Homogeneous spaces

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15a. Quotient spaces

We have seen that the closed subgroups [math]G\subset U_N^+[/math] can be investigated with a variety of techniques, for the most belonging to algebraic geometry and probability theory, and with most of our new, original results concerning the free case, where [math]S_N^+\subset G\subset U_N^+[/math]. All this suggests developing, more generally, a theory of “free geometry”, again of algebraic geometry and probability flavor. And also, why not developing as well, along the same lines, theories like “half-classical geometry”, or “twisted geometry”, and so on.

This is certainly possible, but quite time-consuming, and going well beyond the purposes of the present book. Instead, we will provide in this chapter an introduction to all this. Our purpose, quite modest, will be that of extending some of our quantum group results to certain classes of “quantum homogeneous spaces”. With this being the first step towards constructing the above-mentioned noncommutative geometry theories.

Before starting, a few words on motivations. These come from physics, and more specifically from quantum mechanics, of course. Classical mechanics is described by classical geometry, and so quantum mechanics should be described by some kind of quantum geometry, it's as simple as that. In practice however, all this is quite new, 100 years old, and no one really knows how to do this. In addition, the experts are bitterly split, with Connes and his group believing in differential geometry and smoothness, and with us, meaning me and you, dear reader, and our friends, believing instead in algebraic geometry, or call that Riemannian geometry a la Nash , and probability.

But probably enough talking, let's have something started, and more comments later. Let us begin with some generalities regarding the quotient spaces. We have:

Proposition

Given a quantum subgroup [math]H\subset G[/math], with associated quotient map [math]\rho:C(G)\to C(H)[/math], if we define the quotient space [math]X=G/H[/math] by setting

[[math]] C(X)=\left\{f\in C(G)\Big|(\rho\otimes id)\Delta f=1\otimes f\right\} [[/math]]

then we have a coaction [math]\Phi:C(X)\to C(X)\otimes C(G)[/math], obtained as the restriction of the comultiplication of [math]C(G)[/math]. In the classical case, we obtain the usual space [math]X=G/H[/math].

Show Proof

Observe that [math]C(X)\subset C(G)[/math] is indeed a subalgebra, because it is defined via a relation of type [math]\varphi(f)=\psi(f)[/math], with [math]\varphi,\psi[/math] morphisms. Observe also that in the classical case we obtain the algebra of continuous functions on [math]X=G/H[/math], because:

[[math]] \begin{eqnarray*} (\rho\otimes id)\Delta f=1\otimes f &\iff&(\rho\otimes id)\Delta f(h,g)=(1\otimes f)(h,g),\forall h\in H,\forall g\in G\\ &\iff&f(hg)=f(g),\forall h\in H,\forall g\in G\\ &\iff&f(hg)=f(kg),\forall h,k\in H,\forall g\in G \end{eqnarray*} [[/math]]

Regarding now the construction of [math]\Phi[/math], observe that for [math]f\in C(X)[/math] we have:

[[math]] \begin{eqnarray*} (\rho\otimes id\otimes id)(\Delta\otimes id)\Delta f &=&(\rho\otimes id\otimes id)(id\otimes\Delta)\Delta f\\ &=&(id\otimes\Delta)(\rho\otimes id)\Delta f\\ &=&(id\otimes\Delta)(1\otimes f)\\ &=&1\otimes\Delta f \end{eqnarray*} [[/math]]

Thus [math]f\in C(X)[/math] implies [math]\Delta f\in C(X)\otimes C(G)[/math], and this gives the existence of [math]\Phi[/math], as in the statement. Finally, all the other assertions are clear.

■

As an illustration, in the group dual case we have:

Proposition

Assume that [math]G=\widehat{\Gamma}[/math] is a discrete group dual.

• The quantum subgroups of [math]G[/math] are [math]H=\widehat{\Lambda}[/math], with [math]\Gamma\to\Lambda[/math] being a quotient group.
• For such a quantum subgroup [math]\widehat{\Lambda}\subset\widehat{\Gamma}[/math], we have [math]\widehat{\Gamma}/\widehat{\Lambda}=\widehat{\Theta}[/math], where [math]\Theta=\ker(\Gamma\to\Lambda)[/math].

Show Proof

This is well-known, the idea being as follows:

(1) In one sense, this is clear. Conversely, since the algebra [math]C(G)=C^*(\Gamma)[/math] is cocommutative, so are all its quotients, and this gives the result.

(2) Consider a quotient map [math]r:\Gamma\to\Lambda[/math], and denote by [math]\rho:C^*(\Gamma)\to C^*(\Lambda)[/math] its extension. With [math]f=\sum_{g\in\Gamma}\lambda_g\cdot g\in C^*(\Gamma)[/math] we have:

[[math]] \begin{eqnarray*} f\in C(\widehat{\Gamma}/\widehat{\Lambda}) &\iff&(\rho\otimes id)\Delta(f)=1\otimes f\\ &\iff&\sum_{g\in\Gamma}\lambda_g\cdot r(g)\otimes g=\sum_{g\in\Gamma}\lambda_g\cdot 1\otimes g\\ &\iff&\lambda_g\cdot r(g)=\lambda_g\cdot 1,\forall g\in\Gamma\\ &\iff&supp(f)\subset\ker(r) \end{eqnarray*} [[/math]]

But this means [math]\widehat{\Gamma}/\widehat{\Lambda}=\widehat{\Theta}[/math], with [math]\Theta=\ker(\Gamma\to\Lambda)[/math], as claimed.

