Where's the negative section of a deformation of a Hirzebruch surface?

Question

As in 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.

Answer 1

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, $$ \deg(N_{C_t/S_t}) = 3m - (n + m) = 2m - n. $$