# Normal bundle of a component of a reducible fiber

Suppose that $\pi : X \to S$ is an elliptic fibration, and that the fiber over a point $s \in S$ is reducible, with several components. Call one of them $C$. How can I compute the normal bundle of $C$ in $X$? I'm not assuming the base is a curve.

I'd be happy to hear how to do this in an example, so for concreteness let's say that $S$ is the versal deformation space of a fiber of Kodaira type $I_2$: the central fiber consists of two smooth rational curves meeting at two points, $S$ is two-dimensional, and there are two curves in the base through $s$ over which the fibers are nodal cubics, corresponding to smoothing the two nodes of the singular fiber. What are the normal bundles of the two curves in the singular fiber?

I can choose a smooth curve $\gamma \subset S$ going through $s$, with the preimage a smooth surface $X_\gamma \subset X$. It's easy to compute the normal bundle of $C$ in $X_\gamma$ (they are $(-2)$-curves in the example), and there is $0 \to N_{C/X_\gamma} \to N_{C/X}$, but it's not clear to me that I should expect $N_{C/X_\gamma}$ to be a direct summand, since $C$ isn't obtained as the complete intersection of such surfaces.

• Minor point: I don't think the surface $X_\gamma$ you mention can be smooth. Here's why: if it were, then choosing two general smooth curves $\gamma$, $\delta$ intersecting transversely at $s$, then their preimages would be smooth surfaces intersecting transversely at every point of the fibre. But of course that doesn't happen, since the fibre has 2 singular points. – user5117 Mar 1 '13 at 16:43
• By the way, I like this question, and your answer! – user5117 Mar 1 '13 at 16:44

I'll take a stab an answering my own question here, in case anyone's interested.

It's true that if $C \subset S \subset X$, all smooth, there's a sequence $0 \to N_{C/S} \to N_{C/X} \to N_{S/X}\vert_C \to 0$, but I don't see that it should split in general (though it will if $C$ is a complete intersection). In the example above, $N_{C/X_\gamma} \cong \mathcal O(-2)$ and $N_{X_\gamma/X}\vert_C \cong \mathcal O$, but I think $N_{C/X}$ is the nonsplit extension $\mathcal O(-1) \oplus \mathcal O(-1)$.

Forgive me for re-notating, with $C_1$ and $C_2$ the two curves in the singular fiber, intersecting at $p_1$ and $p_2$, and $C$ their union. There's an exact sequence $0 \to N_{C_1/X} \to (N_{C/X})\vert_{C_1} \to T_{C_2,p_1} \oplus T_{C_2,p_2} \to 0$ This has the property that a first-order deformation (from a section in $H^0(C,N_{C/X})$) smooths $p_i$ if and only if it maps to something nonzero in $T_{C_2,p_i}$. This is written down in these notes by Debarre, page 40.

The group in the middle is $\mathcal O \oplus \mathcal O$. In one direction $p_1$ is smoothed, and in another $p_2$ is, so the map $\mathcal O \oplus \mathcal O \to k \oplus k$ sends $(1,0)$ to $(1,0)$ and $(0,1)$ to $(0,1)$. It follows that the kernel (i.e. the normal bundle in question) is $\mathcal O(-1) \oplus \mathcal O(-1)$.

I guess to answer the question in general we need to know something about how the singularities of the fibers deform as we move in the base.

Let $C_1,\dots,C_k$ be the components. Then $O(C_1+\dots+C_k)_|{C_1} = O$ because this bundle is a pullback from the base. On the other hand, for each $i \ne 1$ one has $O(C_i)_{|C_1} = O_{C_1}(C_i \cap C_1)$. Thus $$N_{C_1/X} = O(C_1)_{|C_1} = O(-C_2 - \dots - C_k)_{|C_1} = O_{C_1}(-C'\cap C_1),$$ where $C' = C_2 + \dots + C_k$.

In the example the normal bundle is $O(-2)$.

• Sasha: the OP said S is not a curve, so the C_i are not divisors in X. – user5117 Feb 27 '13 at 19:11