Q:

$\alpha\wedge\beta = 0$ 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?

A:

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 $\alpha = 0.$ 

