14d. Tannakian results

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

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We discuss here some converses to the above results, which are rather specialized results, of Tannakian nature. We will first prove that any quantum permutation group [math]G\subset S_N^+[/math] produces a planar subalgebra of [math]\mathcal S_N[/math]. In order to do so, we first have:

Theorem

Given a quantum permutation group [math]G\subset S_N^+[/math], consider the associated coaction map on [math]C(X)[/math], where [math]X=\{1,\ldots,N\}[/math],

[[math]] \Phi:C(X)\to C(X)\otimes C(G)\quad,\quad e_j\to\sum_je_j\otimes u_{ji} [[/math]]

and then consider the tensor powers of this coaction, which are the following linear maps:

[[math]] \Phi^k:C(X^k)\to C(X^k)\otimes C(G)\quad,\quad e_{i_1\ldots i_k}\to\sum_{j_1\ldots j_k}e_{j_1\ldots j_k}\otimes u_{j_1i_1}\ldots u_{j_ki_k} [[/math]]

The fixed point spaces of these latter coactions are then given by the formula

[[math]] P_k=Fix(u^{\otimes k}) [[/math]]

and form a planar subalgebra of the spin planar algebra [math]\mathcal S_N[/math].

Show Proof

This can be done in several steps, as follows:

(1) Since the map [math]\Phi[/math] is a coaction, its tensor powers [math]\Phi^k[/math] are coactions too, and at the level of the fixed point algebras we have the following formula, which is standard:

[[math]] P_k=Fix(u^{\otimes k}) [[/math]]

(2) In order to prove now the planar algebra assertion, we use the presentation result for the spin planar algebras established before, involving the multiplications, inclusions, expectations, Jones projections and rotations.

(3) Consider the rotation [math]R_k[/math]. Rotating, then applying [math]\Phi^k[/math], and rotating backwards by [math]R_k^{-1}[/math] is the same as applying [math]\Phi^k[/math], then rotating each [math]k[/math]-fold product of coefficients of [math]\Phi[/math].

(4) Thus the elements obtained by rotating, then applying [math]\Phi^k[/math], or by applying [math]\Phi^k[/math], then rotating, differ by a sum of Dirac masses tensored with commutators in [math]A=C(G)[/math]:

[[math]] \Phi^kR_k(x)-(R_k\otimes id)\Phi^k(x)\in C(X^k)\otimes [A,A] [[/math]]

(5) Now let [math]\int_A[/math] be the Haar functional of [math]A[/math], and consider the conditional expectation onto the fixed point algebra [math]P_k[/math], which is given by the following formula:

[[math]] \phi_k=\left(id\otimes\int_A\right)\Phi^k [[/math]]

The square of the antipode being the identity, the Haar integration [math]\int_A[/math] is a trace, so it vanishes on commutators. Thus [math]R_k[/math] commutes with [math]\phi_k[/math]:

[[math]] \phi_kR_k=R_k\phi_k [[/math]]

(6) The commutation relation [math]\phi_kT=T\phi_l[/math] holds in fact for any [math](l,k)[/math]-tangle [math]T[/math]. These tangles are called annular, and the proof is by verification on generators of the annular category. In particular we obtain, for any annular tangle [math]T[/math]:

[[math]] \phi_kT\phi_l=T\phi_l [[/math]]

(7) We conclude from this that the annular category is contained in the suboperad [math]\mathcal P'\subset\mathcal P[/math] of the planar operad consisting of tangles [math]T[/math] satisfying the following condition, where [math]\phi =(\phi_k)[/math], and where [math]i(.)[/math] is the number of input boxes:

[[math]] \phi T\phi^{\otimes i(T)}=T\phi^{\otimes i(T)} [[/math]]

On the other hand the multiplicativity of [math]\Phi^k[/math] gives [math]M_k\in\mathcal P'[/math]. Since [math]\mathcal P[/math] is generated by multiplications and annular tangles, it follows that we have:

[[math]] \mathcal P'=P [[/math]]

(8) Thus for any tangle [math]T[/math] the corresponding multilinear map between spaces [math]P_k(X)[/math] restricts to a multilinear map between spaces [math]P_k[/math]. In other words, the action of the planar operad [math]\mathcal P[/math] restricts to [math]P[/math], and makes it a subalgebra of [math]\mathcal S_N[/math], as claimed.

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As a second result now, completing our study, we have:

Theorem

Given a subalgebra [math]Q\subset\mathcal S_N[/math], there is a unique quantum group

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

whose associated planar algebra is [math]Q[/math].

Show Proof

The idea is that this will follow by applying Tannakian duality to the annular category over [math]Q[/math]. Let [math]n,m[/math] be positive integers. To any element [math]T_{n+m}\in Q_{n+m}[/math] we can associate a linear map [math]L_{nm}(T_{n+m}):P_n(X)\to P_m(X)[/math] in the following way:

[[math]] L_{nm}\left(\begin{matrix}|\ |\ |\cr T_{n+m}\cr |\ |\ |\end{matrix}\right): \left(\begin{matrix}|\cr a_n\cr |\end{matrix}\right) \to \left(\begin{matrix} \hskip 1.5mm |\hskip 3.0mm |\hskip 3.0mm \cap\cr \ \ T_{n+m}\hskip 0.0mm |\cr \hskip 1.9mm |\hskip 1.2mm |\hskip 3.2mm |\hskip2.2mm |\cr a_n|\hskip 3.2mm |\hskip 2.2mm |\cr \hskip 2.1mm\cup \hskip3.5mm |\hskip 2.2mm | \end{matrix}\right) [[/math]]

That is, we consider the planar [math](n,n+m,m)[/math]-tangle having an small input [math]n[/math]-box, a big input [math]n+m[/math]-box and an output [math]m[/math]-box, with strings as on the picture of the right. This defines a certain multilinear map, as follows:

[[math]] P_n(X)\otimes P_{n+m}(X)\to P_m(X) [[/math]]

Now let us put the element [math]T_{n+m}[/math] in the big input box. We obtain in this way a certain linear map [math]P_n(X)\to P_m(X)[/math], that we call [math]L_{nm}[/math]. Now let us set:

[[math]] Q_{nm}=\left\{ L_{nm}(T_{n+m}):P_n(X)\to P_m(X)\Big| T_{n+m}\in Q_{n+m}\right\} [[/math]]

These spaces form a Tannakian category, and so by we obtain a Woronowicz algebra [math](A,u)[/math], such that the following equalities hold, for any [math]m,n[/math]:

[[math]] Hom(u^{\otimes m},u^{\otimes n})=Q_{mn} [[/math]]

We prove that [math]u[/math] is a magic unitary. We have [math]Hom(1,u^{\otimes 2})=Q_{02}=Q_2[/math], so the unit of [math]Q_2[/math] must be a fixed vector of [math]u^{\otimes 2}[/math]. But [math]u^{\otimes 2}[/math] acts on the unit of [math]Q_2[/math] as follows:

[[math]] u^{\otimes 2}(1)=\sum_{kl}\begin{pmatrix}k&k\\ l&l\end{pmatrix}\otimes (uu^t)_{kl} [[/math]]

From [math]u^{\otimes 2}(1)=1\otimes 1[/math] ve get that [math]uu^t[/math] is the identity matrix, and together with the unitarity of [math]u[/math], this gives [math]u^t=u^*=u^{-1}[/math]. Consider now the Jones projection [math]E_1\in Q_3[/math]. The linear map [math]M=L_{21}(E_1)[/math] is the multiplication [math]\delta_i\otimes\delta_j\to\delta_{ij}\delta_i[/math], and we have:

[[math]] (M\otimes id)u^{\otimes 2}\left(\begin{pmatrix}i&i\\ j&j\end{pmatrix}\otimes 1\right) =\sum_{k}\begin{pmatrix}k\\ k\end{pmatrix}\delta_k\otimes u_{ki}u_{kj} [[/math]]

[[math]] u(M\otimes id)\left(\begin{pmatrix}i&i\\ j&j\end{pmatrix}\otimes 1\right) =\sum_k\begin{pmatrix}k\\ k\end{pmatrix}\delta_k\otimes\delta_{ij}u_{ki} [[/math]]

Thus [math]u_{ki}u_{kj}=\delta_{ij}u_{ki}[/math] for any [math]i,j,k[/math], and we deduce from this that [math]u[/math] is a magic unitary. Now if [math]P[/math] is the planar algebra associated to [math]u[/math], we have [math]Hom(1,v^{\otimes n})=P_n=Q_n[/math], as desired. As for the uniqueness, this is clear from the Peter-Weyl theory from .

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The above results, following old papers from the early 00s, subsequent to , regarding the subgroups [math]G\subset S_N^+[/math], have several generalizations, to the subgroups [math]G\subset O_N^+[/math] and [math]G\subset U_N^+[/math], as well as subfactor versions, going beyond the purely combinatorial level. For the modern story, we refer here to Tarrago-Wahl and related papers.

General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].