Take, for example, the Klein bottle K. Its de Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1. It is in general true that de Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since de Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...


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    $\begingroup$ I very much like this question! $\endgroup$ – B. Bischof Apr 25 '10 at 3:27

You can compute the integer (co)homology groups of a compact manifold from a Morse function $f$ together with a generic Riemannian metric $g$; the metric enters through the (downward) gradient flow equation $$ \frac{d}{dt}x(t)+ \mathrm{grad}_g(f) (x(t)) = 0 $$ for paths $x(t)$ in the manifold.

After choosing further Morse functions and metrics, in a generic way, you can recover the ring structure, Massey products, cohomology operations, Reidemeister torsion, functoriality.

The best-known way to compute the cohomology from a Morse function is to form the Morse cochain complex, generated by the critical points (see e.g. Hutchings's Lecture notes on Morse homology). Poincaré duality is manifest.

Another way, due to Harvey and Lawson, is to observe that the de Rham complex $\Omega^{\ast}(M)$ sits inside the complex of currents $D^\ast(M)$, i.e., distribution-valued forms. The closure $\bar{S}_c$ of the the stable manifold $S_c$ of a critical point $c$ of $f$ defines a Dirac-delta current $[\bar{S}_c]$. As $c$ varies, these span a $\mathbb{Z}$-subcomplex $S_f^\ast$ of $D^*(M)$ whose cohomology is naturally the singular cohomology of $M$.

The second approach could be seen as a "de Rham theorem over the integers", because over the reals, the inclusions of $S_f\otimes_{\mathbb{Z}} \mathbb{R}$ and $\Omega^{\ast}_M$ into $D^\ast(M)$ are quasi-isomorphisms, and the resulting isomorphism of $H^{\ast}_{dR}(M)$ with $H^\ast(S_f\otimes_{\mathbb{Z}}\mathbb{R})=H^\ast_{sing}(X;\mathbb{R})$ is the de Rham isomorphism.

  • $\begingroup$ Thanks a ton! Your second approach was exactly the sort of thing I was looking for, though it is a very useful observation that Morse theory also does the trick. $\endgroup$ – Paul Siegel Apr 26 '10 at 15:16

The Cheeger--Muller theorem (and related results) allow one to get some control over the torsion in the homology using an analysis of the spectrum of the Laplacian. For one application, see the recent paper of Bergeron and Venkatesh (which also contains a reference to the work of Cheeger and Muller).


In your example, the orientation cover is the 2-d torus T. It is a double cover of the Klein bottle with H1(T)=R2, so the answer for this example is definitely yes. However, I am not sure if this example is so good, as the torsion here comes form the orientation cover being connected.


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