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Can anyone give a nonlinear regular function from C^2 to C^2 with a constant nonzero Jacobian? It seems to me that the only such functions are linear.

According to the Jacobian conjecture, a function from C^2 to C^2 with a constant nonzero Jacobian must have an inverse.

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You apparently will have to clarify: did you mean a constant Jacobian matrix or a constant Jacobian determinant? Those two are related, but different. – J. M. Dec 15 '10 at 1:33
I meant Jacobian determinant. – Craig Feinstein Dec 15 '10 at 1:35
up vote 3 down vote accepted

Some trial and error should get it.

I figured $\left[\begin{array}{cc} x & 1\\ x-1 & 1\end{array}\right]$ was a good target Jacobian matrix, and so got $f(x,y)=(\frac 1 2 x^2 + y,\frac 1 2 x^2-x + y)$. Is that what you're looking for? Evidently it has an inverse, $g(a,b)=(a-b,a-\frac 1 2(a-b)^2)$.

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Yes, that's what I am looking for. Evidently, I was wrong when I thought they couldn't exist. Now, I'll ask what is the most general form? – Craig Feinstein Dec 15 '10 at 1:38

If $u(x,y)$ is of the form $af(x-y)+bx$, where $a$ and $b$ are constants and $f$ is a one-variable differentiable function, and if $v(x,y)=y-x$, then $$ \frac{{\partial u(x,y)}}{{\partial x}} \frac{{\partial v(x,y)}}{{\partial y}} - \frac{{\partial u(x,y)}}{{\partial y}}\frac{{\partial v(x,y)}}{{\partial x}} = [af'(x-y)+b]+[-af'(x-y)] = b. $$

EDIT: More simply, letting $u(x,y)=af(x+y)+bx$ and $v(x,y)=x+y$ gives the same result as above.

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Any linear function has constant Jacobian determinant, as does any map of the form $(z_1,z_2) \rightarrow (z_1, z_2 - f(z_1))$ or $(z_1,z_2) \rightarrow (z_1 - f(z_2), z_2)$. As a result, any finite composition of maps of these forms will have constant Jacobian. These include the examples that Shai Covo and Andrew Marshall listed. I forget if there are known examples outside this category.

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