How to calculate the dual vector field of a given vector field? For example, given the vector field $V = x\, \frac  \partial  {\partial z} + y \frac \partial {\partial x} $ in $(x,y,z) \in \mathbb{R}^3$, what is its dual one-form field?
Thanks!
 A: In general one may try to find a dual form $\alpha_X$ of a vector field $X$ by hand, imposing $\alpha_X(X) = 1$ and $\alpha_X(Y)$ for any $Y$ which is not proportional to $X$. In other words, you may try to mimic the property of dual forms as in vector spaces. However, in this way it might happen that $\alpha_X$ is not defined on the entire starting manifold. 
For example: in your case you may consider $g$ as the standard metric on $\mathbb{R}^3$. Endow $\mathbb{R}^3$ with global coordinates $(x^1,x^2,x^3) = (x,y,z)$. If you impose that $\alpha_X = \sum_{i=1}^3 \alpha_i(x^1,x^2,x^3)dx^i$ is 1 when evaluated on your $V$ and is 0 in the complement (with respect to $g$) of Span($V$), you end up with
$$\alpha_X = \frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dz.$$
You immediately see that this is not defined on all of $\mathbb{R}^3$, since it is not defined at the origin.
You can avoid this problem using the musical isomorphism $\flat$, which is a map between the tangent bundle of $\mathbb{R}^3$ and its cotangent. If this does not sound familiar, you may think that $\flat$ takes a vector field on $\mathbb{R}^3$ in input, and returns a dual form defined on $\mathbb{R}^3$. $\flat$ is defined through the metric $g$ as follows: $\flat(X) \mapsto X^{\flat}$, where $X^{\flat}$ is a 1-form now, and it acts on vector fields like this: $X^{\flat}(Y) = g(X,Y)$. 
Using this definition you can compute $V^{\flat} = xdz + ydx$. In fact $V^{\flat}(V) = x^2+y^2 = g(V,V)$. Further, if $U = \sum_{i=1}^3 f_i(x,y,z)\frac{\partial}{\partial x^i}$, then $V^{\flat}(U) = g(V,U)$ (check!).