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Given two quantum spaces [math]X,Y[/math], we say that [math]X[/math] is a quotient space of [math]Y[/math] when we have an embedding of algebras [math]\alpha:C(X)\subset C(Y)[/math]. With this convention, we have:

Definition

We call a quotient space [math]G\to X[/math] homogeneous when

[[math]] \Delta(C(X))\subset C(X)\otimes C(G) [[/math]]

where [math]\Delta:C(G)\to C(G)\otimes C(G)[/math] is the comultiplication map.

In other words, an homogeneous quotient space [math]G\to X[/math] is a quantum space coming from a subalgebra [math]C(X)\subset C(G)[/math], which is stable under the comultiplication. The relation with the quotient spaces from Proposition 15.1 is as follows:

Theorem

The following results hold:

• The quotient spaces [math]X=G/H[/math] are homogeneous.
• In the classical case, any homogeneous space is of type [math]G/H[/math].
• In general, there are homogeneous spaces which are not of type [math]G/H[/math].

Show Proof

Once again these results are well-known, the proof being as follows:

(1) This is clear from Proposition 15.1 above.

(2) Consider a quotient map [math]p:G\to X[/math]. The invariance condition in the statement tells us that we must have an action [math]G\curvearrowright X[/math], given by [math]g(p(g'))=p(gg')[/math]. Thus:

[[math]] p(g')=p(g'')\implies p(gg')=p(gg''),\ \forall g\in G [[/math]]

Now observe that the following subset [math]H\subset G[/math] is a subgroup:

[[math]] H=\left\{g\in G\Big|p(g)=p(1)\right\} [[/math]]

Indeed, [math]g,h\in H[/math] implies [math]p(gh)=p(g)=p(1)[/math], so [math]gh\in H[/math], and the other axioms are satisfied as well. Our claim now, finishing the proof here, is that we have [math]X=G/H[/math], via [math]p(g)\to Hg[/math]. Indeed, the map [math]p(g)\to Hg[/math] is well-defined and bijective, because [math]p(g)=p(g')[/math] is equivalent to [math]p(g^{-1}g')=p(1)[/math], and so to [math]Hg=Hg'[/math], as desired.

(3) Given a discrete group [math]\Gamma[/math] and an arbitrary subgroup [math]\Theta\subset\Gamma[/math], the quotient space [math]\widehat{\Gamma}\to\widehat{\Theta}[/math] is homogeneous. Now by using Proposition 15.2 above, we can see that if [math]\Theta\subset\Gamma[/math] is not normal, the quotient space [math]\widehat{\Gamma}\to\widehat{\Theta}[/math] is not of the form [math]G/H[/math].

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Let us try now to understand the properties of the homogeneous spaces [math]G\to X[/math], in the above sense. We have the following result, which is once again well-known:

Proposition

Assume that a quotient space [math]G\to X[/math] is homogeneous.

• The restriction [math]\Phi:C(X)\to C(X)\otimes C(G)[/math] of [math]\Delta[/math] is a coaction.
• We have [math]\Phi(f)=f\otimes 1\implies f\in\mathbb C1[/math], and [math](id\otimes\int)\Phi f=\int f[/math].
• The restriction of [math]\int[/math] is the unique unital form satisfying [math](\tau\otimes id)\Phi=\tau(.)1[/math].

Show Proof

These results are all elementary, the proof being as follows:

(1) This is clear from definitions, because [math]\Delta[/math] itself is a coaction.

(2) If [math]f\in C(G)[/math] is such that [math]\Delta(f)=f\otimes 1[/math], then by applying the counit we obtain:

[[math]] (\varepsilon\otimes id)\Delta f=(\varepsilon\otimes id)(f\otimes 1) [[/math]]

Thus [math]f=\varepsilon(f)1[/math], as desired. As for the second assertion, this follows from the left invariance property [math](id\otimes\int)\Delta f=\int f[/math] of the Haar functional, by restriction to [math]C(X)[/math].

(3) By using the right invariance property [math](\int\otimes id)\Delta f=\int f[/math] of the Haar functional of [math]C(G)[/math], we obtain that [math]tr=\int_{|C(X)}[/math] is [math]G[/math]-invariant, in the sense that:

[[math]] (tr\otimes id)\Phi f=tr(f)1 [[/math]]

Conversely, assuming that [math]\tau:C(X)\to\mathbb C[/math] satisfies [math](\tau\otimes id)\Phi f=\tau(f)1[/math], we have:

[[math]] \left(\tau\otimes\int\right)\Phi(f) =\int(\tau\otimes id)\Phi(f) =\int(\tau(f)1) =\tau(f) [[/math]]

On the other hand, we can compute the same quantity as follows:

[[math]] \left(\tau\otimes\int\right)\Phi(f) =\tau\left(id\otimes\int\right)\Phi(f) =\tau(tr(f)1) =tr(f) [[/math]]

Thus we have [math]\tau(f)=tr(f)[/math] for any [math]f\in C(X)[/math], and this finishes the proof.

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Let us discuss now an extra issue, of analytic nature. The point is that for one of the most basic examples of actions, [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math], the associated morphism [math]\alpha:C(X)\to C(G)[/math] is not injective. In order to include such examples, we must relax our axioms:

Definition

An extended homogeneous space consists of a morphism of algebras [math]\alpha:C(X)\to C(G)[/math], and a coaction map [math]\Phi:C(X)\to C(X)\otimes C(G)[/math], such that

[[math]] \xymatrix@R=16mm@C=20mm{ C(X)\ar[r]^\Phi\ar[d]_\alpha&C(X)\otimes C(G)\ar[d]^{\alpha\otimes id}\\ C(G)\ar[r]^\Delta&C(G)\otimes C(G) } [[/math]]

commutes, and such that

[[math]] \xymatrix@R=16mm@C=20mm{ C(X)\ar[r]^\Phi\ar[d]_\alpha&C(X)\otimes C(G)\ar[d]^{id\otimes\int}\\ C(G)\ar[r]^{\int(.)1}&C(X) } [[/math]]

commutes as well, where [math]\int[/math] is the Haar integration over [math]G[/math]. We write then [math]G\to X[/math].

When [math]\alpha[/math] is injective we obtain an homogeneous space in the previous sense. The examples with [math]\alpha[/math] being not injective, which motivate the above formalism, include the standard action [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math], and the standard action [math]U_N^+\curvearrowright S^{N-1}_{\mathbb C,+}[/math].

Here are a few general remarks on the above axioms:

Proposition

Assume that we have morphisms of algebras [math]\alpha:C(X)\to C(G)[/math] and [math]\Phi:C(X)\to C(X)\otimes C(G)[/math], satisfying [math](\alpha\otimes id)\Phi=\Delta\alpha[/math].

• If [math]\alpha[/math] is injective on a dense [math]*[/math]-subalgebra [math]A\subset C(X)[/math], and [math]\Phi(A)\subset A\otimes C(G)[/math], then [math]\Phi[/math] is automatically a coaction map, and is unique.
• The ergodicity type condition [math](id\otimes\int)\Phi=\int\alpha(.)1[/math] is equivalent to the existence of a linear form [math]\lambda:C(X)\to\mathbb C[/math] such that [math](id\otimes\int)\Phi=\lambda(.)1[/math].

Show Proof

This is something elementary, the idea being as follows:

(1) Assuming that we have a dense [math]*[/math]-subalgebra [math]A\subset C(X)[/math] as in the statement, satisying [math]\Phi(A)\subset A\otimes C(G)[/math], the restriction [math]\Phi_{|A}[/math] is given by:

[[math]] \Phi_{|A}=(\alpha_{|A}\otimes id)^{-1}\Delta\alpha_{|A} [[/math]]

This restriction and is therefore coassociative, and unique. By continuity, [math]\Phi[/math] itself follows to be coassociative and unique, as desired.

(2) Assuming [math](id\otimes\int)\Phi=\lambda(.)1[/math], we have [math](\alpha\otimes\int)\Phi=\lambda(.)1[/math]. On the other hand, we have as well the following formula:

[[math]] \left(\alpha\otimes\int\right)\Phi=\left(id\otimes\int\right)\Delta\alpha=\int\alpha(.)1 [[/math]]

Thus we obtain [math]\lambda=\int\alpha[/math], as claimed.

■

Given an extended homogeneous space [math]G\to X[/math] as above, with associated map [math]\alpha:C(X)\to C(G)[/math], we can consider the image of this latter map:

[[math]] \alpha:C(X)\to C(Y)\subset C(G) [[/math]]

Equivalently, at the level of the associated noncommutative spaces, we can factorize the corresponding quotient map [math]G\to Y\subset X[/math]. With these conventions, we have:

Proposition

Consider an extended homogeneous space [math]G\to X[/math].

• [math]\Phi(f)=f\otimes 1\implies f\in\mathbb C1[/math].
• [math]tr=\int\alpha[/math] is the unique unital [math]G[/math]-invariant form on [math]C(X)[/math].
• The image space obtained by factorizing, [math]G\to Y[/math], is homogeneous.

Show Proof

We have several assertions to be proved, the idea being as follows:

(1) This follows indeed from [math](id\otimes\int)\Phi(f)=\int\alpha(f)1[/math], which gives:

[[math]] f=\int\alpha(f)1 [[/math]]

(2) The fact that [math]tr=\int\alpha[/math] is indeed [math]G[/math]-invariant can be checked as follows:

[[math]] \begin{eqnarray*} (tr\otimes id)\Phi f &=&(\smallint\alpha\otimes id)\Phi f\\ &=&(\smallint\otimes id)\Delta\alpha f\\ &=&\smallint\alpha(f)1\\ &=&tr(f)1 \end{eqnarray*} [[/math]]

As for the uniqueness assertion, this follows as before.

