8d. Uniformity

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

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Following , we discuss in what follows the structure of the row algebras in the case where the underlying quantum group is easy, which is the case that we are mostly interested in. As in , which was written some time ago, and based on dealing with the orthogonal case, [math]G\subset O_N^+[/math], we will restrict the attention to the orthogonal case. With the remark of course that the unitary extension looks quite straightforward. We will need the following key result, coming from , :

Theorem

For an easy subgroup [math]G_N\subset O_N^+[/math], the following are equivalent:

• [math]G=(G_N)[/math] is uniform, in the sense that we have
  [[math]] G_N\cap O_k^+=G_k [[/math]]
  for any [math]k\leq N[/math], with respect to the standard embedding [math]O_k^+\subset O_N^+[/math].
• The corresponding category of partitions
  [[math]] D=(D(k,l)) [[/math]]
  is stable under the operation which consists in removing blocks.

Show Proof

This can proved in several steps, as follows:

(1) In order to establish the equivalence between the above two conditions, we will prove that [math]G_N\cap O_k^+=G_k'[/math], where [math]G'=(G_N')[/math] is the easy quantum group associated to the category [math]D'[/math] generated by all subpartitions of the partitions in [math]D[/math].

(2) We know that the correspondence between categories of partitions and easy quantum groups comes from Woronowicz's Tannakian duality in , with the quantum group [math]G_N\subset O_N^+[/math] associated to a category of partitions [math]D=(D(s))[/math] obtained by imposing to the fundamental representation of [math]O_N^+[/math] the fact that its [math]s[/math]-th tensor power must fix [math]\xi_\pi[/math], for any [math]s\in\mathbb N[/math] and [math]\pi\in D(s)[/math]. In other words, we have the following formula:

[[math]] C(G_N)=C(O_N^+)\Big/\Big \lt \xi_\pi\in Fix(u^{\otimes s}),\forall s,\,\forall\pi\in D(s)\Big \gt [[/math]]

Now since [math]\xi_\pi\in Fix(u^{\otimes s})[/math] means [math]u^{\otimes s}(\xi_\pi\otimes 1)=\xi_\pi\otimes 1[/math], this condition is equivalent to the following collection of equalities, one for each multi-index [math]i\in\{1,\ldots,N\}^s[/math]:

[[math]] \sum_{j_1\ldots j_s}\delta_\pi(j)u_{i_1j_1}\ldots u_{i_sj_s}=\delta_\pi(i)1 [[/math]]

Summarizing, we have the following presentation result:

[[math]] C(G_N)=C(O_N^+)\Big/\left \lt \sum_{j_1\ldots j_s}\delta_\pi(j)u_{i_1j_1}\ldots u_{i_sj_s}=\delta_\pi(i)1,\forall s,\,\forall\pi\in D(s),\,\forall i\right \gt [[/math]]

(3) Let now [math]k\leq N[/math], assume that we have a compact quantum group [math]K\subset O_k^+[/math], with fundamental representation denoted [math]u[/math], and consider the following [math]N\times N[/math] matrix:

[[math]] \tilde{u}=\begin{pmatrix}u&0\\ 0&1_{N-k}\end{pmatrix} [[/math]]

Our claim is that for any [math]s\in\mathbb N[/math] and any [math]\pi\in P(s)[/math], we have:

[[math]] \xi_\pi\in Fix(\tilde{u}^{\otimes s})\iff\xi_{\pi'}\in Fix(u^{\otimes s'}),\,\forall\pi'\subset\pi [[/math]]

Here [math]\pi'\subset\pi[/math] means that [math]\pi'\in P(s')[/math] is obtained from [math]\pi\in P(s)[/math] by removing some of its blocks. The proof of this claim is standard. Indeed, when making the replacement [math]u\to\tilde{u}[/math] and trying to check the condition [math]\xi_\pi\in Fix(\tilde{u}^{\otimes s})[/math], we have two cases:

-- [math]\delta_\pi(i)=1[/math]. Here the [math] \gt k[/math] entries of [math]i[/math] must be joined by certain blocks of [math]\pi[/math], and we can consider the partition [math]\pi'\in D(s')[/math] obtained by removing these blocks. The point now is that the collection of [math]\delta_\pi(i)=1[/math] equalities to be checked coincides with the collection of [math]\delta_\pi(i)=1[/math] equalities expressing the fact that we have [math]\xi_\pi\in Fix(u^{\otimes s'})[/math], for any [math]\pi'\subset\pi[/math].

