10b. Reflection groups

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

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Summarizing, the problem of conceptually understanding [math]G(\square)[/math] remains open. In order to present now the correct, final solution, the idea will be that to look at the quantum group [math]G^+(|\ |)[/math] instead, which is equal to it, according to Proposition 10.2 (3). We first have the following result, extending Proposition 10.2 (4) above:

Proposition

For a disconnected union of graphs we have

[[math]] G^+(X_1)\;\hat{*}\;\ldots\;\hat{*}\;G^+(X_k)\subset G^+(X_1\sqcup\ldots\sqcup X_k) [[/math]]

and this inclusion is in general not an isomorphism.

Show Proof

The proof of the first assertion is nearly identical to the proof of Proposition 10.2 (4) above. Indeed, the adjacency matrix of the disconnected union is given by:

[[math]] d_{X_1\sqcup\ldots\sqcup X_k}=diag(d_{X_1},\ldots,d_{X_k}) [[/math]]

[[math]] w=diag(u_1,\ldots,u_k) [[/math]]

We have then [math]d_{X_i}u_i=u_id_{X_i}[/math], and this implies [math]dw=wd[/math], which gives the result. As for the last assertion, this is something that we already know, from Proposition 10.6 (2).

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In the case where the graphs [math]X_1,\ldots,X_k[/math] are identical, which is the one that we are truly interested in, we can further build on this. Following Bichon , we have:

Proposition

Given closed subgroups [math]G\subset U_N^+[/math], [math]H\subset S_k^+[/math], with fundamental corepresentations [math]u,v[/math], the following construction produces a closed subgroup of [math]U_{Nk}^+[/math]:

[[math]] C(G\wr_*H)=(C(G)^{*k}*C(H))/ \lt [u_{ij}^{(a)},v_{ab}]=0 \gt [[/math]]

In the case where [math]G,H[/math] are classical, the classical version of [math]G\wr_*H[/math] is the usual wreath product [math]G\wr H[/math]. Also, when [math]G[/math] is a quantum permutation group, so is [math]G\wr_*H[/math].

Show Proof

Consider indeed the matrix [math]w_{ia,jb}=u_{ij}^{(a)}v_{ab}[/math], over the quotient algebra in the statement. It is routine to check that [math]w[/math] is unitary, and in the case [math]G\subset S_N^+[/math], our claim is that this matrix is magic. Indeed, the entries are projections, because they appear as products of commuting projections, and the row sums are as follows:

[[math]] \begin{eqnarray*} \sum_{jb}w_{ia,jb} &=&\sum_{jb}u_{ij}^{(a)}v_{ab}\\ &=&\sum_bv_{ab}\sum_ju_{ij}^{(a)}\\ &=&1 \end{eqnarray*} [[/math]]

As for the column sums, these are as follows:

[[math]] \begin{eqnarray*} \sum_{ia}w_{ia,jb} &=&\sum_{ia}u_{ij}^{(a)}v_{ab}\\ &=&\sum_av_{ab}\sum_iu_{ij}^{(a)}\\ &=&1 \end{eqnarray*} [[/math]]

With these observations in hand, it is routine to check that [math]G\wr_*H[/math] is indeed a quantum group, with fundamental corepresentation [math]w[/math], by constructing maps [math]\Delta,\varepsilon,S[/math] as in section 1, and in the case [math]G\subset S_N^+[/math], we obtain in this way a closed subgroup of [math]S_{Nk}^+[/math]. Finally, the assertion regarding the classical version is standard as well. See .

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We refer to Bichon and to Tarrago-Wahl for more details regarding the above construction. Now with this notion in hand, we have the following result:

Theorem

Given a connected graph [math]X[/math], and [math]k\in\mathbb N[/math], we have the formulae

[[math]] G(kX)=G(X)\wr S_k [[/math]]

[[math]] G^+(kX)=G^+(X)\wr_*S_k^+ [[/math]]

where [math]kX=X\sqcup\ldots\sqcup X[/math] is the [math]k[/math]-fold disjoint union of [math]X[/math] with itself.

Show Proof

The first formula is something well-known, which follows as well from the second formula, by taking the classical version. Regarding now the second formula, it is elementary to check that we have an inclusion as follows, for any finite graph [math]X[/math]:

[[math]] G^+(X)\wr_*S_k^+\subset G^+(kX) [[/math]]

Indeed, we want to construct an action [math]G^+(X)\wr_*S_k^+\curvearrowright kX[/math], and this amounts in proving that we have [math][w,d]=0[/math]. But, the matrices [math]w,d[/math] are given by:

[[math]] w_{ia,jb}=u_{ij}^{(a)}v_{ab}\quad,\quad d_{ia,jb}=\delta_{ij}d_{ab} [[/math]]

With these formulae in hand, we have the following computation:

[[math]] \begin{eqnarray*} (dw)_{ia,jb} &=&\sum_kd_{ik}w_{ka,jb}\\ &=&\sum_kd_{ik}u_{kj}^{(a)}v_{ab}\\ &=&(du^{(a)})_{ij}v_{ab} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} (wd)_{ia,jb} &=&\sum_kw_{ia,kb}d_{kj}\\ &=&\sum_ku_{ik}^{(a)}v_{ab}d_{kj}\\ &=&(u^{(a)}d)_{ij}v_{ab} \end{eqnarray*} [[/math]]

Thus we have [math][w,d]=0[/math], and from this we obtain:

[[math]] G^+(X)\wr_*S_k^+\subset G^+(kX) [[/math]]

Regarding now the reverse inclusion, which requires [math]X[/math] to be connected, this follows by doing some matrix analysis, by using the commutation with [math]u[/math]. To be more precise, let us denote by [math]w[/math] the fundamental corepresentation of [math]G^+(kX)[/math], and set:

[[math]] u_{ij}^{(a)}=\sum_bw_{ia,jb}\quad,\quad v_{ab}=\sum_iv_{ab} [[/math]]

It is then routine to check, by using the fact that [math]X[/math] is indeed connected, that we have here magic unitaries, as in the definition of the free wreath products. Thus we obtain the reverse inclusion, that we were looking for, namely:

[[math]] G^+(kX)\subset G^+(X)\wr_*S_k^+ [[/math]]

To be more precise, the key ingredient is the fact that when [math]X[/math] is connected, the [math]*[/math]-algebra generated by [math]d_X[/math] contains a matrix having nonzero entries.

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We are led in this way to the following result, from :

Theorem

Consider the graph consisting of [math]N[/math] segments.

• Its symmetry group is the hyperoctahedral group [math]H_N=\mathbb Z_2\wr S_N[/math].
• Its quantum symmetry group is the quantum group [math]H_N^+=\mathbb Z_2\wr_*S_N^+[/math].

Show Proof

This comes from the above results, as follows:

(1) This is clear from definitions, with the remark that the relation with the formula [math]H_N=G(\square_N)[/math] comes by viewing the [math]N[/math] segments as being the [math][-1,1][/math] segments on each of the [math]N[/math] coordinate axes of [math]\mathbb R^N[/math]. Indeed, a symmetry of the [math]N[/math]-cube is the same as a symmetry of the [math]N[/math] segments, and so we obtain, as desired:

[[math]] G(\square_N)=\mathbb Z_2\wr S_N [[/math]]

(2) This follows from Theorem 10.11 above, applied to the segment graph. Observe also that (2) implies (1), by taking the classical version.

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Now back to the square, we have [math]G^+(\square)=H_2^+[/math], and our claim is that this is the “good” and final formula. In order to prove this, we must work out the easiness theory for [math]H_N,H_N^+[/math], and prove that [math]H_N\to H_N^+[/math] is an easy quantum group liberation.

Following , we first have the following result:

Proposition

The algebra [math]C(H_N^+)[/math] can be presented in two ways, as follows:

• As the universal algebra generated by the entries of a [math]2N\times 2N[/math] magic unitary having the following “sudoku” pattern, with [math]a,b[/math] being square matrices:
  [[math]] w=\begin{pmatrix}a&b\\b&a\end{pmatrix} [[/math]]
• As the universal algebra generated by the entries of a [math]N\times N[/math] orthogonal matrix which is “cubic”, in the sense that, for any [math]j\neq k[/math]:
  [[math]] u_{ij}u_{ik}=u_{ji}u_{ki}=0 [[/math]]

As for [math]C(H_N)[/math], this has similar presentations, among the commutative algebras.

Show Proof

Here the first assertion follows from Theorem 10.12, via Proposition 10.10, and the last assertion is clear as well, because [math]C(H_N)[/math] is the abelianization of [math]C(H_N^+)[/math]. Thus, we are left with proving that the algebras [math]A_s,A_c[/math] coming from (1,2) coincide. We construct first the arrow [math]A_c\to A_s[/math]. The elements [math]a_{ij},b_{ij}[/math] being self-adjoint, their differences are self-adjoint as well. Thus [math]a-b[/math] is a matrix of self-adjoint elements. We have the following formula for the products on the columns of [math]a-b[/math]:

[[math]] \begin{eqnarray*} (a-b)_{ik}(a-b)_{jk} &=&a_{ik}a_{jk}-a_{ik}b_{jk}-b_{ik}a_{jk}+b_{ik}b_{jk}\\ &=&\begin{cases} 0&{\rm for }\ i\neq j\\ a_{ik}+b_{ik}&{\rm for}\ i=j \end{cases} \end{eqnarray*} [[/math]]

In the [math]i=j[/math] case the elements [math]a_{ik}+b_{ik}[/math] sum up to [math]1[/math], so the columns of [math]a-b[/math] are orthogonal. A similar computation works for rows, so [math]a-b[/math] is orthogonal. Now by using the [math]i\neq j[/math] computation, along with its row analogue, we conclude that [math]a-b[/math] is cubic. Thus we can define a morphism [math]A_c\to A_s[/math] by the following formula:

[[math]] \varphi(u_{ij})=a_{ij}-b_{ij} [[/math]]

We construct now the inverse morphism. Consider the following elements:

[[math]] \alpha_{ij}=\frac{u_{ij}^2+u_{ij}}{2}\quad,\quad \beta_{ij}=\frac{u_{ij}^2-u_{ij}}{2} [[/math]]

These are projections, and the following matrix is a sudoku unitary:

[[math]] M=\begin{pmatrix} (\alpha_{ij})&(\beta_{ij})\\ (\beta_{ij})&(\alpha_{ij}) \end{pmatrix} [[/math]]

Thus we can define a morphism [math]A_s\to A_c[/math] by the following formulae:

[[math]] \psi(a_{ij})=\frac{u_{ij}^2+u_{ij}}{2}\quad,\quad \psi(b_{ij})=\frac{u_{ij}^2-u_{ij}}{2} [[/math]]

We check now the fact that [math]\psi,\varphi[/math] are indeed inverse morphisms:

[[math]] \begin{eqnarray*} \psi\varphi(u_{ij}) &=&\psi(a_{ij}-b_{ij})\\ &=&\frac{u_{ij}^2+u_{ij}}{2}-\frac{u_{ij}^2-u_{ij}}{2}\\ &=&u_{ij} \end{eqnarray*} [[/math]]

As for the other composition, we have the following computation:

[[math]] \begin{eqnarray*} \varphi\psi(a_{ij}) &=&\varphi\left(\frac{u_{ij}^2+u_{ij}}{2}\right)\\ &=&\frac{(a_{ij}-b_{ij})^2+(a_{ij}-b_{ij})}{2}\\ &=&a_{ij} \end{eqnarray*} [[/math]]

A similar computation gives [math]\varphi\psi(b_{ij})=b_{ij}[/math], which completes the proof.

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We can now work out the easiness property of [math]H_N,H_N^+[/math], with respect to the cubic representations, and we are led to the following result, which is fully satisfactory:

Theorem

The quantum groups [math]H_N,H_N^+[/math] are both easy, as follows:

• [math]H_N[/math] corresponds to the category [math]P_{even}[/math].
• [math]H_N^+[/math] corresponds to the category [math]NC_{even}[/math].

Show Proof

This follows indeed from the fact that the cubic relations are implemented by the one-block partition in [math]P(2,2)[/math], which generates [math]NC_{even}[/math]. See .

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General references

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