Free, open-source online mathematics for students, teachers and workers

Let $M$ be a manifold. Then

$$H^\ast(M\times \mathbb{R} )\simeq H^\ast(M)$$ $$H^\ast_c(M\times \mathbb{R} )\simeq H_c^{\ast-1}(M)$$

The case with general cohomology comes directly from the fact that $M$ and $M\times\mathbb{R}$ have the same homotopy type. We may take as homotopic inverses the las projection $p:(x,t)\longmapsto x$ and section $s:x\longmapsto (x,0)$ maps; in this setting

$$s^\ast:(\beta+\mathrm{d} t\wedge\alpha)\longmapsto \beta(0)$$ $$p^\ast:\beta\longmapsto \beta+\mathrm{d} t\wedge 0$$

($\beta$ independient of $t$) provide an isomorphism in cohomology

The compact support scenario is completely different; there is no functoriality and the lemma is indeed different: $H^\ast_c(M\times \mathbb{R} )\simeq H_c^{\ast-1}(M)$ means $H^k_c(M\times \mathbb{R} )\simeq H_c^{k-1}(M)$ for all $k$). We'll have to come up with a different, direct approach

Forms in $M\times \mathbb{R} $ are still of the type $\beta+\mathrm{d} t\wedge\alpha$, but now $\beta$ and $\alpha$ have compact support in $M\times \mathbb{R}$. Let's consider the following homomorphism:

$$\pi:\Omega^k_c(M\times \mathbb{R} )\longrightarrow \Omega_c^{k-1}(M)$$ $$\pi(\beta+\mathrm{d} t\wedge\alpha)=\int_{\mathbb{R} }\alpha$$

The integral in well-defined because $\alpha$ has compact support along the fiber $x\times\mathbb{R}$, and since the support of $\int_{\mathbb{R} }\alpha$ is (contained in) the projection of the support of $\alpha$, the image in indeed in $\Omega_c^{k-1}(M)$. We assert that it induces another map $\pi:H^k_c(M\times \mathbb{R} )\longrightarrow H_c^{k-1}(M)$ that is isomorfism

First of all, $\pi$ is well-defined for cohomology:

$$\mathrm{d} \pi(\beta+\mathrm{d} t\wedge\alpha)=\mathrm{d}\left(\int_{\mathbb{R} }\alpha\right)=\int_{\mathbb{R} }\mathrm{d}\alpha$$ $$\pi\bar{\mathrm{d}}(\beta+\mathrm{d} t\wedge\alpha)=\pi\left(\mathrm{d}\beta+\mathrm{d} t\wedge\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)\right)= \int_{\mathbb{R} }\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)=-\int_{\mathbb{R} }\mathrm{d}\alpha$$

because $\displaystyle\int_{\mathbb{R} }\dfrac{\partial \beta}{\partial t}=\int_{-R}^R\dfrac{\partial \beta}{\partial t}=\beta(R)-\beta(-R)=0$ for $R$ big enough ($\beta$ has compact support in each fiber). So $\mathrm{d} \pi=-\pi\bar{\mathrm{d}}$; it's not the equality $\mathrm{d} \pi=\pi\bar{\mathrm{d}}$ required for $\pi$ to be a complex homomorphism, but it is enough for $\pi$ to be well-defined in cohomology: closed forms map to closed forms, exact forms map to exact forms

On the one hand, we have another homomorfism

$$\sigma:\Omega_c^{k-1}(M)\longrightarrow \Omega_c^k(M\times\mathbb{R} )$$ $$\alpha\longmapsto \rho(t)\mathrm{d} t\wedge\alpha$$

where $\rho$ is some compact support function in $\mathbb{R} $ with $\displaystyle\int_\mathbb{R} \rho=1$. And it happens $\pi\circ\sigma=\text{id}_{\Omega_c^{k-1}(M)}$ and similarly $\pi\circ\sigma=\text{id}_{H_c^{k-1}}(M)$ in cohomology. In particular $\pi$ is surjective

To check that $\pi$ in injective (for cohomology), let's consider some closed form $\beta+\mathrm{d} t\wedge\alpha\longmapsto \displaystyle\int_{\mathbb{R} }\alpha$, and suppose that $\displaystyle\int_{\mathbb{R} }\alpha=\mathrm{d} \gamma$ is exact; is the starting form also exact? Let's study the composition of a hypothetical form whose exterior derivative is $\beta+\mathrm{d} t\wedge\alpha$

$$\bar{\mathrm{d}}(\delta+\mathrm{d} t\wedge \cdots)=\mathrm{d} \delta+\mathrm{d} t\wedge\cdots\Longrightarrow \beta=\mathrm{d} \delta$$

But we shouldn't forget that $\beta+\mathrm{d} t\wedge\alpha$ is closed:

$$\bar{\mathrm{d}}(\beta+\mathrm{d} t\wedge\alpha)=\left(\mathrm{d}\beta+\mathrm{d} t\wedge\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)\right)=0\Longrightarrow \begin{cases} \mathrm{d}\beta=0\\ \dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha=0 \end{cases}$$ $$\mathrm{d}\alpha=\dfrac{\partial \beta}{\partial t}=\dfrac{\partial \mathrm{d} \delta}{\partial t}=\mathrm{d} \dfrac{\partial \delta}{\partial t}$$

and it seems reasonable to think that $\dfrac{\partial \delta}{\partial t}=\alpha$, so we define $\delta=\int_{-\infty}^t\alpha$. For this choice,

$$\bar{\mathrm{d}}\delta=\mathrm{d}\delta+\mathrm{d} t\wedge\dfrac{\partial \delta}{\partial t}=\mathrm{d} \int_{-\infty}^t\alpha+\mathrm{d} t\wedge\dfrac{\partial }{\partial t}\int_{-\infty}^t\alpha=\int_{-\infty}^t\mathrm{d}\alpha+\mathrm{d} t\wedge\alpha=\int_{-\infty}^t\dfrac{\partial \beta}{\partial t}+\mathrm{d} t\wedge\alpha=\beta+\mathrm{d} t\wedge\alpha$$

exactly as required. But: this form has compact support as to the projection on $M$, but as to the $\mathbb{R}$ direction, $\delta$ may be nonzero beyond the support of $\alpha$:

$$\delta(R)=\int_{-\infty}^R\alpha=\mathrm{d} \gamma$$

for $R$ big enough. So we add a correction:

$$\delta-\rho(t)\mathrm{d} \gamma-\mathrm{d} t\wedge\dfrac{\partial \rho(t)}{\partial t}\gamma$$

where $\rho$ is a differentiable function equal to 0 in $(-\infty,-1]$ and equal to 1 in $[1,\infty)$. $\rho(t)\mathrm{d} \gamma$ corrects $\delta$ and $\dfrac{\partial \rho(t)}{\partial t}$ has compact support; moreover

$$\bar{\mathrm{d}}\left(\rho(t)\mathrm{d} \gamma+\mathrm{d} t\wedge\dfrac{\partial \rho(t)}{\partial t}\gamma\right)=\mathrm{d}(\rho(t)\mathrm{d} \gamma)+\mathrm{d} t\wedge\left(\dfrac{\partial \rho(t)}{\partial t}\mathrm{d}\gamma-\mathrm{d}\left(\dfrac{\partial \rho(t)}{\partial t}\gamma\right)\right)=0$$

This argument is not replicable in the non-compact case: $\displaystyle\int_{-\infty}^t\alpha$ may diverge, and if $\displaystyle\delta=\int_{0}^t\alpha$, then $\mathrm{d} \delta=\beta-\beta(0)$, which requires some additional $\delta_0$ satisfying $\mathrm{d} \delta_0=\beta(0)$, i.e. again everything points out the cohomology of $M$ of the same order