# Is this intuitive way of thinking about the implicit function theorem correct?

"Intuitive version": Let $f: \mathbb{R}^n \to \mathbb{R}^m$, $n\ge m$, be a smooth map. If, given $q \in \mathbb{R}^m$, for a fixed point $p \in f^{-1}(q)$ one can define a maximum-dimension vector space using the partial derivatives of $f$ then there exists a diffeomorphism between a neighborhood in $f^{-1}(q)$ of $p$ and $\mathbb{R}^{n-m}$ (i.e. one can "smoothly deform" a neighborhood of $f^{-1}(q)$ into $\mathbb{R}^{n-m}$).

When I say "non-degenerate" I mean of maximum possible dimension, which is the maximum possible number of linearly independent vectors which can be defined by the partial derivatives.

One has to specify the point $q$ before specifying the point $p$, because otherwise one would try to construct a diffeomorphism between a neighborhood in $\mathbb{R}^n$ of $p$ and $\mathbb{R}^{n-m}$, which obviously does not make any sense -- we have to specify that the neighborhood is in $f^{-1}(q)$.

Questions:

1. Is this way of thinking about the implicit function theorem correct?

2. Why is $f^{-1}(q)$ of dimension $n-m$ when the dimension of the space spanned by the partial derivatives of $f$ at every point $p \in f^{-1}(q)$ is $m$? If $f^{-1}(q)$ has $m$-dimensional tangent spaces then why should it be an $n-m$ dimensional manifold, and not an $m-$dimensional one?

3. If for one of the points in $f^{-1}(q)$ the derivative of $f$ doesn't have full rank (i.e. we can only define a degenerate tangent space), then does that make that point a singular point of $f^{-1}(q)$?

4. If $p_0$ is the point from the third question, then is $f^{-1}(q) \setminus \{p_0 \}$ a smooth manifold? The example I have in mind is the singular point of a curve, e.g. $(0,0)$ of $y^2 - x^3=0$.

5. What prevents the following scenario: for every point $p \in f^{-1}(q)$, $Df(p)$ has rank $m-1$, so $f^{-1}(q)$ is a manifold of dimension $n-m-1$? Does the constant rank theorem allow this possibility, explaining why the constant rank theorem is a generalization?

• Another way to state this I think using Riemannian geometry is: the fact that $f$ has maximal rank allows us to define an exponential map from a neighborhood in $\mathbb{R}^{n-m}$ to a neighborhood of $p$ in $f^{-1}(q)$, i.e. that $f$ has maximum rank means that it has maximum rank in a neighborhood of $p$ in $\mathbb{R}^n$, so we can take a neighborhood of the projection of $p$ into the kernel of $f$ (i.e. $\mathbb{R}^{n-m}$) and warp it diffeomorphically into a neighborhood of $p$ in $f^{-1}(q)$, thus we can approximate $f^{-1}(q)$ by tangent spaces near every point, i.e. normal coordinates. – Chill2Macht Oct 30 '16 at 16:14

First, I don't like your use of the word nondegenerate. Why not use the same language the rest of the world does and say the derivative mapping $df_p$ has rank $m$?

$f^{-1}(q)$ does not have $m$-dimensional tangent spaces. Think about linear equations. If you have $m$ independent linear equations, they impose $m$ independent conditions on your $n$ variables, and you're left with a linear subspace of dimension $n-m$.

As far as number (3) is concerned, $p$ need not be a singular point. Consider $f\colon\Bbb R^2\to\Bbb R$ given by $f(x,y)=x^2$. $f^{-1}(0)=\{x=0\}$ is perfectly nice, even though the derivative is $0$ at all $p$ in the set. So the answer to (4) is, "Sure, if the derivative has maximal rank at all other points $p\in f^{-1}(q_0)$."

I believe I already gave you a counterexample to (5). The rank theorem only applies when the rank of $df_p$ is constant over (an open subset of) your whole domain (not a proper subset).

• I edited my post to say "maximum dimension" vector space instead of non-degenerate vector space -- you're right that the latter terminology isn't good, because it doesn't even necessarily suggest that $f$ is an immersion. – Chill2Macht Nov 9 '16 at 8:52

I just understood the answer to (2): the partial derivatives don't span the tangent space, they span the space normal to the tangent space. The standard example to think of is the gradient of a level set -- it's always perpendicular to the hypersurface. Even when thinking about derivations, the tangent space is essentially thought of as the orthogonal complement to the span of the partial derivatives. Thus, if the partial derivatives span an $m$ dimensional space because the derivative $Df$ has maximal rank $m$, then the normal space at the point is $m$-dimensional, so the orthogonal complement to the normal space, which is (isomorphic to) the tangent space is $(n-m)$-dimensional.

Update: also I think this fact might be related to the Gauss Lemma in Riemannian geometry and/or normal coordinates -- this isn't obvious from the Wikipedia article, but at least based on my provisional reading of pp. 285-288 in Kuehnel's Differential Geometry, Chapter 7.