4c. The correspondence

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

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We recall that we want to prove that we have [math]E_C^{(s)'}\subset E_{C_{A_C}}^{(s)'}[/math], for any [math]s\in\mathbb N[/math]. For this purpose, we must first refine Theorem 4.15, in the case [math]A=A_C[/math]. In order to do so, we will use an explicit model for [math]A_C[/math]. In order to construct such a model, let [math] \lt u_{ij} \gt [/math] be the free [math]*[/math]-algebra over [math]\dim(H)^2[/math] variables, with comultiplication and counit as follows:

[[math]] \Delta(u_{ij})=\sum_ku_{ik}\otimes u_{kj} [[/math]]

[[math]] \varepsilon(u_{ij})=\delta_{ij} [[/math]]

Following Malacarne , we can model this [math]*[/math]-bialgebra, in the following way:

Proposition

Consider the following pair of dual vector spaces,

[[math]] F=\bigoplus_kB\left(H^{\otimes k}\right) [[/math]]

[[math]] F^*=\bigoplus_kB\left(H^{\otimes k}\right)^* [[/math]]

and let [math]f_{ij},f_{ij}^*\in F^*[/math] be the standard generators of [math]B(H)^*,B(\bar{H})^*[/math].

• [math]F^*[/math] is a [math]*[/math]-algebra, with multiplication [math]\otimes[/math] and involution [math]f_{ij}\leftrightarrow f_{ij}^*[/math].
• [math]F^*[/math] is a [math]*[/math]-bialgebra, with [math]\Delta(f_{ij})=\sum_kf_{ik}\otimes f_{kj}[/math] and [math]\varepsilon(f_{ij})=\delta_{ij}[/math].
• We have a [math]*[/math]-bialgebra isomorphism [math] \lt u_{ij} \gt \simeq F^*[/math], given by [math]u_{ij}\to f_{ij}[/math].

Show Proof

Since [math]F^*[/math] is spanned by the various tensor products between the variables [math]f_{ij},f_{ij}^*[/math], we have a vector space isomorphism as follows, given by [math]u_{ij}\to f_{ij},u_{ij}^*\to f_{ij}^*[/math]:

[[math]] \lt u_{ij} \gt \simeq F^* [[/math]]

The corresponding [math]*[/math]-bialgebra structure induced on [math]F^*[/math] is the one in the statement.

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Now back to our algebra [math]A_C[/math], we have the following modelling result for it:

Proposition

The smooth part of the algebra [math]A_C[/math] is given by

[[math]] \mathcal A_C\simeq F^*/J [[/math]]

where [math]J\subset F^*[/math] is the ideal coming from the following relations,

[[math]] \begin{eqnarray*} &&\sum_{p_1,\ldots,p_k}T_{i_1\ldots i_l,p_1\ldots p_k}f_{p_1j_1}\otimes\ldots\otimes f_{p_kj_k}\\ &=&\sum_{q_1,\ldots,q_l}T_{q_1\ldots q_l,j_1\ldots j_k}f_{i_1q_1}\otimes\ldots\otimes f_{i_lq_l}\quad,\quad\forall i,j \end{eqnarray*} [[/math]]

one for each pair of colored integers [math]k,l[/math], and each [math]T\in C(k,l)[/math].

Show Proof

Our first claim is that [math]A_C[/math] appears as enveloping [math]C^*[/math]-algebra of the following universal [math]*[/math]-algebra, where [math]u=(u_{ij})[/math] is regarded as a formal corepresentation:

[[math]] \mathcal A_C=\left \lt (u_{ij})_{i,j=1,\ldots,N}\Big|T\in Hom(u^{\otimes k},u^{\otimes l}),\forall k,l,\forall T\in C(k,l)\right \gt [[/math]]

Indeed, this follows from Proposition 4.4 above, because according to the result there, the relations defining [math]C(U_N^+)[/math] are included into those that we impose. With this claim in hand, the conclusion is that we have a formula as follows, where [math]I[/math] is the ideal coming from the relations [math]T\in Hom(u^{\otimes k},u^{\otimes l})[/math], with [math]T\in C(k,l)[/math]:

[[math]] \mathcal A_C= \lt u_{ij} \gt /I [[/math]]

Now if we denote by [math]J\subset F^*[/math] the image of the ideal [math]I[/math] via the [math]*[/math]-algebra isomorphism [math] \lt u_{ij} \gt \simeq F^*[/math] from Proposition 4.16, we obtain an identification as follows:

[[math]] \mathcal A_C\simeq F^*/J [[/math]]

In order to compute [math]J[/math], let us go back to [math]I[/math]. With standard multi-index notations, and by assuming that [math]k,l\in\mathbb N[/math] are usual integers, for simplifying, a relation of type [math]T\in Hom(u^{\otimes k},u^{\otimes l})[/math] inside [math] \lt u_{ij} \gt [/math] is equivalent to the following conditions:

