Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $\deg K=0$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.