Question: Suppose I have some one-form (linear functional), $f = \alpha \; \omega^1 + \beta \; \omega^2 + \gamma \omega^3$ for some basis forms $\{\omega^i\}$ and coefficients $\{\alpha, \beta, \gamma\} \in \mathbb{R}$. How do I find a vector normal to this one-form?
My thought process: Geometrically, one-forms give a sense of how far in a certain direction a vector "pierces" (e.g. how many isograms span the vector). I know the vector perpendicular to this one-form should scale with the number of isograms it pieces (i.e. if my vector $v$ originally pierces 2 isograms, then $2v$ will pierce 4 isograms. I'm not sure how to quantize this thought process using manipulations with the one-form given above.
You want to compute the kernel of $f$. Let $(e_1,e_2,e_3)$ be dual to $(\omega^1,\omega^2,\omega^3)$ and write $v = v^1e_1+v^2e_2+v^3e_3$. Then $f(v)=\alpha v^1+\beta v^2+\gamma v^3$. This vanishes if and only if $$\alpha v^1+\beta v^2+\gamma v^3=0.$$For instance, if $\gamma\neq 0$, then $$v = v^1e_1+v^2e_2+\left(\frac{-\alpha v^1-\beta v^2}{\gamma}\right)e_3 = v^1\left(e_1-\frac{\alpha}{\gamma}e_3\right)+v^2\left(e_2-\frac{\beta}{\gamma}e_3\right).$$So the kernel/normal space of $f$ has $(e_1 - (\alpha/\gamma)e_3, e_2-(\beta/\gamma)e_3)$ as a basis.