8

Let $X$ be a $d$-dimensional smooth projective connected variety with cotangent sheaf $\Omega^1_X$ over $\mathbb C$. Suppose that for any nonsingular complete curve $C$ and non-constant morphism $\gamma:C\to X$, we have that $$\gamma^\ast \Omega^1_X = \mathcal O_C^{\oplus d}.$$

What can we say about $X$?

I know that if $X$ is an abelian variety, it has the above property.

Are there any other varieties which satisfy the above property? For instance, what if the determinant of $\Omega^1_X$ is trivial, i.e., $\omega_X = \mathcal O_X$. Do we have the above property?

What if $X$ is a K3 surface? (This is a special case of the above question.)

Arctic Char
  • 16,972
Tom
  • 647

1 Answers1

2

This is only a partial answer, but it is not true in general even for $K3$ surfaces. Let $S$ be a smooth algebraic $K3$, containing a smooth rational curve $C \stackrel{\gamma}{\hookrightarrow} S$. Then by adjunction, the normal bundle is $\mathcal{N}_{C|S} \cong \mathcal{O}_C(-2)$. There is a short exact sequence $$ 0 \to \mathcal{N}_{C|S}^{\check{}} \to \gamma^*\Omega^1_S \to \Omega^1_C \to 0 ~, $$ which becomes $$ 0 \to \mathcal{O}_C(2) \to \gamma^*\Omega^1_S \to \mathcal{O}_C(-2) \to 0 ~. $$ Since $\mathcal{O}_C(2)$ doesn't inject into $\mathcal{O}_C^{\oplus 2}$, $\gamma^*\Omega^1_S$ must be non-trivial.


As suggested by Tom, the above can be generalised (and it can probably be generalised further by somebody who knows their stuff):

If $X$ is a smooth variety containing a smooth rational curve $\gamma : C \hookrightarrow X$, then $X$ does not have the property given in the OP. To see this, start with the following exact sequence: $$ 0 \to {N}_{C|X}^{\check{}} \to \gamma^*\Omega^1_X \to \Omega^1_C \to 0 $$ where $N_{C|X}^{\check{}}$ is the conormal bundle. Taking determinants, we find $$ \gamma^* \omega_X \cong \omega_C\otimes\det (N_{C|X}^{\check{}}) \cong \mathcal{O}_C(-2)\otimes\det (N_{C|X}^{\check{}}) $$ So if $\gamma^*\Omega^1_X$ is trivial, we must have $\det (N_{C|X}^{\check{}}) \cong \mathcal{O}_C(2)$. As $N_{C|X}^{\check{}}$ is isomorphic to a sum of line bundles, at least one of them must have positive degree. But then it cannot inject into a trivial bundle, so we have a contradiction.

Rhys
  • 4,443
  • How can we generalize this to varieties $S$ containing a smooth rational curve with trivial canonical bundle? It seems to me the same strategy should work. One computes the normal bundle, and writes down a similar exact sequence it seems, no? – Tom May 11 '13 at 19:09
  • Yes, you're right. I'll add it to my answer. – Rhys May 13 '13 at 09:27