guide:23e6904627

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At the level of the really “new” examples now, we have basic liberation constructions, going back to the pioneering work of Wang , , and to the subsequent papers , , as well as several more recent constructions. We first have, following Wang :

Theorem

The following universal algebras are Woronowicz algebras,

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

so the underlying compact quantum spaces [math]O_N^+,U_N^+[/math] are compact quantum groups.

Show Proof

This follows from the elementary fact that if a matrix [math]u=(u_{ij})[/math] is orthogonal or biunitary, as above, then so must be the following matrices:

[[math]] u^\Delta_{ij}=\sum_ku_{ik}\otimes u_{kj}\quad,\quad u^\varepsilon_{ij}=\delta_{ij}\quad,\quad u^S_{ij}=u_{ji}^* [[/math]]

Consider indeed the matrix [math]U=u^\Delta[/math]. We have then:

[[math]] (UU^*)_{ij} =\sum_{klm}u_{il}u_{jm}^*\otimes u_{lk}u_{mk}^* =\sum_{lm}u_{il}u_{jm}^*\otimes\delta_{lm} =\delta_{ij} [[/math]]

In the other sense the computation is similar, as follows:

[[math]] (U^*U)_{ij} =\sum_{klm}u_{kl}^*u_{km}\otimes u_{li}^*u_{mj} =\sum_{lm}\delta_{lm}\otimes u_{li}^*u_{mj} =\delta_{ij} [[/math]]

The verification of the unitarity of [math]\bar{U}[/math] is similar. We first have:

[[math]] (\bar{U}U^t)_{ij} =\sum_{klm}u_{il}^*u_{jm}\otimes u_{lk}^*u_{mk} =\sum_{lm}u_{il}^*u_{jm}\otimes\delta_{lm} =\delta_{ij} [[/math]]

In the other sense the computation is similar, as follows:

[[math]] (U^t\bar{U})_{ij} =\sum_{klm}u_{kl}u_{km}^*\otimes u_{li}u_{mj}^* =\sum_{lm}\delta_{lm}\otimes u_{li}u_{mj}^* =\delta_{ij} [[/math]]

Regarding now the matrix [math]u^\varepsilon=1_N[/math], and also the matrix [math]u^S[/math], their biunitarity its clear. Thus, we can indeed define morphisms [math]\Delta,\varepsilon,S[/math] as in Definition 2.8, by using the universal properties of [math]C(O_N^+)[/math], [math]C(U_N^+)[/math], and this gives the result.

■

Let us study now the above quantum groups, with the techniques that we have. As a first observation, we have embeddings of compact quantum groups, as follows:

[[math]] \xymatrix@R=15mm@C=15mm{ U_N\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&O_N^+\ar[u] } [[/math]]

The basic properties of [math]O_N^+,U_N^+[/math] can be summarized as follows:

Theorem

The quantum groups [math]O_N^+,U_N^+[/math] have the following properties:

• The closed subgroups [math]G\subset U_N^+[/math] are exactly the [math]N\times N[/math] compact quantum groups. As for the closed subgroups [math]G\subset O_N^+[/math], these are those satisfying [math]u=\bar{u}[/math].
• We have liberation embeddings [math]O_N\subset O_N^+[/math] and [math]U_N\subset U_N^+[/math], obtained by dividing the algebras [math]C(O_N^+),C(U_N^+)[/math] by their respective commutator ideals.
• We have as well embeddings [math]\widehat{L}_N\subset O_N^+[/math] and [math]\widehat{F}_N\subset U_N^+[/math], where [math]L_N[/math] is the free product of [math]N[/math] copies of [math]\mathbb Z_2[/math], and where [math]F_N[/math] is the free group on [math]N[/math] generators.

Show Proof

All these assertions are elementary, as follows:

(1) This is clear from definitions, and from Proposition 2.14.

(2) This follows from the Gelfand theorem, which shows that we have presentation results for [math]C(O_N),C(U_N)[/math] as follows, similar to those in Theorem 2.23:

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

(3) This follows from (1) and from Proposition 2.11 above, with the remark that with [math]u=diag(g_1,\ldots,g_N)[/math], the condition [math]u=\bar{u}[/math] is equivalent to [math]g_i^2=1[/math], for any [math]i[/math].

■

As an interesting philosophical conclusion, if we denote by [math]L_N^+,F_N^+[/math] the discrete quantum groups which are dual to [math]O_N^+,U_N^+[/math], then we have embeddings as follows:

[[math]] L_N\subset L_N^+\quad,\quad F_N\subset F_N^+ [[/math]]

Thus [math]F_N^+[/math] is some kind of “free free group”, and [math]L_N^+[/math] is its real counterpart. This is not surprising, since [math]F_N,L_N[/math] are not “fully free”, their group algebras being cocommutative.

The last assertion in Theorem 2.24 suggests the following construction, from :

Proposition

Given a closed subgroup [math]G\subset U_N^+[/math], consider its “diagonal torus”, which is the closed subgroup [math]T\subset G[/math] constructed as follows:

[[math]] C(T)=C(G)\Big/\left \lt u_{ij}=0\Big|\forall i\neq j\right \gt [[/math]]

This torus is then a group dual, [math]T=\widehat{\Lambda}[/math], where [math]\Lambda= \lt g_1,\ldots,g_N \gt [/math] is the discrete group generated by the elements [math]g_i=u_{ii}[/math], which are unitaries inside [math]C(T)[/math].

Show Proof

Since [math]u[/math] is unitary, its diagonal entries [math]g_i=u_{ii}[/math] are unitaries inside [math]C(T)[/math]. Moreover, from [math]\Delta(u_{ij})=\sum_ku_{ik}\otimes u_{kj}[/math] we obtain, when passing inside the quotient:

[[math]] \Delta(g_i)=g_i\otimes g_i [[/math]]

It follows that we have [math]C(T)=C^*(\Lambda)[/math], modulo identifying as usual the [math]C^*[/math]-completions of the various group algebras, and so that we have [math]T=\widehat{\Lambda}[/math], as claimed.

■

With this notion in hand, Theorem 2.24 (3) tells us that the diagonal tori of [math]O_N^+,U_N^+[/math] are the group duals [math]\widehat{L}_N,\widehat{F}_N[/math]. We will be back to this later.

Here is now a more subtle result on [math]O_N^+,U_N^+[/math], having no classical counterpart:

Proposition

Consider the quantum groups [math]O_N^+,U_N^+[/math], with the corresponding fundamental corepresentations denoted [math]v,u[/math], and let [math]z=id\in C(\mathbb T)[/math].

• We have a morphism [math]C(U_N^+)\to C(\mathbb T)*C(O_N^+)[/math], given by [math]u=zv[/math].
• In other words, we have a quantum group embedding [math]\widetilde{O_N^+}\subset U_N^+[/math].
• This embedding is an isomorphism at the level of the diagonal tori.

Show Proof

The first two assertions follow from Proposition 2.19, or simply from the fact that [math]u=zv[/math] is biunitary. As for the third assertion, the idea here is that we have a similar model for the free group [math]F_N[/math], which is well-known to be faithful, [math]F_N\subset\mathbb Z*L_N[/math].

■

We will be back to the above morphism later on, with a proof of its faithfulness, after performing a suitable GNS construction, with respect to the Haar functionals.

Let us construct now some more examples of compact quantum groups. Following , , , , we can introduce some intermediate liberations, as follows:

Proposition

We have intermediate quantum groups as follows,

[[math]] \xymatrix@R=15mm@C=15mm{ U_N\ar[r]&U_N^*\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&O_N^*\ar[r]\ar[u]&O_N^+\ar[u]} [[/math]]

with [math]*[/math] standing for the fact that [math]u_{ij},u_{ij}^*[/math] must satisfy the relations [math]abc=cba[/math].

Show Proof

This is similar to the proof of Theorem 2.23, by using the elementary fact that if the entries of [math]u=(u_{ij})[/math] half-commute, then so do the entries of [math]u^\Delta[/math], [math]u^\varepsilon[/math], [math]u^S[/math].

