# The Unapologetic Mathematician

## 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 $M$.