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Proof Review. A Theorem in Spivak's Book.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB.
I need your help to confirm that I have got the proof right of a very important theorem.

Theorem 2-13 in Spivak's Calculus on Manifolds.

Let $p\leq n$ and $f:\mathbb R^n\to\mathbb R^p$ be a continuously differentiable function in an open set $O$ of $\mathbb R^n$.
Let $a$ be a point in $O$ such that $f(a)=0$ and assume that the $p\times n$ matrix $M$ with the $i,j$-th entry $[M]_{i,j}=D_jf_i(a)$ has rank $p$.
Then there exists an open set $A$ of $\mathbb R^n$ which contains $a$, and a diffeomorphism $h:A\to\mathbb R^n$ such that $f\circ h(x_1,\ldots,x_n)=(x_{p-n+1},\ldots,x_n)$.

In Spivak's book, it says that Theorem 5.1 is immediate using the above theorem. I have tried to prove a slight variation of Theorem 5.1 below.

Notation:
Let $x\in \mathbb R^n$. We write $x_k^+$ as a shorthand for $(x_{k+1},\ldots,x_n)$ and $x_k^-$ as a shorthand for $(x_1,\ldots,x_{k-1})$.

To Prove:
Let $p\leq n$ and $f:\mathbb R^n\to\mathbb R^p$ be a continuously differentiable function such that $Df(x)$ has rank $p$ whenever $f(x)=0$.
Then $f^{-1}(0)$ is a $(n-p)$-dimensional manifold in $\mathbb R^n$.

Proposed Proof:
Let $a=(a_1,\ldots,a_n)$ be in $f^{-1}(0)$.
Then $f(a)=0$ and thus by Spivak's Theorem 2.13 there exists an open set $A$ of $\mathbb R^n$ which contains $a$, and a diffeomorphism $h:A\to\mathbb R^n$ such that $f\circ h(x)=x_{n-p}^+$.
Write $M=\{x\in A:x_{n-p}^+=0\}$.
We now show that $h(M)=f^{-1}(0)$.

Claim 1: $h(M)\subseteq f^{-1}(0)$.
Proof:
Let $y\in M$.
Then $f\circ h(y)=y_{n-p}^+$ and since $y_{n-p}^+=0$, we have $f(h(y))=0$.
Therefore $f(h(M))=\{0\}$.
This gives $h(M)\subseteq f^{-1}(0)$ and the claim is settled.

Claim 2: $f^{-1}(0)\subseteq h(M)$.
Proof:
Let $x\in f^{-1}(0)$.
Since $h$ is a diffeomorphism, it is a bijection and thus there is $y\in A$ such that $h(y)=x$.
Thus $f(h(y))=f(x)=0$.
This means $f\circ h(y)=0$, that is $y_{n-p}^+=0$, meaning $y\in M$.
Hence $x\in h(M)$ and since $x$ was arbitrarily chosen in $f^{-1}(0)$, we conclude that $f^{-1}(0)\subseteq h(M)$ and the claim is settled.

From the above two claims we have shown that $f^{-1}(0)=h(M)$.
Note that $M=A\cap \{x\in \mathbb R^n:x_{n-p}^+=0\}$.
Since $A$ is an open set in $\mathbb R^n$ and $\{x\in \mathbb R^n:x_{n-p}^+=0 \}$ is a $(n-p)$-dimensional manifold in $\mathbb R^n$, we infer that $M$ is a $(n-p)$-dimensional manifold in $\mathbb R^n$.
Now since $h$ was a diffeomorphism, and since diffeomorphisms take manifolds to manifolds and preserve the dimension, we know that $h(M)$ is a $(n-p)$-dimensional manifold in $\mathbb R^n$.
Having already shown that $h(M)=f^{-1}(0)$, our lemma is proved.
___

Can anybody please check the proof and confirm that it's correct or else point point out the errors?

Thanks in advance for taking the time out.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I don't see anything wrong with this. One thing I noticed in the original text was that Spivak did not include the condition "continuously differentiable" but merely said $f$ is differentiable, which may have been an oversight on his part.

As far as your proof goes, the proof of the 2 claims is a bit wordy, but this is just a stylistic objection. I find it obvious that:

$h(M) \subseteq f^{-1}(0)$ as soon as you show that $f(h(y)) = 0$, and likewise that:

$f^{-1}(0) \subseteq h(M)$ as soon as you show $y \in M$, which is also obvious from:

$f(h(y)) = 0$.

