- #1

PhysicsRock

- 116

- 18

- TL;DR Summary
- I am trying to work through the proof for Stokes' generalized theorem given in Lee's Introduction to Smooth Manifolds. I am, however, a little puzzled on how to work out the induced orientation on ##\partial \mathbb{H}^n##.

To my understanding, an orientation can be expressed by choosing a no-where vanishing top form, say ##\eta := f(x^1,...,x^n) dx^1 \wedge ... \wedge dx^n## with ##f \neq 0## everywhere on some manifold ##M##, which is ##\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n \geq 0 \}## here specifically. To determine the induced orientation on the boundary ##\partial\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n = 0 \}## we would now pick any outward pointing vector field, let's call it ##X##, and obtain a new orientation form on ##\partial\mathbb{H}^n##, say ##\eta^\prime##, by first considering an ordered basis ##(v_1,...,v_{n-1})## of ##T_pM## for all ##p \in M \equiv \mathbb{H}^n## and demanding

$$

\eta^\prime(v_1,...,v_{n-1}) := \eta(X(p),v_1,...,v_{n-1}).

$$

If we choose the standard orientation on ##\mathbb{H}^n## by setting ##\eta = dx^1 \wedge ... \wedge dx^n##, we should end up with the induced orientation being ##\eta^\prime = (-1)^n dx^1 \wedge ... \wedge dx^{n-1}##. Of course, in this simple case we can choose ##X = -\partial_n##, resulting in

$$

\eta^\prime(v_1, ..., v_{n-1}) = \eta(-\partial_n \vert_p, v_1, ..., v_{n-1})

$$

This is where I'm kinda stuck. If I do remember correctly, then the relation between the basis for ##T_pM## and ##T^*_pM## is ##dx^i(\partial_j) = {\delta^i}_j##, so the expression above would give some

$$

dx^1(-\partial_n) = -{\delta^1}_n = 0 \text{ if } n \neq 1.

$$

The only way I can make sense of this would be to rearrange the wedge product, i.e. writing

$$

\eta = dx^1 \wedge dx^2 \wedge ... \wedge dx^n = (-1)^{n-1} dx^n \wedge dx^1 \wedge dx^2 \wedge ... \wedge dx^{n-1},

$$

which, together with the factor of ##(-1)## of ##X##, would exactly give the desired ##(-1)^n##. So essentially, my question would be whether this is allowed to do and, of course, correct.

$$

\eta^\prime(v_1,...,v_{n-1}) := \eta(X(p),v_1,...,v_{n-1}).

$$

If we choose the standard orientation on ##\mathbb{H}^n## by setting ##\eta = dx^1 \wedge ... \wedge dx^n##, we should end up with the induced orientation being ##\eta^\prime = (-1)^n dx^1 \wedge ... \wedge dx^{n-1}##. Of course, in this simple case we can choose ##X = -\partial_n##, resulting in

$$

\eta^\prime(v_1, ..., v_{n-1}) = \eta(-\partial_n \vert_p, v_1, ..., v_{n-1})

$$

This is where I'm kinda stuck. If I do remember correctly, then the relation between the basis for ##T_pM## and ##T^*_pM## is ##dx^i(\partial_j) = {\delta^i}_j##, so the expression above would give some

$$

dx^1(-\partial_n) = -{\delta^1}_n = 0 \text{ if } n \neq 1.

$$

The only way I can make sense of this would be to rearrange the wedge product, i.e. writing

$$

\eta = dx^1 \wedge dx^2 \wedge ... \wedge dx^n = (-1)^{n-1} dx^n \wedge dx^1 \wedge dx^2 \wedge ... \wedge dx^{n-1},

$$

which, together with the factor of ##(-1)## of ##X##, would exactly give the desired ##(-1)^n##. So essentially, my question would be whether this is allowed to do and, of course, correct.

Last edited by a moderator: