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Let $U,V$ be two open sets in $R^n$ and $f:U\to V$ proper $C^{\infty}$ map (proper = preimage of compact set is compact). Then we have

$\int f^{*}\omega=\deg(f)\int \omega,$

for $\omega \in \Omega_c^{n}(V)$. How to prove that if $f$ is linear mapping i.e. $f(x)=Ax$ for nonsignular $n\times n$ matrix we have $\deg(A)=sign(\det (A))$?

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    Yes, I have corrected it. Sorry!2012-06-16

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The degree of a linear map is the signum of $\det$, not $\det$ (the degree is integer valued, while $\det$ is real valued -- already this should make you suspicious). (And the answer to your question is, basically, the change of variables formula for the integral, once you replace $\det$ by it's sign).

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    @dmm If you post such a question it is usually hard to answer cause the people reading it do not know how the concepts were introduced to you and what you have available as tools. Maybe you have a standard textbook you are following, then those who know it may be able to help you based on this knowledge. The approaches to differential forms and degree I know all rely on the transformation formula (which reads $\int_{\phi(U)}f(x) d^nx = \int_U f(\phi(x))|\det D\phi(x)|d^n x$, just to make sure we are talking about the same thing).2012-06-16