TITLE: Showing a vector field is tangent to the 2-sphere
QUESTION [11 upvotes]: How would I go about showing that the vector field $X = x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}$ is tangent to the unit sphere in $\mathbb{R}^3$? I can see it pictorially but I'm not sure what would be required to actually prove it. 
(This isn't homework! I understand the definitions of vector field as either a derivation of $C^\infty(M)$ or as a section of the tangent bundle $TM$, but I don't know how to compute anything about it so I'm trying to work through some examples.)

REPLY [11 votes]: Here's how to make the argument from the comments a little more rigorous from basic properties of tangent bundles.  Let $i: S^2 \to \mathbb R^3$ be the inclusion.  For every $x \in S^2$ this induces an injection $i_* T_xS^2 \to T_x \mathbb R^3$.  Then asking if a vector field $X$ on $\mathbb R^3$ is tangent to $S^2$ is the same as asking if for all $x \in S^2, X_x \in i_* T_x S^2$.  Now consider the map $f: \mathbb R^3 \to \mathbb R, (x,y,z) \mapsto x^2 + y^2 + z^2$.  Now $f \circ i : S^2 \to \mathbb R$ is constant so that $f_* \circ i_* = 0$.  In other words, $i_* TS^2 \subset \ker f_*$.  Since $f_* \ne 0$ on $S^2$, by dimensionality reasons we must have that $i_* TS^2 = \ker f_*$.  Finally, a vector field is in $\ker f_*$ is equivalent to $Xf = 0$ since $Xf = df(X) = f_* X$.