Using the standard contact structure of $\mathbb R^{2n+1}$ on $S^{2n+1}$?

Question

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.

Answer 1

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 $\mathcal{L}_X\omega=\omega$.

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 $\mathrm{d}(i_X\omega)=\omega$.

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 $\alpha:=i_X\omega$ is contact form on $M$.

Remark. To be strictly correct, one owes to consider $i^*\alpha$, where $i\colon M\hookrightarrow V$, 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: $$\alpha\wedge(\mathrm{d}\alpha)^n=i_X\omega\wedge w^n=\frac{1}{n+1}i_X\omega^{n+1},$$ 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 $i_X\omega^{n+1}\neq 0$. Therefore, with our computation $\alpha\wedge(\mathrm{d}\alpha)^n\neq 0$, whence the result. $\Box$

Let $V=\mathbb{R}^{2n+2}$, $\omega:=\sum\limits_{i=1}^{n+1}\mathrm{d}x_i\wedge\mathrm{d}y_i$, $X=\frac{1}{2}\sum\limits_{i=1}^{n+1}(x_i\mathrm{d}x_i+y_i\mathrm{d}y_i)$ 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 $x\in S^{2n+1}$, $\langle X_x,x\rangle\neq 0$, so that $X_x\not\in T_xS^{2n+1}$.

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