■

In the same spirit, we have as well intermediate spheres as follows, with the symbol [math]*[/math] standing for the fact that [math]x_i,x_i^*[/math] must satisfy the relations [math]abc=cba[/math]:

[[math]] \xymatrix@R=15mm@C=15mm{ S^{N-1}_\mathbb C\ar[r]&S^{N-1}_{\mathbb C,*}\ar[r]&S^{N-1}_{\mathbb C,+}\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_{\mathbb R,*}\ar[r]\ar[u]&S^{N-1}_{\mathbb R,+}\ar[u] } [[/math]]

At the level of the diagonal tori, we have the following result:

Theorem

The tori of the basic spheres and quantum groups are as follows,

[[math]] \xymatrix@R=15mm@C=15mm{ \widehat{\mathbb Z^N}\ar[r]&\widehat{\mathbb Z^{\circ N}}\ar[r]&\widehat{\mathbb Z^{*N}}\\ \widehat{\mathbb Z_2^N}\ar[r]\ar[u]&\widehat{\mathbb Z_2^{\circ N}}\ar[r]\ar[u]&\widehat{\mathbb Z_2^{*N}}\ar[u]} [[/math]]

with [math]\circ[/math] standing for the half-classical product operation for groups.

Show Proof

The idea here is as follows:

(1) The result on the left is well-known.

(2) The result on the right follows from Theorem 2.24 (3).

(3) The middle result follows as well, by imposing the relations [math]abc=cba[/math].

■

Let us discuss now the relation with the noncommutative spheres. Having the things started here is a bit tricky, and as a main source of inspiration, we have:

Proposition

Given an algebraic manifold [math]X\subset S^{N-1}_\mathbb C[/math], the formula

[[math]] G(X)=\left\{U\in U_N\Big|U(X)=X\right\} [[/math]]

defines a compact group of unitary matrices, or isometries, called affine isometry group of [math]X[/math]. For the spheres [math]S^{N-1}_\mathbb R,S^{N-1}_\mathbb C[/math] we obtain in this way the groups [math]O_N,U_N[/math].

Show Proof

The fact that [math]G(X)[/math] as defined above is indeed a group is clear, its compactness is clear as well, and finally the last assertion is clear as well. In fact, all this works for any closed subset [math]X\subset\mathbb C^N[/math], but we are not interested here in such general spaces.

■

We have the following quantum analogue of the above construction:

Proposition

Given an algebraic manifold [math]X\subset S^{N-1}_{\mathbb C,+}[/math], the category of the closed subgroups [math]G\subset U_N^+[/math] acting affinely on [math]X[/math], in the sense that the formula

[[math]] \Phi(x_i)=\sum_jx_j\otimes u_{ji} [[/math]]

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

[[math]] \Phi:C(X)\to C(X)\otimes C(G) [[/math]]

has a universal object, denoted [math]G^+(X)[/math], and called affine quantum isometry group of [math]X[/math].

Show Proof

Observe first that in the case where [math]\Phi[/math] as above exists, this morphism is automatically a coaction, in the sense that it satisfies the following conditions:

[[math]] (\Phi\otimes id)\Phi=(id\otimes\Delta)\Phi [[/math]]

[[math]] (id\otimes\varepsilon)\Phi=id [[/math]]

In order to prove now the result, assume that [math]X\subset S^{N-1}_{\mathbb C,+}[/math] comes as follows:

[[math]] C(X)=C(S^{N-1}_{\mathbb C,+})\Big/\Big \lt f_\alpha(x_1,\ldots,x_N)=0\Big \gt [[/math]]

Consider now the following variables:

[[math]] X_i=\sum_jx_j\otimes u_{ji}\in C(X)\otimes C(U_N^+) [[/math]]

Our claim is that [math]G=G^+(X)[/math] in the statement appears as follows:

[[math]] C(G)=C(U_N^+)\Big/\Big \lt f_\alpha(X_1,\ldots,X_N)=0\Big \gt [[/math]]

In order to prove this claim, we have to clarify how the relations [math]f_\alpha(X_1,\ldots,X_N)=0[/math] are interpreted inside [math]C(U_N^+)[/math], and then show that [math]G[/math] is indeed a quantum group. So, pick one of the defining polynomials, [math]f=f_\alpha[/math], and write it as follows:

[[math]] f(x_1,\ldots,x_N)=\sum_r\sum_{i_1^r\ldots i_{s_r}^r}\lambda_r\cdot x_{i_1^r}\ldots x_{i_{s_r}^r} [[/math]]

With [math]X_i=\sum_jx_j\otimes u_{ji}[/math] as above, we have the following formula:

