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(This question relates to my incomplete answer at https://math.stackexchange.com/a/892212/168832.)

Is the following true (for all n)?

"If $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable and satisfies $\det(f'(x)) = 0$ for all $x$, then $f$ is not injective."

If so, what's the most elementary proof you can think of? It is clearly true for $n=1$. In lieu of a proof for the general case, I'll accept answers for other small $n$.

I have a simple intuitive argument: pick a path in $\mathbb R^n$ such that at each point of the path, it points along some vector in the kernel of $Df$ at that point. (Remember, $\det(f'(x)) = 0$ for all $x$.) Now take the integral of the directional derivative (along the curve) of $f$ over the curve. It describes a difference between two values in the range of $f$, and it should come out to zero (QED). I have a problem showing that such a path exists and is suitable for the purpose described.

Note: "elementary" means stuff that comes before chapter 3 in Spivak's "Calculus on Manifolds". However, note also that Spivak seems to assume that integrals and elementary facts about functions in one variable are available to the reader. Here's a list of things that were not covered in Spivak at the point where the problem came up: constant rank theorem, implicit function theorem. The inverse function theorem was introduced in the same chapter as the problem was given, so that would be ok to use.

Note: This is not quite the initial problem from Spivak, and does not necessarily need to be proved to solve the original problem (see the link).

Community
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aleino
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  • The constant rank theorem is one way. Elementary, but not necessarily simple. Presumably the inverse/implicit function theorem is on the table? – copper.hat Aug 17 '14 at 06:37
  • My intuition tells me that your intuitive argument will be difficult, mostly because we have no way of knowing how nasty the 'path along the kernel vectors' is. – Christian Chapman Aug 17 '14 at 06:59
  • @enthdegree Yeah. It seems that the results that would allow me to use such a path are not as elementary as I'd want. – aleino Aug 17 '14 at 07:08
  • If your function is smooth, though, or at least has a few more differentials, then your argument works fine. – Amitai Yuval Aug 17 '14 at 07:35

2 Answers2

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Very **non**elementary proof:

If $f$ were injective, it would be an open map by invariance of domain.

But by Sards theorem, the set of critical values (in this case, this is equal to the range of $f$) is a null-set, contradiction.

EDIT: Note that Sard's theorem is indeed applicable, because $f : \Bbb{R}^n \to \Bbb{R}^n$, see http://en.m.wikipedia.org/wiki/Sard's_theorem

PhoemueX
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  • I added an explanation of what I mean by "elementary", but I think you managed to post this answer before my edit appeared. (I also added an intuitive argument that might yield an elementary proof if I can just work out one detail.) – aleino Aug 17 '14 at 06:32
  • Sard's theorem is definitely off the table, but thanks anyway. :) – aleino Aug 17 '14 at 06:33
  • I'm not sure that Sard's theorem works here. It does, of course, when $f$ is smooth, but not if it is only $C^1$. – Amitai Yuval Aug 17 '14 at 07:36
  • @Yuval: It works, because the dimensions match, i.e. $f : \Bbb{R}^n \to \Bbb{R}^n$, see http://en.m.wikipedia.org/wiki/Sard's_theorem – PhoemueX Aug 17 '14 at 08:34
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One way of proving this is to use relevant parts of the constant rank theorem proof.

(Frankly, I find the constant rank theorem proof a little opaque, but the following observation works for me!)

The proof hinges on the following observation: Suppose the matrix $M$ has rank $r$, and $M$ can be partitioned as $M =\begin{bmatrix} A & B \\ C & D \end{bmatrix}$ where $A$ is an invertible $r \times r$ matrix. If we compute $M' = \begin{bmatrix} I & 0 \\ -C A^{-1} & I \end{bmatrix} M = \begin{bmatrix} A & B \\ 0 & D-C A^{-1} B \end{bmatrix}$, we see that $M'$ must still have rank $r$, and hence we must have $D-C A^{-1} B = 0$.

Suppose $r = \max_x \operatorname{rk} Df(x)$. By hypothesis we have $r < n$. Let $\hat{x}$ be such that $\operatorname{rk} Df(\hat{x}) = r$. In particular, $Df(\hat{x})$ must have a $r \times r$ invertible minor, and hence by continuity, we have $\operatorname{rk} Df(x) = r$ for $x$ in some neighbourhood $U$ of $\hat{x}$.

With appropriate choice of permutation matrices $P_1,P_2$, we can have the upper left $r \times r$ block of $P_1 Df(\hat{x}) P_2$ be invertible.

By considering $\tilde{f}(y) = P_1 f(P_2 y)$ at the point $\hat{y} = P_2^{-1} \hat{x}$, we see that $D \tilde{f} (\hat{y}) = P_1 Df(\hat{x}) P_2$, so to reduce notational clutter, I will just assume that the upper left $r \times r$ block of $f$ is invertible. I will also partition $x=(x_1,x_2)$, where $x_1 \in \mathbb{R}^r$. Similarly, I will write $f=(f_1,f_2)$.

