Tannakian Reconstruction

These notes are from the February 23, 2016 lecture of 18.757, Representations of Lie Algebras, taught by Laura Rider.

Fix a field ${k}$ and let ${G}$ be a finite group. In this post we will show that one can reconstruct the group ${G}$ from the monoidal category of ${k[G]}$ -modules (i.e. its ${G}$ -representations).

1. Hopf algebras

We won’t do anything with Hopf algebras per se, but it will be convenient to have the language.

Recall that an associative ${k}$ -algebra is a ${k}$ -vector space ${A}$ equipped with a map ${m : A \otimes A \rightarrow A}$ and ${i : k \hookrightarrow A}$ (unit), satisfying some certain axioms.

Then a ${k}$ -coalgebra is a map

$\displaystyle \Delta : A \rightarrow A \otimes A \qquad \varepsilon : A \rightarrow k$

called comultiplication and counit respectively, which satisfy the dual axioms. See \url{https://en.wikipedia.org/wiki/Coalgebra}.

Now a Hopf algebra ${A}$ is a bialgebra ${A}$ over ${k}$ plus a so-called antipode ${S : A \rightarrow A}$ . We require that the diagram

commutes.

Given a Hopf algebra ${A}$ group-like element in ${A}$ is an element of

$\displaystyle G = \left\{ x \in A \mid \Delta(x) = x \otimes x \right\}.$

Exercise 1

Show that ${G}$ is a group with multiplication ${m}$ and inversion ${S}$ .

Now the example

Example 2 (Group algebra is Hopf algebra)

The group algebra ${k[G]}$ is a Hopf algebra with

${m}$ , ${i}$ as expected.
${\varepsilon}$ the counit is the trivial representation.
${\Delta}$ comes form ${g \mapsto g \otimes g}$ extended linearly.
${S}$ takes ${g \mapsto g^{-1}}$ extended linearly.

Theorem 3

The group-like elements are precisely the basis elements ${1_k \cdot g \in k[g]}$ .

Proof: Assume ${V = \sum_{g \in G} a_g g}$ is grouplike. Then by assumption we should have

$\displaystyle \sum_{g \in G} a_g (g \otimes g) = \Delta(v) = \sum_{g \in G} \sum_{h \in G} a_ga_h (g \otimes h).$

Comparing each coefficient, we get that

$\displaystyle a_ga_h = \begin{cases} a_g & g = h \\ 0 & \text{otherwise}. \end{cases}$

This can only occur if some ${a_g}$ is ${1}$ and the remaining coefficients are all zero. $\Box$

2. Monoidal functors

Recall that monoidal category (or tensor category) is a category ${\mathscr C}$ equipped with a functor ${\otimes : \mathscr C \times \mathscr C \rightarrow \mathscr C}$ which has an identity ${I}$ and satisfies some certain coherence conditions. For example, for any ${A,B,C \in \mathscr C}$ we should have a natural isomorphism

$\displaystyle A \otimes (B \otimes C) \xrightarrow{a_{A,B,C}} (A \otimes B) \otimes C.$

The generic example is of course suggested by the notation: vector spaces over ${k}$ , abelian groups, or more generally modules/algebras over a ring ${R}$ .

Now take two monoidal categories ${(\mathscr C, \otimes_\mathscr C)}$ and ${(\mathscr D, \otimes_\mathscr D)}$ . Then a monoidal functor ${F : \mathscr C \rightarrow \mathscr D}$ is a functor for which we additionally need to select an isomorphism

$\displaystyle F(A \otimes B) \xrightarrow{t_{A,B}} F(A) \otimes F(B).$

We then require that the diagram

commutes, plus some additional compatibility conditions with the identities of the ${\otimes}$ ‘s (see Wikipedia for the list).

We also have a notion of a natural transformation of two functors ${t : F \rightarrow G}$ ; this is just making the squares

commute. Now, suppose ${F : \mathscr C \rightarrow \mathscr C}$ is a monoidal functor. Then an automorphism of ${F}$ is a natural transformation ${t : F \rightarrow F}$ which is invertible, i.e. a natural isomorphism.

3. Application to ${k[G]}$

With this language, we now reach the main point of the post. Consider the category of ${k[G]}$ modules endowed with the monoidal ${\otimes}$ (which is just the tensor over ${k}$ , with the usual group representation). We want to reconstruct ${G}$ from this category.

Let ${U}$ be the forgetful functor

$\displaystyle U : \mathsf{Mod}_{k[G]} \rightarrow \mathsf{Vect}_k.$

It’s easy to see this is in fact an monoidal functor. Now let ${\text{Aut }^{\otimes}(U)}$ be the set of monoidal automorphisms of ${U}$ .

The key claim is the following:

Theorem 4 ( ${G}$ is isomorphic to ${\text{Aut }^\otimes(U)}$ )

Consider the map

$\displaystyle i : G \rightarrow \text{Aut }^\otimes(U) \quad\text{by}\quad g \mapsto T^g.$

Here, the natural transformation ${T^g}$ is defined by the components

$\displaystyle T^g_{(V,\phi)} : (V, \phi) \rightarrow U(V, \phi) = V \quad\text{by}\quad v \mapsto \phi(g) v.$

Then ${i}$ is an isomorphism of groups.

In particular, using only ${\otimes}$ structure this exhibits an isomorphism ${G \cong \text{Aut }^\otimes(U)}$ . Consequently this solves the problem proposed at the beginning of the lecture.

Proof: It’s easy to see ${i}$ is a group homomorphism.

To see it’s injective, we show ${1_G \neq g \in G}$ gives ${T^g}$ isn’t the identity automorphism. i.e. we need to find some representation for which ${g}$ acts nontrivially on ${V}$ . Now just take the regular representation, which is faithful!