Central Z extensions

Here's one of my favorite basic algebra/topology things.  You can read about this in Brown's Cohomology of Groups book.  

A short exact sequence of groups $$0 \to \mathbb{Z} \to H \to G \to 1$$ is called a central $\mathbb{Z}$-extension of $G$ if the $\mathbb{Z}$ subgroup of $H$ sits in the center of $H$.  For example, $$0 \to \mathbb{Z} \to G \oplus \mathbb{Z} \to G \to 1$$ is the trivial central $\mathbb{Z}$-extension of $G$ and $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0$$ is a non-trivial extension, where the first map is multiplication by 2.  

Are there other extensions of $\mathbb{Z}/2$?  How do you classify central $\mathbb{Z}$-extensions?  Given an extension and a homomorphism $f: G' \to G$, can we lift $f$ to $H$?  These seem hard to answer, but actually they are easy!!!  And we can topologize this problem.  

The one thing we need is group cohomology.  Lightning crash course for this post: if $G$ is a group, then $H^*(G)$ is just defined to be $H^*(K(G,1))$.  [This is group cohomology with coefficients in $\mathbb{Z}$.  There's much more interesting group cohomology coefficients, but we won't touch on that here.]  

For example, $H^*(\mathbb{Z}) \cong H^*(S^1) \cong \mathbb{Z}_{(0)} \oplus \mathbb{Z}_{(1)}$ and $H^*(\mathbb{Z}/2) \cong H^*(\mathbb{R}P^\infty)$.  Easy peasy.  (Do people still say that?)  A homomorphism $f: G \to K$ induces a homomorphism $f^*: H^*(K) \to H^*(G)$ which you can think of like map coming from singular cohomology.  Since $K(G*H,1) \simeq K(G,1) \vee K(H,1)$ and $K(G \times H,1) \simeq K(G,1) \times K(H,1)$, there's a lot of things we can compute quickly.  (In fact, you can deduce a Mayer-Vietoris type principle type thing for group cohomology using the Mayer-Vietoris sequence for cohomology, so it's all pretty manageable.)

Now, the awesome fact is that central $\mathbb{Z}$-extensions of a group $G$ (up to a suitable notion of equivalence) are in bijective correspondence with elements in $H^2(G)$.  Want some topological intuition?  Given a principal $S^1$-bundle $E$ over $X$, we get a fibration $S^1 \to E \to X$ and the induced map on $\pi_1$ will actually give a central $\mathbb{Z}$-extension of $\pi_1(X)$.  (Principal is what is yielding central here.)  But topologically, we know that principal $S^1$-bundles correspond to $H^2(X)$.  This tells us that elements of $H^2$ give us central $\mathbb{Z}$-extensions.  Conversely, given a central $\mathbb{Z}$-extension, using the Eilenberg-MacLane spaces, we can find $S^1 \to E \to X$ where $E = K(H,1)$ and $X = K(G,1)$ such that the induced maps on $\pi_1$ give the central extension.  This can be made into a circle bundle, and hence an element of $H^2(G)$.  

So, let's characterize the central $\mathbb{Z}$-extensions of $\mathbb{Z}/2$.  Well, we just want to compute $H^2(\mathbb{R}P^\infty) = \mathbb{Z}/2$.  Therefore, there are exactly two extensions: the trivial one and the non-trivial one which we've already found.  So we've classified them!  Very cool!

What if you have a homomorphism $f: G' \to G$?  When can you lift this to $H$?  Well, let $\rho \in H^2(G)$ correspond to the extension over $G$.  Then, $f$ lifts if and only if $f^*\rho = 0 \in H^2(G')$.  The way to think of this is that $f^*\eta$ is the pullback extension for any $\eta \in H^2(G')$.  (In other words, think of the extension of $G$ as an $S^1$-bundle over $K(G,1)$, and then $f^*\eta$ corresponds to an $S^1$-bundle to $K(G',1)$, and hence a $\mathbb{Z}$-extension.)  Why can we lift if $f^*\rho = 0$?  Well, using the pullback extension, $H'$, we get a map $\tilde{f}$ from $H' \to H$ satisfying $\pi \tilde{f} = f \pi'$, where $\pi : H \to G$ and $\pi': H' \to G'$ are the projections from the short exact sequences.  [Think about the analogous picture for pullback bundles.  There's a map between bundles that covers the map on the base.]  But since the extension is trivial, we have a section, i.e. $H' = G' \oplus \mathbb{Z}$ and there's an obvious map from $\sigma : G' \to H'$ given by $\sigma(g') = (g',0)$.  Hence, $\widetilde{f}\sigma$ is the lift!  

Where have you seen central $\mathbb{Z}$-extensions in low-dimensional topology? 

1) Seifert fibered spaces.  Their fundamental groups are central $\mathbb{Z}$-extensions of the "orbifold" fundamental group of the base orbifold.  For example, the Brieskorn spheres $\Sigma(p,q,r)$ have fundamental group a central $\mathbb{Z}$-extensions of the triangle group $\langle x, y, z \mid x^p = y^q = z^r = xyz = 1 \rangle$.  (If you change the extension, but fix the triangle group, then you will get some Seifert fibered spaces with singular fibers of order $p,q,r$ but with different $H_1$.)  

2) The Lie group $SL_2(\mathbb{R})$ has universal cover $\widetilde{SL_2(\mathbb{R})}$ as a central $\mathbb{Z}$-extension.  Consequently, if you have a homology three-sphere $Y$ which is a $K(\pi,1)$ and a homomorphism $\pi_1(Y)$ to $S_2(\mathbb{R})$, we can always lift it since $H^2(Y) = 0$.  [Other than $S^3$ and $\Sigma(2,3,5)$, any prime homology three-sphere is a $K(\pi_1,1)$, so $H^2(Y) = H^2(\pi_1(Y))$.]  Since $\widetilde{SL_2(\mathbb{R})}$ is a left-orderable group, it follows that a non-trivial homomorphism from $\pi_1(Y)$ to $SL_2(\mathbb{R})$ implies that $\pi_1(Y)$ is left-orderable.  [Boyer-Gordon-Watson proved that if the fundamental group of a prime three-manifold admits a non-trivial homomorphism to a left-orderable group, then the three-manifold has left-orderable fundamental group.]