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In GTM82, I read a model for the relative cohomology of (M,N) with N a submanifold of M.

And in the page: Relative De Rham cohomologies , I got to know that there is another model for relative cohomology by using differential forms of M, like those forms which restrict to 0 on N.

Taladris called this Godbillon model. And Johannes Ebert said this model is used by jost in his book "Riemannian geometry and geometric analysis". But I can't find this model although I have checked that book(the 6th edtion) twice.

I wanna know in which paper,book, or article I can find the definition of the second model.

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The question you link refers to the book "Elements de Topologie Algébrique" by C. Godbillon, have you checked there? –  Mark Grant Feb 25 '13 at 11:42
    
Thank you! Unfortunately, I can not speak french, and I can't find this book. –  Ryan Du Feb 25 '13 at 13:23

1 Answer 1

up vote 5 down vote accepted

Suppose that $C$ is a closed subset of $M$. Denote by $\newcommand{\eO}{\mathscr{O}}$ $\eO$ its complement.

The DeRham cohomology of $M$ is in fact the cohomology associated to a particular soft resolution of the constant sheaf $\newcommand{\ur}{\underline{\mathbb{R}}}$ $\ur$ on $M$.

To any sheaf $\newcommand{\eS}{\mathscr{S}}$ $\eS$ on $M$ we can associate a sheaf $\eS_{\eO}$, also on $M$ whose stalk $\eS_{\eO}(x)$ at $x\in M$ is $\eS(x)$ if $x\in \eO$, and $0$ if $x\in M\setminus \eO$. For any open subset $U\subset M$ the space $\Gamma(U, \eS_{\eO})$ of section of $\eS_{\eO}$ over $U$ consists of the section $s\in \Gamma(\eO\cap U,\eS)$ such that the support of $s$ is closed in $U$. The operation $\eS\mapsto \eS_{\eO}$ is an exact functor. Moreover if $\eS$ is soft, so is $\eS_{\eO}$. Thus the DeRham resolution

$$ 0\to\ur\to\Omega^0\to\\Omega^1\to\cdots, $$

$\Omega^k=$ the sheaf of smooth $k$-forms on $M$, produces a soft resolution of $\ur_{\eO}$

$$ 0\to\ur_{\eO}\to\Omega_{\eO}^0\to\\Omega_{\eO}^1\to\cdots . $$

Thus the cohomology of the sheaf $\ur_{\eO}$ is computed by the the cohomology of the complex

$$ \Gamma(M, \Omega_{\eO}^0)\stackrel{d}{\to}\Gamma(M, \Omega^1_{\eO})\stackrel{d}{\to}\cdots, \tag{1} $$

where, as explained above $\Gamma(M, \Omega^k_{\eO})$ consists of smooth $k$-forms on $\eO$ whose support is a closed subset in $M$. Equivalently $\Gamma(M, \Omega^k_{\eO})$ consists of forms on $M$ whose supports do not intersect $C$. If $M$ is compact, then $\Gamma(M, \Omega^k_{\eO})$ consists of form with compact support contained in $\eO$.

On the other hand, the cohomology of $\ur_{\eO}$ can be identified with the relative cohomology $H^\bullet(M,C;\mathbb{R})$. The complex (1) is the a DeRHam model for this cohomology. For more details I recommend the comprehensive book of Kashiwara and Schapira Sheaves on Manifolds or my notes which are less comprehensive, but may guide you through the literature. In particular, see Remark 2.17 of my notes.

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Thank you very much. I understand what you say. But I want to avoid the use of sheaf theory. I want know the direct construction of the de rham model of relative cohomology although the proof may need sheaf toeory. I want know the direct construction of Godbillon or the one jost use. –  Ryan Du Feb 26 '13 at 3:01
    
You can avoid using the word sheaf, but you cannot avoid using the technologies from sheaf theory. Even Bott and Tu in the book you quoted drop all pretenses and introduce pre-sheaves. In any case if you work with the DeRham complex of forms with supports disjoint from $N$ you get the relative cohomology $H^\bullet(M,N;\mathbb{R})$. It is easy to construct the long exact sequence of the pair $(M,N)$. However, proving that the DeRham cohomoloygy of a manifold is the same as the singular cohomology with real coefficients, is a more challenging proposition. –  Liviu Nicolaescu Feb 26 '13 at 10:54

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