Could someone please clarify this Spivak's definition of the volume element and the example with $\det$? How does the underlined text in red follows? Why is $\det$ an example of such an element?
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Related: Volume of $n$-dimensional parallelepiped as determinant – Winther Apr 01 '18 at 21:26
1 Answers
The parts right before the red underline are the justification for what follows. Consider a fixed orientation $\mu$ on $V$, and a given inner product $T$. We may then speak of orthonormal oriented bases. So let $\{v_1, \ldots, v_n\}$ be an orthonormal basis with orientation $\mu$. Then consider some alternating $n$-linear form $\omega$ on $V$ which does not vanish on this basis. Then, normalizing if necessary, we may assume $\omega(v_1, \ldots, v_n) = 1$.
From page 82, of Spivak, if $A$ is the change of basis matrix from a basis $\{v_1, \ldots, v_n\}$ to a basis $\{w_1 = Av_1, \ldots, w_n = Av_n\}$, then $$ \omega(w_1, \ldots, w_n) = \det A \cdot \omega(v_1, \ldots, v_n). $$ Thus, if we single out a $\mu$-oriented basis $\{v_i\}$, it follows by definition that if $\{w_i\}$ is another $\mu$-oriented basis, the change of basis matrix $A$ has determinant +1, so $\omega(w_i) = \omega(v_i)$. Thus, $\omega$ is 1 on every $\mu$-oriented orthonormal basis. If $\eta$ is another alternating $n$-linear form such that $\eta(v_i) = 1$, then $\omega = \eta$ by $n$-linearity.
The determinant is an example of an alternating $n$-linear form which satisfies this uniqueness criterion, since $\det(e_1, \ldots, e_n) = 1$.
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Thank you! Why does $\eta$ agree with $\omega$ on every orthonormal $\mu$-oriented basis? Also, I thought any linear function is determined by its valued on an arbitrary basis. But seems you deduce that $\omega=\eta$ from the fact that their values agree on every orthonormal $\mu$-oriented basis. Is this unnecessary assumption (that $\eta=\omega$ on all orthonormal $\mu$-oriented bases)? – user557 Apr 01 '18 at 23:08
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Sorry about that - I've edited my answer to address this. You're right in that we only need them to agree on a single basis. – Chris Apr 02 '18 at 00:12
