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What is the universal cover of the wedge sum of the torus and the real projective plane?

I know from Hatcher's Algebraic Topology that the universal cover of $\mathbb{R}P^2 \vee \mathbb{R}P^2 $ is an infinite number of spheres each one of them attached to two other spheres. I tried to mimic this construction somehow for this situation "gluing" together the universal covers of the torus and the projective plane and getting something like $\mathbb{R}^2$ with an infinite number of spheres attached but this doesn't seem to work.

How can I calculate the universal cover of this space?

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$\widetilde{\Bbb{RP}^2 \vee T^2}$ is going to look like a tree with vertices corresponding to either $S^2$ or $\Bbb R^2$ and edges corresponding to one-point union of the two spaces corresponding to the vertices it joins.

The tree is a colored tree, with vertices colored by blue and red, each blue vertex adjacent only to red vertices and each red vertex adjacent only to blue vertices. Neighborhood of a red vertex consist of $\Bbb Z/2$-many vertices and neighborhood of a blue vertex consist of $\Bbb Z^2$-many vertices. This is because the wedge point $x$ in $\Bbb{RP}^2 \vee T^2$ lifts to $\Bbb Z/2$-many points in each $S^2$, and $\Bbb Z^2$-many points in each $\Bbb R^2$. Replacing each red vertex by an $S^2$, each blue vertex by an $\Bbb R^2$ and each edge by one-point union of the two vertex spaces gives me the desired universal cover.

Here is a picture of the part of the graph. While there are infinitely many red vertices adjacent to blue vertices, only finitely many are drawn for obvious reasons and the existence of the rest are dotted. As we see, the graph is a tree with vertex set partitioned into two colors and valence of blue vertices is $|\Bbb Z^2|$ and valence of red vertices is $2$.

enter image description here

Thus, ultimately, the space $\widetilde{\Bbb{RP}^2 \vee T^2}$ is iterative one-point-union of infinitely many $S^2$'s and $\Bbb{R}^2$'s, with each $S^2$ wedged with two $\Bbb R^2$'s, and each $\Bbb R^2$ wedged with $\Bbb Z^2$-many $S^2$'s.

$\text{Explanation}$: To see this, note that $\Bbb R^2$ is the universal cover of $T^2$, hence $\Bbb R^2 \bigvee_{\Bbb Z^2} \Bbb{RP}^2$ ($\Bbb R^2$ with a copy of projective plane attached at each integer lattice) covers $\Bbb{RP}^2 \vee T^2$. Now $S^2$ is the universal cover of $\Bbb{RP}^2$, so you can similarly "unwrap" one of the projective planes from $\Bbb Z^2$-many of them to get the cover $\Bbb R^2 \bigvee_{\Bbb Z^2 - (0, 0)} \Bbb{RP}^2 \vee (S^2 \vee \Bbb R^2 \bigvee_{\Bbb Z^2 - (0, 0)} \Bbb{RP}^2)$. Covering all of the wedged $\Bbb{RP}^2$'s likewise, one will end up with the cover $\Bbb{R}^2 \bigvee_{\Bbb Z^2} (S^2 \vee \Bbb R^2 \bigvee_{\Bbb Z^2} \Bbb{RP}^2)$. "Unwrapping" iteratively in this process will give you a tree-like structure, entirely consisting of $S^2$ and $\Bbb R^2$, hence simply connected and thus a universal cover of your space.


$\text{Remark}$: The reason that you get a much nicer thing for $\Bbb{RP}^2 \vee \Bbb{RP}^2$ is that your tree consists of vertices corresponding only to $S^2$ and the wedge point lifts only to 2 points in each $S^2$. This implies for every $S^2$-vertex, there are only two $S^2$-vertices adjacent to it in the graph, so globally it looks like an infinite string of $S^2$'s, each two of them touching at a point. Note that the graph is still a tree, with each vertex being of valence $2$.

enter image description here

The presence of a space (i.e., $T^2$) with infinite fundamental group ($\pi_1(T^2) \cong \Bbb Z^2$) makes things worse.

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