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 $$\Delta : A \rightarrow A \otimes A \qquad \varepsilon : A \rightarrow k$$ called comultiplication and counit respectively, which satisfy the dual axioms. See 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 $$G = \left\{ x \in A \mid \Delta(x) = x \otimes x \right\}.$$

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 $$\sum_{g \in G} a_g (g \otimes g) = \Delta(v) = \sum_{g \in G} \sum_{h \in G} a_ga_h (g \otimes h).$$ Comparing each coefficient, we get that $$a_ga_h = \begin{cases} a_g & g = h \\ 0 & \text{otherwise}. \end{cases}$$ 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 $$A \otimes (B \otimes C) \xrightarrow{a_{A,B,C}} (A \otimes B) \otimes C.$$ 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 $$F(A \otimes B) \xrightarrow{t_{A,B}} F(A) \otimes F(B).$$ 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 $$U : \mathsf{Mod}_{k[G]} \rightarrow \mathsf{Vect}_k.$$ It’s easy to see this is in fact an monoidal functor. Now let $\operatorname{Aut}^{\otimes}(U)$ be the set of monoidal automorphisms of $U$.

The key claim is the following:

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

Consider the map $$i : G \rightarrow \operatorname{Aut}^\otimes(U) \quad\text{by}\quad g \mapsto T^g.$$ Here, the natural transformation $T^g$ is defined by the components $$T^g_{(V,\phi)} : (V, \phi) \rightarrow U(V, \phi) = V \quad\text{by}\quad v \mapsto \phi(g) v.$$ Then $i$ is an isomorphism of groups.

In particular, using only $\otimes$ structure this exhibits an isomorphism $G \cong \operatorname{Aut}^\otimes(U)$. 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 $T \in \operatorname{Aut}^\otimes(U)$ 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 $$f_v : k[G] \rightarrow (V, \phi) \quad\text{by}\quad 1_{k[G]} \mapsto v.$$ If we apply the forgetful functor to this, we have a diagram

$\Box$

Next, we claim

Lemma 6. $T_{k[G]}(1_{k[G]})$ is a grouplike element of $k[G]$.

Proof: Draw the diagram and note that it implies $$\Delta(T_{k[G]}(1_{k[G]})) = T_{k[G]}(1_{k[G]}) \otimes T_{k[G]}(1_{k[G]}).$$ $\Box$

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