Let $X$ be a projective non-singular complex surface. Let $i$ be an involution on $X$. I am confused about a couple of points.
Suppose $C$ is a curve on $X$ such that $C$ is moved by $i$ i.e. $C\neq i(C)$. Let $C'$ be any curve such that $C'$ is linearly equivalent to $C$.
1) Then is it true that $C'\neq i(C')$ ?
2) Also is $i(C)$ linearly equivalent to $i(C')$? That is i want to check if $O(i(C))=O(i(C'))$. I don't know if this is true.
Any help would be very useful.
For the first question the answer is no. Suppose that $X = \mathbb{P}^2$, $C$ is given by $xy-z^2=0$ and $i(x:y:z) = (x:z:y)$. Note that $i(C)$ is defined by $xz-y^2=0$. We have that $C' = \{ x^2+y^2+z^2 =0\}$ is linearly equivalent to $C$, as every conic is, and invariant by $i$.
For the second question, the answer is yes. Let $i\colon X \rightarrow X$ be any automorphism. If $C$ and $C'$ are linearly equivalent, then there exists a rational function $f$ such that ${\rm div}(f) = C-C'$. Then we see that ${\rm div}(f\circ i^{-1}) = i(C)-i(C')$ i.e. $i(C)$ and $i(C')$ are linearly equivalent.