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I am interested in obtaining injectivity of a $C^1$ map from the nonvanishing minors of its Jacobian matrix. Here is a brief history of the topic.

In 1953, Samuelson asked the following:

If the upper left-hand principal minors of the Jacobian matrix of a map $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ do not vanish, is it true that $F$ must be injective?

In 1965, Gale and Nikaido gave a counterexample in $\mathbb{R}^2$. In their paper the following is proved

Gale-Nikaido theorem: If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ are positive, then $F$ is injective.

Since then, some effort has been made to weaken the assumption in Gale-Nikaido theorem since the assumption seems to be too restrictive in application. A comprehensive dicussion can be found in T. Parthasarathy, On Global Univalence Theorems, Lecture Notes in Mathematics, Vol. 977, 1983. In the case of polynomial map, this is related to the real version of Jacobian conjecture.

A possible generalization I'm interested in is the following, which seems to be open.

Question: If all the principal minors of the Jacobian matrix of $F: \mathbb{R}^n\rightarrow \mathbb{R}^n$ do not vanish, is $F$ necessarily injective?

In Gale and Nikaido's paper, the case of $\mathbb{R}^2$ was answered in affirmative, the case of $\mathbb{R}^3$ was claimed in affirmative (yet no complete proof seems to be known).

My motivation comes from trying to make a change of variables to globally rectify a curved coordinate system so that Plancherel theorem can be applied. Any information would be appreciated : )

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@Syang: You may also want to take a look at Zorich Theorem (en.wikipedia.org/wiki/Zorich's_theorem): If $f: {\mathbb R}^n\to {\mathbb R}^n$ is a locally-injective quasiregular map, then, for $n\ge 3$, the map $f$ is a homeomorphism. A smooth map $f$ is $K$-quasiregular if $||Df(x)||^n\le K |J_f(x)|$ for all $x\in {\mathbb R}^n$. The assumptions are somewhat different from the ones you are asking, but the conclusion is the same. –  Misha May 7 '12 at 4:34
    
@Misha: Thank you. Zorich's theorem and the results on en.wikipedia.org/wiki/Quasiregular_map are very impressive. Unfortunately the maps I am concerned with are not quasiregular. Their Jocobians collapse when two coordinates coincide whereas their gradients do not. –  Syang Chen May 12 '12 at 17:33
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2 Answers

You might be interested by C. Soulé, M.Kaufman, R.Thomas results (search "multistationarity"). This might seem unrelated to your question but it is in fact related.

Briefly, they study various necessary conditions for a differential equation $dx/dt=F(x)$, $x\in\mathbb{R}^n$ to have several non degenerate stationary points $F(x)=0$.

The conditions depend on a signed "interaction graph" $G(x)$ deduced from the signs in the Jacobian matrix of $F$ at $x$.

By taking the contrapositive, applied to $F-c$ or various other simple transform , you obtain sufficient conditions for $F$ to be injective (assuming non-vanishing jacobian determinant).

Hope this helps.

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It is interesting. But their results don't seem to go beyond the Gale-Nikaido theorem. Maybe I'm missing something, it seems all the sufficient conditions they obtained (in term of properties of graphs associated to the Jacobian) guarantee injectivity via G-N theorem. –  Syang Chen Feb 15 '12 at 3:27
    
Sure, but the obtained criteria seem more diverse than the one in G-N. –  BS. Feb 27 '12 at 14:43
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The answer is NO. Consider the two-dimensional example $F=(f,g)$ with $$ f(x,y) = \sqrt{2}e^{x/2}\cos(ye^{-x}) $$ $$ g(x,y) = \sqrt{2}e^{x/2}\sin(ye^{-x}) $$ The determinant of the Jacobian matrix is 1 at all points $(x,y)$ in the plane. However, $f(0, 2\pi)=f(0,0)$ and $g(0,2\pi)=g(0,0)$, hence the map $F=(f,g)$ is not injective. It is still an open question (the Keller Jacobian Conjecture, 1939) if a polynomial map $F:C^n\rightarrow C^n$ whose Jacobian determinant is identically 1 is injective.

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Marc, the first principal minor in the Jacobian of $F$ is just the $x$-derivative of $f(x,y)$, and I don't believe that this quantity is non-vanishing. If you evaluate the derivative as a function of $y$ at $x=0$ for instance, it vanishes whenever $\tan(y) = \frac{1}{2y}$. –  Vidit Nanda Aug 17 '12 at 17:02
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