guide:Fc24d4509f

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

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In order to prove that we have [math]C_{A_C}\subset C[/math], for any Tannakian category [math]C[/math], let us begin with some abstract constructions. Following Malacarne , we have:

Proposition

Given a tensor category [math]C=C((k,l))[/math] over a Hilbert space [math]H[/math],

[[math]] E_C^{(s)} =\bigoplus_{|k|,|l|\leq s}C(k,l) \subset\bigoplus_{|k|,|l|\leq s}B(H^{\otimes k},H^{\otimes l}) = B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) [[/math]]

is a finite dimensional [math]C^*[/math]-subalgebra. Also,

[[math]] E_C =\bigoplus_{k,l}C(k,l) \subset\bigoplus_{k,l}B(H^{\otimes k},H^{\otimes l}) \subset B\left(\bigoplus_kH^{\otimes k}\right) [[/math]]

is a closed [math]*[/math]-subalgebra.

Show Proof

This is clear indeed from the categorical axioms from Definition 4.6, via the standard embeddings and isomorphisms in the statement.

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Now back to our reconstruction question, given a tensor category [math]C=(C(k,l))[/math], we want to prove that we have [math]C=C_{A_C}[/math], which is the same as proving that we have:

[[math]] E_C=E_{C_{A_C}} [[/math]]

Equivalently, we want to prove that we have equalities as follows, for any [math]s\in\mathbb N[/math]:

[[math]] E_C^{(s)}=E_{C_{A_C}}^{(s)} [[/math]]

The problem, however, is that these equalities are not easy to establish directly. In order to solve this question, we will use a standard commutant trick, as follows:

Theorem

For any [math]C^*[/math]-algebra [math]B\subset M_n(\mathbb C)[/math] we have the formula

[[math]] B=B'' [[/math]]

where prime denotes the commutant, given by:

[[math]] A'=\left\{T\in M_n(\mathbb C)\Big|Tx=xT,\forall x\in A\right\} [[/math]]

Show Proof

This is a particular case of von Neumann's bicommutant theorem, which follows as well from the explicit description of [math]B[/math] given in chapter 3 above. To be more precise, let us decompose [math]B[/math] as there, as a direct sum of matrix algebras:

[[math]] B=M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C) [[/math]]

The center of each matrix algebra being reduced to the scalars, the commutant of this algebra is then as follows, with each copy of [math]\mathbb C[/math] corresponding to a matrix block:

[[math]] B'=\mathbb C\oplus\ldots\oplus\mathbb C [[/math]]

By taking once again the commutant, and computing over the matrix blocks, we obtain the algebra [math]B[/math] itself, and this gives the formula in the statement.

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Now back to our questions, we recall that we want to prove that we have [math]C=C_{A_C}[/math], for any Tannakian category [math]C[/math]. By using the bicommutant theorem, we have:

Proposition

Given a Tannakian category [math]C[/math], the following are equivalent:

• [math]C=C_{A_C}[/math].
• [math]E_C=E_{C_{A_C}}[/math].
• [math]E_C^{(s)}=E_{C_{A_C}}^{(s)}[/math], for any [math]s\in\mathbb N[/math].
• [math]E_C^{(s)'}=E_{C_{A_C}}^{(s)'}[/math], for any [math]s\in\mathbb N[/math].

In addition, the inclusions [math]\subset[/math], [math]\subset[/math], [math]\subset[/math], [math]\supset[/math] are automatically satisfied.

Show Proof

This follows from the above results, as follows:

[math](1)\iff(2)[/math] This is clear from definitions.

[math](2)\iff(3)[/math] This is clear from definitions as well.

[math](3)\iff(4)[/math] This comes from the bicommutant theorem. As for the last assertion, we have indeed [math]C\subset C_{A_C}[/math] from Proposition 4.9, and this shows that we have as well:

[[math]] E_C\subset E_{C_{A_C}} [[/math]]

We therefore obtain the following inclusion:

[[math]] E_C^{(s)}\subset E_{C_{A_C}}^{(s)} [[/math]]

By taking now the commutants, this gives:

[[math]] E_C^{(s)}\supset E_{C_{A_C}}^{(s)} [[/math]]

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

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Summarizing, in order to finish, given a tensor category [math]C=(C(k,l))[/math], we would like to prove that we have inclusions as follows, for any [math]s\in\mathbb N[/math]:

[[math]] E_C^{(s)'}\subset E_{C_{A_C}}^{(s)'} [[/math]]

Let us first study the commutant on the right. As a first observation, we have:

Proposition

Given a Woronowicz algebra [math](A,u)[/math], we have

[[math]] E_{C_A}^{(s)}=End\left(\bigoplus_{|k|\leq s}u^{\otimes k}\right) [[/math]]

as subalgebras of the following algebra:

