# Pencils with many completely decomposable fibers

Let $F= \frac{G}{H} : \mathbb P^n \to \mathbb P^1$ be a non-constant rational function ($G$ and $H$ homogenous polynomials of the same degree in $\mathbb C^{n+1})$. The fiber over $(\lambda:\mu) \in \mathbb P^1$ is completely decomposable if all the irreducible factors of the polynomial $\mu G - \lambda H$ are linear polynomials. In a joint work with Sergey Yuzvinsky, we proved that when $n=4$ either (a) $F$, after a linear change of coordinates, belongs to $\mathbb C(x_0,x_1,x_2,x_3)$; or (b) $F$ has at most three completely decomposable fibers.

We couldn't come up with an example not satisfying (a), but with three completely decomposable fibers. Is anyone aware of an example like that ?

Almost one year later: Let me describe the known examples of rational functions on $\mathbb P^3$ with three completely decomposable fibers. For every natural number $d \ge 1$, consider the rational function $F_d : \mathbb P^3 \to \mathbb P^1$ defined as $$F_d(x_0:x_1:x_2:x_3) = \frac{(x_0^d - x_1^d)(x_2^d - x_3^d)}{(x_0^d - x_2^d)(x_1^d - x_3^d)} .$$ Clearly the fibers over $(0:1)$ and $(1:0)$ are completely decomposable. The fiber over $(-1:1)$ is also completely decomposable since $$F^{-1}(1:-1) = \lbrace (x_0^d - x_3^d)(x_1^d - x_2^d) = 0 \rbrace \, .$$

For $d=1$, $F_d$ can be written in fewer variables, but for $d \ge 2$ this is not the case.

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I think the question is fine, but you should probably add the open-problem tag (as I have done for you) if you are reasonably sure that this is an open problem. –  David Speyer Nov 11 '09 at 14:06
From what I recall from your paper, in (a) there should be 3 coordinates, not 4 (the pencil is pulled back from P^2), am I right? Just want to clarify the story (it was some time ago when I read the paper). –  Vladimir Dotsenko Nov 21 '09 at 15:21
You are absolutely right. But we are aware of an example in P^3 with three completely decomposable fibers which does not come from P^2. –  jvp Nov 22 '09 at 10:37