Question

I'm reading the book about moduli spaces by Huybrechts and Lehn, and i'm stuck understanding a proof, it is Theorem 6.1.8.:

Given a K3-surface $X$ and a 2-dimensional space $M$, coherent and torsion free sheafs $F$ on $X$ and $E$ on $M\times X$. We have projections $p,q$ from $M\times X$ to $M$ and $X$ resp.

They claim the class $a:=ch(p_{*}\mathcal{H}om(q^{*}F,E))$ as an element in $H^{*}(M,\mathbb{Q})$, where $p_{*}\mathcal{H}om(q^{*}F,E)=\sum\limits_{i=0}^2 (-1)^i \mathcal{E}xt^i_p(q^{*}F,E)$, only depends on the classes of $ch(q^{*}F)$ and $ch(E)$ as elements in $H^{*}(M\times X,\mathbb{Q})$, where $\mathcal{E}xt_p^i(q^{*}F,E)=R^i(p_{*}\mathcal{H}om(q^{*}F,E))$.

So using Grothendieck-Riemann-Roch as suggested shows: $ch(\sum\limits_{i=0}^2 (-1)^i \mathcal{E}xt^i_p(q^{*}F,E))td(M)=p_{*}(ch(\mathcal{H}om(q^{*}F,E)td(M\times X))$

Here i am stuck. Why does this show that $a$ only depends on $ch(E)$ and $ch(q^{*}F)$. I think one has to show that $ch(\mathcal{H}om(q^{*}F,E))$ only depends on this classes, but i can't see why.

Answer 1

If you apply ${\mathcal H}om(-,E)$ to a resolution of a sheaf $G$, you obtain a complex, the cohomology of which are ${\mathcal E}xt^i(G,E)$, hence by additivity of the Chern character, the alternating sum of Chern characters of the terms of the complex equals the alternating sum of Chern characters of the Ext sheaves.

Answer 2

This was to long for a comment, so i post this as an answer: Using Sasha's answer i tried my best, and here are my computations. Feel free to report any mistakes.

Take a locally free resolution $G_{*} \rightarrow q^{*}F$. Then we have $\mathcal{E}xt_p^i(q^{*}F,E)=h^i(\mathcal{H}om_p(G_{*},E))$, where $h^i = ker(d_i)/im(d_{i-1})$. Now:

$ch(\sum\limits_{i=0}^2(-1)^i \mathcal{E}xt_p^i(q^{*}F,E))=ch(\sum\limits_{i=0}^2 (-1)^i h^i(\mathcal{H}om_p(G_{*},E)))$ $= ch(ker(d_0))-ch(ker(d_1)/im(d_0))+ch(coker(d_1))$ $=ch(ker(d_0))+ch(im(d_0))-ch(ker(d_1))-ch(im(d_1))+ch(\mathcal{H}om_p(G_2,E))$ $ch(ker(d_0)\oplus im(d_0))-ch(ker(d_1)\oplus im(d_1))+ch(\mathcal{H}om_p(G_2,E))$ $ch(\mathcal{H}om_p(G_0,E))-ch(\mathcal{H}om_p(G_1,E))+ch(\mathcal{H}om_p(G_2,E))$ (1)

Now one uses Grothendieck-Riemann-Roch:

$ch(\mathcal{H}om_p(G_i,E))=ch(p_{*}\mathcal{H}om(G_i,E))=p_{*}(ch(\mathcal{H}om(G_i,E))td(X))$

Now the $G_i$ are locally free, so $ch(\mathcal{H}om(G_i,E))=ch(G_i^{*})ch(E)$ and (1) gives:

$p_{*}(ch(G_0^{*})ch(E)td(X))-p_{*}(ch(G_1^{*})ch(E)td(X))+p_{*}(ch(G_2^{*})ch(E)td(X))$

which is $p_{*}((\sum\limits_{i=0}^2 (-1)^i ch(G_i^{*}))ch(E)td(X))$, but since $G_{*}$ was a resolution of $q^{*}F$ this is

$p_{*}(ch((q^{*}F)^{*})ch(E)td(X))$

So we have: $ch(\sum\limits_{i=0}^2(-1)^i \mathcal{E}xt_p^i(q^{*}F,E))=p_{*}(ch((q^{*}F)^{*})ch(E)td(X))$

So i think my assumption, that $ch(\mathcal{H}om(M,N))=ch(M^{*})ch(N)$ is always true, is wrong?