[[math]] f(X_1,\ldots,X_N) =\sum_r\sum_{i_1^r\ldots i_{s_r}^r}\lambda_r\sum_{j_1^r\ldots j_{s_r}^r}x_{j_1^r}\ldots x_{j_{s_r}^r}\otimes u_{j_1^ri_1^r}\ldots u_{j_{s_r}^ri_{s_r}^r} [[/math]]

Since the variables on the right span a certain finite dimensional space, the relations [math]f(X_1,\ldots,X_N)=0[/math] correspond to certain relations between the variables [math]u_{ij}[/math]. Thus, we have indeed a closed subspace [math]G\subset U_N^+[/math], coming with a universal map:

[[math]] \Phi:C(X)\to C(X)\otimes C(G) [[/math]]

In order to show now that [math]G[/math] is a quantum group, consider the following elements:

[[math]] u_{ij}^\Delta=\sum_ku_{ik}\otimes u_{kj}\quad,\quad u_{ij}^\varepsilon=\delta_{ij}\quad,\quad u_{ij}^S=u_{ji}^* [[/math]]

Consider as well the following associated elements, with [math]\gamma\in\{\Delta,\varepsilon,S\}[/math]:

[[math]] X_i^\gamma=\sum_jx_j\otimes u_{ji}^\gamma [[/math]]

From the relations [math]f(X_1,\ldots,X_N)=0[/math] we deduce that we have:

[[math]] f(X_1^\gamma,\ldots,X_N^\gamma) =(id\otimes\gamma)f(X_1,\ldots,X_N) =0 [[/math]]

But this shows that for any exponent [math]\gamma\in\{\Delta,\varepsilon,S\}[/math] we can map [math]u_{ij}\to u_{ij}^\gamma[/math], and it follows that [math]G[/math] is indeed a compact quantum group, and we are done.

■

Following and related papers, we can now formulate:

Theorem

The quantum isometry groups of the basic spheres are

[[math]] \xymatrix@R=15mm@C=17mm{ U_N\ar[r]&U_N^*\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&O_N^*\ar[r]\ar[u]&O_N^+\ar[u]} [[/math]]

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

Show Proof

Let us first construct an action [math]U_N^+\curvearrowright S^{N-1}_{\mathbb C,+}[/math]. We must prove here that the variables [math]X_i=\sum_jx_j\otimes u_{ji}[/math] satisfy the defining relations for [math]S^{N-1}_{\mathbb C,+}[/math], namely:

[[math]] \sum_ix_ix_i^*=\sum_ix_i^*x_i=1 [[/math]]

But this follows from the biunitarity of [math]u[/math]. We have indeed:

[[math]] \begin{eqnarray*} \sum_iX_iX_i^* &=&\sum_{ijk}x_jx_k^*\otimes u_{ji}u_{ki}^*\\ &=&\sum_jx_jx_j^*\otimes1\\ &=&1\otimes1 \end{eqnarray*} [[/math]]

In the other sense the computation is similar, as follows:

[[math]] \begin{eqnarray*} \sum_iX_i^*X_i &=&\sum_{ijk}x_j^*x_k\otimes u_{ji}^*u_{ki}\\ &=&\sum_jx_j^*x_j\otimes1\\ &=&1\otimes1 \end{eqnarray*} [[/math]]

Regarding now [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math], here we must check the extra relations [math]X_i=X_i^*[/math], and these are clear from [math]u_{ia}=u_{ia}^*[/math]. Finally, regarding the remaining actions, the verifications are clear as well, because if the coordinates [math]u_{ia}[/math] and [math]x_a[/math] are subject to commutation relations of type [math]ab=ba[/math], or of type [math]abc=cba[/math], then so are the variables [math]X_i=\sum_jx_j\otimes u_{ji}[/math].

We must prove now that all these actions are universal:

\underline{[math]S^{N-1}_{\mathbb R,+},S^{N-1}_{\mathbb C,+}[/math].} The universality of [math]U_N^+\curvearrowright S^{N-1}_{\mathbb C,+}[/math] is trivial by definition. As for the universality of [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math], this comes from the fact that [math]X_i=X_i^*[/math], with [math]X_i=\sum_jx_j\otimes u_{ji}[/math] as above, gives [math]u_{ia}=u_{ia}^*[/math]. Thus [math]G\curvearrowright S^{N-1}_{\mathbb R,+}[/math] implies [math]G\subset O_N^+[/math], as desired.

