11b. Schur-Weyl twists

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

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In order to discuss these questions, we must first review the above construction of the twists of [math]S,T,U,K[/math], which was something quite ad-hoc, and replace this by something more conceptual. We use easiness. Let us start with something that we know, namely:

Proposition

The intermediate easy quantum groups

[[math]] H_N\subset G\subset U_N^+ [[/math]]

come via Tannakian duality from the intermediate categories of partitions

[[math]] P_{even}\supset D\supset\mathcal {NC}_2 [[/math]]

with [math]P_{even}(k,l)\subset P(k,l)[/math] being the category of partitions whose blocks have even size.

Show Proof

This is something coming from the general easiness theory for quantum groups, discussed in chapter 2. Indeed, as explained there, the easy quantum groups appear as certain intermediate compact quantum groups, as follows:

[[math]] S_N\subset G\subset U_N^+ [[/math]]

To be more precise, such a quantum group is easy when the corresponding Tannakian category comes from an intermediate category of partitions, as follows:

[[math]] P\supset D\supset\mathcal {NC}_2 [[/math]]

Now since this correspondence makes correspond [math]H_N\leftrightarrow P_{even}[/math], once again as explained in chapter 2, we are led to the conclusion in the statement.

■

The idea now will be that the twisting operation [math]G\to\bar{G}[/math], in the easy case, can be implemented, via Tannakian duality as usual, via a signature operation on [math]P_{even}[/math]. Given a partition [math]\tau\in P(k,l)[/math], let us call “switch” the operation which consists in switching two neighbors, belonging to different blocks, in the upper row, or in the lower row. Also, we use the standard embedding [math]S_k\subset P_2(k,k)[/math], via the pairings having only up-to-down strings. With these conventions, we have the following result, from :

Theorem

There is a signature map [math]\varepsilon:P_{even}\to\{-1,1\}[/math], given by

[[math]] \varepsilon(\tau)=(-1)^c [[/math]]

where [math]c[/math] is the number of switches needed to make [math]\tau[/math] noncrossing. In addition:

• For [math]\tau\in S_k[/math], this is the usual signature.
• For [math]\tau\in P_2[/math] we have [math](-1)^c[/math], where [math]c[/math] is the number of crossings.
• For [math]\tau\leq\pi\in NC_{even}[/math], the signature is [math]1[/math].

Show Proof

In order to show that the signature map [math]\varepsilon:P_{even}\to\{-1,1\}[/math] in the statement, given by [math]\varepsilon(\tau)=(-1)^c[/math], is well-defined, we must prove that the number [math]c[/math] in the statement is well-defined modulo 2. It is enough to perform the verification for the noncrossing partitions. More precisely, given [math]\tau,\tau'\in NC_{even}[/math] having the same block structure, we must prove that the number of switches [math]c[/math] required for the passage [math]\tau\to\tau'[/math] is even.

In order to do so, observe that any partition [math]\tau\in P(k,l)[/math] can be put in “standard form”, by ordering its blocks according to the appearence of the first leg in each block, counting clockwise from top left, and then by performing the switches as for block 1 to be at left, then for block 2 to be at left, and so on. Here the required switches are also uniquely determined, by the order coming from counting clockwise from top left.

Here is an example of such an algorithmic switching operation, with block 1 being first put at left, by using two switches, then with block 2 left unchanged, and then with block 3 being put at left as well, but at right of blocks 1 and 2, with one switch:

[[math]] \xymatrix@R=3mm@C=3mm{\circ\ar@/_/@{.}[drr]&\circ\ar@{-}[dddl]&\circ\ar@{-}[ddd]&\circ\\ &&\ar@/_/@{.}[ur]&\\ &&\ar@/^/@{.}[dr]&\\ \circ&\circ\ar@/^/@{.}[ur]&\circ&\circ} \item[a]ymatrix@R=4mm@C=1mm{&\\\to\\&\\& } \item[a]ymatrix@R=3mm@C=3mm{\circ\ar@/_/@{.}[dr]&\circ\ar@{-}[dddl]&\circ&\circ\ar@{-}[dddl]\\ &\ar@/_/@{.}[ur]&&\\ &&\ar@/^/@{.}[dr]&\\ \circ&\circ\ar@/^/@{.}[ur]&\circ&\circ} \item[a]ymatrix@R=4mm@C=1mm{&\\\to\\&\\&} \item[a]ymatrix@R=3mm@C=3mm{\circ\ar@/_/@{.}[r]&\circ&\circ\ar@{-}[dddll]&\circ\ar@{-}[dddl]\\ &&&\\ &&\ar@/^/@{.}[dr]&\\ \circ&\circ\ar@/^/@{.}[ur]&\circ&\circ} \item[a]ymatrix@R=4mm@C=1mm{&\\\to\\&\\& } \item[a]ymatrix@R=3mm@C=3mm{\circ\ar@/_/@{.}[r]&\circ&\circ\ar@{-}[dddll]&\circ\ar@{-}[dddll]\\ &&&\\ &&&\\ \circ&\circ&\circ\ar@/^/@{.}[r]&\circ} [[/math]]

