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Integrating over a Manifold

  1. Oct 17, 2010 #1
    First let me write out the definition of a manifold given in my book:

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

    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 [tex] \mathbb{R}^n [/tex] 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 [tex] \mathbb{R}^n [/tex]?

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

    Note that [tex] dV [/tex] represents the integral with respect to volume and [tex] V(D \alpha_i) = \sqrt{det[(D\alpha_i)^{tr} D\alpha_i]} [/tex]
  2. jcsd
  3. Oct 18, 2010 #2
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