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I was trying to check a variation of theorem 2.2:

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I did the first part of the proof using the standard contact structure on $\mathbb R^{2n + 1}$ which worked out as expected. Then I wanted to do it for a slight modification of the theorem where I replace Euclidean space with the sphere $S^{2n +1}\subseteq \mathbb R^{2n +2}$ endowed with the standard contact structure on the sphere $\displaystyle \theta = \sum_{k=1}^{n+1} x_i dy_i - y_i dx_i$.

But I couldn't get $\theta = 0$ in that case by just substituting the coordinates from the theorem into the form.

But there is a workaround which suddenly makes everything work out but I don't understand why or how.

Here is what one can do: Note that if $(x_1, \dots, x_n, y_1, \dots, y_n)$ are standard coordinates on $\mathbb R^{2n}$ then on the sphere we can assume that one of them is $1$. Say for example, $x_n=1$. This is because locally the sphere has coordinate charts that just project a plane onto it and that plane may be described by the equation $x_n=1$ for example.

Next, let $z=y_{n}$. This is just a renaming of one of the coordinates so no reason to worry about it.

Substituting $x_{n}=1$ and $z=y_n$ into $\theta$ we now have expressed $\theta$ in local coordinates on the sphere.

If we now substitute the coordinates from the theorem into $\theta$ lo and behold, it turns out to be the zero form as desired.

But I am non the wiser.

Why does it need this in between step of expressing $\theta$ in local coordinates? What exactly does it do?

I wonder whether it is even correct because if it was I do not see why I could not substitute the coordinates from the theorem into $\theta$ directly.

Any help is greatly appreciated. The more detailed the better, as I am very confused.

Edit (in response to Hurkyl)

Here are the calculations I did in the case that did not work out:

We have $\displaystyle \theta = \sum_{k=1}^{n+1} x_i dy_i - y_i dx_i$ and

$$ x_i = x_i, x_j = - {\partial f \over \partial y_j}, y_j = y_j, y_i = {\partial f \over \partial x_i}, z = f - \sum_{i\in I}x_i {\partial f \over \partial x_i}$$

So I substituted in to $\theta$:

$$\begin{align} \theta &= \sum_{i \in I}x_i d\left({\partial f\over \partial x_i} \right) - {\partial f \over \partial x_i} dx_i + \sum_{j \in J} -{\partial f \over \partial y_j} dy_j - y_j d\left(-{\partial f \over \partial y_j} \right) \end{align}$$

I then calculated the parts separately:

$$\begin{aligned}&\sum_{i \in I}x_i d\left({\partial f\over \partial x_i} \right) - {\partial f \over \partial x_i} dx_i =\\ &=\sum_{i\in I} x_i \sum_{k \in I} {\partial^2 f\over \partial x_i \partial x_k} dx_k + \sum_{i\in I} x_i \sum_{j \in J} {\partial^2 f\over \partial x_i \partial y_j} dy_j - \sum_{i\in I} {\partial f \over \partial x_i}dx_i \end{aligned}$$

$$\begin{aligned} & \sum_{j \in J} -{\partial f \over \partial y_j} dy_j - y_j d\left(-{\partial f \over \partial y_j} \right)=\\ &- \sum_{j \in J}{\partial f \over \partial y_j} dy_j + \sum_{j\in J} y_j \sum_{i \in I} {\partial^2 f \over \partial y_j \partial x_i} dx_i + \sum_{j\in J} y_j \sum_{k \in J} {\partial^2 f \over \partial y_j \partial y_k } dy_k \end{aligned}$$

It seems to me that there is no factor of two that I am missing.

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I speculate your problem is that you didn't take advantage of the fact that on spheres centered on the origin, we have an identity

$$ \sum_i x_i^2 + y_i^2 = r^2 $$

and consequently

$$ \sum_i 2 x_i \mathrm{d} x_i + 2 y_i \mathrm{d} y_i = 0 $$

is a linear dependence between the differential forms involved.

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  • $\begingroup$ I added my calculations to my question. I don't see where I made a mistake but I don't get the factor of $2$ that you suggest in your answer. $\endgroup$ – self-learner Jun 28 '16 at 3:05

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