The point now is that, under the assumption [math]\tau\in NC_{even}(k,l)[/math], each of the moves required for putting a leg at left, and hence for putting a whole block at left, requires an even number of switches. Thus, putting [math]\tau[/math] is standard form requires an even number of switches. Now given [math]\tau,\tau'\in NC_{even}[/math] having the same block structure, the standard form coincides, so the number of switches [math]c[/math] required for the passage [math]\tau\to\tau'[/math] is indeed even.

Regarding now the remaining assertions, these are all elementary:

(1) For [math]\tau\in S_k[/math] the standard form is [math]\tau'=id[/math], and the passage [math]\tau\to id[/math] comes by composing with a number of transpositions, which gives the signature.

(2) For a general [math]\tau\in P_2[/math], the standard form is of type [math]\tau'=|\ldots|^{\cup\ldots\cup}_{\cap\ldots\cap}[/math], and the passage [math]\tau\to\tau'[/math] requires [math]c[/math] mod 2 switches, where [math]c[/math] is the number of crossings.

(3) Assuming that [math]\tau\in P_{even}[/math] comes from [math]\pi\in NC_{even}[/math] by merging a certain number of blocks, we can prove that the signature is 1 by proceeding by recurrence.

■

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

Definition

Associated to any partition [math]\pi\in P_{even}(k,l)[/math] is the linear map

[[math]] \bar{T}_\pi:(\mathbb C^N)^{\otimes k}\to(\mathbb C^N)^{\otimes l} [[/math]]

given by the following formula, with [math]e_1,\ldots,e_N[/math] being the standard basis of [math]\mathbb C^N[/math],

[[math]] \bar{T}_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j_1\ldots j_l}\bar{\delta}_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

and where [math]\bar{\delta}_\pi\in\{-1,0,1\}[/math] is [math]\bar{\delta}_\pi=\varepsilon(\tau)[/math] if [math]\tau\geq\pi[/math], and [math]\bar{\delta}_\pi=0[/math] otherwise, with:

[[math]] \tau=\ker\binom{i}{j} [[/math]]

In other words, what we are doing here is to add signatures to the usual formula of [math]T_\pi[/math]. Indeed, observe that the usual formula for [math]T_\pi[/math] can be written as folllows:

[[math]] T_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j:\ker(^i_j)\geq\pi}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

Now by inserting signs, coming from the signature map [math]\varepsilon:P_{even}\to\{\pm1\}[/math], we are led to the following formula, which coincides with the one given above:

[[math]] \bar{T}_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{\tau\geq\pi}\varepsilon(\tau)\sum_{j:\ker(^i_j)=\tau}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

We will be back later to this analogy, with more details on what can be done with it. For the moment, we must first prove a key categorical result, as follows:

Proposition

The assignement [math]\pi\to\bar{T}_\pi[/math] is categorical, in the sense that

[[math]] \bar{T}_\pi\otimes\bar{T}_\sigma=\bar{T}_{[\pi\sigma]}\quad,\quad \bar{T}_\pi \bar{T}_\sigma=N^{c(\pi,\sigma)}\bar{T}_{[^\sigma_\pi]}\quad,\quad \bar{T}_\pi^*=\bar{T}_{\pi^*} [[/math]]

where [math]c(\pi,\sigma)[/math] are certain positive integers.

