# Number of singular fibers in families of hypersurfaces

Consider the projection map $$\pi: X = V(t_0 f + t_1 gh) \to \mathbf P^1,$$ where $[t_0: t_1]$ are the homogeneous coordinates on $\mathbf P^1$, $f=f(x_0, \dots, x_n)$ is a homogeneous polynomial of some degree $d$ such that $V(f) \subset \mathbf P^n$ is smooth and $g$ and $h$ satisfy $\deg g+\deg h = d$ and they similarly define smooth hypersurfaces in $\mathbf P^n$ such that $V(gh) \subset X$ is a normal crossings divisor, i.e., $V(g) \cap V(h)$ is again smooth of the expected dimension.

As the fiber $V(f) \cong \pi^{-1}([1:0])$ is smooth by assumption, the map is generically smooth, so has only a finite number of singular fibers.

Question: Is there an upper bound on the number of singular fibers of $\pi$? (In terms of the degree of $f$, for example?)

(If there is an answer to a similar question for more general families of hypersurfaces, I am interested in this, too.)

Thank you!

• In general there is no upper bound for small $d$. Take for example a family of conics $V(aX^2 + bY^2 + cZ^2) \subset Proj ( \mathbf{C}[X,Y,Z])$ parametrized by $[a:b:c] \in \mathbf{P}^2$. The singular locus of this family is the fundamental triangle in $\mathbf{P}^2$. Since there are irreducible rational curves of arbitrary degree in $\mathbf{P}^2$, the intersection number of these rational curves with the singular locus is not bounded. – HYL Dec 11 '14 at 10:39

Definitely there is an upper bound for the degree, namely the degree $(n+1)(d-1)^{n-1}$ of the discriminant hypersurface $\Delta$ in the projective space $\mathbb{P} S_d$ associated to the finite dimensional $k$-vector space $S_d = k[x_0,\dots,x_n]_d$ of degree $d$ homogeneous polynomials in $x_0,\dots,x_n$. If you look at my previous answer (or look this up in Fulton's "Intersection Theory", Eisenbud-Harris's "3264 and All That" or various other sources), you will see that the degree of the discriminant hypersurface is computable (in a more general context). Here it is particularly simple, just $(n+1)(d-1)^{n-1}$.
The point is, your pencil of hypersurfaces gives a morphism $f:\mathbb{P}^1 \to \mathbb{P} S_d$ that is an isomorphism onto a line $\Lambda$ in $\mathbb{P} S_d$. So now you are asking the cardinality of the intersection $\Lambda \cap \Delta$, under your hypothesis that $\Lambda$ is not contained in $\Delta$. As $\Delta$ is just a hypersurface and $\Lambda$ is just a line, the total intersection number is precisely the degree of $\Delta$. The total intersection number is a sum (à la Fulton-MacPherson refined intersection theory) of positive integers (intersection multiplicities) for each connected component of $\Lambda \cap \Delta$.
You could ask for a lower bound on the intersection multiplicity coming from the connected component (i.e., point) $f([1:0])$. This would give a refined upper bound above. This multiplicity should be computable via excess intersection.
• Dear Jason, there seems to be a typo in your response: according to the Eisenbud-Harris reference you mentioned and also according to Gelfand-Kapranov-Zelevinsky, the degree of $\Delta$ is given by $(n+1)(d-1)^n$. – Jakob Jan 5 '15 at 10:26
• @Jakob: "... the degree of $\Delta$ is $(n+1)(d-1)^n$." You are absolutely correct. That was a typo. – Jason Starr Jan 5 '15 at 11:26