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I am reading this paper https://arxiv.org/abs/1608.04797

Let $\Theta$ be the stack $\mathbb A^1/{\mathbb G_m}$. Let $X$ be a smooth projective curve of genus $g$ over a field $k$. Let $Coh_P$ be the stack of coherent sheaves parameterized by $X$ and with Hilbert polynomial $P$.

The paper says : groupoid of maps $f$ from $\mathbb A^1/{\mathbb G_m}$ to $Coh_P$ is equivalent to the groupoid whose objects consist of a coherent sheaf $f(1)=[E] \in Coh_P $ and a $\mathbb Z$ weighted filtration of $E$. Maps $*/ \mathbb G_m \to Coh_P$ correspond to $\mathbb Z$ graded coherent sheaves and restriction of a map $f$ from $\Theta \to Coh_P$ to $0/ \mathbb G_m$ classify gr(E).

I am unable to make sense of this.

  1. What does the stack $\Theta$ look like and what does f(1) mean? More precisely what does 1 mean here?
  1. Where do the filtrations and gradations appear from?
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  • $\begingroup$ It's a nice exercise to show first that quasicoherent sheaves on $B \mathbb{G}_m$ can be identified with graded vector spaces, and second that quasicoherent sheaves on $\mathbb{A}^1/\mathbb{G}_m$ can be identified with filtered vector spaces (using a suitable generalization of filtrations). I don't know what the deal with the fixed Hilbert polynomial is. $1$ refers to the point $1$ on $\mathbb{A}^1$. $\endgroup$ Commented Feb 16, 2023 at 19:36
  • $\begingroup$ In the paper it is mentioned that the stack $\mathbb A^1 / \mathbb G_m$ comes from $\mathbb G_m$ acting by weight -1. Could you please elaborate what this means? $\endgroup$ Commented Feb 16, 2023 at 20:02

1 Answer 1

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Maps $\mathrm{B} G \to \mathcal{X}$ correspond to an object of $\mathcal{X}(k)$ along with a $G$-action. Indeed, the map $* \to \mathrm{B} G \to \mathcal{X}$ selects an object $x$ and for each test scheme $T$ we get a natural map,

$$ \{ T \text{-torsors} \} \to \mathcal{X}(T) $$

so that the trivial torsor maps to $x$. This is determined by the induced $G$-action on $x$ since, after a cover, each torsor becomes trivial and the gluing maps pass to gluing data for an object of $\mathcal{X}$.

Maps $[E/G] \to \mathcal{X}$ correspond to "$G$-equivariant objects over $E$" meaning the data of $x \in \mathcal{X}(E)$ with a $G$-action in $\mathcal{X}$ which is compatible with the $G$-action on $E$ along the functor $\mathcal{X} \to \mathrm{Sch}_k$.

First, $\mathbb{G}_m$-equivariant coherent sheaves are exactly coherent sheaves with a $\mathbb{Z}$-grading. In some sense, this is an incarnation of the fact that the actions of tori split up into a weight space decomposition. In the affine case, this is very clear, we need a comultiplication map,

$$ M \to M \otimes k[t, t^{-1}] $$

which gives a decomposition of $M$ into the sum over $M_n = \{ m \mid m \mapsto m \otimes t^n\}$. The (co)associativity of the action shows that this is a $\mathbb{Z}$-grading.

Now we need to consider $\mathbb{G}_m$-equivariant sheaves over $\mathbb{A}^1$. This means a coherent sheaf $\mathcal{F}$ on $X \times \mathbb{A}^1$ flat over $\mathbb{A}^1$ with a $\mathbb{G}_m$-action over $\mathbb{A}^1$. We get an actual sheaf $\mathcal{F}|_1$ taking the fiber over $1 \in \mathbb{A}^1$ which is what "$f(1)$" should mean. Consider this situation affine-locally. We have a $A[t]$-module $M$ which is $\mathbb{Z}$-graded compatibly with the $\mathbb{Z}$-grading on $A[t]$ (although this grading on $A[t]$ is trivial in negative degrees this need not be true of $M$ e.g. $M = A[t, t^{-1}]$. We think of $f(1) = M/(t-1)$ as the underlying object which messes up the grading but preserves the decreasing filtration

$$ M_{\ge n} = \bigoplus\limits_{k \ge n} M_n $$

because the action of $(t-1)$ preserves this filtration but not preserve the ascending filtration. Therefore, we get a $\mathbb{Z}$-filtered $A$-module $M/(t-1)$.

This is analogous to the "dynamical description of Parabolic, Levi, and Cartan subgroups" where we are asking for the limit of a $\mathbb{G}_m$-parametrized sheaf "as $t \to 0$". This is the sort of condition that gives a parabolic so it may not be so surprising that it recovers filtered objects.

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  • $\begingroup$ Is M assumed to be t-torsion free here? $\endgroup$ Commented Feb 17, 2023 at 7:58
  • $\begingroup$ I forgot to mention that $\mathcal{F}$ is required to be flat over $\mathbb{A}^1$. In particular, this implies that $M$ is $t$-torsion-free. $\endgroup$
    – Ben C
    Commented Feb 17, 2023 at 13:04

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