16c. Kesten measures

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

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Let us compute now the Kesten measure [math]\mu=law(\chi)[/math], in the case where the deformation matrix is generic, as before. Our results here will be a combinatorial moment formula, a geometric interpretation of it, and an asymptotic result. We first have:

Theorem

We have the moment formula

[[math]] \int\chi^p =\frac{1}{|G|\cdot|H|}\#\left\{\begin{matrix}i_1,\ldots,i_p\in G\\ d_1,\ldots,d_p\in H\end{matrix}\Big|\begin{matrix}[(i_1,d_1),(i_2,d_2),\ldots,(i_p,d_p)]\ \ \ \ \\=[(i_1,d_p),(i_2,d_1),\ldots,(i_p,d_{p-1})]\end{matrix}\right\} [[/math]]

where the sets between square brackets are by definition sets with repetition.

Show Proof

According to the various formulae above, the factorization found in Theorem 16.15 is, at the level of standard generators, as follows:

[[math]] \begin{matrix} C(S_{G\times H}^+)&\to&C^*(\Gamma_{G,H})\otimes C(G)&\to&M_{G\times H}(\mathbb C)\\ u_{ia,jb}&\to&\frac{1}{|H|}\sum_cF_{b-a,c}c^{(i)}\otimes v_{ij}&\to&W_{ia,jb} \end{matrix} [[/math]]

Thus, the main character of the quantum permutation group that we found in Theorem 16.15 is given by the following formula:

[[math]] \begin{eqnarray*} \chi &=&\frac{1}{|H|}\sum_{iac}c^{(i)}\otimes v_{ii}\\ &=&\sum_{ic}c^{(i)}\otimes v_{ii}\\ &=&\left(\sum_{ic}c^{(i)}\right)\otimes\delta_1 \end{eqnarray*} [[/math]]

Now since the Haar functional of [math]C^*(\Gamma)\rtimes C(H)[/math] is the tensor product of the Haar functionals of [math]C^*(\Gamma),C(H)[/math], this gives the following formula, valid for any [math]p\geq1[/math]:

[[math]] \int\chi^p=\frac{1}{|G|}\int_{\widehat{\Gamma}_{G,H}}\left(\sum_{ic}c^{(i)}\right)^p [[/math]]

Consider the elements [math]S_i=\sum_cc^{(i)}[/math]. By using the embedding in Proposition 16.11 (2), with the notations there we have:

[[math]] S_i=\sum_c(b_{i0}-b_{ic},c) [[/math]]

Now observe that these elements multiply as follows:

[[math]] S_{i_1}\ldots S_{i_p}=\sum_{c_1\ldots c_p} \begin{pmatrix} b_{i_10}-b_{i_1c_1}+b_{i_2c_1}-b_{i_2,c_1+c_2}&&\\ +b_{i_3,c_1+c_2}-b_{i_3,c_1+c_2+c_3}+\ldots\ldots&,&c_1+\ldots+c_p&\\ \ldots\ldots+b_{i_p,c_1+\ldots+c_{p-1}}-b_{i_p,c_1+\ldots+c_p}&& \end{pmatrix} [[/math]]

In terms of the new indices [math]d_r=c_1+\ldots+c_r[/math], this formula becomes:

[[math]] S_{i_1}\ldots S_{i_p}=\sum_{d_1\ldots d_p} \begin{pmatrix} b_{i_10}-b_{i_1d_1}+b_{i_2d_1}-b_{i_2d_2}&&\\ +b_{i_3d_2}-b_{i_3d_3}+\ldots\ldots&,&d_p&\\ \ldots\ldots+b_{i_pd_{p-1}}-b_{i_pd_p}&& \end{pmatrix} [[/math]]

Now by integrating, we must have [math]d_p=0[/math] on one hand, and on the other hand:

[[math]] [(i_1,0),(i_2,d_1),\ldots,(i_p,d_{p-1})]=[(i_1,d_1),(i_2,d_2),\ldots,(i_p,d_p)] [[/math]]

Equivalently, we must have [math]d_p=0[/math] on one hand, and on the other hand:

[[math]] [(i_1,d_p),(i_2,d_1),\ldots,(i_p,d_{p-1})]=[(i_1,d_1),(i_2,d_2),\ldots,(i_p,d_p)] [[/math]]

Thus, by translation invariance with respect to [math]d_p[/math], we obtain:

[[math]] \int_{\widehat{\Gamma}_{G,H}}S_{i_1}\ldots S_{i_p} =\frac{1}{|H|}\#\left\{d_1,\ldots,d_p\in H\Big|\begin{matrix}[(i_1,d_1),(i_2,d_2),\ldots,(i_p,d_p)]\ \ \ \ \\=[(i_1,d_p),(i_2,d_1),\ldots,(i_p,d_{p-1})]\end{matrix}\right\} [[/math]]

It follows that we have the following moment formula:

[[math]] \int_{\widehat{\Gamma}_{G,H}}\left(\sum_iS_i\right)^p =\frac{1}{|H|}\#\left\{\begin{matrix}i_1,\ldots,i_p\in G\\ d_1,\ldots,d_p\in H\end{matrix}\Big|\begin{matrix}[(i_1,d_1),(i_2,d_2),\ldots,(i_p,d_p)]\ \ \ \ \\=[(i_1,d_p),(i_2,d_1),\ldots,(i_p,d_{p-1})]\end{matrix}\right\} [[/math]]

Now by dividing by [math]|G|[/math], we obtain the formula in the statement.

■

The formula in Theorem 16.16 can be interpreted as follows:

Theorem

With [math]M=|G|,N=|H|[/math] we have the formula

[[math]] law(\chi)=\left(1-\frac{1}{N}\right)\delta_0+\frac{1}{N}law(A) [[/math]]

where the matrix on the right,

[[math]] A\in C(\mathbb T^{MN},M_M(\mathbb C)) [[/math]]

is given by [math]A(q)=[/math] Gram matrix of the rows of [math]q[/math].

Show Proof

According to Theorem 16.16, we have the following formula:

[[math]] \begin{eqnarray*} \int\chi^p &=&\frac{1}{MN}\sum_{i_1\ldots i_p}\sum_{d_1\ldots d_p}\delta_{[i_1d_1,\ldots,i_pd_p],[i_1d_p,\ldots,i_pd_{p-1}]}\\ &=&\frac{1}{MN}\int_{\mathbb T^{MN}}\sum_{i_1\ldots i_p}\sum_{d_1\ldots d_p}\frac{q_{i_1d_1}\ldots q_{i_pd_p}}{q_{i_1d_p}\ldots q_{i_pd_{p-1}}}\,dq\\ &=&\frac{1}{MN}\int_{\mathbb T^{MN}}\sum_{i_1\ldots i_p}\left(\sum_{d_1}\frac{q_{i_1d_1}}{q_{i_2d_1}}\right)\left(\sum_{d_2}\frac{q_{i_2d_2}}{q_{i_3d_2}}\right)\ldots\left(\sum_{d_p}\frac{q_{i_pd_p}}{q_{i_1d_p}}\right)dq \end{eqnarray*} [[/math]]

Consider now the Gram matrix in the statement, namely:

[[math]] A(q)_{ij}= \lt R_i,R_j \gt [[/math]]

Here [math]R_1,\ldots,R_M[/math] are the rows of the following matrix:

[[math]] q\in \mathbb T^{MN}\simeq M_{M\times N}(\mathbb T) [[/math]]

We have then the following computation:

[[math]] \begin{eqnarray*} \int\chi^p &=&\frac{1}{MN}\int_{\mathbb T^{MN}} \lt R_{i_1},R_{i_2} \gt \lt R_{i_2},R_{i_3} \gt \ldots \lt R_{i_p},R_{i_1} \gt \\ &=&\frac{1}{MN}\int_{\mathbb T^{MN}}A(q)_{i_1i_2}A(q)_{i_2i_3}\ldots A(q)_{i_pi_1}\\ &=&\frac{1}{MN}\int_{\mathbb T^{MN}}Tr(A(q)^p)dq\\ &=&\frac{1}{N}\int_{\mathbb T^{MN}}tr(A(q)^p)dq \end{eqnarray*} [[/math]]

But this gives the formula in the statement, and we are done.

■

In general, the moments of the Gram matrix [math]A[/math] are given by a quite complicated formula, and we cannot expect to have a refinement of Theorem 16.17, with [math]A[/math] replaced by a plain, non-matricial random variable, say over a compact abelian group. However, this kind of simplification appears at [math]M=2[/math], and since phenomenon this is quite interesting, we will explain this now. As a first remark, at [math]M=2[/math] we have:

Proposition

For [math]F_2\otimes_QF_H[/math], with [math]Q\in M_{2\times N}(\mathbb T)[/math] generic, we have

[[math]] N\int\left(\frac{\chi}{N}\right)^p=\int_{\mathbb T^N}\sum_{k\geq0}\binom{p}{2k}\left|\frac{a_1+\ldots+a_N}{N}\right|^{2k}da [[/math]]

where the integral on the right is with respect to the uniform measure on [math]\mathbb T^N[/math].

Show Proof

In order to prove the result, consider the following quantity, which appeared in the proof of Theorem 16.17:

[[math]] \Phi(q)=\sum_{i_1\ldots i_p}\sum_{d_1\ldots d_p}\frac{q_{i_1d_1}\ldots q_{i_pd_p}}{q_{i_1d_p}\ldots q_{i_pd_{p-1}}} [[/math]]

We can “half-dephase” the matrix [math]q\in M_{2\times N}(\mathbb T)[/math] if we want to, as follows:

[[math]] q=\begin{pmatrix}1&\ldots&1\\ a_1&\ldots&a_N\end{pmatrix} [[/math]]

Let us compute now the above quantity [math]\Phi(q)[/math], in terms of the numbers [math]a_1,\ldots,a_N[/math]. Our claim is that we have the following formula:

[[math]] \Phi(q)=2\sum_{k\geq0}N^{p-2k}\binom{p}{2k}\left|\sum_ia_i\right|^{2k} [[/math]]

Indeed, the idea is that:

-- The [math]2N^k[/math] contribution will come from [math]i=(1\ldots1)[/math] and [math]i=(2\ldots2)[/math].

-- Then we will have a [math]p(p-1)N^{k-2}|\sum_ia_i|^2[/math] contribution coming from indices of type [math]i=(2\ldots 21\ldots1)[/math], up to cyclic permutations.

-- Then we will have a [math]2\binom{p}{4}N^{p-4}|\sum_ia_i|^4[/math] contribution coming from indices of type [math]i=(2\ldots 21\ldots12\ldots21\ldots1)[/math].

-- And so on.

In practice now, this gives the result. Indeed, in order to prove our claim, in order to find the [math]N^{p-2k}|\sum_ia_i|^{2k}[/math] contribution, we have to count the circular configurations consisting of [math]p[/math] numbers [math]1,2[/math], such that the [math]1[/math] values are arranged into [math]k[/math] non-empty intervals, and the [math]2[/math] values are arranged into [math]k[/math] non-empty intervals as well. Now by looking at the endpoints of these [math]2k[/math] intervals, we have [math]2\binom{p}{2k}[/math] choices, and this gives the above formula. Now by integrating, this gives the formula in the statement.

■

Observe now that the integrals in Proposition 16.18 can be computed as follows:

[[math]] \begin{eqnarray*} \int_{\mathbb T^N}|a_1+\ldots+a_N|^{2k}da &=&\int_{\mathbb T^N}\sum_{i_1\ldots i_k}\sum_{j_1\ldots j_k}\frac{a_{i_1}\ldots a_{i_k}}{a_{j_1}\ldots a_{j_k}}da\\ &=&\#\left\{i_1\ldots i_k,j_1\ldots j_k\Big|[i_1,\ldots,i_k]=[j_1,\ldots,j_k]\right\}\\ &=&\sum_{k=\sum r_i}\binom{k}{r_1,\ldots,r_N}^2 \end{eqnarray*} [[/math]]

We obtain in this way the following “blowup” result, for our measure:

Proposition

For [math]F_2\otimes_QF_H[/math], with [math]Q\in M_{2\times N}(\mathbb T)[/math] generic, we have

[[math]] \mu=\left(1-\frac{1}{N}\right)\delta_0+\frac{1}{2N}\left(\Psi^+_*\varepsilon+\Psi^-_*\varepsilon\right) [[/math]]

where [math]\varepsilon[/math] is the uniform measure on [math]\mathbb T^N[/math], and where the blowup function is:

[[math]] \Psi^\pm(a)=N\pm\left|\sum_ia_i\right| [[/math]]

Show Proof

We use the formula found in Proposition 16.18, along with the following standard identity, coming from the Taylor formula:

[[math]] \sum_{k\geq0}\binom{p}{2k}x^{2k}=\frac{(1+x)^p+(1-x)^p}{2} [[/math]]

By using this identity, Proposition 16.18 reformulates as follows:

[[math]] N\int\left(\frac{\chi}{N}\right)^p=\frac{1}{2}\int_{\mathbb T^N}\left(1+\left|\frac{\sum_ia_i}{N}\right|\right)^p+\left(1-\left|\frac{\sum_ia_i}{N}\right|\right)^p\,da [[/math]]

Now by multiplying by [math]N^{p-1}[/math], we obtain the following formula:

[[math]] \int\chi^k=\frac{1}{2N}\int_{\mathbb T^N}\left(N+\left|\sum_ia_i\right|\right)^p+\left(N-\left|\sum_ia_i\right|\right)^p\,da [[/math]]

But this gives the formula in the statement, and we are done.

■

We can further improve the above result, by reducing the maps [math]\Psi^\pm[/math] appearing there to a single one, and we are led to the following statement:

Theorem

For [math]F_2\otimes_QF_H[/math], with [math]Q\in M_{2\times N}(\mathbb T)[/math] generic, we have

[[math]] \mu=\left(1-\frac{1}{N}\right)\delta_0+\frac{1}{N}\Phi_*\varepsilon [[/math]]

where [math]\varepsilon[/math] is the uniform measure on [math]\mathbb Z_2\times\mathbb T^N[/math], and where the blowup map is:

[[math]] \Phi(e,a)=N+e\left|\sum_ia_i\right| [[/math]]

Show Proof

This is clear indeed from Proposition 16.19.

■

As already mentioned, the above results at [math]M=2[/math] are something quite special. In the general case, [math]M\in\mathbb N[/math], it is not clear how to construct a nice blowup of the measure. All the above results are quite interesting in the general context of subfactor theory, where the blowup question is one of the main open questions, related to the continuations of Jones' planar algebra work in , and to many other things, mainly coming from advanced quantum physics. For more on all this, we refer to and its previous versions, which were more subfactor-centered, and which can be found on the internet.

General references

Banica, Teo (2024). "Invitation to Hadamard matrices". arXiv:1910.06911 [math.CO].