Notation $\mathbb{R}${${\partial _x} + y{\partial _z}, {\partial_y}$} I'm reading Ko Honda's notes on Contact Geometry (course page link; notes link) and ran into a bit of notational confusion. When (page 2, after HW4) talking about the standard contact form on $\mathbb{R}^3$ (which is given by $\alpha_0=dz-ydx$), Honda denotes the kernel of this form by:
$ker(\alpha_0)=\xi_0=\mathbb{R}${${\partial _x} + y{\partial _z}, {\partial_y}$}
Is anyone able to make sense of this notation for me? I'm also a little confused on how to compute the kernel of a form (which I've read is a tangent bundle by this link:Is the kernel of a differential form a subset of $M$ or of $TM$?) which might be helpful as well (if someone could clarify how that kernel was calculated given this form).
Thanks in advance!
 A: $\mathbb{R}\{\partial_x+y\partial_z,\partial_y\}$ is another notation for the span over $\mathbb{R}$ of $\{\partial_x+y\partial_z,\partial_y\}$.
To check this is indeed the kernel lets sub in an arbitrary vector.  At any point $p\in \mathbb{R}^3$, $\alpha_0(p)$ is simply a linear functional on $T_p\mathbb{R}^3$.  Therefore, let us consider the vector $$v=a\partial_x+b\partial_y+c\partial_z\in T_p\mathbb{R}^3$$ where $a,b,c\in \mathbb{R}$.  Then $$\alpha_0(p)(v)=(dz-ydx)(v)=dz(v)-ydx(v).$$
Since the one forms $dx,dy,dz$ are dual to $\partial_x,\partial_y,\partial_z$ they simply pull out the corresponding components. Therefore $$\alpha_0(p)(v) =dz(v)-ydx(v) = c-ya.$$  Finally, if $v\in \text{ker}\,\alpha_0$ then $\alpha_0(v)=0$ which would imply $c=ya$.  Therefore, we can write an arbitrary element of the kernel as $$v=a\partial_x+b\partial_y+ya\partial_z=a(\partial_x+y\partial_z)+b\partial_y$$ for arbitrary constants $a,b\in \mathbb{R}$ which is the same as saying $$\text{ker}\,\alpha_0=\text{span}_{\mathbb{R}}\{{\partial_x+y\partial_z,\partial_y}\}=\mathbb{R}\{\partial_x+y\partial_z,\partial_y\}$$
