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Why is a (not necessarily linear) mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ area- and orientation preserving iff the determinant of its jacobian is $\pm 1$ ?

(I understand by an area-preserving mapping $f$ a mapping $f$ such that the measure $m(f^{-1}(A))=m(A)$, where $m(\cdot)$ denotes the measure of a measurable set $A$.)

I have no idea how to prove this... but I'd also be happy with a reference.
(I'd also be happy for a sketch of the proof for a less general definition of "area-preserving", where $A$ is not just any measurable set, but a polytope - this definition would work easier with the concept of determinant since the volumen of a polytope is just the absolute value of the determinant of the vector that represent it's edges.)
Googling didn't get me anything.

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    Why not $m(f(A)) = m(A)$? – mvw Jan 21 '15 at 11:09
  • @mvw I just followed along this definition http://mathworld.wolfram.com/Area-PreservingMap.html which at first sight seemed reasonably geometric and intuitive. but ours does too. I think the mathworld definitino may be better, because preimages of sets generally work with all kinds of properties imposed on the sets (like being measurable). Though for a complete technical, in which case these definition are equivalent and which is more general, I unfortunately don't have the time right now. –  Jan 21 '15 at 11:19
  • See http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables. – lhf Jan 21 '15 at 11:24
  • @mvw I guess that definition is stated in the mentioned form because $f^{-1}(A)$ is the largest set which is mapped into $A$. – Janko Bracic Jan 21 '15 at 11:24
  • @lhf So you say the change of variable formulas for integrals proves everything ? –  Jan 21 '15 at 11:50
  • @user36772, yes, though there are probably several technical details to attend to. Note the second theorem, which is about Lebesgue measurable functions. I'm proposing to consider integrating characteristic functions. – lhf Jan 21 '15 at 11:51
  • @lhf so depending on which definition I use for "area-preserving" (see the edit of my question) different version of the change-of-variables theorem come to use ? –  Jan 21 '15 at 11:53
  • As the MathWorld page implies ("absolute value $1$"), a determinant of $-1$ is also area-preserving. – TonyK Jan 22 '15 at 07:57
  • @TonyK Yes, I should have been more specific and should have said "orientation preserving". Gonna update the question. –  Jan 22 '15 at 15:36

1 Answers1

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In dynamical systems or ergodic theory it is preferable to call a map $f:\>X\to Y$ measure preserving (or area preserving when $X$ and $Y$ are surfaces) if $$\mu\bigl(f^{-1}(B)\bigr)=\mu(B)\qquad\forall B\subset Y\ .\tag{1}$$ This allows for functions that are many-to-one to be measure preseving nevertheless. But using this definition the Jacobian determinant need not have absolute value $1$. For instance, the map $$f:\quad S^1\to S^1,\qquad e^{it}\mapsto e^{2it}$$ is measure preserving, but its Jacobian determinant is $=2$.

When $f$ is essentially injective then $(1)$ can be replaced by $$\mu\bigl(f(A)\bigr)=\mu(A)\qquad\forall A\subset X\ .$$ Now it is proven in calculus that when $f$ is essentially injective and $f(A)=B$ then for any reasonable function $g:B\to{\mathbb R}$ one has $$\int_B g(x)\ {\rm d}(x)=\int_Ag\bigl(f(u)\bigr)\>|J_f(u)|\>{\rm d}(u)\ .$$ Putting $g(x):\equiv 1$ here gives $$\mu\bigl(f(A)\bigr)=\mu(B)=\int_B 1\ {\rm d}(x)=\int_A |J_f(u)|\>{\rm d}(u)\ .$$ Here the right hand side can only be $=\mu(A)$ for every $A\subset X$ if $|J_f(u)|\equiv1$.

Christian Blatter
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  • Thanks! This is a great answer. I have two questions though: 1) What would be the most general $g$ that would pass as "reasonable" and the most general notation of integral for that the transformation formula holds ? 2) One commenter mentioned that I expressed myself inaccurately, since $\det =+1$ (I corrected my question now) doesn't just mean area-preserving; it moreover means area- and orientation preserving. You perfectly showed that a function satisfying my above definition of area-preserving has determinant $\pm 1$. [...] –  Jan 22 '15 at 15:46
  • Do you know, if there is geometric definition of "orientation-preserving" other than $det=+1$ (just as there is a geometric definition of "area-preserving", different that defining "area-preserving to mean $\det=\pm 1$), so that I could you that definition and then try to show that that implies $\det =+1$ ? –  Jan 22 '15 at 15:47
  • @user36772: I suggest you formulate this as a separate question, because it is an interesting problem in itself. – Christian Blatter Jan 22 '15 at 15:54
  • Done. For those who are interested: The separate question can be found here: http://math.stackexchange.com/questions/1119360/variants-of-the-change-of-variables-formula –  Jan 25 '15 at 20:15