Neighborhoods of the fixed points of a $\Bbb Z_2$ action on the 4-torus

Question

Regard the $4$-torus $T^4$ as a quotient of $\Bbb C^2$ by the $\Bbb Z^4$ action given by translation. Then consider the $\Bbb Z_2$-action on $T^4=\Bbb C^2/\Bbb Z^4$ given by $(z_1,z_2)\mapsto (-z_1,-z_2)$. Let $X$ denote the resulting quotient space $T^4/\Bbb Z_2$. It is easy to see that there are exactly 16 fixed points of this $\Bbb Z_2$ action. Let $x_1,\dots,x_{16}$ denote the images in $X$ of the fixed points. How can we show that each $x_i$ has a neighborhood in $X$ that is homeomorphic to the cone of $\Bbb RP^3$?

Answer 1

The equivariant tubular neighborhood theorem says that if $G$ acts on $M$ and $S \subset M$ is a closed $G$-invariant submanifold, then there are arbitrarily small open neighborhoods of $S$ which are equivariantly diffeomorphic to the normal bundle $NS$. (Note that $NS$ inherits a $G$-action.)

in your case $S$ is a fixed point, and so there is a neighborhood of $S$ equivariantly diffeomorphic to $N(x) = T_x M$. The action at this point is given by the negation action (check at each point separately). Therefore there is a neighborhood of $x$ diffeomorphic in particular to the 4-ball $D^4$ with negation action. Note that this is equivariantly diffeomorphic to the cone on $S^3$ with antipodal action (there is no action on the radial/cone coordinate). Taking the quotient, we see that in $T^4/(\Bbb Z/2)$ each fixed point has a neighborhood diffeomorphic to $C(S^3/[\Bbb Z/2]) = C(\Bbb{RP}^3)$.