(3) The condition [math](\alpha\otimes id)\Phi=\Delta\alpha[/math], together with the fact that [math]i[/math] is injective, allows us to factorize [math]\Delta[/math] into a morphism [math]\Psi[/math], as follows:

[[math]] \xymatrix@R=12mm@C=30mm{ C(X)\ar[r]^\Phi\ar[d]_\alpha&C(X)\otimes C(G)\ar[d]^{\alpha\otimes id}\\ C(Y)\ar@.[r]^\Psi\ar[d]_i&C(Y)\otimes C(G)\ar[d]^{i\otimes id}\\ C(G)\ar[r]^\Delta&C(G)\otimes C(G) } [[/math]]

Thus the image space [math]G\to Y[/math] is indeed homogeneous, and we are done.

■

Finally, we have the following result, which further clarifies our formalism:

Theorem

Let [math]G\to X[/math] be an extended homogeneous space, and construct quotients [math]X\to X'[/math], [math]G\to G'[/math] by performing the GNS construction with respect to [math]\int\alpha,\int[/math]. Then [math]\alpha[/math] factorizes into an inclusion [math]\alpha':C(X')\to C(G')[/math], and we have an homogeneous space.

Show Proof

We factorize [math]G\to Y\subset X[/math] as above. By performing the GNS construction with respect to [math]\int i\alpha,\int i,\int[/math], we obtain a diagram as follows:

[[math]] \xymatrix@R=12mm@C=30mm{ C(X)\ar[r]^p\ar[d]_\alpha&C(X')\ar[d]^{\alpha'}\ar[dr]^{tr'}\\ C(Y)\ar[r]^q\ar[d]_i&C(Y')\ar[d]^{i'}&\mathbb C\\ C(G)\ar[r]^r&C(G')\ar[ur]_{\int'} } [[/math]]

Indeed, with [math]tr=\int\alpha[/math], the GNS quotient maps [math]p,q,r[/math] are defined respectively by:

[[math]] \begin{eqnarray*} \ker p&=&\left\{f\in C(X)\Big|tr(f^*f)=0\right\}\\ \ker q&=&\left\{f\in C(Y)\Big|\smallint(f^*f)=0\right\}\\ \ker r&=&\left\{f\in C(G)\Big|\smallint(f^*f)=0\right\} \end{eqnarray*} [[/math]]

Next, we can define factorizations [math]i',\alpha'[/math] as above. Observe that [math]i'[/math] is injective, and that [math]\alpha'[/math] is surjective. Our claim now is that [math]\alpha'[/math] is injective as well. Indeed:

[[math]] \begin{eqnarray*} \alpha'p(f)=0 &\implies&q\alpha(f)=0\\ &\implies&\int\alpha(f^*f)=0\\ &\implies&tr(f^*f)=0\\ &\implies&p(f)=0 \end{eqnarray*} [[/math]]

We conclude that we have [math]X'=Y'[/math], and this gives the result.

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Summarizing, the basic homogeneous space theory from the classical case extends to the quantum group setting, with a few twists, both of algebraic and analytic nature.

15b. Partial isometries

We discuss now some explicit examples of homogeneous spaces. This can be done at several levels of generality, and there has been quite some work here, starting with . In what follows we discuss the formalism in , which is quite broad, while remaining not very abstract. We will study the spaces of the following type:

[[math]] X=(G_M\times G_N)\big/(G_L\times G_{M-L}\times G_{N-L}) [[/math]]

These spaces cover indeed the quantum groups and the spheres. And also, they are quite concrete and useful objects, consisting of certain classes of “partial isometries”. Our main result will be a verification of the Bercovici-Pata liberation criterion, for certain variables associated [math]\chi\in C(X)[/math], in a suitable [math]L,M,N\to\infty[/math] limit.

We begin with a study in the classical case. Our starting point will be:

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 [[/math]]

[[math]] 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 [[/math]]

[[math]] 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 [[/math]]

[[math]] 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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As an illustration, at [math]L=M=N[/math] we recover in this way the usual matrix description of [math]O_N,U_N[/math]. Also, at [math]L=M=1[/math] we obtain the usual description of [math]S^{N-1}_\mathbb R,S^{N-1}_\mathbb C[/math], as row spaces over the corresponding groups [math]O_N,U_N[/math]. Finally, at [math]L=N=1[/math] we obtain the usual description of [math]S^{N-1}_\mathbb R,S^{N-1}_\mathbb C[/math], as column spaces over the corresponding groups [math]O_N,U_N[/math].

