- #1
caffeinemachine
Gold Member
MHB
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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.
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.