guide:Ee829b2733

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

We recall that we want to prove that we have [math]E_C^{(s)'}\subset E_{C_{G_C}}^{(s)'}[/math], for any [math]s\in\mathbb N[/math]. And for this purpose, we must first refine Theorem 14.11, in the case [math]G=G_C[/math].

Generally speaking, in order to prove anything about [math]G_C[/math], we are in need of an explicit model for this group. 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}\quad,\quad \varepsilon(u_{ij})=\delta_{ij} [[/math]]

Following , 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)\quad,\quad 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 as follows:
  [[math]] f_{ij}\leftrightarrow f_{ij}^* [[/math]]
• [math]F^*[/math] is a [math]*[/math]-bialgebra, with [math]*[/math]-bialgebra operations as follows:
  [[math]] \Delta(f_{ij})=\sum_kf_{ik}\otimes f_{kj}\quad,\quad \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:

[[math]] \lt u_{ij} \gt \simeq F^*\quad,\quad u_{ij}\to f_{ij}\quad,\quad u_{ij}^*\to f_{ij}^* [[/math]]

The corresponding [math]*[/math]-bialgebra structure induced on the vector space [math]F^*[/math] is then the one in the statement, and this gives the result.

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

Proposition

The smooth part of the algebra [math]A_C=C(G_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, for any [math]i,j[/math],

[[math]] \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} [[/math]]

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

Show Proof

As a first observation, [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]]

With this observation 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 14.15, we obtain an identification as follows:

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

With standard multi-index notations, and by assuming now that [math]k,l\in\mathbb N[/math] are usual integers, for simplifying the presentation, the general case being similar, 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]] \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} [[/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 14.11. 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, [math]\pi_v:a\to (a_k)_{kk}[/math].

Show Proof

We know from Proposition 14.13 above that we have an identification of [math]*[/math]-bialgebras [math]\mathcal A_C\simeq F^*/J[/math]. But this gives a quotient map, as follows:

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

At the dual level, this gives [math]\mathcal A_C^*\subset F[/math]. To be more precise, we have:

[[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 14.13, 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, [math]\mathcal A_C^*[/math] is given by the formula in the statement. It remains to compute [math]\pi_v[/math]:

[[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)\quad,\quad 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 then 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 category [math]C[/math] that we have [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 14.14 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_\nu[/math] in Proposition 14.14.

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Following , we can now state and prove our main result, as follows:

Theorem

The Tannakian duality constructions

[[math]] C\to G_C\quad,\quad G\to C_G [[/math]]

are inverse to each other.

Show Proof

According to our various results above, we have to prove that, for any Tannakian category [math]C[/math], and any [math]s\in\mathbb N[/math], we have an inclusion as follows:

[[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]]

In order to do so, we use the following formula, from Proposition 14.15:

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

We know from the above that we have an identification as follows:

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

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

[[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 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. We should mention that there are many other versions of this duality, the story being as follows:

• The original version of the duality, due to Tannaka and Krein, states that any compact group [math]G[/math] can be recovered from the knowledge of its category of representations [math]\mathcal R_G[/math], viewed as subcategory of the category [math]\mathcal H[/math] of the finite dimensional Hilbert spaces, with each [math]v\in\mathcal R_G[/math] corresponding to its Hilbert space [math]H_v\in\mathcal H[/math].
• Regarding the proof of this fact, this is something established long ago by Tannaka and Krein. From a more modern perspective, coming from the work of Grothendieck, Saavedra and others, the group [math]G[/math] simply appears as the group of endomorphisms of the embedding functor [math]\mathcal R_G\subset\mathcal H[/math]. See Chari-Pressley .
• With this understood, comes now a non-trivial result, due independently to Deligne and Doplicher-Roberts , stating that the compact group [math]G[/math] can be recovered from the sole knowledge of the category [math]\mathcal R_G[/math]. This is obviously something more advanced, and the proof is quite tricky.
• Getting back to Earth now, as already explained in the above, for concrete applications it is technically convenient to replace the full category [math]\mathcal R_G[/math] by its subcategory [math]\mathcal R_G^\circ[/math] consisting of the Peter-Weyl representations. And here, we have the analogue of the original result of Tannaka-Krein, explained in the above.
• For more specialized questions, it is convenient to further shrink the Peter-Weyl category [math]\mathcal R_G^\circ[/math], consisting of the spaces [math]C(k,l)=Hom(u^{\otimes k},u^{\otimes l})[/math], to its diagonal algebra [math]P_G=\Delta\mathcal R_G[/math], consisting of the [math]*[/math]-algebras [math]C(k,k)=End(u^{\otimes k})[/math]. This is the idea behind Jones' planar algebras , with [math]P_G[/math] being such an algebra.
• And this is not the end of the story, because one can still try to have Doplicher-Roberts and Deligne type results in the Peter-Weyl category setting, or in the planar algebra setting. We refer here to the modern quantum algebra literature, where Tannakian duality, in all its forms, is something highly valued.

Importantly, Tannakian duality of all sorts is just a way of “linearizing” the group, and there is as well a second method, namely considering the associated Lie algebra. Again, we refer here to the quantum algebra literature, as for instance Chari-Pressley .

General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].