# unique volume element in a vector space

by yifli
Tags: element, space, unique, vector, volume
 P: 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
 Emeritus Sci Advisor PF Gold P: 15,671 Hi yifli! I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes $f(v_1,...,v_n)=1$?
P: 70
 Quote by yifli Here is my question: If $\det(A)=-1$, then $f(u_1,u_2,\cdots,u_n)=-1$, which makes $f$ not unique
 Quote by micromass Hi yifli! I don't quite see why your reasoning implies that f is not unique. What's the other tensor that makes $f(v_1,...,v_n)=1$?
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

Emeritus
PF Gold
P: 15,671

## unique volume element in a vector space

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

What does it mean for $\det(A)$ to be -1?
Emeritus
 Quote by yifli So given a fixed orientation, it's impossible for $\det(A)$ to be -1? What does it mean for $\det(A)$ to be -1?