\underline{[math]S^{N-1}_\mathbb R,S^{N-1}_\mathbb C[/math].} We use here a trick from Bhowmick-Goswami . Assuming first that we have an action [math]G\curvearrowright S^{N-1}_\mathbb R[/math], consider the following variables:

[[math]] w_{kl,ij}=u_{ki}u_{lj} [[/math]]

[[math]] p_{ij}=x_ix_j [[/math]]

In terms of these variables, which can be thought of as being projective coordinates, the corresponding projective coaction map is given by:

[[math]] \Phi(p_{ij})=\sum_{kl}p_{kl}\otimes w_{kl,ij} [[/math]]

We have the following formulae:

[[math]] \begin{eqnarray*} \Phi(p_{ij})&=&\sum_{k \lt l}p_{kl}\otimes(w_{kl,ij}+w_{lk,ij})+\sum_kp_{kk}\otimes w_{kk,ij}\\ \Phi(p_{ji})&=&\sum_{k \lt l}p_{kl}\otimes(w_{kl,ji}+w_{lk,ji})+\sum_kp_{kk}\otimes w_{kk,ji} \end{eqnarray*} [[/math]]

By comparing these two formulae, and then by using the linear independence of the variables [math]p_{kl}=x_kx_l[/math] with [math]k\leq l[/math], we conclude that we must have:

[[math]] w_{kl,ij}+w_{lk,ij}=w_{kl,ji}+w_{lk,ji} [[/math]]

Let us apply the antipode to this formula. For this purpose, observe that we have:

[[math]] S(w_{kl,ij}) =S(u_{ki}u_{lj}) =S(u_{lj})S(u_{ki}) =u_{jl}u_{ik} =w_{ji,lk} [[/math]]

Thus by applying the antipode we obtain:

[[math]] w_{ji,lk}+w_{ji,kl}=w_{ij,lk}+w_{ij,kl} [[/math]]

By relabelling the indices, we obtain from this:

[[math]] w_{kl,ij}+w_{kl,ji}=w_{lk,ij}+w_{lk,ji} [[/math]]

Now by comparing with the original relation, we obtain:

[[math]] w_{lk,ij}=w_{kl,ji} [[/math]]

But, recalling that we have [math]w_{kl,ij}=u_{ki}u_{lj}[/math], this formula reads:

[[math]] u_{li}u_{kj}=u_{kj}u_{li} [[/math]]

We therefore conclude we have [math]G\subset O_N[/math], as claimed. The proof of the universality of the action [math]U_N\curvearrowright S^{N-1}_\mathbb C[/math] is similar.

\underline{[math]S^{N-1}_{\mathbb R,*},S^{N-1}_{\mathbb C,*}[/math].} Assume 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 then that, with [math]p_{ab}=z_a\bar{z}_b[/math], we have:

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

By multiplying these two 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]]

Since the variables [math]p_{ij}p_{kl}=z_i\bar{z}_jz_k\bar{z}_l[/math] depend only on [math]|\{i,k\}|,|\{j,l\}|\in\{1,2\}[/math], and this dependence produces the only relations between them, we are led to [math]4[/math] equations:

(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 [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 actions [math]U_N^*\curvearrowright S^{N-1}_{\mathbb C,*}[/math] and of [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math], and from [math]U_N^*\cap O_N^+=O_N^*[/math].

■

As a conclusion to all this, we have now a simple and reliable definition for the compact quantum groups, in the Lie case, namely [math]G\subset U_N^+[/math], covering all the compact Lie groups, [math]G\subset U_N[/math], covering as well all the duals [math]\widehat{\Gamma}[/math] of the finitely generated groups, [math]F_N\to\Gamma[/math], and allowing the construction of several interesting examples, such as [math]O_N^+,U_N^+[/math].

With respect to the noncommutative geometry questions raised in chapter 1 above, we certainly have here some advances. In order to further advance, however, we would need now representation theory results, in the spirit of Weyl , for our quantum isometry groups. We will develop all this in what follows, in the next few chapters.

General references

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