Understanding vector field(s) on $\mathbb{S}^3$.

Question

I was slving the exercises of John Lee's book "Introduction to Smooth Manifolds", where there is an exercise asking us to prove that $\mathbb{S}^3$ is parallelizable. In the hint, the author asks us to consider the vector fields:

$$X_1 = -x\dfrac{\partial}{\partial w} + w \dfrac{\partial}{\partial x} - z \dfrac{\partial}{\partial y} + y \dfrac{\partial}{\partial z},$$

$$X_2 = -y\dfrac{\partial}{\partial w} + z \dfrac{\partial}{\partial x} + w \dfrac{\partial}{\partial y} - x \dfrac{\partial}{\partial z},$$

$$X_3 = -z\dfrac{\partial}{\partial w} - y \dfrac{\partial}{\partial x} + x \dfrac{\partial}{\partial y} + w \dfrac{\partial}{\partial z}.$$

I get the hint and how to use it. What I don't understand is why are the vector fields $4$-dimensional? Isn't $\mathbb{S}^3$ a $3$-dimensional manifold? This is why the tangent vectors should have only $3$ coordinates! I also searched other places on the internet and more or less, everybody uses $4$ coordinates for a vector field on $\mathbb{S}^3$. Could anybody help me understand this?

Answer 1

$S^3$ is often defined as a subset (submanifold) of $\Bbb{R}^4$, so at a point $p\in S^3\subset \Bbb{R}^4$, the tangent space $T_pS^3$ is a subset of $T_p\Bbb{R}^4$ (or atleast there is a very natural inclusion $T_pS^3\hookrightarrow T_p\Bbb{R}^4$). If you denote the coordinates in $\Bbb{R}^4$ as $(w,x,y,z)$, then there is the "standard basis" of vector fields $\frac{\partial}{\partial w},\dots \frac{\partial}{\partial z}$. We have defined three vector fields $X_1,X_2,X_3$ by taking linear combinations of these coordinate vector fields. Your job is to show that

• For every point $p\in S^3$, $X_1(p),X_2(p),X_3(p)$ actually lie in the subspace $T_pS^3$ (note that a-priori, they only lie in $T_p\Bbb{R}^4$).
• For every $p\in S^3$, $\{X_1(p),X_2(p),X_3(p)\}$ is actually a basis for $T_pS^3$.

Your statement about "Isn't $S^3$ a $3$-dimensional manifold. This is why the tangent vectors should have only 3 coordinates" isn't quite right. The issue is that $(\Bbb{R}^4,(w,x,y,z))$ is a coordinate chart for the $4$-dimensional manifold $\Bbb{R}^4$, it is NOT a chart for the submanifold $S^3$. If you instead use a chart $(V,(\xi_1,\xi_2,\xi_3))$ of $S^3$, then sure an element of $T_pS^3$ can be written as a linear combination \begin{align} a\frac{\partial}{\partial \xi^1}(p)+b\frac{\partial}{\partial \xi^2}(p)+c\frac{\partial}{\partial \xi^3}(p) \end{align}

Answer 2

For what it's worth:

Let $\mathrm{inc}:\mathbb S^3\to\mathbb R^4$ be the inclusion. The $X_i$ as defined are sections of the bundle $T\mathbb R^4$. On the level of sets, for each $i$ there is a unique map $Y_i$ such that the following diagram commutes:

$$\begin{array} A\mathbb S^3 & \stackrel{\mathrm{inc}}{\hookrightarrow} & \mathbb R^4 \\ \downarrow{Y_i} & & \downarrow{X_i} \\ T\mathbb S^3 & \stackrel{d\,\mathrm{inc}}{\hookrightarrow} & T\mathbb R^4 \end{array} $$

With $\mathrm{inc}$ also $d\,\mathrm{inc}$ is an embedding, so the $Y_i$ are smooth since $d\,\mathrm{inc}\circ Y_i=X_i\circ\mathrm{inc}$ is smooth. Furthemore the $Y_i$ are linearly independent, since the $X_i$ are linearly independent and the inclusion is an immersion. Therefore the $Y_i$ form a global frame of $T\mathbb S^3$ and so $T\mathbb S^3$ is trivial.

Answer 3

You can parametrize the sphere $S^3$ by polar coordinates but computations will be painful! You can identify $S^3$ to the Lie group $SU(2)$ and find a basis to its Lie algebra. Lie groups are parallelizable (pick any basis of left invariant vector fields).