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The standard contact structure on $mathbb R^{2n+1}$ is given by

$$ alpha = dz + sum_{i=1}^n x_i dy_i$$

And the standard contact structure on $S^{2n+1}$ is given by

$$ beta = sum_{i=1}^n x_i dy_i - y_i dx_i$$

For $n=1$ we get $alpha = dz + x dy$ and $beta = xdy - y dx + y dz - z dy$.

Is it possible to use the standard contact structure $alpha$ on
$S^{2n+1}$? For example, does $alpha = dz + x dy$ define a contact structure on $S^1$?

It is not clear to me why it would not. But if it did there would be no need to define a different contact form for the sphere.

As mentioned by A. Hwang, there is no way to build the standard contact form $alpha_{textrm{sph}}$ of the $S^{2n+1}$ from the standard contact structure of $mathbb{R}^{2n+1}$, the reason being that $S^{2n+1}$ is not a subset of $mathbb{R}^{2n+1}$.

However, one can construct $alpha_{mathrm{sph}}$ from the symplectic structure of $mathbb{R}^{2n+2}$ and this is a rather general process.

Definition. Let $(V,omega)$ be a symplectic manifold, then a vector field $X$ of $V$ is said to be a Liouville vector field if and only if the Lie derivative of $omega$ along $X$ is equal to $omega$, namely $mathcal{L}_Xomega=omega$.

Geometric interpretation. The vector field $X$ is Liouville if and only if its local flow expands $omega$.

Remark. Since $omega$ is closed, from the Cartan's formula, this is equivalent to $mathrm{d}(i_Xomega)=omega$.

Now, here is the link with contact geometry:

Theorem. Let $(V,omega)$ be a symplectic manifold, $X$ be a Liouville vector field and $M$ be a hypersurface. Assume that $X$ is transverse to $M$, then $alpha:=i_Xomega$ is contact form on $M$.

Remark. To be strictly correct, one owes to consider $i^*alpha$, where $icolon Mhookrightarrow V$, but this is just a waste of notation since the pullback commutes with the exterior product and derivative.

Remark. In this setting, $mathrm{d}alpha=omega_{vert M}$ and $M$ is said to be an hypersurface of contact-type of $(V,omega)$.

Proof. Let $2(n+1)$ be the dimension of $V$, then $M$ has dimension $2n+1$. Notice that one has:
$$alphawedge(mathrm{d}alpha)^n=i_Xomegawedge w^n=frac{1}{n+1}i_Xomega^{n+1},$$
the last equality is derived by induction on the integer $n$.

Since $omega$ is non-degenerate on $V$, then $omega^{n+1}neq 0$ and since $X$ is transverse to $M$, one has $i_Xomega^{n+1}neq 0$. Therefore, with our computation $alphawedge(mathrm{d}alpha)^nneq 0$, whence the result. $Box$

Let $V=mathbb{R}^{2n+2}$, $omega:=sumlimits_{i=1}^{n+1}mathrm{d}x_iwedgemathrm{d}y_i$, $X=frac{1}{2}sumlimits_{i=1}^{n+1}(x_imathrm{d}x_i+y_imathrm{d}y_i)$ and $M=mathbb{S}^{2n+1}$, then:

• $(V,omega)$ is a symplectic manifold,




• $M$ is an hypersurface of $V$,




• $X$ is transverse to $M$, since if $xin S^{2n+1}$, $langle X_x,xrangleneq 0$, so that $X_xnotin T_xS^{2n+1}$.

Furthermore, one has the following equality:
$$i_Xomega=frac{1}{2}sum_{i=1}^{n+1}(x_imathrm{d}y_i-y_imathrm{d}x_i)=frac{1}{2}alpha_{mathrm{sph}},$$
so that $mathcal{L}_Xomega=omega$ and $X$ is a Liouville vector field of $V$. Whence, $frac{1}{2}alpha_{textrm{sph}}$ is a contact structure on $S^{2n+1}$.