Compute the volume element in a differentiable manifold.

Question

Let $A\subset\mathbb{R}^n$ open $g:A\to\mathbb{R}$ of class $C^{1}$ and $g'(x)\not=0$ in each $x\in A$ then I want to compute $dV$ in the differentiable manifold $ M = g^{-1}(0)$.

The thing is that we have the result that says: Let $M\subset\mathbb{R}^n$ be a differentiable manifold of dimension $n-1$ with orientation $\mu$ so $$dV(x)((v_1)_p, \ldots, (v_{n-1})_p) = \det\begin{pmatrix}\nu_x & v_1 & \cdots & v_{n-1}\end{pmatrix}.$$

where $\nu_p\in(M_p)^\perp\subset\mathbb{R}^n$ is the unitary normal vector, this is, $\nu_p$ is the vector such that $[\nu_p, (v_1)_p, \ldots, (v_{n-1})_p]$ is the right-handed orientation in $\mathbb{R}^n$ for any basis $\{(v_1)_p, \ldots, (v_{n-1})_p\}$ of the tangent space $M_p$ with orientation $[(v_1)_p, \ldots, (v_{n-1})_p] = \mu_p$

Then it turns out that the problem is only to prove that $M$ is orientable, but how can I prove this with the hypotheses I got? Or How can I solve this problem?

NOTE: The notation and definitions are from the book Calculus in manifolds by Spivak the problem isn't taken from that book.

THANKS FOR YOUR HELP :)

ATTEMPT:

First we give to our differentiable manifold the orientation that has $\mathbb{R}^n$ then we pick the vector $\nu = \frac{\nabla g}{|\nabla g|}$, since this is well define because $\nabla g$ doesn't vanish this give us a unitary normal vector, this new $\nu$ has a direction, and therefore if it is wrong oriented we take $- \nu$ BUT I DONT UNDERSTAND WHY IF $\mathbb{R}^n$ is orientable then so is $M$ Thnks :)

In case it's helpful, this question (and its answer) may address your remaining doubts (depending on your definition of "orientation"). — Andrew D. Hwang, May 29 '15 at 23:10

score 1 · Answer 1 · answered May 27 '15 at 13:48

1

Since $\nabla g$ is non-vanishing in $A$ by hypothesis, the normalized gradient $\nu = \nabla g/\|\nabla g\|$ is a continuous unit normal field on the hypersurface $M$. Since the ambient space is orientable, $M$ itself is orientable.

answered May 27 '15 at 13:48

Andrew D. Hwang

78,195

thanks a lot for your answer, so,first of all, what is the ambient space? – user162343 May 27 '15 at 13:51
The ambient space is $A$ (or $\mathbf{R}^{n}$). :) – Andrew D. Hwang May 27 '15 at 14:03
Ok, then I use the result that I posted and that is all, right? – user162343 May 27 '15 at 14:05
Yes, that's right. – Andrew D. Hwang May 27 '15 at 15:11
Another question how do you prove thtat $\nu$ is the unit normal vector? – user162343 May 27 '15 at 16:53
The gradient field of a differentiable function is orthogonal to the level sets. – Andrew D. Hwang May 27 '15 at 17:04
Ok so you are considering the equation $g^{-1}=O$ as a level curve right? – user162343 May 27 '15 at 17:10
Well it is because I want to know about the following, do the gradient defines an orientation in the euclidean space because what if the gradient gives the opposite orientation un the manifold? – user162343 May 27 '15 at 18:30
Yes, $M = g^{-1}(0)$ is a level set. (It's a level curve only if $n = 2$.) 2. If the normalized gradient gives the "wrong" orientation, use $-\nabla g/|\nabla g|$. (One of the two must "work" in your formula for the volume element.)

Andrew D. Hwang

May 27 '15 at 19:31

Ok, let me post my way of writting this and can you give me your verification plaese ? Just vive some minutes because I am at the phone jajajaja – user162343 May 27 '15 at 19:39

ready :) I have edited my post – user162343 May 27 '15 at 20:17

I'm no longer sure what you're trying to express: You have an equation for $dV$ that requires a continuous unit normal field on $M$; the gradient (or its negative) gives you a suitable normal field at each point. If you have a basis for $T_{p}M$, you can calculate the value of $dV$ on this basis. Does that not answer the title question? – Andrew D. Hwang May 27 '15 at 20:58

Yes that is right but What I want to prove is that my differentiable manifold is orientable because with that I use the result I posted, so you said it is because we pick un the unitary gradient or its negative and that give us and orientation but I don't know how you prove that indeed the gradient is the vetor we need – user162343 May 27 '15 at 21:02

Compute the volume element in a differentiable manifold.

1 Answers1