Unique volume element in a vector space

Click For Summary

Discussion Overview

The discussion centers on the uniqueness of the volume element defined by an alternating tensor in the context of vector spaces with orthonormal bases. Participants explore the implications of the determinant of the transformation matrix between two bases and its relation to orientation.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that if the determinant of the transformation matrix A is -1, then the alternating tensor f evaluated at the new basis does not yield a unique value, as it could be -1 instead of 1.
  • Another participant questions the reasoning behind the claim of non-uniqueness, asking what other tensor could also satisfy the condition f(v_1,...,v_n)=1.
  • A clarification is made that f(u_1,u_2,...,u_n) may equal -1, even though f(v_1,v_2,...,v_n) is defined to be 1.
  • It is noted that f will yield 1 for any basis with the same orientation and -1 for those with opposite orientation.
  • Participants discuss the implications of having a fixed orientation and whether it is possible for the determinant to be -1.
  • One participant explains that a determinant of 1 indicates bases with the same orientation, while -1 indicates opposite orientations, leading to a discussion about the nature of orthonormal transformation matrices.
  • It is stated that a transformation matrix with determinant 1 represents a rotation, while one with determinant -1 represents a rotation combined with a reflection.

Areas of Agreement / Disagreement

Participants express differing views on the uniqueness of the volume element and the implications of the determinant's sign. The discussion remains unresolved regarding the uniqueness of f and the conditions under which it holds.

Contextual Notes

Participants rely on definitions of orientation and properties of determinants, but the discussion does not resolve the implications of these concepts on the uniqueness of the volume element.

yifli
Messages
68
Reaction score
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
 
Physics news on Phys.org
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]?
 
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
 
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.
 
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?
 
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.
 
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.
 

Similar threads

MHB Orthonormal Basis times a real Matrix
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Proof that if T is Hermitian, eigenvectors form an orthonormal basis
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Confused about small detail in rank-nullity theorem
  • · Replies 1 ·
Replies
1
Views
3K
MHB Vector Space - Proving Associativity
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Wronskian: null space equals the span of independent functions
  • · Replies 23 ·
Replies
23
Views
2K
High School Simple question about the dual space
  • · Replies 17 ·
Replies
17
Views
3K
Undergrad Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
  • · Replies 3 ·
Replies
3
Views
3K
MHB So V is not vector over field \Bbb{R}
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Proving inequality about dimension of quotient vector spaces
  • · Replies 25 ·
Replies
25
Views
4K
High School Is it invalid to redefine the sgn function in this way in a proof?
  • · Replies 15 ·
Replies
15
Views
5K