Polygonal spheres

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14a. Polygonal spheres

In this chapter we build on the findings from the previous chapter, still following , , with the idea in mind that all this material belongs to a new and exciting area of noncommutative algebra, which can help in building an algebraic geometry theory for the free manifolds, and which therefore needs to be prioritarily developed.

As in the previous chapter, due to various technical difficulties with the complex case, at least at this stage of the things, we will basically restrict the attention to the real case. The main objects of study here are the 3 real spheres, which are as follows:

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

We have seen that the study of the relations between the coordinates [math]x_1,\ldots,x_N[/math] of these real spheres naturally leads to the twisted versions of these spheres, namely:

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

More precisely, the study of the algebraic relations between the coordinates [math]x_1,\ldots,x_N[/math] of the real spheres leads to the study of the various intersections between the twisted and untwisted spheres. These [math]3\times3[/math] intersections form a square diagram, as follows:

[[math]] \xymatrix@R=13mm@C=1mm{ S^{N-1}_\mathbb R\ar[rr]&&S^{N-1}_{\mathbb R,*}\ar[rrr]&&&S^{N-1}_{\mathbb R,+}\\ S^{N-1}_\mathbb R\cap\bar{S}^{N-1}_{\mathbb R,*}\ar[rr]\ar[u]&&S^{N-1}_{\mathbb R,*}\cap\bar{S}^{N-1}_{\mathbb R,*}\ar[rrr]\ar[u]&&&\bar{S}^{N-1}_{\mathbb R,*}\ar[u]\\ S^{N-1}_\mathbb R\cap\bar{S}^{N-1}_\mathbb R\ar[rr]\ar[u]&&S^{N-1}_{\mathbb R,*}\cap\bar{S}^{N-1}_\mathbb R\ar[rrr]\ar[u]&&&\bar{S}^{N-1}_\mathbb R\ar[u]} [[/math]]

We have seen as well that these intersections all appear as “polygonal spheres”, which are certain real algebraic manifolds, according to the following result:

Theorem

The [math]5[/math] main spheres, and the intersections between them, are

[[math]] \xymatrix@R=12mm@C=14mm{ S^{N-1}_\mathbb R\ar[r]&S^{N-1}_{\mathbb R,*}\ar[r]&S^{N-1}_{\mathbb R,+}\\ S^{N-1,1}_\mathbb R\ar[r]\ar[u]&S^{N-1,1}_{\mathbb R,*}\ar[r]\ar[u]&\bar{S}^{N-1}_{\mathbb R,*}\ar[u]\\ S^{N-1,0}_\mathbb R\ar[r]\ar[u]&\bar{S}^{N-1,1}_\mathbb R\ar[r]\ar[u]&\bar{S}^{N-1}_\mathbb R\ar[u]} [[/math]]

where [math]\dot{S}^{N-1,d-1}_{\mathbb R,\times}\subset\dot{S}^{N-1}_{\mathbb R,\times}[/math] is obtained by assuming [math]x_{i_0}\ldots x_{i_d}=0[/math], for [math]i_0,\ldots,i_d[/math] distinct.

Show Proof

This is something that we know from chapter 13, the idea being that commutation and anticommutation produces vanishing relations.

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We refer to chapter 13 for more on these spheres, including their algebraic axiomatization and main properties, and the “standard parametrization” result there.

In this chapter we discuss the extension of the axiomatics for abstract noncommutative geometries that we have, from chapter 4, in order to cover both the twisted and untwisted cases, and the intersections. This is a very natural question, in view of our findings from chapter 13. For this purpose, we are in need of some new quantum isometry group computations. In order to deal with the polygonal spheres, we will need:

Proposition

Assume that [math]X\subset S^{N-1}_\mathbb R[/math] is invariant, for any [math]i[/math], under:

[[math]] x_i\to-x_i [[/math]]

• If the coordinates [math]x_1,\ldots,x_N[/math] are linearly independent inside [math]C(X)[/math], then the group
  [[math]] G(X)=G^+(X)\cap O_N [[/math]]
  consists of the usual isometries of [math]X[/math].
• In addition, in the case where the products of coordinates
  [[math]] \left\{x_ix_j\Big|i\leq j\right\} [[/math]]
  are linearly independent inside [math]C(X)[/math], we have [math]G^+(X)=G(X)[/math].

Show Proof

This is a standard trick, that we will heavily use here, which follows from Bhowmick-Goswami and Goswami , the idea being as follows:

(1) The assertion here is well-known, [math]G(X)=G^+(X)\cap O_N[/math] being by definition the biggest subgroup [math]G\subset O_N[/math] acting affinely on [math]X[/math]. We refer to for details, and for a number of noncommutative extensions of this fact, with [math]G(X)[/math] replaced by [math]G^+(X)[/math].

(2) Consider an arbitrary coaction map on the algebra [math]C(X)[/math], as follows:

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

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

In order to establish the result, we must prove that the variables [math]u_{ij}[/math] commute. But this follows by using a strandard trick, from , that we will briefly recall now. We can write the action of [math]\Phi[/math] on the commutators between the coordinates as follows:

[[math]] \Phi([x_i,x_j])=\sum_{k\leq l}\left(1-\frac{\delta_{kl}}{2}\right)x_kx_l\otimes\left([u_{ki},u_{lj}]-[u_{kj},u_{li}]\right) [[/math]]

Now since the variables [math]\{x_kx_l|k\leq l\}[/math] were assumed to be linearly independent, we obtain from this that we have the following formula:

[[math]] [u_{ki},u_{lj}]=[u_{kj},u_{li}] [[/math]]

Moreover, if we apply now the antipode we further obtain:

[[math]] [u_{jl},u_{ik}]=[u_{il},u_{jk}] [[/math]]

By relabelling, this gives the following formula:

[[math]] [u_{ki},u_{lj}]=[u_{li},u_{kj}] [[/math]]

Now by comparing with the original equality of commutators, from above, we conclude from this that we have a commutation relation, as follows:

[[math]] [u_{ki},u_{lj}]=0 [[/math]]

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

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With the above notion in hand, let us investigate the polygonal spheres. We recall that, according to the various computations from the previous chapters, the quantum isometry groups of the 5 main spheres are as follows:

[[math]] \xymatrix@R=18mm@C=18mm{ S^{N-1}_\mathbb R\ar[r]\ar@{~}[d]&S^{N-1}_{\mathbb R,*}\ar[r]\ar@{~}[d]&S^{N-1}_{\mathbb R,+}\ar@{~}[d]&\bar{S}^{N-1}_{\mathbb R,*}\ar[l]\ar@{~}[d]&\bar{S}^{N-1}_\mathbb R\ar[l]\ar@{~}[d]\\ O_N\ar[r]&O_N^*\ar[r]&O_N^+&\bar{O}_N^*\ar[l]&\bar{O}_N\ar[l]} [[/math]]

In the polygonal sphere case now, we begin with the computations of the quantum isometry groups in the classical case. We have here the following result, from :

Theorem

The quantum isometry groups of the classical polygonal spheres

[[math]] S^{N-1,d-1}_\mathbb R=\left\{x\in S^{N-1}_\mathbb R\Big|x_{i_0}\ldots x_{i_d}=0,\forall i_0,\ldots,i_d\ {\rm distinct}\right\} [[/math]]

are as follows:

• At [math]d=1[/math] we obtain the free hyperoctahedral group [math]H_N^+[/math].
• At [math]d=2,\ldots,N-1[/math] we obtain the hyperoctahedral group [math]H_N[/math].
• At [math]d=N[/math] we obtain the orthogonal group [math]O_N[/math].

