# To what extent is the global angular form well-defined?

I am reading Differential Forms in Algebraic Topology by Bott & Tu. The authors constructed the global angular form $\psi$ for an oriented $k$-sphere bundle $E$ over a smooth manifold $M$. It has the properties:

1. $\psi$, when restricted to each fiber, is the generator for the top cohomology of the fiber. To be more precise, $\psi|_{E_p}$ is a closed $k$- form and $\int_{E_p}\psi|_{Ep}=1$. The integral makes sense since the sphere bundle is oriented.

2. $d\psi=-\pi^*(e)$, where $\pi:E\to M$ is the natural projection and $e=e(E)$ is the Euler class of the sphere bundle.

The global angular form is clearly not uniquely characterized by the above two properties. For any closed form $\eta$ on $M$, $\psi+\pi^*(\eta)$ will satisfy 1 and 2 if $\psi$ does. Note that $\psi$ is in general not closed, hence does not represent a cohomology class. Now my question is:

To what extent is the global angular form of an oriented sphere bundle unique, satisfying 1 and 2 above?

Since the global angular form is in some sense choosing a polar coordinate system for each fiber and the Euler class indicates the obstruction against this. (I am sorry for my imprecision at this point. I may need some help here.) So I have conjectured the following proposition:

With notations as above. Let $\omega$ be a $k$-form on $E$ such that

1. $\omega$ restricts to the generator of the top cohomology of each fibre and
2. $d\omega=\pi^*(\eta)$ for some closed form $\eta$ on $M$. Then $\eta$ is the Euler class of the sphere bundle.

Can someone prove of disprove this proposition? Any comment on the intuitive understanding of the global angular forms is appreciated. Thanks!

• Why doesn't LaTeX work in the grey area? Could somebody help me fix it? – YZhou Jan 9 '13 at 12:01