The symmetric case:
Let $w \in \mathbb{C}(x,y)$, and write $w=\frac{u}{v}$, where $u,v \in \mathbb{C}[x,y]$.
Let $\beta: (x,y) \mapsto (x,-y)$; $\beta$ is an involution (= an automorphism of order two) on $\mathbb{C}[x,y]$, and hence an involution on $\mathbb{C}(x,y)$.
Assume that $\beta(w)=w$.
Is it possible to find the 'exact' form of $w$?
My answer: Write $u=s+k$ and $v=s'+k'$, where $s,s' \in \mathbb{C}[x,y]$ are symmetric w.r.t. $\beta$ and $k,k' \in \mathbb{C}[x,y]$ are skew-symmetric w.r.t. $\beta$.
Observe that it is easy to find the general form of a symmetric element w.r.t. $\beta$: $a_{2n}y^{2n}+\cdots+a_{2}y^2+a_0$, where $a_{2i} \in \mathbb{C}[x]$. Similarly, the general form of a skew-symmetric element w.r.t. $\beta$: $b_{2m+1}y^{2m+1}+\cdots+b_{3}y^3+b_1y$, where $b_{2j+1} \in \mathbb{C}[x]$.
Now, $\beta(w)=w$ means that $\frac{s-k}{s'-k'}=\frac{s+k}{s'+k'}$, and then after a direct calculation we obtain $ks'=sk'$, equivalently, $\frac{k}{k'}=\frac{s}{s'}$. Then if we write $\mathbb{C}(x,y) \ni c:=\frac{k}{k'}=\frac{s}{s'}$, we get $u=s+k=cs'+ck'=c(s'+k')=cv$, so $w=\frac{u}{v}=\frac{cv}{v}=c=\frac{k}{k'}=\frac{s}{s'}$.
Therefore, we obtained that $w$ must be a quotient of two symmetric polynomials (w.r.t. $\beta$) or $w$ must be a quotient of two skew-symmetric polynomials (w.r.t. $\beta$).
Actually, a skew-symmetric element is a multiple of $y$ by a symmetric element, so 'the reduced form' of $w$ is as a quotient of two symmetric polynomials. ('reduced form' to symmetric and skew-symmetric elements, but not reduced to irreducible factors).
Am I right?
Thank you very much!
The skew-symmetric case:
Let $w \in \mathbb{C}(x,y)$, and write $w=\frac{u}{v}$, where $u,v \in \mathbb{C}[x,y]$.
Let $\beta: (x,y) \mapsto (x,-y)$; $\beta$ is an involution (= an automorphism of order two) on $\mathbb{C}[x,y]$, and hence an involution on $\mathbb{C}(x,y)$.
Assume that $\beta(w)=-w$.
Is it possible to find the 'exact' form of $w$?
My answer: Write $u=s+k$ and $v=s'+k'$, where $s,s' \in \mathbb{C}[x,y]$ are symmetric w.r.t. $\beta$ and $k,k' \in \mathbb{C}[x,y]$ are skew-symmetric w.r.t. $\beta$.
Now, $\beta(w)=-w$ means that $\frac{s-k}{s'-k'}=-\frac{s+k}{s'+k'}$, and then after a direct calculation we obtain $ss'=kk'$, equivalently, $\frac{s}{k'}=\frac{k}{s'}$. Then if we write $\mathbb{C}(x,y) \ni c:=\frac{s}{k'}=\frac{k}{s'}$, we get $u=s+k=ck'+cs'=c(k'+s')=cv$, so $w=\frac{u}{v}=\frac{cv}{v}=c=\frac{s}{k'}=\frac{k}{s'}$.
Therefore, we obtained that $w$ must be a quotient of a symmetric polynomial with a skew-symmetric polynomial, or vice versa.
Since $k$ is skew-symmetric w.r.t. $\beta$, it has the form $k=s''y$ for some symmetric $s''$. Therefore, $w=\frac{k}{s'}$ becomes $w=\frac{s''y}{s'}=\frac{s''}{s'}y$.