restriction of the cotangent bundle of an elliptic surface

Question

Given an nonisotrivial elliptic fibration $f:X\rightarrow P^1$, where $X$ is smooth and $P^1$ is a projective line. Could anybody provide some information on the restriction of the cotangent bundle $\Omega^1_{X}$ to the smooth fibers of $f$, e.g. does it split to line bundles, (semi)stable...?

Answer 1

It's semi-stable, but not stable on every fiber. Assuming that the characteristic is zero this sheaf does not split. This may be true in positive characteristic, but as Damian Rössler points out the proof below requires characteristic zero.

Claim 1. Let $V\subseteq \mathbb P^1$ be an arbitrary non-empty open subset and $U=f^{-1}V$. Then $$0 \to f^*\Omega^1_{\mathbb P^1}|_U \to \Omega^1_X|_U \to \Omega^1_{X/\mathbb P^1}|_U \to 0\tag{$\star$}$$ is not a split sequence.

Proof. We may obviously assume that $f$ is smooth over $U$. Since $f$ is a non-isotrivial fibration, the associated Kodaira-Spencer morphism of sheaves is non-zero, that is, pushing forward the dual of the original sequence, $$0 \to T_{X/\mathbb P^1} \to T_X \to f^*T_{\mathbb P^1} \to 0,\tag{$\star^\vee$}$$ gives an injective map $$ \kappa: T_{\mathbb P^1}\to R^1f_* T_X. $$ Clearly, this is injective on any non-empty open set $U$ which implies that $(\star^\vee)|_U$ cannot be split and hence the Claim is proven. $\square$

Remark 1 The map $\kappa$ is the sheaf version of the classical Kodaira-Spencer map. That is obtained by tensoring with the residue field of a point on the target to get $T_{\mathbb P^1, t}\to H^1(X_t, T_{X_t})$ for $t\in \mathbb P^1$. That may be zero even for non-isotrivial families

Remark 2 The fact that this sequence does not split does not mean that the sheaf in the middle cannot be the direct sum of two line bundles. Contemplate the Euler sequence of $\mathbb P^1$: $$ 0 \to \omega_{\mathbb P^1} \to \mathscr O_{\mathbb P^1}(-1) \oplus \mathscr O_{\mathbb P^1}(-1)\to \mathscr O_{\mathbb P^1} \to 0. $$

On the other hand, this proof gives a little more:

Corollary (of the proof). $\Omega_X^1|_U\not\simeq \mathscr O_U\oplus \mathscr O_U$.

Proof. Considering the long exact sequence that $\kappa$ is sitting in we get $$ 0\to f_*T_{X/\mathbb P^1}\to f_*T_X \to T_{\mathbb P^1} \to R^1f_*T_{X/\mathbb P^1} \to \dots $$ Using that the fibers are genus $1$ curves we see that $f_*T_{X/\mathbb P^1}$ is a line bundle. Since $\kappa$ is non-trivial, it has to be injective and hence $f_*T_{X/\mathbb P^1}\to f_*T_X$ is surjective. On the other hand, if $T_X|_U$ was trivial, then $f_*T_X$ would have rank $2$ which would lead to contradiction. $\square$

So we need a little more:

Claim 2 Let $V\subseteq \mathbb P^1$ be an arbitrary non-empty open subset and $U=f^{-1}V$. Then the sheaf $\Omega^1_X|_U$ does not split as a non-trivial direct sum.

Remark As I said, the fact that $(\star)|_U$ is not split does not mean that $\Omega_X^1|_U$ is not either, but more particulars about that sequence actually imply that...

Proof. Suppose $\Omega_X^1|_U=\mathscr L\oplus \mathscr M$. If this is a non-trivial decomposition, then both $\mathscr L$ and $\mathscr M$ are line bundles. We may assume that $V\neq \mathbb P^1$ and hence $f^*\Omega^1_{\mathbb P^1}|_U\simeq \mathscr O_U$ and $\Omega^1_{X/\mathbb P^1}|_U\simeq \mathscr O_U$, so we have a short exact sequence $$ 0\to \mathscr O_U \to \mathscr L\oplus \mathscr M\to \mathscr O_U \to 0. $$ If either of the induced maps, say $\mathscr M\to \mathscr O_U$, is trivial, then the other one, $\mathscr L\to \mathscr O_U$, is surjective and hence an isomorphism and hence $\mathscr L\simeq\mathscr M\simeq \mathscr O_U$. However, this is impossible by the Corollary above.

So we may assume that the induced maps $\mathscr L\to \mathscr O_U$ and $\mathscr M\to \mathscr O_U$ are non-trivial. Similarly the induced maps $\mathscr O_U\to \mathscr L$ and $\mathscr O_U\to \mathscr M$ are also non-trivial. These together imply that both $\mathscr L$ and $\mathscr L^{-1}$ as well as both $\mathscr M$ and $\mathscr M^{-1}$ have non-zero global sections and therefore, again, $\mathscr L\simeq\mathscr M\simeq \mathscr O_U$, which is again impossible by the Corollary above. $\square$

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[This part does not require a restriction on the characteristic.]

As far as (semi-)stability goes, since $U$ is not projective, it doesn't make much sense to ask on $U$. You could ask if it is relatively semi-stable, or semi-stable on the fibers. From the short exact sequence $(\star)$ you can see that its degree on any fiber is $0$ and an argument similar to the proof of Claim 2 says that any sub-line bundle would have to have degree at most $0$. (For any sub line bundle, if the induced map to $\mathscr O$ is non-trivial, then this is clear, if it is trivial, then it has to be contained in the kernel, which is also $\mathscr O$, so again clear.) So, this is actually semi-stable on every fiber. Since it contains a copy of $\mathscr O$, it is not stable.

For other properties, you can probably use the above short exact sequences and arguments.

Answer 2

We have the exact sequence $0 \to f^*\Omega^1_{\mathbb P^1} \to \Omega^1_X \to \Omega^1_{X/\mathbb P^1} \to 0$

Because the map is smooth, this is an exact sequence of bundles, i.e. $f^* \Omega^1_{\mathbb P^1}$ and $\Omega^1_{X/\mathbb P^1}$ are both line bundles.

Since you've removed at least one fiber, and both line bundles have divisor classes that are multiples of the class of a fiber, both line bundles will be trivial.

I do not think the exact sequence splits in general, but it might split in a non-canonical way.