Need more help with differential forms The standard contact form on the sphere $S^{2n +1}$ in $\mathbb R^{2n + 2}$ is given by
$$ \omega = \sum_{k=1}^{n+1} x_k dy_k - y_k dx_k$$
(see e.g. here)
Now what I'm confused about is that this form uses all $2n + 2$ coordinates of $\mathbb R^{2n + 2}$ but the sphere is only $2n + 1$-dimensional. 

Question 1: Doesn't one have to restrict this to the sphere?

By that I mean compose with the differential of the inclusion map $i:S^{2n +1}\hookrightarrow \mathbb R^{2n +2}$. It's not clear to me though what $i$ should be: a point $x$ in $S^{2n +1}$ only has $2n+1$ coordinates. 
Question 2:

So do we map an arbitrary coordinte to $0$? But if so, wouldn't then
  $i$ be not defined on all of the sphere (but only chartwise)?

When restricted I expect $\omega$ to become an expression of only $2n+1$ coordinates. Am I on the right track of understanding or is my current understanding all complete nonsense?
I also believe that $\omega$ is globally defined on $S^{2n+1}$. But I don't understand how it's possible since the sphere does not admit a global coordinate system. 
Question 3: 

How is it possible that $\omega$ is nonetheless a globally defined
  differential form?

 A: Consider the standard embedded sphere defined by $S^{2n+1}=\{x\in\mathbb{R^{2n+2}}:\|x\|=1\}$. $\omega$ is obviously globally defined on $\mathbb{R^{2n+2}}$ thus its restriction on $S^{2n+1}$ is a global form.
Note that for a unit normal vector $n=x=\sum x^j\frac\partial {\partial x^j}+y^j\frac\partial {\partial y^j}$, 
$$\omega(n)=\big(\sum x^kdy^k-y^kdx^k\big) \big(\sum x^j\frac\partial {\partial x^j}+y^j\frac\partial {\partial y^j}\big)=\sum x^ky^k-y^kx^k=0$$
That is, $\omega$ is REALLY a form on the tangent space.
A: $\newcommand{\Reals}{\mathbf{R}}$The Cartesian coordinates $x_{1}, \dots, x_{n+1}$, $y_{1}, \dots, y_{n+1}$ are globally-defined smooth functions on the sphere $S^{2n+1}$. As you say, they do not comprise a coordinate system because at each point, one coordinate can be expressed as a smooth function of the others. (Which coordinate is up to you. The only constraint is, if some coordinate is zero at $p$, you cannot solve for that coordinate as a smooth function of the others in a neighborhood of $p$.)
The coordinate differentials $dx_{1}, \dots, dx_{n+1}$, $dy_{1}, \dots, dy_{n+1}$ are therefore smooth $1$-forms. For example, $dx_{k}$ is a smooth $1$-form on $\Reals^{2n+2}$, and by restriction (a.k.a., under pullback by the inclusion map $i$) defines a smooth $1$-form on the sphere $S^{2n+1}$.
Note, incidentally, that while $dx_{k}$ is nowhere vanishing, $i^{*}dx_{k} = 0$ at two points: the standard basis element $e_{k}$ and its negative. That should suffice to convince you that $dx_{k}$ (a $1$-form on $\Reals^{2n+2}$) and $i^{*}dx_{k}$ (a $1$-form on $S^{2n+1}$) are not equal, even though "they're defined by the same formula". The domain matters!
Now, to your questions:


*

*Yes, technically you're letting $i:S^{2n+1} \to \Reals^{2n+2}$ denote the inclusion, and are considering $i^{*}\omega$. The differential of the inclusion satisfies $i_{*}(v) = v$ for every tangent vector to the sphere. That is, if $p \in S^{2n+1}$, then $i_{*}:T_{p}S^{2n+1} \to \Reals_{p}^{2n+2}$ sends each vector $v$ (in the $(2n+1)$-dimensional tangent space $T_{p}S^{2n+1}$) to itself (viewed as an element of $T_{p}\Reals^{2n+2}$).

*No, you would not map an arbitrary coordinate to $0$, but instead would express one coordinate as a function of the others and calculate partial derivatives with respect to those coordinates. When you restrict $\omega$, it is locally dependent on only $(2n+1)$ coordinates. Note carefully, however, that doing so forces you to make an inelegant (and unnecessary) choice, only to get the simple answer of the preceding item.

*As noted at the top, the Cartesian coordinates define smooth, real-valued functions on the sphere, and their differentials define smooth $1$-forms on the sphere. Consequently, the formula
$$
\omega = \sum_{k=1}^{n+1} (x_{k}\, dy_{k} - y_{k}\, dx_{k})
\tag{1}
$$
defines a smooth $1$-form on the sphere. It simply happens not to a representation in local coordinates on the sphere.
The important point is, if you are given an arbitrary point
$$
p = (p_{1}, \dots, p_{n+1}, q_{1}, \dots, q_{n+1}) \in S^{2n+1}
\tag{2a}
$$
and an arbitrary tangent vector
$$
v = (u_{1}, \dots, u_{n+1}, v_{1}, \dots, v_{n+1}) \in T_{p}S^{2n+1},
\tag{2b}
$$
you can compute
$$
\omega(p)(v) = \sum_{k=1}^{n+1} (p_{k}\, v_{k} - q_{k}\, u_{k}).
$$
The following facts are immaterial:


*

*Not every $(2n+2)$-tuple (2a) represents a point of $S^{2n+1}$;

*The $(2n + 2)$ smooth functions in (2a) are not local coordinates on the sphere;

*Not every $(2n+2)$-tuple (2b) represents a tangent vector of $S^{2n+1}$.
A: There seems to be some confusions on this thread.
The point is that $\omega \in \Omega^1 (\mathbb R^{2n + 2})$ is a global differential 1-form on $\mathbb R^{2n + 2}$, but in no case a form on $S^{2n +1}$.
Whatever that form $\omega$ is you can restrict it to $S^{2n +1}$ and obtain a form $i^*\omega \in \Omega^1 (S^{2n +1})$.
But you should not  ask whether $\omega$ "is" a form on $S^{2n +1}$: it is definitely not because a differential form lives on one and only one manifold, and $\omega$ lives on $\mathbb R^{2n + 2},$ period!
It is indeed true that $\omega$ happens to be zero on vectors normal to the sphere but this is irrelevant to the question. 
