Posted by: cnpde | September 12, 2012

## Basics of Riemann Geometry 2

In this post, I will discuss the most brilliant part of differential geometry — Exterior differential calculus. Before that I will make some preparation.

Following last post of this series, we already have tangent and cotangent vector space at point $p$, denoted as $T_p$ and $T_p^*$. Collecting the space for all point $p$ on the manifold, we get bundles. We have known how to define tensor for finite dimensional vector space. Now we are going to define tensor on tangent or cotangent spaces, which are infinite dimensional vector spaces. Define $(r,s)-$ type tensor space of manifold M at p as follows:

$T_s^r(p)=T_p\otimes...\otimes T_p\otimes T_p^* \otimes T_p^*$.

After collecting all tensor space for all point of $M$, we get tensor bundle $T_s^r$. The natual projection $\pi$ from $T_s^r$ to $M$ is the bundle projection, while $T_s^r(p)$ is the fibre of the bundle $T_s^r$ at point $p$.

Assume $f: M\rightarrow T_s^r$ is a smooth mapping, if $\pi\circ f=id: M\rightarrow M$, then we say $f$ is a smooth section of tensor bundle $T_s^r$, or a $(r,s)$ type smooth tensor field. The definition is easy to extend to exterior differential form by anti-symmetrizing.

We can define exterior vector bundles and exterior form bundles as

$\Lambda^r(M)=\bigcup\limits_{p\in M}\Lambda^r(T_p)$

$\Lambda^r(M^*)=\bigcup\limits_{p\in M}\Lambda^r(T_p^*)$.

So exterior differential form of degree r on M is a smooth map $M\rightarrow \Lambda^r(M^*)$ as a section of exterior form bundles. Another view is to look it as a smooth map  $T(M)\times ...\times T(M)\rightarrow C^\infty(M)$, which is multi-linear and alternating.