Now back to the general case, observe 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=15mm@C=15mm{ \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 15.11 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 the following groups:

[[math]] O_L\times O_{M-L}\times O_{N-L} [[/math]]

[[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 both 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 then parametrized by triples [math](x,a,b)[/math] belonging respectively to:

[[math]] O_L\times O_{M-L}\times O_{N-L} [[/math]]

[[math]] U_L\times U_{M-L}\times U_{N-L} [[/math]]

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

Show Proof

This is just a reformulation of Proposition 15.12 above, 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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Once again, the basic examples here come from the cases [math]L=M=N[/math] and [math]L=M=1[/math]. At [math]L=M=N[/math] the quotient spaces at right are respectively:

[[math]] O_N,U_N [[/math]]

At [math]L=M=1[/math] the quotient spaces at right are respectively:

[[math]] O_N/O_{N-1}\quad,\quad U_N/U_{N-1} [[/math]]

In fact, in the general orthogonal [math]L=M[/math] case we obtain the following spaces:

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

Also, in the general unitary [math]L=M[/math] case we obtain the following spaces:

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

Similarly, the examples coming from the cases [math]L=M=N[/math] and [math]L=N=1[/math] are particular cases of the general [math]L=N[/math] case, where we obtain the following spaces:

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

In the unitary case, we obtain the following spaces:

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

Summarizing, we have here some basic homogeneous spaces, unifying the real and complex spheres with the orthogonal and unitary groups.

15c. Free isometries

We can liberate the spaces [math]O_{MN}^L,U_{MN}^L[/math], 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((u_{ij})_{i=1,\ldots,M,j=1,\ldots,N}\Big|u=\bar{u},uu^t={\rm projection\ of\ trace}\ L\right)\\ C(U_{MN}^{L+})&=&C^*\left((u_{ij})_{i=1,\ldots,M,j=1,\ldots,N}\Big|uu^*,\bar{u}u^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}u_{ij}u_{ij}^* =\sum_{ij}u_{ij}^*u_{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, we first have the following result:

Proposition

At [math]L=M=1[/math] we obtain the following diagram,

[[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]} [[/math]]

and at [math]L=N=1[/math] we obtain the following diagram:

[[math]] \xymatrix@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]]

Show Proof

Both the assertions are clear from definitions.

■

We have as well the following result:

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

We recall that the various quantum groups in the statement are constructed as follows, with the symbol [math]\times[/math] standing once again for “commutative” and “free”:

[[math]] \begin{eqnarray*} C(O_N^\times)&=&C^*_\times\left((u_{ij})_{i,j=1,\ldots,N}\Big|u=\bar{u},uu^t=u^tu=1\right)\\ C(U_N^\times)&=&C^*_\times\left((u_{ij})_{i,j=1,\ldots,N}\Big|uu^*=u^*u=1,\bar{u}u^t=u^t\bar{u}=1\right) \end{eqnarray*} [[/math]]

On the other hand, according to Proposition 15.11 and to Definition 15.14 above, we have the following presentation results:

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

[[math]] Tr(\bar{u}u^t)=Tr(u^*u) [[/math]]

We therefore obtain the following formulae:

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

But this leads to the conclusion in the statement.

■

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]] u_{ij}\to\sum_{kl}u_{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]] u_{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 [[/math]]

[[math]] 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=u_{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]uu^*u=u[/math], let us set:

[[math]] U_{ij}=\sum_{kl}u_{kl}\otimes a_{ki}\otimes b_{lj}^* [[/math]]

We have then the following computation:

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

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

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

(2) Assuming [math]uu^*u=u[/math], let us set:

[[math]] V_{ij}=\sum_{r\leq L}a_{ri}\otimes b_{rj}^* [[/math]]

We have then the following computation:

[[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]]

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

[[math]] \begin{eqnarray*} \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 \end{eqnarray*} [[/math]]

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

■

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 15.17.

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}u_{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 some discrete extensions of the above constructions:

Definition

Associated to any partial permutation, [math]\sigma:I\simeq J[/math] with [math]I\subset\{1,\ldots,N\}[/math] and [math]J\subset\{1,\ldots,M\}[/math], is the real/complex partial isometry

[[math]] T_\sigma:span\left(e_i\Big|i\in I\right)\to span\left(e_j\Big|j\in J\right) [[/math]]

given on the standard basis elements by [math]T_\sigma(e_i)=e_{\sigma(i)}[/math].

Let [math]S_{MN}^L[/math] be the set of partial permutations [math]\sigma:I\simeq J[/math] as above, with range [math]I\subset\{1,\ldots,N\}[/math] and target [math]J\subset\{1,\ldots,M\}[/math], and with [math]L=|I|=|J|[/math]. We have:

Proposition

The space of partial permutations signed by elements of [math]\mathbb Z_s[/math],

[[math]] H_{MN}^{sL}=\left\{T(e_i)=w_ie_{\sigma(i)}\Big|\sigma\in S_{MN}^L,w_i\in\mathbb Z_s\right\} [[/math]]

is isomorphic to the quotient space

[[math]] (H_M^s\times H_N^s)/(H_L^s\times H_{M-L}^s\times H_{N-L}^s) [[/math]]

via a standard isomorphism.

Show Proof

This follows by adapting the computations in the proof of Proposition 15.12 above. Indeed, we have an action map as follows, which is transitive:

[[math]] H_M^s\times H_N^s\to H_{MN}^{sL}\quad,\quad (A,B)U=AUB^* [[/math]]

Consider now the following point:

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

The stabilizer of this point follows to be the following group:

[[math]] H_L^s\times H_{M-L}^s\times H_{N-L}^s [[/math]]

To be more precise, this group is embedded via:

[[math]] (x,a,b)\to\left[\begin{pmatrix}x&0\\0&a\end{pmatrix},\begin{pmatrix}x&0\\0&b\end{pmatrix}\right] [[/math]]

But this gives the result.

