14b. Quotient spaces

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We have seen so far that free geometry is a broad and fluffy subject, with countless potential paths to be taken, and interesting ramifications, and no wonder here, because hundreds of books have been written on classical geometry, and it is probably possible to write as many on free geometry. In practice now, this suggests thinking a bit, and making some good choices for the remainder of this chapter. Our choices will be as follows:

(1) We will first explain how Weingarten integration and the Bercovici-Pata bijection work, for a remarkable class of homogeneous spaces, generalizing the spheres.

(2) Then, we will go back to the question of going beyond Bercovici-Pata, and we will discuss here the free hyperspherical laws, and the free hypergeometric laws.

Getting started now, we would like to find a suitable collection of “free homogeneous spaces”, generalizing at the same time the free spheres [math]S[/math], and the free unitary groups [math]U[/math]. This can be done at several levels of generality, and central here is the construction of the free spaces of partial isometries, which can be done in fact for any easy quantum group. In order to explain this, let us start with the classical case. We have here:

Definition

Associated to any integers [math]L\leq M,N[/math] are the spaces

[[math]] O_{MN}^L=\left\{T:E\to F\ {\rm isometry}\Big|E\subset\mathbb R^N,F\subset\mathbb R^M,\dim_\mathbb RE=L\right\} [[/math]]

[[math]] U_{MN}^L=\left\{T:E\to F\ {\rm isometry}\Big|E\subset\mathbb C^N,F\subset\mathbb C^M,\dim_\mathbb CE=L\right\} [[/math]]

where the notion of isometry is with respect to the usual real/complex scalar products.

As a first observation, at [math]L=M=N[/math] we obtain the groups [math]O_N,U_N[/math]:

[[math]] O_{NN}^N=O_N\quad,\quad U_{NN}^N=U_N [[/math]]

Another interesting specialization is [math]L=M=1[/math]. Here the elements of [math]O_{1N}^1[/math] are the isometries [math]T:E\to\mathbb R[/math], with [math]E\subset\mathbb R^N[/math] one-dimensional. But such an isometry is uniquely determined by [math]T^{-1}(1)\in\mathbb R^N[/math], which must belong to [math]S^{N-1}_\mathbb R[/math]. Thus, we have [math]O_{1N}^1=S^{N-1}_\mathbb R[/math]. Similarly, in the complex case we have [math]U_{1N}^1=S^{N-1}_\mathbb C[/math], and so our results here are:

[[math]] O_{1N}^1=S^{N-1}_\mathbb R\quad,\quad U_{1N}^1=S^{N-1}_\mathbb C [[/math]]

Yet another interesting specialization is [math]L=N=1[/math]. Here the elements of [math]O_{1N}^1[/math] are the isometries [math]T:\mathbb R\to F[/math], with [math]F\subset\mathbb R^M[/math] one-dimensional. But such an isometry is uniquely determined by [math]T(1)\in\mathbb R^M[/math], which must belong to [math]S^{M-1}_\mathbb R[/math]. Thus, we have [math]O_{M1}^1=S^{M-1}_\mathbb R[/math]. Similarly, in the complex case we have [math]U_{M1}^1=S^{M-1}_\mathbb C[/math], and so our results here are:

[[math]] O_{M1}^1=S^{M-1}_\mathbb R\quad,\quad U_{M1}^1=S^{M-1}_\mathbb C [[/math]]

In general, the most convenient is to view the elements of [math]O_{MN}^L,U_{MN}^L[/math] as rectangular matrices, and to use matrix calculus for their study. We have indeed:

Proposition

We have identifications of compact spaces

[[math]] O_{MN}^L\simeq\left\{U\in M_{M\times N}(\mathbb R)\Big|UU^t={\rm projection\ of\ trace}\ L\right\} [[/math]]

[[math]] U_{MN}^L\simeq\left\{U\in M_{M\times N}(\mathbb C)\Big|UU^*={\rm projection\ of\ trace}\ L\right\} [[/math]]

with each partial isometry being identified with the corresponding rectangular matrix.

Show Proof

We can indeed identify the partial isometries [math]T:E\to F[/math] with their corresponding extensions [math]U:\mathbb R^N\to\mathbb R^M[/math], [math]U:\mathbb C^N\to\mathbb C^M[/math], obtained by setting [math]U_{E^\perp}=0[/math]. Then, we can identify these latter maps [math]U[/math] with the corresponding rectangular matrices.

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In order to advance, observe now that the isometries [math]T:E\to F[/math], or rather their extensions [math]U:\mathbb K^N\to\mathbb K^M[/math], with [math]\mathbb K=\mathbb R,\mathbb C[/math], obtained by setting [math]U_{E^\perp}=0[/math], can be composed with the isometries of [math]\mathbb K^M,\mathbb K^N[/math], according to the following scheme:

[[math]] \xymatrix@R=17mm@C=17mm{ \mathbb K^N\ar[r]^{B^*}&\mathbb K^N\ar@.[r]^U&\mathbb K^M\ar[r]^A&\mathbb K^M\\ B(E)\ar@.[r]\ar[u]&E\ar[r]^T\ar[u]&F\ar@.[r]\ar[u]&A(F)\ar[u] } [[/math]]

With the identifications in Proposition 14.15 made, the precise statement here is:

Proposition

We have action maps as follows, which are both transitive,

[[math]] O_M\times O_N\curvearrowright O_{MN}^L\quad,\quad (A,B)U=AUB^t [[/math]]

[[math]] U_M\times U_N\curvearrowright U_{MN}^L\quad,\quad (A,B)U=AUB^* [[/math]]

whose stabilizers are respectively [math]O_L\times O_{M-L}\times O_{N-L}[/math] and [math]U_L\times U_{M-L}\times U_{N-L}[/math].

Show Proof

We have indeed action maps as in the statement, which are transitive. Let us compute now the stabilizer [math]G[/math] of the following point:

[[math]] U=\begin{pmatrix}1&0\\0&0\end{pmatrix} [[/math]]

Since [math](A,B)\in G[/math] satisfy [math]AU=UB[/math], their components must be of the following form:

[[math]] A=\begin{pmatrix}x&*\\0&a\end{pmatrix}\quad,\quad B=\begin{pmatrix}x&0\\ *&b\end{pmatrix} [[/math]]

Now since [math]A,B[/math] are unitaries, these matrices follow to be block-diagonal, and so:

[[math]] G=\left\{(A,B)\Big|A=\begin{pmatrix}x&0\\0&a\end{pmatrix},B=\begin{pmatrix}x&0\\ 0&b\end{pmatrix}\right\} [[/math]]

The stabilizer of [math]U[/math] is parametrized by triples [math](x,a,b)[/math] belonging to [math]O_L\times O_{M-L}\times O_{N-L}[/math] and [math]U_L\times U_{M-L}\times U_{N-L}[/math], and we are led to the conclusion in the statement.

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Finally, let us work out the quotient space description of [math]O_{MN}^L,U_{MN}^L[/math]. We have here:

Theorem

We have isomorphisms of homogeneous spaces as follows,

[[math]] \begin{eqnarray*} O_{MN}^L&=&(O_M\times O_N)/(O_L\times O_{M-L}\times O_{N-L})\\ U_{MN}^L&=&(U_M\times U_N)/(U_L\times U_{M-L}\times U_{N-L}) \end{eqnarray*} [[/math]]

with the quotient maps being given by [math](A,B)\to AUB^*[/math], where [math]U=(^1_0{\ }^0_0)[/math].

Show Proof

This is just a reformulation of Proposition 14.16, by taking into account the fact that the fixed point used in the proof there was [math]U=(^1_0{\ }^0_0)[/math].

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Summarizing, we have here some basic homogeneous spaces, unifying the spheres with the rotation groups. The point now is that we can liberate these spaces, as follows:

Definition

Associated to any integers [math]L\leq M,N[/math] are the algebras

[[math]] \begin{eqnarray*} C(O_{MN}^{L+})&=&C^*\left((v_{ij})_{i=1,\ldots,M,j=1,\ldots,N}\Big|v=\bar{v},vv^t={\rm projection\ of\ trace}\ L\right)\\ C(U_{MN}^{L+})&=&C^*\left((v_{ij})_{i=1,\ldots,M,j=1,\ldots,N}\Big|vv^*,\bar{v}v^t={\rm projections\ of\ trace}\ L\right) \end{eqnarray*} [[/math]]

with the trace being by definition the sum of the diagonal entries.

