I am absolutely stuck, reading Bott and Tu isomorphism of de Rham cohomology. Please help. On page 92,


Step 2. $r^{*}$ is injective.

$$ r(\omega)=D\phi^{'}=d\phi^{'}, ~~~\delta \phi^{'}=0 $$

I think the claim that $ \delta \phi^{'}=0 $ is not correct and hence neither is the proof. Can someone check the proof. It starts from page 91.

Edit: Here is why I think the proof is not correct. Let's follow the proof from page 91 to 92 for two columns only. In the injective part, we argue that we can write $\phi$ as a sum:


where $\phi^{'}$ has only top component. This is shown on the previous page as follow. Let $\phi=(\alpha_0, \alpha_1)$ then by exactness there exists $\beta$ such that $\alpha-D\beta$ has only top component. Let's write this explicitly. We set $\beta=\phi^{''}=(\beta_0, 0)$ and $\delta\beta_{0}=\alpha_{1}$ to get the notation on page 92.

$$D\phi^{''}=D(\beta_{0}, 0) = (d\beta_0, \delta \beta_0)$$ and thus:

$$\phi^{'}=\phi-D\phi^{''}=(\alpha_0-d\beta_0, \alpha_1-\delta \beta_0)=(\alpha_0-d\beta_0, 0)$$

so $$\delta\phi^{'}= \delta(\alpha_0-d\beta_0)=\delta\alpha_{0}-d\alpha_1$$ and there is no reason why we should assume $\delta\alpha_0-d\alpha_1=0$.

  • 2
    $\begingroup$ Why do you think the claim is not correct? $\endgroup$
    – Pedro
    Commented Dec 18, 2016 at 14:09
  • $\begingroup$ I took a look and could not see an obvious reason why $\delta\phi'=0$. I noticed they used the fact again for the general proof a few pages later, again without justification. $\endgroup$
    – ziggurism
    Commented Dec 18, 2016 at 14:10
  • 1
    $\begingroup$ I would guess downvotes are because people don't like questions that are not self-contained. You should add all relevant context to your question, instead of asking people to open a PDF and page through 100 pages $\endgroup$
    – ziggurism
    Commented Dec 18, 2016 at 14:11

2 Answers 2


The crucial step is that one can always replace a cochain in $K^{**}$ by one with one with last component zero.

To see that $r*$ is onto, take a cocycle in $K^{**}$. By the first remark, this is represented by a pair of forms $(\eta_1,\eta_2)$, and the cocycle condition means in the vertical direction, that each $\eta_i$ is closed, and second, that $\eta_1=\eta_2$ in $U\cap V$. Thus there is a global form $\omega$ that restricts to $(\eta_1,\eta_2)$.

To see that $r*$ is injective, take a global form $\omega$ and suppose the restrictions $(\omega_1,\omega_2)$ have trivial class in $\operatorname{Tot} K^{**}$, so this is of the form $D\varphi$ for some $\varphi$. By the first remark, one can assume $\varphi$ has zero last component. Now $r^*\omega$ also has zero last component -- it has $(q,0)$ component $r^*\omega$ and $(q-1,1)$ component $0$, and because $D\varphi=r^*\omega$, it follows that $\delta\varphi=0$, so $\varphi$ is a globally defined form. But we also have $d\varphi=r^*\omega$, so $\omega$ is exact.

Their proof can be phrased as follows. There is a double complex $C^{**}$ obtained by looking at the exact sequence of complexes as such a double complex:

$$ \Omega^*(M)\stackrel{r}\longrightarrow \Omega^*(U)\oplus \Omega^*(V) \longrightarrow \Omega^*(U\cap V)$$

This is a first quadrant cohomological double complex, and it has exact rows. Now there is induced a map $$r: \Omega^*(M)\to \operatorname{Tot}(K^{**})$$

as described by Bott and Tu. Now one checks the cone of $r$ is exactly the total complex of $C^{**}$, and because $C^{**}$ has exact rows -- by virtue of the exactness of the Mayer-Vietoris sequnce --- its total complex is acyclic. Because $\operatorname{cone}(r^*)$ is acyclic, it follows that $r$ is a quasi-isomorphism, as desired.

So, let us show that if a first quadrant cohomological complex $C^{**}$ has exact rows, then $\operatorname{Tot}(C^{**})$ is acyclic. Write $d'$ for the vertical differential and $d''$ for the horizontal one.

Consider a cocycle $c=(c_{0,q},c_{1,q-1},\ldots,c_{q,0})$. Then $d''c_{q,0}=0$. Because the rows are exact, there is $b_{q,0}$ such that $d''b_{q,0}=c_{q,0}$. Then $c'=c-Db$ where $b=(0,\ldots,0,b_{q,0})$ has last component $0$, and $Dc'=0$. Continue in this way: inductively assume you have killed the last $j$ components, so now $d''c_{q-j,j}=0$, so there is $b_{q-j,j}$ such that $d''b_{q-j,j}=c_{q-j,j}$. Inductively, we reach $c''$ with only the top component $(0,q)$ nonzero. But if $Dc''=0$, this means that $d''_{0,q}c''_{0,q}=0$, and since $d''$ is injective at $(0,*)$ because our complex has exact rows, it follows that $c''=0$. This is exactly what Bott and Tu are doing.

  • $\begingroup$ So Bott and Tu meant to apply the argument "we can assume the form to only have top component" to $D\phi$, rather than to $\phi$? $\endgroup$
    – ziggurism
    Commented Dec 18, 2016 at 16:03
  • $\begingroup$ @ziggurism No. They mean what they said. $D\varphi$ has only top component by construction, because $r^*$ maps to the first column of $\operatorname{Tot}(K^{**})$, and one can assume $\varphi$ has only top component by their claim. $\endgroup$
    – Pedro
    Commented Dec 18, 2016 at 16:04
  • $\begingroup$ Hm ok so $\varphi\in K^{(q,0)}$ and therefore $D\varphi = d\varphi\pm\delta\varphi\in K^{(q+1,0)}\oplus K^{(q,1)}$ but since it equals $r^*\omega\in K^{(q+1,0)}$ we conclude the second component vanishes. Is that it? $\endgroup$
    – ziggurism
    Commented Dec 18, 2016 at 16:10
  • $\begingroup$ @ziggurism Yes. ${}$ $\endgroup$
    – Pedro
    Commented Dec 18, 2016 at 16:11
  • $\begingroup$ So did we actually need to make this argument about $\varphi$ having only top component? Seems like a red herring. Without it, we have $\varphi=\alpha+\beta\in K^{q,0}\oplus K^{q-1,1}$ and then $D\varphi=(d\pm\delta)(\alpha+\beta)=d\alpha + (\delta\alpha+d\beta) + \delta\beta \in K^{q+1,0}\oplus K^{q,1}\oplus K^{q-1,2}$. Then set it equal to $r^*\omega\in K^{q+1,0}$ and conclude that the components in $K^{q,1}$ and $K^{q-1,2}$ vanish. $\endgroup$
    – ziggurism
    Commented Dec 18, 2016 at 16:20

To address the edit: you're misunderstanding what they are doing. A generic element in the total complex $C^*(\mathfrak U)$ of $K^{**}$ is of the form $(a,b)$ with $a\in \Omega^*(U)\oplus \Omega^*(V)$ and $b\in \Omega^*(U\cap V)$.

The map $r^* : \Omega^*(M)\longrightarrow C^*(\mathfrak U)$ sends $\omega$ to $(a,0)$ where $a=(\omega\mid_U,\omega\mid_V)$. Take a closed form $\omega$, and asume that $r^*(\omega)=(a,0)$ is a boundary in $C^*(\mathfrak U)$. Then $(a,0) = D\varphi$ for some $\varphi$. Now $\varphi$ is an element of $C^*(\mathfrak U)$, and by the argument in the book, there is $\varphi'$ of the form $(\eta',0)$ so that $\varphi-\varphi'=D\nu$ in $C^*(\mathfrak U)$. Thus $(a,0) = D\varphi'$. Now $D\varphi'=D(\eta',0)=(d\eta',\delta \eta')$, and since this equals $(a,0)$, then $$d\eta'=a$$ $$\delta \eta'=0$$

The first equation says that $\omega$ is the boundary of $\eta'$ on each $U,V$, the second says that $\eta'=(\eta_1,\eta_2) \in \Omega^*(U)\oplus \Omega^*(V)$ is a globally defined form, so $\omega$ is a boundary, as desired.

  • $\begingroup$ Actually then if we assume $D\phi=(a,0)$ $\delta\alpha_0-d\alpha_1=0$ has to be zero according even my own construction. Since it is the second term in $D\phi^{'}=D\phi=(a,0)$? right $\endgroup$
    – alireza
    Commented Dec 18, 2016 at 20:50
  • $\begingroup$ Exactly. You don't have to assume it, it is given by the definition of $r^*$. $\endgroup$
    – Pedro
    Commented Dec 18, 2016 at 20:51
  • $\begingroup$ yeahhhhhhh! :). $\endgroup$
    – alireza
    Commented Dec 18, 2016 at 20:52
  • $\begingroup$ would you see this? math.stackexchange.com/questions/2109142/… $\endgroup$
    – alireza
    Commented Jan 22, 2017 at 18:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .