$\alpha\wedge\beta = 0$ for all $\beta$ implies $\alpha = 0$ without using the Hodge dual

Let $\alpha$ be a differential $k$-form on an orientable smooth $n$-dimensional manifold. If $\alpha\wedge\beta = 0$ for every differential $(n - k)$-form $\beta$, then $\alpha = 0$ because we can choose a Riemannian metric from which we can construct the corresponding Hodge dual $\ast$ and obtain $0 = \alpha\wedge\ast\alpha = \|\alpha\|^2dV$.

Can we deduce that $\alpha = 0$ without choosing a metric and considering the Hodge dual?

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If you allow wedging with $0$-forms, this is trivial: wedge with a constant function. If you do not, this becomes false for $n$-forms (assuming you are on an $n$-dimensional manifold), again for trivial reasons. I think you need to put bounds on the degrees of $\alpha$ and $\beta$. – Aaron Dec 6 '12 at 14:50
@Aaron: Of course you are right, I have changed my question accordingly. – Michael Albanese Dec 8 '12 at 7:26

This is pointwise. Choose a basis of the 1-forms $\omega_1, \ldots, \omega_n.$ Let $I$ denote any subset of $\{1,2,\ldots,n \}$ containing $k$ elements. then let $$\omega_I = \omega_{i_1} \wedge \cdots \wedge \omega_{i_k}.$$ Meanwhile, let $I'$ denote the subset consisting of the other $n-k$ indices, that is $$I \cap I' = \{ \}, \; \; I \cup I' = \{1,2,\ldots,n \}.$$ Why not, let $$\Omega = \omega_{1} \wedge \cdots \wedge \omega_{n}.$$ We get $$\omega_I \wedge \omega_{I'} = \pm \Omega,$$ but if $I \neq J,$ then $I \cap J' \neq \{ \},$ and so $$I \neq J \Longrightarrow \omega_I \wedge \omega_{J'} = 0.$$

Your $\alpha$ is a $k$-form, so $$\alpha = \sum_{I} a_I \omega_I$$ with real numbers $a_I.$ As result, given any fixed $J,$ we are just picking out the $J$ coefficient with $$\alpha \wedge \omega_{J'} = \; \pm \, a_{J} \; \Omega.$$ Your hypothesis that any $(n-k)$ form $\beta$ gives $\alpha \wedge \beta = 0$ can be applied with $\beta = \omega_{J'}.$ So the conclusion is that all the coefficients $a_J =0$ and $\alpha = 0.$

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I see, you did not specify 1-forms. Let me type this in, perhaps you can decide the status of the argument for other $p$-forms.

This is Cartan's Lemma, which is about 1-forms. On a smooth manifold (or small open neighborhood in one) of dimension $n,$ let $p \leq n.$ Let $\omega_1, \ldots, \omega_p$ be 1-forms that are (pointwise) linearly independent. Then let t $\theta_1, \ldots, \theta_p$ be 1-forms such that $$\sum_{i=1}^p \theta_i \wedge \omega_i = 0.$$ Then there are smooth functions $A_{ij} = A_{ji}$ such that $$\forall i \leq p, \; \; \; \theta_i = \sum_{j=1}^p A_{ij} \omega_j \; \; .$$

So, extend your 1-form $\alpha$ to a full basis $\omega_1, \ldots, \omega_{n-1}, \omega_n = \alpha.$ You are saying all wedges are zero, so $$\sum_{i=1}^{n-1} \alpha \wedge \omega_i = 0.$$ Then $$\alpha = \sum_{j=1}^{n-1} A_{ij} \omega_j \; \; .$$ However, $\omega_n = \alpha$ is part of the basis, which is contradicted, so this contradicts the ability to form such a basis and the assumption that $\alpha \neq 0.$

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