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Let $V$ be an $n$-dimensional vector subspace of $\mathbb R^N$ and let $(\alpha^1,\dots,\alpha^n )$ be a basis of $V^\vee$ dual to an orthonormal positively oriented basis of $V$. Prove that the $\alpha^1\wedge\dots\wedge\alpha^n$ is the unique $n$-covector such that $\alpha^1\wedge\dots\wedge\alpha^n(v_1,\dots,v_n)=1/n!$ for all positively oriented ordered orthonormal bases.

So it suffices to check that $$\alpha^1\wedge\dots\wedge\alpha^n(v_1,\dots,v_n)=1/n!$$ for some positively oriented orthonormal basis $(v_1,\dots,v_k)$. (The reason is this question.) So we have (for definitions, see this question) $$\alpha^1\wedge\dots\wedge \alpha^n(v_1,\dots,v_n)=\operatorname{Alt}(\alpha^1\otimes\dots\otimes \alpha^n)(v_1,\dots,v_n)=\\\frac{1}{n!}\sum_{\sigma\in S_n}\operatorname{sgn}\sigma \cdot \alpha^1(v_{\sigma(1)})\dots\alpha^n(v_{\sigma(n)})=\frac{1}{n!}\det[\alpha^i(v_j)]=\frac{1}{n!},$$ where the last equality holds because $\alpha^i(v_j)=\delta_{ij}$ (so the matrix is the identity).

But I have one concern: I never used that $(v_1,\dots,v_n)$ is an orthonormal or even positively oriented. Is my solution incorrect? If so, what exactly is incorrect?

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  • $\begingroup$ Seems I figured out what I was doing wrong. I assumed $(\alpha^i)$ is dual to $(v_j)$; but I should have applied $\alpha^1\wedge\dots\wedge\alpha^n$ to an arbitrary positively oriented orthonormal basis, not just to the one which $(\alpha^i)$ is dual to. This requires one more step, but it's easy to do using math.stackexchange.com/questions/2740307/…. Such a proof will use both the fact that the basis is positively oriented and orthonormal (to conclude that the determinant from the question I cited is equal to one). $\endgroup$ – user500094 Apr 17 '18 at 1:59

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