Observe that the above universal algebras are indeed well-defined, as it was previously the case for the free spheres, and this due to the trace conditions, which read:

[[math]] \sum_{ij}v_{ij}v_{ij}^* =\sum_{ij}v_{ij}^*v_{ij} =L [[/math]]

We have inclusions between the various spaces constructed so far, as follows:

[[math]] \xymatrix@R=15mm@C=15mm{ O_{MN}^{L+}\ar[r]&U_{MN}^{L+}\\ O_{MN}^L\ar[r]\ar[u]&U_{MN}^L\ar[u]} [[/math]]

At the level of basic examples now, at [math]L=M=1[/math] and at [math]L=N=1[/math] we obtain the following diagrams, showing that our formalism covers indeed the free spheres:

[[math]] \xymatrix@R=15mm@C=15mm{ S^{N-1}_{\mathbb R,+}\ar[r]&S^{N-1}_{\mathbb C,+}\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_\mathbb C\ar[u]} \qquad\qquad \item[a]ymatrix@R=15mm@C=15mm{ S^{M-1}_{\mathbb R,+}\ar[r]&S^{M-1}_{\mathbb C,+}\\ S^{M-1}_\mathbb R\ar[r]\ar[u]&S^{M-1}_\mathbb C\ar[u]} [[/math]]

We have as well the following result, in relation with the free rotation groups:

Proposition

At [math]L=M=N[/math] we obtain the diagram

[[math]] \xymatrix@R=15mm@C=15mm{ O_N^+\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&U_N\ar[u]} [[/math]]

consisting of the groups [math]O_N,U_N[/math], and their liberations.

Show Proof

According to the above, we have the following presentation results:

[[math]] \begin{eqnarray*} C(O_{NN}^{N\times})&=&C^*_\times\left((v_{ij})_{i,j=1,\ldots,N}\Big|v=\bar{v},vv^t={\rm projection\ of\ trace}\ N\right)\\ C(U_{NN}^{N\times})&=&C^*_\times\left((v_{ij})_{i,j=1,\ldots,N}\Big|vv^*,\bar{v}v^t={\rm projections\ of\ trace}\ N\right) \end{eqnarray*} [[/math]]

We use now the standard fact that if [math]p=aa^*[/math] is a projection then [math]q=a^*a[/math] is a projection too. We use as well the following formulae:

[[math]] Tr(vv^*)=Tr(v^t\bar{v})\quad,\quad Tr(\bar{v}v^t)=Tr(v^*v) [[/math]]

We therefore obtain the following formulae:

[[math]] \begin{eqnarray*} C(O_{NN}^{N\times})&=&C^*_\times\left((v_{ij})_{i,j=1,\ldots,N}\Big|v=\bar{v},\ vv^t,v^tv={\rm projections\ of\ trace}\ N\right)\\ C(U_{NN}^{N\times})&=&C^*_\times\left((v_{ij})_{i,j=1,\ldots,N}\Big|vv^*,v^*v,\bar{v}v^t,v^t\bar{u}={\rm projections\ of\ trace}\ N\right) \end{eqnarray*} [[/math]]

Now observe that, in tensor product notation, the conditions at right are all of the form [math](tr\otimes id)p=1[/math]. Thus, [math]p[/math] must be follows, for the above conditions:

[[math]] p=vv^*,v^*v,\bar{v}v^t,v^t\bar{v} [[/math]]

We therefore obtain that, for any faithful state [math]\varphi[/math], we have [math](tr\otimes\varphi)(1-p)=0[/math]. It follows from this that the following projections must be all equal to the identity:

[[math]] p=vv^*,v^*v,\bar{v}v^t,v^t\bar{v} [[/math]]

But this leads to the conclusion in the statement.