[[math]] \begin{eqnarray*} &&\sum_{p_1,\ldots,p_k}T_{i_1\ldots i_l,p_1\ldots p_k}u_{p_1j_1}\ldots u_{p_kj_k}\\ &=&\sum_{q_1,\ldots,q_l}T_{q_1\ldots q_l,j_1\ldots j_k}u_{i_1q_1}\ldots u_{i_lq_l}\quad,\quad\forall i,j \end{eqnarray*} [[/math]]

Now by recalling that the isomorphism of [math]*[/math]-algebras [math] \lt u_{ij} \gt \to F^*[/math] is given by [math]u_{ij}\to f_{ij}[/math], and that the multiplication operation of [math]F^*[/math] corresponds to the tensor product operation [math]\otimes[/math], we conclude that [math]J\subset F^*[/math] is the ideal from the statement.

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With the above result in hand, let us go back to Theorem 4.15. We have:

Proposition

The linear space [math]\mathcal A_C^*[/math] is given by the formula

[[math]] \mathcal A_C^*=\left\{a\in F\Big|Ta_k=a_lT,\forall T\in C(k,l)\right\} [[/math]]

and the representation

[[math]] \pi_v:\mathcal A_C^*\to B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) [[/math]]

appears diagonally, by truncating, via the following formula:

[[math]] \pi_v:a\to (a_k)_{kk} [[/math]]

Show Proof

We know from Proposition 4.17 that we have:

[[math]] \mathcal A_C\simeq F^*/J [[/math]]

But this gives a quotient map [math]F^*\to\mathcal A_C[/math], and so an inclusion as follows:

[[math]] \mathcal A_C^*\subset F [[/math]]

To be more precise, we have the following formula:

[[math]] \mathcal A_C^*=\left\{a\in\ F\Big|f(a)=0,\forall f\in J\right\} [[/math]]

Now since [math]J= \lt f_T \gt [/math], where [math]f_T[/math] are the relations in Proposition 4.17, we obtain:

[[math]] \mathcal A_C^*=\left\{a\in F\Big|f_T(a)=0,\forall T\in C\right\} [[/math]]

Given [math]T\in C(k,l)[/math], for an arbitrary element [math]a=(a_k)[/math], we have:

[[math]] \begin{eqnarray*} &&f_T(a)=0\\ &\iff&\sum_{p_1,\ldots,p_k}T_{i_1\ldots i_l,p_1\ldots p_k}(a_k)_{p_1\ldots p_k,j_1\ldots j_k}=\sum_{q_1,\ldots,q_l}T_{q_1\ldots q_l,j_1\ldots j_k}(a_l)_{i_1\ldots i_l,q_1\ldots q_l},\forall i,j\\ &\iff&(Ta_k)_{i_1\ldots i_l,j_1\ldots j_k}=(a_lT)_{i_1\ldots i_l,j_1\ldots j_k},\forall i,j\\ &\iff&Ta_k=a_lT \end{eqnarray*} [[/math]]

Thus, the dual space [math]\mathcal A_C^*[/math] is given by the formula in the statement. It remains to compute the representation [math]\pi_v[/math], which appears as follows:

[[math]] \pi_v:\mathcal A_C^*\to B\left(\bigoplus_{|k|\leq s}H^{\otimes k}\right) [[/math]]

With [math]a=(a_k)[/math], we have the following computation:

[[math]] \begin{eqnarray*} \pi_v(a)_{i_1\ldots i_k,j_1\ldots j_k} &=&a(v_{i_1\ldots i_k,j_1\ldots j_k})\\ &=&(f_{i_1j_1}\otimes\ldots\otimes f_{i_kj_k})(a)\\ &=&(a_k)_{i_1\ldots i_k,j_1\ldots j_k} \end{eqnarray*} [[/math]]

Thus, our representation [math]\pi_v[/math] appears diagonally, by truncating, as claimed.

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In order to further advance, consider the following vector spaces:

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

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

We denote by [math]a\to a_s[/math] the truncation operation [math]F\to F_s[/math]. We have:

Proposition

The following hold:

• [math]E_C^{(s)'}\subset F_s[/math].
• [math]E_C'\subset F[/math].
• [math]\mathcal A_C^*=E_C'[/math].
• [math]Im(\pi_v)=(E_C')_s[/math].

Show Proof

These results basically follow from what we have, as follows:

(1) We have an inclusion as follows, as a diagonal subalgebra:

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

The commutant of this algebra is given by:

[[math]] F_s'=\left\{b\in F_s\Big|b=(b_k),b_k\in\mathbb C,\forall k\right\} [[/math]]

On the other hand, we know from the identity axiom for the catrgory [math]C[/math] that this algebra is contained inside [math]E_C^{(s)}[/math]:

[[math]] F_s'\subset E_C^{(s)} [[/math]]

Thus, our result follows from the bicommutant theorem, as follows:

[[math]] F_s'\subset E_C^{(s)}\implies F_s\supset E_C^{(s)'} [[/math]]

(2) This follows from (1), by taking inductive limits.

