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Unique volume element in a vector space 
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#1
Jun1611, 05:14 PM

P: 70

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 


#2
Jun1611, 05:21 PM

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Hi yifli!
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
Jun1611, 09:37 PM

P: 70




#4
Jun1611, 09:39 PM

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P: 18,040

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.



#5
Jun1611, 09:47 PM

P: 70

What does it mean for [itex]\det(A)[/itex] to be 1? 


#6
Jun1611, 09:51 PM

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