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.
Best Answer
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.