# 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?

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Nice question! Why do you consider$N\ge 4$ (and say, not $N=2,3$)? Where does this question come from? – Dmitri 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)! – Dmitri Jul 17 '11 at 21:30
Dmitri, I don't see how it works for $N=2$. Can you include your argument? – Sándor Kovács 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.

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I see I missed the condition that $Q\in\sigma$! – Sándor Kovács 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.

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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, ... – Jérémy Blanc 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? – Jérémy Blanc Aug 13 '12 at 21:22