Let $X$ be a complex smooth projective variety of dimension $d$. Let $K(X) := K(\text{Coh}(X))$ denote the Grothendieck group of coherent sheaves on $X$. For two coherent sheaves $E$ and $F$ on $X$, define their Euler pairing by
$$\chi(E,F) = \sum\_{i=0}^{d} (-1)^i \text{dim Ext}^i(E,F).$$
The Euler pairing descends to $K(X)$ by its additivity on short exact sequences in $\text{Coh}(X)$. Let $\ker\_L \subset K(X)$ denote the left radical of $\chi$, that is, $E \in \ker\_L$ if and only if $\chi(E,F) = 0$ for all $F \in K(X)$. By Serre duality, the left and right radicals of $\chi$ agree with each other. Define the numerical Grothendieck group of $X$ by
$$N(X) = K(X)/\ker\_L.$$
In [Bridgeland's notes on derived categories](http://www.tom-bridgeland.staff.shef.ac.uk/talks/warwick.pdf), he said that it's not clear whether the Chern character descends to $N(X)$, but it's true when $X$ is of dimension $\leq 2$.


Why does the Chern character descend to $N(X)$ when $d \leq 2$?


Suppose $d \leq 2$. I want to prove that if $\chi(E,F) = 0$ for all $F \in \text{Coh}(X)$, then $\text{ch}(E) = 0$. This should be a consequence of the Hirzebruch-Riemann-Roch (HRR) theorem, which says
$$\chi(E,F) = \int\_X \text{ch}(E)^\vee \text{ch}(F) \text{td}(X),$$
where $$\text{ch}(E)^\vee = \sum\_{i=0}^d (-1)^i \text{ch}\_i(E),$$ and $$\text{td}(X) = 1 + \frac{1}{2}c\_1(X) + \frac{1}{12}(c\_1(X)^2+c\_2(X)).$$Here I assume the Chern character takes values in the rational cohomology $H^\*(X) \otimes \mathbb{Q}$. If $d = 1$, i.e., $X$ is a curve, then the third term in $\text{td}(X)$ is zero. In this case, let $g$ be the genus of $X$, let $(r\_1, d\_1)$ be the rank and degree of the coherent sheaf $E$ and $(r\_2, d\_2)$ be those of $F$. Then by the HRR theorem, one obtains the following
$$\chi(E,F) = r\_1d\_2 - r\_2d\_1 + r\_1r\_2(1-g).$$
So choosing $(r\_2, d\_2) = (0, 1)$ kills $r\_1$, and further choosing $(r\_2, d\_2) = (1, 0)$ kills $d\_1$. So the case $d = 1$ is settled.
Now let $X$ be a surface. The question is to choose suitable $F$ to kill each $\text{ch}\_i(E)$. But I don't see how this is always possible.



Why does the Chern character descend to the numerical Grothendieck group for surfaces?
First, taking $F$ to be the structure sheaf of a point you "kill" $\mathrm{ch}\_0(E)$.


Next, since intersection pairing is non-degenerate on
$$
\mathrm{Im}(\mathrm{Pic}(X) \to H^2(X,\mathbb{Q})) = \mathrm{NS}(X)
$$
you can choose $F\_i$ to be a collection of the structure sheaves of curves on $X$ to "kill" $\mathrm{ch}\_1(E)$.


Finally, taking $F$ to be the structure sheaf of $X$ you "kill" $\mathrm{ch}\_2(E)$.


