Induced orientation on boundary of ##\mathbb{H}^n## in ##\mathbb{R}^n##

  • #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.
 
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  • #2
Yes, you can arrange the wedge product. Differential forms are alternating tensors so ##dx^1 \wedge dx^2 = -dx^2 \wedge dx^1##.

When you write ##\eta(-\partial_n, v_1, ..., v_{n-1})##, you should think of it as an alternating map, meaning if you switch adjacent elements you introduce a factor of -1.

It may be good to also review what a form actually is under the hood as a tensor to see how it eats vectors so to speak.

For instance,
$$(dx^1 \wedge dx^2)_p(v,w) = dx^1(v)dx^2(w) - dx^2(v)dx^1(w)$$
Or similar to your case,
$$(dx^1 \wedge dx^2)_p(\partial_2, \partial_1) = dx^1(\partial_2)dx^2(\partial_1) - dx^2(\partial_2)dx^1(\partial_1) = -1 $$
 
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  • #3
jbergman said:
For instance,
$$(dx^1 \wedge dx^2)_p(v,w) = dx^1(v)dx^2(w) - dx^2(v)dx^1(w)$$
Or similar to your case,
$$(dx^1 \wedge dx^2)_p(\partial_2, \partial_1) = dx^1(\partial_2)dx^2(\partial_1) - dx^2(\partial_2)dx^1(\partial_1) = -1 $$
This is a very helpful insight that I've actually never seen. Unfortunately, I haven't been able to take a proper differential geometry class so far, all my current knowledge comes from a very short and not quite detailed basic introduction in last semesters electrodynamics lecture. All the practice problems had the vectors in the right order already, so there hasn't been any need for rearrangements so far.

Also, just for clarity, does this rule extend to higher order wedge products by taking all possible permutations and multiplying with their sign? The way you've written it seems to imply a rule like

$$
(dx^1 \wedge dx^2 \wedge ... \wedge dx^n)(\partial_{i_1},\partial_{i_2}, ..., \partial_{i_n}) = \sum_\sigma \text{sgn}(\sigma) dx^{\sigma(1)}(\partial_{i_1}) dx^{\sigma(2)}(\partial_{i_2})...
$$
 
  • #4
PhysicsRock said:
This is a very helpful insight that I've actually never seen. Unfortunately, I haven't been able to take a proper differential geometry class so far, all my current knowledge comes from a very short and not quite detailed basic introduction in last semesters electrodynamics lecture. All the practice problems had the vectors in the right order already, so there hasn't been any need for rearrangements so far.

Also, just for clarity, does this rule extend to higher order wedge products by taking all possible permutations and multiplying with their sign? The way you've written it seems to imply a rule like

$$
(dx^1 \wedge dx^2 \wedge ... \wedge dx^n)(\partial_{i_1},\partial_{i_2}, ..., \partial_{i_n}) = \sum_\sigma \text{sgn}(\sigma) dx^{\sigma(1)}(\partial_{i_1}) dx^{\sigma(2)}(\partial_{i_2})...
$$
Yes, though, people typically permute the arguments (vectors) as opposed to permuting the forms that make up the wedge products.
$$
(dx^1 \wedge dx^2 \wedge ... \wedge dx^n)(\partial_{i_1},\partial_{i_2}, ..., \partial_{i_n}) = \sum_\sigma \text{sgn}(\sigma) dx^{1}(\partial_{\sigma(1)) dx^{2}(\partial_{\sigma(2)})...
$$
 
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  • #5
jbergman said:
Yes, though, people typically permute the arguments (vectors) as opposed to permuting the forms that make up the wedge products.
Does that make a difference? Probably easy to check, but I know myself very well and I tend to make the worst mistakes with the easiest problems.
 
  • #6
PhysicsRock said:
Does that make a difference? Probably easy to check, but I know myself very well and I tend to make the worst mistakes with the easiest problems.
No.
 
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  • #7
PhysicsRock said:
This is a very helpful insight that I've actually never seen. Unfortunately, I haven't been able to take a proper differential geometry class so far, all my current knowledge comes from a very short and not quite detailed basic introduction in last semesters electrodynamics lecture. All the practice problems had the vectors in the right order already, so there hasn't been any need for rearrangements so far.

Also, just for clarity, does this rule extend to higher order wedge products by taking all possible permutations and multiplying with their sign? The way you've written it seems to imply a rule like

$$
(dx^1 \wedge dx^2 \wedge ... \wedge dx^n)(\partial_{i_1},\partial_{i_2}, ..., \partial_{i_n}) = \sum_\sigma \text{sgn}(\sigma) dx^{\sigma(1)}(\partial_{i_1}) dx^{\sigma(2)}(\partial_{i_2})...
$$
Yes, this is why/how, the determinant comes into place.
 
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