I have a basic question about a certain action on product of projective lines $\mathbb{P}^1 \times \mathbb{P}^1$ by group $\mathbb{Z}/2$ which Sasha introduced in this MO discussion.
How does it here exactly work? Sasha noted that 'the group acts diagonally, and nontrivially on each factor'.
My first guess was that this action swaps just the both factors a la $(a,b) \mapsto (b,a)$ but it is described the be diagonal, that is it acts as $(a, b) \mapsto (ga, gb)$ with $g \in \mathbb{Z}/2$.
Which action is here meant? simply the 'minus sign'-action $(a,b) \mapsto (-a,-b)$ or something else?
#update: As Prof Mosher remarked that's definitely diagonal action and therefore is given by $(ga, gb)$. Now the question is what is $ga$ or in other words how $\mathbb{Z}/2$ acts on $\mathbb{P}^1$? I think that my first idea that it acts by multiply a minus sign fails, because we would obtain a trivial action since every 'point' in $\mathbb{P}^1$ is given in homogeneous coordinates as $[x:y] \in \mathbb{P}^1$ up to multiplicative factor therefore multiplying it by a minus sign doesn't change it. Therefore the action is trivial.
Otherwise possibly the action by $\mathbb{Z}/2$ on $\mathbb{P}^1$ is defined here with 'minus sign' in 'affine' sense, namely that if we take $\mathbb{P}^1 = Proj(k[X_0, X_1])$ then it acts on affine pieces on rings $k[X_0/X_1]$ and $k[X_1/X_0]$ by multiplication by $-1$ eg $X_0/X_1 \mapsto -X_0/X_1$?
You've written the action in your post: the action of each $g \in \mathbb Z / 2$ on each factor of $\mathbb P^1 \times \mathbb P^1$ is already given, and then $g$ acts on $(a,b) \in \mathbb P^1 \times \mathbb P^1$ by taking it to $(ga,gb) \in \mathbb P^1 \times \mathbb P^1$.