Show Proof

Observe first that the sphere [math]S^{N-1,d-1}_\mathbb R[/math] appears by definition as a union on [math]\binom{N}{d}[/math] copies of the sphere [math]S^{d-1}_\mathbb R[/math], one for each choice of [math]d[/math] coordinate axes, among the coordinate axes of [math]\mathbb R^N[/math]. We can write this decomposition as follows, with [math]I_N=\{1,\ldots,N\}[/math]:

[[math]] S^{N-1,d-1}_\mathbb R=\bigcup_{I\subset I_N,|I|=d}(S^{d-1}_\mathbb R)^I [[/math]]

With this observation in hand, the proof goes as follows:

(1) At [math]d=1[/math] our sphere is given by the following formula:

[[math]] S^{N-1,0}_\mathbb R=\mathbb Z_2^{\oplus N} [[/math]]

To be more precise, what we have here is the set formed by the endpoints of the [math]N[/math] copies of [math][-1,1][/math] on the coordinate axes of [math]\mathbb R^N[/math]. Thus by the free wreath product results in the corresponding quantum isometry group is [math]H_N^+[/math]:

[[math]] \begin{eqnarray*} G^+(S^{N-1,0}_\mathbb R) &=&G^+(\mathbb Z_2^{\oplus N})\\ &=&G^+(|\ |\ldots |\ |)\\ &=&\mathbb Z_2\wr_*S_N^+\\ &=&H_N^+ \end{eqnarray*} [[/math]]

(2) In order to discuss now the case [math]d\geq2[/math], the idea is to use Proposition 14.2 (2). Our claim is that the following elements are linearly independent:

[[math]] \left\{x_ix_j\Big|i\leq j\right\} [[/math]]

Since [math]S^{N-1,1}_\mathbb R\subset S^{N-1,d}_\mathbb R[/math], we can restrict attention to the case [math]d=2[/math]. Here the above decomposition is as follows, where [math]\mathbb T^{\{i,j\}}[/math] denote the various copies of [math]\mathbb T[/math]:

[[math]] S^{N-1,d-1}_\mathbb R=\bigcup_{i \lt j}\mathbb T^{\{i,j\}} [[/math]]

Now observe that the following elements are linearly independent over [math]\mathbb T\subset\mathbb R^2[/math]:

[[math]] \left\{x^2,y^2,xy\right\} [[/math]]

We deduce that the following elements are linearly independent over [math]S^{N-1,d-1}_\mathbb R[/math]:

[[math]] \left\{x_ix_j\Big|i\leq j\right\} [[/math]]

Thus, our claim is proved, and so Proposition 14.2 (2) applies, and gives:

[[math]] G^+(X)=G(X) [[/math]]

Thus, we are left with proving the following formula, for any [math]d\in\{2,\ldots,N-1\}[/math]:

[[math]] G(X)=H_N [[/math]]

-- Let us first discuss the case [math]d=2[/math]. By using the decomposition formula from the beginning of the proof, here any affine isometric action [math]U\curvearrowright S^{N-1,1}_\mathbb R[/math] must permute the [math]\binom{N}{2}[/math] circles [math]\mathbb T^I[/math], so we can write, for a certain permutation of the indices [math]I\to I'[/math]:

[[math]] U(\mathbb T^I)=\mathbb T^{I'} [[/math]]

Now since [math]U[/math] is bijective, we deduce that for any [math]I,J[/math] we have:

[[math]] U\left(\mathbb T^I\cap\mathbb T^J\right)=\mathbb T^{I'}\cap\mathbb T^{J'} [[/math]]

The point now is that for [math]|I\cap J|=0,1,2[/math] we have:

[[math]] \mathbb T^I\cap\mathbb T^J\simeq\emptyset,\{-1,1\},\mathbb T [[/math]]

By taking now the union over [math]I,J[/math] with [math]|I\cap J|=1[/math], we deduce that:

[[math]] U(\mathbb Z_2^{\oplus N})=\mathbb Z_2^{\oplus N} [[/math]]

Thus we must have [math]U\in H_N[/math], and we are done with the case [math]d=2[/math].

-- In the general case now, [math]d\in\{2,\ldots,N-1\}[/math], we can proceed similarly, by recurrence. Indeed, for any subsets [math]I,J\subset I_N[/math] with [math]|I|=|J|=d[/math] we have:

[[math]] (S^{d-1}_\mathbb R)^I\cap(S^{d-1}_\mathbb R)^J=(S^{|I\cap J|-1}_\mathbb R)^{I\cap J} [[/math]]

By using [math]d\leq N-1[/math], we deduce that we have the following formula:

[[math]] S^{N-1,d-2}_\mathbb R=\bigcup_{|I|=|J|=d,|I\cap J|=d-1}(S^{|I\cap J|-1}_\mathbb R)^{I\cap J} [[/math]]

On the other hand, by using exactly the same argument as in the [math]d=2[/math] case, we deduce that the space on the right is invariant, under any affine isometric action on [math]S^{N-1,d-1}_\mathbb R[/math]. Thus by recurrence we obtain, as desired, that we have:

[[math]] G(S^{N-1,d-1}_\mathbb R) =G(S^{N-1,d-2}_\mathbb R) =H_N [[/math]]

(3) At [math]d=N[/math] the result is known since , with the proof coming from the equality [math]G^+(X)=G(X)[/math], deduced from Proposition 14.2 (2), as explained above.

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The study in the twisted case is considerably more difficult than in the classical case, and we have complete results only at [math]d=1,2,N[/math], as follows:

Theorem

The quantum isometry groups of twisted polygonal spheres, given by

[[math]] C(\bar{S}^{N-1,d-1}_\mathbb R)=C(\bar{S}^{N-1}_\mathbb R)\Big/\Big \lt x_{i_0}\ldots x_{i_d}=0,\forall i_0,\ldots,i_d\ {\rm distinct}\Big \gt [[/math]]

are as follows:

• At [math]d=1[/math] we obtain the free hyperoctahedral group [math]H_N^+[/math].
• At [math]d=2[/math] we obtain the hyperoctahedral group [math]H_N[/math].
• At [math]d=N[/math] we obtain the twisted orthogonal group [math]\bar{O}_N[/math].

Show Proof

The idea is to adapt the proof of Theorem 14.3:

(1) At [math]d=1[/math] the situation is very simple, because we have:

[[math]] \bar{S}^{N-1,0}_\mathbb R=S^{N-1,0}_\mathbb R=\mathbb Z_2^{\oplus N} [[/math]]

By Theorem 14.3 (1), coming from the free wreath product computations in , the corresponding quantum isometry group is indeed [math]H_N^+[/math].

(2) In order to deal now with the case [math]d=2[/math], in analogy with what was done before in the classical case, as a first ingredient, we will need the twisted analogue of the trick from , explained in the proof of Proposition 14.2 (2).

This twisted trick is known to work for the twisted sphere [math]\bar{S}^{N-1}_\mathbb R[/math] itself, and the situation is in fact similar for any closed subset [math]X\subset\bar{S}^{N-1}_\mathbb R[/math], having the property that the following variables are linearly indepedent:

[[math]] \left\{x_ix_j\Big|i\leq j\right\} [[/math]]

More presisely, our claim is that under this linear independence assumption, if a quantum group [math]G\subset O_N^+[/math] acts on [math]X[/math], then we must have:

[[math]] G\subset\bar{O}_N [[/math]]

Indeed, consider a coaction map, written as follows:

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

By making products, we have the following formula:

[[math]] \Phi(x_ix_j)=\sum_k x_k^2\otimes u_{ki}u_{kj}+\sum_{k \lt l}x_kx_l\otimes(u_{ki}u_{lj}-u_{li}u_{kj}) [[/math]]

We deduce that with [math][[a,b]]=ab+ba[/math] we have the following formula:

[[math]] \Phi([[x_i,x_j]])=\sum_kx_k^2\otimes [[u_{ki},u_{kj}]]+\sum_{k \lt l}x_kx_l\otimes ([u_{ki},u_{lj}]-[u_{li},u_{kj}]) [[/math]]

Now assuming [math]i\neq j[/math], we have [math][[x_i,x_j]]=0[/math], and we therefore obtain, for any [math]k[/math]:

[[math]] [[u_{ki},u_{kj}]]=0 [[/math]]

We also have, for any [math]k \lt l[/math], the following formula:

[[math]] [u_{ki},u_{lj}]=[u_{li},u_{kj}] [[/math]]

By applying the antipode and then by relabelling, the latter relation gives:

[[math]] [u_{ki},u_{lj}]=0 [[/math]]

Thus we have reached to the defining relations for the quantum group [math]\bar{O}_N[/math], from chapter 11, and so we have [math]G\subset\bar{O}_N[/math], as claimed.

