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.
 A: 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.
A: 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 $O(-2)$.
