I have the one form : $dz + x\, dy$.
If I want to evaluate this on a vector field, say $-\partial_{z}$, how do I do it?
I don't understand what 1-forms defined with an exterior derivative do to vector fields. An answer for general one-form and/or a resolution for the mentioned case would be very helpful.
Thanks in advance.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$First, $(dz + x\, dy)(-\dd_{z}) = -1$.
Generally, working in $\Reals^{3}$ for concreteness, if $(x, y, z)$ denotes some coordinate system, $(\dd_{x}, \dd_{y}, \dd_{z})$ denotes the associated coordinate vector fields, and $(dx, dy, dz)$ denotes the coordinate $1$-forms, then $$ dx(\dd_{x}) = \frac{\dd x}{\dd x} = 1,\qquad dy(\dd_{y}) = \frac{\dd y}{\dd y} = 1,\qquad dz(\dd_{z}) = \frac{\dd z}{\dd z} = 1, $$ and all other evaluations are zero.
To handle general forms and/or vector fields, "extend using bilinearity over smooth functions". Concretely, if $\omega = f_{1}\, dx + f_{2}\, dy + f_{3}\, dz$ is a general $1$-form, then $$ \omega(\dd_{x}) = f_{1},\qquad \omega(\dd_{y}) = f_{2},\qquad \omega(\dd_{z}) = f_{3}. $$ If $v = v_{1}\, \dd_{x} + v_{2}\, \dd_{y} + v_{3}\, \dd_{z}$ is a general vector field, then $$ \omega(v) = f_{1} v_{1} + f_{2} v_{2} + f_{3} v_{3}. $$