10c. Complex reflections

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Following , the basic algebraic results regarding [math]S_N,S_N^+[/math] and [math]H_N,H_N^+[/math] appear in fact as the [math]s=1,2[/math] particular cases of the following result:

Theorem

The complex reflection groups [math]H_N^s=\mathbb Z_s\wr S_N[/math] and their free analogues [math]H_N^{s+}=\mathbb Z_s\wr_*S_N^+[/math], defined for any [math]s\in\mathbb N[/math], have the following properties:

• They have [math]N[/math]-dimensional coordinates [math]u=(u_{ij})[/math], subject to the relations:
  [[math]] u_{ij}u_{ij}^*=u_{ij}^*u_{ij} [[/math]]
  [[math]] p_{ij}=u_{ij}u_{ij}^*={\rm magic} [[/math]]
  [[math]] u_{ij}^s=p_{ij} [[/math]]
• They are easy, the corresponding categories [math]P^s\subset P,NC^s\subset NC[/math] being given by the fact that we have [math]\#\circ-\#\bullet=0(s)[/math], as a weighted sum, in each block.

Show Proof

We already know that the results hold at [math]s=1,2[/math], and the proof in general is similar. With respect to the above proof at [math]s=2[/math], the situation is as follows:

(1) Observe first that the result holds at [math]s=1[/math], where we obtain the magic condition, and at [math]s=2[/math] as well, where we obtain something equivalent to the cubic condition. In general, this follows from a [math]\mathbb Z_s[/math]-analogue of Proposition 10.13. See .

(2) Once again, the result holds at [math]s=1[/math], trivially, and at [math]s=2[/math] as well, where our condition is equivalent to [math]\#\circ+\#\bullet=0(2)[/math], in each block. In general, this follows as in the proof of Theorem 10.14, by using the one-block partition in [math]P(s,s)[/math]. See .

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We have as well a result at [math]s=\infty[/math], which is of particular interest, as follows:

Theorem

The pure complex reflection groups [math]K_N=\mathbb T\wr S_N[/math] and their free analogues [math]K_N^+=\mathbb T\wr_*S_N^+[/math] have the following properties:

• They have [math]N[/math]-dimensional coordinates [math]u=(u_{ij})[/math], subject to the relations:
  [[math]] u_{ij}u_{ij}^*=u_{ij}^*u_{ij} [[/math]]
  [[math]] p_{ij}=u_{ij}u_{ij}^*={\rm magic} [[/math]]
• They are easy, the corresponding categories [math]\mathcal P_{even}\subset P,\mathcal{NC}_{even}\subset NC[/math] being given by the fact that we have [math]\#\circ=\#\bullet[/math], as a weighted equality, in each block.

Show Proof

The assertions here appear as an [math]s=\infty[/math] extension of (1,2) in Theorem 10.15 above, and their proof can be obtained along the same lines, as follows:

(1) This follows indeed by working out a [math]\mathbb T[/math]-analogue of the computations in the proof of Proposition 10.13 above. Again, for details we refer here to .

(2) This result appears too as a [math]s=\infty[/math] extension of the results that we already have, and for details here, we refer once again to .

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We can now focus on [math]H_N,H_N^+,K_N,K_N^+[/math], with the idea in mind of completing the continuous quantum group picture from chapter 7. Before doing this, however, we have two more quantum groups to be introduced, namely [math]H_N^*,K_N^*[/math]. We have here:

Theorem

We have quantum groups [math]H_N^*,K_N^*[/math], both easy, as follows,

• [math]H_N^*=H_N^+\cap O_N^*[/math], corresponding to the category [math]P_{even}^*[/math],
• [math]K_N^*=K_N^+\cap U_N^*[/math], corresponding to the category [math]\mathcal P_{even}^*[/math],

with the symbol [math]*[/math] standing for the fact that the corresponding partitions, when relabelled clockwise [math]\circ\bullet\circ\bullet\ldots[/math], must contain the same number of [math]\circ,\bullet[/math], in each block.

Show Proof

This is standard, from the results that we already have, regarding the various quantum groups involved, because the intersection operations at the quantum group level correspond to generation operations, at the category of partitions level.

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We can now complete the “continuous” picture from chapter 7 above, as follows:

Theorem

The basic orthogonal and unitary quantum groups are related to the basic real and complex quantum reflection groups as follows,

[[math]] \xymatrix@R=15mm@C=17mm{ U_N\ar[r]&U_N^*\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&O_N^*\ar[r]\ar[u]&O_N^+\ar[u]} \ \ \ \ \ \xymatrix@R=8mm@C=5mm{\\ \leftrightarrow&\\&\\}\ \ \item[a]ymatrix@R=15mm@C=17mm{ K_N\ar[r]&K_N^*\ar[r]&K_N^+\\ H_N\ar[r]\ar[u]&H_N^*\ar[r]\ar[u]&H_N^+\ar[u]} [[/math]]

the connecting operations [math]U\leftrightarrow K[/math] being given by [math]K=U\cap K_N^+[/math] and [math]U=\{K,O_N\}[/math].

