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I am trying to do the following problem, taken from Iitaka's "Algebraic Geometry".

Let C be a smooth, geometrically connected curve of genus g over a field k. Assume that $g \geq 2$. If $f \in Aut(C)$ satisfies that $f^\ast w = w$ for all $w \in \Omega^1_{C/k}$, show that $f=id$.

I have no idea so far, so any hint would be nice (or solution).

TottenLeiber
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    Is it not the case (in the non-hyperelliptic case) that $f$ leaves the canonical curve in $\mathbb P^{g-1}$ pointwise fixed? – Ted Shifrin May 28 '13 at 16:52

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I am not sure if the following method works, but for the case when C is non-hyperelliptic, consider the following: The canonical map $g:C \rightarrow \mathbb{P}^{g-1}_k$ is a closed immersion since C is not hyperelliptic, so assume that $f \in Aut(C)$ acts as the identity on $\Omega^1_{C/k}$. Then we have a commutative diagram $$\require{AMScd} \begin{CD} C @>{g}>> \mathbb{P}^{g-1}_k;\\ @VfVV @VidVV \\ C @>{g}>> \mathbb{P}^{g-1}_k; \end{CD}.$$

This gives us $g= g \circ f$, and since g is a monomorphism (closed immersion) we get that $f=id_C$. However, I am a bit unsure if this is true, since the result seem a bit too strong and I would like people to tell me if it is wrong, so don't trust me all too much!

Dedalus
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  • So if anyone has the time, I'd be very happy if someone could say if my reasoning breaks down or not, and in that case where. – Dedalus May 28 '13 at 19:22
  • I think the key point is that any automorphism of the canonical embedding must be induced by a linear automorphism of $\mathbb P^{g-1}$, so having the identity on $g(C)$ means that we have the identity on $C$. I guess the hyperelliptic case takes a bit of analysis, taking into account the hyperelliptic involution. – Ted Shifrin May 29 '13 at 01:30
  • Ted Shifrin: You are probably right. Please tell me if you think my reasoning breaks down here for the non-hyperelliptic case. – Dedalus May 29 '13 at 13:09
  • In the hyperelliptic case, the canonical curve is the rational normal curve in $\mathbb P^{g-1}$ and the mapping factors through the $2:1$ projection to $\mathbb P^1$. Indeed, the image is invariant under the hyperelliptic involution $\iota$, but $\iota^*\omega = -\omega$ for any $\omega\in H^0(C,\Omega^1)$, so it should follow that the only automorphism $f=\text{id}$ is the only option. – Ted Shifrin May 29 '13 at 14:14
  • Ted Shifrin: Okay, thanks for that case. Does the posted answer work for the non-hyperelliptic? – TottenLeiber May 31 '13 at 12:02