Let $S$ be a K3 surface and $h:=c\_1(i^\*\mathcal{O}\_{\mathbb{P^3}}(1))$, then we can compute that $c\_1(S)=0,c\_2(S)=6h^2$. Hence
\begin{align}
\sqrt{\text{td}(S)}=1+\frac{c\_2(S)}{24}=1+\frac{1}{4}h^2
\end{align}
According to materials e.g. *Lectures on K3 Surfaces*, the Mukai vector of a vector bundle $E$ on $S$ is defined by
$$\nu(E):=\text{ch}(E).\sqrt{\text{td}(S)}=(\text{rk}(E),c\_1(E),\text{ch}\_2(E)+\text{rk}(E))$$
I am a bit confused about the operation $.$ between $\text{ch}(E)$ and $\sqrt{\text{td}(S)}$. By the appearance, I guess that it was given by
$$v\_k=\sum\_{i+j=k}\text{ch}\_i(E)\bullet(\sqrt{\text{td}(S)})\_j$$
by some operation $\bullet$. But then it turns out that $\text{rk}(E)\bullet\frac{1}{4}h^2=\text{rk}(E)$. Here I get $h=4$ and $c\_2(S)=24$.



> 
> But $h\in H^2(S,\mathbb{Z})$ is a cohomology class, how should I understand $h=4$?
> 
> 
> 



Understand the Mukai vector
$c\_2(S)$ should not depend on the choice of polarization. See Corollary 3.3 of Huybrecht's Lectures on K3 surfaces.


$c\_2(S) = 24 c\_s$ where $c\_s$ is the generator of $H^2(S, \mathbb{Z})$. Then note that $\sqrt{td(S)} = (1+2c\_S)^{\frac{1}{2}} = 1 + c\_S$.