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Regarding now the homogeneous space structure of [math]O_{MN}^{L\times},U_{MN}^{L\times}[/math], the situation here is a bit more complicated in the free case than in the classical case, due to a number of algebraic and analytic issues. We first have the following result:

Proposition

The spaces [math]U_{MN}^{L\times}[/math] have the following properties:

• We have an action [math]U_M^\times\times U_N^\times\curvearrowright U_{MN}^{L\times}[/math], given by [math]v_{ij}\to\sum_{kl}v_{kl}\otimes a_{ki}\otimes b_{lj}^*[/math].
• We have a map [math]U_M^\times\times U_N^\times\to U_{MN}^{L\times}[/math], given by [math]v_{ij}\to\sum_{r\leq L}a_{ri}\otimes b_{rj}^*[/math].

Similar results hold for the spaces [math]O_{MN}^{L\times}[/math], with all the [math]*[/math] exponents removed.

Show Proof

In the classical case, consider the following action and quotient maps:

[[math]] U_M\times U_N\curvearrowright U_{MN}^L\quad,\quad U_M\times U_N\to U_{MN}^L [[/math]]

The transposes of these two maps are as follows, where [math]J=(^1_0{\ }^0_0)[/math]:

[[math]] \begin{eqnarray*} \varphi&\to&((U,A,B)\to\varphi(AUB^*))\\ \varphi&\to&((A,B)\to\varphi(AJB^*)) \end{eqnarray*} [[/math]]

But with [math]\varphi=v_{ij}[/math] we obtain precisely the formulae in the statement. The proof in the orthogonal case is similar. Regarding now the free case, the proof goes as follows:

(1) Assuming [math]vv^*v=v[/math], let us set [math]U_{ij}=\sum_{kl}v_{kl}\otimes a_{ki}\otimes b_{lj}^*[/math]. We have then:

[[math]] \begin{eqnarray*} (UU^*U)_{ij} &=&\sum_{pq}\sum_{klmnst}v_{kl}v_{mn}^*v_{st}\otimes a_{ki}a_{mq}^*a_{sq}\otimes b_{lp}^*b_{np}b_{tj}^*\\ &=&\sum_{klmt}v_{kl}v_{ml}^*v_{mt}\otimes a_{ki}\otimes b_{tj}^*\\ &=&\sum_{kt}v_{kt}\otimes a_{ki}\otimes b_{tj}^*\\ &=&U_{ij} \end{eqnarray*} [[/math]]

Also, assuming that we have [math]\sum_{ij}v_{ij}v_{ij}^*=L[/math], we obtain:

[[math]] \begin{eqnarray*} \sum_{ij}U_{ij}U_{ij}^* &=&\sum_{ij}\sum_{klst}v_{kl}v_{st}^*\otimes a_{ki}a_{si}^*\otimes b_{lj}^*b_{tj}\\ &=&\sum_{kl}v_{kl}v_{kl}^*\otimes1\otimes1\\ &=&L \end{eqnarray*} [[/math]]

(2) Assuming [math]vv^*v=v[/math], let us set [math]V_{ij}=\sum_{r\leq L}a_{ri}\otimes b_{rj}^*[/math]. We have then:

[[math]] \begin{eqnarray*} (VV^*V)_{ij} &=&\sum_{pq}\sum_{x,y,z\leq L}a_{xi}a_{yq}^*a_{zq}\otimes b_{xp}^*b_{yp}b_{zj}^*\\ &=&\sum_{x\leq L}a_{xi}\otimes b_{xj}^*\\ &=&V_{ij} \end{eqnarray*} [[/math]]

Finally, assuming that we have [math]\sum_{ij}u_{ij}u_{ij}^*=L[/math], we obtain:

[[math]] \sum_{ij}V_{ij}V_{ij}^* =\sum_{ij}\sum_{r,s\leq L}a_{ri}a_{si}^*\otimes b_{rj}^*b_{sj} =\sum_{l\leq L}1 =L [[/math]]

By removing all the [math]*[/math] exponents, we obtain as well the orthogonal results.

