claudio

I have a compact manifold M without boundary. This contains a proper
submanifold N which is not contractible (i.e. not homotopy equivalent
to a point). Can I conclude that M is not contractible as well?

david l. johnson

You can conclude that even without the submanifold. For any compact
manifold w/o boundary, the top-degree homology will not vanish (over Z for
orientable manifolds, and over Z/2Z for non-orientable ones).

I*presume you are asking about finite-dimensional manifolds.

--

David L. Johnson

__o | When you are up to your ass in alligators, it's hard to remember
_`\(,_ | that your initial objective was to drain the swamp. -- LBJ
(_)/ (_) |

anton deitmar

Well, it depends.
Is the empty set a manifold?
If so, it is not homotopy equivalent to a point, so it is not contractible.
Then choose M to be a point and N to be the empty set.

If you insist that N is non-empty, then M bust consist of more than one point, so either M is not connected, hence not contractible, or it is of positive dimension, in which case we can assume that M is connected. There are two possibilities:

- M is orientable: then, by Poincare-duality, it has non-trivial homology, hence is not contractible.

- M is not orientable: then it has a double covering which is orientable, hence it has a non-trivial fundamental group, hence is not contractible.

Cheers,
Anton

claudio

The question was not formulated competely. Here is the real question
I am interested: If I puncture my larger compact manifold without
boundary M (thus it gets a boundary) but I know it still contains a
submanifold N (compact, without boundary) not "touched" by the
puncturing of M and hence noncontractible, can I still conclude that
the punctured M is noncotractible or can I conclude nothing?

gavin wraith

In message <efj8tb$833$1@news.ks.uiuc.edu>


No. M a 2-sphere, N its equator.

--
Gavin Wraith (gavin@wra1th.plus.com)
Home page: http://www.wra1th.plus.com/

lrudolph

claudio <altafini@sissa.it> writes:


You can't possibly mean what you're saying, can you? If you
puncture the n-sphere M, then you get the magnificently contractible
n-space R^n, but of course R^n (for n > 1) contains loads and loads
of compact, boundaryless, non-contractible submanifolds N!

Lee Rudolph