Show Proof

According to the general results in chapter 7, in terms of categories of partitions, the operations introduced in the statement reformulate as follows:

[[math]] D_K= \lt D_U,\mathcal{NC}_{even} \gt \quad,\quad D_U=D_K\cap P_2 [[/math]]

On the other hand, by putting together the various easiness results that we have, the categories of partitions for the quantum groups in the statement are as follows:

[[math]] \xymatrix@R=16mm@C=15mm{ \mathcal P_2\ar[d]&\mathcal P_2^*\ar[l]\ar[d]&\mathcal{NC}_2\ar[l]\ar[d]\\ P_2&P_2^*\ar[l]&NC_2\ar[l]} \ \ \ \ \ \xymatrix@R=8mm@C=5mm{\\ :&\\&\\}\ \ \item[a]ymatrix@R=16mm@C=12mm{ \mathcal P_{even}\ar[d]&\mathcal P_{even}^*\ar[l]\ar[d]&\mathcal{NC}_{even}\ar[l]\ar[d]\\ P_{even}&P_{even}^*\ar[l]&NC_{even}\ar[l]} [[/math]]

It is elementary to check that these categories are related by the above intersection and generation operations, and we conclude that the correspondence holds indeed.

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Our purpose now will be that of showing that a twisted analogue of the above result holds. It is convenient to include in our discussion two more quantum groups, coming from , and denoted [math]H_N^{[\infty]},K_N^{[\infty]}[/math], which are constructed as follows:

Theorem

We have intermediate liberations [math]H_N^{[\infty]},K_N^{[\infty]}[/math] as follows, constructed by using the relations [math]\alpha\beta\gamma=0[/math], for any [math]a\neq c[/math] on the same row or column of [math]u[/math],

[[math]] \xymatrix@R=15mm@C=17mm{ K_N\ar[r]&K_N^*\ar[r]&K_N^{[\infty]}\ar[r]&K_N^+\\ H_N\ar[r]\ar[u]&H_N^*\ar[r]\ar[u]&H_N^{[\infty]}\ar[r]\ar[u]&H_N^+\ar[u]} [[/math]]

with the convention [math]\alpha=a,a^*[/math], and so on. These quantum groups are easy, the corresponding categories [math]P_{even}^{[\infty]}\subset P_{even}[/math] and [math]\mathcal P_{even}^{[\infty]}\subset\mathcal P_{even}[/math] being generated by [math]\eta=\ker(^{iij}_{jii})[/math].

Show Proof

This is routine, by using the fact that the relations [math]\alpha\beta\gamma=0[/math] in the statement are equivalent to the following condition, with [math]|k|=3[/math]:

[[math]] \eta\in End(u^{\otimes k}) [[/math]]

For further details on these quantum groups, we refer to , .

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In order to discuss the twisting, we will need the following technical result:

Proposition

We have the following equalities,

[[math]] \begin{eqnarray*} P_{even}^*&=&\left\{\pi\in P_{even}\Big|\varepsilon(\tau)=1,\forall\tau\leq\pi,|\tau|=2\right\} \\ P_{even}^{[\infty]}&=&\left\{\pi\in P_{even}\Big|\sigma\in P_{even}^*,\forall\sigma\subset\pi\right\}\\ P_{even}^{[\infty]}&=&\left\{\pi\in P_{even}\Big|\varepsilon(\tau)=1,\forall\tau\leq\pi\right\} \end{eqnarray*} [[/math]]

where [math]\varepsilon:P_{even}\to\{\pm1\}[/math] is the signature of even permutations.

Show Proof

This is routine combinatorics, the idea being as follows:

(1) Given [math]\pi\in P_{even}[/math], we have [math]\tau\leq\pi,|\tau|=2[/math] precisely when [math]\tau=\pi^\beta[/math] is the partition obtained from [math]\pi[/math] by merging all the legs of a certain subpartition [math]\beta\subset\pi[/math], and by merging as well all the other blocks. Now observe that [math]\pi^\beta[/math] does not depend on [math]\pi[/math], but only on [math]\beta[/math], and that the number of switches required for making [math]\pi^\beta[/math] noncrossing is [math]c=N_\bullet-N_\circ[/math] modulo 2, where [math]N_\bullet/N_\circ[/math] is the number of black/white legs of [math]\beta[/math], when labelling the legs of [math]\pi[/math] counterclockwise [math]\circ\bullet\circ\bullet\ldots[/math] Thus [math]\varepsilon(\pi^\beta)=1[/math] holds precisely when [math]\beta\in\pi[/math] has the same number of black and white legs, and this gives the result.

