Half-liberation

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

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9a. Spheres and tori

We have seen in chapter 4 that the quadruplets of type [math](S,T,U,K)[/math] can be axiomatized, and that at the level of basic examples we have 4 such quadruplets, corresponding to the usual real and complex geometries [math]\mathbb R^N,\mathbb C^N[/math], and to the free versions of these:

[[math]] \xymatrix@R=50pt@C=50pt{ \mathbb R^N_+\ar[r]&\mathbb C^N_+\\ \mathbb R^N\ar[u]\ar[r]&\mathbb C^N\ar[u] } [[/math]]

Our purpose in what follows will be that of extending the above diagram, with the construction of some supplementary examples. There are two methods here:

(1) Look for intermediate geometries [math]\mathbb R^N\subset\mathcal X\subset\mathbb R^N_+[/math], and their complex analogues.

(2) Look for intermediate geometries [math]\mathbb R^N\subset\mathcal X\subset \mathbb C^N[/math], and their free analogues.

We will see that, in each case, there is a “standard” solution, and that these solutions can be combined. Thus, we will end up with a total of [math]3\times3=9[/math] solutions, as follows:

[[math]] \xymatrix@R=40pt@C=40pt{ \mathbb R^N_+\ar[r]&\mathbb T\mathbb R^N_+\ar[r]&\mathbb C^N_+\\ \mathbb R^N_*\ar[u]\ar[r]&\mathbb T\mathbb R^N_*\ar[u]\ar[r]&\mathbb C^N_*\ar[u]\\ \mathbb R^N\ar[u]\ar[r]&\mathbb T\mathbb R^N\ar[u]\ar[r]&\mathbb C^N\ar[u] } [[/math]]

There is quite some work to be done here, and the construction of these 9 geometries will take us the whole present chapter, and most of the next chapter as well. We will see also, at the end of the next chapter, that under certain strong axioms, of combinatorial type, these 9 geometries are conjecturally the only ones. Finally, in chapters 11-12 below we will discuss a number of related topics, such as twisting, and matrix models.

Observe that all this is in direct continuation of what we did in Part I, with no obvious relation with Part II. However, and here comes our point, once these intermediate geometries constructed, we will also have to “develop” them, meaning looking at various homogeneous spaces [math]X=G/H[/math], and other manifolds [math]X[/math], and here the theory developed in Part II, while mainly designed for being of help with free geometry, will be of great use. By the way, let us mention too that the intermediate geometries to be developed here, in Part III, will be quite close to the classical geometries, of [math]\mathbb R^N,\mathbb C^N[/math], and so our manifolds [math]X[/math] will start having interesting geometric features, that we will explore as well. Finally, for our outline to be complete, later in Part IV we will go back to the free geometries, of [math]\mathbb R^N_+,\mathbb C^N_+[/math], and develop more theory there, based on all this knowledge.

A few words on our motivations, too. There are many of them, as follows:

(1) The real half-classical geometry, of [math]\mathbb R^N_*[/math], is something very interesting in quantum group theory, due to the fact that the corresponding orthogonal group, [math]O_N^*[/math], is conjecturally the unique intermediate subgroup [math]O_N\subset G\subset O_N^+[/math]. Thus, regardless of our precise axioms here, the geometry of [math]\mathbb R^N_*[/math] can only be, at least conjecturally, the only intermediate geometry [math]\mathbb R^N\subset\mathcal X\subset\mathbb R^N_+[/math], so is definitely worth a study, mathematically speaking.

(2) Still talking [math]\mathbb R^N_*[/math], the geometry here is not that far from the geometries of [math]\mathbb R^N,\mathbb C^N[/math], so the study here can potentially lead into many things not available in the free case, and not discussed so far in this book, such as differential geometry, Lie theory, K-theory, and many more. Thus, in a certain sense, [math]\mathbb R^N_*[/math] is the “bridge” between our free geometry and more traditional visions of noncommutative geometry, such as Connes' .

(3) And pretty much the same goes for the other geometries to be investigated in this Part III, and particularly for the complex half-classical geometry, of [math]\mathbb C^N_*[/math], for reasons similar to those in the real case, and also for the twisted geometries, of [math]\bar{\mathbb R}^N,\bar{\mathbb C}^N[/math], making a link between our free quantum groups and free geometry with the more traditional vision of quantum groups and noncommutative geometry of Drinfeld-Jimbo , .

(4) Thus, plenty of good reasons for looking into such things, be them philosophical, or more concrete. And also, talking now physics, an interesting discovery, due to Bhowmick-D'Andrea-Dabrowski , and fine-tuned in their later paper with Das , is that the computations for the free gauge group of the Standard Model, in its Chamseddine-Connes formulation , , crucially involve the quantum group [math]U_N^*[/math].

And that is all, for the moment, more on this later. Getting to work now, our starting point will be the general axioms found in chapter 4, which are as follows:

Definition

An abstract noncommutative geometry is described by a quadruplet [math](S,T,U,K)[/math], formed of intermediate objects as follows,

[[math]] S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb C,+} [[/math]]

[[math]] T_N\subset T\subset\mathbb T_N^+ [[/math]]

[[math]] O_N\subset U\subset U_N^+ [[/math]]

[[math]] H_N\subset K\subset K_N^+ [[/math]]

subject to a set of connecting formulae between them, as follows,

[[math]] \begin{matrix} S&=&S_U\\ S\cap\mathbb T_N^+&=&T&=&K\cap\mathbb T_N^+\\ G^+(S)&=& \lt O_N,T \gt &=&U\\ K^+(T)&=&U\cap K_N^+&=&K \end{matrix} [[/math]]

with the usual convention that all this is up to the equivalence relation.

All this is of course quite tricky, and a bit simplified too, in the above form, and for full details on this, and comments, we refer to chapter 4. Now with this in hand, let us get into our first question, namely finding intermediate geometries as follows:

[[math]] \mathbb R^N\subset\mathcal X\subset\mathbb R^N_+ [[/math]]

Since such a geometry is given by a quadruplet [math](S,T,U,K)[/math], as above, forgetting about correspondences, we are led to 4 different intermediate object questions, as follows:

[[math]] S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+} [[/math]]

[[math]] T_N\subset T\subset T_N^+ [[/math]]

[[math]] O_N\subset U\subset O_N^+ [[/math]]

[[math]] H_N\subset K\subset H_N^+ [[/math]]

At the sphere and torus level, there are obviously uncountably many solutions, without supplementary assumptions, and it is hard to get beyond this, with bare hands. Thus, our hopes will basically come from the unitary and reflection quantum groups, where things are more rigid than for spheres and tori. Let us record, however, the following fact regarding the spheres, from , which will appear to be relevant, later on:

Theorem

The algebraic manifold [math]S^{(k)}\subset S^{N-1}_{\mathbb R,+}[/math] obtained by imposing the relations [math]a_1\ldots a_k=a_k\ldots a_1[/math] to the standard coordinates of [math]S^{N-1}_{\mathbb R,+}[/math] is as follows:

• At [math]k=1[/math] we have [math]S^{(k)}=S^{N-1}_{\mathbb R,+}[/math].
• At [math]k=2,4,6,\ldots[/math] we have [math]S^{(k)}=S^{N-1}_\mathbb R[/math].
• At [math]k=3,5,7,\ldots[/math] we have [math]S^{(k)}=S^{(3)}[/math].

