Let $E$ be your favorite elliptic curve, and let $Tor^m$ be the moduli stack of torsion sheaves of degree $m$ on $E$. This sounds horrible, but it's not so bad; it's a global quotient of a smooth Quot scheme (the space of rank m subbundles of degree -m in $\mathcal{O}_E^{\oplus m}$) modulo the obvious action of GL(m).

For various nefarious reasons of my own, I would like to have an explicit presentation of the cohomology ring of this stack (i.e. the equivariant cohomology of this Quot scheme for GL(m)).

Does anyone know of such a presentation?

EDIT: Since Torsten asked in comments, I'd be perfectly happy just to understand the rational cohomology, though obviously integral would be even better. I would also mention that I really want the ring structure, not just the vector spaces.

  • $\begingroup$ I'm a bit confused. Isn't the stack $Tor^m$ the quotient of an open subset of the Quot scheme? Namely one considers the subscheme $Q^0$ of quotients $h:\mathcal{O}_E^{\oplus m}\to T$ where $T$ is torsion of length 2 and such that $H^0(h)$ is an isomorphism... Then $Tor^m = Q^0/GL_m$. $\endgroup$ Jan 11, 2013 at 16:35
  • $\begingroup$ I meant of length $m$ not of length 2. $\endgroup$ Jan 11, 2013 at 17:40

2 Answers 2


It seems that one can obtain the additive structure of rational cohomology without too much effort (in no way have I checked this carefully so caveat lector applies). As Allen noticed, for rational cohomology it is enough to compute $T$-equivariant cohomology and then take $\Sigma_m$-invariants (if this is to work also for integral cohomology a more careful analysis would have to be made I think). Now, the space $Tor^m_C$ of length $m$ quotients of $\mathcal O^m_C$ (I use $C$ as everything I say will work for any smooth and proper curve $C$) is smooth and proper so we may use the Bialinski-Birula analysis (choosing a general $\rho\colon \mathrm{G}_m \to T$) and we first look at the fixed point locus of $T$.

Now, for every sequence $(k_1,\ldots,k_m)$ of non-negative integers with $k_1+\ldots+k_m=m$ gives a map $S^{k_1}C\times\cdots\times S^{k_m}C \to Tor^m_C$, where $S^kC$ is the symmetric product interpreted as a Hilber scheme, and the map takes $(\mathcal I_1,\ldots,\mathcal I_m)$ to $\bigoplus_i\mathcal I_i \hookrightarrow \mathcal O^m_C$. It is clear that this lands in the $T$-fixed locus and almost equally clear that this is the whole $T$-fixed locus (any $T$-invariant submodule must be the direct sum of its weight-spaces).

We can now use $\rho$ to get a stratification parametrised by the sequences $(k_1,\ldots,k_m)$. Concretely, the tangent space of $Tor^m_C$ to a point of $S^{k_1}C\times\cdots\times S^{k_m}C$ has character $(k_1\alpha_1+\cdots+k_m\alpha_m)\beta$, where $\beta=\sum_i\alpha_i^{-1}$ (and we think of characters as elements of the group ring of the character group of $T$ and the $\alpha_i$ are the natural basis elements of the character group). This shows that $\rho$ can for instance be chosen to be $t \mapsto (1,t,t^2,\ldots,t^{m-1})$. In any case the stratum corresponding to $(k_1,\ldots,k_m)$ has as character for its tangent space the characters on the tangent space on which $T'$ is non-negative and its normal bundle consists of those on which $T'$ is non-negative (hence with the above choice, the character on the tangent space is $\sum_{i\geq j}k_i\alpha_i\alpha_j^{-1}$ and in particular the dimension of the stratum is $\sum_iik_i$). We now have that each stratum is a vector bundle of the corresponding fixed point locus so in particular the equivariant cohomology of it is the equivariant cohomology of the fixed point locus and in particular is free over the cohomology ring of $T$. Furthermore, if we build up the cohomology using the stratification (and the Gysin isomorphism), at each stage a long exact sequence splits once we have shown that the top Chern class of the normal bundle is a non-zero divisor in the cohomology of the equivariant cohomology of the component of the fixed point locus (using the Atiyah-Bott criterion).

However, the non-zero divisor condition seems more or less automatic (and that fact should be well-known): The normal bundle $\mathcal N$ of the $(k_1,\ldots,k_m)$-part, $F$ say, of the fixed point locus splits up as direct sum $\bigoplus_\alpha \mathcal N_\alpha$, where $T$ acts by the character $\alpha$ on $\mathcal N_\alpha$. Then the equivariant total Chern class of $\mathcal N_\alpha$ inside of $H^\ast_T(F)=H^\ast(F)\bigotimes H^\ast_T(pt)$ is the Chern polynomial of $\mathcal N_\alpha$ as ordinary vector bundle evaluated at $c_1(\alpha)=\alpha\in H^2_T(pt)$. Hence, if we quasi-order the characters of $T$ by using $\beta \mapsto -\rho(\beta)$ (``quasi'' as many characters get the same size), then as $-\rho(\alpha)>0$ (because $\alpha$ appears in the normal bundle) we get that $1\otimes\alpha^{n_\alpha}\in H^\ast_T(F)$, where $n_\alpha$ is the rank of $\mathcal N_\alpha$, is the term of $c_n(\mathcal N_\alpha)$ of largest order. Hence, we get that for the top Chern class of $\mathcal N$ which is the product $\prod_\alpha c_{n_\alpha}(\mathcal N_\alpha)$ its term of largest order has $1$ as $H^\ast(F)$-coefficient and hence is a non-zero divisor.

As the cohomology of $S^nC$ is torsion free we get that all the involved cohomology is also torsion free and everything works over the integers but as I've said to go integrally from $T$-equivariant cohomology to $\mathrm{GL}_m$-equivariant cohomology is probably non-trivial.

If one wants to get a hold on the multiplicative structure one could use that the fact that the Atiyah-Bott criterion works implies that that $H^\ast_T(Tor^m_C)$ injects into the equivariant cohomology of the fixed point locus. The algebra structure of the cohomology of $S^nC$ is clear (at least rationally) so we get an embedding into something with known multiplicative structure. The tricky thing may be to determine the image. We do get a lower bound for the image by looking at the ring generated by the Chern classes of of the tautological bundle but I have no idea how close that would get us to the actual image. (It is a well-known technique anyway used in for instance equivariant Schubert calculus so there could be known tricks.)

There is another source of elements, namely we have the map $Tor^m_C \to S^mC$. This map becomes even better if one passes to the $\Sigma_m$-invariants.

[Later] Upon further thought I realise that the relation between $T$- and $G=GL_m$-equivariant cohomology is simpler than I thought. The point is (and more knowledgeable people certainly know this) that the map $EG\times_TX \to EG\times_GX$ is a $G/T$ bundle and $G$ being special $H^*_T(X)$ is free as a $H^*_G(X)$-module (with $1$ being one of the basis elements). That means that $H^*_G(X) \to H^*_T(X)$ is injective but more precisely $H^*_T(X)/H^*_G(X)$ is torsion free with $\Sigma_m$-action without invariants so that $H^*_G(X)$ is the ring of $\Sigma_m$-invariants of $H^*_T(X)$.


How about your coauthor's paper http://arxiv.org/abs/math.AG/0602161 ? That computes T^m-equivariant cohomology, in which the GL(m)-equivariant sits as the S_m-invariant part.

EDIT: As Ben points out, he asked for elliptic curves, whereas the Braden-Chen-Sottile paper is about rational curves. Oops.

  • 1
    $\begingroup$ As far as I can tell, that paper is about the P^1 version of this question. Which is of course still useful, just different (in particular, life is easier when your basis is toric, since then you have an extra action to cut things down with). $\endgroup$
    – Ben Webster
    Mar 12, 2010 at 1:59
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    $\begingroup$ Also, in my defense, that paper is from before I had met Tom. $\endgroup$
    – Ben Webster
    Mar 12, 2010 at 2:00
  • $\begingroup$ To begin with are you interested in rational or integral cohomology? Secondly, the distance to the $\mathrm{P}^1$ case is perhaps not that great. For a general smooth curve $C$ you have a map from your moduli stack to the $m$'th symmetric power of $C$ associating to a sheaf its cycle. Étale locally this map is independent of the curve and you might try to use the Leray spectral sequence for it. $\endgroup$ Mar 12, 2010 at 6:31

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