as far as I know, there are two main ways to have a relative version of De Rham Cohomology for a pair (M,N), where M and N are smooth manifolds and N is a closed (as a topological subspace) submanifold of M:

1) Godbillon, Elements de topologie algébrique: $\Omega^p(M,N)$ is the space of all forms on $M$ whose restriction to $N$ is zero. This is a subalgebra of $\Omega^p(M)$ so it defines a cohomological space $H^p(M,N)$.

2) Bott-Tu, Differential forms in algebraic topology: this times, $\Omega^p(M,N)=\Omega^p(M)\oplus \Omega^{p-1}(N)$ with differential $d(\omega,\theta)=(d\omega,i^*(\omega)-d\theta)$, where $i:N\to M$ is the inclusion.

Does these two cohomologies give the same results? Otherwise, are they related and how are they related?

Bott and Tu's paragraph on the relative De Rham cohomology is very short. Does someone know a good reference on this subject?

Thank you in advance.

  • $\begingroup$ the otimes should be an oplus... $\endgroup$
    – Xandi Tuni
    Oct 30, 2012 at 8:52
  • 1
    $\begingroup$ Yes, thanks Johannes for editing. And for your answer. I just seen it right now (because of Sandy), so I need time to check it and ask details. $\endgroup$
    – Taladris
    Nov 1, 2012 at 0:17

4 Answers 4


A chain map $\Theta$ from the Godbillon theory to the Bott-Tu version is given by $\omega \mapsto (\omega,0)$ (note that is a chain map only on $\Omega^{p} (M;N)_{G}$). I claim that this induces an isomorphism on cohomology. A couple of special cases is obvious: if $N=\emptyset$, then both theories agree with absolute de Rham theory. If $N \to M$ is a homotopy equivalence, both theories are trivial by long exact sequences and homotopy invariance of the absolute theory.

For the general case, pick a tubular neighborhood $U$ of $N$. You get short exact sequences of chain complexes (in both cases)

$$ 0\to \Omega (M;N) \to \Omega(U;N) \oplus \Omega (M-N) \to \Omega (U-N) \to 0 $$

(exactness is checked by means of a partition of unity), and $\Theta$ compares the both short exact sequences. The associated (Mayer-Vietoris) exact sequence and the $5$-lemma concludes the proof.

  • 1
    $\begingroup$ What is $G$ here, and what does it mean for $\Theta$ to "compare" two short exact sequences? $\endgroup$ Oct 30, 2012 at 10:08
  • 1
    $\begingroup$ I assume $G$=Godbillon. That is $\Omega(M;N)$ with a $G$ subscript is the Godbillon definition and without a $G$ is the Bott-Tu definition. $\endgroup$ Oct 30, 2012 at 10:27
  • 1
    $\begingroup$ @Michael: absence of the G subscript means that the above exact sequence of chain complexes can be defined for the two kinds of relative cohomology (with different maps in the sequences of course). $\endgroup$
    – Taladris
    Nov 1, 2012 at 3:10
  • 3
    $\begingroup$ @Vel: "Compare" means that we have a commutative diagram connecting the exact sequences $ 0\to \Omega(M;N)_{G} \to \Omega(U;N)_{G}\oplus\Omega(M\setminus N)\to \Omega(U\setminus N)\to 0$ and $ 0\to \Omega(M;N)_{BT} \to \Omega(U;N)_{BT}\oplus\Omega(M\setminus N)\to \Omega(U\setminus N)\to 0$ (where the $G$ and $BT$ respectively mean Godbillon and Bott-Tu) and the vertical maps are $\Theta$, $\Theta\times id$ and $id$. These maps induce isomorphisms in cohomology (as Johannes claimed). @Johannes: thank you! I considered the chain map $\Theta$ but unsuccessfully. $\endgroup$
    – Taladris
    Nov 1, 2012 at 3:12

The fact that the DeRham cohomology of $M$ computes the singular cohomology is a reflection of the fact that the cohomology of a sheaf (in this case the constant sheaf $\newcommand{\bR}{\mathbb{R}}$ $\underline{\bR}$ over $M$) can be computed by using any (soft, flabby) resolution of this sheaf. The geometry appears when we produce a concrete such resolution, in your case the DeRham resolution.

Thus, an appropriate question would be if the relative cohomology could be given a sheaffy description.

The answer is yes. If $\newcommand{\eS}{\mathscr{S}}$ $\eS$ is a sheaf of Abelian groups on $M$ and $Z\subset M$ is a closed subset, then we denote by $\eS_Z$ the ``restriction'' of $\eS$ to $Z$ extended by $0$ outside. I want to emphasize that $\eS_Z$ is a sheaf on $M$: its stalk at $x\in M$ is $\eS_x$ if $x\in Z$, and $0$ otherwise. We have a natural surjective morphism

$$ r:\eS\to\eS_Z $$

We denote its kernel by $\eS_{M\setminus Z}$.We thus get a short exact sequence of sheaves

$$ 0\to\eS_{M\setminus Z} \to \eS\to\eS_Z\to 0. $$

This leads to an exact sequence in cohomology

$$ \cdots \to H^\bullet(M,\eS_{M\setminus Z})\to H^\bullet(M,\eS)\to H^\bullet(M,\eS_Z) \stackrel{+1}{\to}\cdots $$

When $\eS=\underline{\bR}$ is the constant sheaf on $M$ with stalk $\bR$ then $H^\bullet(M,\underline{\bR})$ is the singular cohomology of $M$, $H^\bullet(M,\underline{\bR}_Z)$ is the cohomology of $Z$ and $H^\bullet(M,\underline{\bR}_{M\setminus Z})$ is the (relative) cohomology of the pair $(M,Z)$.

You can get the two descriptions by from this fact alone (the one in Bott-Tu requires additionally the cone of construction). I refer to Iversen's book Cohomology of sheaves.

  • 1
    $\begingroup$ For the benefit of anyone else learning about relative de Rham cohomology, Liviu explains this in another answer here and gives a link to his notes. $\endgroup$
    – ಠ_ಠ
    Mar 31, 2016 at 8:30

The morphism from Bott-Tu cohomolgy to Godbillon cohomology can be made explicit: $$\Psi:H^*(M,N)_{BT}\to H^*(M,N)_G$$

$$(\omega,\theta)\mapsto \omega-d(\pi^*\theta\wedge \eta)\tag{1}\label{1}$$

Here we take a tubular neighborhood $T$ of $N$ in $M$, and let $\pi:T\to N$ be the projection. $\eta$ is a bump function defined in the following way: Take a sequence of tubular neighborhoods $N\subseteq T’’\subseteq T’\subseteq T$ of $N$, such that the closure of the previous one is contained in the next. Then $\eta$ is a smooth function which is constant $1$ in $T’’$ and is zero on $M\setminus T’$.

First, $\Psi$ is well-defined: If $(\omega,\theta)$ is a closed form, then $i^*\omega=d\theta$, so the expression $(\ref{1})$ vanishes on $N$ as a result of $\eta$ being constant $1$ near $N$. Also, one can easily verify that $\Psi$ sends closed forms to closed forms. When $(\omega,\theta)=d(\omega',\theta')$ is exact, we have $$\Psi d(\omega',\theta')=\Psi(d\omega',i^*\omega'-d\theta')=d\omega'-d(\pi^*(i^*\omega'-d\theta')\wedge \eta).$$ One should note that $\omega'-\pi^*i^*\omega'\wedge\eta$ vanishes on $N$, and $d\theta'\wedge\eta=d(\theta'\wedge\eta)\pm\theta'\wedge d\eta$. Since $d\eta$ vanishes on $N$ and $dd(\theta'\wedge\eta)=0$, we conclude $\Psi d(\omega',\theta')=d\lambda$, where $\lambda$ is a differential form on $X$ that vanishes on $N$, so $\Psi$ send exact forms to exact forms. (However, one should note that $\Psi$ is not well-defined on the level of non-closed forms in general.)

Second, one note that $\Psi$ is independent of choice of tubular neighborhoods and the bump function $\eta$, since any of such two will lead to an expression in $(\ref{1})$ differing by an exact form.

Finally, let’s show $\Psi$ is inverse to the map $$\Theta: H^*(M,N)_{G}\to H^*(M,N)_{BT}$$ $$\alpha\mapsto (\alpha,0).$$

Obviously, $\Psi\circ\Theta=Id$. On the other side,

$$\Theta\circ\Psi(\omega,\theta)=\big (\omega-d(\pi^*\theta\wedge \eta),0\big ),$$

which differ to $(\omega,\theta)$ by an exact form $d(\pi^*\theta\wedge \eta,0)$, so $\Theta\circ\Psi=Id$ on the level of cohomology.

Therefore, the two relative cohomolgy theories are isomorphic.


AG learner's is almost the answer I wanted, but really bugged me that that $\Psi$ wasn't a chain map that was homotopic to the identity. I think I figured out how to make that happen. Let $T$ be an open tubular neighborhood of $N$, and $r\colon R\to N$ a smooth deformation retract. Let $f$ be a smooth bump function whose support is in $T$ and which is 1 on $N$. We can find the formula for $\Psi$ by considering the homotopy $h(\alpha, \beta)=(fr^*\beta,0)$. We have that $$(dh+hd)(\alpha,\beta)=(d(fr^*\beta)+fr^*(\alpha|_N-d\beta),\beta)$$ so $$(1-dh-hd)(\alpha,\beta)=(\alpha-fr^*(\alpha|_N)-df\wedge r^*\beta,0).$$ This coincides with AG learner's formula on closed forms, and is a chain map.


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.