Need more help with differential forms

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\Reals}{\mathbf{R}}$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:

1. 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 $T_{p}\Reals^{2n+2}$).

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.

3. 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 $S^{2n+1}$.

Answer 3

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.