Riemann-Roch and dim of deformation space.

2012-06-18T16:46:30Z

Let's consider curve $C\subset \mathbb P^n$ of degree $d$ and genus $g$. We want to calculate dimension of deformation space of $C$, i.e. $h^0(C,L)$ where $L$ is the normal bundle.

We can decompose $L$ as $L_1\subset L_2\dots \subset L$, such as $dim L_{i+1}/L_i =1 \ $ and apply Riemann-Roch to each $L_{i+1}/L_i$.

I heard that this gives us $(n-1)(d-g+1)+2g-2 + 2d\ $ (as expected to be dimension of deformation space of $C$ or the number of points need to fix to count curves degree $d$ and genus $g$ passing through them).

But I can't calculate it ! Could you help me with this or give me a reference ?

Answer by David Speyer for Riemann-Roch and dim of deformation space.

2012-06-18T17:15:54Z

Let's start by stating Riemmann-Roch for vector bundles: If $C$ is a smooth projective curve of genus $g$, and $E$ is a vector bundle of rank $r$ and degree $\delta$, then $h^0(C,E) - h^1(C,E) = \delta - r(g-1)$. If $E$ has a filtration such as you describe, you can prove this by using Riemann-Roch on each quotient line bundle, and then adding up the contributions in the long exact sequence of cohomology.

Now, we want to consider the case where $E$ is the normal bundle $N$ to $C \subset \mathbb{P}^n$. The rank of $E$ is $n-1$. We have the short exact sequence $0 \to T_C \to T_{\mathbb{P}^n}|_C \to N \to 0$ . The tangent bundle to $C$ has degree $-(2g-2)$. The tangent bundle to $\mathbb{P}^n$ has first chern class $(n+1)$ times the hyperplane class; the hyperclane class restricts to $d$ times the fundamental class of $C$. So the degree of $N$ is $d(n+1) + (2g-2)$ and Riemann-Roch gives $$d(n+1)+(2g-2)-(n-1)(g-1)$$ which matches your formula.

Riemann-Roch and dim of deformation space. - MathOverflow

Riemann-Roch and dim of deformation space.

Answer by David Speyer for Riemann-Roch and dim of deformation space.