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Let $R$ be a reasonable ring (maybe I mean a PID, or $\mathbb{Z}$, and when sufficiently desperate, a field). Now consider fixed sequences $C_n$ and $H_n$ of $R$-modules, which are tame in every possible sense -- finite rank, zero for large enough $n$, etc. Here's the problem:

What is the set (space? category?) of all $R$-linear maps $d_n:C_n \to C_{n-1}$ so that $(C_\bullet,d_\bullet)$ forms a chain complex with homology groups $H_n$?

I've tried to figure things out in baby cases without much luck in terms of seeing the bigger picture. You can easily create situations where no $d_\bullet$s will suffice, but most fascinating are the cases where several different sequences of boundary operators will work!

A very general thing to do is fix a basis of all chain groups in sight and examine conditions that entries of various $d_\bullet$-representing matrices have to satisfy. Just requiring the $d_\bullet$s to produce a chain complex yields an algebraic variety generated by a system of multivariate quadratic equations (which sounds nightmarish). So unless forcing the homology to be prescribed by $H_n$s greatly simplifies things, I should not expect a very explicit answer. However, I'd like to know what the obstructions to getting such an answer might be.

In the cases of interest, all my $d_\bullet$s can be assumed to have an upper-triangular form as $R$-matrices, although I'd also be interested in a solution that ignores such structural constraints.

Update: Thanks to David Speyer's answer, which provides a wealth of information on the problem when $R$ is a field. I'm still seeking answers for more general choices of $R$.

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  • $\begingroup$ I speak from memory so I may not be that precise. You should look at the derived category of finite complexes of $R$-modules. If $R=\mathbb{Z}$, then any complex of Abelian groups is quasi-isomorphic to direct sums of trivial complexes $\cdots 0\to C_n\stackrel{0}{\to} C_{n+1}\to 0\cdots$, $n\in \mathbb{Z}$. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 16, 2015 at 16:45
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    $\begingroup$ Am I misunderstanding you? If $H_n=C_n$ for all $n$, doesn't this force all $d_n$ to vanish? $\endgroup$
    – Dan Petersen
    Commented Sep 16, 2015 at 16:52
  • $\begingroup$ @DanPetersen No, I just wrote something stupid. Fixing it now. $\endgroup$
    – Vidit Nanda
    Commented Sep 16, 2015 at 17:03
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    $\begingroup$ The complex you wrote is quasi-isomorphic to a trivial complex. Two complexes are quasi-isomorphic if there exists a morphism of complexes between them that induces an isomorphism in homology. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 16, 2015 at 17:14
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    $\begingroup$ "when sufficiently desperate, a field" I have often wondered about this question over $\mathbb{C}$, even just for exact sequences. For short exact sequences, we recover (both forms of the) Grassmannian, which is the primordial moduli space, and an endless supply of fascinating questions. I guess it would be fun to work over $\mathbb{Z}$ one day, but I don't think any desperation comes into it :) $\endgroup$
    – John Wiltshire-Gordon
    Commented Sep 16, 2015 at 18:43

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Over a field $k$, this is a special case of quiver loci. The general framework is that you are given a triangular array $(r_{ij})_{1 \leq i \leq j \leq n}$ of nonnegative integers. Consider a list of $n-1$ matrices $M_1$, $M_2$, …, $M_{n-1}$ of size $r_{ii} \times r_{(i+1)(i+1)}$. The quiver locus is the set where $\mathrm{rank}(M_i M_{i+1} \cdots M_{j-1}) \leq r_{ij}$. Your case is where $r_{ij}=0$ for $j-i \geq 2$.

See Knutson, Miller and Shimozono, and the many citations therein.

To be a bit more explicit, if you want $\dim V_i = d_i$ and $\dim H_i(V_{\bullet})=h_i$, then take $r_{ii} = d_i$, compute $r_{i(i+1)}$ by solving the linear equations $r_{(i-1)i} + r_{i(i+1)} + h_i= d_i$, and put all other $r_{ij}=0$.

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  • $\begingroup$ This certainly gives a chain complex, but how does one go about constraining kernel mod image so that we get the desired homology? $\endgroup$
    – Vidit Nanda
    Commented Sep 17, 2015 at 1:46
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    $\begingroup$ I've edited in an answer to that question. $\endgroup$
    – David E Speyer
    Commented Sep 17, 2015 at 1:55

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