guide:203101203b

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In connection with our axiomatization questions for the quadruplets [math](S,T,U,K)[/math], we can construct now the correspondences [math]S\to U[/math], in the following way:

Theorem

The quantum isometry groups of the basic spheres are

[[math]] \xymatrix@R=15mm@C=14mm{ S^{N-1}_{\mathbb R,+}\ar[r]&S^{N-1}_{\mathbb C,+}\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_\mathbb C\ar[u] } \qquad \item[a]ymatrix@R=8mm@C=15mm{\\ \to} \qquad \item[a]ymatrix@R=16mm@C=18mm{ O_N^+\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&U_N\ar[u]} [[/math]]

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

Show Proof

We have 4 results to be proved, and following , and related papers, where this result was established in its above form, we can proceed as follows:

\underline{[math]S^{N-1}_{\mathbb C,+}[/math]}. 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]]

By using the biunitarity of [math]u[/math], we have the following computation:

[[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]]

Once again by using the biunitarity of [math]u[/math], we have as well:

[[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]]

Thus we have an action [math]U_N^+\curvearrowright S^{N-1}_{\mathbb C,+}[/math], which gives [math]G^+(S^{N-1}_{\mathbb C,+})=U_N^+[/math], as desired.

\underline{[math]S^{N-1}_{\mathbb R,+}[/math]}. Let us first construct an action [math]O_N^+\curvearrowright S^{N-1}_{\mathbb R,+}[/math]. We already know 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], so we just have to check that these variables are self-adjoint. But this is clear from [math]u=\bar{u}[/math], as follows:

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

Conversely, assume that we have an action [math]G\curvearrowright S^{N-1}_{\mathbb R,+}[/math], with [math]G\subset U_N^+[/math]. The variables [math]X_i=\sum_jx_j\otimes u_{ji}[/math] must be then self-adjoint, and the above computation shows that we must have [math]u=\bar{u}[/math]. Thus our quantum group must satisfy [math]G\subset O_N^+[/math], as desired.

\underline{[math]S^{N-1}_\mathbb C[/math]}. The fact that we have an action [math]U_N\curvearrowright S^{N-1}_\mathbb C[/math] is clear. Conversely, assume that we have an action [math]G\curvearrowright S^{N-1}_\mathbb C[/math], with [math]G\subset U_N^+[/math]. We must prove that this implies [math]G\subset U_N[/math], and we will use a standard trick of Bhowmick-Goswami . We have:

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

By multiplying this formula with itself we obtain:

[[math]] \Phi(x_ix_k)=\sum_{jl}x_jx_l\otimes u_{ji}u_{lk} [[/math]]

[[math]] \Phi(x_kx_i)=\sum_{jl}x_lx_j\otimes u_{lk}u_{ji} [[/math]]

Since the variables [math]x_i[/math] commute, these formulae can be written as:

[[math]] \Phi(x_ix_k)=\sum_{j \lt l}x_jx_l\otimes(u_{ji}u_{lk}+u_{li}u_{jk})+\sum_jx_j^2\otimes u_{ji}u_{jk} [[/math]]

[[math]] \Phi(x_ix_k)=\sum_{j \lt l}x_jx_l\otimes(u_{lk}u_{ji}+u_{jk}u_{li})+\sum_jx_j^2\otimes u_{jk}u_{ji} [[/math]]

Since the tensors at left are linearly independent, we must have:

[[math]] u_{ji}u_{lk}+u_{li}u_{jk}=u_{lk}u_{ji}+u_{jk}u_{li} [[/math]]

By applying the antipode to this formula, then applying the involution, and then relabelling the indices, we succesively obtain:

[[math]] u_{kl}^*u_{ij}^*+u_{kj}^*u_{il}^*=u_{ij}^*u_{kl}^*+u_{il}^*u_{kj}^* [[/math]]

[[math]] u_{ij}u_{kl}+u_{il}u_{kj}=u_{kl}u_{ij}+u_{kj}u_{il} [[/math]]

[[math]] u_{ji}u_{lk}+u_{jk}u_{li}=u_{lk}u_{ji}+u_{li}u_{jk} [[/math]]

Now by comparing with the original formula, we obtain from this:

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

In order to finish, it remains to prove that the coordinates [math]u_{ij}[/math] commute as well with their adjoints. For this purpose, we use a similar method. We have:

[[math]] \Phi(x_ix_k^*)=\sum_{jl}x_jx_l^*\otimes u_{ji}u_{lk}^* [[/math]]

[[math]] \Phi(x_k^*x_i)=\sum_{jl}x_l^*x_j\otimes u_{lk}^*u_{ji} [[/math]]

Since the variables on the left are equal, we deduce from this that we have:

[[math]] \sum_{jl}x_jx_l^*\otimes u_{ji}u_{lk}^*=\sum_{jl}x_jx_l^*\otimes u_{lk}^*u_{ji} [[/math]]

Thus we have [math]u_{ji}u_{lk}^*=u_{lk}^*u_{ji}[/math], and so [math]G\subset U_N[/math], as claimed.

\underline{[math]S^{N-1}_\mathbb R[/math]}. The fact that we have an action [math]O_N\curvearrowright S^{N-1}_\mathbb R[/math] is clear. In what regards the converse, this follows by combining the results that we already have, as follows:

[[math]] \begin{eqnarray*} G\curvearrowright S^{N-1}_\mathbb R &\implies&G\curvearrowright S^{N-1}_{\mathbb R,+},S^{N-1}_\mathbb C\\ &\implies&G\subset O_N^+,U_N\\ &\implies&G\subset O_N^+\cap U_N=O_N \end{eqnarray*} [[/math]]

Thus, we conclude that we have [math]G^+(S^{N-1}_\mathbb R)=O_N[/math], as desired.

■

Let us discuss now the construction [math]U\to S[/math]. In the classical case the situation is very simple, because the sphere [math]S=S^{N-1}[/math] appears by rotating the point [math]x=(1,0,\ldots,0)[/math] by the isometries in [math]U=U_N[/math]. Moreover, the stabilizer of this action is the subgroup [math]U_{N-1}\subset U_N[/math] acting on the last [math]N-1[/math] coordinates, and so the sphere [math]S=S^{N-1}[/math] appears from the corresponding rotation group [math]U=U_N[/math] as an homogeneous space, as follows:

[[math]] S^{N-1}=U_N/U_{N-1} [[/math]]

In functional analytic terms, all this becomes even simpler, the correspondence [math]U\to S[/math] being obtained, at the level of algebras of functions, as follows:

[[math]] C(S^{N-1})\subset C(U_N)\quad,\quad x_i\to u_{1i} [[/math]]

In general now, the homogeneous space interpretation of [math]S[/math] as above fails, due to a number of subtle algebraic and analytic reasons, explained in and related papers. However, we can have some theory going by using the functional analytic viewpoint, with an embedding [math]x_i\to u_{1i}[/math] as above. Let us start with the following observation:

Proposition

For the basic spheres, we have a diagram as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ C(S)\ar[r]^\Phi\ar[d]^\alpha&C(S)\otimes C(U)\ar[d]^{\alpha\otimes id}\\ C(U)\ar[r]^\Delta&C(U)\otimes C(U) } [[/math]]

where the map on top is the affine coaction map,

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

and the map on the left is given by [math]\alpha(x_i)=u_{1i}[/math].

Show Proof

The diagram in the statement commutes indeed on the standard coordinates, the corresponding arrows being as follows, on these coordinates:

[[math]] \xymatrix@R=50pt@C=50pt{ x_i\ar[r]\ar[d]&\sum_jx_j\otimes u_{ji}\ar[d]\\ u_{1i}\ar[r]&\sum_ju_{1j}\otimes u_{ji} } [[/math]]

Thus by linearity and multiplicativity, the whole the diagram commutes.

■

We therefore have the following result:

Theorem

We have a quotient map and an inclusion as follows,

[[math]] U\to S_U\subset S [[/math]]

with [math]S_U[/math] being the first row space of [math]U[/math], given by

[[math]] C(S_U)= \lt u_{1i} \gt \subset C(U) [[/math]]

at the level of the corresponding algebras of functions.

Show Proof

At the algebra level, we have an inclusion and a quotient map as follows:

[[math]] C(S)\to C(S_U)\subset C(U) [[/math]]

Thus, we obtain the result, by transposing.

■

We will prove in what follows that the inclusion [math]S_U\subset S[/math] constructed above is an isomorphism. This will produce the correspondence [math]U\to S[/math] that we are currently looking for. In order to do so, we will use the uniform integration over [math]S[/math], which can be introduced, in analogy with what happens in the classical case, in the following way:

Definition

We endow each of the algebras [math]C(S)[/math] with its integration functional

[[math]] \int_S:C(S)\to C(U)\to\mathbb C [[/math]]

obtained by composing the morphism of algebras given by

[[math]] x_i\to u_{1i} [[/math]]

with the Haar integration functional of the algebra [math]C(U)[/math].

In order to efficiently integrate over the sphere [math]S[/math], and in the lack of some trick like spherical coordinates, we need to know how to efficiently integrate over the corresponding quantum isometry group [math]U[/math]. There is a long story here, going back to the papers of Weingarten , then Collins-\'Sniady in the classical case, and to the more recent papers , , and then , in the quantum group case. Following , we have:

Theorem

Assuming that a compact quantum group [math]G\subset U_N^+[/math] is easy, coming from a category of partitions [math]D\subset P[/math], we have the Weingarten formula

[[math]] \int_Gu_{i_1j_1}^{e_1}\ldots u_{i_kj_k}^{e_k}=\sum_{\pi,\sigma\in D(k)}\delta_\pi(i)\delta_\sigma(j)W_{kN}(\pi,\sigma) [[/math]]

for any indices [math]i_r,j_r\in\{1,\ldots,N\}[/math] and any exponents [math]e_r\in\''ptyset,*\''[/math], where [math]\delta[/math] are the usual Kronecker type symbols, and where

[[math]] W_{kN}=G_{kN}^{-1} [[/math]]

is the inverse of the matrix [math]G_{kN}(\pi,\sigma)=N^{|\pi\vee\sigma|}[/math].

Show Proof

Let us arrange indeed all the integrals to be computed, at a fixed value of the exponent [math]k=(e_1\ldots e_k)[/math], into a single matrix, of size [math]N^k\times N^k[/math], as follows:

[[math]] P_{i_1\ldots i_k,j_1\ldots j_k}=\int_Gu_{i_1j_1}^{e_1}\ldots u_{i_kj_k}^{e_k} [[/math]]

According to the construction of the Haar measure of Woronowicz , explained in chapter 2, this matrix [math]P[/math] is the orthogonal projection onto the following space:

[[math]] Fix(u^{\otimes k})=span\left(\xi_\pi\Big|\pi\in D(k)\right) [[/math]]

In order to compute this projection, consider the following linear map:

[[math]] E(x)=\sum_{\pi\in D(k)} \lt x,\xi_\pi \gt \xi_\pi [[/math]]

Consider as well the inverse [math]W[/math] of the restriction of [math]E[/math] to the following space:

[[math]] span\left(T_\pi\Big|\pi\in D(k)\right) [[/math]]

By a standard linear algebra computation, it follows that we have:

[[math]] P=WE [[/math]]

But the restriction of [math]E[/math] is the linear map corresponding to [math]G_{kN}[/math], so [math]W[/math] is the linear map corresponding to [math]W_{kN}[/math], and this gives the result. See .

■

Following , we can now integrate over the spheres [math]S[/math], as follows:

Proposition

The integration over the basic spheres is given by

[[math]] \int_Sx_{i_1}^{e_1}\ldots x_{i_k}^{e_k}=\sum_\pi\sum_{\sigma\leq\ker i}W_{kN}(\pi,\sigma) [[/math]]

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

Show Proof

According to our conventions, the integration over [math]S[/math] is a particular case of the integration over [math]U[/math], via [math]x_i=u_{1i}[/math]. By using the formula in Theorem 3.9, we obtain:

[[math]] \begin{eqnarray*} \int_Sx_{i_1}^{e_1}\ldots x_{i_k}^{e_k} &=&\int_Uu_{1i_1}^{e_1}\ldots u_{1i_k}^{e_k}\\ &=&\sum_{\pi,\sigma\in D(k)}\delta_\pi(1)\delta_\sigma(i)W_{kN}(\pi,\sigma)\\ &=&\sum_{\pi,\sigma\in D(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma) \end{eqnarray*} [[/math]]

Thus, we are led to the formula in the statement.

■

Again following , , we have the following key result:

Theorem

The integration functional of [math]S[/math] has the ergodicity property

[[math]] \left(id\otimes\int_U\right)\Phi(x)=\int_Sx [[/math]]

where [math]\Phi:C(S)\to C(S)\otimes C(U)[/math] is the universal affine coaction map.

Show Proof

In the real case, [math]x_i=x_i^*[/math], it is enough to check the equality in the statement on an arbitrary product of coordinates, [math]x_{i_1}\ldots x_{i_k}[/math]. The left term is as follows:

[[math]] \begin{eqnarray*} \left(id\otimes\int_U\right)\Phi(x_{i_1}\ldots x_{i_k}) &=&\sum_{j_1\ldots j_k}x_{j_1}\ldots x_{j_k}\int_Uu_{j_1i_1}\ldots u_{j_ki_k}\\ &=&\sum_{j_1\ldots j_k}\ \sum_{\pi,\sigma\in D(k)}\delta_\pi(j)\delta_\sigma(i)W_{kN}(\pi,\sigma)x_{j_1}\ldots x_{j_k}\\ &=&\sum_{\pi,\sigma\in D(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma)\sum_{j_1\ldots j_k}\delta_\pi(j)x_{j_1}\ldots x_{j_k} \end{eqnarray*} [[/math]]

Let us look now at the last sum on the right. The situation is as follows:

(1) In the free case we have to sum quantities of type [math]x_{j_1}\ldots x_{j_k}[/math], over all choices of multi-indices [math]j=(j_1,\ldots,j_k)[/math] which fit into our given noncrossing pairing [math]\pi[/math], and just by using the condition [math]\sum_ix_i^2=1[/math], we conclude that the sum is 1.

(2) The same happens in the classical case. Indeed, our pairing [math]\pi[/math] can now be crossing, but we can use the commutation relations [math]x_ix_j=x_jx_i[/math], and the sum is again 1.

Thus the sum on the right is 1, in all cases, and we obtain:

[[math]] \left(id\otimes\int_U\right)\Phi(x_{i_1}\ldots x_{i_k}) =\sum_{\pi,\sigma\in D(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma) [[/math]]

On the other hand, another application of the Weingarten formula gives:

[[math]] \begin{eqnarray*} \int_Sx_{i_1}\ldots x_{i_k} &=&\int_Uu_{1i_1}\ldots u_{1i_k}\\ &=&\sum_{\pi,\sigma\in D(k)}\delta_\pi(1)\delta_\sigma(i)W_{kN}(\pi,\sigma)\\ &=&\sum_{\pi,\sigma\in D(k)}\delta_\sigma(i)W_{kN}(\pi,\sigma) \end{eqnarray*} [[/math]]

Thus, we are done with the proof of the result, in the real case. In the complex case the proof is similar, by adding exponents everywhere. See , .

■

Still following , , we can now deduce a useful abstract characterization of the integration over the spheres, as follows:

Theorem

There is a unique positive unital trace [math]tr:C(S)\to\mathbb C[/math] satisfying

[[math]] (tr\otimes id)\Phi(x)=tr(x)1 [[/math]]

where [math]\Phi[/math] is the coaction map of the corresponding quantum isometry group,

[[math]] \Phi:C(S)\to C(S)\otimes C(U) [[/math]]

and this is the canonical integration, as constructed in Definition 3.8.

Show Proof

First of all, it follows from the Haar integral invariance condition for [math]U[/math] that the canonical integration has indeed the invariance property in the statement, namely:

[[math]] (tr\otimes id)\Phi(x)=tr(x)1 [[/math]]

In order to prove now the uniqueness, let [math]tr[/math] be as in the statement. We have:

[[math]] \begin{eqnarray*} tr\left(id\otimes\int_U\right)\Phi(x) &=&\int_U(tr\otimes id)\Phi(x)\\ &=&\int_U(tr(x)1)\\ &=&tr(x) \end{eqnarray*} [[/math]]

On the other hand, according to Theorem 3.11, we have as well:

[[math]] tr\left(id\otimes\int_U\right)\Phi(x) =tr\left(\int_Sx\right) =\int_Sx [[/math]]

We therefore conclude that [math]tr[/math] equals the standard integration, as claimed.

■

Getting back now to our axiomatization questions, we have:

Theorem

The operation [math]S\to S_U[/math] produces a correspondence as follows,

[[math]] \xymatrix@R=15mm@C=15mm{ S^{N-1}_{\mathbb R,+}\ar[r]&S^{N-1}_{\mathbb C,+}\\ S^{N-1}_\mathbb R\ar[r]\ar[u]&S^{N-1}_\mathbb C\ar[u]} \qquad \item[a]ymatrix@R=8mm@C=15mm{\\ \to} \qquad \item[a]ymatrix@R=17mm@C=16mm{ O_N^+\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&U_N\ar[u] } [[/math]]

between basic unitary groups and the basic noncommutative spheres.

Show Proof

We use the ergodicity formula from Theorem 3.11, namely:

[[math]] \left(id\otimes\int_U\right)\Phi=\int_S [[/math]]

We know that [math]\int_U[/math] is faithful on [math]\mathcal C(U)[/math], and that we have:

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

The coaction map [math]\Phi[/math] follows to be faithful as well. Thus for any [math]x\in\mathcal C(S)[/math] we have:

[[math]] \int_Sxx^*=0\implies x=0 [[/math]]

Thus [math]\int_S[/math] is faithful on [math]\mathcal C(S)[/math]. But this shows that we have:

[[math]] S=S_U [[/math]]

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

■

General references

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