Show Proof

We have to go back to the proof from the untwisted case, from chapter 2, and insert signs. We have to check three conditions, as follows:

\underline{1. Concatenation}. In the untwisted case, this was based on the following formula:

[[math]] \delta_\pi\begin{pmatrix}i_1\ldots i_p\\ j_1\ldots j_q\end{pmatrix} \delta_\sigma\begin{pmatrix}k_1\ldots k_r\\ l_1\ldots l_s\end{pmatrix} =\delta_{[\pi\sigma]}\begin{pmatrix}i_1\ldots i_p&k_1\ldots k_r\\ j_1\ldots j_q&l_1\ldots l_s\end{pmatrix} [[/math]]

In the twisted case, it is enough to check the following formula:

[[math]] \varepsilon\left(\ker\begin{pmatrix}i_1\ldots i_p\\ j_1\ldots j_q\end{pmatrix}\right) \varepsilon\left(\ker\begin{pmatrix}k_1\ldots k_r\\ l_1\ldots l_s\end{pmatrix}\right)= \varepsilon\left(\ker\begin{pmatrix}i_1\ldots i_p&k_1\ldots k_r\\ j_1\ldots j_q&l_1\ldots l_s\end{pmatrix}\right) [[/math]]

Let us denote by [math]\tau,\nu[/math] the partitions on the left, so that the partition on the right is of the form [math]\rho\leq[\tau\nu][/math]. Now by switching to the noncrossing form, [math]\tau\to\tau'[/math] and [math]\nu\to\nu'[/math], the partition on the right transforms into [math]\rho\to\rho'\leq[\tau'\nu'][/math]. Now since the partition [math][\tau'\nu'][/math] is noncrossing, we can use Theorem 11.7 (3), and we obtain the result.

\underline{2. Composition}. In the untwisted case, this was based on the following formula:

[[math]] \sum_{j_1\ldots j_q}\delta_\pi\begin{pmatrix}i_1\ldots i_p\\ j_1\ldots j_q\end{pmatrix} \delta_\sigma\begin{pmatrix}j_1\ldots j_q\\ k_1\ldots k_r\end{pmatrix} =N^{c(\pi,\sigma)}\delta_{[^\pi_\sigma]}\begin{pmatrix}i_1\ldots i_p\\ k_1\ldots k_r\end{pmatrix} [[/math]]

In order to prove now the result in the twisted case, it is enough to check that the signs match. More precisely, we must establish the following formula:

[[math]] \varepsilon\left(\ker\begin{pmatrix}i_1\ldots i_p\\ j_1\ldots j_q\end{pmatrix}\right) \varepsilon\left(\ker\begin{pmatrix}j_1\ldots j_q\\ k_1\ldots k_r\end{pmatrix}\right) =\varepsilon\left(\ker\begin{pmatrix}i_1\ldots i_p\\ k_1\ldots k_r\end{pmatrix}\right) [[/math]]

Let [math]\tau,\nu[/math] be the partitions on the left, so that the partition on the right is of the form [math]\rho\leq[^\tau_\nu][/math]. Our claim is that we can jointly switch [math]\tau,\nu[/math] to the noncrossing form. Indeed, we can first switch as for [math]\ker(j_1\ldots j_q)[/math] to become noncrossing, and then switch the upper legs of [math]\tau[/math], and the lower legs of [math]\nu[/math], as for both these partitions to become noncrossing. Now observe that when switching in this way to the noncrossing form, [math]\tau\to\tau'[/math] and [math]\nu\to\nu'[/math], the partition on the right transforms into [math]\rho\to\rho'\leq[^{\tau'}_{\nu'}][/math]. Now since the partition [math][^{\tau'}_{\nu'}][/math] is noncrossing, we can apply Theorem 11.7 (3), and we obtain the result.

\underline{3. Involution}. Here we must prove the following formula:

[[math]] \bar{\delta}_\pi\begin{pmatrix}i_1\ldots i_p\\ j_1\ldots j_q\end{pmatrix}=\bar{\delta}_{\pi^*}\begin{pmatrix}j_1\ldots j_q\\ i_1\ldots i_p\end{pmatrix} [[/math]]

But this is clear from the definition of [math]\bar{\delta}_\pi[/math], and we are done.

■

As a conclusion, our twisted construction [math]\pi\to\bar{T}_\pi[/math] has all the needed properties for producing quantum groups, via Tannakian duality, and we can now formulate:

Theorem

Given a category of partitions [math]D\subset P_{even}[/math], the construction

[[math]] Hom(u^{\otimes k},u^{\otimes l})=span\left(\bar{T}_\pi\Big|\pi\in D(k,l)\right) [[/math]]

produces via Tannakian duality a quantum group [math]\bar{G}_N\subset U_N^+[/math], for any [math]N\in\mathbb N[/math].

