Notational issue: Difference between $dH$ and $\nabla H$ Below is a passage from the book I am reading on Geometric Mechanics (Foundation of Mechanics by Abraham and Marsden) to brush myself up on the subject before beginning research in Hybrid Dynamical Systems (I thought it would be best to directly insert the passage instead of retyping, let me know if this is problematic). If anyone feels this is better suited for physics stack, I will place it there instead, though my question is mathematical in nature.


I believe I understand the material presented here just fine, except for one little issue: I see no difference between $\nabla H(\xi)$ and $dH(\xi)$, and am confused by the change in notation.
From my understanding, $$\nabla H(\xi) = (\partial H/ \partial q^i, \partial H/ \partial p_i)$$
so that $$X_H(\xi) = J \nabla H(\xi) = (\partial H/ \partial p_i, -\partial H/ \partial q^i)$$
thus 
$$\omega(X_H, v) = (\partial H/ \partial p_i, -\partial H/ \partial q^i) J v = <(\partial H/ \partial q^i, \partial H/ \partial p_i), v> = <dH, v>$$
(sorry for poor usage of inner product here, but I believe it is clear what I mean).
Is this correct that, in fact, $\nabla H = dH$. If so, does anyone see a reason for the dual notation here?
 A: The difference between $\nabla H$ and $dH$ is that the former is a vector field, while the latter is covector field (1-form).
The fact that $\nabla H$ is a vector field is transparent when you calculate directional derivative: $D_vH(\xi) = \nabla H(\xi)\cdot v.$ The right hand side wouldn't even be well defined if $\nabla H(\xi)$ weren't a vector. When you write $\nabla H(\xi) = (\frac{\partial H}{\partial x^1}(\xi),\ldots, \frac{\partial H}{\partial x^n}(\xi))$ you are just writing it down in some basis for tangent space at $\xi$ coming from coordinate map $(x^1,\ldots,x^n)$.
On the other hand, $dH$ is a covector field, which means that at given point $\xi$, $dH(\xi)$ is a linear functional acting on tangent space at $\xi$, defined by $dH(\xi)v = v(H)$, or given some coordinate map $(x^1,\ldots,x^n)$ we have $dH(\xi)(\frac\partial{\partial x^i}|_\xi) = \frac{\partial H}{\partial x^i}(\xi).$
So, what's deal here?
Well, you can notice that $\nabla H(\xi)\cdot (\frac{\partial}{\partial x^i}|_\xi) = dH(\xi)(\frac{\partial}{\partial x^i}|_\xi)$ and the difference is the same as when you consider any (real) finite dimensional inner product space $V$ and its dual $V^*$: for every linear functional $f$ there is unique vector $v$ such that $f(w) = \langle v, w\rangle$. When you fix some basis for $V$, you get corresponding dual basis for $V^*$. The same thing happens here: for a given point $\xi$ fix some coordinates $(x^1,\ldots,x^n)$, get a corresponding base $\{\frac{\partial}{\partial x^1}|_\xi,\ldots,\frac{\partial}{\partial x^n}|_\xi\}$ for tangent space at $\xi$, and get its dual basis $\{dx^1|_\xi,\ldots,dx^n|_\xi\}$.