Our second claim is that the above trick applies to any [math]\bar{S}^{N-1,d-1}_\mathbb R[/math] with [math]d\geq2[/math]. Consider indeed the following maps, obtained by setting [math]x_k=0[/math] for [math]k\neq i,j[/math]:

[[math]] \pi_{ij}:C(\bar{S}^{N-1,d-1}_\mathbb R)\to C(\bar{S}^1_\mathbb R) [[/math]]

By using these maps, we conclude that the following variables are indeed linearly independent over [math]\bar{S}^{N-1,d-1}_\mathbb R[/math], as desired:

[[math]] \left\{x_ix_j\Big|i\leq j\right\} [[/math]]

Summarizing, we have proved so far that if a compact quantum group [math]G\subset O_N^+[/math] acts on a polygonal sphere [math]\bar{S}^{N-1,d-1}_\mathbb R[/math] with [math]d\geq2[/math], then we must have:

[[math]] G\subset\bar{O}_N [[/math]]

In order to finish, we must now adapt the second part of the proof of Proposition 14.2, and since this is quite unobvious at [math]d\geq3[/math], due to various technical reasons, we will restrict now attention to the case [math]d=2[/math], as in the statement.

So, consider a compact quantum group [math]G\subset\bar{O}_N[/math]. The question is that of understanding when we have a coaction map, as follows:

[[math]] \Phi:C(\bar{S}^{N-1,1}_\mathbb R)\to C(G)\otimes C(\bar{S}^{N-1,1}_\mathbb R) [[/math]]

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

In order for this to happen, the elements [math]X_i=\sum_jx_j\otimes u_{ji}[/math] must satisfy the relations [math]X_iX_jX_k=0[/math], for any [math]i,j,k[/math] distinct.

So, let us compute [math]X_iX_jX_k[/math] for [math]i,j,k[/math] distinct. We have:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\sum_{abc}x_ax_bx_c\otimes u_{ai}u_{bj}u_{ck}\\ &=&\sum_{a,b,c\ not \ distinct}x_ax_bx_c\otimes u_{ai}u_{bj}u_{ck}\\ &=&\sum_{a\neq b}x_a^2x_b\otimes u_{ai}u_{aj}u_{bk}+\sum_{a\neq b}x_ax_bx_a\otimes u_{ai}u_{bj}u_{ak}\\ &+&\sum_{a\neq b}x_bx_a^2\otimes u_{bi}u_{aj}u_{ak}+\sum_ax_a^3\otimes u_{ai}u_{aj}u_{ak} \end{eqnarray*} [[/math]]

By using [math]x_ax_bx_a=-x_a^2x_b[/math] and [math]x_bx_a^2=x_a^2x_b[/math], we deduce that we have:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\sum_{a\neq b}x_a^2x_b\otimes(u_{ai}u_{aj}u_{bk}-u_{ai}u_{bj}u_{ak}+u_{bi}u_{aj}u_{ak})\\ &+&\sum_ax_a^3\otimes u_{ai}u_{aj}u_{ak}\\ &=&\sum_{ab}x_a^2x_b\otimes(u_{ai}u_{aj}u_{bk}-u_{ai}u_{bj}u_{ak}+u_{bi}u_{aj}u_{ak}) \end{eqnarray*} [[/math]]

By using now the defining relations for [math]\bar{O}_N[/math], which apply to the variables [math]u_{ij}[/math], this formula can be written in a cyclic way, as follows:

[[math]] X_iX_jX_k=\sum_{ab}x_a^2x_b\otimes(u_{ai}u_{aj}u_{bk}+u_{aj}u_{ak}u_{bi}+u_{ak}u_{ai}u_{bj}) [[/math]]

We use now the fact that the variables on the left, namely [math]x_a^2x_b[/math], are linearly independent. We conclude that, in order for our quantum group [math]G\subset\bar{O}_N[/math] to act on [math]\bar{S}^{N-1,1}_\mathbb R[/math], its coordinates must satisfy the following relations, for any [math]i,j,k[/math] distinct:

[[math]] u_{ai}u_{aj}u_{bk}+u_{aj}u_{ak}u_{bi}+u_{ak}u_{ai}u_{bj}=0 [[/math]]

By multiplying to the right by [math]u_{kb}[/math] and then by summing over [math]b[/math], we deduce from this that we have, for any [math]i,j[/math]:

[[math]] u_{ai}u_{aj}=0 [[/math]]

Now since the quotient of [math]C(\bar{O}_N)[/math] by these latter relations is the algebra [math]C(H_N)[/math], we conclude that we have, as claimed:

[[math]] G^+(\bar{S}^{N-1,1}_\mathbb R)=H_N [[/math]]

(3) At [math]d=N[/math] the result is already known, and its proof follows in fact from the “twisted trick” explained in the proof of (2) above, applied to [math]\bar{S}^{N-1}_\mathbb R[/math].

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14b. Quantum groups

In general now, the idea will be that the quantum isometry groups of the intersections of the spheres will basically appear as intersections of the corresponding quantum isometry groups. To start with, we must compute the intersections between the quantum orthogonal groups and their twists. The result here, which is similar to the one for the corresponding spheres, established in chapter 13, is as follows:

Proposition

The [math]5[/math] orthogonal groups and their twists, and the intersections between them, are as follows, at any [math]N\geq3[/math]:

[[math]] \xymatrix@R=12mm@C=16mm{ O_N\ar[r]&O_N^*\ar[r]&O_N^+\\ H_N\ar[r]\ar[u]&H_N^*\ar[r]\ar[u]&\bar{O}_N^*\ar[u]\\ H_N\ar[r]\ar[u]&H_N\ar[r]\ar[u]&\bar{O}_N\ar[u]} [[/math]]

At [math]N=2[/math] the same holds, with the lower left square being replaced by:

[[math]] \xymatrix@R=12mm@C=16mm{ O_2\ar[r]&O_2^+\\ H_2\ar[u]\ar[r]&\bar{O}_2\ar[u]} [[/math]]

Show Proof

We have to study 4 quantum group intersections, as follows:

(1) [math]O_N\cap\bar{O}_N[/math]. Here an element [math]U\in O_N[/math] belongs to the intersection when its entries satisfy [math]ab=0[/math] for any [math]a\neq b[/math] on the same row or column of [math]U[/math]. But this means that our matrix [math]U\in O_N[/math] must be monomial, and so we get [math]U\in H_N[/math], as claimed.

(2) [math]O_N\cap\bar{O}_N^*[/math]. At [math]N=2[/math] the defining relations for [math]\bar{O}_N^*[/math] dissapear, and so we have the following computation, which leads to the conclusion in the statement:

[[math]] O_2\cap\bar{O}_2^*=O_2\cap O_2^+=O_2 [[/math]]

At [math]N\geq3[/math] now, the following inclusion is clear:

[[math]] H_N\subset O_N\cap\bar{O}_N^* [[/math]]

In order to prove the converse inclusion, pick [math]U\in O_N[/math] in the intersection, and assume that [math]U[/math] is not monomial. By permuting the entries we can further assume:

[[math]] U_{11}\neq0\quad,\quad U_{12}\neq0 [[/math]]

From [math]U_{11}U_{12}U_{i3}=0[/math] for any [math]i[/math] we deduce that the third column of [math]U[/math] is filled with [math]0[/math] entries, a contradiction. Thus we must have [math]U\in H_N[/math], as claimed.

(3) [math]O_N^*\cap\bar{O}_N[/math]. At [math]N=2[/math] we have the following computation, as claimed:

[[math]] O_2^*\cap\bar{O}_2=O_2^+\cap\bar{O}_2=\bar{O}_2 [[/math]]

At [math]N\geq3[/math] now, the best is to use the result in (4) below. Indeed, knowing that we have [math]O_N^*\cap\bar{O}_N^*=H_N^*[/math], our intersection is then:

[[math]] G=H_N^*\cap\bar{O}_N [[/math]]

Now since the standard coordinates on [math]H_N^*[/math] are known to satisfy [math]ab=0[/math] for [math]a\neq b[/math] on the same row or column of [math]u[/math], the commutation/anticommutation relations defining [math]\bar{O}_N[/math] reduce to plain commutation relations. Thus [math]G[/math] follows to be classical, [math]G\subset O_N[/math], and by using (1) above we obtain the following formula, as claimed:

[[math]] \begin{eqnarray*} G &=&H_N^*\cap\bar{O}_N\cap O_N\\ &=&H_N^*\cap H_N\\ &=&H_N \end{eqnarray*} [[/math]]

(4) [math]O_N^*\cap\bar{O}_N^*[/math]. The result here is non-trivial, and we must use the half-liberation technology from . The quantum group [math]H_N^\times=O_N^*\cap\bar{O}_N^*[/math] is indeed half-classical in the sense of , and since we have [math]H_N^*\subset H_N^\times[/math], this quantum group is not classical. Thus the main result in applies, and shows that [math]H_N^\times\subset O_N^*[/math] must come, via the crossed product construction there, from an intermediate compact group, as follows:

[[math]] \mathbb T\subset G\subset U_N [[/math]]

Now observe that the standard coordinates on [math]H_N^\times[/math] are by definition subject to the conditions [math]abc=0[/math] when [math](r,s)=(\leq2,3),(3,\leq2)[/math], with the notations and conventions from chapter 11 above. It follows that the standard coordinates on [math]G[/math] are subject to the conditions [math]\alpha\beta\gamma=0[/math] when [math](r,s)=(\leq2,3),(3,\leq2)[/math], where [math]r,s=span(a,b,c)[/math], and [math]\alpha=a,a^*,\beta=b,b^*,\gamma=c,c^*[/math]. Thus we have an inclusion as follows:

[[math]] G\subset\bar{U}_N^* [[/math]]

We deduce that we have an inclusion as follows, with [math]K_N^\circ=U_N\cap\bar{U}_N^*[/math]:

[[math]] G\subset K_N^\circ [[/math]]

But this intersection can be computed exactly as in the real case, in the proof of (2) above, and we obtain [math]K_2^\circ=U_2[/math], and [math]K_N^\circ=\mathbb T\wr S_N[/math] at [math]N\geq 3[/math].

But the half-liberated quantum groups obtained from [math]U_2[/math] and [math]\mathbb T\wr S_N[/math] via the half-liberation construction in are well-known, these being [math]O_2^*=O_2^+[/math] and [math]H_N^*[/math]. Thus by functoriality we have [math]H_2^\times\subset O_2^+[/math] and [math]H_N^\times\subset H_N^*[/math] at [math]N\geq 3[/math], and since the reverse inclusions are clear, we obtain [math]H_2^\times=O_2^+[/math] and [math]H_N^\times=H_N^*[/math] at [math]N\geq 3[/math], as claimed.

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Let us go back now to the sphere left, namely [math]S^{N-1,1}_{\mathbb R,*}[/math]. Things are quite tricky here, and we will need the following technical result, from Raum-Weber :

Proposition

Let [math]H_N^{[\infty]}\subset O_N^+[/math] be the compact quantum group obtained via the relations [math]abc=0[/math], whenever [math]a\neq c[/math] are on the same row or column of [math]u[/math].

• We have inclusions [math]H_N^*\subset H_N^{[\infty]}\subset H_N^+[/math].
• We have [math]ab_1\ldots b_rc=0[/math], whenever [math]a\neq c[/math] are on the same row or column of [math]u[/math].
• We have [math]ab^2=b^2a[/math], for any two entries [math]a,b[/math] of [math]u[/math].

Show Proof

We briefly recall the proof in , for future use in what follows. Our first claim is that [math]H_N^{[\infty]}[/math] comes, as an easy quantum group, from the following diagram:

[[math]] \xymatrix@R=5mm@C=0.1mm{&\\\pi\ \ =\\&}\xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[rr]&&\\ \circ&\circ&\circ} [[/math]]

Indeed, this diagram acts via the following linear map:

[[math]] T_\pi(e_{ijk})=\delta_{ik}e_{ijk} [[/math]]

We therefore have the following formula:

[[math]] \begin{eqnarray*} T_\pi u^{\otimes 3}e_{abc} &=&T_\pi\sum_{ijk}e_{ijk}\otimes u_{ia}u_{jb}u_{kc}\\ &=&\sum_{ijk}e_{ijk}\otimes\delta_{ik}u_{ia}u_{jb}u_{kc} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} u^{\otimes 3}T_\pi e_{abc} &=&u^{\otimes 3}\delta_{ac}e_{abc}\\ &=&\sum_{ijk}e_{ijk}\otimes\delta_{ac}u_{ia}u_{jb}u_{kc} \end{eqnarray*} [[/math]]

Thus the condition [math]T_\pi\in End(u^{\otimes 3})[/math] is equivalent to the following relations:

[[math]] (\delta_{ik}-\delta_{ac})u_{ia}u_{jb}u_{kc}=0 [[/math]]

The non-trivial cases are [math]i=k,a\neq c[/math] and [math]i\neq k,a=c[/math], and these produce the relations [math]u_{ia}u_{jb}u_{ic}=0[/math] for any [math]a\neq c[/math], and [math]u_{ia}u_{jb}u_{ka}=0[/math], for any [math]i\neq k[/math]. Thus, we have reached to the standard relations for the quantum group [math]H_N^{[\infty]}[/math].

(1) We have the following formula:

[[math]] \xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[ddrr]&\circ\ar@{-}[dd]&\circ\ar@{-}[ddll]\ar@/^/@{.}[r]&\circ\ar@{-}[dd]&\circ\ar@{-}[dd]\\ &&&\ar@{-}[r]&\\ \circ&\circ&\circ\ar@/_/@{.}[r]&\circ&\circ}\ \ \xymatrix@R=5mm@C=1mm{&\\=\\&}\xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[rr]&&\\ \circ&\circ&\circ} [[/math]]

We have as well the following formula:

[[math]] \xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]\ar@/^/@{.}[r]&\circ\ar@{-}[dd]\\ \ar@{-}[rr]&&\\ \circ&\circ\ar@/_/@{.}[r]&\circ}\ \ \xymatrix@R=5mm@C=1mm{&\\=\\&}\xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[r]&&\\ \circ&\circ} [[/math]]

Thus, we obtain inclusions as desired, namely:

[[math]] H_N^*\subset H_N^{[\infty]}\subset H_N^+ [[/math]]

(2) At [math]r=2[/math], the relations [math]ab_1b_2c=0[/math] come indeed from the following diagram:

[[math]] \xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\ar@/^/@{.}[r]&\circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[rr]&&&\ar@{-}[rr]&&\\ \circ&\circ&\circ\ar@/_/@{.}[r]&\circ&\circ&\circ} \ \ \xymatrix@R=5mm@C=1mm{&\\=\\&} \item[a]ymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[rrr]&&&\\ \circ&\circ&\circ&\circ} [[/math]]

In the general case [math]r\geq2[/math] the proof is similar, see for details.

(3) We use here an idea from . By rotating [math]\pi[/math], we obtain:

[[math]] \xymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\circ\ar@{-}[dd]\\ \ar@{-}[rr]&&\\ \circ&\circ&\circ} \ \ \xymatrix@R=5mm@C=0.1mm{&\\ \to\\&}\ \item[a]ymatrix@R=5mm@C=5mm{ \circ\ar@{-}[dd]&\circ\ar@{.}[dd]&\\ \ar@{-}[rrr]&&\ar@{-}[d]&\ar@{-}[d]\\ \circ&\circ&\circ&\circ} \ \ \xymatrix@R=5mm@C=0.1mm{&\\ \to\\&}\ \item[a]ymatrix@R=5mm@C=5mm{ \circ\ar@{-}[d]&\circ\ar@{-}[dd]&\circ\ar@{.}[ddll]\\ \ar@{-}[rr]&&\ar@{-}[d]\\ \circ&\circ&\circ} [[/math]]

Let us denote by [math]\sigma[/math] the partition on the right. Since [math]T_\sigma(e_{ijk})=\delta_{ij}e_{kji}[/math], we obtain:

[[math]] \begin{eqnarray*} T_\sigma u^{\otimes 3}e_{abc} &=&T_\sigma\sum_{ijk}e_{ijk}\otimes u_{ia}u_{jb}u_{kc}\\ &=&\sum_{ijk}e_{kji}\otimes\delta_{ij}u_{ia}u_{jb}u_{kc} \end{eqnarray*} [[/math]]

On the other hand, we obtain as well the following formula:

[[math]] \begin{eqnarray*} u^{\otimes 3}T_\sigma e_{abc} &=&u^{\otimes 3}\delta_{ab}e_{cba}\\ &=&\sum_{ijk}e_{kji}\otimes\delta_{ab}u_{kc}u_{jb}u_{ia} \end{eqnarray*} [[/math]]

Thus [math]T_\sigma\in End(u^{\otimes 3})[/math] is equivalent to the following relations:

[[math]] \delta_{ij}u_{ia}u_{jb}u_{kc}=\delta_{ab}u_{kc}u_{jb}u_{ia} [[/math]]

Now by setting [math]j=i,b=a[/math] in this formula we obtain the commutation relations in the statement, namely [math]u_{ia}^2u_{kc}=u_{kc}u_{ia}^2[/math], and this finishes the proof.

■

The relation of [math]H_N^{[\infty]}[/math] with the polygonal spheres comes from the following fact:

Proposition

Let [math]X\subset S^{N-1}_{\mathbb R,+}[/math] be closed, let [math]d\geq2[/math], and set:

[[math]] X^{d-1}=X\cap S^{N-1,d-1}_{\mathbb R,+} [[/math]]

Then for a quantum group [math]G\subset H_N^{[\infty]}[/math] the following are equivalent:

• [math]x_i\to\sum_jx_j\otimes u_{ji}[/math] defines a coaction map, as follows:
  [[math]] \Phi:C(X^{d-1})\to C(X^{d-1})\otimes C(G) [[/math]]
• [math]x_i\to\sum_jx_j\otimes u_{ji}[/math] defines a morphism, as follows:
  [[math]] \widetilde{\Phi}:C(X)\to C(X^{d-1})\otimes C(G) [[/math]]

In particular, [math]G^+(X)\cap H_N^{[\infty]}[/math] acts on [math]X^{d-1}[/math], for any [math]d\geq2[/math].

Show Proof

The idea here is to use the relations in Proposition 14.6 (2):

[math](1)\implies(2)[/math] This is clear, by composing [math]\Phi[/math] with the following projection map:

[[math]] \pi:C(X)\to C(X^{d-1}) [[/math]]

[math](2)\implies(1)[/math] In order to prove this implication, we must understand when a coaction map as follows exists:

[[math]] C(X^{d-1})\to C(X^{d-1})\otimes C(G) [[/math]]

In order for this to happen, the variables [math]X_i=\sum_jx_j\otimes u_{ji}[/math] must satisfy the relations defining [math]X[/math], which hold indeed by (2), and must satisfy as well the following relations, with [math]i_0,\ldots,i_d[/math] distinct, which define the polygonal spheres [math]S^{N-1,d-1}_{\mathbb R,+}[/math]:

[[math]] X_{i_0}\ldots X_{i_d}=0 [[/math]]

The point now is that, under the assumption [math]G\subset H_N^{[\infty]}[/math], these latter relations are automatic. Indeed, by using Proposition 14.6 (2), for [math]i_0,\ldots,i_d[/math] distinct we obtain:

[[math]] \begin{eqnarray*} X_{i_0}\ldots X_{i_d} &=&\sum_{j_0\ldots j_d}x_{j_0}\ldots x_{j_d}\otimes u_{j_0i_0}\ldots u_{j_di_d}\\ &=&\sum_{j_0\ldots j_d\ distinct}0\otimes u_{j_0i_0}\ldots u_{j_di_d}\ \ +\!\!\!\!\sum_{j_0\ldots j_d\ not\ distinct}x_{j_0}\ldots x_{j_d}\otimes0\\ &=&0+0=0 \end{eqnarray*} [[/math]]

Thus the coaction in (1) exists precisely when (2) is satisfied, and we are done.

Finally, the last assertion is clear from [math](2)\implies(1)[/math], because the universal coaction of [math]G=G^+(X)[/math] gives rise to a map [math]\widetilde{\Phi}:C(X)\to C(X^{d-1})\otimes C(G)[/math] as in (2).

■

As an illustration, we have the following result:

Theorem

The compact quantum groups

[[math]] H_N,H_N,H_N^*,H_N^*,H_N^{[\infty]} [[/math]]

act respectively on the spheres

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

at any [math]d\geq2[/math].

Show Proof

We use Proposition 14.7. We know that the quantum isometry groups at [math]d=N[/math] are respectively equal to the following quantum groups:

[[math]] O_N,\bar{O}_N,O_N^*,\bar{O}_N^*,O_N^+ [[/math]]

Our claim is that, by intersecting these quantum groups with [math]H_N^{[\infty]}[/math], we obtain the quantum groups in the statement. Indeed:

(1) [math]O_N\cap H_N^{[\infty]}=H_N[/math] is clear from definitions.

(2) [math]\bar{O}_N\cap H_N^{[\infty]}=H_N[/math] follows from [math]\bar{O}_N\cap H_N^+\subset O_N[/math], which in turn follows from the computation (3) in the proof of Proposition 14.5, with [math]H_N^*[/math] replaced by [math]H_N^+[/math].

(3) [math]O_N^*\cap H_N^{[\infty]}=H_N^*[/math] follows from [math]O_N^*\cap H_N^+=H_N^*[/math].

(4) [math]\bar{O}_N^*\cap H_N^{[\infty]}\supset H_N^*[/math] is clear, and the reverse inclusion can be proved by a direct computation, similar to the computation (3) in the proof of Proposition 14.5.

(5) [math]O_N^+\cap H_N^{[\infty]}=H_N^{[\infty]}[/math] is clear from definitions.

■

Observe that the above result is “sharp”, in the sense that the actions there are the universal ones, in the classical case at any [math]d\in\{2,\ldots,N-1\}[/math], as well as in the twisted case at [math]d=2[/math]. Indeed, this follows from the various results established above. We refer to for more comments on all this, and for some related open problems.

14c. Middle computation

Let us discuss now the computation of the quantum isometry group for the polygonal sphere which is left, namely [math]S^{N-1,1}_{\mathbb R,*}[/math], appearing in the middle of our square diagram of polygonal spheres, from the beginning of this chapter. We have seen in the above that the quantum group [math]H_N^*[/math] acts on [math]S^{N-1,1}_{\mathbb R,*}[/math]. This action, however, is not universal, due to:

Proposition

The discrete group dual

[[math]] G=\widehat{\mathbb Z_2^{*N}} [[/math]]

acts on the polygonal sphere [math]S^{N-1,1}_{\mathbb R,*}[/math].

