Notes from my last meeting with Pete

Posted byayushkhaitan3437Posted inUncategorized

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

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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