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I'm trying to work out Proposition (6.16) from Bott and Tu which states that $H^*_{cv}(M\times\mathbb{R}^n)$ is isomorphic to $H^{*-n}(M)$. The first group being forms compactly supported along the fibre.

The idea is to define chain maps from one space (of forms) to the other. One of the maps, $\pi_*$, is integration along the vertical direction and the other, $e_*$ is (presumably) given by wedging with $n$ many bump functions $e$ of integral $1$, one in each of the fibre variables.

It is quite clear that $\pi_*\circ{e_*}=id$. To show that the other composition is identity (on cohomology), I follow what the book does in earlier proofs and try to construct a homotopy operator $K:\Omega^*_{cv}(M\times\mathbb{R}^n)\rightarrow\Omega^{*-1}_{cv}(M\times\mathbb{R}^n)$ as follows.

$$K(\phi f(x,t)dt_I):=\sum_{i}^{}(\phi\int_{-\infty}^{t_i}f(x,t)dt_i(-1)^{i-1}dt_{I-{i}}-\phi\int_{-\infty}^{\infty}f(x,t)dt_i(-1)^{i-1}.\int_{-\infty}^{t_i}e(s)ds.dt_{I-i})$$ Where $\phi$ is a form pulled back from $M$, $I$ is a multi-index and $x$, $t$ denote dependencies on the horizontal and vertical directions respectively. (Note, this is indeed compactly supported along the fibre)

I've tried many times now to show $Kd-dK=1-e_*\circ{\pi_*}$ (up to a constant) without luck and I'm starting to think some special care needs to be taken in defining the maps. My calculations allow me to cancel almost everything in $dK-Kd$ to get $c(1-e_*\circ{\pi_*})$ and one other term. I can't get rid of that unwanted term and another problem is that the constant $c$ sometimes vanishes depending on how many terms there are in $I$ (When there are half as many as $n$, the dimension of the fibre).

I don't expect a complete computation ($dK$ and $Kd$ have 8 terms each) as an answer as that would be very tedious and messy (But it would be great if someone could do so!). But I hope someone who's done this before can tell me if this is indeed the approach. The books says this proposition 'carries over verbatim from (4.7)' which I find very upsetting.

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    $\begingroup$ I have done this before, but I might have been less conscientious than you; still, my guess (as yours) is that your $K$ is incorrect. Have you tried working this out for $n=1$? I see your $*-1$, but you are using the $n$ dimensions to define $K$. Perhaps, you should view $\pi_* = (\pi_1)_*\cdots (\pi_n)_*$, and have a $K_i$ to deal with one variable at a time. (After all, 4.6 has $M\times \mathbb R^1$.) By the way: 'Very upsetting' - I'm sure! And your gnashing of teeth and weeping, but commendably, no profanity or obscenity, have been heard as far as in northern NJ, where I am.... $\endgroup$ – peter a g May 31 '16 at 14:17
  • $\begingroup$ Well - my previous comment seems (even more) stupid given your edit... Hmn. what is your constant $c$? $\endgroup$ – peter a g Jun 1 '16 at 0:31
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    $\begingroup$ My point in my first comment was: Suppose $A, B$ and $C$ are differential objects (e.g., complexes), and pairs $$ f: A \to B \text { and } \phi: B \to A$$ and $$ g : B \to C \text { and } \gamma: C \to B$$ of maps commuting with $d$ - e.g. $df=fd$, where $d=d_B$ and $d=d_A$ as relevant, and so on, for $\phi, g$, and $\gamma$ . Suppose there are homotopies $$ Kd - dK = 1 - \phi f \text { and } Ld - dL = 1 - \gamma g,$$ with $K: A\to A$, and $L:B\to B$. Then there is a homotopy $$ Md - dM = 1- \phi\gamma g f, $$ where $ M= \phi L f + K$. (cont.) $\endgroup$ – peter a g Jun 2 '16 at 11:25
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    $\begingroup$ (cont.) Did you construct your $K$ (as in your question - corresponding to my $M$ above for the case $n=2$ ) using that method? If so, I don't see how it can be wrong, at least in principle, as the abstract (toy) principle seems pretty clear. I'll try to go through the construction (in this case) this weekend if need be - as I wrote before, you are more conscientious than I... $\endgroup$ – peter a g Jun 2 '16 at 11:35
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    $\begingroup$ Yes, you're not alone... I've wasted I don't know how much time by misreading (and, reading, at least in the case of election coverage). I'll put the comment down as an answer over the weekend... in the meantime, I'm glad it helped. $\endgroup$ – peter a g Jun 2 '16 at 15:59
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To take this off of the un-answered queue - but (basically verbatim) as in the comments above.

Lemma Suppose $A,B$ and $C$ are differential objects (e.g., complexes), and pairs $$f:A\to B \text{ and } \phi:B\to A$$ and $$g:B\to C \text { and } \gamma:C→B$$

of maps commuting with $d$ - e.g., $df=fd$, where $d=d_B$ and $d=d_A$ as relevant, and so on, for $\phi$,$g$ and $\gamma$. Suppose there are homotopies $K:A→A$, and $L:B→B$: $$ Kd−dK=1−\phi f \text { and }Ld−dL= 1−\gamma g.$$

Then there is a homotopy $M:A\to A$: $$Md−dM=1−\phi\gamma gf,$$ where $M=\phi Lf+K.$

The proof of the lemma is immediate....

Using the lemma, one can collate homotopies, 'killing' one dimension at a time, i.e., by using the homotopy for one dimension of the book:

$$K_N:\Omega^*_{cv}(N\times\mathbb{R}^1)\rightarrow\Omega^{*-1}_{cv}(N\times\mathbb{R}^1),$$ with $N= M \times \mathbb R^k$, and $k = 0, \cdots,\ n-1$, where, in the notation of the original question, $K_N\phi f =0$, and (up to sign, depending on the weight of the form, if one wants to use the lemma above, exactly as stated), $$K_N \phi f(x,t)\,dt = \phi\left ( \int^{t}_{-\infty} f(x,s)\,ds - \int^t_{-\infty} e(s)\,ds \cdot \int^\infty_{-\infty} f(x,s)\,ds\right).$$

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