guide:A7bd00c3c4

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

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Following , let us discuss now some more subtle examples of stationary models, related to the Pauli matrices, and Weyl matrices, and physics. We first have:

Definition

Given a finite abelian group [math]H[/math], the associated Weyl matrices are

[[math]] W_{ia}:e_b\to \lt i,b \gt e_{a+b} [[/math]]

where [math]i\in H[/math], [math]a,b\in\widehat{H}[/math], and where [math](i,b)\to \lt i,b \gt [/math] is the Fourier coupling [math]H\times\widehat{H}\to\mathbb T[/math].

As a basic example, consider the simplest cyclic group, namely:

[[math]] H=\mathbb Z_2=\{0,1\} [[/math]]

Here the Fourier coupling is [math] \lt i,b \gt =(-1)^{ib}[/math], and the Weyl matrices act as follows:

[[math]] W_{00}:e_b\to e_b\qquad,\qquad W_{10}:e_b\to(-1)^be_b [[/math]]

[[math]] W_{11}:e_b\to(-1)^be_{b+1}\qquad,\qquad W_{01}:e_b\to e_{b+1} [[/math]]

Thus, we have the following formulae for the Weyl matrices:

[[math]] W_{00}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\quad,\quad W_{10}=\begin{pmatrix}1&0\\0&-1\end{pmatrix} [[/math]]

[[math]] W_{11}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\quad,\quad W_{01}=\begin{pmatrix}0&1\\1&0\end{pmatrix} [[/math]]

We recognize here, up to some multiplicative factors, the four Pauli matrices. Now back to the general case, we have the following well-known result:

Proposition

The Weyl matrices are unitaries, and satisfy:

• [math]W_{ia}^*= \lt i,a \gt W_{-i,-a}[/math].
• [math]W_{ia}W_{jb}= \lt i,b \gt W_{i+j,a+b}[/math].
• [math]W_{ia}W_{jb}^*= \lt j-i,b \gt W_{i-j,a-b}[/math].
• [math]W_{ia}^*W_{jb}= \lt i,a-b \gt W_{j-i,b-a}[/math].

Show Proof

The unitary follows from (3,4), and the rest of the proof goes as follows:

(1) We have indeed the following computation:

[[math]] \begin{eqnarray*} W_{ia}^* &=&\left(\sum_b \lt i,b \gt E_{a+b,b}\right)^*\\ &=&\sum_b \lt -i,b \gt E_{b,a+b}\\ &=&\sum_b \lt -i,b-a \gt E_{b-a,b}\\ &=& \lt i,a \gt W_{-i,-a} \end{eqnarray*} [[/math]]

(2) Here the verification goes as follows:

[[math]] \begin{eqnarray*} W_{ia}W_{jb} &=&\left(\sum_d \lt i,b+d \gt E_{a+b+d,b+d}\right)\left(\sum_d \lt j,d \gt E_{b+d,d}\right)\\ &=&\sum_d \lt i,b \gt \lt i+j,d \gt E_{a+b+d,d}\\ &=& \lt i,b \gt W_{i+j,a+b} \end{eqnarray*} [[/math]]

(3,4) By combining the above two formulae, we obtain:

[[math]] \begin{eqnarray*} W_{ia}W_{jb}^* &=& \lt j,b \gt W_{ia}W_{-j,-b}\\ &=& \lt j,b \gt \lt i,-b \gt W_{i-j,a-b} \end{eqnarray*} [[/math]]

We obtain as well the following formula:

[[math]] \begin{eqnarray*} W_{ia}^*W_{jb} &=& \lt i,a \gt W_{-i,-a}W_{jb}\\ &=& \lt i,a \gt \lt -i,b \gt W_{j-i,b-a} \end{eqnarray*} [[/math]]

But this gives the formulae in the statement, and we are done.

■

Observe that, with [math]n=|H|[/math], we can use an isomorphism [math]l^2(\widehat{H})\simeq\mathbb C^n[/math] as to view each [math]W_{ia}[/math] as a usual matrix, [math]W_{ia}\in M_n(\mathbb C)[/math], and hence as a usual unitary, [math]W_{ia}\in U_n[/math].

Given a vector [math]\xi[/math], we denote by [math]Proj(\xi)[/math] the orthogonal projection onto [math]\mathbb C\xi[/math]. We have:

Proposition

Given a closed subgroup [math]E\subset U_n[/math], we have a representation

[[math]] \pi_H:C(S_N^+)\to M_N(C(E)) [[/math]]

[[math]] w_{ia,jb}\to[U\to Proj(W_{ia}UW_{jb}^*)] [[/math]]

where [math]n=|H|,N=n^2[/math], and where [math]W_{ia}[/math] are the Weyl matrices associated to [math]H[/math].

Show Proof

The Weyl matrices being given by [math]W_{ia}:e_b\to \lt i,b \gt e_{a+b}[/math], we have:

[[math]] tr(W_{ia})=\begin{cases} 1&{\rm if}\ (i,a)=(0,0)\\ 0&{\rm if}\ (i,a)\neq(0,0) \end{cases} [[/math]]

Together with the formulae in Proposition 16.14, this shows that the Weyl matrices are pairwise orthogonal with respect to the following scalar product on [math]M_n(\mathbb C)[/math]:

[[math]] \lt x,y \gt =tr(x^*y) [[/math]]

Thus, these matrices form an orthogonal basis of [math]M_n(\mathbb C)[/math], consisting of unitaries:

[[math]] W=\left\{W_{ia}\Big|i\in H,a\in\widehat{H}\right\} [[/math]]

Thus, each row and each column of the matrix [math]\xi_{ia,jb}=W_{ia}UW_{jb}^*[/math] is an orthogonal basis of [math]M_n(\mathbb C)[/math], and so the corresponding projections form a magic unitary, as claimed.

■

We will need the following well-known result:

Proposition

With [math]T=Proj(x_1)\ldots Proj(x_p)[/math] and [math]||x_i||=1[/math] we have

[[math]] \lt T\xi,\eta \gt = \lt \xi,x_p \gt \lt x_p,x_{p-1} \gt \ldots \lt x_2,x_1 \gt \lt x_1,\eta \gt [[/math]]

for any [math]\xi,\eta[/math]. In particular, we have:

[[math]] Tr(T)= \lt x_1,x_p \gt \lt x_p,x_{p-1} \gt \ldots \lt x_2,x_1 \gt [[/math]]

Show Proof

For [math]||x||=1[/math] we have [math]Proj(x)\xi= \lt \xi,x \gt x[/math]. This gives:

[[math]] \begin{eqnarray*} T\xi &=&Proj(x_1)\ldots Proj(x_p)\xi\\ &=&Proj(x_1)\ldots Proj(x_{p-1}) \lt \xi,x_p \gt x_p\\ &=&Proj(x_1)\ldots Proj(x_{p-2}) \lt \xi,x_p \gt \lt x_p,x_{p-1} \gt x_{p-1}\\ &=&\ldots\\ &=& \lt \xi,x_p \gt \lt x_p,x_{p-1} \gt \ldots \lt x_2,x_1 \gt x_1 \end{eqnarray*} [[/math]]

Now by taking the scalar product with [math]\eta[/math], this gives the first assertion. As for the second assertion, this follows from the first assertion, by summing over [math]\xi=\eta=e_i[/math].

■

Now back to the Weyl matrix models, let us first compute [math]T_p[/math]. We have:

Proposition

We have the formula

[[math]] \begin{eqnarray*} &&(T_p)_{ia,jb}\\ &=&\frac{1}{N} \lt i_1,a_1-a_p \gt \ldots \lt i_p,a_p-a_{p-1} \gt \lt j_1,b_1-b_2 \gt \ldots \lt j_p,b_p-b_1 \gt \\ &&\int_Etr(W_{i_1-i_2,a_1-a_2}UW_{j_2-j_1,b_2-b_1}U^*)\ldots tr(W_{i_p-i_1,a_p-a_1}UW_{j_1-j_p,b_1-b_p}U^*)dU \end{eqnarray*} [[/math]]

with all the indices varying in a cyclic way.

