Draw a cover of $S^1 \vee S^1$ whose $\pi_1$ is isomorphic to $\langle a^2,b^3,aba^{-1}b^{-1} \rangle \leq F_2$.

This is a follow-up to my post from yesterday regarding the kernel K of a map $\Phi: F_2\to \mathbb{Z}_2 \bigoplus \mathbb{Z}_3$ which sends $a\mapsto ([0]_2,[1]_3)$ and $b\mapsto ([1]_2, [0]_3)$.

Having obtained $K=\ker{\Phi}=\langle a^2,b^3,aba^{-1}b^{-1} \rangle \leq F_2$ I am now tasked with drawing a covering of $S^1 \vee S^1$ such that the fundamental group of the covering space maps isomorphically to $K$ under the homomorphism induced by $\Phi$.

I know that any covering of $S^1 \vee S^1$ will be a 4-regular graph (each vertex will have 4 half-edges), and I have seen many examples of interesting covers of $S^1 \vee S^1$ such as in this post.

Using these as inspiration my first attempt at drawing a covering is this:

enter image description here

However, I then noticed that the outer vertices do not have 4 half-edges. So my next thought is just to extend the graph to infinity, making a fractal of the pattern (i.e. each outer 'red' vertex would be joined to a vertex of another 'blue' triangle). Though I don't know how to TeK that just yet.

I reckon the fundamental group of such an infinite graph would be (choosing any base point) $\langle a^2, ba^2b^{-1}=a^2, b^{-1}a^2b=a^2, b^3 \rangle$ assuming we have a commutativity relation.

Also, I'm not sure how $aba^{-1}b^{-1}$ could be a loop in the covering space, maybe I am confused about that generator. Is it effectively the same as a commutivity relation?

Is this correct, or am I missing something?

edit: Was able to finish up the graph correctly in Tikz thanks to the answer below. Here it is: enter image description here

  • $\begingroup$ I may be completely wrong. But what if you just join the three outer vertices with one blue triangle? $\endgroup$
    – Zeekless
    Feb 22, 2022 at 13:57
  • $\begingroup$ I don't think that would work since it would create 3 new holes, and I don't believe when you trace them out on the graph that they would be in the kernel. $\endgroup$ Feb 22, 2022 at 14:06
  • $\begingroup$ I haven't checked your construction in detail, but it's not wrong just because you have chosen not to give the graph a minimal set of vertices. you can just treat each of the three red loop as comprising a single edge. $\endgroup$
    – Rob Arthan
    Feb 22, 2022 at 16:28
  • 1
    $\begingroup$ Having had another look, it looks to me like you are getting $\langle a^2, b^3\rangle$, but I can't see $aba^{-1}b^{-1}$: have a think about the cell structure of a torus. $\endgroup$
    – Rob Arthan
    Feb 22, 2022 at 16:30

1 Answer 1


I'm pretty sure the diagram below (which was suggested by Zeekless) is the covering space you're looking for. Firstly, it's clearly a covering space since each vertex has the required pairs of arrows coming in and out. Call the covering space $X$, and let $p:X \to S^1 \vee S^1$ be the covering map.

Secondly, all the generators you want are in the image of the fundamental group. For example, if you start at any vertex and trace out the paths "$a$", "$b$", "$a^{-1}$", and then "$b^{-1}$", you get back where you started. Therefore, the word $aba^{-1}b^{-1}$ is in $p_*\pi_1(X)$. Similarly, following "$a$" twice and "$b$" three times (respectively) gets you back where you started, so we have $\langle a^2, b^3, aba^{-1}b^{-1}\rangle \subseteq p_*\pi_1(X)$.

We claim that actually this inclusion is an equality. To show this, we can consider $p_*\pi_1(X)/K$, where $K = \langle a^2, b^3, aba^{-1}b^{-1}\rangle$, and show that the quotient is trivial. Note that you already showed that $K$ is a kernel, hence a normal subgroup of $F_2$, and therefore also of $\pi_1(X)$, so we are allowed to quotient by it. Let $[w] \in p_*\pi_1(X)/K$, where $w$ is a word in $a$ and $b$. Since $aba^{-1}b^{-1}\in K$, the quotient group is abelian, so we may assume that $w = a^ib^j$ for some $i,j$. Since $a^2,b^3\in K$, we may assume that $0 \leq i \leq 1$ and $0 \leq j \leq 2$. There are then six possibilities for $w$. Of these six, only the identity is actually in $p_*\pi_1(X)$, which means that $[w]$ is the identity in $p_*\pi_1(X)/K$. It follows that $p_*\pi_1(X)/K$ is trivial, so $K = p_*\pi_1(X)$.

enter image description here

  • $\begingroup$ Ah! I had just figured this out, but you beat me to it. $\endgroup$ Feb 22, 2022 at 17:24
  • $\begingroup$ Haha, I mean you could cheat and use Tikz-cd with some bullets for vertices and some "bend left" and "bend right" for the arrows. That wouldn't look perfect but it would be pretty quick using something like tikzcd.yichuanshen.de $\endgroup$ Feb 22, 2022 at 17:27
  • $\begingroup$ Indeed. I used Tikz for my graph above, so maybe I'll finish it out 'officially'. Thank you $\endgroup$ Feb 22, 2022 at 17:29
  • $\begingroup$ You’re welcome :) $\endgroup$ Feb 22, 2022 at 17:33
  • $\begingroup$ Thank you for making this rigour! $\endgroup$
    – Zeekless
    Feb 24, 2022 at 11:51

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