# why do automorphisms preserve ample divisors?

Let $X \hookrightarrow \mathbb{P}$ be a smooth hypersurface inside some projective space $\mathbb{P}$ and let $H$ be a smooth hyperplane section of $X$. Now let $\varphi$ be an automorphism of $X$.

Why is it true that $\varphi^\ast$ acts trivially on the cohomology class of $H$?

-
There is no reason, except in the case where the embedding is canonical or anticanonical. Think for example to a cubic curve in $\mathbb P^2$ – Jérémy Blanc Mar 6 '14 at 15:41
Consider $\mathbb{P}^1\times \mathbb{P}^1$ embedded by an "unbalanced" divisor. The involution does not act trivially on the divisor. – Jack Huizenga Mar 6 '14 at 15:42
Sorry I forgot to say $X$ is a hypersurface – beginigeb Mar 6 '14 at 15:54

It is indeed true unless $X$ is a cubic plane curve or a quartic surface. The reason is the following: $\varphi ^*$ preserve the canonical class, which is $(d-n-1)H$ (for $X$ a smooth hypersurface of degree $d$ in $\mathbb{P}^n$). Since $H^2(X,\mathbb{Z})$ has no torsion, this implies $\varphi ^*H=H$ if $d\neq n+1$. Now if $n\geq 4$, we have $H^2(X,\mathbb{Z})=\mathbb{Z}$ by Lefschetz theorem, which implies again $\varphi ^*H=H$. The remaining cases are cubic plane curves and quartic surfaces, and indeed in those cases there are examples of $\varphi$ with $\varphi ^*H\neq H$.
Yes, same argument. The exceptions are : curves $(2,2)$ in $\mathbb{P}^3$, surfaces $(2,3)$ in $\mathbb{P}^4$ and $(2,2,2)$ in $\mathbb{P}^5$. – abx Mar 6 '14 at 16:46
Is there really an example among cubic plane curves? Shouldn't they all have $H^2(X,\mathbb Z)=\mathbb Z$? – Will Sawin Mar 22 '14 at 22:42