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Given two modules $M$ and $N$ there is a nice scheme parametrizing extensions

$0 \rightarrow M \rightarrow E \rightarrow N \rightarrow 0$

namely $Ext^1(N,M)$ or, leaving out the trivial extension, the projective space $P(Ext^1(N,M))$.

There are (at least) two natrual generalizations

  1. n-step extensions

    $M \rightarrow E_1 \rightarrow E_2 \rightarrow \dots \rightarrow E_n \rightarrow N$

    between $N$ and $M$.

  2. Filtered modules: Parametrize modules $E$ which admit a filtration

    $0 \subset F_1 \subset F_2 \dots F_n=E$

    with fixed graded objects $E_i=F_i/F_{i-1}$.

I suppose in the first case one can use the group Ext^n(M,N), although I never saw a construction of a universal family. Is there a good reference?

In the second case, I do not have a clue. So my main question is:

Is there a nice moduli space of filtered objects?

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3  
As you probably know $\mathrm{Ext}^n$ does not parameterise $n$-fold extensions directly but rather certain equivalence classes of them. The equivalence relation is the smallest with the property that $M\to E_1\to E_2\to\cdots \to N$ and $M\to F_1\to F_2\to\cdots \to N$ are equivalent when there's a chain map between them which is the identity on $M$ and $N$. When $n\ge2$ onw can add on an arbitrary module to $E_1$ and $E_2$ and get an equivalent $n$-step extension, so the equivalence classes are proper classes :-) –  Robin Chapman May 21 '10 at 13:21

3 Answers 3

I. In the first case, you can use the fact that if $$\cdots\to P_n\to P_{n-1}\to\cdots\to P_1\to P_0$$ is a projective resolution of $N$, then every Yoneda $n$-extension of $M$ by $N$ can be represented an extension of the form $$ 0\to M\to E\to P_{n-2}\to P_{n-3}\to\cdots \to P_1\to P_0\to N\to 0$$ and where $E$ is a module which is constructed as a pushout of a diagram of the form $$M\leftarrow P_n\rightarrow P_1$$ This gets you a sensible set of representatives of $n$-extensions (the isomorphism classes of $n$-extensions, as opposed to equivalece classes, do not form a set, so one needs to do something like this) which you can probably make into a scheme. You next want to quotient by equivalence---I do not see immediately how that'll work.

II. For the second case, and if you are considering finite dimensional modules over a finitely generated algebra $A$, you can construct an analogue of the representation variety $\mathrm{Rep}_d(A)$ for filtered modules with specific subquotients. For example, suppose you want a variety of modules $M$ of total dimension $d$ with a filtration $0=F_0\subseteq F_1\subseteq F_2\subseteq F_3=M$ such that $F_1/F_0\cong N_1$, $F_2/F_2\cong N_2$ and $F_3/F_2\cong N_3$. Up to isomorphism, you can suppose that $M=k^d$, and that the $F_i$ are a standard partial flag (so that $F_i$ is the subspace of $k^d$ of vectors whose last $d-\dim F_i$ coordinates vanish)

The action of $M$ is then completely given by $n$ $d$-by-$d$ matrices, where $n$ is the size of a generating set of $A$, and the fact that $M$ is an actual module, that chosen filtration is a module filtration, and that the subquotients are what they should be can be expressed in terms of polynomial equations involving the coefficients of those $n$ matrices.

This determines a scheme, whose points are $A$-module structures on $k^d$ which satisfy the desired conditions, and which contain representatives of all isoclasses of modules satisfying those conditions. Of course, the points of this scheme are not in correspondence with isoclasses: to do that, you need to pass to the quotient by the appropriate change-of-basis group (but that will kill the scheme structure, I guess...)

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Thanks for your answer. Your hint for the first point is really helpful. Ad II) I have seen people (e.g. Reineke arXiv:0802.2147) doing the moduli of quiver representations using a similar construction and GIT quotients. So I am confident that one can produce a sheme in the way you described. However, the application I have in mind goes as follows. Suppose $E_i$ are sheaves on an algebraic variety and $E_i$ are stable in some stability cndition. How many sheaves $E$ have a Hader-Narasimhan filtration with semi-stable factors $E_i$? In this case Artin-Algebras are only of limited help. –  Heinrich Hartmann May 22 '10 at 9:42
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I can almost guarantee that a scheme structure does NOT exist on such a quotient. In fact, unless I misunderstood something, this is more or less equivalent to the problem of describing a moduli space of $n\times n$-matrices up to conjugacy. And this can never exist as a moduli space, even in a coarse sense. See the paper by Mumford and Suominen "An introduction to the theory of moduli", 1970. –  Daniel Larsson May 22 '10 at 14:18
    
@Daniel, that really depends on the starting data. For example, if the algebra $A$ is the group algebra of a finite group (over a field of characteristic zero) you do get a (zero-dimensional) scheme. –  Mariano Suárez-Alvarez May 24 '10 at 15:50

Hi Heinrich,

in the situation you have in mind (sheaves on an algebraic variety), such spaces are not too difficult to construct as Artin stacks. If you omit the condition that the i-th filtration quotient is isomorphic to a given one, then such a universal Artin stack is e.g. constructed in Bridgeland's introduction to Hall-algebras (arXiv:1002.4372, he calls them $\mathcal M^{(n)}$), but of course also in earlier articles by Joyce. Basically it follows from the existence of relative quot schemes.

These universal extension stacks have evaluation morphisms to $\mathcal M$, the stack of all sheaves, sending the filtration to its i-th quotients, so you can take a base change via the map from $\Spec k \to \mathcal M \times \dots \times \mathcal M$ given by your set of objects $E_i$, and the fiber product will be the Artin stack you are looking for.

If you want a scheme instead of an Artin stack - then I would ask back "why?" :) Nevertheless, it would be useful to understand this fiber product better when $n > 2$.

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Hi Heinrich!

I don't know. But you could take a look at Carlos Simpson's general definition of "filtered object" on pages 24/25 of his paper The Hodge filtration on non-abelian cohomology: Roughly, a filtered X is a $\mathbb{G}_m$-equivariant map from an X to $\mathbb{A}^1$, so you could get your moduli space as a mapping space, or as an object of the "comma site" of maps of objects of your site into $\mathbb{A}^1$...

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Dear Peter, thanks for your answer. Unfortunately I do not see how to apply this construction to the question. It seems to me that this gives an answer to the "inverse" problem of parametrizing filtrations of a given object. –  Heinrich Hartmann May 21 '10 at 14:50
    
Well, I had thought that maybe you could look at the fibered category of all modules over the comma site/A^1. Since the fiber over 0 gives the graded object, you could maybe form a pullback where you fix the desired quotients of the filtration... But all of this is wild speculation from a layman and may be nonsense! If you like, I delete this answer to attract others! –  Peter Arndt May 21 '10 at 15:41
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At least it's abstract nonsense :) –  Lars May 21 '10 at 18:19

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