Timeline for $\pi_1$ of 4-manifolds that "look like" disk bundles
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2016 at 13:42 | comment | added | Danny Ruberman | I'm not clear on exactly what you're asking at this point. If you want to continue by email, I'm sure you can find me somewhere out there. | |
| Sep 30, 2016 at 12:59 | comment | added | PVAL | Connect summing with a homology 4-sphere (which isn't a homotopy 4-sphere) automatically rules out the surjectivity of $\pi_1(L) \to \pi_1(X)$. | |
| Sep 30, 2016 at 12:39 | comment | added | Danny Ruberman | Yes, as I pointed out in the comments on the original question; you could always take something simply connected and connected and connect sum with a homology 4-sphere (which are plentiful; every homology 3-sphere gives rise to one (and sometimes two) by a spinning construction. | |
| Sep 30, 2016 at 0:47 | comment | added | PVAL | 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.). | |
| Sep 29, 2016 at 23:54 | comment | added | Danny Ruberman | 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 $\pi_1$. | |
| Sep 29, 2016 at 23:07 | vote | accept | PVAL | ||
| Sep 29, 2016 at 23:06 | comment | added | PVAL | 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? | |
| Sep 29, 2016 at 1:36 | history | answered | Danny Ruberman | CC BY-SA 3.0 |