■

In the free case now, the idea is similar, by using inspiration from the construction of the quantum group [math]H_N^{s+}=\mathbb Z_s\wr_*S_N^+[/math] in . The result here is as follows:

Proposition

The compact quantum space [math]H_{MN}^{sL+}[/math] associated to the algebra

[[math]] C(H_{MN}^{sL+})=C(U_{MN}^{L+})\Big/\left \lt u_{ij}u_{ij}^*=u_{ij}^*u_{ij}=p_{ij}={\rm projections},u_{ij}^s=p_{ij}\right \gt [[/math]]

has an action map, and is the target of a quotient map, as in Theorem 15.18 above.

Show Proof

We must show that if the variables [math]u_{ij}[/math] satisfy the relations in the statement, then these relations are satisfied as well for the following variables:

[[math]] U_{ij}=\sum_{kl}u_{kl}\otimes a_{ki}\otimes b_{lj}^*\quad,\quad V_{ij}=\sum_{r\leq L}a_{ri}\otimes b_{rj}^* [[/math]]

We use the fact that the standard coordinates [math]a_{ij},b_{ij}[/math] on the quantum groups [math]H_M^{s+},H_N^{s+}[/math] satisfy the following relations, for any [math]x\neq y[/math] on the same row or column of [math]a,b[/math]:

[[math]] xy=xy^*=0 [[/math]]

We obtain, by using these relations:

[[math]] \begin{eqnarray*} U_{ij}U_{ij}^* &=&\sum_{klmn}u_{kl}u_{mn}^*\otimes a_{ki}a_{mi}^*\otimes b_{lj}^*b_{mj}\\ &=&\sum_{kl}u_{kl}u_{kl}^*\otimes a_{ki}a_{ki}^*\otimes b_{lj}^*b_{lj} \end{eqnarray*} [[/math]]

We have as well the following formula:

[[math]] \begin{eqnarray*} V_{ij}V_{ij}^* &=&\sum_{r,t\leq L}a_{ri}a_{ti}^*\otimes b_{rj}^*b_{tj}\\ &=&\sum_{r\leq L}a_{ri}a_{ri}^*\otimes b_{rj}^*b_{rj} \end{eqnarray*} [[/math]]

In terms of the projections [math]x_{ij}=a_{ij}a_{ij}^*[/math], [math]y_{ij}=b_{ij}b_{ij}^*[/math], [math]p_{ij}=u_{ij}u_{ij}^*[/math], we have:

[[math]] U_{ij}U_{ij}^*=\sum_{kl}p_{kl}\otimes x_{ki}\otimes y_{lj} [[/math]]

[[math]] V_{ij}V_{ij}^*=\sum_{r\leq L}x_{ri}\otimes y_{rj} [[/math]]

By repeating the computation, we conclude that these elements are projections. Also, a similar computation shows that [math]U_{ij}^*U_{ij},V_{ij}^*V_{ij}[/math] are given by the same formulae. Finally, once again by using the relations of type [math]xy=xy^*=0[/math], we have:

[[math]] \begin{eqnarray*} U_{ij}^s &=&\sum_{k_rl_r}u_{k_1l_1}\ldots u_{k_sl_s}\otimes a_{k_1i}\ldots a_{k_si}\otimes b_{l_1j}^*\ldots b_{l_sj}^*\\ &=&\sum_{kl}u_{kl}^s\otimes a_{ki}^s\otimes(b_{lj}^*)^s \end{eqnarray*} [[/math]]

We have as well the following formula:

[[math]] \begin{eqnarray*} V_{ij}^s &=&\sum_{r_l\leq L}a_{r_1i}\ldots a_{r_si}\otimes b_{r_1j}^*\ldots b_{r_sj}^*\\ &=&\sum_{r\leq L}a_{ri}^s\otimes(b_{rj}^*)^s \end{eqnarray*} [[/math]]

Thus the conditions of type [math]u_{ij}^s=p_{ij}[/math] are satisfied as well, and we are done.

■

Let us discuss now the general case. We have the following result:

Proposition

The various spaces [math]G_{MN}^L[/math] constructed so far appear by imposing to the standard coordinates of [math]U_{MN}^{L+}[/math] the relations

[[math]] \sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\delta_\pi(i)\delta_\sigma(j)u_{i_1j_1}^{e_1}\ldots u_{i_sj_s}^{e_s}=L^{|\pi\vee\sigma|} [[/math]]

with [math]s=(e_1,\ldots,e_s)[/math] ranging over all the colored integers, and with [math]\pi,\sigma\in D(0,s)[/math].

