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Exercise from Guillemin and Pollack:

Let $T$ be the volume element of $\mathbb R^2$ (i.e., the unique $2$-covector $\mathbb R^2\to \mathbb R^2$ that equals $\frac{1}{2!}$ on all positively oriented orthonormal bases. Prove that for any vectors $v_1,v_2\in\mathbb R^2$, $T(v_1,v_2)$ is $\pm$ one half the volume of the parallelogram spanned by $v_1$ and $v_2$. Furthermore, when $v_1$ and $v_2$ are independent, then the sign equals the sign of the basis $(v_1,v_2)$ in the standard orientation of $\mathbb R^2$. Generalize to $\mathbb R^3$. Now how would you define the volume of a parallelepiped in $\mathbb R^k$?

I think I understand how to do this exercise, but still have a couple of qustions.

Is it correct that it is assumed in this exercise that it is known that $\det [v_1 | v_2]$ and is $\pm$ the volume of the parallelepiped spanned by $v_1,v_2$ (and that $\det [w_1|w_2|w_3]$ for $w_i\in \mathbb R^3$ is $\pm$ the volume of the parallelepiped spanned by $w_1,w_2,w_3$)?

What is the definition that the authors want? I can define the volume of the parallelepiped spanned by $u_1,\dots,u_k$ in $\mathbb R^k$ to be the modulus of the volume form on $\mathbb R^k$ evaluated at $(u_1,\dots,u_k)$ times $k!$. Is that what they want?

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  • $\begingroup$ Why does it need to multiply k! ? $\endgroup$ – OrdinaryWitch Apr 17 '18 at 4:52
  • $\begingroup$ @YuanweiChen Because according to Guillemin and Pollack, the volume element of $\mathbb R^k$ equals $1/k!$ (and not just $1$) on all positively oriented orthonormal bases. $\endgroup$ – user500094 Apr 17 '18 at 18:49

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