Unique volume element in a vector space

[yifli]

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

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

Here is my question:
If \det(A)=-1, then f(u_1,u_2,\cdots,u_n)=-1, which makes f 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 f(v_1,...,v_n)=1?

 

[yifli]

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

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

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

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.

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

What does it mean for \det(A) to be -1?

 

[micromass]

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

What does it mean for \det(A) to be -1?

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.