Regular values are useful because of the generalization of the first part of the implicit function theorem: if $q$ is a regular value of $f:M \to N$ (with dimension $m$ and $n$ respectively), then $$A = f^{-1}(q) \subseteq M$$ is a topological manifold of dimension $m - n$.

I am and have been stuck for quite some time on the implicit function theorem as say we have a point $x \in M$, $x$ then has a dimension (or rank?) of $m$. Now if $m \geq n$, $f(x)$ has a dimension, or rank of $n$. Now is this always the case? I am a bit confused here. Either way if we let $f(x) = q$ have a dimension $n$ - would it then be a regular value? But furthermore say $m = 5$ and $n = 3$, why would we have $A$ having a rank of $5 - 3 = 2$ as while $2$ of the "basis" in $$A \subset M$$ would collapse, but then we would still have some $$B \subseteq N$$ for which $$f(A) = B.$$ Hence why wouldn't $A$ have a rank of $3$ in $M$?

I found : What is the 'implicit function theorem'? useful - especially DBr's answer. The mathematics for the Rank Theorem make some sense, I suppose I don't understand what the Rank and hence the Implicit function is really telling us. I know that somehow, due to the implicit function theorem I am able to write a function of $m$ variables in $m-1$ variables or maybe more (I just see, for example $x + y = 1$ is also written as $y = f(x) = 1 - x$, and thus we have reduced a single variable). I don't understand how this is related to the Rank theorem and the Rank of the image being less.

Now with regular values, I understand that they are not the image of critical points, but don't understand how the critical points play in to the Rank and hence Implicit Function Theorem.

I know I probably massacred the above with several things that might be wrong - but I can't seem to figure out how this works together - especially with the implicit function theorem and why we get these strange $m - n$ dimensions. This confusion is at the root of a large mis-understanding of things that I hope to fix. I guess in the end I am really hoping to understand "what is the point" of these two Theorems.

Thanks for any insights.


  • $\begingroup$ Dimensions of manifolds (as in the first paragraph of your question) and ranks of maps (as in DBr's answer that you mentioned) make good sense to me, but I don't understand the material in your second paragraph about a point $x\in M$ having "a dimension (or rank?) of $m$". A point has dimension $0$; rank of a point isn't defined. $\endgroup$ – Andreas Blass Apr 29 '13 at 14:40

I think you have confused rank, nullity and dimension. $M$ has a dimension $m$ (meaning it is locally $\mathbb{R}^m$), $N$ has a dimension $n$. If you have a point $p \in M$ with $f(p) = q \in N$, you can look at the linearization $Df(p)$. This is a linear map on the tangent spaces $T_p M \to T_q N$. You can think of this as being conjugate to a linear map from $\mathbb{R}^m \to \mathbb{R}^n$. The dimension of its kernel is the nullity of $Df(p)$ and the dimension of its image is the rank of $Df(p)$. We also say it is the rank of $f$ at $p$ since there is no ambiguity.

Now, let me give some (IMO excellent) unsolicited advice. If you are stuck with a nonlinear problem, think about the linear case.

What is the inverse function theorem trying to do? if we have a map $f \colon M \to N$ and have $f(p) = q$, we then want to understand $f^{-1}(q)$. In other words, we are looking to solve the nonlinear system of equations given by $f(x) = q$, once we already know that we have a solution ($x=p$).

What is this in the linear case? if $f$ is a linear map from $M = \mathbb{R}^m$ to $N = \mathbb{R}^n$, we are trying to solve a linear system of equations \begin{align} a_{11} x_1 + a_{12} x_2 + \dots + a_{1m} x_m &= q_1 \\ a_{21} x_1 + a_{22} x_2 + \dots + a_{2m} x_m &= q_2 \\ \dots \\ a_{n1} x_1 + a_{n2} x_2 + \dots + a_{nm} x_m &= q_n. \end{align} I will let $A$ denote the matrix representing $f$ in this basis, so we are really solving $A \mathbf{x} = \mathbf{q}$, where we already know that $A \mathbf{p} = \mathbf{q}$. Therefore, we can rewrite this as solving $A(\mathbf{x} - \mathbf{p}) = 0$, and thus this problem is equivalent to finding the null space of $A$.

Now, the rank theorem comes in: $nullity + rank = m$. The null-space of this matrix $A$ then has dimension given by the nullity, which is the same as $m - rank$. The largest you can hope for is that $rank = n$, and this is the scenario that guarantees that for any $p \in N$, you can solve the linear equation.

OK, let's move on to the inverse function theorem. If you have a map $f \colon M \to N$, with additionally $Df(p)$ of rank $n$, then the theorem says that the nonlinear problem is "essentially the same" as the linear problem. If the rank is smaller, then you don't really know what is going on just by looking at the linear equation.

The main point you don't seem to understand is that you must check some property of $Df(p)$ in order to apply the theorem. There are many situations in which this breaks down. Consider, for instance, the map \begin{align} f \colon \mathbb{R} &\to \mathbb{R} \\ x &\mapsto x^2 \end{align} We have $f(0) = 0$, but $Df(0) = 0$ also, so the map does not have maximal rank at $x = 0$, and thus we can't apply the theorem. This is good because we can't locally solve the equation $f(x) = -\epsilon$ for any $\epsilon > 0$.

If, however, we look at $f(2) = 4$ (i.e. in the notation I had, $p = 2$, $q=4$). Then, $Df(2) = 4$, where we see $4$ as the linear operator $\mathbb{R} \to \mathbb{R}$ that multiplies by $4$. This has full rank, so the inverse function theorem tells us that we can solve $x^2 = 4 + y$ for $y$ sufficiently small. (As you well know by easier means.)

This is one of a large class of examples where the following general principle holds: if you are studying a nonlinear problem in a region where the linearized problem is sufficiently non-degenerate, then the linear problem tells you lots about the nonlinear problem. (This is clearly a very vague statement, but is surprisingly useful.)

One example you have seen already in your studies is the second derivative test. Suppose you have a critical point of a function (on $\mathbb{R}$ for simplicity) and you want to tell if it is a local max or a local min. If the second derivative has a sign at the critical point, this allows you to tell, but if the second derivative vanishes, you can't conclude directly.


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