18
$\begingroup$

On pp. 166 of Scorpan's "The Wild World of 4--manifolds", he gives an example of a parallelizable 4-manifold ($S^1\times S^3$) and then asserts: "there are no simply connected examples". It's confusing to me whether he means that there are no such examples in general, or examples that are 4-manifolds. In any case, here is my question:

Do there exist (compact, smooth, oriented) simply connected manifolds that are parallelizable? 4-manifolds?

(Recall that a manifold is called parallelizable if it has a trivial tangent bundle.)

$\endgroup$
2
  • 3
    $\begingroup$ At the very least "compact" should be added, otherwise just take $\mathbb{R}^n$. In fact, any contractible manifold will be both simply connected and parallelizable. $\endgroup$
    – Matt
    Jun 13, 2013 at 18:34
  • $\begingroup$ @Matt you're absolutely right. I've edited the question. $\endgroup$
    – Aru Ray
    Jun 13, 2013 at 18:59

2 Answers 2

21
$\begingroup$

A simply connected smooth compact 4-manifold has its homology concentrated in even degrees, by Poincare duality. Thus its Euler characteristic is positive. By the Poincare-Hopf index theorem, such a manifold can have no nowhere vanishing vector field, and so is certainly not parallelizable.

$\endgroup$
5
$\begingroup$

At least $S^3$ is a lie group so parallelizable.

$\endgroup$
2
  • 2
    $\begingroup$ More generally, the orthogonal groups have simply connected compact universal covering spaces. $\endgroup$ Jun 14, 2013 at 5:51
  • 2
    $\begingroup$ The question is specifically about 4-manifolds. We could say now that, as a corollary, that there are no 4-dimensional simply connected Lie groups. $\endgroup$
    – Turion
    May 5, 2016 at 18:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .