Non-zero tangent vectors in Hermitian manifolds

Question

I'm a physicist trying to learn about Chern's class and forms by myself. To that respect I feel the Chern's writings far more pleasant than esoteric literature written by and for physicists, which are more than usual neither elegant nor to the point.

I'm actually reading Chern's work, especially [C1 = Characteristic Classes of Hermitian Manifolds, 1946] and [C2 = Characteristic Classes and Characteristic Forms, 1991], both reprinted in

A Mathematician and His Mathematical Works, Selected papers of S.S. Chern, Editors S.Y. Cheng, P. Li and G. Tian. World Scientific, 1996.

but otherwise not difficult to find. I'll try making the question intelligible without access to these articles. The question is anyway naive I think.

I'm a bit puzzled by the following: In [C1] Chern defines an equivalence class between non-zero tangent vectors $Z$ and $W$ when their components $Z^i$ and $W^i$ verifies $W^i=\rho Z^i$, with $\rho$ a real positive quantity. Then he goes on for the algebraic topological construction.

I'm just wondering why he excludes the zero tangent vectors ? I feel this is a really trivial argument, but I can not understand it. For me, the zero vectors would belong to their own class or equivalence composed of a unique element, and so would not violate any classification. I must be wrong here, but I do not understand why.

In [C2] Chern states the Poincaré-Hopf theorem, and proposes to answer the question : Is such a local product structure always a product globally ? The local product stucture is nothing but the one in the definition of the vector bundle, in the sense of Steenrod now widely used. Then Chern says a product bundle has sections which are nowhere zero in order to relate his question to the Poincaré-Hopf theorem. One more time I accept this statement, but do not appreciate it. Is there an easy way to understand it ? I think I'm missing some easy details in the definition of the section and/or tangent bundle, but I can not figure out which one.

Any help is welcome, either to improve this question, or to answer it :-)

Answer 1

I don't have the paper with me right now, so I don't understand the context of your first question. I will come back once I go in to the office and find my copy.

Let $M$ be a compact smooth manifold of dimension $n$. If the tangent bundle of $M$ is trivial (i.e. $M$ is parallelisable), there is a vector bundle isomorphism $\varphi : TM \to M \times \mathbb{R}^n$. The trivial bundle has $n$ natural nowhere sections given by $s_i : M \to M\times\mathbb{R}^n$, $s_i(x) = (x, e_i)$, where $i = 1, \dots, n$. Using the isomorphism $\varphi$, we can obtain $n$ nowhere zero sections of $TM$, namely $\varphi^{-1}\circ s_i$, $i = 1, \dots, n$. Applying the Poincaré-Hopf theorem to any one of these sections, we see that $\chi(M) = 0$.

Note, the sections $\varphi^{-1}\circ s_i$ above are actually linearly independent. The existence of such a collection is equivalent to the bundle being trivial. That is, if $E$ is a rank $n$ bundle on $M$ and $\sigma_1, \dots, \sigma_n$ are linearly independent sections of $E$, then any $e \in E_p$ can be written as $e = a_1\sigma_1(p) + \dots + a_n\sigma_n(p)$, so we can define $\varphi : E \to M\times\mathbb{R}^n$ by $\varphi(e) = (p, (a_1, \dots, a_n))$. This is a vector bundle isomorphism. Moreover, in terms of the above construction, $\varphi^{-1}\circ s_i = \sigma_i$.