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Jul 3, 2014 at 19:12 comment added user27920 By the way, this is all quite classical for Riemannian manifolds to equip the line bundle of top-degree differential forms with an induced metric tensor and thereby define in a coordinate-free manner the volume form associated to an orientation (as the unique fibral unit section in the half-line singled out by the orientation).
Jul 3, 2014 at 19:09 comment added user27920 Using the dictionary between quadratic and symmetric bilinear (or hermitian and "conjugate-symmetric" semi-bilinear) forms, the issue of induced form on a tensor or exterior power becomes one for bilinear or semi-bilinear forms, which in turn is given by habitual formulas on elementary tensors or elementary wedge products. I don't know a reference since I figured these things out for myself when the need arose and I always assumed that everyone else did likewise; must be somewhere in Bourbaki (?). Try "tensor algebra, tensor pairings" in Google.
Jul 3, 2014 at 18:37 comment added Mikhail Borovoi @user52824: Thank you! As I expected, my questions reduce to linear algebra. Now what is the induced quadratic (resp., Hermitian) form on $\det(V)$? Could you please either explain or give references? I would be grateful if you could write an answer based on your comment. I will accept it immediately.
Jul 3, 2014 at 15:26 comment added user27920 The answer to Question 1 is: isomorphism classes of triples $(V,F,e)$ where $(V,F)$ is a non-degenerate quadratic space of rank $n$ over $\mathbf{R}$ and $e$ is a vector in $\det(V)$ on which the induced quadratic form has value $(-1)^{q}$; this $e$ is a kind of "orientation". Similar story for the other one. This works similarly over any field (not of char. 2) at all, replacing $(-1)^q$ with whatever value you like among those attained by the induced quadratic form on $\det(V)$ for the initially given space.
Jul 3, 2014 at 12:40 history edited Mikhail Borovoi CC BY-SA 3.0
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Jul 3, 2014 at 12:33 history asked Mikhail Borovoi CC BY-SA 3.0