Show Proof

This follows indeed from the Tannakian results from chapter 2, exactly as in the easy case, by using this time Proposition 11.9 as technical ingredient. To be more precise, Proposition 11.9 shows that the linear spaces on the right form a Tannakian category, and so the results in chapter 2 apply, and give the result.

■

We can unify the easy quantum groups, or at least the examples coming from categories [math]D\subset P_{even}[/math], with the quantum groups constructed above, as follows:

Definition

A closed subgroup [math]G\subset U_N^+[/math] is called [math]q[/math]-easy, or quizzy, with deformation parameter [math]q=\pm1[/math], when its tensor category appears as follows,

[[math]] Hom(u^{\otimes k},u^{\otimes l})=span\left(\dot{T}_\pi\Big|\pi\in D(k,l)\right) [[/math]]

for a certain category of partitions [math]D\subset P_{even}[/math], where, for [math]q=-1,1[/math]:

[[math]] \dot{T}=\bar{T},T [[/math]]

The Schur-Weyl twist of [math]G[/math] is the quizzy quantum group [math]\bar{G}\subset U_N^+[/math] obtained via [math]q\to-q[/math].

We will see later on that the easy quantum group associated to [math]P_{even}[/math] itself is the hyperochahedral group [math]H_N[/math], and so that our assumption [math]D\subset P_{even}[/math], replacing [math]D\subset P[/math], simply corresponds to [math]H_N\subset G[/math], replacing the usual condition [math]S_N\subset G[/math].

For the moment, our most pressing task is that of checking that, when applying the Schur-Weyl twisting to the basic unitary quantum groups, we obtain the ad-hoc twists that we previously constructed. This is indeed the case:

Theorem

The twisted unitary quantum groups introduced before,

[[math]] \xymatrix@R=12mm@C=12mm{ O_N^+\ar[r]&\mathbb TO_N^+\ar[r]&U_N^+\\ \bar{O}_N^*\ar[r]\ar[u]&\mathbb T\bar{O}_N^*\ar[r]\ar[u]&\bar{U}_N^*\ar[u]\\ \bar{O}_N\ar[r]\ar[u]&\mathbb T\bar{O}_N\ar[r]\ar[u]&\bar{U}_N\ar[u]} [[/math]]

appear as Schur-Weyl twists of the basic unitary quantum groups.

Show Proof

This is something routine, in several steps, as follows:

(1) The basic crossing, [math]\ker\binom{ij}{ji}[/math] with [math]i\neq j[/math], comes from the transposition [math]\tau\in S_2[/math], so its signature is [math]-1[/math]. As for its degenerated version [math]\ker\binom{ii}{ii}[/math], this is noncrossing, so here the signature is [math]1[/math]. We conclude that the linear map associated to the basic crossing is:

[[math]] \bar{T}_{\slash\!\!\!\backslash}(e_i\otimes e_j) =\begin{cases} -e_j\otimes e_i&{\rm for}\ i\neq j\\ e_j\otimes e_i&{\rm otherwise} \end{cases} [[/math]]

For the half-classical crossing, namely [math]\ker\binom{ijk}{kji}[/math] with [math]i,j,k[/math] distinct, the signature is once again [math]-1[/math], and by examining the signatures of the various degenerations of this half-classical crossing, we are led to the following formula:

[[math]] \bar{T}_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}(e_i\otimes e_j\otimes e_k) =\begin{cases} -e_k\otimes e_j\otimes e_i&{\rm for}\ i,j,k\ {\rm distinct}\\ e_k\otimes e_j\otimes e_i&{\rm otherwise} \end{cases} [[/math]]

(2) Our claim now if that for an orthogonal quantum group [math]G[/math], the following holds, with the quantum group [math]\bar{O}_N[/math] being the one in Theorem 11.2:

[[math]] \bar{T}_{\slash\!\!\!\backslash}\in End(u^{\otimes 2})\iff G\subset\bar{O}_N [[/math]]

Indeed, by using the formula of [math]\bar{T}_{\slash\!\!\!\backslash}[/math] found in (1) above, we obtain:

[[math]] \begin{eqnarray*} (\bar{T}_{\slash\!\!\!\backslash}\otimes1)u^{\otimes 2}(e_i\otimes e_j\otimes1) &=&\sum_ke_k\otimes e_k\otimes u_{ki}u_{kj}\\ &-&\sum_{k\neq l}e_l\otimes e_k\otimes u_{ki}u_{lj} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} u^{\otimes 2}(\bar{T}_{\slash\!\!\!\backslash}\otimes1)(e_i\otimes e_j\otimes1) &=&\begin{cases} \sum_{kl}e_l\otimes e_k\otimes u_{li}u_{ki}&{\rm if}\ i=j\\ -\sum_{kl}e_l\otimes e_k\otimes u_{lj}u_{ki}&{\rm if}\ i\neq j \end{cases} \end{eqnarray*} [[/math]]

