unique volume element in a vector space

yifli

Given two orthonormal bases [itex]v_1,v_2,\cdots,v_n[/itex] and [itex]u_1,u_2,\cdots,u_n[/itex] for a vector space [itex]V[/itex], we know the following formula holds for an alternating tensor [itex]f[/itex]:
[tex]f(u_1,u_2,\cdots,u_n)=\det(A)f(v_1,v_2,\cdots,v_n)[/tex]
where A is the orthogonal matrix that changes one orthonormal basis to another orthonormal basis.

The volume element is the unique [itex]f[/itex] such that [itex]f(v_1,v_2,\cdots,v_n)=1[/itex] for any orthonormal basis [itex]v_1,v_2,\cdots,v_n[/itex] given an orientation for [itex]V[/itex].

Here is my question:
If [itex]\det(A)=-1[/itex], then [itex]f(u_1,u_2,\cdots,u_n)=-1[/itex], which makes [itex]f[/itex] not unique

micromass

Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes [itex]f(v_1,...,v_n)=1[/itex]?

yifli

Sorry for the confusion. Actually I mean [itex]f(u_1,u_2,\cdots,u_n)[/itex] is not necessarily equal to 1, it may be equal to -1 also, even though [itex]f(v_1,v_2,\cdots,v_n)[/itex] is made to be 1

micromass

Yes, but the point is that f of every basis with the same orientation will give you 1. If you take the opposite orientation, then you will get -1 of course.

yifli

So given a fixed orientation, it's impossible for [itex]\det(A)[/itex] to be -1?

What does it mean for [itex]\det(A)[/itex] to be -1?

micromass

The transition matrix between two orthonormal bases has det(A)=1 if and only if the bases have the same orientation and has det(A)=-1 if they have opposite orientation.

I like Serena

A orthonormal transformation matrix with det(A)=1 is a rotation.
A orthonormal transformation matrix with det(A)=-1 is a rotation combined with a reflection.

So the volume sign would be invariant.