6 edited body

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 $\mbox{PSL}(2,\mathbb \mbox{PGL}(2,\mathbb C(\mathbb P^{n-1}))$ 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.

Let $n\ge 3$. If \mbox{Inv}(\mathbb P^n)$be the subgroup of$\mbox{Bir}(\mathbb{P}^n)$is generated by involutions and$\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$you would need at least there are uncountably many involutions of degree$> \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$\mbox{PSL}(2,\mathbb C(\mathbb P^{n-1}))$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. 4 edited body Let$n\ge 3$. If$\mbox{Bir}(\mathbb{P}^n)$is generated by involutions then for any$d >1$you would need at least uncountably many involutions of degree$> d$. This is a direct consequece 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. 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 $$1 \to \mbox{PSL}(2,\mathbb mbox{PGL}(2,\mathbb C(\mathbb P^{n-1})) \to \mbox{St}_p(\mathbb P^n) \to \mbox{Bir}(\mathbb P^{n-1}) \to 1$$ 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$\mbox{PSL}(2,\mathbb C(\mathbb P^{n-1}))$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.