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Let $\mathcal{E}$ be a coherent sheaf on a smooth variety $X$ of dimension $n$. We know that in such a case there is a locally-free resolution of length $n$:

$$E_{n} \to \ldots \to E_{1} \to \mathcal{E}.$$

This allows us to define the Chern character of $\mathcal{E}$ by:

$$\text{ch}(\mathcal{E}) = \sum_{i=1}^{n} (-1)^{i} \text{ch}(E_{i}) \in H^{2*}(X, \mathbb{Q}),$$

where one uses the well-known notion of the Chern character of the locally-free sheaf $E_{i}$. We typically denote the Chern character $\big(\text{ch}_{0}(\mathcal{E}), \ldots, \text{ch}_{n}(\mathcal{E})\big)$. (Although the higher Chern characters don't generally vanish, they integrate to zero against $X$, so we omit them.)

Let's say the support of $\mathcal{E}$ is $d$ dimensional. I am struggling to prove that $\text{ch}_{k}(\mathcal{E})=0$ for $k < n-d$. Can someone please help me see this rigorgously, and/or provide sources? Moreover, I feel like it should be true that $\text{ch}_{n-d}(\mathcal{E})$ should be very closely related to the (Poincare dual of the) homology class of $\text{Supp}(\mathcal{E})$. Is this true?

I'm worried that one might have to use something very powerful like Grothendieck-Riemann-Roch, but I've been trying to think of it in terms of the topological interpretation of the Chern classes $c_{k}(E_{i})$ from the locally-free resolution above. However, the necessary cancellations are not at all obvious to me and seem almost too miraculous.

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If you are happy to work with Chow groups - then here's a way of seeing this:

Recall that given a variety $X$ and an open dense subset $U$, with $Z=X\backslash U$, we have that for every $k\in \mathbb{Z}_{\ge 0}$ the following exact sequence

$$A_k(Z) \rightarrow A_k(X) \stackrel{i^*}{\rightarrow} A_k(U) \rightarrow 0$$ where $i:U\rightarrow X$ is the inclusion map and $i^*$ is given by flat pullback.

Now we let $Z = Supp(\mathcal{E})$. By assumption this is $d$-dimensional, and so for any $k>d$, we have that $i^*: A_k(X) \rightarrow A_k(U)$ is an isomorphism. In this setting because we are working with the chern character, we should really tensor everything by $\mathbb{Q}$, so we have an isomorphism $i^*: A_k(X) \otimes \mathbb{Q} \rightarrow A_k(U) \otimes \mathbb{Q}$ for $k>d$.

Now on $U$, we have that $$0\rightarrow E_n|_U \rightarrow E_{n-1}|_U \rightarrow \ldots \rightarrow E_1|_U \rightarrow 0$$ is exact, hence using the additivity of Chern character we have that $\sum_{i=1}^n (-1)^i ch(E_i|_U) = 0$. On the other hand, we know that $i^*$ commutes with $ch$, so we deduce that $i^* (\sum_{i=1}^n (-1)^i ch(E_i)) = 0$, i.e. $i^* ch(\mathcal{E}) = 0$. By the fact that $i^*$ is an isomorphism for $k>d$, we deduce that $ch_k(\mathcal{E}))= 0\in A_k(X)\otimes \mathbb{Q}$ for $k>d$.

Now since $X$ is smooth, we have that $A_k(X) \cong A^{n-k}(X)$, and therefore we have that $ch^{k}(\mathcal{E}) = ch_{n-k}(\mathcal{E}) = 0$ for $k<n-d$. (Here I'm using upper indices rather than lower indices to distinguish the grading on $A^*$ and $A_*$ (cohomological/homological)).

Finally the cycle map $A^*(X) \rightarrow H^{2*}(X)$ maps $ch(\mathcal{E})$ (viewed as an element of Chow) to $ch(\mathcal{E})$ (viewed as an element in cohomology), so we are done.

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  • $\begingroup$ Dear loch, why do you invoke smoothness for the equality $A_k(X) =A^{n-k}(X)$ ? This is just the definition of $A^{n-k}(X)$, as you seem to imply yourself further down your answer. $\endgroup$ – Georges Elencwajg Jul 19 '18 at 10:34
  • $\begingroup$ Hi @GeorgesElencwajg, my impression is that people don't really talk about $A^*(X) $ when $X$ is not smooth (well - at least not defined by $A^*(X) = A_{n-*}(X)$), because there is no natural ring structure on this. Perhaps another way of phrasing the above is that our cycle map $cl: A_*(X) \rightarrow H_*(X)$ satisfies $cl(c_i(E) \cap \alpha) = c_i(E) \cap cl(\alpha)$, and I'm using Poincare duality to go from $H_*(X)$ to $H^*(X)$. Perhaps smoothness is not needed for the original statement though? I'm not quite sure.. $\endgroup$ – loch Jul 19 '18 at 12:28
  • $\begingroup$ @loch Thanks a lot! That's a nice proof. Do you happen to know any references (perhaps citable) where they prove this? $\endgroup$ – Benighted Jul 19 '18 at 15:26
  • $\begingroup$ @Benighted everything related to Chow groups and cycle maps can be found in Fulton's intersection theory. I don't know a reference specifically for this though.. $\endgroup$ – loch Jul 19 '18 at 16:31
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    $\begingroup$ @Benighted Sorry for the late reply! You are right that if we're using ordinary homology then we do need compactness to apply Poincare duality (since we're not just concerned with cohomology with compact support). In the general case though we are really working with Borel-Moore homology (which agrees with usual homology, when everything is compact), where PD holds more generally. $\endgroup$ – loch Aug 17 '18 at 9:27

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