For [math]i=j[/math] the conditions are [math]u_{ki}^2=u_{ki}^2[/math] for any [math]k[/math], and [math]u_{ki}u_{li}=-u_{li}u_{ki}[/math] for any [math]k\neq l[/math]. For [math]i\neq j[/math] the conditions are [math]u_{ki}u_{kj}=-u_{kj}u_{ki}[/math] for any [math]k[/math], and [math]u_{ki}u_{lj}=u_{lj}u_{ki}[/math] for any [math]k\neq l[/math]. Thus we have exactly the relations between the coordinates of [math]\bar{O}_N[/math], and we are done.

(3) Our claim now if that for an orthogonal quantum group [math]G[/math], the following holds, with the quantum group [math]\bar{O}_N^*[/math] being the one in Theorem 11.3:

[[math]] \bar{T}_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}\in End(u^{\otimes 3})\iff G\subset\bar{O}_N^* [[/math]]

Indeed, by using the formula of [math]\bar{T}_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}[/math] found in (1) above, we obtain:

[[math]] \begin{eqnarray*} (\bar{T}_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}\otimes1)u^{\otimes 2}(e_i\otimes e_j\otimes e_k\otimes1) &=&\sum_{abc\ not\ distinct}e_c\otimes e_b\otimes e_a\otimes u_{ai}u_{bj}u_{ck}\\ &-&\sum_{a,b,c\ distinct}e_c\otimes e_b\otimes e_a\otimes u_{ai}u_{bj}u_{ck} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} &&u^{\otimes 2}(\bar{T}_{\slash\hskip-1.6mm\backslash\hskip-1.1mm|\hskip0.5mm}\otimes1)(e_i\otimes e_j\otimes e_k\otimes1)\\ &&=\begin{cases} \sum_{abc}e_c\otimes e_b\otimes e_a\otimes u_{ck}u_{bj}u_{ai}&{\rm for}\ i,j,k\ {\rm not\ distinct}\\ -\sum_{abc}e_c\otimes e_b\otimes e_a\otimes u_{ck}u_{bj}u_{ai}&{\rm for}\ i,j,k\ {\rm distinct} \end{cases} \end{eqnarray*} [[/math]]

For [math]i,j,k[/math] not distinct the conditions are [math]u_{ai}u_{bj}u_{ck}=u_{ck}u_{bj}u_{ai}[/math] for [math]a,b,c[/math] not distinct, and [math]u_{ai}u_{bj}u_{ck}=-u_{ck}u_{bj}u_{ai}[/math] for [math]a,b,c[/math] distinct. For [math]i,j,k[/math] distinct the conditions are [math]u_{ai}u_{bj}u_{ck}=-u_{ck}u_{bj}u_{ai}[/math] for [math]a,b,c[/math] not distinct, and [math]u_{ai}u_{bj}u_{ck}=u_{ck}u_{bj}u_{ai}[/math] for [math]a,b,c[/math] distinct. Thus we have the relations between the coordinates of [math]\bar{O}_N^*[/math], as desired.

(4) Now with the above in hand, we obtain that the Schur-Weyl twists of [math]O_N,O_N^*[/math] are indeed the quantum groups [math]\bar{O}_N,\bar{O}_N^*[/math] from Theorem 11.2 and Theorem 11.3.

(5) The proof in the unitary case is similar, by adding signs in the above computations (2,3), the conclusion being that the Schur-Weyl twists of [math]U_N,U_N^*[/math] are [math]\bar{U}_N,\bar{U}_N^*[/math].

■

Let us clarify now the relation between the maps [math]T_\pi,\bar{T}_\pi[/math]. We recall that the Möbius function of any lattice, and in particular of [math]P_{even}[/math], is given by:

[[math]] \mu(\sigma,\pi)=\begin{cases} 1&{\rm if}\ \sigma=\pi\\ -\sum_{\sigma\leq\tau \lt \pi}\mu(\sigma,\tau)&{\rm if}\ \sigma \lt \pi\\ 0&{\rm if}\ \sigma\not\leq\pi \end{cases} [[/math]]

With this notation, we have the following result:

Proposition

For any partition [math]\pi\in P_{even}[/math] we have the formula

[[math]] \bar{T}_\pi=\sum_{\tau\leq\pi}\alpha_\tau T_\tau [[/math]]

where [math]\alpha_\sigma=\sum_{\sigma\leq\tau\leq\pi}\varepsilon(\tau)\mu(\sigma,\tau)[/math], with [math]\mu[/math] being the Möbius function of [math]P_{even}[/math].

Show Proof

The linear combinations [math]T=\sum_{\tau\leq\pi}\alpha_\tau T_\tau[/math] acts on tensors as follows:

[[math]] \begin{eqnarray*} T(e_{i_1}\otimes\ldots\otimes e_{i_k}) &=&\sum_{\tau\leq\pi}\alpha_\tau T_\tau(e_{i_1}\otimes\ldots\otimes e_{i_k})\\ &=&\sum_{\tau\leq\pi}\alpha_\tau\sum_{\sigma\leq\tau}\sum_{j:\ker(^i_j)=\sigma}e_{j_1}\otimes\ldots\otimes e_{j_l}\\ &=&\sum_{\sigma\leq\pi}\left(\sum_{\sigma\leq\tau\leq\pi}\alpha_\tau\right)\sum_{j:\ker(^i_j)=\sigma}e_{j_1}\otimes\ldots\otimes e_{j_l} \end{eqnarray*} [[/math]]

Thus, in order to have [math]\bar{T}_\pi=\sum_{\tau\leq\pi}\alpha_\tau T_\tau[/math], we must have [math]\varepsilon(\sigma)=\sum_{\sigma\leq\tau\leq\pi}\alpha_\tau[/math], for any [math]\sigma\leq\pi[/math]. But this problem can be solved by using the Möbius inversion formula, and we obtain the numbers [math]\alpha_\sigma=\sum_{\sigma\leq\tau\leq\pi}\varepsilon(\tau)\mu(\sigma,\tau)[/math] in the statement.

■

With the above results in hand, let us go back now to the question of twisting the quantum reflection groups. It is convenient to include in our discussion two more quantum groups, coming from and denoted [math]H_N^{[\infty]},K_N^{[\infty]}[/math], constructed as follows:

Proposition

We have quantum groups [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]]

These quantum groups are both easy, with the corresponding categories of partitions, denoted [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 condition [math]\eta\in End(u^{\otimes k})[/math], with [math]|k|=3[/math]. For details here, and for more on these two quantum groups, which are very interesting objects, and that we have actually already met in chapter 4 above, we refer to the paper of Raum-Weber .

■

In order to discuss now the Schur-Weyl twisting of the various quantum reflection groups that we have, 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, from , , 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 Proposition 11.14, by computing [math] \lt \eta \gt [/math], and for the complete proof here we refer to .

(3) We use the fact, also from , that the relations [math]g_ig_ig_j=g_jg_ig_i[/math] are trivially satisfied for real reflections. Thus, we have:

[[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 we have [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

The basic quantum reflection groups, namely

[[math]] \xymatrix@R=13mm@C=13mm{ H_N^+\ar[r]&\mathbb TH_N^+\ar[r]&K_N^+\\ H_N^*\ar[r]\ar[u]&\mathbb TH_N^*\ar[r]\ar[u]&K_N^*\ar[u]\\ H_N\ar[r]\ar[u]&\mathbb TH_N\ar[r]\ar[u]&K_N\ar[u]} [[/math]]

equal their own Schur-Weyl twists.

Show Proof

This result, from , 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 Theorem 11.7 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 signatures found in Proposition 11.15.

-- [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.

(2) In the complex case the proof is similar, and we refer here to .

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In relation now with the tori, we have the following result:

Theorem

The diagonal tori of the twisted quantum groups are

[[math]] \xymatrix@R=13.5mm@C=13.5mm{ T_N^+\ar[r]&\mathbb TT_N^+\ar[r]&\mathbb T_N^+\\ T_N^*\ar[r]\ar[u]&\mathbb TT_N^*\ar[r]\ar[u]&\mathbb T_N^*\ar[u]\\ T_N\ar[r]\ar[u]&\mathbb TT_N\ar[r]\ar[u]&\mathbb T_N\ar[u]} [[/math]]

exactly as in the untwisted case.

Show Proof

This is clear for the quantum reflection groups, which are not twistable, and for the quantum unitary groups this is elementary as well, coming from definitions.

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General references

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