Show Proof

The standard coordinates on the polygonal sphere [math]S^{N-1,1}_{\mathbb R,*}[/math] are by definition subject to the following relations:

[[math]] x_ix_jx_k=\begin{cases} 0&{\rm for}\ i,j,k\ {\rm distinct}\\ x_kx_jx_i&{\rm otherwise} \end{cases} [[/math]]

Let us try now to understand under which assumptions on a compact quantum group [math]G[/math], and in particular on a group dual, we have a coaction map, as follows:

[[math]] \Phi:C(S^{N-1,1}_{\mathbb R,*})\to C(S^{N-1,1}_{\mathbb R,*})\otimes C(G) [[/math]]

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

In order for this to happen, the following variables must satisfy the above relations:

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

For the group dual [math]G=\widehat{\mathbb Z_2^{*N}}[/math] we have by definition [math]u_{ij}=\delta_{ij}g_i[/math], where [math]g_1,\ldots,g_N[/math] are the standard generators of [math]\mathbb Z_2^{*N}[/math]. We therefore have:

[[math]] X_iX_jX_k=x_ix_jx_k\otimes g_ig_jg_k [[/math]]

[[math]] X_kX_jX_i=x_kx_jx_i\otimes g_kg_jg_i [[/math]]

Thus the formula [math]X_iX_kX_k=0[/math] for [math]i,j,k[/math] distinct is clear, and the formula [math]X_iX_jX_k=X_kX_jX_i[/math] for [math]i,j,k[/math] not distinct requires [math]g_ig_jg_k=g_kg_jg_i[/math] for [math]i,j,k[/math] not distinct, which is clear as well. Indeed, at [math]i=j[/math] this latter relation reduces to [math]g_k=g_k[/math], at [math]i=k[/math] this relation is trivial, [math]g_ig_jg_i=g_ig_jg_i[/math], and at [math]j=k[/math] this relation reduces to [math]g_i=g_i[/math].

■

More generally now, we have the following result:

Proposition

The intermediate liberation of the hyperoctahedral group

[[math]] G=H_N^{[\infty]} [[/math]]

acts on the polygonal sphere [math]S^{N-1,1}_{\mathbb R,*}[/math].

Show Proof

We proceed as in the proof of Theorem 14.4. By expanding the formula of [math]X_iX_jX_k[/math] and by using the relations for the sphere [math]S^{N-1,1}_{\mathbb R,*}[/math], we have:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\sum_{abc}x_ax_bx_c\otimes u_{ai}u_{bj}u_{ck}\\ &=&\sum_{a,b,c\ not \ distinct}x_ax_bx_c\otimes u_{ai}u_{bj}u_{ck}\\ &=&\sum_{a\neq b}x_a^2x_b\otimes(u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak})\\ &+&\sum_{a\neq b}x_ax_bx_a\otimes u_{ai}u_{bj}u_{ak}\\ &+&\sum_a x_a^3\otimes u_{ai}u_{aj}u_{ak} \end{eqnarray*} [[/math]]

Now by assuming [math]G=H_N^{[\infty]}[/math], as in the statement, and by using the various formulae in Proposition 14.6, we obtain, for any [math]i,j,k[/math] distinct:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\sum_{a\neq b}x_a^2x_b\otimes(0\cdot u_{bk}+u_{bi}\cdot 0)\\ &+&\sum_{a\neq b}x_ax_bx_a\otimes0\\ &+&\sum_ax_a^3\otimes(0\cdot u_{ak})\\ &=&0 \end{eqnarray*} [[/math]]

It remains to prove that we have [math]X_iX_jX_k=X_kX_jX_i[/math], for [math]i,j,k[/math] not distinct. By replacing [math]i\leftrightarrow k[/math] in the above formula of [math]X_iX_jX_k[/math], we obtain:

[[math]] \begin{eqnarray*} X_kX_jX_i &=&\sum_{a\neq b}x_a^2x_b\otimes(u_{ak}u_{aj}u_{bi}+u_{bk}u_{aj}u_{ai})\\ &+&\sum_{a\neq b}x_ax_bx_a\otimes u_{ak}u_{bj}u_{ai}\\ &+&\sum_ax_a^3\otimes u_{ak}u_{aj}u_{ai} \end{eqnarray*} [[/math]]

Let us compare this formula with the above formula of [math]X_iX_jX_k[/math]. The last sum being 0 in both cases, we must prove that for any [math]i,j,k[/math] not distinct and any [math]a\neq b[/math] we have:

[[math]] u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=u_{ak}u_{aj}u_{bi}+u_{bk}u_{aj}u_{ai} [[/math]]

[[math]] u_{ai}u_{bj}u_{ak}=u_{ak}u_{bj}u_{ai} [[/math]]

By symmetry the three cases [math]i=j,i=k,j=k[/math] reduce to two cases, [math]i=j[/math] and [math]i=k[/math]. The case [math]i=k[/math] being clear, we are left with the case [math]i=j[/math], where we must prove:

[[math]] u_{ai}u_{ai}u_{bk}+u_{bi}u_{ai}u_{ak}=u_{ak}u_{ai}u_{bi}+u_{bk}u_{ai}u_{ai} [[/math]]

[[math]] u_{ai}u_{bi}u_{ak}=u_{ak}u_{bi}u_{ai} [[/math]]

By using [math]a\neq b[/math], the first equality reads:

[[math]] u_{ai}^2u_{bk}+0\cdot u_{ak}=u_{ak}\cdot 0+u_{bk}u_{ai}^2 [[/math]]

Since by Proposition 14.6 (3) we have [math]u_{ai}^2u_{bk}=u_{bk}u_{ai}^2[/math], this formula is satisfied, and we are done. As for the second equality, this reads:

[[math]] 0\cdot u_{ak}=u_{ak}\cdot 0 [[/math]]

But this is true as well, and this ends the proof.

■

We will prove now that the action in Proposition 14.10 is universal. In order to do so, we need to convert the formulae of type [math]X_iX_jX_k=0[/math] and [math]X_iX_jX_k=X_kX_jX_i[/math] into relations between the quantum group coordinates [math]u_{ij}[/math], and this requires a good knowledge of the linear relations between the variables [math]x_a^2x_b,x_ax_bx_a,x_a^3[/math] over the sphere [math]S^{N-1,1}_{\mathbb R,*}[/math]. So, we must first study these variables. The answer here is given by:

Proposition

The variables

[[math]] \left\{x_a^2x_b,x_ax_bx_a,x_a^3\Big|a\neq b\right\} [[/math]]

are linearly independent over the sphere [math]S^{N-1,1}_{\mathbb R,*}[/math].

