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Before getting into the spheres, let us discuss integration questions. The result here, valid for any Schur-Weyl twist in our sense, is as follows:

Theorem

We have the Weingarten type formula

[[math]] \int_{\bar{G}}u_{i_1j_1}^{e_1}\ldots u_{i_kj_k}^{e_k}=\sum_{\pi,\sigma\in D(k)}\bar{\delta}_\pi(i_1\ldots i_k)\bar{\delta}_\sigma(j_1\ldots j_k)W_{kN}(\pi,\sigma) [[/math]]

where [math]W_{kN}=G_{kN}^{-1}[/math], with [math]G_{kN}(\pi,\sigma)=N^{|\pi\vee\sigma|}[/math], for [math]\pi,\sigma\in D(k)[/math].

Show Proof

This follows exactly as in the untwisted case, the idea being that the signs will cancel. Let us recall indeed from Definition 11.8 and the comments afterwards that the twisted vectors [math]\bar{\xi}_\pi[/math] associated to the partitions [math]\pi\in P_{even}(k)[/math] are as follows:

[[math]] \bar{\xi}_\pi=\sum_{\tau\geq\pi}\varepsilon(\tau)\sum_{i:\ker(i)=\tau}e_{i_1}\otimes\ldots\otimes e_{i_k} [[/math]]

Thus, the Gram matrix of these vectors is given by:

[[math]] \begin{eqnarray*} \lt \xi_\pi,\xi_\sigma \gt &=&\sum_{\tau\geq\pi\vee\sigma}\varepsilon(\tau)^2\left|\left\{(i_1,\ldots,i_k)\Big|\ker i=\tau\right\}\right|\\ &=&\sum_{\tau\geq\pi\vee\sigma}\left|\left\{(i_1,\ldots,i_k)\Big|\ker i=\tau\right\}\right|\\ &=&N^{|\pi\vee\sigma|} \end{eqnarray*} [[/math]]

Thus the Gram matrix is the same as in the untwisted case, and so the Weingarten matrix is the same as well as in the untwisted case, and this gives the result.

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In relation now with the spheres, we have the following result:

Theorem

The twisted spheres have the following properties:

• They have affine actions of the twisted unitary quantum groups.
• They have unique invariant Haar functionals, which are ergodic.
• Their Haar functionals are given by Weingarten type formulae.
• They appear, via the GNS construction, as first row spaces.

Show Proof

The proofs here are similar to those from the untwisted case, via a routine computation, by adding signs where needed, and with the main technical ingredient, namely the Weingarten formula, being available from Theorem 11.18. See .

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As a conclusion now, we have shown that the various quadruplets [math](S,T,U,K)[/math] constructed in chapters 1-10 have twisted counterparts [math](\bar{S},T,\bar{U},K)[/math]. The question that we would like to solve now is that of finding correspondences, as follows:

[[math]] \xymatrix@R=50pt@C=50pt{ \bar{S}\ar[r]\ar[d]\ar[dr]&T\ar[l]\ar[d]\ar[dl]\\ \bar{U}\ar[u]\ar[ur]\ar[r]&K\ar[l]\ar[ul]\ar[u] } [[/math]]

In order to discuss this, let us get back to the axioms from chapter 4. We have seen there that the 12 correspondences come in fact from 7 correspondences, as follows:

[[math]] \xymatrix@R=50pt@C=50pt{ S\ar[r]\ar[d]&T\ar[d]\ar[dl]\\ U\ar[u]\ar[r]&K\ar[u] } [[/math]]

In the twisted case, 6 of these correspondences hold as well, but the remaining one, namely [math]S\to T[/math], definitely does not hold as stated, and must be modified. Let us begin our discussion with the quantum isometry group results. We have here:

Theorem

We have the quantum isometry group formula

[[math]] \bar{U}=G^+(\bar{S}) [[/math]]

in all the [math]9[/math] main twisted cases.

Show Proof

The proofs here are similar to those from the untwisted case, via a routine computation, by adding signs where needed, which amounts in replacing the usual commutators [math][a,b]=ab-ba[/math] by twisted commutators, given by:

[[math]] [[a,b]]=ab+ba [[/math]]

There is one subtle point, however, coming from the fact that the linear independence of various products of coordinates of length 1,2,3, which was something clear in the untwisted case, is now a non-trivial question. But this can be solved via a technical application of the Weingarten formula, from Theorem 11.18. See .

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Regarding now the [math]K=G^+(T)\cap K_N^+[/math] axiom, this is something that we already know. However, regarding the correspondence [math]S\to T[/math], things here fail in the twisted case. Our “fix” for this, or at least the best fix that we could find, is as follows:

Theorem

Given an algebraic manifold [math]X\subset S^{N-1}_{\mathbb C,+}[/math], define its toral isometry group as being the biggest subgroup of [math]\mathbb T_N^+[/math] acting affinely on [math]X[/math]:

[[math]] \mathcal G^+(X)=G^+(X)\cap\mathbb T_N^+ [[/math]]

With this convention, for the [math]9[/math] basic spheres [math]S[/math], and for their twists as well, the toral isometry group equals the torus [math]T[/math].

Show Proof

We recall from chapter 3 that the affine quantum isometry group [math]G^+(X)\subset U_N^+[/math] of a noncommutative manifold [math]X\subset S^{N-1}_{\mathbb C,+}[/math] coming from certain polynomial relations [math]P[/math] is constructed according to the following procedure:

[[math]] P(x_i)=0\implies P\left(\sum_jx_j\otimes u_{ji}\right)=0 [[/math]]

Similarly, the toral isometry group [math]\mathcal G^+(X)\subset\mathbb T_N^+[/math] is constructed as follows:

[[math]] P(x_i)=0\implies P\left(x_i\otimes u_i\right)=0 [[/math]]

In the easy case one can prove that the following formula holds:

[[math]] G^+(\bar{S})=\overline{G^+(S)} [[/math]]

By intersecting with [math]\mathbb T_N^+[/math], we obtain from this that we have:

[[math]] \mathcal G^+(\bar{S})=\mathcal G^+(S) [[/math]]

The result can be of course be proved as well directly. For [math]\bar{S}^{N-1}_\mathbb R[/math] we have:

[[math]] \Phi(x_ix_j)=x_ix_j\otimes u_iu_j [[/math]]

[[math]] \Phi(x_jx_i)=x_jx_i\otimes u_ju_i [[/math]]

Thus we obtain [math]u_iu_j=-u_ju_i[/math] for [math]i\neq j[/math], and so the quantum group is [math]T_N[/math]. The proof in the complex, half-liberated and hybrid cases is similar.

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Regarding the hard liberation axiom, this seems to hold indeed in all the cases under consideration, but this is non-trivial, and not known yet. As a conclusion, we conjecturally have an extension of our [math](S,T,U,K)[/math] formalism, with the [math]S\to T[/math] axiom needing a modification as above, which covers the twisted objects [math](\bar{S},T,\bar{U},K)[/math] as well.

General references

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