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, 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.
