1
$\begingroup$

I read that both the zeroth homology and cohomology both count the connected components of a finite simplicial complex. (I am restricting to finite simplicial complexes so that connected=path connected and direct sum=direct product.)

However, by Poincare duality, $H^k\cong H_{n-k}$.

Hence, does that mean that actually $H^n$ and $H_n$ also count the connected components of a simplicial complex? (Since $H^0\cong H_{n}$ and $H^n\cong H_{0}$ by Poincare duality).

However that doesn't seem right? I recall that $H_n$ counts the number of $n$-dimensional "holes", which can be different from the number of connected components.

Thanks for any enlightenment.

Update: I realize Poincare duality only works for manifold without boundary. Is this the reason?

$\endgroup$
  • $\begingroup$ A form of Poincare duality does indeed hold for manifolds with boundary. A triangulation of a given manifold actually gives rise to a dual cell structure. In this way counting the 0-dimensional holes is the same as counting the n-dimensional holes of the dual structure. I'd suggest the wiki article en.wikipedia.org/wiki/Poincar%C3%A9_duality for some background. $\endgroup$ – Tyrone Dec 7 '17 at 14:49
2
$\begingroup$

A few preliminary notes:

  • To simplify things I will assume the coefficient ring $R = \mathbb Z.$ The same arguments also go through for general $R,$ provided we restrict to $\mathbb Z$-orientable surfaces.

  • The usual formulation of Poincaré duality holds for compact, oriented manifolds without boundary (to see where this can fail, consider the unit ball in $\mathbb R^n$). There is a relative version for manifolds with boundary, but I won't go into that here.

Recall that if $M$ is a (closed oriented) manifold, we can decompose it into its path components as $M = \bigsqcup_{\alpha \in A} M_{\alpha}.$ Then we get an isomorphism, $$ H_k(M) \cong \bigoplus_{\alpha \in A} H_k(M_{\alpha}) $$ and similarly for cohomology. The important point is that if $M$ is orientable, then so are all of its path components. So assuming $M$ has dimension $n$ we get,

$$ H_n(M_{\alpha}) \cong \mathbb Z. $$

From this you see that for each path component, each path component gives rise to a (co)homology class in grading $n.$


Now what you're probably wondering is why the last claim I stated holds. To simplify things I will further assume $M$ is connected and explain why $H_n(M) \cong \mathbb Z.$

Formally we can prove this assuming $M$ is triangulable ($M$ is homeomorphic to a simplicial complex, where all the top dimensional simplicies have dimension $n$), which is always true albeit annoying to prove. One can check that if you sum the $n$-simplicies arising from the triangulation (with appropriate orientations), you get a cycle and hence a non-trivial class in the top homology. The idea here is we can partition $M$ into these $n$-dimensional solid triangles, which fit together nicely in the sense that each $(n-1)$-simplex is a face of two $n$-simplicies. This already gives you some idea of the non-triviality of the top homology,since these simplicies somewhat loop back together.

The idea of $n$-dimensional holes does somewhat work to give some intuition also, where we restrict to the case where $M$ is smooth and embedded in $\mathbb R^{n+1}.$ I quite like this explanation which basically considers inserting an $(n+1)$-dimensional solid object inside the space; in this simplified setting it will be an open subset of $\mathbb R^{n+1}.$ Of course I don't think this will hold in general, but by considering the compact orientable surfaces (as illustrated in the link above) you can convince yourself that it works for $n=2.$ It's instructive to think about how this relates to the more formal viewpoint of adding simplicies.

The case of cohomology is a very different matter, since it doesn't measure $n$-dimensional holes. In that case I think the best way to get intuition is via the de Rham cohomology for smooth manifolds.

$\endgroup$

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.