16c. Measure blowup

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

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All the above, which was quite nice, was about index [math]N\in[1,4][/math], where the Jones annular theory result from does not apply. In higher index now, [math]N\in(4,\infty)[/math], where the Jones result does apply, the precise correct “blowup” manipulation on the spectral measure is not known yet. The known results here are as follows:

• One one hand, there is as a computation for some basic Hadamard subfactors, with nice blowup, on a certain noncommutative manifold .
• On the other hand, there are many computations by Evans-Pugh, with quite technical blowup results, on some suitable real algebraic manifolds .

We will discuss in what follows (1), and to be more precise the computation of the spectral measure, and then the blowup problem, for the subfactors coming from the deformed Fourier matrices. Let us start with the following definition:

Definition

Given two finite abelian groups [math]G,H[/math], we consider the corresponding deformed Fourier matrix, given by the formula

[[math]] (F_G\otimes_Q F_H)_{ia,jb}=Q_{ib}(F_G)_{ij}(F_H)_{ab} [[/math]]

and we factorize the associated representation [math]\pi_Q[/math] of the algebra [math]C(S_{G\times H}^+)[/math],

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C(G_Q)\ar[ur]_\pi&} [[/math]]

with [math]C(G_Q)[/math] being the Hopf image of this representation [math]\pi_Q[/math].

Explicitely computing the above quantum permutation group [math]G_Q\subset S_{G\times H}^+[/math], as function of the parameter matrix [math]Q\in M_{G\times H}(\mathbb T)[/math], will be our main purpose, in what follows. In order to do so, we first have the following elementary result:

Proposition

We have a factorization as follows,

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C(H\wr_*G)\ar[ur]_\pi&} [[/math]]

given on the standard generators by the formulae

[[math]] U_{ab}^{(i)}=\sum_jW_{ia,jb}\quad,\quad V_{ij}=\sum_aW_{ia,jb} [[/math]]

independently of [math]b[/math], where [math]W[/math] is the magic matrix producing [math]\pi_Q[/math].

Show Proof

With [math]K=F_G,L=F_H[/math] and [math]M=|G|,N=|H|[/math], the formula of the magic matrix [math]W\in M_{G\times H}(M_{G\times H}(\mathbb C))[/math] associated to [math]H=K\otimes_QL[/math] is as follows:

[[math]] \begin{eqnarray*} (W_{ia,jb})_{kc,ld} &=&\frac{1}{MN}\cdot\frac{Q_{ic}Q_{jd}}{Q_{id}Q_{jc}}\cdot\frac{K_{ik}K_{jl}}{K_{il}K_{jk}}\cdot\frac{L_{ac}L_{bd}}{L_{ad}L_{bc}}\\ &=&\frac{1}{MN}\cdot\frac{Q_{ic}Q_{jd}}{Q_{id}Q_{jc}}\cdot K_{i-j,k-l}L_{a-b,c-d} \end{eqnarray*} [[/math]]

Our claim now is that the representation [math]\pi_Q[/math] constructed in Definition 16.21 can be factorized in three steps, up to the factorization in the statement, as follows:

[[math]] \xymatrix@R=60pt@C=50pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[d]&&M_{G\times H}(\mathbb C)\\ C(S_H^+\wr_*S_G^+)\ar[r]\ar@{. \gt }[rru]&C(S_H^+\wr_*G)\ar[r]\ar@{. \gt }[ur]&C(H\wr_*G)\ar@{. \gt }[u]} [[/math]]

Indeed, the construction of the map on the left is standard. Regarding the second factorization, this comes from the fact that since the elements [math]V_{ij}[/math] depend on [math]i-j[/math], they satisfy the defining relations for the quotient algebra [math]C(S_G^+)\to C(G)[/math]. Finally, regarding the third factorization, observe that [math]W_{ia,jb}[/math] depends only on [math]i,j[/math] and on [math]a-b[/math]. By summing over [math]j[/math] we obtain that the elements [math]U_{ab}^{(i)}[/math] depend only on [math]a-b[/math], and we are done.

■

We have now all needed ingredients for refining Proposition 16.22, as follows:

Proposition

We have a factorization as follows,

[[math]] \xymatrix@R=40pt@C=30pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C^*(\Gamma_{G,H})\rtimes C(G)\ar[ur]_\rho&} [[/math]]

where the group on the bottom is given by:

[[math]] \Gamma_{G,H}=H^{*G}\Big/\left \lt [c_1^{(i_1)}\ldots c_s^{(i_s)},d_1^{(j_1)}\ldots d_s^{(j_s)}]=1\Big|\sum_rc_r=\sum_rd_r=0\right \gt [[/math]]

Show Proof

Assume that we have a representation, as follows:

[[math]] \pi:C^*(\Gamma)\rtimes C(G)\to M_L(\mathbb C) [[/math]]

