Unique volume element in a vector space

In summary, 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.
  • #1
yifli
70
0
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
 
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  • #2
Hi yifli! :smile:
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]?
 
  • #3
yifli said:
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 said:
Hi yifli! :smile:
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]?

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
 
  • #4
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
micromass said:
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 [itex]\det(A)[/itex] to be -1?

What does it mean for [itex]\det(A)[/itex] to be -1?
 
  • #6
yifli said:
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?

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
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.
 

What is a unique volume element in a vector space?

A unique volume element in a vector space is a mathematical concept that represents the smallest unit of volume in a given vector space. It is also known as the determinant or Jacobian determinant.

Why is a unique volume element important in vector spaces?

A unique volume element is important because it allows us to measure the volume of a given region in a vector space. It also helps us to understand the effects of linear transformations on the volume of a space.

How is a unique volume element calculated?

A unique volume element is calculated by taking the determinant of the basis vectors of a given vector space. This can be done using various methods such as the cofactor expansion or the Gaussian elimination method.

Can a unique volume element be negative?

Yes, a unique volume element can be negative. This happens when the basis vectors of a vector space are oriented in a way that results in a negative determinant. This does not affect the magnitude of the volume, but it indicates a change in orientation.

How does a unique volume element relate to linear independence?

A unique volume element is closely related to linear independence. In fact, a set of vectors is linearly independent if and only if their determinant is non-zero, which means that they span a unique volume element in the vector space.

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