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Let us examine now the relation between the above maps. In the classical case, given a quotient space [math]X=G/H[/math], the associated action and quotient maps are given by:

[[math]] \begin{cases} a:X\times G\to X&:\quad (Hg,h)\to Hgh\\ p:G\to X&:\quad g\to Hg \end{cases} [[/math]]

Thus we have [math]a(p(g),h)=p(gh)[/math]. In our context, a similar result holds:

Theorem

With [math]G=G_M\times G_N[/math] and [math]X=G_{MN}^L[/math], where [math]G_N=O_N^\times,U_N^\times[/math], we have

[[math]] \xymatrix@R=15mm@C=30mm{ G\times G\ar[r]^m\ar[d]_{p\times id}&G\ar[d]^p\\ X\times G\ar[r]^a&X } [[/math]]

where [math]a,p[/math] are the action map and the map constructed in Proposition 14.20.

Show Proof

At the level of the associated algebras of functions, we must prove that the following diagram commutes, where [math]\Phi,\alpha[/math] are morphisms of algebras induced by [math]a,p[/math]:

[[math]] \xymatrix@R=15mm@C=25mm{ C(X)\ar[r]^\Phi\ar[d]_\alpha&C(X\times G)\ar[d]^{\alpha\otimes id}\\ C(G)\ar[r]^\Delta&C(G\times G) } [[/math]]

When going right, and then down, the composition is as follows:

[[math]] \begin{eqnarray*} (\alpha\otimes id)\Phi(u_{ij}) &=&(\alpha\otimes id)\sum_{kl}v_{kl}\otimes a_{ki}\otimes b_{lj}^*\\ &=&\sum_{kl}\sum_{r\leq L}a_{rk}\otimes b_{rl}^*\otimes a_{ki}\otimes b_{lj}^* \end{eqnarray*} [[/math]]

On the other hand, when going down, and then right, the composition is as follows, where [math]F_{23}[/math] is the flip between the second and the third components:

[[math]] \begin{eqnarray*} \Delta\pi(u_{ij}) &=&F_{23}(\Delta\otimes\Delta)\sum_{r\leq L}a_{ri}\otimes b_{rj}^*\\ &=&F_{23}\left(\sum_{r\leq L}\sum_{kl}a_{rk}\otimes a_{ki}\otimes b_{rl}^*\otimes b_{lj}^*\right) \end{eqnarray*} [[/math]]

Thus the above diagram commutes indeed, and this gives the result.

■

Let us discuss now the integration over the above spaces [math]G_{MN}^L[/math]. We first have:

Definition

The integration functional of [math]G_{MN}^L[/math] is the composition

[[math]] \int_{G_{MN}^L}:C(G_{MN}^L)\to C(G_M\times G_N)\to\mathbb C [[/math]]

of the representation [math]v_{ij}\to\sum_{r\leq L}a_{ri}\otimes b_{rj}^*[/math] with the Haar functional of [math]G_M\times G_N[/math].

As an illustration here, observe that in the case [math]L=M=N[/math] we obtain the integration over [math]G_N[/math]. Also, at [math]L=M=1[/math], or at [math]L=N=1[/math], we obtain the integration over the sphere. In the general case now, we first have the following result:

Proposition

The integration functional of [math]G_{MN}^L[/math] has the invariance property

[[math]] \left(\int_{G_{MN}^L}\!\otimes\ id\right)\Phi(x)=\int_{G_{MN}^L}x [[/math]]

with respect to the coaction map [math]\Phi(v_{ij})=\sum_{kl}v_{kl}\otimes a_{ki}\otimes b_{lj}^*[/math].

Show Proof

We can restrict the attention to the orthogonal case, the proof in the unitary case being similar. We must check the following formula:

[[math]] \left(\int_{G_{MN}^L}\!\otimes\ id\right)\Phi(v_{i_1j_1}\ldots v_{i_sj_s})=\int_{G_{MN}^L}v_{i_1j_1}\ldots v_{i_sj_s} [[/math]]

Let us compute the left term. This is given by:

