4
$\begingroup$

The question as in the title:

Is there a simple example of a compact orientable smooth finite-dimensional manifold whose singular cohomology groups with integer coefficients are not isomorphic to those of some smooth homogeneous space? My heart tells me that the two holed torus is such an example but I left my algebraic topology skills in my other pair of pants.

(I would also probably satisfied if this were shown only for compact homogeneous spaces)

Similarly, is there a simple example of a nice manifold whose integral singular cohomology groups are not isomorphic to the singular cohomology of any double coset space of any finite dimensional Lie group? (I believe these double coset space do not have to be smooth manifolds right?)

I am of course assuming that such examples exist.

$\endgroup$
  • $\begingroup$ mathoverflow.net/questions/89345/… is related. $\endgroup$ – Ryan Budney Aug 21 '13 at 18:45
  • $\begingroup$ @RyanBudney: It is related, but I think I am asking a weaker question, unless all cohomology groups being isomorphic means homotopy equivalence for nice manifolds. $\endgroup$ – BebopButUnsteady Aug 21 '13 at 19:05
  • $\begingroup$ Also, Qiachou Yuan's comment to that question indicates that all the surfaces of genus 2 or more are double coset spaces of $SL_2(\mathbb{R})$ $\endgroup$ – BebopButUnsteady Aug 21 '13 at 19:10
  • $\begingroup$ the genus 2 surface is not a homogeneous space, though. Double cosets of Lie groups are sometimes called "geometric" manifolds. $\endgroup$ – Ryan Budney Aug 21 '13 at 19:11
  • $\begingroup$ @Bebopunsteady: I would guess you need more than the group structure, probably the ring structure of cohomology. $\endgroup$ – Ryan Budney Aug 21 '13 at 19:12
3
$\begingroup$

The most compelling answer to the first question is that compact homogenous spaces have non-negative Euler characteristic. The proof for compact $G$ is quite short and is outlined in the comments to this answer to the similar question on MO, while the answer provides a reference for general case. Euler characteristic can be calculated from dimensions of cohomology groups, so there could be no homogenous spaces with cohomology like that of two-holed torus.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is sufficient for my needs I think. Thank you! $\endgroup$ – BebopButUnsteady Aug 22 '13 at 16:57
1
$\begingroup$

The first part has an easy partial answer: the cohomology groups of the surfaces genus $\geq$ 2 are not isomorphic to any compact oriented homogenous space , since it would have to be two dimensional, and therefore a torus or sphere. These have different cohomology groups by direct computation.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.