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Let $X=(S^1\times S^1)\vee S^1$. What are all connected 3-sheeted covering spaces of $X$ up to homeomorphism.

I found one which is three squares with three circles from the universal covering. But I am not sure it is right. My guess about the question is that I need to show there exists several types first then draw them but I don't know how. Thanks for your help!

Ling
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  • As a starting point to help your thinking, what are the three sheeted covers of $S^1$ and $S^1 \times S^1$? – Aaron Jul 18 '17 at 16:41
  • The three sheeted cover of $S^1$ is a three circle spring? And the cover for $S^1\times S^1$ is three squares with the opposite sides identified? – Ling Jul 18 '17 at 16:54
  • You could call it a three circle spring, but it's probably more correct to call it a circle which goes around the original circle 3 times as fast. If you view the circle as the unit circle in the complex plane, it is the map $z\mapsto z^3$ And for the torus, it's that map cross the identity map, one way or the other. But in both case, as spaces, the covering space is just the original space, it's just the map that determines the covering which makes it something different. – Aaron Jul 18 '17 at 17:08

1 Answers1

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I have been able to find two 3-sheeted coverings. For one of them, take a torus, and every third of the way around, attach a circle. For the other, take a circle, and every third of the way around, attack a torus.

The easiest way to prove that these are all the coverings should be to attack the problem algebraically. Covering spaces correspond to subgroups of the fundamental group, with the number of sheets of the cover being the same as the index of the subgroup. Therefore, we want to examine the index 3 subgroups of $\pi_1((S^1\times S^1)\vee S^1)\cong \mathbb Z^2 * \mathbb Z$, which has presentation $(a,b,c\mid ab=ba)$. I want to say that the index $3$ subgroups are the ones generated by $a^3,b,c$; by $a,b^3, c$; and by $a,b,c^3$, and that the first two yield the same covering space. However, I am unsure of myself here.

Aaron
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  • Many thanks!!!! – Ling Jul 21 '17 at 21:01
  • Sorry but I have one more question. Are they normal covering spaces? I don't really understand how to tell whether they are "symmetric" or not. I am not quite sure they need to be symmetric about what? Thanks a lot a lot! – Ling Jul 21 '17 at 21:18
  • Yes, they are normal. The group of deck transformations in both cases is generated by rotations by $2\pi/3$ (assuming you have embedded the spaces in $\mathbb R^3$ such that the central circle/torus is symmetric about the $z$-axis, and placed the outer circles/tori with $3$-fold symmetry). It is easy enough to look at the orbit of a point and see that it has cardinality 3, which is the same as the rank (is that the right term?) of the cover. – Aaron Jul 21 '17 at 22:06
  • Although I shouldn't say "easy enough" because I have a picture in my head that you may not, and until you have that picture it will not be obvious. The question you have to ask is, "You said what the covering space was, but you didn't say what the map was....what is the map?" Just like the 3 fold cover of the circle sends $(r,\theta)$ to $(r,3\theta)$, ours sends $(r,\theta,z)$ to $(r,3\theta,z)$ (using the embedding in my previous comment). You should verify that the rotations I mention actually commute with this map. – Aaron Jul 21 '17 at 22:12
  • Thanks very much! This is really helpful! :) – Ling Jul 22 '17 at 19:40
  • I am sorry but I have one more question. How many different homeomorphism types are there between the two covering spaces. To be honest, I don't understand what a homeomorphism type is. Many thanks!! – Ling Jul 25 '17 at 20:27