Orbit space for a nice action

Question

Consider the action of $\mathbb Z_2$ on $\mathbb P^5$ by $1.x_1=-x_6, 1.x_2=-x_5, 1.x_3=-x_3, 1.x_4=-x_4, 1.x_5=-x_2, 1.x_6=-x_1$ where $1$ is the non-identity element of $\mathbb Z_2$. I need to understand the orbit space for the action of $\mathbb Z_2$ on ($\mathbb P^5 -(\mathbb P^5)^{\mathbb Z_2}$) in this case. I computed the fixed locus and it is isomorphic to $\mathbb P^1$ but I don't have a clear picture how the orbit space looks like. Is there a nice realization of the orbit space ? May be a bundle over something ?

Answer 1

Consider the change of variables

$$ \begin{array}{ll} u_1 & = x_1-x_6 \\ u_2 & = x_2-x_5 \\ u_3 & =x_3 \\ u_4 & =x_4 \\ u_5 & =x_2+x_5 \\ u_6 & =x_1+x_6 \end{array} $$

With respect to this basis, the lifted action of the nontrivial element $g$ of $\mathbb{Z}_2$ is given by the diagonal matrix $\mathrm{diag}(+1,+1,-1,-1,-1,-1)$. Thus, the fixed point set in $\mathbb{A}^6$ (in $u$ coordinates) is given by those vectors of the form $(u_1,u_2,0,0,0,0)$, while there are additional fixed points in $\mathbb{P}^5$ of the form $[0,0,u_3,u_4,u_5,u_6]$. The first parametrizes a copy of $\mathbb{P}^1$ in $\mathbb{P}^5$, and the second parametrizes a copy of $\mathbb{P}^3$ inside $\mathbb{P}^5$. The complement $C$ is comprsied of all $[u_1,u_2,u_3,u_4,u_5,u_6]$ in which both of $(u_1,u_2)$ and $(u_3,u_4,u_5,u_6)$ are nonzero. We then have a surjection $A\times\mathbb{P}^1\times\mathbb{P}^3\to C/\mathbb{Z}_2$,

$$ (a,[u_1,u_2],[u_3,u_4,u_5,u_6])\mapsto [au_1,au_2,u_3,u_4,u_5,u_6], $$

where $A=(\mathbb{A}^1\setminus\{0\})/\mathbb{Z}_2$ (the action being $a\leftrightarrow -a$).

If the field is $\mathbb{R}$, this is $\simeq\mathbb{R}\times S^1\times \mathbb{RP}^3$. If it's $\mathbb{C}$, this is $\simeq \mathbb{R}\times S^1\times S^2\times\mathbb{CP}^3$.