MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $H$?

share|cite|improve this question
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 $\varphi ^*H\neq H$.

share|cite|improve this answer
Thanks! Is this also true for smooth complete intersections? – beginigeb Mar 6 '14 at 16:16
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

Your Answer


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.