Show Proof

We use a trick of Bichon-Dubois-Violette . Consider the 1-dimensional polygonal version of the complex sphere [math]S^{N-1}_\mathbb C[/math], which is by definition given by:

[[math]] S^{N-1,1}_\mathbb C=\left\{z\in S^{N-1}_\mathbb C\Big|z_iz_jz_k=0,\forall i,j,k\ {\rm distinct}\right\} [[/math]]

We have then a [math]2\times2[/math] matrix model for the coordinates of [math]S^{N-1,1}_{\mathbb R,*}[/math], as follows:

[[math]] x_i\to\gamma_i=\begin{pmatrix}0&z_i\\ \bar{z}_i&0\end{pmatrix} [[/math]]

Indeed, the matrices [math]\gamma_i[/math] on the right are all self-adjoint, their squares sum up to [math]1[/math], they half-commute, and they satisfy, for [math]i,j,k[/math] distinct:

[[math]] \gamma_i\gamma_j\gamma_k=0 [[/math]]

Thus we have indeed a morphism as follows, as claimed:

[[math]] C(S^{N-1,1}_{\mathbb R,*})\to M_2(C(S^{N-1,1}_\mathbb C))\quad,\quad x_i\to\gamma_i [[/math]]

We can use this model in order to prove the linear independence. Consider indeed the variables that we are interested in, namely:

[[math]] x_a^2x_b\quad,\quad x_ax_bx_a\quad,\quad x_a^3 [[/math]]

In the model, these variables are mapped to the following variables:

[[math]] \gamma_a^2\gamma_b=\begin{pmatrix}0&|z_a|^2z_b\\ |z_a|^2\bar{z}_b&0\end{pmatrix} [[/math]]

[[math]] \gamma_a\gamma_b\gamma_a=\begin{pmatrix}0&z_a^2\bar{z}_b\\ \bar{z}_a^2z_b&0\end{pmatrix} [[/math]]

[[math]] \gamma_a^3=\begin{pmatrix}0&|z_a|^2z_a\\ |z_a|^2\bar{z}_a&0\end{pmatrix} [[/math]]

Now observe that the following variables are linearly independent over [math]S^1_\mathbb C[/math]:

[[math]] |z_1|^2z_2,|z_2|^2z_1,z_1^2\bar{z}_2,z_2^2\bar{z}_1,|z_1|^2z_1,|z_2|^2z_2 [[/math]]

Thus the upper right entries of the above matrices are linearly independent over [math]S^{N-1,1}_\mathbb C[/math]. Thus the matrices themselves are linearly independent, and this proves our result.

■

With the above result in hand, we can now reformulate the coaction problem into a purely quantum group-theoretical problem, as follows:

Proposition

A quantum group [math]G\subset O_N^+[/math] acts on [math]S^{N-1,1}_{\mathbb R,*}[/math] precisely when its standard coordinates [math]u_{ij}[/math] satisfy the following relations:

• [math]u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=0[/math] for any [math]i,j,k[/math] distinct.
• [math]u_{ai}u_{bj}u_{ak}=0[/math] for any [math]i,j,k[/math] distinct.
• [math]u_{ai}^2u_{bk}=u_{bk}u_{ai}^2[/math].
• [math]u_{ak}u_{ai}u_{bi}=u_{bi}u_{ai}u_{ak}[/math].
• [math]u_{ai}u_{bi}u_{ak}=u_{bk}u_{bi}u_{ai}[/math].

Show Proof

We use notations from the beginning of the proof of Proposition 14.10, along with the following formula, also established there:

[[math]] \begin{eqnarray*} X_iX_jX_k &=&\sum_{a\neq b}x_a^2x_b\otimes(u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak})\\ &+&\sum_{a\neq b}x_ax_bx_a\otimes u_{ai}u_{bj}u_{ak}\\ &+&\sum_ax_a^3\otimes u_{ai}u_{aj}u_{ak} \end{eqnarray*} [[/math]]

In order to have an action as in the statement, these quantities must satisfy [math]X_iX_kX_k=0[/math] for [math]i,j,k[/math] distinct, and [math]X_iX_kX_k=X_kX_jX_i[/math] for [math]i,j,k[/math] not distinct. Now by using Proposition 14.11, we conclude that the relations to be satisfied are as follows:

(A) For [math]i,j,k[/math] distinct, the following must hold:

[[math]] u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=0,\forall a\neq b [[/math]]

[[math]] u_{ai}u_{bj}u_{ak}=0,\forall a\neq b [[/math]]

[[math]] u_{ai}u_{aj}u_{ak}=0,\forall a [[/math]]

(B) For [math]i,j,k[/math] not distinct, the following must hold:

[[math]] u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=u_{ak}u_{aj}u_{bi}+u_{bk}u_{aj}u_{ai},\forall a\neq b [[/math]]

[[math]] u_{ai}u_{bj}u_{ak}=u_{ak}u_{bj}u_{ai},\forall a\neq b [[/math]]

[[math]] u_{ai}u_{aj}u_{ak}=u_{ak}u_{aj}u_{ai},\forall a [[/math]]

The last relations in (A,B) can be merged with the other ones, and we are led to:

(A') For [math]i,j,k[/math] distinct, the following must hold:

[[math]] u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=0,\forall a,b [[/math]]

[[math]] u_{ai}u_{bj}u_{ak}=0,\forall a,b [[/math]]

(B') For [math]i,j,k[/math] not distinct, the following must hold:

[[math]] u_{ai}u_{aj}u_{bk}+u_{bi}u_{aj}u_{ak}=u_{ak}u_{aj}u_{bi}+u_{bk}u_{aj}u_{ai},\forall a,b [[/math]]

[[math]] u_{ai}u_{bj}u_{ak}=u_{ak}u_{bj}u_{ai},\forall a,b [[/math]]

Observe that the relations (A') are exactly the relations (1,2) in the statement.

Let us further process the relations (B'). In the case [math]i=k[/math] the relations are automatic, and in the cases [math]j=i,j=k[/math] the relations that we obtain coincide, via [math]i\leftrightarrow k[/math]. Thus (B') reduces to the set of relations obtained by setting [math]j=i[/math], which are as follows:

[[math]] u_{ai}u_{ai}u_{bk}+u_{bi}u_{ai}u_{ak}=u_{ak}u_{ai}u_{bi}+u_{bk}u_{ai}u_{ai} [[/math]]

[[math]] u_{ai}u_{bi}u_{ak}=u_{ak}u_{bi}u_{ai} [[/math]]

Observe that the second relation is the relation (5) in the statement. Regarding now the first relation, with the notation [math][x,y,z]=xyz-zyx[/math], this is as follows:

[[math]] [u_{ai},u_{ai},u_{bk}]=[u_{ak},u_{ai},u_{bi}] [[/math]]

By applying the antipode, we obtain from this:

[[math]] [u_{kb},u_{ia},u_{ia}]=[u_{ib},u_{ia},u_{ka}] [[/math]]

By relabelling [math]a\leftrightarrow i[/math] and [math]b\leftrightarrow k[/math], this relation becomes:

[[math]] [u_{bk},u_{ai},u_{ai}]=[u_{ak},u_{ai},u_{bi}] [[/math]]

Now since we have [math][x,y,z]=-[z,y,x][/math], by comparing this latter relation with the original one, a simplification occurs, and the resulting relations are as follows:

[[math]] [u_{ai},u_{ai},u_{bk}]=[u_{ak},u_{ai},u_{bi}]=0 [[/math]]

But these are exactly the relations (3,4) in the statement, and we are done.

■

Now by solving the quantum group problem raised by Proposition 14.12, we obtain:

Proposition

We have the folowing formula:

[[math]] G^+(S^{N-1,1}_{\mathbb R,*})=H_N^{[\infty]} [[/math]]

Show Proof

The inclusion [math]\supset[/math] is clear from Proposition 14.10. For the converse, we already have the result at [math]N=2[/math], so assume [math]N\geq3[/math]. We will use many times the conditions (1-5) in Proposition 14.12. By using (2), for [math]i\neq j[/math] we have:

[[math]] \begin{eqnarray*} &&u_{ai}u_{bj}u_{ak}=0,\forall k\neq i,j\\ &\implies&u_{ai}u_{bj}u_{ak}^2=0,\forall k\neq i,j\\ &\implies&u_{ai}u_{bj}\left(\sum_{k\neq i,j}u_{ak}^2\right)=0,\forall i\neq j\\ &\implies&u_{ai}u_{bj}(1-u_{ai}^2-u_{aj}^2)=0,\forall i\neq j \end{eqnarray*} [[/math]]

Now by using (3), we can move the variable [math]u_{bj}[/math] to the right. By further multiplying by [math]u_{bj}[/math] to the right, and then summing over [math]b[/math], we obtain:

[[math]] \begin{eqnarray*} &&u_{ai}u_{bj}(1-u_{ai}^2-u_{aj}^2)=0,\forall i\neq j\\ &\implies&u_{ai}(1-u_{ai}^2-u_{aj}^2)u_{bj}=0,\forall i\neq j\\ &\implies&u_{ai}(1-u_{ai}^2-u_{aj}^2)u_{bj}^2=0,\forall i\neq j\\ &\implies&u_{ai}(1-u_{ai}^2-u_{aj}^2)=0,\forall i\neq j \end{eqnarray*} [[/math]]

We can proceed now as follows, by summing over [math]j\neq i[/math]:

[[math]] \begin{eqnarray*} &&u_{ia}(1-u_{ai}^2-u_{aj}^2)=0,\forall i\neq j\\ &\implies&u_{ai}u_{aj}^2=u_{ai}-u_{ai}^3,\forall i\neq j\\ &\implies&u_{ai}(1-u_{ai}^2)=(N-1)(u_{ai}-u_{ai}^3)\\ &\implies&u_{ai}=u_{ai}^3 \end{eqnarray*} [[/math]]

Thus the standard coordinates are partial isometries, and so [math]G\subset H_N^+[/math]. On the other hand, we know from the proof of Proposition 14.6 (3) that the quantum subgroup [math]G\subset H_N^+[/math] obtained via the relations [math][a,b^2]=0[/math] is [math]H_N^{[\infty]}[/math], and this finishes the proof.

■

We have now complete results for the 9 main spheres, as follows:

Theorem

The quantum isometry groups of the [math]9[/math] polygonal spheres are

[[math]] \xymatrix@R=12mm@C=17mm{ O_N\ar[r]&O_N^*\ar[r]&O_N^+\\ H_N\ar[r]\ar[u]&H_N^{[\infty]}\ar[r]\ar[u]&\bar{O}_N^*\ar[u]\\ H_N^+\ar[r]\ar[u]&H_N\ar[r]\ar[u]&\bar{O}_N\ar[u]} [[/math]]

where [math]H_N^+,H_N^{[\infty]}[/math] and [math]\bar{O}_N,O_N^*,\bar{O}_N^*,O_N^*[/math] are quantum versions of [math]H_N,O_N[/math].

Show Proof

This follows indeed by putting together the above results.

■

As a conclusion, we have a key computation available, but there are still many questions left, regarding the extension of our [math](S,T,U,K)[/math] formalism, as to cover the intersections between the twisted and untwisted geometries. And here, things are quite complicated, because as explained in chapter 11, even in the plain untwisted case, there is one axiom which needs to be modified, with questions which are currently open.

As explained in the present and in the previous chapter, on several occasions, all this is not exactly something theoretical, but is rather something of practical interest, in connection with the algebraic geometry of the free spheres, and other free manifolds. Finally, let us mention too that none of the questions to be solved is difficult, so to say. It's just that there is so much work to be done, and that hasn't been done yet.

14d. Complex extension

Still following , let us discuss now a straightforward complex extension of the above results. Our starting point will be the following definition:

Definition

The complex polygonal spheres, denoted

[[math]] S^{N-1,d-1}_{\mathbb C},\bar{S}^{N-1,d-1}_{\mathbb C},\bar{S}^{N-1,d-1}_{\mathbb C,*},S^{N-1,d-1}_{\mathbb C,+} [[/math]]

are constructed from [math]S^{N-1}_{\mathbb C,+}[/math] in the same way as their real versions, namely

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

are constructed from [math]S^{N-1}_{\mathbb R,+}[/math], namely by assuming that the corresponding vanishing relations hold between the variables [math]x_i=z_i,z_i^*[/math].

As in the real case, we will restrict now the attention to the 5 main spheres, coming from , and to their intersections. We have 9 such spheres here, as follows:

[[math]] \xymatrix@R=13mm@C=3mm{ S^{N-1}_\mathbb C\ar[rr]&&S^{N-1}_{\mathbb C,*}\ar[rrr]&&&S^{N-1}_{\mathbb C,+}\\ S^{N-1}_\mathbb C\cap\bar{S}^{N-1}_{\mathbb C,*}\ar[rr]\ar[u]&&S^{N-1}_{\mathbb C,*}\cap\bar{S}^{N-1}_{\mathbb C,*}\ar[rrr]\ar[u]&&&\bar{S}^{N-1}_{\mathbb C,*}\ar[u]\\ S^{N-1}_\mathbb C\cap\bar{S}^{N-1}_\mathbb C\ar[rr]\ar[u]&&S^{N-1}_{\mathbb C,*}\cap\bar{S}^{N-1}_\mathbb C\ar[rrr]\ar[u]&&&\bar{S}^{N-1}_\mathbb C\ar[u]} [[/math]]

The intersections can be computed as in the real case, and we have:

Proposition

The [math]5[/math] main spheres, and the intersections between them, are

[[math]] \xymatrix@R=12mm@C=12mm{ S^{N-1}_\mathbb C\ar[r]&S^{N-1}_{\mathbb C,*}\ar[r]&S^{N-1}_{\mathbb C,+}\\ S^{N-1,1}_\mathbb C\ar[r]\ar[u]&S^{N-1,1}_{\mathbb C,*}\ar[r]\ar[u]&\bar{S}^{N-1}_{\mathbb C,*}\ar[u]\\ S^{N-1,0}_\mathbb C\ar[r]\ar[u]&\bar{S}^{N-1,1}_\mathbb C\ar[r]\ar[u]&\bar{S}^{N-1}_\mathbb C\ar[u]} [[/math]]

with all the maps being inclusions.

Show Proof

This is similar to the proof from the real case, from chapter 13, by replacing in all the computations there the variables [math]x_i[/math] by the variables [math]x_i=z_i,z_i^*[/math]. For full details here, and for more on these spheres, we refer as before to the paper .

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As before in the real case, all this raises the question of updating our noncommutative geometry axioms from chapter 4, as to cover such intersections, which are quite interesting objects. And here, we are mostly in need of quantum isometry group results. The computation, from , is very similar to the one from the real case, the result being:

Theorem

The quantum isometry groups of the [math]9[/math] main complex spheres are

[[math]] \xymatrix@R=13mm@C=17mm{ U_N\ar[r]&U_N^*\ar[r]&U_N^+\\ K_N\ar[r]\ar[u]&K_N^{[\infty]}\ar[r]\ar[u]&\bar{U}_N^*\ar[u]\\ K_N^+\ar[r]\ar[u]&K_N\ar[r]\ar[u]&\bar{U}_N\ar[u]} [[/math]]

where [math]K_N[/math] and its versions are the complex analogues of [math]H_N[/math] and its versions.

Show Proof

The idea here is that the proof here is quite similar to the proof in the real case, by replacing [math]H_N,O_N[/math] with their complex analogues [math]K_N,U_N[/math]. See .

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As a conclusion, we have many technical results available, but there are still many questions left, regarding the extension of our [math](S,T,U,K)[/math] formalism, as to cover the intersections between the twisted and untwisted geometries.

So long for free algebraic geometry. These things are quite recent, going back to the 10s, and there is still an enormous amount of work to be done, in order to have something serious started. And what is quite puzzling here, but not really puzzling if you're a bit familiar with algebraic geometry, and its story, is that everything is somehow “trivial”, but the massive accumulation of such trivialities eventually leads to difficulties.

Which can only make us think at Grothendieck, the problems that he met when doing algebraic geometry, and his strategy for dealing with them. The story goes that if Grothendieck was to be on top of a mountain, and had to go to the mountain next to it, he would rather fill the valley in between with small rocks, no matter how long it takes, and walk straight, instead of going downhill and then uphill.

But hey, there should be a trick somewhere, in order to avoid such Grothendieck type things, the mathematics that we developed so far is dangerously pushing us into. Up to you to find it, either based on other things discussed in this book, or on some new idea of yours. Plus of course, and obviously, if you feel like doing some Grothendieck type work in free algebraic geometry, just go for that. No fuss, only trivialities, can only work.

General references

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