The Unapologetic Mathematician

Mathematics for the interested outsider

Compactly Supported De Rham Cohomology

Since we’ve seen that all contractible spaces have trivial de Rham cohomology, we can’t use that tool to tell them apart. Instead, we introduce de Rham cohomology with compact support. This is just like the regular version, except we only use differential forms with compact support. The space of compactly supported k-forms on M is \Omega_c^k(M); closed and exact forms are denoted by Z_c^k(M) and B_c^k(M), respectively. And the cohomology groups themselves are H_c^k(M).

To see that these are useful, we’ll start slowly and compute H_c^n(\mathbb{R}^n). Obviously, if \omega is an n-form on \mathbb{R}^n its exterior derivative must vanish, so Z_c^n(\mathbb{R}^n)=\Omega_c^n(\mathbb{R}^n). If \omega\in B_c^n(\mathbb{R}^n), then we write \omega=d\eta for some compactly-supported n-1-form \eta. The support of both \omega and \eta is contained in some large n-dimensional parallelepiped R, so we can use Stokes’ theorem to write

\displaystyle\int\limits_{\mathbb{R}^n}\omega=\int\limits_Rd\eta=\int\limits_{\partial R}\eta=0

I say that the converse is also true: if \omega integrates to zero over all of \mathbb{R}^n — the integral is defined because \omega is compactly supported — then \omega=d\eta for some compactly-supported \eta. We’ll actually prove an equivalent statement; if U is a connected open subset of \mathbb{R}^n containing the support of \omega we pick some parallelepiped Q_0\subseteq U and an n-form \omega_0 supported in Q_0 with integral 1. If \omega is any compactly supported n-form with support in U and integral c, then \omega-c\omega_0=d\eta for some compactly-supported \eta. It should be clear that our assertion is a special case of this one.

To prove this, let Q_i\subseteq U be a sequence of parallelepipeds covering the support of \omega. Another partition of unity argument tells us that it suffices to prove this statement within each of the Q_i, so we can assume that \omega is supported within some parallelepiped Q. I say that we can connect Q to Q_0 by a sequence of N parallelepipeds contained in U, each of which overlaps the next. This follows because the set of points in U we can reach with such a sequence of parallelepipeds is open, as is the set of points we can’t; since U is connected, only one of these can be nonempty, and since we can surely reach any point in Q_0, the set of points we can’t reach must be empty.

So now for each i we can pick \nu_i supported in the intersection of the ith and i+1st parallelepipeds and with integral 1. The difference \nu_i-\nu_{i-1} is supported in the ith parallelepiped and has integral 0; since the parallelepiped is contractible, we can conclude that \nu_{i-1} and \nu_i differ by an exact form. Similarly, \omega_0-\nu_1 has integral 0, as does \omega-c\nu_N, so these also give us exact forms. And thus putting them all together we find that


is a finite linear combination of a bunch of exact n-forms, and so it’s exact as well.

The upshot is that the map sending an n-form \omega to its integral over \mathbb{R}^n is a linear surjection whose kernel is exactly B_c^n(\mathbb{R}^n). This means that H_c^n(\mathbb{R}^n)=Z_c^n(\mathbb{R}^n)/B_c^n(\mathbb{R}^n)\cong\mathbb{R}.

December 6, 2011 - Posted by | Differential Topology, Topology |


  1. […] time we saw that compactly-supported de Rham cohomology is nonvanishing in the top degree for . I say that this is true for any oriented, connected […]

    Pingback by Nonvanishing Compactly-Supported de Rham Cohomology « The Unapologetic Mathematician | December 8, 2011 | Reply

  2. […] will be a nice, simple definition. As we saw last time, the top-degree compactly-supported de Rham cohomology is easy to calculate for a connected, oriented manifold . Specifically, we know that […]

    Pingback by The Degree of a Map « The Unapologetic Mathematician | December 9, 2011 | Reply

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