# Tannakian description of a semi-direct product

Let $H$ and $K$ be affine group schemes over a field $k$ of caracteristic zero. Let $\varphi:H\to Aut(K)$ be a group action. Then we can form the semi-direct product $G = K\ltimes H$.

Problem: Describe the tannakian category $Rep_k(G)$ in terms of $Rep_k(K)$, $Rep_k(H)$ and $\varphi$.

If $\varphi$ is the trivial action, $G$ is the direct product $K\times H$. In this case, $Rep_k(G) = Rep_k(K) \boxtimes Rep_k(G)$ is just the Deligne tensor product of the tannakian categories.

Question: What happens in the opposite scenario where the action $\varphi$ is faithful? Can one characterize this situation in terms of Ext groups of simple objects in $Rep_k(K)$ and $Rep_k(H)$?

I'm mostly insterested in the case where $K$ and $H$ are extensions of $\mathbb{G}_m$ by a pro-unipotent group but I have no idea where to start.

-
You might want to look in Weintraub: Representation Theory of Finite Groups on page 125. Construction 2.13 describes how you can compute the irreducible representation of a semidirect product with $H$ abelian. You have to consider a direct sum over the $H$ orbits on $\widehat{K}$ and then the tannakian category of the stabilizer in $H$ of a character, I guess. That includes your example, if $K$ is abelian. –  Marc Palm May 3 '11 at 8:46

The algebraic stack BG is the quotient of BK by the action of H, induced by its action on K. So we can describe coherent sheaves on it (aka reps of G) via descent from BK - i.e. reps of G are H-equivariant sheaves on BK, or H-equivariant objects in Rep K. Now this is not yet the answer you want, since it involves Rep K and H, rather than Rep H.

One indirect (and maybe slightly imprecise) answer is to use the Morita equivalence between Rep H-module categories and categories with H action: the Rep H-module category Rep G corresponds under this equivalence to Rep K as a category with H action. One direction takes a category with H action to its H-equivariant objects, which carry a natural Rep H action. The other direction takes a Rep H category to the category of its eigenobjects. Here an eigenobject is an object $M\in C$ with a functorial identification $$V\ast M \simeq \underline{V}\otimes M$$ of the module action of $V\in Rep K$ with the naive tensor product by the underlying vector space of $V$. (see e.g. this paper) -- again this is just playing with monadic formalism.. So I would then characterize Rep G as the Rep H-module (via induction of representations) whose eigenobject category is Rep K as an H-category..

This Morita equivalence for finite groups appears in papers of Mueger and Ostrik cited in here. For affine group schemes I proved a derived version of this result with Nadler and Francis, but it's not available sadly. I haven't thought this through in the usual Tannakian setting, so maybe I'm missing something obvious, but it should be a fairly straightforward (though 2-categorical!) application of Barr-Beck I would think: we're simply claiming that categories over BH are described via descent from a point, and the descent data is an action of H.

-
A more Tannakian answer along the same lines is relative Tannakian theory. Restriction to $H$ gives a fibre functor $Rep_k(G) \to Rep_k(H)$ and the surjection $G\to H$ gives a "constant object" functor $Rep_k(H) \to Rep_k(G)$. One can imitate the usual Tannaka theory replacing $Rep_k(1)=Vec_k$ by $Rep_k(H)$ getting a description of $Rep_k(G)$ as the category of modules in $Rep_k(H)$ of some group scheme in $Rep_k(H)$. .... –  Torsten Ekedahl May 14 '10 at 4:18
... In this case the group scheme is just $K$ with the given action of $H$ (of course it may not be a finite group scheme so its affine algebra may not lie in $Rep_k(H)$ but rather in the category of ind-objects for $Rep_k(H)$. –  Torsten Ekedahl May 14 '10 at 4:18
Torsten - thanks, that is more Tannakian! (I guess the description I gave should extend to nonsplit group extensions, but in the split case may be overkill) –  David Ben-Zvi May 14 '10 at 13:32
@David: Yes, the non-split case is a problem for the relative Tannakian which will no doubt result in me actually reading through your answer :-) –  Torsten Ekedahl May 14 '10 at 15:47

This is just a reiteration of David's answer: in this situation the group $H$ acts on the category $Rep(K)$ (an element $h\in H$ sends a representation of $K$ into its twist by $\phi(h)$; I should add that the notion of an action of affine group on abelian category is a bit subtle..). In this language, the category $Rep(G)$ is just an "equivariantization" of the category $Rep(K)$ (in other words the objects of $Rep(G)$ are just $H-$equivariant objects of $Rep(K)$).

-

Even when viewed as an additive category, Rep(G) is not semisimple, so it's not really clear to me what such a description would entail... But simple objects in it are (in a different language) irreducible representations of a semidirect product of groups and Mackey theory was invented precisely with the goal of determining them. The answer, in short, is that each simple G-module has the form $\operatorname{Ind}_{L}^{G}(\sigma^{\prime})$, where $\sigma$ is a simple $H$-module, $K_\sigma$ is the stabilizer of $\sigma$ in $K$ (i.e. consists of all elements $k\in K$ s.t. the action of $k$ on $H$ conjugates $\sigma$ into an isomorphic module), $L=K_{\sigma}H<G$ and $\sigma^{\prime}$ is the natural extension of $\sigma$ to $L$. The catch is that the usual Mackey theory applies to unitary representations (which can be infinite-dimensional) of topological groups. Nevertheless, induction functor is defined in the algebraic setting and I think that the "Mackey machine" works for formal reasons (this must be described in Jantzen, but I don't have it close at hand to check).

Returning to the full category Rep(G), there are various filtrations of finite-length representations with simple quotients, and one can take tensor products of filtered objects. However, in general one doesn't expect a manageable description of Rep(G) even in the special situation of a semidirect product of a unipotent group and a torus: this already includes the case when G is the Borel subgroup of a semisimple algebraic group, which has been studied but is not completely understood. Good luck!

Addendum Of course, if G is a solvable connected algebraic group, by Lie – Kolchin every irreducible representation is one-dimensional, so simple G-modules are the same as simple G/[G,G]-modules, which have an easy description.

-