-- [math]\delta_\pi(i)=0[/math]. In this case the situation is quite similar. Indeed, the collection of [math]\delta_\pi(i)=0[/math] equalities to be checked coincides, modulo some [math]0=0[/math] identities, which hold automatically, with the collection of [math]\delta_\pi(i)=0[/math] equalities expressing the fact that we have [math]\xi_\pi\in Fix(u^{\otimes s'})[/math], for any [math]\pi'\subset\pi[/math].

(4) Our second claim is that given a quantum group [math]K\subset O_N^+[/math], with fundamental representation denoted [math]v[/math], the algebra of functions on [math]H=K\cap O_k^+[/math] is given by:

[[math]] C(H)=C(O_k^+)\Big/\Big \lt \xi\in Fix(\tilde{u}^{\otimes s}),\,\forall\xi\in Fix(v^{\otimes s})\Big \gt [[/math]]

But this follows indeed from Woronowicz's results in , because the algebra on the right comes from the Tannakian formulation of the intersection operation.

(5) Now with the above two claims in hand, we can conclude that we have the following formula, where [math]G'=(G_N')[/math] is the easy quantum group associated to the category [math]D'[/math] generated by all the subpartitions of the partitions in [math]D[/math]:

[[math]] G_N\cap U_k^+=G_k' [[/math]]

In particular we see that the condition [math]G_N\cap U_k^+=G_k^+[/math] for any [math]k\leq N[/math] is equivalent to [math]D=D'[/math], and this gives the result.

■

Let us study now the following inclusions of algebras, constructed in Definition 8.15, where [math]G=(G_n)[/math] is a uniform easy quantum group:

[[math]] C_\times(G_N/G_k)\subset C(G_N/G_k) [[/math]]

For classification purposes the uniformity axiom is something very natural and useful, substantially cutting from complexity, and we have the following result, from :

Theorem

The classical and free uniform orthogonal easy quantum groups, with inclusions between them, are as follows:

[[math]] \xymatrix@R=20pt@C=20pt{ &H_N^+\ar[rr]&&O_N^+\\ S_N^+\ar[rr]\ar[ur]&&B_N^+\ar[ur]\\ &H_N\ar[rr]\ar@.[uu]&&O_N\ar@.[uu]\\ S_N\ar@.[uu]\ar[ur]\ar[rr]&&B_N\ar@.[uu]\ar[ur] } [[/math]]

Moreover, this is an intersection/easy generation diagram, in the sense that for any of its square subdiagrams [math]P\subset Q,R\subset S[/math] we have [math]P=Q\cap R[/math] and [math]\{Q,R\}=S[/math].

Show Proof

In this statement all the quantum groups are objects that we are familiar with, and that we know to be easy, except for [math]B_N\subset O_N[/math] and [math]B_N^+\subset O_N^+[/math], which are the bistochastic group and its free analogue, constructed via the relation [math]\xi\in Fix(u)[/math], where [math]\xi[/math] is the all-one vector. Since this all-one vector corresponds to the singleton partition, the quantum groups [math]B_N,B_N^+[/math] follow to be easy too, coming from the categories [math]P_{12},NC_{12}[/math] of singletons and pairings. Thus, the quantum groups in the statement are all easy, and clearly uniform too, the corresponding categories of partitions being as follows:

[[math]] \xymatrix@R=20pt@C6pt{ &NC_{even}\ar[dl]\ar@.[dd]&&NC_2\ar[dl]\ar[ll]\ar@.[dd]\\ NC\ar@.[dd]&&NC_{12}\ar@.[dd]\ar[ll]\\ &P_{even}\ar[dl]&&P_2\ar[dl]\ar[ll]\\ P&&P_{12}\ar[ll] } [[/math]]

Since this latter diagram is an intersection and generation diagram, we conclude that we have an intersection and easy generation diagram of quantum groups, as stated.

