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How to find all involutions on $\mathbb{C}(x,y)$, or at least all involutions $\delta$ on $\mathbb{C}(x,y)$ such that $\delta(x)=x$?

Remarks:

(1) An involution on $\mathbb{C}[x,y]$ is either conjugate to $\beta: (x,y) \mapsto (x,-y)$ or to $\epsilon: (x,y) \mapsto (-x,-y)$ (since the group of automorphisms of $k[x,y]$ is free amalgamated, and by a result of Serre about trees).

(2) In $\mathbb{C}[x,y]$ if we wish to find all involutions $\delta$ with $\delta(x)=x$, just write $\delta(y)=q$. There are two cases: (a) The Jacobian of $\delta$ is $1$, so $q_y=1$, and then $q=y+H(x)$, for some $H(t) \in \mathbb{C}[t]$. Since $\delta$ is of order $2$, by $\delta^2=1$ (=order $\leq 2$), we get $h(x)=0$, so $\delta=1$, which is not an involution. (b) The Jacobian of $\delta$ is $-1$, so $q_y=-1$, and then $q=-y+H(x)$, for some $H(t) \in \mathbb{C}[t]$. Since $\delta$ is of order $2$, by $\delta^2=1$ (=order $\leq 2$), we get that there is no restriction on $h(x)$, so $\delta: (x,y) \mapsto (x,-y+H(x))$. (Another way to obtain this is by using Theorem 1).

(3) Of course, in $\mathbb{C}(x,y)$ there exist more involutions, for example, $(x,y) \mapsto (x,\frac{1}{y})$. Is it true that the $\delta$'s I am looking for are those of (2)(b), $(x,y) \mapsto (x,\frac{1}{y})$ and $(x,y) \mapsto (x,-\frac{1}{y})$?

(4) I have asked this question, in which commenters suggested the following references: Serge Cantat's notes and Julie Deserti's list. (Perhaps I can find an answer to my recent question in those references).

Thank you very much.

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  • $\begingroup$ It seems that webusers.imj-prg.fr/~julie.deserti/biblio/… answers my question (Theorem 2.6). $\endgroup$
    – user237522
    Commented Apr 9, 2017 at 2:48
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    $\begingroup$ However, I am not familiar with Geiser and Bertini involutions, so I will be thankful if someone can elaborate a little on those involutions. $\endgroup$
    – user237522
    Commented Apr 9, 2017 at 2:52

1 Answer 1

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As you explain, the classification of birational involutions of $\mathbb{P}^2$ given by Bayle and Beauville in the article you cite gives the answer.

The Jonquières involutions preserve a rational fibration. They are in fact conjugate to $(x,y)\mapsto (x,p(x)/y)$ for some polynomial $p\in \mathbb{C}$ without any multiple root and of even degree. The fixed locus is given by the curve $y^2=p(x)$, which is rational if $\deg(p)=2$, elliptic is $\deg(p)=4$ and hyperelliptic of genus $g$ when $\deg(p)=2g+2\ge 6$. The birational type of the curve determines the conjugacy class in the Cremona group $\mathrm{Aut}_{\mathbb{C}}(\mathbb{C}(x,y))$. In particular, for $\deg(p)=2$ you have only one conjugacy class, in fact also conjugate to $(x,y)\mapsto (x,-y)$ in $\mathrm{Aut}_{\mathbb{C}}(\mathbb{C}(x,y))$ (but not by an element that fixes $x$).

The other involutions (Geiser and Bertini) do not preserve any rational fibration. They preserve elliptic fibrations, but this does not help so much to write explicitely the map. This can be done in practice but the formulas are quite horrible. In particular, the degree of a Bertini involution is $17$ and the degree of a Geiser Bertini involution is $8$. So $x$ and $y$ are sent onto rational functions, quotients of polynomials of degree $17$ or $8$. Choosing the coordinates such that the point $[1:0:0]$ or $[0:1:0]$ (pencils of $x/z$ and $y/z$ corresponding to $x=cst$ and $y=cst$) you obtain polynomials of degree slightly less ($11$ and $5$) but it not so good for expliciting the maps. The geometric description given by Del Pezzo surfaces of degree $1$ and $2$ that you can find in the article of Bayle and Beauville is definitely better, I think.

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  • $\begingroup$ Just to understand, are you saying that there are only 2 (mod conjugacy) non-Jonquières involutions? Also, in the Jonquières case with $p$ of degree $2n$, one obtains a parameterization by the set $P_{2n}$ of monic polynomials $p$ of degree $2n$ with no multiple root, which has dimension $2n$, modulo identification $\sim$ of $p,p'$ when $y^2=p(x)$ and $y^2=p'(x)$ are equivalent. The quotient $P_{2n}/\sim$ has dimension $\le 2n-2$, but is the value known? (Probably, since moduli spaces of curves are well-known...) For $2n=4$ is it positive? $\endgroup$
    – YCor
    Commented Jan 30, 2019 at 14:54
  • $\begingroup$ No, there are 2 "types" of non-Jonquières involutions: Bertini and Geiser involutions. But the conjugacy classes are given by isomorphism classes of smooth quartic curves for Geiser and non-hyperelliptic curves of genus 4 for Bertini, so you get infinitely many in both cases (but finite dimensional, not like Jonquières). I have added some "s" to make it clearer. For the Jonquières, it is simply up to $PGL_2=\mathrm{Aut}(\mathbb{P}^1)$, so you simply remove $4$. For $2n=4$, this gives a dimension $1$ family: the moduli space of elliptic curves (given by the $j$-invariant). $\endgroup$
    – Jérémy Blanc
    Commented Jan 30, 2019 at 20:34
  • $\begingroup$ I guess you mean "remove 3" (for $2n\ge 4$), yielding a parameter space of dimension $\max(0,2n-3)$. $\endgroup$
    – YCor
    Commented Jan 30, 2019 at 20:35
  • $\begingroup$ Absolutely. For $2n=2$ you have only one conjugacy class and for $2n\ge4$ you get dimension $2n-3$. $\endgroup$
    – Jérémy Blanc
    Commented Jan 30, 2019 at 21:05

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