3
$\begingroup$

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.

enter image description here

$p$ 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?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 31 at 8:37
  • $\begingroup$ 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? $\endgroup$
    – MEG
    Jan 31 at 23:29
  • $\begingroup$ 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. $\endgroup$ Feb 1 at 1:12
5
$\begingroup$

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 $\mathbb Z / 2 \mathbb Z$.

$\endgroup$
3
  • $\begingroup$ . 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 $2^n-1$. $\endgroup$
    – MEG
    Jan 31 at 18:49
  • $\begingroup$ 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? $\endgroup$
    – MEG
    Jan 31 at 18:49
  • $\begingroup$ Yes, this all sounds correct. In particular your list of index 2 subgroups of the $S^1 \vee \mathbb RP^2$ group looks good. $\endgroup$
    – Lee Mosher
    Jan 31 at 19:43

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.