# Unique volume element in a vector space

1. ### yifli

70
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

2. ### micromass

20,075
Staff Emeritus
Hi yifli!
I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes $f(v_1,...,v_n)=1$?

3. ### yifli

70
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

4. ### micromass

20,075
Staff Emeritus
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.

5. ### yifli

70
So given a fixed orientation, it's impossible for $\det(A)$ to be -1?

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

6. ### micromass

20,075
Staff Emeritus
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.

7. ### I like Serena

6,183
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.

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