If $C$ is a smooth elliptic curve and $f: C \to \mathbb P^n$, then 
$H^1(C,f^\*T\_{\mathbb P^n}) = 0.$ How do I prove this? The implication is that map
from $C$ to $\mathbb P^n$ is unobstructed.



$H^1$ of the pull back of the tangent bundle.
I am a beginner but here is my attempt:


the Euler sequence on $\mathbb{P}^n$ pulls back to $0 \to O\_C \to O\_C(1)^{n+1} \to f^\*T\_{\mathbb{P}^n} \to 0$ and so from the associated long exact sequence we want to show $H^1(O\_C(1))=0$ (as $H^2(O\_C)=0$ since $\dim C = 1 <2$). Let $D$ be an effective divisor on $C$ so that $O\_C(1)=O\_C(D)$. By Serre duality we have $H^1(O\_C(D))=H^0(O\_C(K-D))=0$ because $\deg K-D = \deg -D <0 $ since on elliptic curve $\deg K=0$.