[[math]] \begin{eqnarray*} X &=&\left(\int_{G_{MN}^L}\!\otimes\ id\right)\sum_{k_xl_x}v_{k_1l_1}\ldots v_{k_sl_s}\otimes a_{k_1i_1}\ldots a_{k_si_s}\otimes b_{l_1j_1}^*\ldots b_{l_sj_s}^*\\ &=&\sum_{k_xl_x}\sum_{r_x\leq L}a_{k_1i_1}\ldots a_{k_si_s}\otimes b_{l_1j_1}^*\ldots b_{l_sj_s}^*\int_{G_M}a_{r_1k_1}\ldots a_{r_sk_s}\int_{G_N}b_{r_1l_1}^*\ldots b_{r_sl_s}^*\\ &=&\sum_{r_x\leq L}\sum_{k_x}a_{k_1i_1}\ldots a_{k_si_s}\int_{G_M}a_{r_1k_1}\ldots a_{r_sk_s} \otimes\sum_{l_x}b_{l_1j_1}^*\ldots b_{l_sj_s}^*\int_{G_N}b_{r_1l_1}^*\ldots b_{r_sl_s}^* \end{eqnarray*} [[/math]]

By using now the invariance property of the Haar functionals of [math]G_M,G_N[/math], we obtain:

[[math]] \begin{eqnarray*} X &=&\sum_{r_x\leq L}\left(\int_{G_M}\!\otimes\ id\right)\Delta(a_{r_1i_1}\ldots a_{r_si_s}) \otimes\left(\int_{G_N}\!\otimes\ id\right)\Delta(b_{r_1j_1}^*\ldots b_{r_sj_s}^*)\\ &=&\sum_{r_x\leq L}\int_{G_M}a_{r_1i_1}\ldots a_{r_si_s}\int_{G_N}b_{r_1j_1}^*\ldots b_{r_sj_s}^*\\ &=&\left(\int_{G_M}\otimes\int_{G_N}\right)\sum_{r_x\leq L}a_{r_1i_1}\ldots a_{r_si_s}\otimes b_{r_1j_1}^*\ldots b_{r_sj_s}^* \end{eqnarray*} [[/math]]

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

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We will prove now that the above functional is in fact the unique positive unital invariant trace on [math]C(G_{MN}^L)[/math]. For this purpose, we will need the Weingarten formula:

Theorem

We have the Weingarten type formula

[[math]] \int_{G_{MN}^L}v_{i_1j_1}\ldots v_{i_sj_s}=\sum_{\pi\sigma\tau\nu}L^{|\pi\vee\tau|}\delta_\sigma(i)\delta_\nu(j)W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) [[/math]]

where the matrices on the right are given by [math]W_{sM}=G_{sM}^{-1}[/math], with [math]G_{sM}(\pi,\sigma)=M^{|\pi\vee\sigma|}[/math].

Show Proof

By using the Weingarten formula for [math]G_M,G_N[/math], we obtain:

[[math]] \begin{eqnarray*} \int_{G_{MN}^L}v_{i_1j_1}\ldots v_{i_sj_s} &=&\sum_{l_1\ldots l_s\leq L}\int_{G_M}a_{l_1i_1}\ldots a_{l_si_s}\int_{G_N}b_{l_1j_1}^*\ldots b_{l_sj_s}^*\\ &=&\sum_{l_1\ldots l_s\leq L}\sum_{\pi\sigma}\delta_\pi(l)\delta_\sigma(i)W_{sM}(\pi,\sigma)\sum_{\tau\nu}\delta_\tau(l)\delta_\nu(j)W_{sN}(\tau,\nu)\\ &=&\sum_{\pi\sigma\tau\nu}\left(\sum_{l_1\ldots l_s\leq L}\delta_\pi(l)\delta_\tau(l)\right)\delta_\sigma(i)\delta_\nu(j)W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) \end{eqnarray*} [[/math]]

The coefficient being [math]L^{|\pi\vee\tau|}[/math], we obtain the formula in the statement.

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We can now derive an abstract characterization of the integration, as follows:

Theorem

The integration of [math]G_{MN}^L[/math] is the unique positive unital trace

[[math]] C(G_{MN}^L)\to\mathbb C [[/math]]

which is invariant under the action of the quantum group [math]G_M\times G_N[/math].