Show Proof

According to the various constructions above, the relations defining [math]G_{MN}^L[/math] can be written as follows, with [math]\sigma[/math] ranging over a family of generators, with no upper legs, of the corresponding category of partitions [math]D[/math]:

[[math]] \sum_{j_1\ldots j_s}\delta_\sigma(j)u_{i_1j_1}^{e_1}\ldots u_{i_sj_s}^{e_s}=\delta_\sigma(i) [[/math]]

We therefore obtain the relations in the statement, as follows:

[[math]] \begin{eqnarray*} \sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\delta_\pi(i)\delta_\sigma(j)u_{i_1j_1}^{e_1}\ldots u_{i_sj_s}^{e_s} &=&\sum_{i_1\ldots i_s}\delta_\pi(i)\sum_{j_1\ldots j_s}\delta_\sigma(j)u_{i_1j_1}^{e_1}\ldots u_{i_sj_s}^{e_s}\\ &=&\sum_{i_1\ldots i_s}\delta_\pi(i)\delta_\sigma(i)\\ &=&L^{|\pi\vee\sigma|} \end{eqnarray*} [[/math]]

As for the converse, this follows by using the relations in the statement, by keeping [math]\pi[/math] fixed, and by making [math]\sigma[/math] vary over all the partitions in the category.

■

In the general case now, where [math]G=(G_N)[/math] is an arbitary uniform easy quantum group, we can construct spaces [math]G_{MN}^L[/math] by using the above relations, and we have:

Theorem

The spaces [math]G_{MN}^L\subset U_{MN}^{L+}[/math] constructed by imposing the relations

[[math]] \sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\delta_\pi(i)\delta_\sigma(j)u_{i_1j_1}^{e_1}\ldots u_{i_sj_s}^{e_s}=L^{|\pi\vee\sigma|} [[/math]]

with [math]\pi,\sigma[/math] ranging over all the partitions in the associated category, having no upper legs, are subject to an action map/quotient map diagram, as in Theorem 15.18.

Show Proof

We proceed as in the proof of Proposition 15.17. We must prove that, if the variables [math]u_{ij}[/math] satisfy the relations in the statement, then so do the following variables:

[[math]] U_{ij}=\sum_{kl}u_{kl}\otimes a_{ki}\otimes b_{lj}^*\quad,\quad V_{ij}=\sum_{r\leq L}a_{ri}\otimes b_{rj}^* [[/math]]

Regarding the variables [math]U_{ij}[/math], the computation here goes as follows:

[[math]] \begin{eqnarray*} &&\sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\delta_\pi(i)\delta_\sigma(j)U_{i_1j_1}^{e_1}\ldots U_{i_sj_s}^{e_s}\\ &=&\sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\sum_{k_1\ldots k_s}\sum_{l_1\ldots l_s}u_{k_1l_1}^{e_1}\ldots u_{k_sl_s}^{e_s}\otimes \delta_\pi(i)\delta_\sigma(j)a_{k_1i_1}^{e_1}\ldots a_{k_si_s}^{e_s}\otimes(b_{l_sj_s}^{e_s}\ldots b_{l_1j_1}^{e_1})^*\\ &=&\sum_{k_1\ldots k_s}\sum_{l_1\ldots l_s}\delta_\pi(k)\delta_\sigma(l)u_{k_1l_1}^{e_1}\ldots u_{k_sl_s}^{e_s}\\ &=&L^{|\pi\vee\sigma|} \end{eqnarray*} [[/math]]

For the variables [math]V_{ij}[/math] the proof is similar, as follows:

[[math]] \begin{eqnarray*} &&\sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\delta_\pi(i)\delta_\sigma(j)V_{i_1j_1}^{e_1}\ldots V_{i_sj_s}^{e_s}\\ &=&\sum_{i_1\ldots i_s}\sum_{j_1\ldots j_s}\sum_{l_1,\ldots,l_s\leq L}\delta_\pi(i)\delta_\sigma(j)a_{l_1i_1}^{e_1}\ldots a_{l_si_s}^{e_s}\otimes(b_{l_sj_s}^{e_s}\ldots b_{l_1j_1}^{e_1})^*\\ &=&\sum_{l_1,\ldots,l_s\leq L}\delta_\pi(l)\delta_\sigma(l)\\ &=&L^{|\pi\vee\sigma|} \end{eqnarray*} [[/math]]

Thus we have constructed an action map, and a quotient map, as in Proposition 15.17 above, and the commutation of the diagram in Theorem 15.18 is then trivial.

■

15d. Integration theory

Still following , let us discuss now the integration over [math]G_{MN}^L[/math]. We 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]u_{ij}\to\sum_{r\leq L}a_{ri}\otimes b_{rj}^*[/math] with the Haar functional of [math]G_M\times G_N[/math].

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(u_{ij})=\sum_{kl}u_{kl}\otimes a_{ki}\otimes b_{lj}^* [[/math]]

Show Proof

We 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(u_{i_1j_1}\ldots u_{i_sj_s})=\int_{G_{MN}^L}u_{i_1j_1}\ldots u_{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}u_{k_1l_1}\ldots u_{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.

■

We 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}u_{i_1j_1}\ldots u_{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}u_{i_1j_1}\ldots u_{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.

■

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

The idea, from , will be that of proving the following ergodicity formula:

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

We restrict the attention to the orthogonal case, the proof in the unitary case being similar. We must verify that the following holds:

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

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}u_{k_1l_1}\ldots u_{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}u_{k_1l_1}\ldots u_{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} \end{eqnarray*} [[/math]]

By using now the summation formula in Theorem 15.23, we obtain:

[[math]] X=\sum_{\pi\sigma\tau\nu}L^{|\pi\vee\tau|}\delta_\sigma(i)\delta_\nu(j)W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) [[/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:

[[math]] \begin{eqnarray*} \tau\left(id\otimes\int_{G_M}\otimes\int_{G_N}\right)\Phi(x) &=&\left(\tau\otimes\int_{G_M}\otimes\int_{G_N}\right)\Phi(x)\\ &=&\left(\int_{G_M}\otimes\int_{G_N}\right)(\tau\otimes id)\Phi(x)\\ &=&\left(\int_{G_M}\otimes\int_{G_N}\right)(\tau(x)1)\\ &=&\tau(x) \end{eqnarray*} [[/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 integration technology, still following , we have the following result, extending previous computations for quantum group characters:

Proposition

For a sum of coordinates

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

which do not overlap on rows and columns we have

[[math]] \int_{G_{MN}^L}\chi_E^s=\sum_{\pi\sigma\tau\nu}K^{|\pi\vee\tau|}L^{|\sigma\vee\nu|}W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) [[/math]]

where [math]K=|E|[/math] is the cardinality of the indexing set.

Show Proof

With [math]K=|E|[/math], we can write [math]E=\{(\alpha(i),\beta(i))\}[/math], for certain embeddings:

[[math]] \alpha:\{1,\ldots,K\}\subset\{1,\ldots,M\} [[/math]]

[[math]] \beta:\{1,\ldots,K\}\subset\{1,\ldots,N\} [[/math]]

In terms of these maps [math]\alpha,\beta[/math], the moment in the statement is given by:

[[math]] M_s=\int_{G_{MN}^L}\left(\sum_{i\leq K}u_{\alpha(i)\beta(i)}\right)^s [[/math]]

By using the Weingarten formula, we can write this quantity as follows:

[[math]] \begin{eqnarray*} &&M_s\\ &=&\int_{G_{MN}^L}\sum_{i_1\ldots i_s\leq K}u_{\alpha(i_1)\beta(i_1)}\ldots u_{\alpha(i_s)\beta(i_s)}\\ &=&\sum_{i_1\ldots i_s\leq K}\sum_{\pi\sigma\tau\nu}L^{|\sigma\vee\nu|}\delta_\pi(\alpha(i_1),\ldots,\alpha(i_s))\delta_\tau(\beta(i_1),\ldots,\beta(i_s))W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu)\\ &=&\sum_{\pi\sigma\tau\nu}\left(\sum_{i_1\ldots i_s\leq K}\delta_\pi(i)\delta_\tau(i)\right)L^{|\sigma\vee\nu|}W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) \end{eqnarray*} [[/math]]

But, as explained before, the coefficient on the left in the last formula is:

[[math]] C=K^{|\pi\vee\tau|} [[/math]]

We therefore obtain the formula in the statement.

■

We can further advance in the classical/twisted and free cases, where the Weingarten theory for the corresponding quantum groups is available from , , , :

Theorem

In the context of the liberation operations

[[math]] O_{MN}^L\to O_{MN}^{L+}\quad,\quad U_{MN}^L\to U_{MN}^{L+}\quad,\quad H_{MN}^{sL}\to H_{MN}^{sL+} [[/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]]

regime and [math]N\to\infty[/math] limit.

Show Proof

We use general theory from , , , . According to Proposition 15.28, in terms of [math]K=|E|[/math], the moments of the variables in the statement are:

[[math]] M_s=\sum_{\pi\sigma\tau\nu}K^{|\pi\vee\tau|}L^{|\sigma\vee\nu|}W_{sM}(\pi,\sigma)W_{sN}(\tau,\nu) [[/math]]

We use now two standard facts, namely:

(1) The fact that in the [math]N\to\infty[/math] limit the Weingarten matrix [math]W_{sN}[/math] is concentrated on the diagonal. This is indeed something very standard.

(2) The fact that we have an inequality as follows, with equality when [math]\pi=\sigma[/math]:

[[math]] |\pi\vee\sigma|\leq\frac{|\pi|+|\sigma|}{2} [[/math]]

Again, this is standard, and for details on all this, we refer to .

Let us discuss now what happens in the regime from the statement, namely:

[[math]] K=\kappa N,L=\lambda N,M=\mu N,N\to\infty [[/math]]

In this regime, we obtain:

[[math]] \begin{eqnarray*} M_s &\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,\bar{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,\bar{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.

(3) Finally, for [math]G_N=H_N^s/H_N^{s+}[/math], the variables [math]\chi_E[/math] are asymptotically Bessel/free Bessel of parameter [math]\frac{\kappa\lambda}{\mu}[/math], and once again in Bercovici-Pata bijection.

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There are several possible extensions of the above result, for instance by using quantum reflection groups instead of unitary quantum groups, and by using twisting operations as well. Finally, there are many interesting questions in relation with Connes' noncommutative geometry , and more specifically with the quantum extension of the Nash embedding theorem . We refer here to , , and related papers.

General references

Banica, Teo (2024). "Introduction to quantum groups". arXiv:1909.08152 [math.CO].