Having shown the existence of $h$, you're done, because that is all that is required to show $f^{-1}(0)$ is an $(n-p)$-dimensional manifold according to the definition at the beginning of chapter 5 (with $M$ playing the role of the set $V \cap (\Bbb R^k \times \{0\}) \subseteq \Bbb R^n$).
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
I don't see anything wrong with this. One thing I noticed in the original text was that Spivak did not include the condition "continuously differentiable" but merely said $f$ is differentiable, which may have been an oversight on his part.
Actually at the beginning of chapter 5 Spivak had said that 'differentiable' henceforth will mean $C^\infty$ so he didn't make a mistake. I wanted to prove the theorem using the much weaker hypothesis of continuous differentiability only and that's why I changed the statement.

As far as your proof goes, the proof of the 2 claims is a bit wordy, but this is just a stylistic objection. I find it obvious that:

$h(M) \subseteq f^{-1}(0)$ as soon as you show that $f(h(y)) = 0$, and likewise that:

$f^{-1}(0) \subseteq h(M)$ as soon as you show $y \in M$, which is also obvious from:

$f(h(y)) = 0$.
I am new to these things so I am writing the proofs in full. :eek:

Having shown the existence of $h$, you're done, because that is all that is required to show $f^{-1}(0)$ is an $(n-p)$-dimensional manifold according to the definition at the beginning of chapter 5 (with $M$ playing the role of the set $V \cap (\Bbb R^k \times \{0\}) \subseteq \Bbb R^n$).
Yes. Thanks. :)
 

Semillon

New member
May 30, 2014
3
Hi there all,

My first post, so apologies if this turns out to be cr*p. I think I am probably at about the stage caffeinemachine was when he posted - i.e. struggling to wade through Spivak chapter 5. I am not at an educational institution, so I'm on my own, which can be a lonely place in Calculus on Manifolds!

Anyway, I also struggle with Theorem 5.1, not least because I don't think the proof of Theorem 2.13 referenced from chapter 2 is valid. In that proof a claim is made about a cont. diff. function defined on an open set A containing a point a. For the general case in paragraph 2, the author first permutes the coordinates of a via a function g, then proceeds to apply the inverse function theorem to the composite f o g. My issue with this is that surely we don't even know this composite exists at a, let alone in an open set containing a!

That aside, with respect caffeinemachine, I have some trouble following your proof of Thm 5.1:
In showing that h(M)=f-1(0), should this not be h(M) = f-1(0) intersected with h(A)? This notwithstanding, do we not require h(M) to be open? You have not shown this. I know these are details, but I'm trying to be rigorous. No offence is meant.
 
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caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi there all,

My first post, so apologies if this turns out to be cr*p. I think I am probably at about the stage caffeinemachine was when he posted - i.e. struggling to wade through Spivak chapter 5. I am not at an educational institution, so I'm on my own, which can be a lonely place in Calculus on Manifolds!

Anyway, I also struggle with Theorem 5.1, not least because I don't think the proof of Theorem 2.13 referenced from chapter 2 is valid. In that proof a claim is made about a cont. diff. function defined on an open set A containing a point a. For the general case in paragraph 2, the author first permutes the coordinates of a via a function g, then proceeds to apply the inverse function theorem to the composite f o g. My issue with this is that surely we don't even know this composite exists at a, let alone in an open set containing a!

That aside, with respect caffeinemachine, I have some trouble following your proof of Thm 5.1:
In showing that h(M)=f-1(0), should this not be h(M) = f-1(0) intersected with h(A)? This notwithstanding, do we not require h(M) to be open? You have not shown this. I know these are details, but I'm trying to be rigorous. No offence is meant.
Hey Semillon!

I wrote this post quite some time ago.

I will review my proof keeping in mind the possible mistakes you have mentioned and reply by the end of the day.

EDIT: I am sorry that I could not reply because I have been quite busy today. I will try to study this post tomorrow and post my response.
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
@ Semillon... You don't need to worry about the "auto-save" function on here. It just stores what you've written as a handy back-up. If you then delete, change, etc your text, the next 'save' will keep a record of your changes. In short, the auto-save is your friend; you don't need to rush. Really... :cool:
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
So I finally reviewed my proof. Sorry for the very late reply.

That aside, with respect caffeinemachine, I have some trouble following your proof of Thm 5.1:
In showing that h(M)=f-1(0), should this not be h(M) = f-1(0) intersected with h(A)?
Isn't $h(A)=\mathbb R^n$? In this light, $f^{-1}(0)\cap h(A)=f^{-1}(0)\cap\mathbb R^n=f^{-1}(0)$.

I know these are details, but I'm trying to be rigorous. No offence is meant.
None taken. :)