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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.

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up vote 8 down vote accepted

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 $\deg K=0$.

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In general, for any non-constant morphism $f:C \to \mathbb P^n$, from a $1$-dimensional Cohen-Macaulay (for instance reduced) curve $C$, one has that $$H^1(C,f^*T_{\mathbb P^n}\otimes \omega_C)=0.$$

Indeed, (as already pointed out by Al e) considering the pull-back of the Euler sequence $$0 \to f^*\mathscr O_{\mathbb P^n} \to f^*\mathscr O_{\mathbb P^n}(1)^{\oplus (n+1)} \to f^*T_{\mathbb{P}^n} \to 0$$ and using the fact that $H^2(C,\mathscr O_C)=0$ automatically by dimension considerations, it is enough to prove that $$H^1(C, f^*\mathscr O_{\mathbb P^n}(1)\otimes \omega_C)=0.$$ By Serre duality this is dual to $H^0(C,f^*\mathscr O_{\mathbb P^n}(-1))$ and since $f$ is non-constant, this is an anti-ample line bundle on $C$ and hence has no global sections.

Remark: One needs the Cohen-Macaulay condition for Serre duality and so that $\omega_C$ is sensible.

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