The Unapologetic Mathematician

Mathematics for the interested outsider

Compact Oriented Manifolds without Boundary have Nontrivial Homology

If we take M to be a manifold equipped with an orientation given by an orientation form \omega. Then \omega is nowhere zero, and \omega(v_1,\dots,v_n)>0 for any positively oriented basis \{v_i\} of \mathcal{T}_pM at any point p\in M.

Next we take c:[0,1]^n\to M to be an orientation-preserving embedding — a singular cube of top dimension. Then the pullback c^*\omega=fdu^1\wedge\dots\wedge du^n for some strictly-positive function f. We conclude that

\displaystyle\int\limits_c\omega=\int\limits_{[0,1]^n}fdu^1\wedge\dots\wedge du^n=\int\limits_{[0,1]^n}f(u^1,\dots,u^n)\,d(u^1,\dots,u^n)>0

The integral of \omega over all of M must surely be even greater than the integral over the image of c, since we can cover M by orientation-preserving singular n-cubes, and none of them can ever contribute a negative to the integral.

If we further suppose that M is compact, we can cover M by finitely many such singular cubes, and the integral on each is well-defined. Using a partition of unity as usual this shows us that the integral over all of M exists and, further, must be strictly positive. In particular it’s not zero.

But now suppose that M also has an empty boundary. Since \omega is a top form, we know that d\omega=0 — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if \omega=d\eta for some n-1-form \eta then Stokes’ theorem would tell us that

\displaystyle\int\limits_M\omega=\int\limits_Md\eta=\int\limits_{\partial M}\eta=0

since \partial M is empty.

And so if M is a compact, oriented n-manifold without boundary, then there must be some n-forms which do not arise from taking the exterior derivatives of n-1-forms. If M is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on M which are not the divergence of any vector field on M.

About these ads

November 24, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. [...] — we see that any compact, oriented manifold without boundary cannot be contractible, since we know that they have some nontrivial homology! LD_AddCustomAttr("AdOpt", "1"); [...]

    Pingback by Homotopic Maps Induce Identical Maps On Homology « The Unapologetic Mathematician | December 6, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 386 other followers

%d bloggers like this: