6b. Free isometries

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

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We can now 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]]

Show Proof

We recall that the various spheres involved are constructed as follows, with the symbol [math]\times[/math] standing for “commutative” and “free”, respectively:

[[math]] \begin{eqnarray*} C(S^{N-1}_{\mathbb R,\times})&=&C^*_\times\left(z_1,\ldots,z_N\Big|z_i=z_i^*,\sum_iz_i^2=1\right)\\ C(S^{N-1}_{\mathbb C,\times})&=&C^*_\times\left(z_1,\ldots,z_N\Big|\sum_iz_iz_i^*=\sum_iz_i^*z_i=1\right) \end{eqnarray*} [[/math]]

Now by comparing with the definition of [math]O_{1N}^{1\times},U_{1N}^{1\times}[/math], this proves our claim.

■

Similarly, we have as well the following result:

Proposition

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

This is similar to the proof of Proposition 6.6, coming from the definition of the various spheres involved, via some standard identifications.

■

Finally, again regarding examples, we have as well the following result:

Theorem

At [math]L=M=N[/math] we obtain the following 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 6.2 and to Definition 6.5, 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 the conditions on the right are all of the form [math](tr\otimes id)p=1[/math]. To be more precise, [math]p[/math] must be as 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]]

But this shows 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]]

Thus, we are led 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 more complicated in the free case than in the classical case, due to a number of reasons, of both algebraic and analytic nature. 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 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 6.9.

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.

■

Summarizing, we have so far free analogues of the spaces of partial isometries [math]O_{MN}^L[/math] and [math]U_{MN}^L[/math], along with some information about their homogeneous space structure, which looks quite axiomatic, as formulated in Theorem 6.10. There are many things to be done, as a continuation of this, and we will do this slowly, our plan being as follows:

(1) In the remainder of this chapter we will discuss as well discrete versions of the above constructions, and then we will go into the thing to be done, namely study of the Haar functional, and verification of the Bercovici-Pata bijection.

(2) And then, in chapters 7-8 below, we will discuss more abstract or more concrete versions of these constructions, following and related papers, the idea being that both generalizing and particularizing are interesting topics to be discussed.

As a general comment now, I can feel that you are a bit puzzled by our strategy, because we are talking here about homogeneous spaces, without knowing what an homogeneous space is, in the quantum setting. To which I would answer, please relax, there is absolutely no hurry with that. We have our spaces, which is a good thing, our study is on the way, another good thing, and the discussion of the homogeneous space structure, which will be something abstract, inspired from what we found in Theorem 6.10, and which will bring 0 advances on our problems to be solved, will be surely done at some point, and more specifically in chapter 7 below, but absolutely no hurry with that.

Hope you got my point, the quantum homogeneous spaces are not the same thing as the classical homogeneous spaces, and their study is quite tricky, following a different path. That's how the quantum world is, sometimes similar to the classical one, but sometimes very different. In case you are not convinced, pick a sphere [math]S[/math], as those studied so far in this book, and try writing that as a quotient space, and deducing from this results which are better than those established so far in this book, about such spheres [math]S[/math].

That will not work. And you will join here cohorts of mathematicians, having tried to develop theories of quantum homogeneous spaces, in nice and gentle analogy with the theory of classical homogeneous spaces, with quite average results. In fact, it is the paper , dealing with noncommutative spheres [math]S[/math] in a radical new way, via basic algebra and probability, that launched the modern theory, that we are explaining here.

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].