Boundary Orientation

Question

Edits to the original post are in bold.

I've been going through @Ted Shifrin's lectures on Stokes's Theorem, and I had a question relating to his choice of orientation of the tangent space as it comes to boundary orientation. In his example (at https://youtu.be/F80elc8JIQk?t=1005), he takes $\mathbb{R}^2_+$ with the standard orientation and says that such choice of orientation on the manifold's interior necessarily gives its boundary the orientation $\{-\mathbf{e}_2, \mathbf{e}_1\}$. Similarly, he said that taking $\mathbb{R}^3_+$ with the standard orientation necessarily gives its boundary the orientation $\{-\mathbf{e}_3, \mathbf{e}_1, \mathbf{e}_2\}$. I understand the choice of the outward normal, but I don't understand the orientation of the tangent space.For instance, I don't understand why Professor Shifrin chose $\{-\mathbf{e}_2, \mathbf{e}_1\}$ instead of $\{-\mathbf{e}_2, -\mathbf{e}_1\}$; such choice would change the boundary orientation.

After some help from @Deane and @peek-a-boo, I finally understand we give the boundary an orientation such that "a basis $(t_1,…,t_k)$ of vectors tangent to the boundary is positively oriented if $(n,t_1,…,t_k)$ is positively oriented on the interior." Compounding this confusion, I misquoted the video. Taking $\mathbb{R}^3_+$ with the standard orientation necessarily gives its boundary the orientation $\{\mathbf{e}_2, \mathbf{e}_1\}$.

Answer 1

I don’t actually understand your confusion. The comment made by Deane above is a definition. You also write

the bases $\{-e_2,e_1\}$ and $\{-e_2,-e_1\}$ have opposite orientations even though $e_1$ and $-e_1$ both are both bases of vectors tangent to the boundary.

Ok… but that is to be expected. The vectors $e_1,-e_1$ are indeed both tangent to the boundary and a basis for its tangent space, but the minus sign means they define opposite orientations for the boundary. Which is why again, I’m not really sure what it is you’re confused about.

Anyway, what’s the ‘real reason’ for why we make this definition? The ‘meta’ is simply that we want Stokes’ formula to say $\int_Md\omega=\int_{\partial M}\omega$, without any rogue minus signs (such as $\int_Md\omega=(-1)^m\int_{\partial M}\omega$ where $m=\dim M$).

To satisfy the definition of boundary orientation, we note that $\{e_1\}$ is a basis for the boundary $\Bbb{R}\times \{0\}=\partial(\Bbb{R}^2_+)$ such that $\{-e_2,e_1\}$ has positive orientation in $\Bbb{R}^2$ (since $\{-e_2,e_1\}\sim -\{e_2,e_1\}\sim (-1)^2\{e_1,e_2\}$, where by $\sim$ I mean ‘is equivalent to’ under the equivalence relation of the change of basis having positive determinant).

On the other hand, $\{e_1,e_2\}$ is a basis for the boundary $\Bbb{R}^2\times\{0\}=\partial(\Bbb{R}^3_+)$ which unfortunately produces the wrong orientation to the one we want, because $\{-e_3,e_1,e_2\}\sim -\{e_3,e_1,e_2\}\sim(-1)^2\{e_1,e_3,e_2\}=(-1)^3\{e_1,e_2,e_3\}$ (the first minus sign is because of the minus in $-e_3$, the next two minus signs come from swapping). Thus, we should orient the boundary using the ordered basis $\{-e_1,e_2\}$ (or equivalently $\{e_1,-e_2\}$ or $\{e_2,e_1\}$), so that the result $\{-e_3,-e_1,e_2\}$ has positive orientation in $\Bbb{R}^3$.

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Here’s something I’ll nitpick about your phrasing (and I’m sure @Ted Shifrin never phrased it as such either)

he takes $\mathbb{R}^2_+$ with the standard orientation and says that such choice of orientation on the manifold's interior necessarily gives its boundary the orientation $\{-\mathbf{e}_2, \mathbf{e}_1\}$.

The correct way to phrase it is:

If we give $\Bbb{R}^2_+$ the orientation defined by the ordered basis $\{e_1,e_2\}$, then we give the boundary the orientation defined by the ordered basis $\{e_1\}$

You wrote $\{-e_2,e_1\}$ in place of $\{e_1\}$ here, but this is poor mathematical grammar because it is $\{e_1\}$ which is a basis for the boundary; $\{-e_2,e_1\}$ is a basis for $\Bbb{R}^2$. It just so happens that this funny basis defines the same orientation as the usual one.

We want the basis of the tangent space to be boundary $\{b_1,\dots, b_{n-1}\}$ to be such that when we look at the funny basis $\{\nu_{\text{out}},b_1,\dots, b_{n-1}\}$, it must be such that it has the same orientation as the original $\{e_1,\dots, e_n\}$.

If what we’re saying is still not clear, then consider the following statements:

• $\{e_1\}$ is a basis for the tangent to the boundary of $\Bbb{R}^2_+$.
• $\{e_1,e_2\}$ is by definition the positively oriented basis for $\Bbb{R}^2$.
• $\{-e_2,e_1\}$ is a basis for $\Bbb{R}^2$.
• $\{-e_2,e_1\}$ and $\{e_1,e_2\}$ are both bases which give rise to the same orientation for $\Bbb{R}^2$.
• Hence, we define $\{e_1\}$ to be the positively oriented basis for the boundary.

I hope you agree with these 5 bullet points, because 1,2,3,5 are simply definition/ extremely trivial linear algebra. The only calculational step is bullet point (4) (that too a simple check that a certain $2\times 2$ matrix has positive determinant). The only ‘unease’ I can expect is why the definition in the fifth bullet point is the way it is; and I’ve already addressed it above (it’s to make Stokes’ formula nice looking).