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



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!

The hard part is showing that it’s surjective. For this we want to reduce it to the regular representation.

Lemma 5

Any {T \in \text{Aut }^\otimes(U)} is completely determined by {T_{k[G]}(1_{k[G]}) \in k[G]}.


Proof: Let {(V, \phi)} be a representation of {G}. Then for all {v \in V}, we have a unique morphism of representations

\displaystyle f_v : k[G] \rightarrow (V, \phi) \quad\text{by}\quad 1_{k[G]} \mapsto v.

If we apply the forgetful functor to this, we have a diagram


Next, we claim

Lemma 6

{T_{k[G]}(1_{k[G]})} is a grouplike element of {k[G]}.


Proof: Draw the diagram

rep-tan-5and note that it implies

\displaystyle \Delta(T_{k[G]}(1_{k[G]})) = T_{k[G]}(1_{k[G]}) \otimes T_{k[G]}(1_{k[G]}).

This implies surjectivity, by our earlier observation that grouplike elements in {k[G]} are exactly the elements of {G}. \Box


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