Understanding some definitions of smooth manifolds and the inverse function theorem

Question

I am an electrical engineering grad student trying to learn some of the basics of differential geometry to better understand some work on signal processing and learning on Grassmanian manifolds. Pretty much I want to understand this paper pretty completely. My relevant background includes a formal undergraduate analysis course, a graduate measure theoretic probability course, and of course a bunch of random engineering mathematics/mathematical physics courses. I'm using the book "Introduction to Differential Geometry for Engineers" by Doolin/Martin to try and get up to speed quickly.

The first "definition" given is the following: A subset $M$ of $\mathbb{R}^n$ is a k-dimensional manifold if for each $\mathbf{x} \in M$ there are: open subsets $U$ and $V$ of $\mathbb{R}^n$ with $x \in U$ and a diffeomorphism $f$ from $U$ to $V$ such that $f(U \cap M) = \{ \mathbf{y} \in V : y^{k+1}=y^{k+2}=...= y^n = 0 \}$. Thus, each point $y$ in the image of $f$ has a representation like $y = (y^1(x), y^2(x), ... y^k(x),0,...,0)$.

The book also quotes an implicit function theorem, which I'll summarize as: If $F:\mathbb{R}^{n-k}\times\mathbb{R}^{k}\rightarrow\mathbb{R}^k$ is continuously differentiable in an open set containing $(\mathbf{a},\mathbf{b})$ and $F(\mathbf{a},\mathbf{b}) = \mathbf{0}$ then if the Jacobian of F is of rank $k$ then there is an open set $A\in \mathbb{R}^{n-k}$ containing $\mathbf{a}$ and an open set $B \in \mathbb{R}^k$ containing $\mathbf{b}$ such that for each $\mathbf{x}\in A$ there is a unique $g(\mathbf{x})\in B$ such that $F(\mathbf{x}, g(\mathbf{x}))=0$ Furthermore the function $g$ is differentiable. Furthermore, there is a change of coordinates such that assigns $g(\mathbf{x}) = \mathbf{0}$.

My question is in regards to the following example from the text:

Consider a $C^{\infty}$ function $F$ with the domain $A \in \mathbb{R}^{n}$ and a range in $\mathbb{R}^k$, $k<n$. Consider the set $M = \{\mathbf{x}\in \mathbb{R}^n:F(\mathbf{x}) =\mathbf{0}\}$ Let the rank of the Jacobian of $F$ be equal to $k$ $\forall$ $\mathbf{x}\in M$. $M$ is an $n-k$ dimensional manifold.

They indicate that this is true precisely because in the inverse function theorem there exists the change of coordinates that allows us to set $g(\mathbf{x}) = 0$ above. I don't see why that fact leads to the conclusion that $M$ is a manifold. Am I missing some obvious construction or something?

I tried to construct an alternative argument:

If I consider the function $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^n$ such that $f( \mathbf{x} ) = [x^{1},x^2,x^3,...,x^n, F(\mathbf{x})]$. I have $f\in C^{\infty}$ and defined on $A$, the domain of $F$, and I think its easy to see that for an open subset of $A$ we have the diffeomorphism as in the "definition" above".

Does this argument work?

Answer 1

There is nothing hidden there, all you need is to put the pieces together:

First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.

If $M = \{\mathbf{x}\in \mathbb{R}^n:F(\mathbf{x}) =\mathbf{0}\}$ where $\operatorname{rk}\operatorname{Jac}_{\mathbf{x}}(F)=k$ for all $\mathbf{x}$, then by the implicit function theorem, for every $(\mathbf{a}, \mathbf{b})\in M$ there are open subsets $A\subset \Bbb R^{n-k}$ and $B\subset \Bbb R^k$ such that $\mathbf{a}\in A$ and $\mathbf{b}\in B$ such that:

$M\cap (A\times B)$ is the graph of a differentiable function $g:A\to B$.

Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $\forall x\in A,g(x)=0$. What this really means is that for each $x\in A$ you change the coordinates of $\{x\}\times B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $A\times B$, in other words

There is a diffeomorphism $f:A\times B\to A\times B'$ (for some open subset $B'\subset \Bbb R^k$ which has to contain $0$), such that $f(M\cap (A\times B))=A\times\{0\}$.

Now to translate this into the notations used for submanifolds:

Let $U=A\times B$, $V=A\times B'$, and let $f:A\times B\to A\times B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(U\cap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.