Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The technique of faithfully flat descent, and, in the case of vector spaces, Galois descent has been used quite a bit in Algebraic Geometry. However, the question of whether, say, a given $k$-vector space $V$ arises from some $L$-vector space $W$ seems like it could be asked in a wide variety of settings. I'm wondering, in particular, if anyone has seen descent in modular representation theory.

share|improve this question
add comment

3 Answers 3

up vote 4 down vote accepted

The question is a little too vague to have a really satisfactory answer, but obviously descent is omnipresent in (modular) representation theory.

For example, the basic result that two representations of a group $G$ on some $k$-vector spaces are isomorphic iff they become isomorphic on some extension $L$ of $k$ is an instance of faithfully flat descent.

Or to take a less trivial example, let $R$ be a d.v.r with residual field $k$ algebraically closed (or just separably closed, or finite, or...) and of field of fraction $K$. Let $\rho$ be a representation of a group $G$ on some extension $L$ of $K$, whose characteristic polynomials of element of $g$ are in $R[x]$. (In characteristic zero, you can replace that condition by "character takes values in $R$"). A natural question in this context is if you can "descend" $\rho$ to a representation over $R$. For example, the answer is yes when there exists a representation of $G$ over $k$ which is absolutely irreducible and whose characteristic polynomial agree with the reduction of those of $\rho$. The proof uses a lot of faithfully flat descent via the theory of Azumaya algebras.

Edit: to answer Martin's comment, I precise how the faithful flatness plays a role in the first example.

Lemma: Let $G$ be a group. Let $k$ be a commutative noetherian ring, $k'$ a flat $k$-algebra. Let $M$, $N$, be $k[G]$-module, finitely generated as $k$-modules. let $M'=M \otimes_k k'$ and $N'=N\otimes_k k'$. Then $Hom_{k[G]}(M,N) \otimes_k k' = Hom_{k'[G]}(M',N')$. Moreover if $k'$ is a faithfully flat $k$-algebra, $Hom_{k[G]}(M,N)$ is non-zero if and only if $Hom_{k'[G]}(M',N')$ is non-zero.

Pf: Let $g_1,\dots,g_n$ be a finite family of elements of $g$ generating the image of $k[G]$ in $End_{k}(N)$. Let $t_i$ be the linear endomorphism of $H:=Hom_{k}(M,N)$ given by $\phi \mapsto \phi g_i - g_i \phi$. Consider the map $H \rightarrow H^n$ given by the $t_i$. The kernel of this map is $Hom_{k[G]}(M,N)$. the kernel of the same map tensorized by $k'$ is $Hom_{k'[G]}(M',N')$. Hence the first result by flatness. The second is clear by faithful flatness.

So for example, if $k$, $k'$ are field, $M$, $N$ simple $k[G]$-modules, one retrieve the classical assertion that $M \simeq N$ off $M' \simeq N'$. But I hope I have shown that the essence of this assertion is faithfully flat descent.

share|improve this answer
    
Sorry for this dumb question, but why does the first example work? We don't have commutative rings here, and even if $G$ is abelian I don't know why $k[G] \hookrightarrow L[G]$ is faithfully flat. –  Martin Brandenburg Apr 9 '12 at 20:21
    
If $G$ is abelian, $k[G] \hookrightarrow L[G]=k[G] \otimes_k L$ is faithfully flat because $k \rightarrow L$ is and to be faithfully felt is preserved by arbitrary base change (see e.g. Matsumura, commutative algebra, page 27). To explain why the first example works, I need more space so I am going to put it in an edit of my answer. –  Joël Apr 9 '12 at 22:01
add comment

Probably not quite what you're looking for, but it certainly involves modular representation theory, and it's really neat! A baby version of descent is used in a proof of Quillen's stratification theorem.

Suppose we have a compact Lie group, $G$, and a smooth $G$-manifold $X$ (if you'd like to feel modular, take $G$ to be finite and $X$ to be a point). We'd like to show that the map $H_G^*X \rightarrow \lim H_G^*A$ is an F-isomorphism, where cohomology has coefficients in some fixed prime, the limit is taken over a certain category involving components of fixed point subspaces and the elementary abelian subgroups of $G$, and an "F-isomorphism" is a map of rings such that all elements of the kernel are nilpotent and every element, $s$, in the codomain satisfies $s^{p^n} \in \text{image}$ for some $n$. (This implies, in particular, that the Krull dimensions of each ring are the same.) In the case of the point, the right hand side is the limit over elementary abelian subgroups of $G$.

Here's one of the ways that Quillen does this. First he constructs a space $\mathcal{F}$ as follows: Choose a faithful unitary representation of $G$, say on $U(n)$, and let $\mathcal{F} = U(n)/S$ where $S$ is the subgroup of the maximal torus consisting of elements of order $p$ (i.e. matrices with $p$-roots of unity on the diagonal and zeros elsewhere). This space has all sorts of nice properties as a $G$-manifold (in particular, all of the isotropy groups are elementary abelian subgroups.)

Now, Quillen proves the theorem for the $G$-manifold $X \times \mathcal{F}$ (notice that even if all you cared about was when $X$ is a point, you'd still need the general statement of the theorem) and then uses a fun argument using a baby version of faithfully flat descent to deduce the theorem for $X$!

I won't give the argument here, but basically he uses the sequence $X \times \mathcal{F} \times \mathcal{F} \Rightarrow X \times \mathcal{F} \rightarrow X$ (I don't know how to make a double arrow here...) and shows that applying H^* gives an equalizer sequence (this is the "descent" part), and applying the functor appearing on the right hand side of the theorem also gives an exact sequence. Then one does a 3-lemma esque argument to conclude the result. It's really neat! For a much more detailed and entertaining account, see

http://www.math.washington.edu/~mitchell/Quillen/qnew.pdf

share|improve this answer
add comment

Descent does appear in modular representation theory (of finite groups) in a slightly surprising form. Indeed, descent techniques allow us to discuss extension from $Stab(kH)$ to $Stab(kG)$, where $H$ is a subgroup of $G$, of index prime to the characteristic of the field $k$, and where $Stab(-)$ denotes the stable category of modules modulo projectives.

To understand how an extension problem turns out to be a descent problem, one should start by the following "monadicity" observation: There exists a commutative separable ring object (i.e. monoid) $A$ in the symmetric monoidal category $C:=Stab(kG)$ such that the category $A-Mod_C$ of $A$-modules in that category $C$ is equivalent to the stable category $Stab(kH)$, and this in such a way that restriction from $G$ to $H$ coincides with extension-of-scalars with respect to $A$.

So restriction to a subgroup is in fact a extension-of-scalars!

The condition that the index of $H$ in $G$ is prime to the characteristic exactly expresses the faithfulness of the ring object $A$. (Flatness is sort of built-in when you deal with tensor triangulated categories.) Hence extension from $H$ to $G$ is really faithful(ly-flat) descent!

One can then reason rather geometrically, pretty much as with (finite) etale extensions. There is even a Grothendieck topology and a stack hiding in the bushes... This is all explained in the preprint

"Stacks of group representations"

available on my publication page.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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