Proof Review. A Theorem in Spivak's Book.

In summary, the conversation discusses a theorem in Spivak's Calculus on Manifolds and the proof of a related theorem. The main theorem states that for a continuously differentiable function $f:\mathbb{R}^n \rightarrow \mathbb{R}^p$ with a certain rank condition, there exists a diffeomorphism that maps the function to a simpler form. The proposed proof uses this theorem to show that the set $f^{-1}(0)$ is an $(n-p)$-dimensional manifold. The proof has been checked and confirmed to be correct.
  • #1
caffeinemachine
Gold Member
MHB
816
15
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.
 
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  • #2
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 $V \cap (\Bbb R^k \times \{0\}) \subseteq \Bbb R^n$).
 
  • #3
Deveno said:
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.

Deveno said:
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:

Deveno said:
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. :)
 
  • #4
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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  • #5
Semillon said:
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.
 
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  • #6
@ 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:
 
  • #7
So I finally reviewed my proof. Sorry for the very late reply.

Semillon said:
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, $f^{-1}(0)\cap h(A)=f^{-1}(0)\cap\mathbb R^n=f^{-1}(0)$.

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

What is a proof review?

A proof review is a process of examining and evaluating the logical steps and reasoning used to prove a mathematical theorem or statement. It involves analyzing the structure and validity of the proof, as well as identifying any errors or gaps in the argument.

Why is proof review important?

Proof review is important because it helps ensure the correctness and rigor of mathematical arguments. By carefully examining a proof, mathematicians can verify the validity of a theorem and build upon it to make further advancements in the field.

What is the Theorem in Spivak's Book?

The Theorem in Spivak's Book refers to a specific theorem or statement presented in Michael Spivak's book "Calculus". This theorem is often used as a foundational concept in calculus and is commonly referred to as the "Fundamental Theorem of Calculus".

What is the process of reviewing a proof?

The process of reviewing a proof involves carefully reading through the proof, checking each step for accuracy and logical progression, and identifying any potential errors or gaps in the argument. It may also involve consulting other sources and discussing the proof with colleagues to gain a deeper understanding of the theorem.

What are some common mistakes made in proofs?

Some common mistakes made in proofs include incorrect assumptions or axioms, errors in logic or reasoning, and misuse of mathematical notation. Other common mistakes include missing steps or skipping important details, as well as unintentional errors in calculations or algebraic manipulations.

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