### Notes from my last meeting with Pete

$b_{i,i+2}=h^1(\bigwedge^{i+1} M_L(1))$

$\bigwedge^i E=\bigwedge^{n-i}E^*\otimes \text{det }E=h^1(\bigwedge^{n-i-1}M^{*}_L)=$(by Serre duality) $h^0(K_C\otimes \bigwedge^{n-i-1}M_L)$

$0\to M_L\to \Gamma\to L\to 0$

$0\to M_L\otimes K\to \Gamma\otimes K\to L\otimes K\to 0$

Estimate $h^0(M_L\otimes K)\approx h^0(\Gamma\otimes K)-h^0(L\otimes K)$ (we can refine this later)

$h^0(\Gamma\otimes K)=(\text{dim}\Gamma)h^0(K)=(n+1)g$ (Riemann’s result, $h^0(K)=g$)

$h^0(L\otimes K)=$(Riemann-Roch)$(d+2g-2)-g+1=d+g-1$

$h^0(L\otimes K)\approx (n+1)g-(d+g-1)$

As described in Hirzebruch’s book, from $0\to M_L\to\Gamma\to L\to 0$, we get $0\to \bigwedge^2 M_L\to \bigwedge^2 \Gamma\to M_L\otimes L\to 0$, and then $0\to \bigwedge^2 M_L\otimes K\to \bigwedge^2 \Gamma\otimes K\to M_L\otimes L\otimes K$

There are two approaches that we can take:

1. Tensor $(0\to M_L\otimes K\to\Gamma\otimes K\to L\otimes K\to 0)$ by $L$ and compute
2. Riemann-Roch for vector bundles: we have $d+r(1-g)$ instead of $d+1-g$. Here, $r$ is the rank.

$0\to A\to B\to C$

$\text{deg}(B)=\text{deg} (A)+\text{deg} (C)$

$\text{deg}(E\otimes L)=\text{deg} (E)+r.\text{deg} (L)$

$\text{deg}(\Gamma)=0$

Advertisements