1
$\begingroup$

What is the universal cover of a torus minus one point?

Same as this question but I did not find the answer to be understandable.

I know a universal covering must be simply connected, i.e. have trivial $\pi_1$.

I also know that the fundamental group of the torus minus a point is $\mathbb{Z} \ast \mathbb{Z} \cong F_2$.

The wedge $S^1 \vee S^1$ has the same $\pi_1$, and universal cover this thing

enter image description here

A comment on the post linked above mentioned thinking of a "fattened up" version of the above Cayley graph, but I'm not sure what this refers to.

When I try to imaging the graph thickened, it seems like it would become a plane, or some sort of snowflake looking thing?

The answer given in the same post was not understandable to me since I'm not familiar with holomorphisms or Riemann uniformization.

Is there an explanation for how to compute this universal cover that relies primarily upon basic Algebraic Topology (e.g. Hatcher)?

$\endgroup$
6
  • 1
    $\begingroup$ Any simply connected two-dimensional boundary-less manifold(may or may not be compact) is either plane $\Bbb R^2$ or sphere $\Bbb S^2$, upto homeomorphism. If sphere $\Bbb S^2$ covers any surface, then it's either $\Bbb S^2$ or projective plane $\Bbb RP^2$. In other words, the universal cover of any two-dimensional boundary-less manifold $X$ ($X$ may or may not be compact, may or may not be orientable) is homeomorphic to the plane $\Bbb R^2$; provided $X\not \cong \Bbb S^2,\Bbb RP^2$. $\endgroup$
    – Sumanta
    Commented Mar 9, 2022 at 4:54
  • $\begingroup$ The answer, as people are telling you, is (up to homeomorphism) the open unit disk. Unfortunately, I don't know of any "elementary" proof of this fact (though there probably is one). Is this an exercise you've set for yourself, or was it assigned in a class? $\endgroup$ Commented Mar 9, 2022 at 5:06
  • $\begingroup$ This is for a Topology class working out of Hatcher. I was thinking as an explanation, since the punctured torus is homotopic to $S^1 \vee S^1$ and the homotopy "goes up" a dimension (2 to 3) then maybe there is some necessary relation that says there's a homotopy from the 4-regular Cayley graph univ. cover of $S^1 \vee S^1$ and the univ. cover of the punctured torus, that goes up from 1 dimension to 2? I don't know if that necessarily follows from the spaces being homotopic though. $\endgroup$ Commented Mar 9, 2022 at 5:09
  • $\begingroup$ Well, this seems to be at least true for path-connected, locally path-connected spaces and universal covers, see here $\endgroup$ Commented Mar 9, 2022 at 5:16
  • 1
    $\begingroup$ It's like the infinite tree in your picture but each vertex is replaced by a small disk and each edge is replaced by a narrow strip. I think it is better for visualization to think of all the edges to be of the same size. There is no need to embed it in the plane. $\endgroup$ Commented Mar 9, 2022 at 6:41

1 Answer 1

7
$\begingroup$

Consider the torus as a square with the sides identified:

a square with the sides identified

We puncture this

a punctured square

and then cut along a diagonal into our puncture. This is OK as long as we remember to glue it back together at the end.

a punctured square with two marked diagonals

Now, we cut

cut along the marked diagonals

and glue back together along $a$ (since really the two edges marked $a$ are the same edge)

glue along a

Now we can straighten this out to get an octagon

a decorated octagon

If we glue the marked edges together, then we'll be left with a punctured torus. Since we want a simply connected covering space, though, we'll glue copies of this fundamental domain together instead! This is analogous to finding a simply connected covering space for the torus by tiling squares. I'll leave it to you to figure out how to glue a bunch of copies of this octagon together in order to tile the (open) upper half plane.

As a hint, the punctured torus is a hyperbolic object, so you'll need to distort these octagons pretty seriously in order to make things fit. In particular, it's not worth trying to make anything "loop back on itself". In the same way that a $4$-regular tree gets smaller the further out you go, and its loops never close, you'll just have to stick more and more of these octagons (which necessarily get smaller and smaller) together wherever the marked edges indicate.


I hope this helps ^_^

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .