15b. Reflection groups

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

In view of the above, the notion of easiness is a quite interesting one, deserving a full, systematic investigation. As a first natural question that we would like to solve, we would like to compute the easy group associated to the category of all partitions [math]P[/math] itself. And here, no surprise, we are led to the most basic, but non-trivial, classical group that we know, namely the symmetric group [math]S_N[/math]. To be more precise, we have the following Brauer type theorem for [math]S_N[/math], which answers our question formulated above:

Theorem

The symmetric group [math]S_N[/math], regarded as group of unitary matrices,

[[math]] S_N\subset O_N\subset U_N [[/math]]

via the permutation matrices, is easy, coming from the category of all partitions [math]P[/math].

Show Proof

Consider indeed the group [math]S_N[/math], regarded as a group of unitary matrices, with each permutation [math]\sigma\in S_N[/math] corresponding to the associated permutation matrix:

[[math]] \sigma(e_i)=e_{\sigma(i)} [[/math]]

Consider as well the easy group [math]G\subset O_N[/math] coming from the category of all partitions [math]P[/math]. Since [math]P[/math] is generated by the one-block “fork” partition [math]Y\in P(2,1)[/math], we have:

[[math]] C(G)=C(O_N)\Big/\Big \lt T_Y\in Hom(u^{\otimes 2},u)\Big \gt [[/math]]

The linear map associated to [math]Y[/math] is given by the following formula:

[[math]] T_Y(e_i\otimes e_j)=\delta_{ij}e_i [[/math]]

In order to do the computations, we use the following formulae:

[[math]] u=(u_{ij})_{ij}\quad,\quad u^{\otimes 2}=(u_{ij}u_{kl})_{ik,jl}\quad,\quad T_Y=(\delta_{ijk})_{i,jk} [[/math]]

We therefore obtain the following formula:

[[math]] (T_Yu^{\otimes 2})_{i,jk} =\sum_{lm}(T_Y)_{i,lm}(u^{\otimes 2})_{lm,jk} =u_{ij}u_{ik} [[/math]]

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

[[math]] (uT_Y)_{i,jk} =\sum_lu_{il}(T_Y)_{l,jk} =\delta_{jk}u_{ij} [[/math]]

Thus, the relation defining [math]G\subset O_N[/math] reformulates as follows:

[[math]] T_Y\in Hom(u^{\otimes 2},u)\iff u_{ij}u_{ik}=\delta_{jk}u_{ij},\forall i,j,k [[/math]]

In other words, the elements [math]u_{ij}[/math] must be projections, which must be pairwise orthogonal on the rows of [math]u=(u_{ij})[/math]. We conclude that [math]G\subset O_N[/math] is the subgroup of matrices [math]g\in O_N[/math] having the property [math]g_{ij}\in\{0,1\}[/math]. Thus we have [math]G=S_N[/math], as desired.

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As a continuation of this, and following the hierarchy of finite groups explained in chapters 9-12, with the original motivations there being actually probabilistic, but with these motivations leading to the same hierarchy that we need here, let us discuss now the hyperoctahedral group [math]H_N[/math]. The result here, from , is quite similar, as follows:

Theorem

The hyperoctahedral group [math]H_N[/math], regarded as a group of matrices,

[[math]] S_N\subset H_N\subset O_N [[/math]]

is easy, coming from the category of partitions with even blocks [math]P_{even}[/math].

Show Proof

This follows as usual from Tannakian duality. To be more precise, consider the following one-block partition, which, as the name indicates, looks like a [math]H[/math] letter:

[[math]] H\in P(2,2) [[/math]]

The linear map associated to this partition is then given by:

[[math]] T_H(e_i\otimes e_j)=\delta_{ij}e_i\otimes e_i [[/math]]

By using this formula, we have the following computation:

[[math]] \begin{eqnarray*} (T_H\otimes id)u^{\otimes 2}(e_a\otimes e_b) &=&(T_H\otimes id)\left(\sum_{ijkl}e_{ij}\otimes e_{kl}\otimes u_{ij}u_{kl}\right)(e_a\otimes e_b)\\ &=&(T_H\otimes id)\left(\sum_{ik}e_i\otimes e_k\otimes u_{ia}u_{kb}\right)\\ &=&\sum_ie_i\otimes e_i\otimes u_{ia}u_{ib} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} u^{\otimes 2}(T_H\otimes id)(e_a\otimes e_b) &=&\delta_{ab}\left(\sum_{ijkl}e_{ij}\otimes e_{kl}\otimes u_{ij}u_{kl}\right)(e_a\otimes e_a)\\ &=&\delta_{ab}\sum_{ij}e_i\otimes e_k\otimes u_{ia}u_{ka} \end{eqnarray*} [[/math]]

