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If we have an orientable bundle $E\rightarrow M$, then the transition maps can be adjusted by Gram-Schmidt process to be in $SO(n,\mathbb{R})$. So the determinant bundle $\det E$ is isomorphic to $M\times \mathbb{R}$ with a trivial section $m\rightarrow (m,1)$.

However in the reversed direction it seems not so clear. Suppose we have a global trivialization $f:\det(E)\rightarrow M\times\mathbb{R}$ that associates $e\rightarrow (m,r)$. Consider a local chart at $M$ and the trivialization $\phi_{U}\rightarrow U\times \mathbb{R}^{n}$. $\phi_{U}$ would induce a local trivialization $\psi_{U}:\det(E)_{U}\rightarrow U\times \mathbb{R}$. Then $f\circ \psi_{U}^{-1}$ can be written as $(m,r)\rightarrow (m,f(r))$. However, we still need to show $f(r)=\{\pm 1\}$(in the previous equation). And this is not so clear to me. (Why doesn't any other real number work?) Taubes then asserts that we have $$f_{U'}\circ \det(g_{U',U})\circ f_{U}^{-1}=1, \forall U\cap U'\not=\emptyset$$

I am wondering why this is true (intuitively). We can of course use $\det_{g_{U,U'}}=\psi_{U'}\circ \psi_{U}$, and the above equation becomes nothing but a combination of coordinate transformations. But this does not really tell us anything intuitively.

My thoughts are to associate $E$ with $E^{*}$, and then $\det(E)$ and $\det(E^{*})$ would have coordinate transformation functions that are inverse to each other. But this does not really show why $f=\pm 1$, since any nonzero real number must be invertible.

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Fix a trivialization of the determinant bundle. If $p\in M$, this allows you to divide the set of ordered bases of $E_p$, the fiber of $E$ over $p$ in two subsets: if $(x_1,\dot,x_n)$ is such an ordered basis, we can say it is positive if the image of $(p,x_1\wedge\cdots\wedge x_n)\in \det(E)$ in $M\times\mathbb R$ has positive second coordinate, and negative in the other case (notice the second coordinate will not be zero)

Show that the set of positive bases gives an orientation of $E$.

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Thanks, this helps. –  Bombyx mori Sep 15 '12 at 18:53

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