Since the rank is constant in a neighbourhood of $\hat{x}$, the above observation shows that ${\partial f_2 (x) \over \partial x_2 } - {\partial f_2 (x) \over \partial x_1 } {\partial f_1 (x) \over \partial x_1 }^{-1} {\partial f_1 (x) \over \partial x_2 } = 0$ for all $x$ in the neighbourhood $U$.

Now let $b = f(\hat{x})$ and consider the equation $f_1((x_1,x_2)) = b_1$. The implicit function theorem gives the existence of a locally defined $\lambda$ such that $f_1((\lambda(x_2), x_2)) = b_1$ for $x_2$ in a connected neighbourhood $U_2$ of $\hat{x}_2$. We have ${ \partial \lambda( x_2) \over \partial x_2} = -{\partial f_1 ((\lambda(x_2), x_2)) \over \partial x_1 }^{-1} {\partial f_1 ((\lambda(x_2), x_2)) \over \partial x_2 }$.

Now consider $g(x_2) = f_2((\lambda(x_2), x_2))$, we have ${ \partial g( x_2) \over \partial x_2} = { \partial f_2((\lambda(x_2), x_2)) \over \partial x_1} { \partial \lambda( x_2) \over \partial x_2} + { \partial f_2((\lambda(x_2), x_2)) \over \partial x_2}$. Combining the above two equations gives ${ \partial g( x_2) \over \partial x_2} = 0 $. Hence $g$ is constant in $U_2$, and so we have $f((\lambda(x_2), x_2)) = b$ for $x_2 \in U_2$.

In particular, $f$ is not injective.

copper.hat
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  • Thanks a lot! I'll need to read trough this more carefully in a few days (a bit short on time at the moment).

    I'll say, though, that the amount of effort (and tricks, like the observation) seems a bit incommensurate with most of the solutions to previous problems in Spivak. (Note also that the problem I stated isn't really in Spivak, and might not be necessary to prove what I was trying to prove initially.)

    Furthermore, the problem arose in the chapter just before the implicit function theorem was introduced, so I was hoping to avoid using that as well. :)

     – aleino Aug 19 '14 at 15:03
  • I hope I'm not being unreasonable here. At this point, I'm just trying to figure out if I missed a chapter in Spivak, or if he jumbled the placement of the problem I was initially trying to solve. (http://math.stackexchange.com/a/892212/168832)

    If no other elementary solutions come up, I'll accept your answer (as soon as I've read it more carefully).

     – aleino Aug 19 '14 at 15:07
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    I think it will be hard to avoid some version of the inverse/implicit function theorem. The earliest relevant exercise I can find in Spivak's "Calculus on Manifolds" (1968 printing) is Problem 2.37 (b) (Show a $C^1$ $f:\mathbb{R}^n \to \mathbb{R}^m$, with $n>m$ is not 1-1). This occurs after the chapter on the inverse function theorem. It is straightforward to modify the above proof to use the inverse function theorem instead (in this context they are essentially equivalent), but I think it obscures what is happening. – copper.hat Aug 19 '14 at 15:42
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    (Also, the observation about matrices above is implicit in all proofs of the constant rank theorem I have read, I think it helps to have it explicitly stated.) – copper.hat Aug 19 '14 at 21:00
  • copper.hat: Yes, the inverse function theorem is available to use.

    I find that using the constant rank theorem would be a very natural way to go, and there is a pretty short proof for it using the inverse function theorem, but I have trouble showing that the rank of the map is continuous. I just noticed that your proof depends on this assumption as well (as you point out).

    Also, if I have the constant rank theorem at my disposal, then there'd be no need to reduce my original problem (in the link), to the problem in this thread.

     – aleino Sep 25 '14 at 14:15
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    @aleino: The rank is not continuous, but if the rank is $r$ at some point, then by continuity of $\det$, the rank must be $\ge r$ in some neighbourhood. In the above proof, we have assumed that $r$ is the $\max$ rank, hence the rank is exactly $r$ in this neighbourhood. Personally, I think of the implicit function theorem (actually Newton's method) as basic from which I derive the rest... – copper.hat Sep 25 '14 at 14:49
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    Aha! So the reasoning goes something like: if the rank is r at x, then there is an r-by-r minor of f'(x) with positive determinant. Since the determinant is continuous, the determinant of this minor must be positive in a neighbourhood, and so the rank cannot decrease in this neighbourhood. (?) – aleino Sep 25 '14 at 15:39
  • @aleino: That's it exactly. – copper.hat Sep 25 '14 at 15:41
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    I think the key observation in the constant rank theorem is the observation connecting the rank and the fact that $D-C A^{-1}B = 0$. I have never seen this made explicit in texts. – copper.hat Sep 25 '14 at 15:43