Let [math]\Lambda[/math] be a [math]G[/math]-stable normal subgroup of [math]\Gamma[/math], so that [math]G[/math] acts on [math]\Gamma/\Lambda[/math], and we can form the product [math]C^*(\Gamma/\Lambda)\rtimes C(G)[/math], and assume that [math]\pi[/math] is trivial on [math]\Lambda[/math]. Then [math]\pi[/math] factorizes as:

[[math]] \xymatrix@R=40pt@C=30pt {C^*(\Gamma)\rtimes C(G)\ar[rr]^\pi\ar[rd]&&M_L(\mathbb C)\\&C^*(\Gamma/\Lambda)\rtimes C(G)\ar[ur]_\rho} [[/math]]

With [math]\Gamma=H^{*G}[/math], this gives the result.

■

We have now all the needed ingredients for proving a main result, as follows:

Theorem

When [math]Q[/math] is generic, the minimal factorization for [math]\pi_Q[/math] is

[[math]] \xymatrix@R=45pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C^*(\Gamma_{G,H})\rtimes C(G)\ar[ur]_\pi&} [[/math]]

where on the bottom

[[math]] \Gamma_{G,H}\simeq\mathbb Z^{(|G|-1)(|H|-1)}\rtimes H [[/math]]

is the discrete group constructed above.

Show Proof

Consider the factorization in Proposition 16.23, which is as follows, where [math]L[/math] denotes the Hopf image of [math]\pi_Q[/math]:

[[math]] \theta : C^*(\Gamma_{G,H})\rtimes C(G)\to L [[/math]]

To be more precise, this morphism produces the following commutative diagram:

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+) \ar[rr]^{\pi_Q} \ar[dr]_{} \ar@/_/[ddr]_{}& & M_{G\times H}(\mathbb C) \\ & L \ar[ur]_{} & \\ & C^*(\Gamma_{G,H})\rtimes C(G) \ar@{-- \gt }[u]_\theta \ar@/_/[uur]_{\pi}& } [[/math]]

The first observation is that the injectivity assumption on [math]C(G)[/math] holds by construction, and that for [math]f \in C(G)[/math], the matrix [math]\pi(f)[/math] is “block scalar”. Now for [math]r \in \Gamma_{G,H}[/math] with [math]\theta(r\otimes 1)=\theta(1 \otimes f)[/math] for some [math]f \in C(G)[/math], we see, using the commutative diagram, that [math]\pi(r \otimes 1)[/math] is block scalar. Thus, modulo some standard algebra, we are done.

■

Summarizing, we have computed the quantum permutation groups associated to the Di\c t\u a deformations of the tensor products of Fourier matrices, in the case where the deformation matrix [math]Q[/math] is generic. For some further computations, in the case where the deformation matrix [math]Q[/math] is no longer generic, we refer to the follow-ups of .

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 16.24 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 16.24 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]. With standard notations, 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 16.25 can be further 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

[[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 16.25, 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 16.26, with [math]A[/math] replaced by a plain, non-matricial random variable, say over a compact abelian group.

However, this kind of simplification does appear at [math]M=2[/math], and since this phenomenon is quite interesting, we will explain this now. We first 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 16.26:

[[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:

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

(2) 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.

(3) 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].

(4) And so on.

In practice now, 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 16.27 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 16.27, 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 16.27 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 16.28.

■

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.

Asymptotically, things are however quite simple. Let us go back indeed to the general case, where [math]M,N\in\mathbb N[/math] are both arbitrary. The problem that we would like to solve now is that of finding the good regime, of the following type, where the measure in Theorem 16.25 converges, after some suitable manipulations:

[[math]] M=f(K)\quad,\quad N=g(K)\quad,\quad K\to\infty [[/math]]

In order to do so, we have to do some combinatorics. Let [math]NC(p)[/math] be the set of noncrossing partitions of [math]\{1,\ldots,p\}[/math], and for [math]\pi\in P(p)[/math] we denote by [math]|\pi|\in\{1,\ldots,p\}[/math] the number of blocks. With these conventions, we have the following result:

Proposition

With [math]M=\alpha K,N=\beta K[/math], [math]K\to\infty[/math] we have:

[[math]] \frac{c_p}{K^{p-1}}\simeq\sum_{r=1}^p\#\left\{\pi\in NC(p)\Big||\pi|=r\right\}\alpha^{r-1}\beta^{p-r} [[/math]]

In particular, with [math]\alpha=\beta[/math] we have:

[[math]] c_p\simeq\frac{1}{p+1}\binom{2p}{p}(\alpha K)^{p-1} [[/math]]

Show Proof

We use the combinatorial formula in Theorem 16.25. Our claim is that, with [math]\pi=\ker(i_1,\ldots,i_p)[/math], the corresponding contribution to [math]c_p[/math] is:

[[math]] C_\pi\simeq \begin{cases} \alpha^{|\pi|-1}\beta^{p-|\pi|}K^{p-1}&{\rm if}\ \pi\in NC(p)\\ O(K^{p-2})&{\rm if}\ \pi\notin NC(p) \end{cases} [[/math]]

As a first observation, the number of choices for a multi-index [math](i_1,\ldots,i_p)\in X^p[/math] satisfying [math]\ker i=\pi[/math] is:

[[math]] M(M-1)\ldots (M-|\pi|+1)\simeq M^{|\pi|} [[/math]]

Thus, we have the following estimate:

[[math]] C_\pi\simeq M^{|\pi|-1}N^{-1}\#\left\{d_1,\ldots,d_p\in Y\Big|[d_\alpha|\alpha\in b]=[d_{\alpha-1}|\alpha\in b],\forall b\in\pi\right\} [[/math]]

Consider now the following partition:

[[math]] \sigma=\ker d [[/math]]

The contribution of [math]\sigma[/math] to the above quantity [math]C_\pi[/math] is then given by:

[[math]] \Delta(\pi,\sigma)N(N-1)\ldots(N-|\sigma|+1)\simeq\Delta(\pi,\sigma)N^{|\sigma|} [[/math]]

Here the quantities on the right are as follows:

[[math]] \Delta(\pi,\sigma)=\begin{cases} 1&{\rm if}\ |b\cap c|=|(b-1)\cap c|,\forall b\in\pi,\forall c\in\sigma\\ 0&{\rm otherwise} \end{cases} [[/math]]

We use now the standard fact that for [math]\pi,\sigma\in P(p)[/math] satisfying [math]\Delta(\pi,\sigma)=1[/math] we have:

[[math]] |\pi|+|\sigma|\leq p+1 [[/math]]

In addition, the equality case is well-known to happen when [math]\pi,\sigma\in NC(p)[/math] are inverse to each other, via Kreweras complementation. This shows that for [math]\pi\notin NC(p)[/math] we have:

[[math]] C_\pi=O(K^{p-2}) [[/math]]

Also, this shows that for [math]\pi\in NC(p)[/math] we have:

[[math]] \begin{eqnarray*} C_\pi &\simeq&M^{|\pi|-1}N^{-1}N^{p-|\pi|-1}\\ &=&\alpha^{|\pi|-1}\beta^{p-|\pi|}K^{p-1} \end{eqnarray*} [[/math]]

Thus, we have obtained the result.

■

We denote by [math]D[/math] the dilation operation, given by:

[[math]] D_r(law(X))=law(rX) [[/math]]

With this convention, we have the following result:

Theorem

With [math]M=\alpha K,N=\beta K[/math], [math]K\to\infty[/math] we have:

[[math]] \mu=\left(1-\frac{1}{\alpha\beta K^2}\right)\delta_0+\frac{1}{\alpha\beta K^2}D_{\frac{1}{\beta K}}(\pi_{\alpha/\beta}) [[/math]]

In particular with [math]\alpha=\beta[/math] we have:

[[math]] \mu=\left(1-\frac{1}{\alpha^2K^2}\right)\delta_0+\frac{1}{\alpha^2K^2}D_{\frac{1}{\alpha K}}(\pi_1) [[/math]]

Show Proof

At [math]\alpha=\beta[/math], this follows from Proposition 16.30. In general now, we have:

[[math]] \begin{eqnarray*} \frac{c_p}{K^{p-1}} &\simeq&\sum_{\pi\in NC(p)}\alpha^{|\pi|-1}\beta^{p-|\pi|}\\ &=&\frac{\beta^p}{\alpha}\sum_{\pi\in NC(p)}\left(\frac{\alpha}{\beta}\right)^{|\pi|}\\ &=&\frac{\beta^p}{\alpha}\int x^pd\pi_{\alpha/\beta}(x) \end{eqnarray*} [[/math]]

When [math]\alpha\geq\beta[/math], where [math]d\pi_{\alpha/\beta}(x)=\varphi_{\alpha/\beta}(x)dx[/math] is continuous, we obtain:

[[math]] \begin{eqnarray*} c_p &=&\frac{1}{\alpha K}\int(\beta Kx)^p\varphi_{\alpha/\beta}(x)dx\\ &=&\frac{1}{\alpha\beta K^2}\int x^p\varphi_{\alpha/\beta}\left(\frac{x}{\beta K}\right)dx \end{eqnarray*} [[/math]]

But this gives the formula in the statement. When [math]\alpha\leq\beta[/math] the computation is similar, with a Dirac mass as 0 dissapearing and reappearing, and gives the same result.

■

We refer to and related papers for more on the above.

General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].