# Rational map having a birational restriction.

Let $\sigma\subset |\mathcal{O} _{\mathbb{P} _{\mathbb{C}}^N}(2)|$ be an $N-$dimensional linear systems of quadrics on $\mathbb{P} _{\mathbb{C}}^N$ ($N\geq 4$), $F:\mathbb{P} _{\mathbb{C}} ^N\dashrightarrow \mathbb{P} _ {\mathbb{C}}^ N$ the rational map associated to $\sigma$, and $Q\in \sigma$ a fixed smooth quadric such that $F|_{Q}:Q\dashrightarrow \mathbb{P} _{\mathbb{C}}^{N-1}$ is birational.

Is $F$ birational?

• Nice question! Why do you consider$N\ge 4$ (and say, not $N=2,3$)? Where does this question come from? Jul 17 '11 at 14:19
• actually $N\geq 4$ can be ignored...
– gio
Jul 17 '11 at 15:57
• Ok, $N=2$ works, I guess you know this. Do you know if for $N=3$ the answer to your question is positive? (I have not checked yet). The more information you give on the question, better it is (at least for me)! Jul 17 '11 at 21:30
• Dmitri, I don't see how it works for $N=2$. Can you include your argument? Jul 17 '11 at 21:42
• Note that resolving the rational maps $F$ and $F| _ Q$, is easy to see that the answer to my question is positive if the base scheme $X$ of $F$ is smooth and irreducible. Also, with less assumptions on $X$, my question is essentially equivalent to this mathoverflow.net/questions/70312/…
– gio
Jul 18 '11 at 8:21

This is a bit too long for a comment, but since it was requested, below a positive answer in the case $N=2$ is given. This might also help to understand the question in the simplest case.

So we start with a smooth quadric $Q=0$ in $\mathbb P^2$ and two more quadrics $Q_1=0$, $Q_2=0$ such that the non-fixed locus of intersection of $Q$ with the linear system $Q_1+tQ_2=0$ is a single point (so the corresponding map $Q\to \mathbb P^1\supset t$ is degree one, i.e. birational). Such situation can happen only if both $Q_1$ and $Q_2$ intersect $Q$ at the same set consisting of three points, or two points of which one is with multiplicity 2, or at one point with multiplicity $3$. Let us denote this set by $x$. Then it is clear that for any generic pencil in the family generated by $Q,Q_1,Q_2$ the fixed locus is $x$ plus one point. If you translate this into the notations of the original problem you see that the map from $\mathbb P^2$ is birational.

• I see I missed the condition that $Q\in\sigma$! Jul 17 '11 at 22:29

Nice question. I think that the answer is yes and more general. I have never seen this but seems natural.

${\bf Lemma}$ Let $f\colon \mathbb{P}^n_{\mathbb{C}}\to \mathbb{P}^n_{\mathbb{C}}$ be a rational map of degree $d$, given by $(x_0:\dots:x_n)\to (f_0:\dots:f_n)$, where the $f_i$ are homogeneous of degree $d$.

Suppose that the hypersurface $H\subset \mathbb{P}^n_{\mathbb{C}}$ given $f_0=0$ is irreducible and that the map from $H$ to $\mathbb{P}^{n-1}_{\mathbb{C}}$ given by the restriction (i.e. $(x_0:\dots:x_n)\to (f_1:\dots:f_n)$ on $H$) is birational.

Then, the map $f$ is birational.

${\bf Proof}$ Let $\sigma$ be the linear system associated, which corresponds to hypersurfaces of $\mathbb{P}^n$ of equation $\sum_{i=0}^n a_i f_i=0$. The restriction being birational, the intersection of $n-1$ general elements of $\sigma$ with $H$ give one mobile point, and a non-mobile part that we call $R$. Since every member of $\sigma$ contains $R$ and because elements have all the same degree, the intersection of $n$ general elements of $\sigma$ gives $R$, plus exactly one mobile point (the last part follows from Bézout). This shows that $f$ is birational.

${\bf Remark:}$ We could also view $R$ as a set of points, curves,... with some multiplicities and use intersection form on the blow-up.

• I do not understand your proof, but certainly the statement "every member of $\sigma$ contains $H$" is not true.
– gio
Aug 13 '12 at 6:02
• Yes, sorry for the two missprints. I exchanged $H$ with $R$. It is almost the same argument as Dmitri, but in dimension $n$. In dimension $2$ and for quadrics, the set $R$ consists of $3$ points. In dimension $3$ for quadrics, it consists of $7$ points, or maybe a line and $5$ points, ... Aug 13 '12 at 6:49
• To me this shows that $f$ is birational only if $H$ is general in $\sigma$, but I have requested that it is special. Please let me know if you agree.
– gio
Aug 13 '12 at 10:42
• I dont see why you mean this. I just assumed $H$ to be irreducible. Where is the problem in the argument? Aug 13 '12 at 21:22