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

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:

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:

• I may be completely wrong. But what if you just join the three outer vertices with one blue triangle? Feb 22, 2022 at 13:57
• 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. Feb 22, 2022 at 14:06
• 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. Feb 22, 2022 at 16:28
• 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. Feb 22, 2022 at 16:30

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 $$K = p_*\pi_1(X)$$.