(2) This simply follows from the equality [math]P_{even}^{[\infty]}= \lt \eta \gt [/math] coming from Theorem 10.19, by computing [math] \lt \eta \gt [/math], and for the complete proof here we refer to Raum-Weber .

(3) We use here the fact, also from , that the relations [math]g_ig_ig_j=g_jg_ig_i[/math] are trivially satisfied for real reflections. This leads to the following conclusion:

[[math]] P_{even}^{[\infty]}(k,l)=\left\{\ker\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}\Big|g_{i_1}\ldots g_{i_k}=g_{j_1}\ldots g_{j_l}\ {\rm inside}\ \mathbb Z_2^{*N}\right\} [[/math]]

In other words, the partitions in [math]P_{even}^{[\infty]}[/math] are those describing the relations between free variables, subject to the conditions [math]g_i^2=1[/math]. We conclude that [math]P_{even}^{[\infty]}[/math] appears from [math]NC_{even}[/math] by “inflating blocks”, in the sense that each [math]\pi\in P_{even}^{[\infty]}[/math] can be transformed into a partition [math]\pi'\in NC_{even}[/math] by deleting pairs of consecutive legs, belonging to the same block. Now since this inflation operation leaves invariant modulo 2 the number [math]c\in\mathbb N[/math] of switches in the definition of the signature, it leaves invariant the signature [math]\varepsilon=(-1)^c[/math] itself, and we obtain in this way the inclusion “[math]\subset[/math]” in the statement. Conversely, given [math]\pi\in P_{even}[/math] satisfying [math]\varepsilon(\tau)=1[/math], [math]\forall\tau\leq\pi[/math], our claim is that:

[[math]] \rho\leq\sigma\subset\pi,|\rho|=2\implies\varepsilon(\rho)=1 [[/math]]

Indeed, let us denote by [math]\alpha,\beta[/math] the two blocks of [math]\rho[/math], and by [math]\gamma[/math] the remaining blocks of [math]\pi[/math], merged altogether. We know that the partitions [math]\tau_1=(\alpha\wedge\gamma,\beta)[/math], [math]\tau_2=(\beta\wedge\gamma,\alpha)[/math], [math]\tau_3=(\alpha,\beta,\gamma)[/math] are all even. On the other hand, putting these partitions in noncrossing form requires respectively [math]s+t,s'+t,s+s'+t[/math] switches, where [math]t[/math] is the number of switches needed for putting [math]\rho=(\alpha,\beta)[/math] in noncrossing form. Thus [math]t[/math] is even, and we are done. With the above claim in hand, we conclude, by using the second equality in the statement, that we have [math]\sigma\in P_{even}^*[/math]. Thus [math]\pi\in P_{even}^{[\infty]}[/math], which ends the proof of “[math]\supset[/math]”.

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With the above result in hand, we can now prove:

Theorem

We have the following results:

• The quantum groups from Theorem 10.19 are equal to their own twists.
• With input coming from this, a twisted version of Theorem 10.18 holds.

Show Proof

This result basically comes from the results that we have.

(1) In the real case, the verifications are as follows:

-- [math]H_N^+[/math]. We know from chapter 7 above that for [math]\pi\in NC_{even}[/math] we have [math]\bar{T}_\pi=T_\pi[/math], and since we are in the situation [math]D\subset NC_{even}[/math], the definitions of [math]G,\bar{G}[/math] coincide.

-- [math]H_N^{[\infty]}[/math]. Here we can use the same argument as in (1), based this time on the description of [math]P_{even}^{[\infty]}[/math] involving the signature found in Proposition 10.20.

-- [math]H_N^*[/math]. We have [math]H_N^*=H_N^{[\infty]}\cap O_N^*[/math], so [math]\bar{H}_N^*\subset H_N^{[\infty]}[/math] is the subgroup obtained via the defining relations for [math]\bar{O}_N^*[/math]. But all the [math]abc=-cba[/math] relations defining [math]\bar{H}_N^*[/math] are automatic, of type [math]0=0[/math], and it follows that [math]\bar{H}_N^*\subset H_N^{[\infty]}[/math] is the subgroup obtained via the relations [math]abc=cba[/math], for any [math]a,b,c\in\{u_{ij}\}[/math]. Thus we have [math]\bar{H}_N^*=H_N^{[\infty]}\cap O_N^*=H_N^*[/math], as claimed.

-- [math]H_N[/math]. We have [math]H_N=H_N^*\cap O_N[/math], and by functoriality, [math]\bar{H}_N=\bar{H}_N^*\cap\bar{O}_N=H_N^*\cap\bar{O}_N[/math]. But this latter intersection is easily seen to be equal to [math]H_N[/math], as claimed.

In the complex case the proof is similar, by using the same arguments.

(2) This can be proved by proceeding as in the proof of Theorem 10.18 above, with of course some care when formulating the result.

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General references

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