Godbillon-Vey class is independent from the choices involved

Question

In the section 2.3 of these notes the Godbillon-Vey class is constructed. It is shown that this class does not depend from the choices involved (lemma 2.11). I have troubles understanding the presented proof-to be more precise, with the last steps of it. Here are some unclear steps:

1. Why the last two summands belongs to $\mathcal{S}(E)^{q+1}$
2. Why this space is $0$? (I guess that this should follow from the fact that the bundle $Q$ is of rank $q$)

3. Is this even correct? $E$ and $Q$ are bundles coming from $TM$ thus sections of these bundles are rather vector fields and not differential form. Shouldn't be the $q+1$-th power of the one forms vanishing on sectionf of $E$ instead?

4. It also seems to me that the Leibniz rule in the final calculation is used wrongly.

Is it possible to correct this proof?

Of course I would be happy if somebody could point me the reference where another proof can be found-but I'm interested in general (codimension $q$ instead of codimension $1$) foliation unless the proof for codimension $1$ can be generalized easily.

Answer 1

I had a hard time trying to read these notes so lets sate from the begining. We have $\Omega = \omega_1 \wedge \dots \wedge\omega_q $ section of $\det Q$ and $d\Omega = \alpha \wedge \Omega$ for some $\alpha \in \Gamma(\Omega^1 \otimes \mbox{End}(E))$. We define the Godbillon class by $$ [\eta] = [\alpha\wedge(d\alpha)^q] \in \mbox{H}^{2q+1}_{dR}(M,\mathbb{R}) $$ We need to check that the choices made so far don't change this class.

1. If we change the basis $\{\omega_1, \dots , \omega_q\}$ to $\{\omega_1', \dots , \omega_q'\}$ we'll have a base change matrix (-valued function) $A$ then $\Omega' = \det(A) \Omega$. Hence $$ d\Omega' = d\det(A)\wedge \Omega + \det(A)d\Omega = (d\det(A)+\det(A)\alpha) \wedge \Omega = (d\log(\det(A))+\alpha)\wedge\Omega' $$ and it follows that $\alpha'\wedge(d\alpha')^q = \alpha\wedge(d\alpha)^q + d(\log(\det(A)(d\alpha)^q)$ wich define the same class.
2. If $d\Omega = \alpha' \wedge \Omega$ for some $\alpha' \neq \alpha$. From $(\alpha'-\alpha) \wedge \Omega=0$ and de Rham division lemma, $\beta:=\alpha'-\alpha$ is a section of $Q$, ie $$ \beta = \sum\limits_{j=1}^q g_j \omega_j. $$ Since these are $1$-forms, $$ (d\alpha')^q = (d\alpha +d\beta)^q = \sum\limits_{j=0}^q \binom{q}{j}(d\alpha)^{q-j}\wedge(d\beta)^j = (d\alpha)^q + \Theta. $$ Then $$ \alpha'\wedge(d\alpha')^q = \alpha\wedge(d\alpha)^q + \beta\wedge(d\alpha)^q+(\alpha+\beta)\wedge\Theta $$ and it remains to show that $\beta\wedge(d\alpha)^q+(\alpha+\beta)\wedge\Theta$ is exact.

From $d\alpha\wedge \Omega =0$ we know that (completing the frame to $\{\omega_1, \dots, \omega_n\}$) $$ d\alpha = \sum\limits_{i<j} f_{ij}\omega_i\wedge \omega_j $$ with $i\leq q$. Then we may write $$ d\alpha = \sum\limits_{i=1}^q \gamma_i\wedge \omega_i $$ and it follows that $ (d\alpha)^q = \pm \gamma_1 \wedge \dots \gamma_q \wedge \Omega $ so $\beta\wedge(d\alpha)^q =0$. By Frobenius Theorem, $d\beta $ depends only on $\omega_i\wedge \omega_j$, $i<j$ with $i \leq q$. Hence the same same argument shows that $\beta\wedge \Theta = 0$. It remains $\alpha\wedge\Theta$.

We have that $\Theta = \sum\limits_{j=1}^q \binom{q}{j}(d\alpha)^{q-j}\wedge(d\beta)^j$. Define $$ \sigma = \sum\limits_{j=1}^q \binom{q}{j}(d\alpha)^{q-j}\wedge(d\beta)^{j-1} $$ Note that $\sigma$ is closed and $d\beta\wedge\sigma = \Theta$. Then $$ d(\beta\wedge\alpha\wedge\sigma) = d\beta\wedge \alpha\wedge\sigma - \beta\wedge d\alpha\wedge\sigma = \alpha \wedge \Theta - \beta\wedge d\alpha\wedge\sigma $$ The previous argument also shows that $\beta\wedge d\alpha\wedge\sigma=0$ and we are done.