guide:776d839156

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

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Before getting into more about inner faithfulness, let us first go back to the stationary models. These models are quite restrictive, because [math]G[/math] must be coamenable. However, there are many interesting examples of coamenable compact quantum groups, and in order to better understand these examples, and also in order to construct some new examples, our idea will be that of looking for stationary models for them. We first have:

Theorem

For [math]\pi:C(G)\to M_K(C(T))[/math], the following are equivalent:

• [math]Im(\pi)[/math] is a Hopf algebra, and [math](tr\otimes\int_T)\pi[/math] is the Haar integration on it.
• [math]\psi=(tr\otimes\int_X)\pi[/math] satisfies the idempotent state property [math]\psi*\psi=\psi[/math].
• [math]T_e^2=T_e[/math], [math]\forall p\in\mathbb N[/math], [math]\forall e\in\{1,*\}^p[/math], where:
  [[math]] (T_e)_{i_1\ldots i_p,j_1\ldots j_p}=\left(tr\otimes\int_T\right)(U_{i_1j_1}^{e_1}\ldots U_{i_pj_p}^{e_p}) [[/math]]

If these conditions are satisfied, we say that [math]\pi[/math] is stationary on its image.

Show Proof

Given a matrix model [math]\pi:C(G)\to M_K(C(T))[/math] as in the statement, we can factorize it via its Hopf image, as in Definition 16.6 above:

[[math]] \pi:C(G)\to C(H)\to M_K(C(T)) [[/math]]

Now observe that the conditions (1,2,3) in the statement depend only on the factorized representation:

[[math]] \nu:C(H)\to M_K(C(T)) [[/math]]

Thus, we can assume in practice that we have [math]G=H[/math], which means that we can assume that [math]\pi[/math] is inner faithful. With this assumption made, the general integration formula from Theorem 16.8 applies to our situation, and the proof of the equivalences goes as follows:

[math](1)\implies(2)[/math] This is clear from definitions, because the Haar integration on any compact quantum group satisfies the idempotent state equation, namely:

[[math]] \psi*\psi=\psi [[/math]]

[math](2)\implies(1)[/math] Assuming [math]\psi*\psi=\psi[/math], we have, for any [math]r\in\mathbb N[/math]:

[[math]] \psi^{*r}=\psi [[/math]]

Thus Theorem 16.8 gives [math]\int_G=\psi[/math], and by using Theorem 16.3, we obtain the result.

In order to establish now [math](2)\Longleftrightarrow(3)[/math], we use the following elementary formula, which comes from the definition of the convolution operation:

[[math]] \psi^{*r}(u_{i_1j_1}^{e_1}\ldots u_{i_pj_p}^{e_p})=(T_e^r)_{i_1\ldots i_p,j_1\ldots j_p} [[/math]]

[math](2)\implies(3)[/math] Assuming [math]\psi*\psi=\psi[/math], by using the above formula at [math]r=1,2[/math] we obtain that the matrices [math]T_e[/math] and [math]T_e^2[/math] have the same coefficients, and so they are equal.

[math](3)\implies(2)[/math] Assuming [math]T_e^2=T_e[/math], by using the above formula at [math]r=1,2[/math] we obtain that the linear forms [math]\psi[/math] and [math]\psi*\psi[/math] coincide on any product of coefficients [math]u_{i_1j_1}^{e_1}\ldots u_{i_pj_p}^{e_p}[/math]. Now since these coefficients span a dense subalgebra of [math]C(G)[/math], this gives the result.

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As a first illustration, we will apply this criterion to certain models for the quantum groups [math]O_N^*,U_N^*[/math]. We first have the following result:

Proposition

We have a matrix model as follows,

[[math]] C(O_N^*)\to M_2(C(U_N))\quad,\quad u_{ij}\to\begin{pmatrix}0&v_{ij}\\ \bar{v}_{ij}&0\end{pmatrix} [[/math]]

where [math]v[/math] is the fundamental corepresentation of [math]C(U_N)[/math], as well as a model as follows,

[[math]] C(U_N^*)\to M_2(C(U_N\times U_N))\quad,\quad u_{ij}\to\begin{pmatrix}0&v_{ij}\\ w_{ij}&0\end{pmatrix} [[/math]]

where [math]v,w[/math] are the fundamental corepresentations of the two copies of [math]C(U_N)[/math].

Show Proof

It is routine to check that the matrices on the right are indeed biunitaries, and since the first matrix is also self-adjoint, we obtain models as follows:

[[math]] C(O_N^+)\to M_2(C(U_N)) [[/math]]

[[math]] C(U_N^+)\to M_2(C(U_N\times U_N)) [[/math]]

Consider now antidiagonal [math]2\times2[/math] matrices, with commuting entries, as follows:

[[math]] X_i=\begin{pmatrix}0&x_i\\ y_i&0\end{pmatrix} [[/math]]

We have then the following computation:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\begin{pmatrix}0&x_i\\ y_i&0\end{pmatrix} \begin{pmatrix}0&x_j\\ y_j&0\end{pmatrix} \begin{pmatrix}0&x_k\\ y_k&0\end{pmatrix}\\ &=&\begin{pmatrix}0&x_iy_jx_k\\ y_ix_jy_k&0\end{pmatrix} \end{eqnarray*} [[/math]]

Since this quantity is symmetric in [math]i,k[/math], we obtain from this:

[[math]] X_iX_jX_k=X_kX_jX_i [[/math]]

Thus, our models above factorize as claimed.

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We can now formulate our first concrete modelling theorem, as folllows:

Theorem

The above antidiagonal models, namely

[[math]] C(O_N^*)\to M_2(C(U_N)) [[/math]]

[[math]] C(U_N^*)\to M_2(C(U_N\times U_N)) [[/math]]

are both stationary.

Show Proof

We first discuss the case of [math]O_N^*[/math]. We use Theorem 16.9 (3). Since the fundamental representation is self-adjoint, the matrices [math]T_e[/math] with [math]e\in\{1,*\}^p[/math] are all equal. We denote this common matrix by [math]T_p[/math]. According to the definition of [math]T_p[/math], we have:

[[math]] (T_p)_{i_1\ldots i_p,j_1\ldots j_p} =\left(tr\otimes\int_H\right)\left[\begin{pmatrix}0&v_{i_1j_1}\\\bar{v}_{i_1j_1}&0\end{pmatrix}\ldots\ldots\begin{pmatrix}0&v_{i_pj_p}\\\bar{v}_{i_pj_p}&0\end{pmatrix}\right] [[/math]]

Since when multipliying an odd number of antidiagonal matrices we obtain an atidiagonal matrix, we have [math]T_p=0[/math] for [math]p[/math] odd. Also, when [math]p[/math] is even, we have:

[[math]] \begin{eqnarray*} (T_p)_{i_1\ldots i_p,j_1\ldots j_p} &=&\left(tr\otimes\int_H\right)\begin{pmatrix}v_{i_1j_1}\ldots\bar{v}_{i_pj_p}&0\\0&\bar{v}_{i_1j_1}\ldots v_{i_pj_p}\end{pmatrix}\\ &=&\frac{1}{2}\left(\int_Hv_{i_1j_1}\ldots\bar{v}_{i_pj_p}+\int_H\bar{v}_{i_1j_1}\ldots v_{i_pj_p}\right)\\ &=&\int_HRe(v_{i_1j_1}\ldots\bar{v}_{i_pj_p}) \end{eqnarray*} [[/math]]

We have [math]T_p^2=T_p=0[/math] when [math]p[/math] is odd, so we are left with proving that we have [math]T_p^2=T_p[/math], when [math]p[/math] is even. For this purpose, we use the following formula:

[[math]] Re(x)Re(y)=\frac{1}{2}\left(Re(xy)+Re(x\bar{y})\right) [[/math]]

By using this identity for each of the terms which appear in the product, and multi-index notations in order to simplify the writing, we obtain:

[[math]] \begin{eqnarray*} &&(T_p^2)_{ij}\\ &=&\sum_{k_1\ldots k_p}(T_p)_{i_1\ldots i_p,k_1\ldots k_p}(T_p)_{k_1\ldots k_p,j_1\ldots j_p}\\ &=&\int_H\int_H\sum_{k_1\ldots k_p}Re(v_{i_1k_1}\ldots\bar{v}_{i_pk_p})Re(w_{k_1j_1}\ldots\bar{w}_{k_pj_p})dvdw\\ &=&\frac{1}{2}\int_H\int_H\sum_{k_1\ldots k_p}Re(v_{i_1k_1}w_{k_1j_1}\ldots\bar{v}_{i_pk_p}\bar{w}_{k_pj_p})+Re(v_{i_1k_1}\bar{w}_{k_1j_1}\ldots\bar{v}_{i_pk_p}w_{k_pj_p})dvdw\\ &=&\frac{1}{2}\int_H\int_HRe((vw)_{i_1j_1}\ldots(\bar{v}\bar{w})_{i_pj_p})+Re((v\bar{w})_{i_1j_1}\ldots(\bar{v}w)_{i_pj_p})dvdw \end{eqnarray*} [[/math]]

Now since [math]vw\in H[/math] is uniformly distributed when [math]v,w\in H[/math] are uniformly distributed, the quantity on the left integrates up to [math](T_p)_{ij}[/math]. Also, since [math]H[/math] is conjugation-stable, [math]\bar{w}\in H[/math] is uniformly distributed when [math]w\in H[/math] is uniformly distributed, so the quantity on the right integrates up to the same quantity, namely [math](T_p)_{ij}[/math]. Thus, we have:

[[math]] \begin{eqnarray*} (T_p^2)_{ij} &=&\frac{1}{2}\Big((T_p)_{ij}+(T_p)_{ij}\Big)\\ &=&(T_p)_{ij} \end{eqnarray*} [[/math]]

Summarizing, we have obtained that for any [math]p[/math], the condition [math]T_p^2=T_p[/math] is satisfied. Thus Theorem 16.9 applies, and shows that our model is stationary, as claimed. As for the proof of the stationarity for the model for [math]U_N^*[/math], this is similar.

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As a second illustration, regarding [math]H_N^*,K_N^*[/math], we have:

Theorem

We have a stationary matrix model as follows,

[[math]] C(H_N^*)\to M_2(C(K_N))\quad,\quad u_{ij}\to\begin{pmatrix}0&v_{ij}\\ \bar{v}_{ij}&0\end{pmatrix} [[/math]]

where [math]v[/math] is the fundamental corepresentation of [math]C(K_N)[/math], as well as a stationary model

[[math]] C(K_N^*)\to M_2(C(K_N\times K_N))\quad,\quad u_{ij}\to\begin{pmatrix}0&v_{ij}\\ w_{ij}&0\end{pmatrix} [[/math]]

where [math]v,w[/math] are the fundamental corepresentations of the two copies of [math]C(K_N)[/math].

Show Proof

This follows by adapting the proof of Proposition 16.10 and Theorem 16.11 above, by adding there the [math]H_N^+,K_N^+[/math] relations. All this is in fact part of a more general phenomenon, concerning half-liberation in general, and we refer here to , .

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Summarizing, we have some interesting theory and examples for both the stationary models, and for the general inner faithful models.

General references

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