(3) With the present notations, the formula of [math]\mathcal A_C^*[/math] from Proposition 4.18 reads:

[[math]] \mathcal A_C^*=F\cap E_C' [[/math]]

Now since by (2) we have [math]E_C'\subset F[/math], we obtain from this [math]\mathcal A_C^*=E_C'[/math].

(4) This follows from (3), and from the formula of [math]\pi_v[/math] in Proposition 4.18.

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Still following the paper of Malacarne , we can now state and prove our main result, originally due to Woronowicz , as follows:

Theorem

The Tannakian duality constructions

[[math]] C\to A_C\quad,\quad A\to C_A [[/math]]

are inverse to each other, modulo identifying full and reduced versions.

Show Proof

According to Proposition 4.9, Proposition 4.12, Theorem 5.15 and Proposition 4.19, we have to prove that, for any Tannakian category [math]C[/math], and any [math]s\in\mathbb N[/math]:

[[math]] E_C^{(s)'}\subset(E_C')_s [[/math]]

By taking duals, this is the same as proving that we have:

[[math]] \left\{f\in F_s^*\Big|f_{|(E_C')_s}=0\right\}\subset\left\{f\in F_s^*\Big|f_{|E_C^{(s)'}}=0\right\} [[/math]]

For this purpose, we use the following formula, coming from Proposition 4.19:

[[math]] \mathcal A_C^*=E_C' [[/math]]

We know as well that we have the following formula:

[[math]] \mathcal A_C=F^*/J [[/math]]

We conclude that the ideal [math]J[/math] is given by:

[[math]] J=\left\{f\in F^*\Big|f_{|E_C'}=0\right\} [[/math]]

Our claim is that we have the following formula, for any [math]s\in\mathbb N[/math]:

[[math]] J\cap F_s^*=\left\{f\in F_s^*\Big|f_{|E_C^{(s)'}}=0\right\} [[/math]]

Indeed, let us denote by [math]X_s[/math] the spaces on the right. The categorical axioms for [math]C[/math] show that these spaces are increasing, that their union [math]X=\cup_sX_s[/math] is an ideal, and that:

[[math]] X_s=X\cap F_s^* [[/math]]

We must prove that we have [math]J=X[/math], and this can be done as follows:

“[math]\subset[/math]” This follows from the following fact, for any [math]T\in C(k,l)[/math] with [math]|k|,|l|\leq s[/math]:

[[math]] \begin{eqnarray*} (f_T)_{|\{T\}'}=0 &\implies&(f_T)_{|E_C^{(s)'}}=0\\ &\implies&f_T\in X_s \end{eqnarray*} [[/math]]

“[math]\supset[/math]” This follows from our description of [math]J[/math], because from [math]E_C^{(s)}\subset E_C[/math] we obtain:

[[math]] f_{|E_C^{(s)'}}=0\implies f_{|E_C'}=0 [[/math]]

Summarizing, we have proved our claim. On the other hand, we have:

[[math]] \begin{eqnarray*} J\cap F_s^* &=&\left\{f\in F^*\Big|f_{|E_C'}=0\right\}\cap F_s^*\\ &=&\left\{f\in F_s^*\Big|f_{|E_C'}=0\right\}\\ &=&\left\{f\in F_s^*\Big|f_{|(E_C')_s}=0\right\} \end{eqnarray*} [[/math]]

Thus, our claim is exactly the inclusion that we wanted to prove, and we are done.

■

Summarizing, we have proved Tannakian duality. As already mentioned in the beginning of this chapter, there are many other forms of Tannakian duality for the compact quantum groups, and we refer here to Woronowicz's original paper , which contains a full discussion of the subject, and to the subsequent literature.

As we will see in a moment, Tannakian duality in the above form is something quite powerful, enabling us to recover the Brauer theorems for [math]O_N,U_N[/math], going back to Brauer's paper , and for their free versions [math]O_N^+,U_N^+[/math] as well, following , .

Later on, in chapter 7 below and afterwards, we will further build on Tannakian duality, with a subsequent notion of “easiness” coming from it. Let us also mention, for the concerned reader, that all this escalation of algebraic methods will eventually lead into very concrete applications, of analytic and probabilistic nature.

As a first application now, let us record the following theoretical fact, from :

Theorem

Each closed subgroup [math]G\subset U_N^+[/math] appears as an algebraic manifold of the free complex sphere,

[[math]] G\subset S^{N^2-1}_{\mathbb C,+} [[/math]]

the embedding being given by the following formula, in double indices:

[[math]] x_{ij}=\frac{u_{ij}}{\sqrt{N}} [[/math]]

Show Proof

This follows from Theorem 4.20, by using the following inclusions:

[[math]] G\subset U_N^+\subset S^{N^2-1}_{\mathbb C,+} [[/math]]

Indeed, both these inclusions are algebraic, and this gives the result.

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General references

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