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]:(adsbygoogle = window.adsbygoogle || []).push({});

[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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# Unique volume element in a vector space

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