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I was trying to solve a problem in differential geometry that I realized the following statement is the core of my argument

Let $f: U\subseteq \mathbb{R}^n \to \mathbb{R}^m$ be a $C^1$ function on an open set $U$ where $n \leqslant m$ such that $\mathrm{rank}{Df}=n$ at some $p\in U$. Show that $f$ is injective in a neighborhood of $p$.

After thinking about it, I think that it can be proven using the constant rank theorem. Firstly, since $f$ is $C^1$, we have $\mathrm{rank}Df\geq n$ in a neighborhood of $p$. Since $n$ is the maximum possible rank, we have $\mathrm{rank}Df = n$ near $p$. So, the constant rank theorem applies.

Now the constant rank theorem says that I can find two open sets $V \subseteq U$ and $W\subseteq \mathbb{R}^m$ such that $f(V) \subseteq W$ and two diffeomorphisms $\psi:\mathbb{R}^n \to V$ and $\varphi:\mathbb{R}^m \to W$ such that $\varphi^{-1}\circ f\circ \psi: \mathbb{R}^n \to \mathbb{R}^m$ has the canonical form $(x_1,\cdots,x_n) \mapsto (x_1,\cdots,x_n,0,\cdots,0)$.

Since $\varphi^{-1}\circ f\circ \psi$ is clearly injective, and $\varphi$ and $\psi$ are diffeomorphisms, $f = \varphi \circ \big(\varphi^{-1}\circ f\circ \psi \big) \circ \psi^{-1}$ is injective on $V$.

Assuming my proof is correct (well, is it?) I still think it's overkill. Is there a proof that is more elementary? Ideally, a proof without using the Inverse Function Theorem. Or if it uses the Inverse Function Theorem, it should not be longer than this one since the constant rank theorem can be proven using the Inverse Function Theorem and hence, it's obvious that a longer proof exists.

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    $\begingroup$ To apply the constant rank theorem the rank needs to be ... constant. What if $D_pf$ fails to have rank $n$ at points inside every neighborhood of $p$? $\endgroup$ – Umberto P. Mar 1 at 8:45
  • $\begingroup$ @UmbertoP. Good point. Then let's assume that $D_pf=n$ in a neighborhood of $p$. I will edit my question accordingly. $\endgroup$ – stressed out Mar 1 at 8:46
  • $\begingroup$ I pretty much thought the same thing. It's all right. $\endgroup$ – Sou Mar 1 at 8:47
  • $\begingroup$ @UmbertoP. By the way, I just remembered that if $\mathrm{rank}D_p f = n$ then at least in some neighborhood $\mathrm{rank}D_p f \geq n$. Isn't this true? Because I can find an invertible $n\times n$ minor in the matrix and if its determinant is non-zero, I still have some space to wiggle before it gets $0$? $\endgroup$ – stressed out Mar 1 at 8:52
  • $\begingroup$ Well, you have to assume $f$ is $C^1$ anyway, and then your last comment applies. $\endgroup$ – Amitai Yuval Mar 1 at 9:16
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I don't think you need all that. The differential $Df_p$ is injective, and the question is how to measure this injectivity and transfer that to $f(x) -f(p)$ which is only approximated by $Df$ near $p$. Here is one way:

Take ball around $p$ such that for any $\vec{v}$ with $\|\vec{v}\|=1$ and any $q$ in this ball we have $Df_p(\vec{v})\cdot Df_q(\vec{v})>0$ (Such neighborhood exists since $Df_p(\vec{v})\cdot Df_p(\vec{v})>m$ for some positive $m$, the sphere $\|\vec{v}\|=1$ is compact, and the map $(\vec{v}, q) \to Df_p(\vec{v})\cdot Df_q(\vec{v})$ is continuous; this implies that the function $q\mapsto \min_{\vec{v}}\big\{\, Df_p(\vec{v})\cdot Df_q(\vec{v})\,\big| \,\,\|v\|=1\,\big\}$ is continuous in $q$, and hence positive near $p$.)

Now suppose $p_1$ and $p_2$ are in this ball. We will show that $f(p_1)\neq f(p_2)$ by showing that $f(p_2)\cdot \vec{w}> f(p_1)\cdot \vec{w}$ for a well-chosen $\vec{w}$.

In fact, let $\frac{p_2-p_1}{|p_2-p_1|}=\vec{v}$. We take $\vec{w}=D_p f (\vec{v})$.

Now, take a unit speed straight line segment $\gamma(t)$ from $p_1$ to $p_2$, so that $\gamma'(t)=\vec{v}$. It is enough to show $(f(\gamma(t))\cdot \vec{w})'= (f(\gamma(t))'\cdot \vec{w}>0$ for all $t$.

By chain rule $D(f\cdot\gamma)= D f_{ \gamma(t) } (\vec{v})$. On the other hand, the path lies inside the ball neighborhood (the ball is convex), so $D f_{ \gamma(t) } (\vec{v}) \cdot \vec{w}=D f_{ \gamma(t) } (\vec{v})\cdot D f_{p} (\vec{v}) >0$. This completes the proof.

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  • $\begingroup$ Beautiful answer. (+1) I understand the argument, but I need more time to get comfortable with the geometric intuition. Now we can use the inverse function theorem to further show that $f: U \to f(U)$ is locally a diffeomorphism. Am I right? This seems true from the constant rank theorem, I wonder if it's true here too. $\endgroup$ – stressed out Mar 1 at 21:59
  • $\begingroup$ I edited your answer to make some changes to your notations. Please check that I haven't made any mistake. Feel free to roll back your answer if you didn't like the edit. $\endgroup$ – stressed out Mar 1 at 22:05
  • $\begingroup$ @stressedout Geometric intuition is not clear from the answer because as usual in math, it is written partially from unprocessed intuition and partially "backwards". The idea I think is as follows: pretend the map is $Df_p$ throughout. Then $f(p_2)-f(p_1)=Df_p(p_1-p_2)$; to detect this is non-zero take it's "component" in the direction of itself. This suggests that even for $f$ itself we should have near $p$ the inequality $[f(p_2)-f(p_1)]\cdot Df_p(p_1-p_2) >0$; and we do. $\endgroup$ – Max Mar 2 at 12:40
  • $\begingroup$ @stressedout As for diffeomorphism, you'd need the fact that the inverse is smooth (you can perhaps bootstrap from differentiable), which I don't think follows in any direct way (and would be reproving that part of the inverse function theorem in any case). The inverse function theorem does not apply directly either, but only after projection, which is the content of Christian Blatter's answer (and of my comment on the surface case in another question) and is indeed the standard way to deal with this. Injectivity just happens to be a part one can prove directly. $\endgroup$ – Max Mar 2 at 12:51
  • $\begingroup$ Yeah. I meant homeomorphism actually.My bad. I was rather thinking about using something like invariance of domain to conclude that surjectivity holds too. $\endgroup$ – stressed out Mar 2 at 15:42
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You cannot do without some form of the inverse function theorem.

As ${\rm rank}\bigl(Df(p)\bigr)=n$ the matrix $\bigl[Df(p)\bigr]$ has an $(n\times n)$-submatrix with nonvanishing determinant; say $$\det \left[{\partial f_i\over\partial x_k}(p)\right]_{1\leq i\leq n, \>1\leq k\leq n}\ \ne0\ .$$ Let $f(p)=:q\in{\mathbb R}^m$, and let $\pi$ be the projection of the full $y$-space ${\mathbb R}^m$ onto its $(y_1,\ldots, y_n)$ coordinate plane $Y'$. Then the map $$f':=\pi\circ f:\quad{\mathbb R}^n\to Y',\qquad (x_1,\ldots, x_n)\mapsto\bigl(f_1(x),\ldots, f_n(x)\bigr)$$ has Jacobian matrix $$\left[{\partial f_i\over\partial x_k}(p)\right]_{1\leq i\leq n, \>1\leq k\leq n}$$ at $p$. The inverse function theorem then implies that $f'$ maps a neighborhood $V$ of $p$ injectively onto a neighborhood $V'$ of $q':=\bigl(f_1(p),\ldots, f_n(p)\bigr)$. This immediately implies that $f$ is injective on $V$.

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  • $\begingroup$ Thank you.This is concise and clear. But can you also do the calculation for the chain rule to show that $Df'$ is invertible near $p$? I tried to do it, but I don't see how the invertible submatrix comes into play. I know that such a submatrix exists, but I think $\pi$ must project $\mathbb{R}^m$ onto the columns of this submatrix. Or am I mistaken? Also @Max has proposed a solution which looks correct to me but I don't see where it uses the inverse function theorem. Can you check it to see if it uses the inverse function theorem or not? $\endgroup$ – stressed out Mar 1 at 21:11
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    $\begingroup$ See my edit. Hope it's clearer now. Max has reproved the injectivity part of the inverse function theorem. That's fine. $\endgroup$ – Christian Blatter Mar 2 at 9:10
  • $\begingroup$ Thank you. So, you're just taking the first $n\times n$ sub-matrix? Doesn't ${\rm rank}\bigl(Df(p)\bigr)=n$ tell us that there must be a submatrix of size $n\times n$? Or does it say that every submatrix of size $n\times n$ must be invertible? My confusion arises from the fact that your indexes range over the first $n$ rows and columns so it seems like you're choosing the top-left submatrix no matter what. $\endgroup$ – stressed out Mar 2 at 15:45
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    $\begingroup$ That's why I said "say". There is a submatrix consisting of $n$ rows of the full $Df(p)$ matrix. For simplicity I assumed these are the first $n$ rows. $\endgroup$ – Christian Blatter Mar 2 at 16:01
  • $\begingroup$ Thank you. I totally missed "say". $\endgroup$ – stressed out Mar 2 at 16:14

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