Show Proof

As a first observation, the commutation relations [math]ab=ba[/math] imply the following relations, for any [math]k\geq2[/math]:

[[math]] a_1\ldots a_k=a_k\ldots a_1 [[/math]]

Thus, for any [math]k\geq2[/math], we have an inclusion [math]S^{(2)}\subset S^{(k)}[/math]. It is also elementary to check that the relations [math]abc=cba[/math] imply the following relations, for any [math]k\geq3[/math] odd:

[[math]] a_1\ldots a_k=a_k\ldots a_1 [[/math]]

Thus, for any [math]k\geq3[/math] odd, we have an inclusion [math]S^{(3)}\subset S^{(k)}[/math]. Our claim now is that we have an inclusion as follows, for any [math]k\geq2[/math]:

[[math]] S^{(k+2)}\subset S^{(k)} [[/math]]

In order to prove this, we must show that the relations [math]a_1\ldots a_{k+2}=a_{k+2}\ldots a_1[/math] between the coordinates [math]x_1,\ldots,x_N[/math] imply the relations [math]a_1\ldots a_k=a_k\ldots a_1[/math] between these coordinates [math]x_1,\ldots,x_N[/math]. But this holds indeed, because of the following implications:

[[math]] \begin{eqnarray*} x_{i_1}\ldots x_{i_{k+2}}=x_{i_{k+2}}\ldots x_{i_1} &\implies&x_{i_1}\ldots x_{i_k}x_j^2=x_j^2x_{i_k}\ldots x_{i_1}\\ &\implies&\sum_jx_{i_1}\ldots x_{i_k}x_j^2=\sum_jx_j^2x_{i_k}\ldots x_{i_1}\\ &\implies&x_{i_1}\ldots x_{i_k}=x_{i_k}\ldots x_{i_1} \end{eqnarray*} [[/math]]

Summing up, we have proved that we have inclusions as follows:

[[math]] S^{(2)}\subset\ldots\ldots\subset S^{(6)}\subset S^{(4)}\subset S^{(2)} [[/math]]

[[math]] S^{(3)}\subset\ldots\ldots\subset S^{(7)}\subset S^{(5)}\subset S^{(3)} [[/math]]

Thus, we are led to the conclusions in the statement.

■

As a conclusion, the “privileged” intermediate sphere [math]S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+}[/math] that we are looking for can only be the sphere [math]S^{(3)}[/math], obtained via the following relations:

[[math]] abc=cba [[/math]]

We should mention that, following , it is possible to go further in this direction, with a study of the spheres given by relations of the following type, with [math]\sigma\in S_k[/math]:

[[math]] a_1\ldots a_k=a_{\sigma(1)}\ldots a_{\sigma(k)} [[/math]]

But this leads to a similar conclusion, namely that the sphere [math]S^{(3)}[/math] constructed above is the only new solution. We will discuss this, which is a bit technical, later, in chapter 13 below. All this remains, however, quite ad-hoc. In short, we have constructed so far a new real sphere, [math]S^{(3)}[/math], and we some evidence for the fact that this sphere might be the only new one, under some extra combinatorial axioms, which are quite technical.

9b. Quantum groups

Let us focus now on the quantum groups. We will see that there is a lot more rigidity here, with regards to what happens for the spheres and tori, which makes things simpler. Our goal will be that of finding the intermediate quantum groups as follows:

[[math]] O_N\subset U\subset O_N^+ [[/math]]

[[math]] H_N\subset K\subset H_N^+ [[/math]]

Quite surprisingly, these two questions are of quite different nature. Indeed, regarding [math]O_N\subset U\subset O_N^+[/math], there is a solution here, denoted [math]O_N^*[/math], coming via the relations [math]abc=cba[/math], and conjecturally nothing more. Regarding however [math]H_N\subset K\subset H_N^+[/math], here it is possible to use for instance crossed products, for constructing uncountably many solutions.

In short, in connection with our intermediate geometry question, we do have in principle our solution, coming via the relations [math]abc=cba[/math], and this is compatible with our above [math]S^{(3)}[/math] guess for the spheres. In order to get started, let us recall that we have:

Theorem

The basic quantum unitary and reflection groups, namely

[[math]] \xymatrix@R=16pt@C=16pt{ &K_N^+\ar[rr]&&U_N^+\\ H_N^+\ar[rr]\ar[ur]&&O_N^+\ar[ur]\\ &K_N\ar[rr]\ar[uu]&&U_N\ar[uu]\\ H_N\ar[uu]\ar[ur]\ar[rr]&&O_N\ar[uu]\ar[ur] } [[/math]]

are all easy, coming from certain categories of partitions.

Show Proof

This is something that we already discussed, in chapter 2, the corresponding categories of partitions being as follows:

[[math]] \xymatrix@R=16pt@C5pt{ &\mathcal{NC}_{even}\ar[dl]\ar[dd]&&\ \ \mathcal {NC}_2\ar[dl]\ar[ll]\ar[dd]\\ NC_{even}\ar[dd]&&NC_2\ar[dd]\ar[ll]\\ &\mathcal P_{even}\ar[dl]&&\ \ \mathcal P_2\ar[dl]\ar[ll]\\ P_{even}&&P_2\ar[ll] } [[/math]]

Thus, we are led to the conclusion in the statement.

■

Getting back now to the half-liberation question, let us start by constructing the solutions. The result here, which is well-known as well, is as follows:

Theorem

We have quantum groups as follows, obtained via the half-commutation relations [math]abc=cba[/math], which fit into the diagram of basic quantum groups:

[[math]] \xymatrix@R=55pt@C=55pt{ K_N^*\ar[r]&U_N^*\\ H_N^*\ar[u]\ar[r]&O_N^*\ar[u] } [[/math]]

These quantum groups are all easy, and the corresponding categories of partitions fit into the diagram of categories of partitions for the basic quantum groups.

Show Proof

This is standard, from , the idea being that the half-commutation relations [math]abc=cba[/math] come from the map [math]T_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}[/math] associated to the half-classical crossing:

[[math]] \slash\hskip-1.9mm\backslash\hskip-1.7mm|\hskip0.5mm\in P(3,3) [[/math]]

Thus, the quantum groups in the statement are indeed easy, obtained by adding the half-classical crossing [math]\slash\hskip-1.9mm\backslash\hskip-1.7mm|\hskip0.5mm[/math] to the corresponding categories of noncrossing partitions. We obtain the following categories, with [math]*[/math] standing for the fact that, when relabelling clockwise the legs [math]\circ\bullet\circ\bullet\ldots[/math], the formula [math]\#\circ=\#\bullet[/math] must hold in each block:

[[math]] \xymatrix@R=50pt@C=50pt{ \mathcal P_{even}^*\ar[d]&\mathcal P_2^*\ar[l]\ar[d]\\ P_{even}^*&P_2^*\ar[l] } [[/math]]

Finally, the fact that our new quantum groups and categories fit well into the previous diagrams of quantum groups and categories is clear from this. See .

■

The point now is that we have the following result:

Theorem

There is only one proper intermediate easy quantum group

[[math]] O_N\subset G\subset O_N^+ [[/math]]

namely the half-classical orthogonal group [math]O_N^*[/math].

Show Proof

According to our definition for the easy quantum groups, we must compute here the intermediate categories of pairings, as follows:

[[math]] NC_2\subset D\subset P_2 [[/math]]

But this can be done via some standard combinatorics, in three steps, as follows:

(1) Let [math]\pi\in P_2-NC_2[/math], having [math]s\geq 4[/math] strings. Our claim is that:

-- If [math]\pi\in P_2-P_2^*[/math], there exists a semicircle capping [math]\pi'\in P_2-P_2^*[/math].

-- If [math]\pi\in P_2^*-NC_2[/math], there exists a semicircle capping [math]\pi'\in P_2^*-NC_2[/math].

Indeed, both these assertions can be easily proved, by drawing pictures.

(2) Consider now a partition [math]\pi\in P_2(k,l)-NC_2(k,l)[/math]. Our claim is that:

-- If [math]\pi\in P_2(k, l)-P_2^*(k,l)[/math] then [math] \lt \pi \gt =P_2[/math].

-- If [math]\pi\in P_2^*(k,l)-NC_2(k,l)[/math] then [math] \lt \pi \gt =P_2^*[/math].

This can be indeed proved by recurrence on the number of strings, [math]s=(k+l)/2[/math], by using (1), which provides us with a descent procedure [math]s\to s-1[/math], at any [math]s\geq4[/math].

(3) Finally, assume that we are given an easy quantum group [math]O_N\subset G\subset O_N^+[/math], coming from certain sets of pairings [math]D(k,l)\subset P_2(k,l)[/math]. We have three cases:

-- If [math]D\not\subset P_2^*[/math], we obtain [math]G=O_N[/math].

-- If [math]D\subset P_2,D\not\subset NC_2[/math], we obtain [math]G=O_N^*[/math].

-- If [math]D\subset NC_2[/math], we obtain [math]G=O_N^+[/math].

Thus, we are led to the conclusion in the statement.

■

It is in fact conjectured that the above result holds without the easiness assumption, and we refer here to . Thus, we have now an answer to our questions, with the half-classical real geometry being most likely unique, between classical and free real.

In practice now, what we have to do is to construct this geometry, and its complex analogue as well, and check the axioms from chapter 4. Let us begin by constructing the corresponding quadruplets. We have here the following result:

Theorem

We half-classical real and complex quadruplets, as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb R,*}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&T_N^*\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ O_N^*\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&H_N^*\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } \qquad\qquad \item[a]ymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb C,*}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&\mathbb T_N^*\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ U_N^*\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&K_N^*\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

obtained via [math]abc=cba[/math], imposed to the standard coordinates and their adjoints.

Show Proof

This is more of an empty statement, with the real quantum groups being those above, and with the other objects, namely complex quantum groups, and then spheres and tori, being constructed in a similar way, by starting with the free objects, and imposing the relations [math]abc=cba[/math] to the standard coordinates, and their adjoints.

■

We should mention here that, while the above constructions look trivial, the story with them was not trivial at all. Indeed, while things are certainly clear in the real case, in the complex case there are several possible ways of imposing the half-commutation relations [math]abc=cba[/math] to the standard coordinates and their adjoints, as follows:

(1) The above way, imposing [math]abc=cba[/math] to everything, both the standard coordinates and their adjoints, is the strongest such way, producing the smallest half-liberations, and in particular the smallest half-classical unitary quantum group, denoted [math]U_N^*[/math].

(2) In an opposite direction, imposing only the relations [math]ab^*c=cb^*a[/math] to the standard coordinates is something reasonable too, and this produces the biggest unitary quantum group which can be reasonably called “half-classical”, denoted [math]U_N^\times[/math].

(3) And then, there are all sorts of intermediate objects in between, [math]U_N^*\subset U_N^\circ\subset U_N^\times[/math], and notably the quantum group [math]U_N^{**}[/math] obtained by stating that the variables [math]\{ab^*,a^*b\}[/math] with [math]a,b[/math] standard coordinates should all commute, which is something natural too.

All this is quite technical, related to all sorts of advanced quantum group considerations, and there has been fierce debate all over the 10s, often between certain authors and their inner selves, on which relations to use, and more specifically, on which of the quantum groups [math]U_N^*\subset U_N^{**}\subset U_N^\times[/math] is the “correct” one. And with the literature on the subject, consisting notably of , , , , , , , , , , , being often confusing, with [math]U_N^*[/math] usually denoting the “correct” quantum group at the time of the paper, from the viewpoint of the paper, in a somewhat reckless way.

The solution to these questions came quite recently, first from the paper of Mang-Weber , who classified all the easy quantum groups [math]U_N^*\subset U_N^\circ\subset U_N^\times[/math], which allows one to have a more relaxed, complete perspective on all this, and then with the present noncommutative geometry considerations, coming as a continuation of , the point being that by Mang-Weber the only “good” quantum group among [math]U_N^*\subset U_N^{**}\subset U_N^\times[/math], which produces a noncommutative geometry in our sense, is [math]U_N^*[/math]. We will back to this later, when discussing , and classification for our noncommutative geometries.

9c. Matrix models

In order to check now our noncommutative geometry axioms, we are in need of a better understanding of the half-liberation operation, via some supplementary results. Let us start with the following simple observation, regarding the real spheres:

Proposition

We have a morphism of [math]C^*[/math]-algebras as follows,

[[math]] C(S^{N-1}_{\mathbb R,*})\to M_2(C(S^{N-1}_\mathbb C))\quad,\quad x_i\to\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix} [[/math]]

where [math]z_i[/math] are the standard coordinates of [math]S^{N-1}_\mathbb C[/math].

Show Proof

We have to prove that the matrices [math]X_i[/math] on the right satisfy the defining relations for [math]S^{N-1}_{\mathbb R,*}[/math]. But these matrices are self-adjoint, and we have:

[[math]] \begin{eqnarray*} \sum_iX_i^2 &=&\sum_i\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix}^2\\ &=&\sum_i\begin{pmatrix}|z_i|^2&0\\0&|z_i|^2\end{pmatrix}\\ &=&\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{eqnarray*} [[/math]]

As for the half-commutation relations, these follow from the following formula:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix}\begin{pmatrix}0&z_j\\ \bar{z}_j&0\end{pmatrix}\begin{pmatrix}0&z_k\\ \bar{z}_k&0\end{pmatrix}\\ &=&\begin{pmatrix}0&z_i\bar{z}_jz_k\\ \bar{z}_iz_j\bar{z}_k&0\end{pmatrix} \end{eqnarray*} [[/math]]

Indeed, the quantities on the right being symmetric in [math]i,k[/math], this gives the result.

■

Regarding the complex spheres, the result here is similar, as follows:

Proposition

We have a morphism of [math]C^*[/math]-algebras as follows,

[[math]] C(S^{N-1}_{\mathbb C,*})\to M_2(C(S^{N-1}_\mathbb C\times S^{N-1}_\mathbb C))\quad,\quad x_i\to\begin{pmatrix}0&z_i\\ y_i&0\end{pmatrix} [[/math]]

where [math]y_i,z_i[/math] are the standard coordinates of [math]S^{N-1}_\mathbb C\times S^{N-1}_\mathbb C[/math].

Show Proof

We have to prove that the matrices [math]X_i[/math] on the right satisfy the defining relations for [math]S^{N-1}_{\mathbb C,*}[/math]. We have the following computation:

[[math]] \begin{eqnarray*} \sum_iX_iX_i^* &=&\sum_i\begin{pmatrix}0&z_i\\ y_i&0\end{pmatrix}\begin{pmatrix}0&\bar{y}_i\\ \bar{z}_i&0\end{pmatrix}\\ &=&\sum_i\begin{pmatrix}|z_i|^2&0\\0&|y_i|^2\end{pmatrix}\\ &=&\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{eqnarray*} [[/math]]

We have as well the following computation:

[[math]] \begin{eqnarray*} \sum_iX_i^*X_i &=&\sum_i\begin{pmatrix}0&\bar{y}_i\\ \bar{z}_i&0\end{pmatrix}\begin{pmatrix}0&z_i\\ y_i&0\end{pmatrix}\\ &=&\sum_i\begin{pmatrix}|y_i|^2&0\\0&|z_i|^2\end{pmatrix}\\ &=&\begin{pmatrix}1&0\\0&1\end{pmatrix} \end{eqnarray*} [[/math]]

As for the half-commutation relations, these follow from the following formula:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\begin{pmatrix}0&z_i\\ y_i&0\end{pmatrix}\begin{pmatrix}0&z_j\\ y_j&0\end{pmatrix}\begin{pmatrix}0&z_k\\ y_k&0\end{pmatrix}\\ &=&\begin{pmatrix}0&z_iy_jz_k\\ y_iz_jy_k&0\end{pmatrix} \end{eqnarray*} [[/math]]

Indeed, the quantities on the right being symmetric in [math]i,k[/math], this gives the result.

■

Our goal now will be that of proving that the morphisms constructed above are faithful, up to the usual equivalence relation for the quantum algebraic manifolds. For this purpose, we will use some projective geometry arguments, the idea being that of proving that the above morphisms are indeed isomorphisms, at the projective version level, and then lifting these isomorphism results, to the affine setting. Let us recall that:

(1) The real projective space [math]P^{N-1}_\mathbb R[/math] is the space of lines in [math]\mathbb R^N[/math] passing through the origin. We have a quotient map [math]S^{N-1}_\mathbb R\to P^{N-1}_\mathbb R[/math], producing an embedding [math]C(P^{N-1}_\mathbb R)\subset C(S^{N-1}_\mathbb R)[/math], whose image is the algebra generated by the variables [math]p_{ij}=x_ix_j[/math].

(2) The complex projective space [math]P^{N-1}_\mathbb C[/math] has a similar description, namely is the space of complex lines in [math]\mathbb C^N[/math] passing through the origin, and we have an embedding [math]C(P^{N-1}_\mathbb C)\subset C(S^{N-1}_\mathbb C)[/math], whose image is generated by the variables [math]p_{ij}=x_i\bar{x}_j[/math].

The spaces [math]P^{N-1}_\mathbb R,P^{N-1}_\mathbb C[/math] have the following functional analytic description:

Theorem

We have presentation results as follows,

[[math]] \begin{eqnarray*} C(P^{N-1}_\mathbb C)&=&C^*_{comm}\left((p_{ij})_{i,j=1,\ldots,N}\Big|p=p^*=p^2,Tr(p)=1\right)\\ C(P^{N-1}_\mathbb R)&=&C^*_{comm}\left((p_{ij})_{i,j=1,\ldots,N}\Big|p=\bar{p}=p^*=p^2,Tr(p)=1\right) \end{eqnarray*} [[/math]]

where by [math]C^*_{comm}[/math] we mean as usual universal commutative [math]C^*[/math]-algebra.

Show Proof

We use the elementary fact that the spaces [math]P^{N-1}_\mathbb C,P^{N-1}_\mathbb R[/math], as defined above, are respectively the spaces of rank one projections in [math]M_N(\mathbb C),M_N(\mathbb R)[/math]. With this picture in mind, it is clear that we have arrows [math]\leftarrow[/math]. In order to construct now arrows [math]\to[/math], consider the universal algebras on the right, [math]A_C,A_R[/math]. These algebras being both commutative, by the Gelfand theorem we can write, with [math]X_C,X_R[/math] being certain compact spaces:

[[math]] A_C=C(X_C)\quad,\quad A_R=C(X_R) [[/math]]

Now by using the coordinate functions [math]p_{ij}[/math], we conclude that [math]X_C,X_R[/math] are certain spaces of rank one projections in [math]M_N(\mathbb C),M_N(\mathbb R)[/math]. In other words, we have embeddings:

[[math]] X_C\subset P^{N-1}_\mathbb C\quad,\quad X_R\subset P^{N-1}_\mathbb R [[/math]]

Bsy transposing we obtain arrows [math]\to[/math], as desired.

■

The above result suggests constructing free projective spaces [math]P^{N-1}_{\mathbb R,+},P^{N-1}_{\mathbb C,+}[/math], simply by lifting the commutativity conditions between the variables [math]p_{ij}[/math]. However, there is something wrong with this, and more specifically with [math]P^{N-1}_{\mathbb R,+}[/math], coming from the fact that if certain noncommutative coordinates [math]x_1,\ldots,x_N[/math] are self-adjoint, then the corresponding projective coordinates [math]p_{ij}=x_ix_j[/math] are not necessarily self-adjoint:

[[math]] x_i=x_i^*\ \,\not\!\!\!\!\implies x_ix_j=(x_ix_j)^* [[/math]]

In short, our attempt to construct free projective spaces [math]P^{N-1}_{\mathbb R,+},P^{N-1}_{\mathbb C,+}[/math] as above is not exactly correct, with the space [math]P^{N-1}_{\mathbb R,+}[/math] being rather “irrelevant”, and with the space [math]P^{N-1}_{\mathbb C,+}[/math] being probably the good one, but being at the same time “real and complex”. Observe that there is some similarity here with the following key result, from chapter 4:

[[math]] PO_N^+=PU_N^+ [[/math]]

To be more precise, we have good evidence here for the fact that, in the free setting, the projective geometry is at the same time real and complex. We will be back to this later, but in the meantime, in view of this, let us formulate the following definition:

Definition

Associated to any [math]N\in\mathbb N[/math] is the following universal algebra,

[[math]] C(P^{N-1}_+)=C^*\left((p_{ij})_{i,j=1,\ldots,N}\Big|p=p^*=p^2,Tr(p)=1\right) [[/math]]

whose abstract spectrum is called “free projective space”.

