Covering space action

• Oct 12th 2009, 01:04 AM
harbottle
Covering space action
I am having problems with 1.3.24 in Hatcher's Algebraic Topology:

24. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H in G determines a composition of covering spaces X -> X/H -> G. Show:

a. Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H in G

The best I can do here is say that since we have a covering space action then pi_1(X/H1) = pi_1(X/H2); not sure how to proceed.

b. Two such covering spaces X/H1 and H/H2 of X/G are isomorphic iff H1 and H2 are conjugate subgroups of G.

c. The covering space X/H -> X/G is normal iff H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H.

(definition of covering space actionat (*) on page 72 in this pdf http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf)
• Oct 12th 2009, 04:03 PM
aliceinwonderland
Quote:

Originally Posted by harbottle
I am having problems with 1.3.24 in Hatcher's Algebraic Topology:

24. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H in G determines a composition of covering spaces X -> X/H -> G. Show:

a. Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H in G

The best I can do here is say that since we have a covering space action then pi_1(X/H1) = pi_1(X/H2); not sure how to proceed.

b. Two such covering spaces X/H1 and X/H2 of X/G are isomorphic iff H1 and H2 are conjugate subgroups of G.

c. The covering space X/H -> X/G is normal iff H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H.

(definition of covering space actionat (*) on page 72 in this pdf http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf)

For (a), Since X is a path-connected, locally path-connected space and the action of G is a covering space action, $\displaystyle \pi_1(X)$ is a subgroup of $\displaystyle \pi_1(X/G)$. By using a Galois's correspondence, there exists a covering space X/H of X/G such that $\displaystyle \pi_1(X/H)$ is a subgroup of $\displaystyle \pi_1(X/G)$. In a similar vein, X is the covering space of X/H.

For (b), an isomorphism between covering spaces $\displaystyle p_1:\tilde{X_1} \rightarrow X$ and $\displaystyle p_2:\tilde{X_2} \rightarrow X$ is a homeomorphism $\displaystyle f:\tilde{X_1} \rightarrow \tilde{X_2}$ such that $\displaystyle p_1 = p_2f$. This implies that for each $\displaystyle x \in X$, f preserves the covering space structure, taking $\displaystyle p_1^{-1}(x)$ to $\displaystyle p_2^{-1}(x)$.

If H_1 and H_2 are conjugate subgroups of G, then so are $\displaystyle p_{1*}\pi_1(X/H_1, x_1)$ and $\displaystyle p_{2*}\pi_1(X/H_2, x_2)$ in $\displaystyle \pi_1(X/G, x_0)$, where $\displaystyle x_1 \in p_1^{-1}(x_0)$ and $\displaystyle x_2 \in p_2^{-1}(x_0)$. The remaining step is to show that there exists a bijection between each point $\displaystyle x_1=p_1^{-1}(x_0)$ and $\displaystyle x_2=p_2^{-1}(x_0)$ for each point $\displaystyle x_0 \in X/G$.

For (c), if $\displaystyle X/H \rightarrow X/G$ is normal, then $\displaystyle L=p_*\pi_1(X/H, x_1)$ is a normal subgroup of $\displaystyle M=\pi_1(X/G, x_0)$. The group of deck transformation is N(L)/L (Hatcher p71, proposition 1.39b). Since L is a normal subgroup of $\displaystyle M=\pi_1(X/G, x_0)$, we have $\displaystyle N(L)/L=M/L$. It remains to show that $\displaystyle M/L=G/H$. This equation intuitively makes sense, but I can't think of the rigorous proof for now. Can you proceed from here?
• Oct 13th 2009, 01:25 AM
harbottle
Quote:

Originally Posted by aliceinwonderland
For (a), Since X is a path-connected, locally path-connected space and the action of G is a covering space action, $\displaystyle \pi_1(X)$ is a subgroup of $\displaystyle \pi_1(X/G)$. By using a Galois's correspondence, there exists a covering space X/H of X/G such that $\displaystyle \pi_1(X/H)$ is a subgroup of $\displaystyle \pi_1(X/G)$. In a similar vein, X is the covering space of X/H.

For (b), an isomorphism between covering spaces $\displaystyle p_1:\tilde{X_1} \rightarrow X$ and $\displaystyle p_2:\tilde{X_2} \rightarrow X$ is a homeomorphism $\displaystyle f:\tilde{X_1} \rightarrow \tilde{X_2}$ such that $\displaystyle p_1 = p_2f$. This implies that for each $\displaystyle x \in X$, f preserves the covering space structure, taking $\displaystyle p_1^{-1}(x)$ to $\displaystyle p_2^{-1}(x)$.

If H_1 and H_2 are conjugate subgroups of G, then so are $\displaystyle p_{1*}\pi_1(X/H_1, x_1)$ and $\displaystyle p_{2*}\pi_1(X/H_2, x_2)$ in $\displaystyle \pi_1(X/G, x_0)$, where $\displaystyle x_1 \in p_1^{-1}(x_0)$ and $\displaystyle x_2 \in p_2^{-1}(x_0)$. The remaining step is to show that there exists a bijection between each point $\displaystyle x_1=p_1^{-1}(x_0)$ and $\displaystyle x_2=p_2^{-1}(x_0)$ for each point $\displaystyle x_0 \in X/G$.

For (c), if $\displaystyle X/H \rightarrow X/G$ is normal, then $\displaystyle L=p_*\pi_1(X/H, x_1)$ is a normal subgroup of $\displaystyle M=\pi_1(X/G, x_0)$. The group of deck transformation is N(L)/L (Hatcher p71, proposition 1.39b). Since L is a normal subgroup of $\displaystyle M=\pi_1(X/G, x_0)$, we have $\displaystyle N(L)/L=M/L$. It remains to show that $\displaystyle M/L=G/H$. This equation intuitively makes sense, but I can't think of the rigorous proof for now. Can you proceed from here?

Thank you very much for this. Some of the concepts are still new to me so I need to take a little while to work out what everything means; I'll work through your solution and see what I can do.