We conclude from this that we have the following equivalence:

[[math]] T_H\in End(u^{\otimes 2})\iff \delta_{ik}u_{ia}u_{ib}=\delta_{ab}u_{ia}u_{ka},\forall i,k,a,b [[/math]]

But the relations on the right tell us that the entries of [math]u=(u_{ij})[/math] must satisfy [math]\alpha\beta=0[/math] on each row and column of [math]u[/math], and so that the corresponding closed subgroup [math]G\subset O_N[/math] consists of the matrices [math]g\in O_N[/math] which are permutation-like, with [math]\pm1[/math] nonzero entries. Thus, the corresponding group is [math]G=H_N[/math], and as a conclusion to this, we have:

[[math]] C(H_N)=C(O_N)\Big/\Big \lt T_H\in End(u^{\otimes 2})\Big \gt [[/math]]

According now to our conventions for easiness, this means that the hyperoctahedral group [math]H_N[/math] is easy, coming from the following category of partitions:

[[math]] D= \lt H \gt [[/math]]

But the category on the right can be computed by drawing pictures, and we have:

[[math]] \lt H \gt =P_{even} [[/math]]

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

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More generally now, we have in fact the following result, from , regarding the series of complex reflection groups [math]H_N^s[/math], which covers both the groups [math]S_N,H_N[/math]:

Theorem

The complex reflection group [math]H_N^s=\mathbb Z_s\wr S_N[/math] is easy, the corresponding category [math]P^s[/math] consisting of the partitions satisfying the condition

[[math]] \#\circ=\#\bullet(s) [[/math]]

as a weighted sum, in each block. In particular, we have the following results:

• [math]S_N[/math] is easy, coming from the category [math]P[/math].
• [math]H_N=\mathbb Z_2\wr S_N[/math] is easy, coming from the category [math]P_{even}[/math].
• [math]K_N=\mathbb T\wr S_N[/math] is easy, coming from the category [math]\mathcal P_{even}[/math].

Show Proof

This is something that we already know at [math]s=1,2[/math], from Theorems 15.10 and 15.11. In general, the proof is similar, based on Tannakian duality. To be more precise, in what regards the main assertion, the idea here is that the one-block partition [math]\pi\in P(s)[/math], which generates the category of partitions [math]P^s[/math] in the statement, implements the relations producing the subgroup [math]H_N^s\subset S_N[/math]. As for the last assertions, these are all elementary:

(1) At [math]s=1[/math] we know that we have [math]H_N^1=S_N[/math]. Regarding now the corresponding category, here the condition [math]\#\circ=\#\bullet(1)[/math] is automatic, and so [math]P^1=P[/math].

(2) At [math]s=2[/math] we know that we have [math]H_N^2=H_N[/math]. Regarding now the corresponding category, here the condition [math]\#\circ=\#\bullet(2)[/math] reformulates as follows:

[[math]] \#\circ+\,\#\bullet=0(2) [[/math]]

Thus each block must have even size, and we obtain, as claimed, [math]P^2=P_{even}[/math].

(3) At [math]s=\infty[/math] we know that we have [math]H_N^\infty=K_N[/math]. Regarding now the corresponding category, here the condition [math]\#\circ=\#\bullet(\infty)[/math] reads:

[[math]] \#\circ=\#\bullet [[/math]]

But this is the condition defining [math]\mathcal P_{even}[/math], and so [math]P^\infty=\mathcal P_{even}[/math], as claimed.

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Summarizing, we have many examples. In fact, our list of easy groups has currently become quite big, and here is a selection of the main results that we have so far:

Theorem

We have a diagram of compact groups as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ K_N\ar[r]&U_N\\ H_N\ar[u]\ar[r]&O_N\ar[u]} [[/math]]

where [math]H_N=\mathbb Z_2\wr S_N[/math] and [math]K_N=\mathbb T\wr S_N[/math], and all these groups are easy.

Show Proof

This follows from the above results. To be more precise, we know that the above groups are all easy, the corresponding categories of partitions being as follows:

[[math]] \xymatrix@R=16mm@C=18mm{ \mathcal P_{even}\ar[d]&\mathcal P_2\ar[l]\ar[d]\\ P_{even}&P_2\ar[l]} [[/math]]

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

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Summarizing, most of the groups that we investigated in this book are covered by the easy group formalism. One exception is the symplectic group [math]Sp_N[/math], but this group is covered as well, by a suitable extension of the easy group formalism. See .

General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].