Observe that we have embeddings of noncommutative spaces, as follows:

[[math]] P^{N-1}_\mathbb R\subset P^{N-1}_\mathbb C\subset P^{N-1}_+ [[/math]]

Let us compute now the projective versions of the noncommutative spheres that we have, including the half-classical ones. We use the following formalism here:

Definition

The projective version of a closed subspace [math]S\subset S^{N-1}_{\mathbb C,+}[/math] is the quotient space [math]S\to PS[/math] determined by the fact that

[[math]] C(PS)\subset C(S) [[/math]]

is the subalgebra generated by [math]p_{ij}=x_ix_j^*[/math], called projective coordinates.

In the classical case, this fits with the usual definition. We will be back with more details in chapter 15 below, which is dedicated to the study of projective geometry. We have the following result, coming from , , :

Theorem

The projective versions of the basic spheres are as follows,

[[math]] \xymatrix@R=10mm@C=20mm{ 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]\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_\mathbb C\ar[u]} \qquad \item[a]ymatrix@R=14mm@C=20mm{\\ \to} \qquad \item[a]ymatrix@R=10mm@C=20mm{ P^{N-1}_+\ar[r]&P^{N-1}_+\\ P^{N-1}_\mathbb C\ar[r]\ar[u]&P^{N-1}_\mathbb C\ar[u]\\ P^{N-1}_\mathbb R\ar[r]\ar[u]&P^{N-1}_\mathbb C\ar[u]} [[/math]]

modulo, in the free case, a GNS construction with respect to the uniform integration.

Show Proof

The formulae on the bottom are true by definition. For the formulae on top, we have to prove first that the variables [math]p_{ij}=x_ix_j^*[/math] over the free sphere [math]S^{N-1}_{\mathbb C,+}[/math] satisfy the defining relations for [math]C(P^{N-1}_+)[/math]. In order to check this, we first have:

[[math]] (p^*)_{ij} =p_{ji}^* =(x_jx_i^*)^* =x_ix_j^* =p_{ij} [[/math]]

We have as well the following computation:

[[math]] (p^2)_{ij} =\sum_kp_{ik}p_{kj} =\sum_kx_ix_k^*x_kx_j^* =x_ix_j^* =p_{ij} [[/math]]

Finally, we have as well the following computation:

[[math]] Tr(p) =\sum_kp_{kk} =\sum_kx_kx_k^* =1 [[/math]]

Thus, we have embeddings of algebraic manifolds, as follows:

[[math]] PS^{N-1}_{\mathbb R,+}\subset PS^{N-1}_{\mathbb C,+}\subset P^{N-1}_+ [[/math]]

Regarding now the GNS construction assertion, this follows by reasoning as in the case of the free spheres, the idea being that the uniform integration on these projective spaces comes from the uniform integration over the following quantum group:

[[math]] PO_N^+=PU_N^+ [[/math]]

All this is quite technical, and we will not need this result, in what follows. We refer here to , and we will back to this in chapter 15 below. Finally, regarding the middle assertions, concerning the projective versions of the half-classical spheres, it is enough to prove here that we have inclusions as follows:

[[math]] P^{N-1}_\mathbb C\subset PS^{N-1}_{\mathbb R,*}\subset PS^{N-1}_{\mathbb C,*}\subset P^{N-1}_\mathbb C [[/math]]

But this can be done in 3 steps, as follows:

(1) [math]P^{N-1}_\mathbb C\subset PS^{N-1}_{\mathbb R,*}[/math]. In order to prove this, we recall from Proposition 9.7 that we have a morphism as follows, where [math]z_i[/math] are the standard coordinates of [math]S^{N-1}_\mathbb C[/math]:

[[math]] C(S^{N-1}_{\mathbb R,*})\to M_2(C(S^{N-1}_\mathbb C))\quad,\quad x_i\to\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix} [[/math]]

Now observe that this model maps the projective coordinates as follows:

[[math]] p_{ij}\to P_{ij}=\begin{pmatrix}z_i\bar{z}_j&0\\0&\bar{z}_iz_j\end{pmatrix} [[/math]]

Thus, at the level of generated algebras, our model maps:

[[math]] \lt p_{ij} \gt \to \lt P_{ij} \gt =C(P^{N-1}_\mathbb C) [[/math]]

We conclude from this that we have a quotient map as follows:

[[math]] C(PS^{N-1}_{\mathbb R,*})\to C(P^{N-1}_\mathbb C) [[/math]]

Thus at the level of corresponding spaces, we have, as desired, an inclusion:

[[math]] P^{N-1}_\mathbb C\subset PS^{N-1}_{\mathbb R,*} [[/math]]

(2) [math]PS^{N-1}_{\mathbb R,*}\subset PS^{N-1}_{\mathbb C,*}[/math]. This is something trivial, coming by functoriality of the operation [math]S\to PS[/math], from the inclusion of spheres:

[[math]] S^{N-1}_{\mathbb R,*}\subset S^{N-1}_{\mathbb C,*} [[/math]]

(3) [math]PS^{N-1}_{\mathbb C,*}\subset P^{N-1}_\mathbb C[/math]. This follows from the half-commutation relations, which imply:

[[math]] ab^*cd^* =cb^*ad^* =cd^*ab^* [[/math]]

Indeed, this shows that the projective version [math]PS^{N-1}_{\mathbb C,*}[/math] is classical, and so:

[[math]] PS^{N-1}_{\mathbb C,*} \subset(P^{N-1}_+)_{class} =P^{N-1}_\mathbb C [[/math]]

Thus, we are led to the conclusion in the statement.

■

We can go back now to the spheres, and we have the following result:

Theorem

We have a morphism of [math]C^*[/math]-algebras as follows,

[[math]] C(S^{N-1}_{\mathbb R,*})\subset M_2(C(S^{N-1}_\mathbb C))\quad,\quad x_i\to\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix} [[/math]]

where [math]z_i[/math] are the standard coordinates of [math]S^{N-1}_\mathbb C[/math].

Show Proof

We know from Proposition 9.7 that we have a morphism as above, and the injectivity follows from Theorem 9.12, by using a standard grading trick. See .

■

In the case of the complex spheres we have a similar result, as follows:

Theorem

We have a morphism of [math]C^*[/math]-algebras as follows,

[[math]] C(S^{N-1}_{\mathbb C,*})\to M_2(C(S^{N-1}_\mathbb C\times S^{N-1}_\mathbb C))\quad,\quad x_i\to\begin{pmatrix}0&z_i\\ y_i&0\end{pmatrix} [[/math]]

where [math]y_i,z_i[/math] are the standard coordinates of [math]S^{N-1}_\mathbb C\times S^{N-1}_\mathbb C[/math].

Show Proof

Again, we know from Proposition 9.8 that we have a morphism as above, and the injectivity follows from Theorem 9.12, via a grading trick, as explained in .

■

Summarizing, we have some interesting affine and projective geometry results regarding the half-classical case. The point now is that the same arguments apply to the tori, and to the quantum groups. We first have the following result:

Proposition

The real half-classical quadruplet, namely

[[math]] \xymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb R,*}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&T_N^*\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ O_N^*\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&H_N^*\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

and the complex real half-classical quadruplet, namely

[[math]] \xymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb C,*}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&\mathbb T_N^*\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ U_N^*\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&K_N^*\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

have [math]2\times2[/math] matrix models, constructed by using antidiagonal matrices, as for the spheres.

Show Proof

This is something that we already know from the spheres, from the various results established above. For the other objects which form the quadruplets, this follows by suitably adapting the proof of Proposition 9.7 and Proposition 9.8.

■

Next, once again in analogy with the sphere theory, we have the following result:

Theorem

The real and complex half-classical quadruplets have the same projective version, which is as follows:

[[math]] \xymatrix@R=50pt@C=50pt{ P^{N-1}_\mathbb C\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&P\mathbb T_N\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ PU_N\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&PK_N\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

Show Proof

As before, this is something that we already know from the spheres, from the various results established above. For the other objects which form the quadruplets, this follows from Proposition 9.15, by suitably adapting the proof of Theorem 9.12.

■

Finally, completing our study, we have the following result:

Theorem

The [math]2\times2[/math] antidiagonal matrix models for the real and complex half-classical quadruplets, constructed above, are faithful.

Show Proof

This is something that we already know from the spheres. For the other objects, this follows by suitably adapting the proof of Theorem 9.13 and Theorem 9.14.

■

Let us mention that the above results are part of a series of more general results, regarding matrix models for half-liberations. We will be back to this later.

9d. Axiom check

Let us check now the axioms, for our real and complex half-classical quadruplets. We first need some quantum isometry group results, which are available from , , , , for which we refer for the full details. First, we have the following result:

Theorem

The quantum isometry groups of the basic spheres are

[[math]] \xymatrix@R=10mm@C=20mm{ 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]\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_\mathbb C\ar[u]} \qquad \item[a]ymatrix@R=14.5mm@C=20mm{\\ \to} \qquad \item[a]ymatrix@R=11.5mm@C=23mm{ O_N^+\ar[r]&U_N^+\\ O_N^*\ar[r]\ar[u]&U_N^*\ar[u]\\ O_N\ar[r]\ar[u]&U_N\ar[u]} [[/math]]

modulo identifying, as usual, the various [math]C^*[/math]-algebraic completions.

Show Proof

We already know this, from chapter 3, for the spheres on top and bottom, so we just have to prove the results in the middle. So, assume as in chapter 3 that we have an action [math]G\curvearrowright S^{N-1}_{\mathbb C,*}[/math]. From [math]\Phi(x_a)=\sum_ix_i\otimes u_{ia}[/math] we obtain, with [math]p_{ab}=z_a\bar{z}_b[/math]:

[[math]] \Phi(p_{ab})=\sum_{ij}p_{ij}\otimes u_{ia}u_{jb}^* [[/math]]

By multiplying two such arbitrary formulae, we obtain:

[[math]] \begin{eqnarray*} \Phi(p_{ab}p_{cd})&=&\sum_{ijkl}p_{ij}p_{kl}\otimes u_{ia}u_{jb}^*u_{kc}u_{ld}^*\\ \Phi(p_{ad}p_{cb})&=&\sum_{ijkl}p_{il}p_{kj}\otimes u_{ia}u_{ld}^*u_{kc}u_{jb}^* \end{eqnarray*} [[/math]]

The left terms being equal, and the first terms on the right being equal too, we deduce that, with [math][a,b,c]=abc-cba[/math], we must have the following equality:

[[math]] \sum_{ijkl}p_{ij}p_{kl}\otimes u_{ia}[u_{jb}^*,u_{kc},u_{ld}^*]=0 [[/math]]

Now observe that the products of projective variables [math]p_{ij}p_{kl}=z_i\bar{z}_jz_k\bar{z}_l[/math] depend only on the following two cardinalities:

[[math]] |\{i,k\}|,|\{j,l\}|\in\{1,2\} [[/math]]

The point now is that this dependence produces the only relations between our variables, we are led in this way to [math]4[/math] equations, as follows:

(1) [math]u_{ia}[u_{jb}^*,u_{ka},u_{lb}^*]=0[/math], [math]\forall a,b[/math].

(2) [math]u_{ia}[u_{jb}^*,u_{ka},u_{ld}^*]+u_{ia}[u_{jd}^*,u_{ka},u_{lb}^*]=0[/math], [math]\forall a[/math], [math]\forall b\neq d[/math].

(3) [math]u_{ia}[u_{jb}^*,u_{kc},u_{lb}^*]+u_{ic}[u_{jb}^*,u_{ka},u_{lb}^*]=0[/math], [math]\forall a\neq c[/math], [math]\forall b[/math].

(4) [math]u_{ia}([u_{jb}^*,u_{kc},u_{ld}^*]+[u_{jd}^*,u_{kc},u_{lb}^*])+u_{ic}([u_{jb}^*,u_{ka},u_{ld}^*]+[u_{jd}^*,u_{ka},u_{lb}^*])=0,\forall a\neq c,b\neq d[/math].

From (1,2) we conclude that (2) holds with no restriction on the indices. By multiplying now this formula to the left by [math]u_{ia}^*[/math], and then summing over [math]i[/math], we obtain:

[[math]] [u_{jb}^*,u_{ka},u_{ld}^*]+[u_{jd}^*,u_{ka},u_{lb}^*]=0 [[/math]]

By applying now the antipode, then the involution, and finally by suitably relabelling all the indices, we successively obtain from this formula:

[[math]] \begin{eqnarray*} [u_{dl},u_{ak}^*,u_{bj}]+[u_{bl},u_{ak}^*,u_{dj}]=0 &\implies&[u_{dl}^*,u_{ak},u_{bj}^*]+[u_{bl}^*,u_{ak},u_{dj}^*]=0\\ &\implies&[u_{ld}^*,u_{ka},u_{jb}^*]+[u_{jd}^*,u_{ka},u_{lb}^*]=0 \end{eqnarray*} [[/math]]

Now by comparing with the original relation, above, we conclude that we have:

[[math]] [u_{jb}^*,u_{ka},u_{ld}^*]=[u_{jd}^*,u_{ka},u_{lb}^*]=0 [[/math]]

Thus we have reached to the formulae defining the quantum group [math]U_N^*[/math], and we are done. Finally, in what regards the universality of the action [math]O_N^*\curvearrowright S^{N-1}_{\mathbb R,*}[/math], this follows from the universality of the following two actions:

[[math]] U_N^*\curvearrowright S^{N-1}_{\mathbb C,*}\quad,\quad O_N^+\curvearrowright S^{N-1}_{\mathbb R,+} [[/math]]

Indeed, we obtain from this that we have [math]U_N^*\cap O_N^+=O_N^*[/math], as desired.

■

Regarding now the quantum isometry groups of the tori, the computation here, again form , , , , and that we partly know from chapter 3, is as follows:

Theorem

The quantum isometry groups of the basic tori are

[[math]] \xymatrix@R=10mm@C=20mm{ T_N^+\ar[r]&\mathbb T_N^+\\ T_N^*\ar[r]\ar[u]&\mathbb T_N^*\ar[u]\\ T_N\ar[r]\ar[u]&\mathbb T_N\ar[u]} \qquad \item[a]ymatrix@R=13.5mm@C=20mm{\\ \to} \qquad \item[a]ymatrix@R=10mm@C=20mm{ H_N^+\ar[r]&K_N^+\\ H_N^*\ar[r]\ar[u]&K_N^*\ar[u]\\ \bar{O}_N\ar[r]\ar@{.}[u]&\bar{U}_N\ar@{.}[u]} [[/math]]

with all arrows being inclusions, and with no vertical maps at bottom right.