Show Proof

This is something very standard, from . Our claim is that we have:

[[math]] \left(id\otimes\int_{G_M}\otimes\int_{G_N}\right)\Phi(v_{i_1j_1}\ldots v_{i_sj_s})=\int_{G_{MN}^L}v_{i_1j_1}\ldots v_{i_sj_s} [[/math]]

Indeed, by using the Weingarten formula, the left term can be written as follows:

[[math]] \begin{eqnarray*} X &=&\sum_{k_1\ldots k_s}\sum_{l_1\ldots l_s}v_{k_1l_1}\ldots v_{k_sl_s}\int_{G_M}a_{k_1i_1}\ldots a_{k_si_s}\int_{G_N}b_{l_1j_1}^*\ldots b_{l_sj_s}^*\\ &=&\sum_{k_1\ldots k_s}\sum_{l_1\ldots l_s}v_{k_1l_1}\ldots v_{k_sl_s}\sum_{\pi\sigma}\delta_\pi(k)\delta_\sigma(i)W_{sM}(\pi,\sigma)\sum_{\tau\nu}\delta_\tau(l)\delta_\nu(j)W_{sN}(\tau,\nu)\\ &=&\sum_{\pi\sigma\tau\nu}\delta_\sigma(i)\delta_\nu(j)W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu)\sum_{k_1\ldots k_s}\sum_{l_1\ldots l_s}\delta_\pi(k)\delta_\tau(l)u_{k_1l_1}\ldots u_{k_sl_s}\\ &=&\sum_{\pi\sigma\tau\nu}L^{|\pi\vee\tau|}\delta_\sigma(i)\delta_\nu(j)W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) \end{eqnarray*} [[/math]]

Now by comparing with the Weingarten formula for [math]G_{MN}^L[/math], this proves our claim. Assume now that [math]\tau:C(G_{MN}^L)\to\mathbb C[/math] satisfies the invariance condition. We have then:

[[math]] \tau\left(id\otimes\int_{G_M}\otimes\int_{G_N}\right)\Phi(x) =\left(\int_{G_M}\otimes\int_{G_N}\right)(\tau(x)1) =\tau(x) [[/math]]

On the other hand, according to the formula established above, we have as well:

[[math]] \tau\left(id\otimes\int_{G_M}\otimes\int_{G_N}\right)\Phi(x) =\tau(tr(x)1) =tr(x) [[/math]]

Thus we obtain [math]\tau=tr[/math], and this finishes the proof.

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As a main application of the above results, we have the following quite conceptual statement, making the link with the Bercovici-Pata bijection :

Theorem

In the context of the liberation operations [math]G_{MN}^L\to G_{MN}^{L+}[/math], the laws of the sums of non-overlapping coordinates,

[[math]] \chi_E=\sum_{(ij)\in E}u_{ij} [[/math]]

are in Bercovici-Pata bijection, in the [math]|E|=\kappa N,L=\lambda N,M=\mu N[/math] and [math]N\to\infty[/math] regime.

Show Proof

We use various formulae from , , . In terms of [math]K=|E|[/math], the moments of the variables in the statement are given by:

[[math]] \begin{eqnarray*} M_s &=&\sum_{\pi\sigma\tau\nu}K^{|\pi\vee\tau|}L^{|\sigma\vee\nu|}W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu)\\ &\simeq&\sum_{\pi\tau}K^{|\pi\vee\tau|}L^{|\pi\vee\tau|}M^{-|\pi|}N^{-|\tau|}\\ &\simeq&\sum_\pi K^{|\pi|}L^{|\pi|}M^{-|\pi|}N^{-|\pi|}\\ &=&\sum_\pi\left(\frac{\kappa\lambda}{\mu}\right)^{|\pi|} \end{eqnarray*} [[/math]]

In order to interpret this formula, we use general theory from , , :

(1) For [math]G_N=O_N/O_N^+[/math], the above variables [math]\chi_E[/math] follow to be asymptotically Gaussian/semicircular, of parameter [math]\frac{\kappa\lambda}{\mu}[/math], and hence in Bercovici-Pata bijection.

(2) For [math]G_N=U_N/U_N^+[/math] the situation is similar, with [math]\chi_E[/math] being asymptotically complex Gaussian/circular, of parameter [math]\frac{\kappa\lambda}{\mu}[/math], and in Bercovici-Pata bijection.

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There are several possible extensions of the above result, to the discrete case, and by using twisting operations as well. We refer here to , and related papers.

General references

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