[[math]] B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) [[/math]]

Show Proof

The category [math]C_A[/math] is by definition given by:

[[math]] C_A(k,l)=Hom(u^{\otimes k},u^{\otimes l}) [[/math]]

Thus, according to the various identifications in Proposition 4.10 above, the corresponding algebra [math]E_{C_A}^{(s)}[/math] appears as follows:

[[math]] \begin{eqnarray*} E_{C_A}^{(s)} &=&\bigoplus_{|k|,|l|\leq s}Hom(u^{\otimes k},u^{\otimes l})\\ &\subset&\bigoplus_{|k|,|l|\leq s}B(H^{\otimes k},H^{\otimes l})\\ &=&B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) \end{eqnarray*} [[/math]]

On the other hand, the algebra of intertwiners of [math]\bigoplus_{|k|\leq s}u^{\otimes k}[/math] is given by:

[[math]] \begin{eqnarray*} End\left(\bigoplus_{|k|\leq s}u^{\otimes k}\right) &=&\bigoplus_{|k|,|l|\leq s}Hom(u^{\otimes k},u^{\otimes l})\\ &\subset&\bigoplus_{|k|,|l|\leq s}B(H^{\otimes k},H^{\otimes l})\\ &=&B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) \end{eqnarray*} [[/math]]

Thus we have indeed the same algebra, and we are done.

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In practice now, we have to compute the commutant of the above algebra. And for this purpose, we can use the following general result:

Proposition

Given a corepresentation [math]v\in M_n(A)[/math], we have a representation

[[math]] \pi_v:A^*\to M_n(\mathbb C) [[/math]]

[[math]] \varphi\to (\varphi(v_{ij}))_{ij} [[/math]]

whose image is given by the following formula:

[[math]] Im(\pi_v)=End(v)' [[/math]]

Show Proof

The first assertion is clear, with the multiplicativity claim coming from:

[[math]] \begin{eqnarray*} (\pi_v(\varphi*\psi))_{ij} &=&(\varphi\otimes\psi)\Delta(v_{ij})\\ &=&\sum_k\varphi(v_{ik})\psi(v_{kj})\\ &=&\sum_k(\pi_v(\varphi))_{ik}(\pi_v(\psi))_{kj}\\ &=&(\pi_v(\varphi)\pi_v(\psi))_{ij} \end{eqnarray*} [[/math]]

Let us first prove the inclusion [math]\subset[/math]. Given [math]\varphi\in A^*[/math] and [math]T\in End(v)[/math], we have:

[[math]] \begin{eqnarray*} [\pi_v(\varphi),T]=0 &\iff&\sum_k\varphi(v_{ik})T_{kj}=\sum_kT_{ik}\varphi(v_{kj}),\forall i,j\\ &\iff&\varphi\left(\sum_kv_{ik}T_{kj}\right)=\varphi\left(\sum_kT_{ik}v_{kj}\right),\forall i,j\\ &\iff&\varphi((vT)_{ij})=\varphi((Tv)_{ij}),\forall i,j \end{eqnarray*} [[/math]]

But this latter formula is true, because [math]T\in End(v)[/math] means that we have:

[[math]] vT=Tv [[/math]]

As for the converse inclusion [math]\supset[/math], the proof is quite similar. Indeed, by using the bicommutant theorem, this is the same as proving that we have:

[[math]] Im(\pi_v)'\subset End(v) [[/math]]

But, by using the above equivalences, we have the following computation:

[[math]] \begin{eqnarray*} T\in Im(\pi_v)' &\iff&[\pi_v(\varphi),T]=0,\forall\varphi\\ &\iff&\varphi((vT)_{ij})=\varphi((Tv)_{ij}),\forall\varphi,i,j\\ &\iff&vT=Tv \end{eqnarray*} [[/math]]

Thus, we have obtained the desired inclusion, and we are done.

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By combining now the above results, we obtain:

Theorem

Given a Woronowicz algebra [math](A,u)[/math], we have

[[math]] E_{C_A}^{(s)'}=Im(\pi_v) [[/math]]

as subalgebras of the following algebra,

[[math]] B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) [[/math]]

where the corepresentation [math]v[/math] is the sum

[[math]] v=\bigoplus_{|k|\leq s}u^{\otimes k} [[/math]]

and where [math]\pi_v:A^*\to M_n(\mathbb C)[/math] is given by [math]\varphi\to(\varphi(v_{ij}))_{ij}[/math].

Show Proof

This follows indeed from Proposition 4.13 and Proposition 4.14.

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Summarizing, we have some advances on the duality question, with the whole problem tending to become something quite concrete, which can be effectively solved.

General references

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