Regarding now the classification, consider an easy quantum group [math]S_N\subset G_N\subset O_N[/math]. This most come from a category [math]P_2\subset D\subset P[/math], and if we assume [math]G=(G_N)[/math] to be uniform, then [math]D[/math] is uniquely determined by the subset [math]L\subset\mathbb N[/math] consisting of the sizes of the blocks of the partitions in [math]D[/math]. Our claim is that the admissible sets are as follows:

• [math]L=\{2\}[/math], producing [math]O_N[/math].
• [math]L=\{1,2\}[/math], producing [math]B_N[/math].
• [math]L=\{2,4,6,\ldots\}[/math], producing [math]H_N[/math].
• [math]L=\{1,2,3,\ldots\}[/math], producing [math]S_N[/math].

In one sense, this follows indeed from our easiness results for [math]O_N,B_N,H_N,S_N[/math]. In the other sense now, assume that [math]L\subset\mathbb N[/math] is such that the set [math]P_L[/math] consisting of partitions whose sizes of the blocks belong to [math]L[/math] is a category of partitions. We know from the axioms of the categories of partitions that the semicircle [math]\cap[/math] must be in the category, so we have [math]2\in L[/math]. Our claim is that the following conditions must be satisfied as well:

[[math]] k,l\in L,\,k \gt l\implies k-l\in L [[/math]]

[[math]] k\in L,\,k\geq 2\implies 2k-2\in L [[/math]]

Indeed, we will prove that both conditions follow from the axioms of the categories of partitions. Let us denote by [math]b_k\in P(0,k)[/math] the one-block partition, namely:

[[math]] b_k=\left\{\begin{matrix}\sqcap\hskip-0.7mm \sqcap&\ldots&\sqcap\\ 1\hskip2mm 2&\ldots&k\end{matrix} \right\} [[/math]]

For [math]k \gt l[/math], we can write the partition [math]b_{k-l}[/math] in the following way:

[[math]] b_{k-l}=\left\{\begin{matrix}\sqcap\hskip-0.7mm \sqcap&\ldots&\ldots&\ldots&\ldots&\sqcap\\ 1\hskip2mm 2&\ldots&l&l+1&\ldots&k\\ \sqcup\hskip-0.7mm \sqcup&\ldots&\sqcup&|&\ldots&|\\ &&&1&\ldots&k-l\end{matrix}\right\} [[/math]]

In other words, we have the following formula:

[[math]] b_{k-l}=(b_l^*\otimes |^{\otimes k-l})b_k [[/math]]

Since all the terms of this composition are in [math]P_L[/math], we have [math]b_{k-l}\in P_L[/math], and this proves our first claim. As for the second claim, this can be proved in a similar way, by capping two adjacent [math]k[/math]-blocks with a [math]2[/math]-block, in the middle. Now, we can conclude as follows:

\underline{Case 1}. Assume [math]1\in L[/math]. By using the first condition with [math]l=1[/math] we get:

[[math]] k\in L\implies k-1\in L [[/math]]

This condition shows that we must have [math]L=\{1,2,\ldots,m\}[/math], for a certain number [math]m\in\{1,2,\ldots,\infty\}[/math]. On the other hand, by using the second condition we get:

[[math]] \begin{eqnarray*} m\in L &\implies&2m-2\in L\\ &\implies&2m-2\leq m\\ &\implies&m\in\{1,2,\infty\} \end{eqnarray*} [[/math]]

The case [math]m=1[/math] being excluded by the condition [math]2\in L[/math], we reach to one of the two sets producing the groups [math]S_N,B_N[/math].

\underline{Case 2}. Assume [math]1\notin L[/math]. By using the first condition with [math]l=2[/math] we get:

[[math]] k\in L\implies k-2\in L [[/math]]

This condition shows that we must have [math]L=\{2,4,\ldots,2p\}[/math], for a certain number [math]p\in\{1,2,\ldots,\infty\}[/math]. On the other hand, by using the second condition we get:

[[math]] \begin{eqnarray*} 2p\in L &\implies&4p-2\in L\\ &\implies&4p-2\leq 2p\\ &\implies&p\in\{1,\infty\} \end{eqnarray*} [[/math]]

Thus [math]L[/math] must be one of the two sets producing [math]O_N,H_N[/math], and we are done. In the free case, [math]S_N^+\subset G_N\subset O_N^+[/math], the situation is quite similar, the admissible sets being once again the above ones, producing this time [math]O_N^+,B_N^+,H_N^+,S_N^+[/math]. See .

■

Let us go back now to the inclusions [math]C_\times(G_N/G_k)\subset C(G_N/G_k)[/math]. Following , we first work out a few simple cases, where these inclusions are isomorphisms:

Proposition

For the basic easy quantum groups, the inclusion of algebras

[[math]] C_\times(G_N/G_k)\subset C(G_N/G_k) [[/math]]

is an isomorphism at [math]N=1[/math], at [math]k=0[/math], at [math]k=N[/math], as well as in the following cases:

• [math]G=B^+[/math]: at [math]k=1[/math].
• [math]G=S^+[/math]: at [math]k=1[/math], and at [math]k=2,N=3[/math].

Show Proof

First, the results at [math]N=1[/math], at [math]k=0[/math], and at [math]k=N[/math] are clear from definitions. Regarding now the special cases, the situation here is as follows:

(1) Since the coordinates of [math]B_N^+[/math] sum up to 1 on each column, we have:

[[math]] u_{1j}=1-\sum_{i \gt 1}u_{ij} [[/math]]

Thus the following inclusion is an isomorphism:

[[math]] C_\times(B_N^+/B_1^+)\subset C(B_N^+) [[/math]]

Thus the inclusion [math]C_\times(B_N^+/B_1^+)\subset C(B_N^+/B_1^+)[/math] must be as well an isomorphism.

(2) By using the same argument as above we obtain that the following inclusion is as well an isomorphism:

[[math]] C_\times(S_N^+/S_1^+)\subset C(S_N^+/S_1^+) [[/math]]

In the remaining case [math]k=2,N=3[/math], or more generally at any [math]k\in\mathbb N[/math] and [math]N \lt 4[/math], it is known from Wang that we have [math]S_N=S_N^+[/math], so the inclusion in the statement is:

[[math]] C(S_N/S_k)\subset C(S_N/S_k) [[/math]]

Thus, in this case we are done again.

■

The axiomatization of the algebras [math]C_\times(G_N/G_k)[/math] is a quite tricky task. However, following , we can axiomatize some bigger algebras, as follows:

Definition

Associated to [math]k\leq N[/math] is the universal [math]C^*[/math]-algebra [math]C_+(G_N/G_k)[/math] generated by the entries of a matrix [math]p=(p_{ij})_{i \gt k,j \gt 0}[/math] subject to the following conditions:

• [math]G=O_N^+[/math]: [math]p[/math] is a transposed “orthogonal isometry”, in the sense that its entries [math]p_{ij}[/math] are self-adjoint, and [math]pp^t=1[/math].
• [math]G=S_N^+[/math]: [math]p[/math] is a transposed “magic isometry”, in the sense that [math]p^t[/math] is an orthogonal isometry, and [math]p_{ij}[/math] are projections, orthogonal on columns.
• [math]G=H_N^+[/math]: [math]p[/math] is a transposed “cubic isometry”, in the sense that [math]p^t[/math] is an orthogonal isometry, with [math]xy=0[/math] for any [math]x\neq y[/math] on the same row of [math]p[/math]
• [math]G=B_N^+[/math]: [math]p[/math] is a transposed “stochastic isometry”, in the sense that [math]p^t[/math] is an orthogonal isometry, with sum [math]1[/math] on rows.

Observe that we have surjective maps, as follows:

[[math]] C_+(G_N/G_k)\to C_\times(G_N/G_k) [[/math]]

Still following , we have the following result:

Theorem

The algebras [math]C_+(G_N/G_k)[/math] and [math]C_\times(G_N/G_k)[/math] are as follows:

• They have coactions of [math]G_N[/math], given by [math]\alpha(p_{ij})=\sum_sp_{is}\otimes u_{sj}[/math].
• They have unique [math]G_N[/math]-invariant states, which are tracial.
• Their reduced algebra versions are isomorphic.
• Their abelianized versions are isomorphic.

Show Proof

This is something quite long, the idea being as follows:

(1) For the algebra [math]C_\times(G_N/G_k)[/math] this is clear, because as explained in , this algebra is “embeddable”, and the coaction of [math]G_N[/math] is simply the restriction of the comultiplication map. For the algebra [math]C_+(G_N/G_k)[/math], consider the following elements:

[[math]] P_{ij}=\sum_{s=1}^Np_{is}\otimes u_{sj} [[/math]]

We have to check that these elements satisfy the same relations as those in Definition 8.21, presenting the algebra [math]C_+(G_n/G_k)[/math], and the proof here goes as follows:

\underline{[math]O^+[/math] case}. First, since [math]p_{ij},u_{ij}[/math] are self-adjoint, so is [math]P_{ij}[/math]. Also, we have:

[[math]] \begin{eqnarray*} \sum_jP_{ij}P_{rj} &=&\sum_{jst}p_{is}p_{rt}\otimes u_{sj}u_{tj}\\ &=&\sum_{st}p_{is}p_{rt}\otimes\delta_{st}\\ &=&\sum_sp_{is}p_{rs}\otimes 1\\ &=&\delta_{ir} \end{eqnarray*} [[/math]]

\underline{[math]H^+[/math] case}. The condition [math]xy=0[/math] on rows is checked as follows ([math]j\neq r[/math]):

[[math]] \begin{eqnarray*} P_{ij}P_{ir} &=&\sum_{st}p_{is}p_{it}\otimes u_{sj}u_{tr}\\ &=&\sum_sp_{is}\otimes u_{sj}u_{sr}\\ &=&0 \end{eqnarray*} [[/math]]

\underline{[math]B^+[/math] case}. The sum 1 condition on rows is checked as follows:

[[math]] \begin{eqnarray*} \sum_jP_{ij} &=&\sum_{js}p_{is}\otimes u_{sj}\\ &=&\sum_sp_{is}\otimes 1\\ &=&1 \end{eqnarray*} [[/math]]

\underline{[math]S^+[/math] case}. Since [math]P^t[/math] is cubic and stochastic, we just check the projection condition:

[[math]] \begin{eqnarray*} P_{ij}^2 &=&\sum_{st}p_{is}p_{it}\otimes u_{sj}u_{tj}\\ &=&\sum_sp_{is}\otimes u_{sj}\\ &=&P_{ij} \end{eqnarray*} [[/math]]

Summmarizing, the matrix [math]P[/math] satisfies the same conditions as [math]p[/math], so we can define a morphism of [math]C^*[/math]-algebras, as follows:

[[math]] \alpha:C_+(G_N/G_k)\to C_+(G_N/G_k)\otimes C(G_N)\quad,\quad \alpha(p_{ij})=P_{ij} [[/math]]

We have the following computation:

[[math]] \begin{eqnarray*} (\alpha\otimes id)\alpha (p_{ij}) &=&\sum_s\alpha(p_{is})\otimes u_{sj}\\ &=&\sum_{st}p_{it}\otimes u_{ts}\otimes u_{sj} \end{eqnarray*} [[/math]]

On the other hand, we have as well the following computation:

[[math]] \begin{eqnarray*} (id\otimes\Delta)\alpha(p_{ij}) &=&\sum_tp_{it}\otimes\Delta(u_{ij})\\ &=&\sum_{st}p_{it}\otimes u_{ts}\otimes u_{sj} \end{eqnarray*} [[/math]]

Thus our map [math]\alpha[/math] is coassociative. The density conditions can be checked by using dense subalgebras generated by [math]p_{ij}[/math] and [math]u_{st}[/math], and we are done.

(2) For the existence part we can use the following composition, where the first two maps are the canonical ones, and the map on the right is the integration over [math]G_N[/math]:

[[math]] C_+(G_N/G_k)\to C_\times(G_N/G_k)\subset C(G_N)\to\mathbb C [[/math]]

Also, the uniqueness part is clear for the algebra [math]C_\times(G_N/G_k)[/math]. Regarding now the uniqueness for [math]C_+(G_N/G_k)[/math], let [math]\int[/math] be the Haar state on [math]G_N[/math], and [math]\varphi[/math] be the [math]G_N[/math]-invariant state constructed above. We claim that [math]\alpha[/math] is ergodic:

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

Indeed, let us recall that the Haar state is given by the following Weingarten formula, where [math]W_{sN}=G_{sN}^{-1}[/math], with [math]G_{sN}(\pi,\sigma)=N^{|\pi\vee\sigma|}[/math]:

[[math]] \int u_{i_1j_1}\ldots u_{i_sj_s}=\sum_{\pi,\sigma\in D(s)}\delta_\pi(i)\delta_\sigma(j)W_{sN}(\pi,\sigma) [[/math]]

Now, let us go back now to our claim. By linearity it is enough to check the above equality on a product of basic generators [math]p_{i_1j_1}\ldots p_{i_sj_s}[/math]. The left term is as follows:

[[math]] \begin{eqnarray*} \left(id\otimes\int\right)\alpha(p_{i_1j_1}\ldots p_{i_sj_s}) &=&\sum_{l_1\ldots l_s}p_{i_1l_1}\ldots p_{i_sl_s}\int u_{l_1j_1}\ldots u_{l_sj_s}\\ &=&\sum_{l_1\ldots l_s}p_{i_1l_1}\ldots p_{i_sl_s}\sum_{\pi,\sigma\in D(s)}\delta_\pi(l)\delta_\sigma(j)W_{sN}(\pi,\sigma)\\ &=&\sum_{\pi,\sigma\in D(s)}\delta_\sigma(j)W_{sN}(\pi,\sigma)\sum_{l_1\ldots l_s}\delta_\pi(l)p_{i_1l_1}\ldots p_{i_sl_s} \end{eqnarray*} [[/math]]

Let us look now at the sum on the right. We have to sum the elements of type [math]p_{i_1l_1}\ldots p_{i_sl_s}[/math], over all multi-indices [math]l=(l_1,\ldots,l_s)[/math] which fit into our partition [math]\pi\in D(s)[/math]. In the case of a one-block partition this sum is simply [math]\sum_lp_{i_1l}\ldots p_{i_sl}[/math], and we claim that:

[[math]] \sum_lp_{i_1l}\ldots p_{i_sl}=\delta_\pi(i) [[/math]]

Indeed, the proof of this formula goes as follows:

\underline{[math]O^+[/math] case}. Here our one-block partition must be a semicircle, [math]\pi=\cap[/math], and the formula to be proved, namely [math]\sum_lp_{il}p_{jl}=\delta_{ij}[/math], follows from [math]pp^t=1[/math].

\underline{[math]S^+[/math] case}. Here our one-block partition can be any [math]s[/math]-block, [math]1_s\in P(s)[/math], and the formula to be proved, namely [math]\sum_lp_{i_1l}\ldots p_{i_sl}=\delta_{i_1,\ldots,i_s}[/math], follows from orthogonality on columns, and from the fact that the sum is 1 on rows.

\underline{[math]B^+[/math] case}. Here our one-block partition must be a semicircle or a singleton. We are already done with the semicircle, and for the singleton the formula to be proved, namely [math]\sum_lp_{il}=1[/math], follows from the fact that the sum is 1 on rows.

\underline{[math]H^+[/math] case}. Here our one-block partition must have an even number of legs, [math]s=2r[/math], and due to the cubic condition the formula to be proved reduces to [math]\sum_lp_{il}^{2r}=1[/math]. But since [math]p_{il}^{2r}=p_{il}^2[/math], independently on [math]r[/math], the result follows from the orthogonality on rows.

In the general case now, since [math]\pi[/math] noncrossing, the computations over the blocks will not interfere, and we will obtain the same result, namely:

[[math]] \sum_lp_{i_1l}\ldots p_{i_sl}=\delta_\pi(i) [[/math]]

Now by plugging this formula into the computation that we have started, we get:

[[math]] \begin{eqnarray*} \left(id\otimes\int\right)\alpha(p_{i_1j_1}\ldots p_{i_sj_s}) &=&\sum_{\pi,\sigma\in D(s)}\delta_\pi(i)\delta_\sigma(j)W_{sN}(\pi,\sigma)\\ &=&\int u_{i_1j_1}\ldots u_{i_sj_s}\\ &=&\varphi(p_{i_1j_1}\ldots p_{i_sj_s}) \end{eqnarray*} [[/math]]

This finishes the proof of our claim. So, let us get back now to the original question. Let [math]\tau:C_+(G_N/G_k)\to\mathbb C[/math] be a linear form as in the statement. We have:

[[math]] \begin{eqnarray*} \tau\left(id\otimes\int\right)\alpha(x) &=&\left(\tau\otimes\int\right)\alpha(x)\\ &=&\int(\tau\otimes id)\alpha(x)\\ &=&\int(\tau(x)1)\\ &=&\tau(x) \end{eqnarray*} [[/math]]

On the other hand, according to our above claim, we have as well:

[[math]] \tau\left(id\otimes\int\right)\alpha(x) =\tau(\varphi(x)1) =\varphi(x) [[/math]]

Thus we get [math]\tau=\varphi[/math], which finishes the proof of the uniqueness assertion.

(3) This follows from the uniqueness assertions in (2), and from some standard facts regarding the reduced versions with respect to Haar states, from Woronowicz .

(4) We denote by [math]G^-[/math] the classical version of [math]G[/math], given by [math]G^-=O,S,H,B[/math] in the cases [math]G=O^+,S^+,H^+,B^+[/math]. We have surjective morphisms of algebras, as follows:

[[math]] \begin{eqnarray*} C_+(G_N/G_k) &\to&C_\times(G_k/G_k)\\ &\to&C_\times(G_N^-/G_k^-)\\ &=&C(G_N^-/G_k^-) \end{eqnarray*} [[/math]]

Thus at the level of abelianized versions, we have surjective morphisms as follows:

[[math]] \begin{eqnarray*} C_+(G_N/G_k)_{comm} &\to&C_\times(G_N/G_k)_{comm}\\ &\to&C(G_N^-/G_k^-) \end{eqnarray*} [[/math]]

In order to prove our claim, namely that the first surjective morphism is an isomorphism, it is enough to prove that the above composition is an isomorphism.

Let [math]r=N-k[/math], and denote by [math]A_{N,r}[/math] the algebra on the left. This is by definition the algebra generated by the entries of a transposed [math]N\times r[/math] isometry, whose entries commute, and which is respectively orthogonal, magic, cubic, bistochastic.

We have a surjective morphism [math]A_{N,r}\to C(G_N^-/G_k^-)[/math], and we must prove that this is an isomorphism.