Show Proof

As before, we just have to prove the results in the middle. In the real case, we must find the conditions on [math]G\subset O_N^+[/math] such that [math]g_a\to\sum_ig_a\otimes u_{ia}[/math] defines a coaction. In order for this map to be a coaction, the variables [math]G_a=\sum_ig_a\otimes u_{ia}[/math] must satisfy the following relations, which define the groups dual to the tori in the statement:

[[math]] G_a^2=1\quad,\quad G_aG_bG_c=G_cG_bG_a [[/math]]

In what regards the squares, we have the following formula:

[[math]] \begin{eqnarray*} G_a^2 &=&\sum_{ij}g_ig_j\otimes u_{ia}u_{ja}\\ &=&1+\sum_{i\neq j}g_ig_j\otimes u_{ia}u_{ja} \end{eqnarray*} [[/math]]

As for the products, with the notation [math][x,y,z]=xyz-zyx[/math], we have:

[[math]] \left[G_a,G_b,G_c\right]=\sum_{ijk}g_ig_jg_k\otimes [u_{ia},u_{jb},u_{kc}] [[/math]]

From the first relations, [math]G_a^2=1[/math], we obtain [math]G\subset H_N^+[/math]. In order to process now the second relations, [math]G_aG_bG_c=G_cG_bG_a[/math], we can split the sum over [math]i,j,k[/math], as follows:

[[math]] \begin{eqnarray*} \left[G_a,G_b,G_c\right] &=&\sum_{i,j,k\ distinct}g_ig_jg_k\otimes[u_{ia},u_{jb},u_{kc}]\\ &+&\sum_{i\neq j}g_ig_jg_i\otimes[u_{ia},u_{jb},u_{ic}]\\ &+&\sum_{i\neq j}g_i\otimes[u_{ia},u_{jb},u_{jc}]\\ &+&\sum_{i\neq k}g_k\otimes[u_{ia},u_{ib},u_{kc}]\\ &+&\sum_ig_i\otimes[u_{ia},u_{ib},u_{ic}] \end{eqnarray*} [[/math]]

Our claim is that the last three sums vanish. Indeed, observe that we have:

[[math]] [u_{ia},u_{ib},u_{ic}]=\delta_{abc}u_{ia}-\delta_{abc}u_{ia}=0 [[/math]]

Thus the last sum vanishes. Regarding now the fourth sum, we have:

[[math]] \begin{eqnarray*} \sum_{i\neq k}[u_{ia},u_{ib},u_{kc}] &=&\sum_{i\neq k}u_{ia}u_{ib}u_{kc}-u_{kc}u_{ib}u_{ia}\\ &=&\sum_{i\neq k}\delta_{ab}u_{ia}^2u_{kc}-\delta_{ab}u_{kc}u_{ia}^2\\ &=&\delta_{ab}\sum_{i\neq k}[u_{ia}^2,u_{kc}]\\ &=&\delta_{ab}\left[\sum_{i\neq k}u_{ia}^2,u_{kc}\right]\\ &=&\delta_{ab}[1-u_{ka}^2,u_{kc}]\\ &=&0 \end{eqnarray*} [[/math]]

The proof for the third sum is similar. Thus, we are left with the first two sums. By using [math]g_ig_jg_k=g_kg_jg_i[/math] for the first sum, the formula becomes:

[[math]] \begin{eqnarray*} \left[G_a,G_b,G_c\right] &=&\sum_{i \lt k,j\neq i,k}g_ig_jg_k\otimes([u_{ia},u_{jb},u_{kc}]+[u_{ka},u_{jb},u_{ic}])\\ &+&\sum_{i\neq j}g_ig_jg_i\otimes[u_{ia},u_{jb},u_{ic}] \end{eqnarray*} [[/math]]

In order to have a coaction, the above coefficients must vanish. Now observe that, when setting [math]i=k[/math] in the coefficients of the first sum, we obtain twice the coefficients of the second sum. Thus, our vanishing conditions can be formulated as follows:

[[math]] [u_{ia},u_{jb},u_{kc}]+[u_{ka},u_{jb},u_{ic}]=0,\forall j\neq i,k [[/math]]

Now observe that at [math]a=b[/math] or [math]b=c[/math] this condition reads [math]0+0=0[/math]. Thus, we can formulate our vanishing conditions in a more symmetric way, as follows:

[[math]] [u_{ia},u_{jb},u_{kc}]+[u_{ka},u_{jb},u_{ic}]=0,\forall j\neq i,k,\forall b\neq a,c [[/math]]

We use now the trick from . We apply the antipode to this formula, and then we relabel the indices [math]i\leftrightarrow c,j\leftrightarrow b,k\leftrightarrow a[/math]. We succesively obtain in this way:

[[math]] [u_{ck},u_{bj},u_{ai}]+[u_{ci},u_{bj},u_{ak}]=0,\forall j\neq i,k,\forall b\neq a,c [[/math]]

[[math]] [u_{ia},u_{jb},u_{kc}]+[u_{ic},u_{jb},u_{ka}]=0,\forall b\neq a,c,\forall j\neq i,k [[/math]]

Since we have [math][x,y,z]=-[z,y,x][/math], by comparing the last formula with the original one, we conclude that our vanishing relations reduce to a single formula, as follows:

[[math]] [u_{ia},u_{jb},u_{kc}]=0,\forall j\neq i,k,\forall b\neq a,c [[/math]]

Our first claim is that this formula implies [math]G\subset H_N^{[\infty]}[/math], where [math]H_N^{[\infty]}\subset O_N^+[/math] is defined via the relations [math]xyz=0[/math], for any [math]x\neq z[/math] on the same row or column of [math]u[/math]. In order to prove this, we will just need the [math]c=a[/math] particular case of this formula, which reads:

[[math]] u_{ia}u_{jb}u_{ka}=u_{ka}u_{jb}u_{ia},\forall j\neq i,k,\forall a\neq b [[/math]]

It is enough to check that the assumptions [math]j\neq i,k[/math] and [math]a\neq b[/math] can be dropped. But this is what happens indeed, because at [math]j=i[/math] we have:

[[math]] \begin{eqnarray*} \left[u_{ia},u_{ib},u_{ka}\right] &=&u_{ia}u_{ib}u_{ka}-u_{ka}u_{ib}u_{ia}\\ &=&\delta_{ab}(u_{ia}^2u_{ka}-u_{ka}u_{ia}^2)\\ &=&0 \end{eqnarray*} [[/math]]

Also, at [math]j=k[/math] we have:

[[math]] \begin{eqnarray*} \left[u_{ia},u_{kb},u_{ka}\right] &=&u_{ia}u_{kb}u_{ka}-u_{ka}u_{kb}u_{ia}\\ &=&\delta_{ab}(u_{ia}u_{ka}^2-u_{ka}^2u_{ia})\\ &=&0 \end{eqnarray*} [[/math]]

