# Finding all Covering Spaces of a given index.

I am reviewing for an upcoming topology qualifying exam, and I have a question regarding a specific type of question:

Find all connected two-sheeted covering spaces of $$S^1\vee \mathbb{R}P^2$$ up to equivalence.

The covering space that I was able to think of is picture below.

$$p$$ sends $$a_1$$ and $$a_2$$ to $$a$$ and $$p$$ is the anti-podal map on $$S^2$$ (the universal cover). I have that $$\pi_1(\tilde X,\tilde x_0)=\langle a_1,b_1a_2\overline{b_1}|\rangle$$ where $$\tilde x_0$$ is the point where $$a_1$$ intersects $$S^2$$, $$b_1$$ is the half of the equator pictured and $$x_0$$ is the point at which $$a$$ intersects $$\mathbb{R}P^2$$. Then $$p_*(\pi_1(\tilde X,\tilde x_0))=\langle a, \gamma a \gamma^{-1}|\gamma^2=1\rangle =\langle a, \gamma a \gamma|\gamma^2=1\rangle$$ where $$\gamma$$ is the generator of $$\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$$. Are there other subgroups of index 2 of $$\pi_1(S^1\vee \mathbb{R}P^2)=\langle a,\gamma|\gamma^2=1\rangle$$? I could try to argue that the only nontrivial covering space of $$\mathbb{R}P^2$$ is $$S^2$$, and thus the covering above is all, but this isn't very rigorous. In general, given a space, how might one go about finding all covering spaces (up to equivalence) of a certain index? I know that given a space $$X$$ (p-conn., locally p-conn., semi-locally simply connected), there is a bijection between the path-connected basepoint preserving covering spaces $$p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$$ (up to equivalence) and the subgroups of $$\pi_1(X,x_0)$$ where the correspondence is given by $$p_* \pi_1(\tilde X,\tilde x_0)$$. I also know that if $$p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$$ is an $$n$$-sheeted covering, then $$p_* \pi_1(\tilde X,\tilde x_0)$$ has index $$n$$ in $$\pi_1(X,x_0)$$. So, if I can find all subgroups of a given index, then I at least know how many covering spaces I need to find. However, finding all subgroups of a given index on a free group isn't always easy, and it still doesn't tell me how to actually construct these spaces. For example, given my covering space above, How can I prove that I have found all $$2$$ sheeted covering spaces without calculating all subgroups of a given index? In general, can I find all $$n$$ sheeted covering spaces without calculating the number of subgroups of index $$n$$ and just guessing as to the actual construction of the covering space ? Also, as an aside question, what would the universal cover of $$S^1\vee \mathbb{R}P^2$$ be?

• Are you familiar with the universal cover of $S^1\vee S^1$? The universal cover of $S^1\vee\mathbb RP^2$ should be a similarly complex network of line segments and $2$-spheres. You have one node for every word of the form $a^{\epsilon_1}b^{n_1}ab^{n_2}a\cdots ab^{n_k}a^{\epsilon_2}$ with $n_j\in\mathbb Z$ and $\epsilon_i\in\left\{0,1\right\}$. Each node is the meeting point for two line segments and a $2$-sphere, and each $2$-sphere has exactly two (antipodal) nodes. And none of these things ever connect up, so a drawing would be some kind of fractal graph with spheres in place of some edges. Jan 31 at 8:37
• Yes I am. Each sphere would have two points that are mapped to the base point of the wedge sum under the the projection. Each of these two points points on a sphere would have 2 line segments (opposite orientation) ending/beginning at it. We can see that the universal cover of $S^1\vee S^1$ is simply connected because it is contractible. Is there a similar way to see that this cover of $S^1\vee\mathbb{R} P^2$ is simply connected?
– MEG
Jan 31 at 23:29
• I mean, the universal cover is simply connected by definition. The local description applies to all covers of $S^1 \vee \mathbb RP^2$ so that doesn't help. In terms of what I wrote, the relevant part is that nothing ever connects up, i.e. there are no cycles. Of course this gadget is not contractible because of the spheres, but it's a similar idea. Feb 1 at 1:12

You've found an example which contains a 2-sheeted connected covering of $$\mathbb RP^2$$. You can stretch that idea to try to imagine other examples: maybe one which contains a 2-sheeted connected covering of $$S^1$$; maybe one which contains 2-sheeted covering spaces of each of $$\mathbb R P^2$$ and $$S^1$$.
Also, finding all index 2 subgroups of a finitely presented group $$G$$ is easier than you think: every index 2 subgroup of $$G$$ is normal and has quotient is isomorphic to $$\mathbb Z / 2 \mathbb Z$$. From this, with a bit more thought, you can deduce that the "kernel" operation induces a bijection between the set of surjective homomorphisms $$G \mapsto \mathbb Z / 2 \mathbb Z$$ and the set of index 2 subgroups of $$G$$. So, if you use Van Kampen's theorem to write down a presentation of the fundamental group of your $$S^1 \vee \mathbb R P^2$$ then you should be able to pretty easily write down a list of all possible surjective homomorphisms to $$\mathbb Z / 2 \mathbb Z$$.
• . We have that the subgroups of index two of $\pi_1(S^1\vee \mathbb{R}P^2)=\langle a,\gamma|\gamma^2=1\rangle$ are $\langle a^2,\gamma a,\gamma a^{-1}|\gamma^2=1\rangle$, $\langle a,\gamma a\gamma|\gamma^2=1\rangle$ and $\langle a^2,\gamma, a\gamma a|\gamma^2=1\rangle$. When constructing a homomorphism from the free group on $n$ generators to $\mathbb{Z}_2$, we can send each generator to either 0 or 1, with the only restriction being at least one generator to $\overline{1}$ so that our homomorphism is surjective. Thus we get $2^n-1$.
• For example, the number of 2 sheeted covering of $S^1\vee S^1\vee S^1$ will be $8-1=7$. Is reasoning correct? If we have relations on some of our generators (i.e. $a^2=1$), then is it possible that there could be fewer subgroups of index 2?
• Yes, this all sounds correct. In particular your list of index 2 subgroups of the $S^1 \vee \mathbb RP^2$ group looks good. Jan 31 at 19:43