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Describing the group of birational automorphisms of $\mathbb{P}^n$, $\mbox{Bir}(\mathbb{P}^n)$, for $n\ge 3$ is a fundamental open problem in birational geometry. For $n=2$, the classical theorem of Noether says that this group is generated by linear transformations and the Cremona transformation, which is given by $$ \phi:(x_0 : x_1 : x_2) \to (x_0^{-1} : x_1^{-1} : x_2^{-1}). $$For $n\ge 3$, there is an analogous Cremona transformation, but it is known that the group $\mbox{Bir}(\mathbb{P}^n)$ is no longer generated by this and $\mbox{PGL}_{n+1}(\mathbb{C})$. My question is therefore

Are there examples of other birational involutions of $\mathbb{P}^3$?

In case the answer is yes, have these been classified? I'm also interested in the analog of Noether's theorem in this case: Are there examples of birational transformations of $\mathbb{P}^3$ that can not be written as a composition of birational involutions?

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up vote 9 down vote accepted

Let $\mbox{Inv}(\mathbb P^n)$ be the subgroup of $\mbox{Bir}(\mathbb{P}^n)$ generated by involutions and $\mbox{PGL}(n+1, \mathbb C)$.

If $n\ge 3$ then any generating set of $\mbox{Inv}(\mathbb P^n)$ contains uncountably many involutions. More precisely for any $d >1$ there are uncountably many involutions of degree $\ge d$ in any generating set of $\mbox{Inv}(\mathbb P^n)$.

This is a direct consequece of the proof of Hudson-Pan Theorem which asserts that $\mbox{Bir}(\mathbb{P}^n)$, $n\ge 3$, needs uncountably many generators of degree greater than $d$ for any $d>1$.

Let me briefly review Pan's proof adapting it to prove the remark above. It is achivied in two steps:

  1. for each hypersurface $H$ of $\mathbb P^{n-1}$ there is a birational transformation of $\mathbb P^n$ which contracts a cone over $H$;

  2. a hypersurface contracted by a product of birational transformations $f_1 \circ f_2 \circ \cdots \circ f_k$ is birational to an hypersurface contracted by one of the birational transformations $f_1, \ldots, f_k$.

The proof of 2 is straight-forward. To prove 1, fix $p \in \mathbb P^n$ and consider the subgroup $\mbox{St}_p(\mathbb P^n) \subset \mbox{Bir}(\mathbb P^n)$ which sends lines through $p$ to lines through $p$. One can show that $\mbox{St}_p(\mathbb P^n)$ fits into the split exact sequence $$ 1 \to \mbox{PGL}(2,\mathbb C(\mathbb P^{n-1})) \to \mbox{St}_p(\mathbb P^n) \to \mbox{Bir}(\mathbb P^{n-1}) \to 1 $$ where the rightmost map is defined by the action on the space of lines through $p$, and the leftmost is defined by the action on the fibers of the $\mathbb P^1$-bundle obtained from $\mathbb P^n$ after blowing up $p$.

Now given $h \in \mathbb C(x_0, \ldots, x_{n-1})$, we can consider the element of $\mbox{PGL}(2,\mathbb C(\mathbb P^{n-1}))$ defined by $$ (s:t) \mapsto (t:h\cdot s) \, . $$ Clearly, this defines a birational involution of $\mathbb P^{n-1} \times \mathbb P^1$ which contracts the divisor associated to $h$. Of course, we can realize it as an element of $\mbox{St}_p(\mathbb P^n)$ inducing the identity on $\mbox{Bir}(\mathbb P^{n-1})$ and contracting a cone over $H=h^{-1}(0)$ with vertex at $p$.

Notice that we can choose uncountably many $H$ in uncountably many distinct birational equivalence classes. Putting this observation together with 1 and 2 allow us to conclude.


You are probably aware of it, but there is classification of birational involutions of $\mathbb P^2$ by Bayle-Beauville which completes and clarifies previous works by Bertini and others. Composing the involutions of $\mathbb P^2$ with elements of $PGL(2,\mathbb C(\mathbb P^2))$ gives many examples of birational involutions of $\mathbb P^3$ which are also in $\mbox{St}_p(\mathbb P^3)$.

For more on $\mbox{St}_p(\mathbb P^n)$ see this other paper of Pan.

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Here is an example of a birational involution of $\mathbb P^n$ that is at least not obviously a combination of Cremona transformations:

Suppose $H\subset \mathbb P^{n+1}$ is a hypersurface of degree $d$ in $\mathbb P^{n+1}$ such that it contains two points $P_1,P_2\in H$ such that the multiplicity of $H$ at $P_i$ is exactly $d-1$ for $i=1,2$. Such hypersurfaces can be constructed with a little work.

Now the projection from $P_i$ defines a birational isomorphism between $H$ and $\mathbb P^n$ and the composition of these defines a birational involution on $\mathbb P^n$.

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You mean $H\subseteq P^{n+1}$? –  Alexander Woo Jan 23 '11 at 22:31
    
Yes, of course. Thanks. –  Sándor Kovács Jan 23 '11 at 23:09
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There are plenty of other examples of birational involutions of $\mathbb{P}^3$ which are known. For the moment, no precise classification as in dimension $2$. However, the following preprint of Yuri Prokhorov gives a first nice step toward a classification:

Yuri Prokhorov, "On birational involutions of $\mathbb{P}^3$" http://arxiv.org/abs/1206.4985

In dimension $2$, we can say if two involutions are conjugate only by looking at fixed points. When the curves pointwise fix a curve of positive genus, the curve has to be the same (easy argument), but this condition is in fact sufficient (hard argument, only follows from classification). The same works for any order, looking carefully at the fixed points of all powers (see J. Blanc, "Elements and cyclic subgroups of finite order of the Cremona group." Comment. Math. Helv. 86 (2011), no. 2, 469-497, http://arxiv.org/abs/0809.4673).

In dimension $3$, we can have the same discussion, replacing fixed curves of positive genus with fixed surfaces which are not birationally ruled (other surfaces can be contracted by birational maps). The article of Yuri Prokhorov studies the case where such fixed surface exists.

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