Finally, at [math]a=b[/math] we have:

[[math]] \begin{eqnarray*} \left[u_{ia},u_{ja},u_{ka}\right] &=&u_{ia}u_{ja}u_{ka}-u_{ka}u_{ja}u_{ia}\\ &=&\delta_{ijk}(u_{ia}^3-u_{ia}^3)\\ &=&0 \end{eqnarray*} [[/math]]

Our second claim now is that, due to [math]G\subset H_N^{[\infty]}[/math], we can drop the assumptions [math]j\neq i,k[/math] and [math]b\neq a,c[/math] in the original relations [math][u_{ia},u_{jb},u_{kc}]=0[/math]. Indeed, at [math]j=i[/math] we have:

[[math]] \begin{eqnarray*} [u_{ia},u_{ib},u_{kc}] &=&u_{ia}u_{ib}u_{kc}-u_{kc}u_{ib}u_{ia}\\ &=&\delta_{ab}(u_{ia}^2u_{kc}-u_{kc}u_{ia}^2)\\ &=&0 \end{eqnarray*} [[/math]]

The proof at [math]j=k[/math] and at [math]b=a[/math], [math]b=c[/math] being similar, this finishes the proof of our claim. We conclude that the half-commutation relations [math][u_{ia},u_{jb},u_{kc}]=0[/math] hold without any assumption on the indices, and so we obtain [math]G\subset H_N^*[/math], as claimed. As for the proof in the complex case, this is similar, and we refer here to and related papers.

■

By intersecting now with [math]K_N^+[/math], as required by our [math](S,T,U,K)[/math] axioms, we obtain:

Theorem

The quantum reflection groups of the basic tori are

[[math]] \xymatrix@R=10mm@C=20mm{ T_N^+\ar[r]&\mathbb T_N^+\\ T_N^*\ar[r]\ar[u]&\mathbb T_N^*\ar[u]\\ T_N\ar[r]\ar[u]&\mathbb T_N\ar[u]} \qquad \item[a]ymatrix@R=13.5mm@C=20mm{\\ \to} \qquad \item[a]ymatrix@R=10mm@C=20mm{ H_N^+\ar[r]&K_N^+\\ H_N^*\ar[r]\ar[u]&K_N^*\ar[u]\\ H_N\ar[r]\ar[u]&K_N\ar[u]} [[/math]]

with all the arrows being inclusions.

Show Proof

We already know that the results on the left and on the right hold indeed. As for the results in the middle, these follow from Theorem 9.19.

■

We can now formulate our extension result, as follows:

Theorem

We have basic noncommutative geometries, as follows,

[[math]] \xymatrix@R=30pt@C=80pt{ \mathbb R^N_+\ar[r]&\mathbb C^N_+\\ \mathbb R^N_*\ar[u]\ar[r]&\mathbb C^N_*\ar[u]\\ \mathbb R^N\ar[u]\ar[r]&\mathbb C^N\ar[u] } [[/math]]

with each [math]\mathbb K^N_\times[/math] symbol standing for the corresponding [math](S,T,U,K)[/math] quadruplet.

Show Proof

We have to check the axioms from chapter 4, for the half-classical geometries. The algebraic axioms are all clear, and the quantum isometry axioms follow from the above computations. Next in line, we have to prove the following formulae:

[[math]] O_N^*= \lt O_N,T_N^* \gt [[/math]]

[[math]] U_N^*= \lt U_N,\mathbb T_N^* \gt [[/math]]

By using standard generation results, it is enough to prove the first formula. Moreover, once again by standard generation results, it is enough to check that:

[[math]] H_N^*= \lt H_N,T_N^* \gt [[/math]]

The inclusion [math]\supset[/math] being clear, we are left with proving the inclusion [math]\subset[/math]. But this follows from the formula [math]H_N^*=T_N^*\rtimes S_N[/math], established by Raum-Weber in , as follows:

[[math]] \begin{eqnarray*} H_N^* &=&T_N^*\rtimes S_N\\ &=& \lt S_N,T_N^* \gt \\ &\subset& \lt H_N,T_N^* \gt \end{eqnarray*} [[/math]]

Alternatively, these formulae can be established by using the technology in , or by using categories and easiness. Finally, the axiom [math]S=S_U[/math] can be proved as in the classical and free cases, by using the Weingarten formula, and the following ergodicity property:

[[math]] \left(id\otimes\int_U\right)\Phi(x)=\int_Sx [[/math]]

Our claim, which will finish the proof, is that this holds as well in the half-classical case. Indeed, in the real case, where [math]x_i=x_i^*[/math], it is enough to check the above equality on an arbitrary product of coordinates, [math]x_{i_1}\ldots x_{i_k}[/math]. The left term is as follows:

[[math]] \begin{eqnarray*} \left(id\otimes\int_{O_N^*}\right)\Phi(x_{i_1}\ldots x_{i_k}) &=&\sum_{j_1\ldots j_k}x_{j_1}\ldots x_{j_k}\int_{O_N^*}u_{j_1i_1}\ldots u_{j_ki_k}\\ &=&\sum_{j_1\ldots j_k}\ \sum_{\pi,\sigma\in P_2^*(k)}\delta_\pi(j)\delta_\sigma(i)W_{kN}(\pi,\sigma)x_{j_1}\ldots x_{j_k}\\ &=&\sum_{\pi,\sigma\in P_2^*(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma)\sum_{j_1\ldots j_k}\delta_\pi(j)x_{j_1}\ldots x_{j_k} \end{eqnarray*} [[/math]]

Let us look now at the last sum on the right. We have to sum there quantities of type [math]x_{j_1}\ldots x_{j_k}[/math], over all choices of multi-indices [math]j=(j_1,\ldots,j_k)[/math] which fit into our given pairing [math]\pi\in P_2^*(k)[/math]. But by using the relations [math]x_ix_jx_k=x_kx_jx_i[/math], and then [math]\sum_ix_i^2=1[/math] in order to simplify, we conclude that the sum of these quantities is 1. Thus, we obtain:

[[math]] \left(id\otimes\int_{O_N^*}\right)\Phi(x_{i_1}\ldots x_{i_k}) =\sum_{\pi,\sigma\in P_2^*(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma) [[/math]]

On the other hand, another application of the Weingarten formula gives:

[[math]] \begin{eqnarray*} \int_{S^{N-1}_{\mathbb R,*}}x_{i_1}\ldots x_{i_k} &=&\int_{O_N^*}u_{1i_1}\ldots u_{1i_k}\\ &=&\sum_{\pi,\sigma\in P_2^*(k)}\delta_\pi(1)\delta_\sigma(i)W_{kN}(\pi,\sigma)\\ &=&\sum_{\pi,\sigma\in P_2^*(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma) \end{eqnarray*} [[/math]]

Thus, we are done. In the complex case the proof is similar, by adding exponents. For further details, we refer to for the real case, and to for the complex case.

■

Summarizing, in relation with the plan made in the beginning of this chapter, we have done so far half of our extension program, for the noncommutative geometries that we have. The second half, along with some classification work, is for the next chapter.

General references

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