40
$\begingroup$

The Tannakian formalism says you can recover a complex algebraic group from its category of finite dimensional representations, the tensor structure, and the forgetful functor to Vect. Intuitively, why should this be enough information to recover the group? And does this work for other base fields (or rings?)?

$\endgroup$

4 Answers 4

68
$\begingroup$

A very nice and very general version of Tannakian formalism is in Jacob Lurie's paper, Tannaka duality for geometric stacks, arXiv:math/0412266.

I like to think of Tannaka duality as recovering a scheme or stack from its category of coherent (or quasicoherent) sheaves, considered as a tensor category. From this POV the intuition is quite clear: having a faithful fiber functor to Vect (or more generally to R-modules) means your stack is covered (in the flat sense) by a point (or by Spec R). This is why you get (if you're over an alg closed field) that having a faithful fiber functor to Vect_k means you're sheaves on the quotient BG of a point by some group G, i.e. Rep G.. Over a more general base, you only locally look like a quotient of Spec R (or Spec k for k non-alg closed) by a group ---- ie you're a BG-bundle over Spec R, aka a G-gerbe. Even more generally, the kind of Tannakian theorem Jacob explains basically says that any stack with affine diagonal can be recovered from its tensor category of quasicoherent sheaves..

Actually the construction of the stack from the tensor category is just a version of the usual functor Spec from rings to schemes. Recall that as a functor, Spec R (k) = homomorphisms from R to k. So given a tensor category C let's define Spec C as the stack with functor of points Spec C(k) = tensor functors from C to k-modules (for any ring k, or algebra over the ground field etc). The Tannakian theorems then say for X reasonable (ie a quasicompact stack with affine diagonal), we have X= Spec Quasicoh(X) --- so X is "affine in a quasicoherent-sheaf sense". Again, the usual Tannakian story is the case X=BG or more generally a G-gerbe.

$\endgroup$
3
  • 1
    $\begingroup$ Not all reconstruction theorems could possibly fit into such a neat interpretation: while some important spectra have functoriality in the direct image direction, some others have in the inverse image direction. Some important spectra are even not compatible with localizations, what is expected in nice commutative sheaf theoretic situations. $\endgroup$ Mar 3, 2010 at 20:17
  • 2
    $\begingroup$ This is only meant to cover reconstruction theorems for symmetric monoidal categories (or their homotopical analogues), not eg the Gabriel-Rosenberg type reconstructions for abelian categories or the Bondal-Orlov reconstructions for derived categories. $\endgroup$ Mar 3, 2010 at 23:30
  • 2
    $\begingroup$ Just to make it clear: the category of quasi-coherent sheaves determines the scheme in all cases, not just in the quasi-seperated, quasi-compact case (as for example in the Gabriel-Rosenberg theorem)? $\endgroup$
    – M Turgeon
    Sep 20, 2012 at 2:52
14
$\begingroup$

This doesn't answer your question directly, but I think does explain why such theorems exist: if your notion of "representations" of some algebraic gadget are insufficient to reconstruct the gadget from the 'collection' of representations, then you better go think up better "representations"! I mean this rather loosely, so for example I'm claiming that thinking about representations of complex algebraic groups without also thinking about how they tensor would be lame.

This for example is why the "representation theory of a subfactor N < M" is all about the bimodules between N and M.

The forgetful functor to Vect I don't think of as so essential -- it lets you construct the original gadget in the category you expect, but if you don't have this function it shouldn't prevent you from believing that there's a "gaget object" in some other category.

$\endgroup$
10
$\begingroup$

I'm not sure how to answer the question about intuition. In regards to your second question, the Tannakian formalism works over any field and in fact works for more general categories than Rep(G). A category C is called a neutral Tannakian category over k (k a field) if it is a rigid abelian tensor category together with a k-linear tensor exact functor from C to k-Vect. This latter functor is known as the fiber functor (and is just the forgetful functor referenced in the original post when C = Rep(G)). In this case, there's a theorem that says any neutral Tannakian category is equivalent to the category of representations of an affine group scheme G, where G is given by the tensor automorphisms of the fiber functor.

One of the original references for this is a 1982 paper by Deligne and Milne titled 'Tannkian Categories.' Deligne also has an article in the more recent FGA explained.

My understanding is that sometimes k can even be replaced by an arbitrary commutative ring. I don't know general conditions under which this holds. However, a great example of the Tannakian formalism in action where k is a general commutative ring is the Mirkovic-Vilonen paper on the geometric Satake correspondence, which they prove in great generality (Ginzburg also has a nice paper on this topic, but only in characteristic zero). The Mirkovic-Vilonen paper can be found here http://arxiv.org/abs/math/0401222

In this paper, they prove that their fiber functor is represented by a k-module which is free over k (k any commutative ring here) and hence the Tannakian formalism still works. I don't know if this condition is also a necessary condition.

There's also a version of the Tannakian formalism over a scheme, so to speak. If we view a group G as a principal G-bundle on a point and the category of vector spaces as the category of vector bundles on a point, then we can try to generalize to a scheme. Namely, if we fix a G-bundle on a scheme X, then this induces an exact tensor functor from Rep(G) to the category of vector bundles on X using the associated bundle construction. It turns out that the converse holds: given an exact tensor functor from Rep(G) to vector bundles on X, this is equivalent to giving a G-bundle on X. There's a proof of this fact in a set of notes on the webpage for the seminar that Dennis Gaitsgory is currently running.

$\endgroup$
4
$\begingroup$

Some time ago I was also puzzled by this same question, and only now, after seeing yours, I start to think maybe I understand the idea. These are my own thoughts though, so you're encouraged to re-check them.

Consider for simplicity an affine algebraic group (on the opposite side would be a complex compact one). Then we know that the regular representation k[G] (respectively, L2(G)) should decompose as \sum R^*\otimes R.

Now what does knowing all the whole tensor category of representation imply? This means we can reconstruct the space k[G] as a G-representation by the above formula, so we know the functions on G! The multiplicatication of functions on G should be embedded in the product above, though I can't yet figure exactly how.

E.g. for a linear group G there is some representation V which contains linear functions. Then functions of the form x^n can be obtained, up to conjugacy, as the highest vectors in V\otimes ... \otimes V, taken n times.

By the correspondence between functions and spaces, this would allow us to reconstruct the original space of group G. Now the action on the space of functions translates into the action of G.

So, I think if on this road one tries to define what the point of G is, one would arrive exactly at the defintion of it as the automorphisms of the fiber functor, the way it is in the Tannakian formalism.

I wonder if somebody could give a reference to a text explaining this point of view?

$\endgroup$
2
  • $\begingroup$ The decomposition of L^2(G) is Peter-Weyl, right? I've never heard of or seen any algebraic version of Peter-Weyl. You seem to be asserting that such a theorem exists. Do you have a reference for this? $\endgroup$ Oct 30, 2009 at 21:23
  • 2
    $\begingroup$ After some thought I posted a question about this: mathoverflow.net/questions/3474/decomposition-of-kg $\endgroup$ Oct 30, 2009 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.