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I want to solve the following exercise from M. Spivak's Calculus on Manifolds:

If $M \subseteq \mathbb{R}^n$ is an orientable $(n-1)$-dimensional manifold, show that there is an open set $A \subseteq \mathbb{R}^n$ and a differentiable $g:A \to \mathbb{R}^1$ so that $M=g^{-1}(0)$ and $g'(x)$ has rank 1 for $x \in M$. Hint: Problem 5-4 does this locally. Use the orientation to choose consistent local solutions and use partitions of unity.

I don't understand why the orientation part is necessary. Once we express $M$ locally as the zero set of a function, why can't we just extend directly with partitions of unity?

EDIT:

Looking at Ted Shifrin's answer here we can represent $M$ as a zero set of some differentiable function without using any orientation argument.

Thank you!

EDIT2:

Now, I understand why the orientability part is important: If we have a differentiable function $g$ which defines $M$ as $g^{-1}(0)$ which has rank $1$ on $M$, then $\nabla g/\|g\|$ would give a continuous unit normal field, which can induce an orientation for $M$.

What I'm still having difficulties understanding is how to actually use a partition of unity to get a global result, and in particular, how can you guarantee that $M=g^{-1}(0)$ exactly? (the inclusion $g^{-1}(0) \subseteq M$ doesn't seem to be trivial)

Thanks again!

Community
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user1337
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  • Ted Shifrin's answer was incomplete. (He was responding to a different question, so he didn't bother to write out the details of constructing a global defining function.) See my comment on his question. The upshot is that his argument doesn't work unless you assume the manifold is closed in $\mathbb R^n$, in which case it's automatically orientable. – Jack Lee Jun 02 '15 at 00:15
  • The natural solution to this is the function "signed distance". You need some tubular neighborhoods and such. Take a look at the book of Guillemin and Pollack on Differential Topology. – orangeskid Jun 12 '15 at 11:13

2 Answers2

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Let $\{U_\alpha\}$ be an open covering of $M$ and $\{g_\alpha:U_\alpha\to\mathbb{R}\}$ functions defining $M$ locally. The orientability guarantees that all $g_\alpha$'s can be chosen so that their gradients point to the same "side" of $M$. Without this condition, once you take partition of unity and sum all $g_\alpha$, you may get non-regular points.

More precisely: If $M$ is oriented there is a smooth non-vanishing normal vector field $N$ on the whole of $M$. When taking $g_\alpha$ like in the previous paragraph, we make sure that for each $\alpha$ we have $$\langle \nabla g_\alpha,N\rangle>0.$$Then we use a partition of unity and get a function $g$, whose gradient satisfies $$\langle \nabla g,N\rangle>0.$$In particular, the gradient does not vanish anywhere on $M$.

Edit: The use of the partition of unity is as follows. Let $\{\varphi_\alpha\}$ be a partition of unity subordinate to the covering $\{U_\alpha\}$. For every $\alpha$ we define$$h_\alpha:M\to\mathbb{R},\qquad h_\alpha(x)=\left\{\begin{array}{cl}\varphi_\alpha(x)g_\alpha(x)&x\in U_\alpha\\0&x\not\in U_\alpha\end{array}\right..$$It is not hard to verify that every $h_\alpha$ is differentiable. Then set $h=\sum h_\alpha$, and you got the function you desire.

The orientation allows us to choose all the $g_\alpha$'s such that whenever $p\in U_{\alpha_1}\cap U_{\alpha_2}$, the values $g_{\alpha_1}(p)$ and $g_{\alpha_2}(p)$ have the same sign. Hence, for any $p\not\in M$, $h(p)$ is a convex combination of non-zero numbers, all having the same sign. This shows $h^{-1}(0)\subset M$.

Amitai Yuval
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  • Would you mind showing me what a complete proof looks like? Intuitively I get what you're saying, but I can't write it down. This is not a homework problem, I'm just reading through the textbook. – user1337 May 31 '15 at 12:23
  • @user1337 I added some details, have a look. – Amitai Yuval May 31 '15 at 18:39
  • Thank you. How would you exactly use the partition of unity to glue these $g_\alpha s$ all together? – user1337 Jun 10 '15 at 16:02
  • @user1337 Please have a look at the edit. – Amitai Yuval Jun 10 '15 at 20:06
  • Thanks again. If possible, could you please extend on why "the orientation allows us to choose all the $g_\alpha$'s..."? – user1337 Jun 11 '15 at 13:07
  • @user1337 This is actually equivalent to the gradient pointing to the same side all along $M$. Every $U_\alpha$ is divided anyway into a positive half and a negative one, separated by $M$ (assuming all sets in the covering are diffeomorphic to a ball, for example). Once all the gradients point to the same side, all positive halves are compatible with one another. – Amitai Yuval Jun 11 '15 at 23:01
  • I'm sorry to keep coming up with questions, but how do you know that the sets $U_\alpha \setminus M$ have two connected components? I can imagine really complicated surfaces in my mind. – user1337 Jun 13 '15 at 13:47
  • @user1337 Smooth submanifolds can't be that complicated. In fact, it is a basic result that for any $p\in M$ there is an open neighborhood $U$ and a diffeomorphism $\varphi:\mathbb{R}^n\to U$ identifying $U\cap M$ with the hyperplane ${x_1=0}$. Consider now a function $g:\mathbb{R}^n\to \mathbb{R}$ which vanishes only on ${x_1=0}$ and whose gradient doesn't vanish on this hyperplane. Then $\mathbb{R}^n\setminus{x_1=0}$ is split into two halves, in any of which $g$ has constant sign. – Amitai Yuval Jun 14 '15 at 10:23
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As he says in the hint, you use orientation to choose CONSISTENT local solutions on which you can use partitions of unity.

The reason we need a consistent choice can be seen by thinking of an example. Consider the Moebius Band. If you make a local choice and use partitions of unity to pass the local choice around the band, you will eventually reach a point where you have made a local choice that does not agree with a local choice you have already made. This image, I think, clarifies the point I am trying to make.

Robert Short
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