# Integrating over a Manifold

1. Oct 17, 2010

### JG89

First let me write out the definition of a manifold given in my book:

Let $$k > 0$$. A k-manifold in $$\mathbb{R}^n$$ of class $$C^r$$ is a subspace $$M$$ of $$\mathbb{R}^n$$ having the following property: For each $$p \in M$$, there is an open set $$V \subset M$$ containing $$p$$, a set $$U$$ that is open in either $$\mathbb{R}^k$$ or $$\mathbb{H}^k$$ (upper half space), and a continuous bijection $$\alpha : U \rightarrow V$$ such that 1) $$\alpha$$ is of class $$C^r$$, 2) $$\alpha^{-1} : V \rightarrow U$$ is continuous, 3) $$D\alpha(x)$$ has rank k for each $$x \in U$$. The map $$\alpha$$ is called a coordinate patch on $$M$$ about $$p$$.

In my text I am reading the chapter on integrating a scalar map over a compact manifold. My question is this: Suppose M is a compact-manifold. As a subset of $$\mathbb{R}^n$$ it is bounded. So instead of going through all the mess of defining a manifold and defining the integral of a continuous function f over a manifold, why not just integrate f over M as one usually would? Using Riemann sums in $$\mathbb{R}^n$$?

Surely this would give the same result as integrating f over M using the definition of integral over a manifold. So what's so special about using a manifold M for integration when we could just consider M as a regular bounded subset of Euclidean space and integrate it how we usually would?

In case you're wondering, here is the definition of the integral of a scalar map over a compact manifold M:

Let M be a compact k-manifold in $$\mathbb{R}^n$$, of class $$C^r$$. Let $$f: M \rightarrow \mathbb{R}$$ be a continuous function. Suppose that $$\alpha_i: A_i \rightarrow M_i$$, for i = 1, ..., N, is a coordinate patch on M, such that $$A_i$$ is open in $$\mathbb{R}^k$$ and M is the disjoint union of the open sets $$M_1, M_2, ..., M_N$$ of M and a set K of measure zero in M. Then $$\int_M f dV = \sum_{i = 1}^N \int_{A_i} f(\alpha_i) V(D \alpha_i)$$.

Note that $$dV$$ represents the integral with respect to volume and $$V(D \alpha_i) = \sqrt{det[(D\alpha_i)^{tr} D\alpha_i]}$$

2. Oct 18, 2010

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