How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?


1 Answer 1


"If part:" Suppose that $\omega^1\wedge ...\wedge \omega^k$ is not zero. Suppose that $a_i\in\mathbb{R}$ where $1\leq i\leq k$ such that $$\tag{1}a_1\omega^1+\cdots+a_k\omega^k=0,$$ the zero covector. Take the wedge product with $\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k$ with $(1)$, we get $$a_i\omega_i\wedge\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k=(-1)^{i-1}a_i\omega^1\wedge ...\wedge \omega^k=0,$$ where $1\leq i\leq k$. Therefore, if $\omega^1\wedge ...\wedge \omega^k\neq 0$, we must have $a_i=0$ for $1\leq i\leq k$. By definition, $\omega^1, ..., \omega^k$ are linearly independent.

After typing the "if part", I find that I forgot about the proof of "only if part". Hope that someone can give the full proof.

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    $\begingroup$ for the only if part: if they are independent, extend them to a basis, take $X_i$'s ($i=1\dots n$) to be the dual basis, and compute $\omega^1, ..., \omega^k(X_1,X_2,...,X_k)=1$ as you suggest $\endgroup$
    – user8268
    Dec 30, 2011 at 14:14
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    $\begingroup$ @user8268 would this work for vector spaces of arbitrary dimensions? I encountered this problem in Tu's book in which the finite dimensionality of $V$ has not been assumed. I believe that in the general case, one would first have to extend $\{\omega_{1},\dots, \omega_{k}\}$ to a basis $\{\omega_{j}\}_{j\in J}$ for $V^{*}$, construct the linearly independent set $\{\Phi_{j}\}_{j\in J}$ in $V^{**}$ such that $\Phi_{j}(\alpha_{k}) = \delta_{jk}$, and then obtain $\{v_{j}\}_{j\in J}\subseteq V$ such that $\alpha_{j}(v_{k}) = \delta_{jk}$ using the canonical injection $V\hookrightarrow V^{**}$. $\endgroup$ Jan 27, 2023 at 10:53

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