To Prove that The Level Set Of A Constant Rank Map is a Manifold

[caffeinemachine]

Let $f:\mathbf R^n\to\mathbf R^m$ be a smooth function of constant rank $r$.
Let $\mathbf a\in \mathbf R^n$ be such that $f(\mathbf a)=\mathbf 0$.
Then $f^{-1}(\mathbf 0)$ is a manifold of dimension $n-r$ in $\mathbf R^n$.

We imitate the proof of Lemma 1 on pg 11 in Topology From A Differentiable Viewpoint by Milnor.

Let $S\in\mathcal L(\mathbf R^{m-r},\mathbf R^m)$ be such that $S(\mathbf R^{m-r})=(df_{\mathbf a}(\mathbf R^n))^\perp$, and $T\in \mathcal L(\mathbf R^n,\mathbf R^{n-r})$ be such that $\ker T\oplus \ker(df_{\mathbf a})=\{\mathbf 0\}$.
Such $S$ and $T$ can be chosen.

Define a function $F:\mathbf R^n\times \mathbf R^{m-r}\to \mathbf R^m\times \mathbf R^{n-r}$ as
$$
F(\mathbf x,\mathbf y)=(f(\mathbf x)+S\mathbf y,T\mathbf x)
$$
for all $(\mathbf x,\mathbf y)\in \mathbf R^n\times \mathbf R^{m-r}$.


Claim 1: $dF_{(\mathbf x,\mathbf y)}(\mathbf p,\mathbf q)=(df_{\mathbf x}(\mathbf p)+S(\mathbf q),T(\mathbf p))$.

Proof: Note that
$$
F(\mathbf x+\mathbf h,\mathbf y+\mathbf k)-F(\mathbf x,\mathbf y)-(df_{\mathbf x}(\mathbf h)+S\mathbf k,T\mathbf h)=(f(\mathbf x+\mathbf h)-f(\mathbf x)-df_{\mathbf x}(\mathbf h),\mathbf 0)
$$
From here it is clear that the claim holds.Claim 2: $dF_{(\mathbf a,\mathbf 0)}$ is invertible.

Proof: Suppose $dF_{(\mathbf a,\mathbf 0)}(\mathbf p,\mathbf q)=\mathbf 0$.
Then $(df_{\mathbf a}(\mathbf p)+S\mathbf q,T\mathbf p)=\mathbf 0$.
Therefore, $df_{\mathbf a9}\mathbf p+S\mathbf q=\mathbf 0$ and $T\mathbf p=\mathbf 0$.
Since $S(\mathbf R^{m-r})=(df_{\mathbf a}(\mathbf R^n))^\perp$, we have $df_{\mathbf a}(\mathbf p)=\mathbf 0$ and $S\mathbf q=\mathbf 0$.
Note that $S$ is injective.
Thus $\mathbf q=\mathbf 0$.
Also, since $\ker T\oplus\ker (df_{\mathbf a})=\mathbf R^n$, we also infer from $df_{\mathbf a}=\mathbf 0$ and $T\mathbf p=\mathbf 0$ that $\mathbf p=\mathbf 0$.
This proves that $dF_{(\mathbf a,\mathbf 0)}$ is injective and hence invertible.

Now by the inverse function theorem, we know that there exists a neighborhood $U$ of $(\mathbf a,\mathbf 0)$ such that $F(U)$ is open in $\mathbf R^m\times \mathbf R^{n-r}$ and $F|U:U\to F(U)$ is a diffeomorphism.

Therefore $M:=F^{-1}(\mathbf 0\times\mathbf R^{n-r})\cap U$ is an $n-r$ dimensional manifold in $\mathbf R^n\times\mathbf R^{m-r}$.
From here if I could conclude that $(\mathbf x,\mathbf y)\in M$ implies that $\mathbf y=0$, then I'd be done.
But I am unable to do so.
Can anybody help?

 

[Fallen Angel]

Hi,

I think you can simply take the open $U$ given by the IFT and intersect it with $\Bbb{R^{n}}$, $U\cap (\Bbb{R}^{n}\times \{0\})$ and still having a diffeomorphism and conclude from here.

 

[caffeinemachine]

Hi,

I think you can simply take the open $U$ given by the IFT and intersect it with $\Bbb{R^{n}}$, $U\cap (\Bbb{R}^{n}\times \{0\})$ and still having a diffeomorphism and conclude from here.

I cannot see how this can work. Can you please elaborate?

 

[Fallen Angel]

Oh sorry caffeinmachine, I was misunderstanding the post, ignore my previous post.

Regarding at this I noticed that I don't remember almost anything about differential geometry, I would look at my notes later, but I can't promise a result :/.

Sorry again

 

[blue_raver22]

I would first commend the author for their thorough and rigorous approach to proving the statement. The use of the inverse function theorem and the careful construction of the function $F$ are key components in the proof.

Regarding the final step, I would suggest considering the definition of $F$ and the fact that $\mathbf a$ is a point in the level set of $f$ at which $f(\mathbf a) = \mathbf 0$. This means that the first component of $F(\mathbf a, \mathbf y)$ is always $\mathbf 0$, regardless of the value of $\mathbf y$. Therefore, if $(\mathbf a, \mathbf y)$ is in the level set $M$, then $\mathbf y$ must be $\mathbf 0$ in order for the first component of $F(\mathbf a, \mathbf y)$ to be $\mathbf 0$. This implies that $F^{-1}(\mathbf 0 \times \mathbf R^{n-r}) \cap U$ is actually a subset of $f^{-1}(\mathbf 0)$, and therefore $M$ is a subset of $f^{-1}(\mathbf 0)$.

In conclusion, the proof provided is valid and successfully demonstrates that the level set of a constant rank map is a manifold.