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This question asks if every involution on $\mathbb{C}[[x_1,\ldots,x_n]]$, after a change of variables/coordinates, is given by $x_i \mapsto \pm x_i$, and some commenters gave a positive answer.

For simplicity, let us concentrate on the case $n=2$, namely, $\mathbb{C}[[x,y]]$.

What is exactly meant by "a formal change of coordinates"? Does it mean just taking the original involution $f$ and composing it (on the right or on the left) by any automorphisms of $\mathbb{C}[[x,y]]$?

For example, $f: (x,y) \mapsto (x,-y+x+x^2+x^3+\cdots)$ is an involution on $\mathbb{C}[[x,y]]$ and if we take the automorphism $h: (x,y) \mapsto (x,y+x+x^2+x^3+\cdots)$, then we get that $hf: (x,y) \mapsto (x,-y)$ is of the claimed form (notice that $h$ and $f$ commute).

Can one please give an example of an involution $f$ such that $f(x)$ or $f(y)$ involves monomials of the form $\lambda x^iy^j$, where $\lambda \in \mathbb{C}-\{0\}$ and $ij \neq 0$.

Of course, in my above example there are no such monomials.

(Remark: I understand that, for a given involution $f$, we can apply the Lemma mentioned in the first comment to $G:=\{1,f\}$, and get that the values of the characters are $\{\pm1\}$ since $f$ is of order $2$).

Edit: If it happens that the given involution $f:(x,y) \mapsto (A,B)$ satisfies $A,B \in \mathbb{C}[x,y]$, then by this it is an involution on $\mathbb{C}[x,y]$; in this case I can show (by quite lengthy calculations+ Serre's theorem about trees) that any involution on $\mathbb{C}[x,y]$ is either conjugate to $(x,y) \mapsto (x,-y)$ or to $(x,y) \mapsto (-x,-y)$.

user237522
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  • Note about $\LaTeX$ symbolization: you can use [\![ instead of [[what is the way is used in many textbooks. – Masacroso Apr 19 '17 at 23:11
  • It doesn't mean composing on the left or right (that would be trivial, you could just take $h=f$), it means conjugating. ie you want $h^{-1}fh$ to be of the desired form. For your example you could take $h(x)=x$, $h(y)=2y-x/(1-x)$. – stewbasic Apr 19 '17 at 23:23
  • @stewbasic, interesting!! I did not think about that option (conjugating). How did you know that? Have you read the article that is mentioned in MO's first comment? Or is it a known fact? – user237522 Apr 19 '17 at 23:28
  • For the benefit of some users: Denote $P=x+x^2+x^3+\cdots$. Then, $f: (x,y) \mapsto (x,-y+P)$, $h: (x,y) \mapsto (x,2y-P)$, $h^{-1}: (x,y) \mapsto (x,\frac{y}{2}+\frac{P}{2})$. A direct computation shows that $h^{-1}fh: (x,y) \mapsto (x,-y)$. – user237522 Apr 19 '17 at 23:50
  • @stewbasic, please, any thoughts/hints about my second question? Is it true that it has a negative answer? (there is no such involution), or one can find such involution? – user237522 Apr 20 '17 at 00:23
  • If you conjugate $(x,y)\mapsto(x,2x^2-y)$ by a linear transformation such as $(x,y)\mapsto(x+y,x-y)$, you get $(x,y)\mapsto(y+(x+y)^2,x-(x+y)^2)$. – stewbasic Apr 20 '17 at 00:37
  • But if one conjugates by an automorphism of $\mathbb{C}[[x,y]]$ which is not linear? – user237522 Apr 20 '17 at 00:42
  • Am I missing something trivial? – user237522 Apr 20 '17 at 00:47

1 Answers1

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An example for the second question:

Denote $w=1-y+y^2-y^3+y^4+\cdots$, which is the inverse of $1+y$, and denote $u=1+y+y^2+y^3+y^4+\cdots$, which is the inverse of $1-y$.

Let $h: (x,y) \mapsto (x+xy,y)$, so its inverse is $h^{-1}: (x,y) \mapsto (xw,y)$.

Conjugate the involution $\beta: (x,y) \mapsto (x,-y)$ by $h$, $f:=h^{-1} \beta h$.

$f$ is the involution that answers the second question; indeed, a direct calculation shows that $f: (x,y) \mapsto (x(1-y)w, -y)$.

user237522
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