# Which surface is homotopy equivalent to $\Bbb{R}^4$ minus the planes $x=y=0$, $z=w=0$?

In completing an exercise I have shown that $\Bbb{R}^3$ minus the axes $x=0$, $y=0$, and $z=0$ is homotopic equivalent to the cube graph $Q_3$. To visualize this, $\Bbb{R}^3-0$ is homotopy equivalent to $S^2$; then, remove the $x=0$ axis and we have a space equivalent to "a cylinder"; finally,remove the $y=0$ and $x=0$ axes and we have a punctured cylinder: expanding the punctures gives the graph.

The next part of the question asks me to consider $\Bbb{R}^4$ minus the planes $x=y=0$, $z=w=0$. I tried a similar approach, however $\Bbb{R}^4$ is much more difficult to imagine. Start with $\Bbb{R}^4-0\simeq S^3$. I think that removing the $x=y=0$ axis gives a surface in $\Bbb{R}^4$ somewhat like a cylinder but "enclosing" the $x=y=0$ plane. The next cut/puncture I have no idea with, however. I cannot think how to remove $z=w=0$, without leaving the resulting surface disconnected.

Another idea I had was by trying to construct something analogous to $Q_3$. Where $Q_3$ had eight vertices, one corresponding to each octant, perhaps my surface in $\Bbb{R}^4$ will have sixteen edges with thirty-two surfaces corresponding to the adjacent orthants (sharing a hyperplane).

• It might be easier to think in terms of what points you're removing from $S^3$: for example, removing the plane $x=y=0$ is the same as removing points on $S^3$ with $x=y=0$, that is, you're removing a set of points equivalent to $z^2+w^2=1$ (which is a circle). Apr 19, 2016 at 22:11
• @Steve D: Hmm and then removing $x^2+y^2=1$ would be another circle. Perhaps then, the surface is a torus (a subset of $\Bbb{C}^2$), or am I way off track? Apr 19, 2016 at 22:35
• I would say you are right on track! Apr 19, 2016 at 22:36

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$To come at Steve D's comment from a slightly different angle, it may help to view the complement of orthogonal coordinate planes in $\Reals^{4}$ as $\Cpx^{2}$ with the "axes" $\Cpx \times \{0\}$ and $\{0\} \times \Cpx$ removed. What remains is $\Cpx^{\times} \times \Cpx^{\times}$.