JG89
- 724
- 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]
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]