# Bott and Tu Proposition 12.1

Question about the proof of Bott and Tu's Proposition 12.1: Given any double complex $$K$$, if $$H_\delta H_d(K)$$ has entries only in one row, then $$H_\delta H_d$$ is isomorphic to $$H_D$$.

Remark: $$\delta$$ is the horizontal, and $$d$$ the vertical, delta map. I do not understand the highlighted statement. From what I understand, if $$H^{p+2,q-1}_\delta H_d = 0$$ this means the following $$\ker(\delta: H^{p+2,q-1} \to H^{p+3, q-1}) = \text{Im}(\delta:H^{p+1,q-1} \to H^{p+2,q-1})$$. However, it is clear that $$\phi_1$$ is not $$d$$ closed, so it does not represent a $$H_d$$ cohomology class, so it surely cannot represent a $$H_\delta H_d$$ cohomology class, right? Any help would be appreciated.

• For context, I added the statement of the proposition (with explanation of notation). Aug 24, 2019 at 14:29

If I understand your question (and I am not screwing up!):

(Up to sign errors... ) $$D''\delta \phi_1 = \delta D''\phi_1 = \delta (-\delta \phi ) = 0$$. Hence $$\delta \phi_1$$ is $$D''$$-closed.

Addendum: Therefore $$\delta\phi_1$$ is both $$d$$ and (visibly) $$\delta$$ closed, and thus represents a class in $$H_\delta H_d$$, which is, by assumption, $$0$$ in the appropriate bi-degree. Hence there exists $$\hat \phi_1$$, such that $$d \hat \phi_1 =0$$, and $$\phi_2$$, such that $$\delta \phi_1 = \delta \hat \phi_1 \pm d \phi_2,$$ or $$\delta (\phi_1 -\hat\phi_1) = \pm d\phi_2.$$ Since $$\hat\phi_1$$ is $$d$$-closed, we can replace $$\phi_1$$ with $$\phi_1 -\hat \phi_1$$ to keep $$\delta \phi + D'' \phi_1 = 0$$, and also to obtain the desired $$\delta \phi_1= \pm d\phi_2$$.

• Wow, i dont know how i didnt see this. Thank you so much. Aug 25, 2019 at 16:00
• I'm v. glad to have helped: B and T was / is my favorite math book. Aug 25, 2019 at 17:28
• One thing im still confused about though, $H_{\delta} H_d = 0$ doesnt necessarily imply $H_d =0$ right? So why should $\delta \phi_1$ being $D''$ closed imply that it is $D''$ exact? Aug 25, 2019 at 22:30
• I think I have an idea of what's going on, correct me if I'm wrong but $H_\delta H_d = 0 \Leftrightarrow H_d H_\delta = 0$, since the differentials commute. Thus since $\delta \phi_1$ is clearly $\delta$ closed it represents a class in $H_d H_\delta$, and since $H_d H_\delta = 0$ and it is $d$ closed as you pointed out, it is therefore $d$ exact. Aug 26, 2019 at 4:19
• See my addendum - how does that look? What do you think? Assuming you buy it: it's been quite a while since I looked at the book but I wonder whether I missed this when I did. If so, shame on me ... esp. given the subsequent stuff on spectral sequences. Aug 26, 2019 at 5:25

Ok, thanks to peter a g and some of my colleagues, I now understand this problem, and I am going to post a full solution, starting from scratch. I believe that the proof in Bott and Tu is slightly misleading. I am not really going to worry about the big $$D$$ notation and just stick to $$\delta$$ and $$d$$ for sanity's sake. The point of this proof is to start with a $$H_\delta H_d$$ cocycle, do some diagram chasing and arrive at a $$H_D$$ cocycle.

We start with an element $$\phi$$ in the bicomplex $$K$$ such that $$\phi \in K^{p,q}$$. If $$\phi$$ were to represent a cohomology class in $$H_\delta H_d$$, it would have to be the case that first it represented a class in $$H_d$$, which means that $$d \phi = 0$$. Now for $$[\phi]_d \in H_d$$ to also represent a cohomology class in $$H_\delta H_d$$ it must be the case that $$\delta [\phi]_d = [\delta \phi]_d = _d$$, which means that $$\delta \phi = d \phi_1$$ for some $$\phi_1$$. Now it is clear that $$\phi_1$$ cannot represent a $$H_\delta H_d$$ cohomology class because it is not $$d$$ closed.

Thus we look at $$\delta \phi_1$$. This is $$d$$ closed because $$d \delta \phi_1 = \delta d \phi_1 = \delta \delta \phi = 0$$. Thus $$[\delta \phi_1]_d$$ represents a cohomology class in $$H_d$$. Also $$\delta [\delta \phi_1]_d = [\delta \delta \phi_1]_d = _d$$ so $$\delta \phi_1$$ also represents a cohomology class in $$H_\delta H_d$$. But $$\delta \phi_1 \in (H_\delta H_d)^{p+2,q-1} = 0$$, which means that since it is a $$\delta$$ closed cohomology class it is also a $$\delta$$ exact cohomology class, i.e. $$[\delta \phi_1]_d = \delta [\phi'_1]_d = [\delta \phi'_1]_d$$ for some $$\phi'_1$$ that represents a $$d$$ cohomology class, i.e. $$d \phi'_1 = 0$$. Now $$[\delta(\phi_1 - \phi'_1)]_d = _d$$ means that $$\delta(\phi_1 - \phi'_1) = d \phi_2$$ for some $$\phi_2$$. Thus $$\delta(\phi_1 - \phi'_1)$$ represents a $$H_d$$ class and it is clear that it also represents a $$H_\delta H_d$$ class. We also see that $$d (\phi_1 - \phi'_1) = \delta \phi$$.

Now we can do the same process to $$\phi_2$$, namely look at $$\delta \phi_2$$, notice it represents a $$H_\delta H_d$$ cohomology class, but in the position cohomology is trivial and thus $$[\delta \phi_2]_d = \delta [\phi'_2]_d$$, and replace $$\phi_2$$ with $$\phi_2 - \phi'_2$$. Keep doing this process until you get to the end of the bicomplex where we have $$\phi_n$$ and $$\delta \phi_n = 0$$. Now we want to replace $$[\phi]_{\delta d}$$ with $$[\phi + (\phi_1 - \phi'_1) + (\phi_2 - \phi'_2) + \dots + \phi_n]_{\delta d}$$, which will now be a $$D$$ cocycle, which is easy to check. Thus we have crafted a map from $$H_\delta H_d$$ to $$H_D$$.

• I agree that the book was misleading. Aug 27, 2019 at 15:43