# $\pi_1$ of 4-manifolds that “look like” disk bundles

Let $X$ be a smooth compact oriented 4-manifold with $\partial X=L(p,1)$, $H_2(X;\Bbb Z)=\Bbb Z$, $H_3(X; \Bbb Z)=0$ and the induced map $\pi_1(L(p,1)) \to X$ surjective. What are the possibilities for $\pi_1(X)$? In particular, are there examples where $\pi_1 \ne 0$?

• Nice question! Where does this come from? Anyway, here is one (maybe silly?) example where $\pi_1(X)\neq 0$: you can take the rational homology ball $W$ bounded by $L(4,1)$ and blow it up. I will try to think about more interesting examples. – Marco Golla Sep 29 '16 at 0:53
• Also, do you want $X$ to be smooth? – Marco Golla Sep 29 '16 at 0:57
• @MarcoGolla Yes, X should be smooth. I've seen constructions before of $W$, but haven't went through them carefully. Is it easy to see the map $\pi_1(L(4,1)) \to \pi_1(W)$ is really surjective (maybe its forced to be and I am forgetting something silly)? If you want details to where this comes up, I'd feel more comfortable emailing you (is that okay?). I'd rather not share preliminary results on a public forum. – PVAL Sep 29 '16 at 1:33
• Sure, feel free to drop me an email. As for the surjectivity, one can see it directly from the surgery diagram, or, less directly, from the long exact sequence for the pair $(W,\partial W)$ (and the fact that $H_1(\partial W) = \pi_1(\partial W)$). – Marco Golla Sep 29 '16 at 7:50
• There are of course lots of rational homology balls in this setting, and while the map on $H_1$ is surjective as Marco says, there's no reason that the map on $\pi_1$ should be surjective in general. For a silly example, take the $W$ referred to (with $\pi_1 = Z_2$) and connect sum in the interior with a homology 4-sphere. – Danny Ruberman Sep 29 '16 at 10:38

Let's take your $p$ to be prime. Then $X$ has to be simply connected, even without all of the hypotheses. Here is the argument.
From the map on $\pi_1(L) \to Z_p$ you get a map $L \to BZ_p$. This map is clearly $0$ in $H_3$, since it factors through the inclusion of $L$ into the $4$-manifold $X$. On the other hand, it's well-known that $L$ generates $H_3(BZ_p) \cong Z_p$. (A quick explanation: you can build $BZ_p$ by attaching a 4-cell by a map of degree $p$ (the universal covering $S^3 \to L$) and then higher cells.)
I think a similar proof works if $p=mn$ with $m$ and $n$ relatively prime, and $\pi_1(X) = Z_n$. You have to know a little more about the map $H_3(BZ_p) \to H_3(BZ_n)$ induced by a surjection. I'm pretty sure it's given by multiplication by $m^2$.
• Thanks for the answer. This helps a lot in the application I had in mind. I'm guessing if you replace $L$ by p-surgery on some knot (for $p$ prime), I'm guessing the results aren't nearly so nice, though I am still interested in exactly what about $X$ you can conclude. In that case you still have a map $L \to BZ_p$, but $L$ might not generate $H_3(BZ_p)$ in any reasonable way. Is this correct? Is there any reasonable criterion for $\pi_1(L)$ (with $H_1(L)=Z_p$) so that if $\pi_1(L) \to \pi_1(X)$ is surjective with $H_2(X)=\Bbb Z, H_3(X)=0$ then $X$ is simply connected? – PVAL Sep 29 '16 at 23:06
• For any 3-manifold M with $H_1(M) = Z_p$, then for some $q$, there's a degree-one map $M \to L(p,q)$ inducing the abelianization $\pi_1(M) \to Z_p$. (The linking form of $M$ determines what $q$ to use.) See for instance Hayat-Legrand, Wang, and Zieschang, Pac. J. Math. 176, No. 1, 1996. So in this case, the same argument works; it only depends on $H_1$, not on $\pi_1$. – Danny Ruberman Sep 29 '16 at 23:54
• So this implies (in the case that $p$ is prime), that $H_1(X)=0$ (since it implies the induced map on abelianizations is zero by your argument), but it seems that $\pi_1(X)$ could still be some interesting perfect group (or is there something I missed.). – PVAL Sep 30 '16 at 0:47
• Connect summing with a homology 4-sphere (which isn't a homotopy 4-sphere) automatically rules out the surjectivity of $\pi_1(L) \to \pi_1(X)$. – PVAL Sep 30 '16 at 12:59