Show Proof

By using the trace formula in Proposition 16.16 above, we obtain:

[[math]] \begin{eqnarray*} &&(T_p)_{ia,jb}\\ &=&\left(tr\otimes\int_E\right)\left(Proj(W_{i_1a_1}UW_{j_1b_1}^*)\ldots Proj(W_{i_pa_p}UW_{j_pb_p}^*)\right)\\ &=&\frac{1}{N}\int_E \lt W_{i_1a_1}UW_{j_1b_1}^*,W_{i_pa_p}UW_{j_pb_p}^* \gt \ldots \lt W_{i_2a_2}UW_{j_2b_2}^*,W_{i_1a_1}UW_{j_1b_1}^* \gt dU \end{eqnarray*} [[/math]]

In order to compute now the scalar products, observe that we have:

[[math]] \begin{eqnarray*} \lt W_{ia}UW_{jb}^*,W_{kc}UW_{ld}^* \gt &=&tr(W_{jb}U^*W_{ia}^*W_{kc}UW_{ld}^*)\\ &=&tr(W_{ia}^*W_{kc}UW_{ld}^*W_{jb}U^*)\\ &=& \lt i,a-c \gt \lt l,d-b \gt tr(W_{k-i,c-a}UW_{j-l,b-d}U^*) \end{eqnarray*} [[/math]]

By plugging these quantities into the formula of [math]T_p[/math], we obtain the result.

■

Consider now the Weyl group [math]W=\{W_{ia}\}\subset U_n[/math], that we already met in the proof of Proposition 16.15 above. We have the following result, from :

Theorem

For any compact group [math]W\subset E\subset U_n[/math], the model

[[math]] \pi_H:C(S_N^+)\to M_N(C(E)) [[/math]]

[[math]] w_{ia,jb}\to[U\to Proj(W_{ia}UW_{jb}^*)] [[/math]]

constructed above is stationary on its image.

Show Proof

We must prove that we have [math]T_p^2=T_p[/math]. We have:

[[math]] \begin{eqnarray*} &&(T_p^2)_{ia,jb}\\ &=&\sum_{kc}(T_p)_{ia,kc}(T_p)_{kc,jb}\\ &=&\frac{1}{N^2}\sum_{kc} \lt i_1,a_1-a_p \gt \ldots \lt i_p,a_p-a_{p-1} \gt \lt k_1,c_1-c_2 \gt \ldots \lt k_p,c_p-c_1 \gt \\ && \lt k_1,c_1-c_p \gt \ldots \lt k_p,c_p-c_{p-1} \gt \lt j_1,b_1-b_2 \gt \ldots \lt j_p,b_p-b_1 \gt \\ &&\int_Etr(W_{i_1-i_2,a_1-a_2}UW_{k_2-k_1,c_2-c_1}U^*)\ldots tr(W_{i_p-i_1,a_p-a_1}UW_{k_1-k_p,c_1-c_p}U^*)dU\\ &&\int_Etr(W_{k_1-k_2,c_1-c_2}VW_{j_2-j_1,b_2-b_1}V^*)\ldots tr(W_{k_p-k_1,c_p-c_1}VW_{j_1-j_p,b_1-b_p}V^*)dV \end{eqnarray*} [[/math]]

By rearranging the terms, this formula becomes:

[[math]] \begin{eqnarray*} &&(T_p^2)_{ia,jb}\\ &=&\frac{1}{N^2} \lt i_1,a_1-a_p \gt \ldots \lt i_p,a_p-a_{p-1} \gt \lt j_1,b_1-b_2 \gt \ldots \lt j_p,b_p-b_1 \gt \\ &&\int_E\int_E\sum_{kc} \lt k_1-k_p,c_1-c_p \gt \ldots \lt k_p-k_{p-1},c_p-c_{p-1} \gt \\ &&tr(W_{i_1-i_2,a_1-a_2}UW_{k_2-k_1,c_2-c_1}U^*)tr(W_{k_1-k_2,c_1-c_2}VW_{j_2-j_1,b_2-b_1}V^*)\\ &&\hskip50mm\ldots\ldots\\ &&tr(W_{i_p-i_1,a_p-a_1}UW_{k_1-k_p,c_1-c_p}U^*)tr(W_{k_p-k_1,c_p-c_1}VW_{j_1-j_p,b_1-b_p}V^*)dUdV \end{eqnarray*} [[/math]]

Let us denote by [math]I[/math] the above double integral. By using [math]W_{kc}^*= \lt k,c \gt W_{-k,-c}[/math] for each of the couplings, and by moving as well all the [math]U^*[/math] variables to the left, we obtain:

[[math]] \begin{eqnarray*} I &=&\int_E\int_E\sum_{kc}tr(U^*W_{i_1-i_2,a_1-a_2}UW_{k_2-k_1,c_2-c_1})tr(W_{k_2-k_1,c_2-c_1}^*VW_{j_2-j_1,b_2-b_1}V^*)\\ &&\hskip50mm\ldots\ldots\\ &&tr(U^*W_{i_p-i_1,a_p-a_1}UW_{k_1-k_p,c_1-c_p})tr(W_{k_1-k_p,c_1-c_p}^*VW_{j_1-j_p,b_1-b_p}V^*)dUdV \end{eqnarray*} [[/math]]

In order to perform now the sums, we use the following formula:

[[math]] \begin{eqnarray*} tr(AW_{kc})tr(W_{kc}^*B) &=&\frac{1}{N}\sum_{qrst}A_{qr}(W_{kc})_{rq}(W^*_{kc})_{st}B_{ts}\\ &=&\frac{1}{N}\sum_{qrst}A_{qr} \lt k,q \gt \delta_{r-q,c} \lt k,-s \gt \delta_{t-s,c}B_{ts}\\ &=&\frac{1}{N}\sum_{qs} \lt k,q-s \gt A_{q,q+c}B_{s+c,s} \end{eqnarray*} [[/math]]

If we denote by [math]A_x,B_x[/math] the variables which appear in the formula of [math]I[/math], we have:

[[math]] \begin{eqnarray*} &&I\\ &=&\frac{1}{N^p}\int_E\int_E\sum_{kcqs} \lt k_2-k_1,q_1-s_1 \gt \ldots \lt k_1-k_p,q_p-s_p \gt \\ &&(A_1)_{q_1,q_1+c_2-c_1}(B_1)_{s_1+c_2-c_1,s_1}\ldots (A_p)_{q_p,q_p+c_1-c_p}(B_p)_{s_p+c_1-c_p,s_p}\\ &=&\frac{1}{N^p}\int_E\int_E\sum_{kcqs} \lt k_1,q_p-s_p-q_1+s_1 \gt \ldots \lt k_p,q_{p-1}-s_{p-1}-q_p+s_p \gt \\ &&(A_1)_{q_1,q_1+c_2-c_1}(B_1)_{s_1+c_2-c_1,s_1}\ldots (A_p)_{q_p,q_p+c_1-c_p}(B_p)_{s_p+c_1-c_p,s_p} \end{eqnarray*} [[/math]]

Now observe that we can perform the sums over [math]k_1,\ldots,k_p[/math]. We obtain in this way a multiplicative factor [math]n^p[/math], along with the condition:

[[math]] q_1-s_1=\ldots=q_p-s_p [[/math]]

Thus we must have [math]q_x=s_x+a[/math] for a certain [math]a[/math], and the above formula becomes:

[[math]] I=\frac{1}{n^p}\int_E\int_E\sum_{csa}(A_1)_{s_1+a,s_1+c_2-c_1+a}(B_1)_{s_1+c_2-c_1,s_1}\ldots(A_p)_{s_p+a,s_p+c_1-c_p+a}(B_p)_{s_p+c_1-c_p,s_p} [[/math]]

Consider now the variables [math]r_x=c_{x+1}-c_x[/math], which altogether range over the set [math]Z[/math] of multi-indices having sum 0. By replacing the sum over [math]c_x[/math] with the sum over [math]r_x[/math], which creates a multiplicative [math]n[/math] factor, we obtain the following formula:

[[math]] I=\frac{1}{n^{p-1}}\int_E\int_E\sum_{r\in Z}\sum_{sa}(A_1)_{s_1+a,s_1+r_1+a}(B_1)_{s_1+r_1,s_1}\ldots(A_p)_{s_p+a,s_p+r_p+a}(B_p)_{s_p+r_p,s_p} [[/math]]

For an arbitrary multi-index [math]r[/math] we have:

[[math]] \delta_{\sum_ir_i,0}=\frac{1}{n}\sum_i \lt i,r_1 \gt \ldots \lt i,r_p \gt [[/math]]

Thus, we can replace the sum over [math]r\in Z[/math] by a full sum, as follows:

[[math]] \begin{eqnarray*} I &=&\frac{1}{n^p}\int_E\int_E\sum_{rsia} \lt i,r_1 \gt (A_1)_{s_1+a,s_1+r_1+a}(B_1)_{s_1+r_1,s_1}\\ &&\hskip40mm\ldots\ldots\\ &&\hskip20mm \lt i,r_p \gt (A_p)_{s_p+a,s_p+r_p+a}(B_p)_{s_p+r_p,s_p} \end{eqnarray*} [[/math]]

In order to “absorb” now the indices [math]i,a[/math], we can use the following formula:

[[math]] \begin{eqnarray*} &&W_{ia}^*AW_{ia}\\ &=&\left(\sum_b \lt i,-b \gt E_{b,a+b}\right)\left(\sum_{bc}E_{a+b,a+c}A_{a+b,a+c}\right)\left(\sum_c \lt i,c \gt E_{a+c,c}\right)\\ &=&\sum_{bc} \lt i,c-b \gt E_{bc}A_{a+b,a+c} \end{eqnarray*} [[/math]]

Thus we have:

[[math]] (W_{ia}^*AW_{ia})_{bc}= \lt i,c-b \gt A_{a+b,a+c} [[/math]]

Our formula becomes:

[[math]] \begin{eqnarray*} &&I\\ &=&\frac{1}{n^p}\int_E\int_E\sum_{rsia}(W_{ia}^*A_1W_{ia})_{s_1,s_1+r_1}(B_1)_{s_1+r_1,s_1}\ldots(W_{ia}^*A_pW_{ia})_{s_p,s_p+r_p}(B_p)_{s_p+r_p,s_p}\\ &=&\int_E\int_E\sum_{ia}tr(W_{ia}^*A_1W_{ia}B_1)\ldots\ldots tr(W_{ia}^*A_pW_{ia}B_p) \end{eqnarray*} [[/math]]

Now by replacing [math]A_x,B_x[/math] with their respective values, we obtain:

[[math]] \begin{eqnarray*} I &=&\int_E\int_E\sum_{ia}tr(W_{ia}^*U^*W_{i_1-i_2,a_1-a_2}UW_{ia}VW_{j_2-j_1,b_2-b_1}V^*)\\ &&\hskip30mm\ldots\ldots\\ &&tr(W_{ia}^*U^*W_{i_p-i_1,a_p-a_1}UW_{ia}VW_{j_1-j_p,b_1-b_p}V^*)dUdV \end{eqnarray*} [[/math]]

By moving the [math]W_{ia}^*U^*[/math] variables at right, we obtain, with [math]S_{ia}=UW_{ia}V[/math]:

[[math]] \begin{eqnarray*} I &=&\sum_{ia}\int_E\int_Etr(W_{i_1-i_2,a_1-a_2}S_{ia}W_{j_2-j_1,b_2-b_1}S_{ia}^*)\\ &&\hskip30mm\ldots\ldots\\ &&tr(W_{i_p-i_1,a_p-a_1}S_{ia}W_{j_1-j_p,b_1-b_p}S_{ia}^*)dUdV \end{eqnarray*} [[/math]]

Now since [math]S_{ia}[/math] is Haar distributed when [math]U,V[/math] are Haar distributed, we obtain:

[[math]] I=N\int_E\int_Etr(W_{i_1-i_2,a_1-a_2}UW_{j_2-j_1,b_2-b_1}U^*)\ldots tr(W_{i_p-i_1,a_p-a_1}UW_{j_1-j_p,b_1-b_p}U^*)dU [[/math]]

But this is exactly [math]N[/math] times the integral in the formula of [math](T_p)_{ia,jb}[/math], from Proposition 16.17 above. Since the [math]N[/math] factor cancels with one of the two [math]N[/math] factors that we found in the beginning of the proof, when first computing [math](T_p^2)_{ia,jb}[/math], we are done.

■

The above computation was of course quite tricky, and there are several possible generalizations of it, and some open questions as well, of both algebraic and analytic nature, all quite interesting. We refer to are related papers for more on all this.

As an illustration for the above result, which is something known for a long time, and quite fundamental, going back to the paper , we have:

Theorem

We have a stationary matrix model

[[math]] \pi:C(S_4^+)\subset M_4(C(SU_2)) [[/math]]

given on the standard coordinates by the formula

[[math]] \pi(u_{ij})=[x\to Proj(c_ixc_j)] [[/math]]

where [math]x\in SU_2[/math], and [math]c_1,c_2,c_3,c_4[/math] are the Pauli matrices.

Show Proof

As already explained in the comments following Definition 16.13, the Pauli matrices appear as particular cases of the Weyl matrices. To be more precise, consider the group [math]H=\mathbb Z_2=\{0,1\}[/math], with standard Fourier coupling, as follows:

[[math]] \lt i,b \gt =(-1)^{ib} [[/math]]

The Weyl matrices, as defined in the above, act then as follows:

[[math]] W_{00}:e_b\to e_b\qquad,\qquad W_{10}:e_b\to(-1)^be_b [[/math]]

[[math]] W_{11}:e_b\to(-1)^be_{b+1}\qquad,\qquad W_{01}:e_b\to e_{b+1} [[/math]]

Thus, we have the following formulae for the Weyl matrices:

[[math]] W_{00}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\quad,\quad W_{10}=\begin{pmatrix}1&0\\0&-1\end{pmatrix} [[/math]]

[[math]] W_{11}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\quad,\quad W_{01}=\begin{pmatrix}0&1\\1&0\end{pmatrix} [[/math]]

We recognize here, up to some multiplicative factors, the four Pauli matrices. By working out now the details of the various constructions above, we conclude that Theorem 16.18 produces in this case the model in the statement.

■

Observe that, since the matrix [math]Proj(c_ixc_j)[/math] depends only on the image of [math]x[/math] in the quotient group [math]SU_2\to SO_3[/math], we can replace the model space [math]SU_2[/math] by the smaller space [math]SO_3[/math], if we want to, and so we have a matrix model as follows:

[[math]] \pi:C(S_4^+)\subset M_4(C(SO_3)) [[/math]]

This is something that can be used in conjunction with the isomorphism [math]S_4^+\simeq SO_3^{-1}[/math] from chapter 9 above, and as explained in , our model becomes in this way something quite conceptual, algebrically speaking, as follows:

[[math]] \pi:C(SO_3^{-1})\subset M_4(C(SO_3)) [[/math]]

In general, going beyond stationarity is a difficult task, and among the results here, let us mention the universal modelling questions for quantum permutations and quantum reflections, and various results on the flat models for the discrete groups, from , and related papers, questions regarding the Hadamard matrix models , and the related fine analytic study on the compact and discrete quantum groups , .

General references

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