\underline{[math]S^+[/math] case}. Since [math]\#(S_N/S_k)=N!/k![/math], it is enough to prove that we have:

[[math]] \dim(A_{N,r})=\frac{N!}{k!} [[/math]]

Let [math]p_{ij}[/math] be the standard generators of [math]A_{N,r}[/math]. By using the Gelfand theorem, we can write [math]p_{ij}=\chi(X_{ij})[/math], where [math]X_{ij}\subset X[/math] are certain subets of a given set [math]X[/math]. Now at the level of sets the magic isometry condition on [math](p_{ij})[/math] tells us that the matrix of sets [math](X_{ij})[/math] has the property that its entries are disjoint on columns, and form partitions of [math]X[/math] on rows.

So, let us try to understand this property for [math]N[/math] fixed, and [math]r=1,2,3,\ldots[/math]

-- At [math]r=1[/math] we simply have a partition [math]X=X_1\sqcup\ldots\sqcup X_N[/math]. So, the universal model can be any such partition, with [math]X_i\neq 0[/math] for any [math]i[/math].

-- At [math]r=2[/math] the universal model is best described as follows: [math]X[/math] is the [math]N\times N[/math] square in [math]\mathbb R^2[/math], regarded as a union of [math]N^2[/math] unit tiles, minus the diagonal, the sets [math]X_{1i}[/math] are the disjoint unions on rows, and the sets [math]X_{2i}[/math] are the disjoint unions on columns.

-- At [math]r\geq 3[/math], the universal solution is similar: we can take [math]X[/math] to be the [math]N^r[/math] cube in [math]\mathbb R^r[/math], with all tiles having pairs of equal coordinates removed, and say that the sets [math]X_{si}[/math] for [math]s[/math] fixed are the various “slices” of [math]X[/math] in the direction of the [math]s[/math]-th coordinate of [math]\mathbb R^r[/math].

Summarizing, the above discussion tells us that [math]\dim(A_{N,r})[/math] equals the number of tiles in the above set [math]X\subset\mathbb R^r[/math]. But these tiles correspond by definition to the various [math]r[/math]-tuples [math](i_1,\ldots,i_r)\in\{1,\ldots,N\}^r[/math] with all [math]i_k[/math] different, and since there are exactly [math]N!/k![/math] such [math]r[/math]-tuples, we obtain, as desired:

[[math]] \dim(A_{N,r})=\frac{N!}{k!} [[/math]]

\underline{[math]H^+[/math] case}. We can use here the same method as for [math]S_N^+[/math]. This time the functions [math]p_{ij}[/math] take values in [math]\{-1,0,1\}[/math], and the algebra generated by their squares [math]p_{ij}^2[/math] coincides with the one computed above for [math]S_N^+[/math], having dimension [math]N!/k![/math]. Now by taking into account the [math]N-k[/math] possible signs we obtain the following estimate, which gives the result:

[[math]] \dim(A_{N,r})\leq\frac{2^{N-k}N!}{k!}=\#(H_N/H_k) [[/math]]

\underline{[math]O^+[/math] case}. We can use the same method, namely a straightforward application of the Gelfand theorem. However, instead of performing a dimension count, which is no longer possible, we have to complete here any transposed [math]N\times r[/math] isometry whose entries commute to a [math]N\times N[/math] orthogonal matrix. But this is the same as completing a system of [math]r[/math] orthogonal norm 1 vectors in [math]\mathbb R^N[/math] into an orthonormal basis of [math]\mathbb R^N[/math], which is of course possible.

\underline{[math]B^+[/math] case}. Since we have a surjective map [math]C(O_N^+)\to C(B_N^+)[/math], we obtain a surjective map [math]C_+(O_N^+/O_k^+)\to A_{N,r}[/math], and hence surjective maps as follows:

[[math]] C(O_N/O_k)\to A_{N,r}\to C(B_N/B_k) [[/math]]

The point now is that this composition is the following canonical map:

[[math]] C(O_N/O_k)\to C(B_N/B_k) [[/math]]

Now by looking at the column vector [math]\xi=(1,\ldots,1)^t[/math], which is fixed by the stochastic matrices, we conclude that the map on the right is an isomorphism, and we are done.

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

General references

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