [math]K\subset_vU_{N-1}[/math], consider the matrix [math]u=diag(v,1)[/math]. Our claim is that for any [math]\pi\in P(k)[/math] we have:

[[math]] \xi_\pi\in Fix(u^{\otimes k})\iff\xi_{\pi'}\in Fix(u^{\otimes k'}),\,\forall\pi'\in P(k'),\pi'\subset\pi [[/math]]

In order to prove this claim, we must study the condition on the left. We have:

[[math]] \begin{eqnarray*} \item[a]i_\pi\in Fix(v^{\otimes k}) &\iff&(u^{\otimes k}\xi_\pi)_{i_1\ldots i_k}=(\xi_\pi)_{i_1\ldots i_k},\forall i\\ &\iff&\sum_j(u^{\otimes k})_{i_1\ldots i_k,j_1\ldots j_k}(\xi_\pi)_{j_1\ldots j_k}=(\xi_\pi)_{i_1\ldots i_k},\forall i\\ &\iff&\sum_j\delta_\pi(j_1,\ldots,j_k)u_{i_1j_1}\ldots u_{i_kj_k}=\delta_\pi(i_1,\ldots,i_k),\forall i \end{eqnarray*} [[/math]]

Now let us recall that our representation has the special form [math]u=diag(v,1)[/math]. We conclude from this that for any index [math]a\in\{1,\ldots,k\}[/math], we have:

[[math]] i_a=N\implies j_a=N [[/math]]

With this observation in hand, if we denote by [math]i',j'[/math] the multi-indices obtained from [math]i,j[/math] obtained by erasing all the above [math]i_a=j_a=N[/math] values, and by [math]k'\leq k[/math] the common length of these new multi-indices, our condition becomes:

[[math]] \sum_{j'}\delta_\pi(j_1,\ldots,j_k)(u^{\otimes k'})_{i'j'}=\delta_\pi(i_1,\ldots,i_k),\forall i [[/math]]

Here the index [math]j[/math] is by definition obtained from the index [math]j'[/math] by filling with [math]N[/math] values. In order to finish now, we have two cases, depending on [math]i[/math], as follows:

\underline{Case 1}. Assume that the index set [math]\{a|i_a=N\}[/math] corresponds to a certain subpartition [math]\pi'\subset\pi[/math]. In this case, the [math]N[/math] values will not matter, and our formula becomes:

[[math]] \sum_{j'}\delta_\pi(j'_1,\ldots,j'_{k'})(u^{\otimes k'})_{i'j'}=\delta_\pi(i'_1,\ldots,i'_{k'}) [[/math]]

\underline{Case 2}. Assume now the opposite, namely that the set [math]\{a|i_a=N\}[/math] does not correspond to a subpartition [math]\pi'\subset\pi[/math]. In this case the indices mix, and our formula reads [math]0=0[/math]. Thus we have [math]\xi_{\pi'}\in Fix(u^{\otimes k'})[/math] in both cases, for any subpartition [math]\pi'\subset\pi[/math], as desired.

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Now back to the laws of truncated characters, we have the following result:

Theorem

For a uniform easy group [math]G=(G_N)[/math], we have the formula

[[math]] \lim_{N\to\infty}\int_{G_N}\chi_t^k=\sum_{\pi\in D(k)}t^{|\pi|} [[/math]]

with [math]D\subset P[/math] being the associated category of partitions.

Show Proof

We use Proposition 4.42. With [math]s=[tN][/math], the formula there becomes:

[[math]] \int_{G_N}\chi_t^k=Tr(W_{kN}G_{k[tN]}) [[/math]]

The point now is that in the uniform case the Gram matrix, and so the Weingarten matrix too, is asymptotically diagonal. Thus, we obtain the following estimate:

[[math]] \begin{eqnarray*} \int_{G_N}\chi_t^k &\simeq&\sum_{\pi\in D(k)}W_{kN}(\pi,\pi)G_{k[tN]}(\pi,\pi)\\ &\simeq&\sum_{\pi\in D(k)}N^{-|\pi|}(tN)^{|\pi|}\\ &=&\sum_{\pi\in D(k)}t^{|\pi|} \end{eqnarray*} [[/math]]

Thus, we are led to the formula in the statement.

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We can now enlarge our collection of truncated character results, and we have:

Theorem

With [math]N\to\infty[/math], the laws of truncated characters are as follows:

• For [math]O_N[/math] we obtain the Gaussian law [math]g_t[/math].
• For [math]U_N[/math] we obtain the complex Gaussian law [math]G_t[/math].
• For [math]S_N[/math] we obtain the Poisson law [math]p_t[/math].
• For [math]H_N[/math] we obtain the Bessel law [math]b_t[/math].
• For [math]H_N^s[/math] we obtain the generalized Bessel law [math]b_t^s[/math].
• For [math]K_N[/math] we obtain the complex Bessel law [math]B_t[/math].

Show Proof

We already know these results at [math]t=1[/math]. In the general case, [math]t \gt 0[/math], these follow via some standard combinatorics, from the formula in Theorem 4.44.

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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].