As in [Deformations of Hirzebruch surfaces and toric action](https://mathoverflow.net/questions/69339/deformations-of-hirzebruch-surfaces-and-toric-action),
the Hirzebruch surface $F\_n$ can be deformed into $F\_{n-2m}$ ($0<2m\leq n$) under the fibration given by
$$
M=\{([x\_0:x\_1],[y\_0:y\_1:y\_2],t)\in \mathbb{P}^1\times \mathbb{P}^2 \times \mathbb{C} \mid x^n\_0y\_1−x^n\_1y\_0+tx^{n−m}\_0x^m\_1y\_2=0\}
$$
over $\mathbb{C}$. What I would like to know is the explicit equations that give the negative section of each Hirzebruch surface in this situation. Of course that of the central fiber $M\_0$, which is a ($-n$)-section, is given by $y\_0=y\_1=0$ in $M\_0$. On the other hand, for a noncentral fiber $M\_t$ I cannot find the way to determine the equations of its ($-n+2m$)-section.


I would be grateful for any comments.



Where's the negative section of a deformation of a Hirzebruch surface?
You can take $y\_0 = 0$, $y\_1 = -tx\_1^m$, $y\_2 = x\_0^m$.


EDIT. Here is a simple computation of the normal bundle of the curve
$$
C\_t = \{([x\_0:x\_1],[0:-tx\_1^m:x\_0^m])\} \subset \mathbb{P}^1 \times \mathbb{P}^2
$$
in the surface
$$
S\_t = \{x\_0^ny\_1−x\_1^ny\_0+tx\_0^{n−m}x\_1^my\_2=0\}
$$
for $t \ne 0$. First, the normal bundle fits into the exact sequence
$$
0 \to N\_{C\_t/S\_t} \to N\_{C\_t/\mathbb{P}^1 \times \mathbb{P}^2} \to N\_{S\_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert\_{C\_t} \to 0.
$$
Since $C\_t$ is a graph of a map $\mathbb{P}^1 \to \mathbb{P}^2$ (of degree $m$), its normal bundle is the pullback of the tangent bundle of $\mathbb{P}^2$, hence its degree is
$$
\deg(N\_{C\_t/\mathbb{P}^1 \times \mathbb{P}^2}) = 3m.
$$
On the other hand, $S\_t$ is a divisor of type $(n,1)$ on $\mathbb{P}^1 \times \mathbb{P}^2$, hence its normal bundle restricted to $C\_t$ has the degree
equation which has degree
$$
\deg(N\_{S\_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert\_{C\_t}) = n + m.
$$
Therefore,
$$
\deg(N\_{C\_t/S\_t}) = 3m - (n + m) = 2m - n.
$$


