This is a question from an old qualifying exam in topology.

Let $S_g$ be the compact orientable surface of genus $g$. Show that $S_g$ has an irregular 3-fold cover when $g>1$.

While the question does not explicitly state it, I am pretty sure we are meant to assume we are looking for a connected cover.

I understand that this amounts to finding a non-normal index 3 subgroup of $$\pi_1(S_g)=\langle a_1,b_1,\dots,a_g,b_g\big|[a_1,b_1]\cdots[a_g,b_g]\rangle.$$

I'm sure there is a group theoretic way to show such a subgroup exists, and I'm certainly interested in hearing those arguments. However, I'm much more interested in a geometric argument or at least understanding of this covering space.

I've toyed around with it and haven't really gotten anywhere. I've tried to construct a non-normal index 3 subgroup by hand, and I've also been looking at the regular covering spaces of $S_g$ (Example 1.41 in Hatcher) and trying to tweak them in some way.

  • $\begingroup$ Does "irregular" mean "branched"? Something else...? $\endgroup$ – Andrew D. Hwang Aug 23 '16 at 21:51
  • 1
    $\begingroup$ Presumably "irregular" means "not regular", where "regular" has the standard meaning that the deck group acts transitively on point pre-images, equivalently the image of the induced map on fundamental groups is normal. $\endgroup$ – Lee Mosher Aug 23 '16 at 21:56
  • 1
    $\begingroup$ The existence of an irregular connected degree 3 cover is quite easy to check: it's equivalent to the existence of a surjection $\pi_1 \to S_3$. But I don't even have a picture of such a cover for $\Sigma_2$. $\endgroup$ – user98602 Aug 23 '16 at 23:05

In case anyone is interested in my solution, here it is. I talked to a few other people to get some ideas, and used the feedback I got here as well.

Using the idea to take a surjection $\pi_1(S_g)\rightarrow\Sigma_3$, I found an index 3 subgroup of $\pi_1(S_g)$. The map is $$\phi:\langle a_1,b_1,\dots,a_g,b_g\Big|[a_1,b_1]\cdots[a_g,b_g]\rangle\rightarrow \Sigma_3$$ $$a_1,a_2\mapsto (12)$$ $$b_1\mapsto (13)$$ $$b_2\mapsto (23)$$

(all other generators map to $e$, but I'll continue with just the case for $g=2$ for simplicity).

Then $\langle (12)\rangle$ is a non normal index 3 subgroup of $\Sigma_3$, so $$\phi^{-1}(\langle(12)\rangle)=\langle a_1,a_2,b_1^2, b_2^2, b_1b_2b_1^{-1},b_2b_1b_2^{-1}\rangle$$

is a non normal index 3 subgroup of $\pi_1(S_g)$.

So far this isn't really adding anything to the responses I got, but I was really interested in a geometric picture of what was going on. After a bit of fiddling around I ended up with this picture.

Geometric realization of 3-sheeted covering space of $S_2$, closed orientable surface of genus 2.

Thanks for the input, everyone!

  • 3
    $\begingroup$ What a lovely picture. I think you should accept your own answer. $\endgroup$ – user98602 Sep 10 '16 at 18:28

This probably isn't the sort of picture you're looking for, but eh.

A 3-fold covering of $M$ canonically (once you pick a bijection $p^{-1}(x) \cong \{1,2,3\}$) gives you a homomorphism $\pi_1(M) \to S_3$, sending a loop to the way its lifts permute the fiber. A regular 3-fold covering is one such that the image of this homomorphism is either trivial or a copy of $\Bbb Z/3$, and a connected one comes from any homomorphism whose image is either $S_3$ or $\Bbb Z/3$. (See if you can prove these facts.)

How do you go backwards from this? If I have a homomorphism $f: \pi_1(M) \to S_3$, I can actually construct a degree 3 covering space corresponding to the action on the fiber. Thinking of $\pi_1(M)$ as the deck transformations on the universal cover, define $$M_f = (\tilde M \times \{1, 2, 3\})/((x,n) \sim(\varphi(x),f(\varphi)(n)).$$ This is a degree three covering space of $M$, with the covering map just the natural covering map in the first factor. So we just need to find a surjective homomorphism $\pi_1(M) \to S_3$ to construct your cover corresponding to a non-normal index 3 subgroup (which would literally be $f^{-1}(\{(1), (12)\})$ eg.)

You could probably try to use this explicit construction to construct the covering space explicitly, though I didn't try; we have a composition of covering maps $\bar{M_f} \to M_f \to M$, where $\bar{M_f}$ is the covering space corresponding to $\text{ker}(f)$, and the first map is a normal covering (as is the composition). You could figure out what $\bar{M_f}$ is, and then find what the $\Bbb Z/2$ action is. But I had trouble pulling this off.

For an easy actual example of such a thing, just take $$f(g_1) = (12), f(h_1) = (13), f(g_2) = (13), f(h_2) = (12),$$ and $f(g_i) = f(h_i) = (1)$ for $i>2$. This also makes it clear why every degree 3 covering of the torus is normal; every homomorphism to $S_3$ must have abelian image!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.