📂Geometry

Differentiable Vector Fields on a Differentiable Manifold

Definition¹

Let’s call $M$ a differentiable manifold. The set of all differentiable vector fields on $M$ is denoted as $\frak{X}(M)$.

$$\frak{X}(M) := \left\{ \text{all vector fileds of calss } C^{\infty} \text{ on } M \right\}$$

Explanation

$\frak{X}(M)$ is a module over the ring $\mathcal{D}(M)$ of $\mathcal{D}(M)$. In other words, for a differentiable function $f \in \mathcal{D}(M)$ and a vector field $X \in \frak{X}(M)$, $fX$ is (pointwise) well defined.

$$\begin{align*} (X + Y)(p) &= X(p) + Y(p) \\ fX(p) &= f(p)X(p) \end{align*} \qquad \forall X, Y \in \frak{X}(M),\quad \forall f \in \mathcal{D}(M)$$

Both $X(p)$ and $Y(p)$ are elements of the vector space $T_{p}M$, so their sum is well defined. Since $f(p) \in \mathbb{R}$ and $X(p) \in T_{p}M$, their product is also well defined.

Moreover, a vector field is itself a differential operator, hence the following product rule applies. For $X, Y \in \frak{X}(M), f \in \mathcal{D}(M)$,

$$X(fY) = X(f)Y + fXY$$

This can be easily shown by direct calculation. If $X = a_{i}\dfrac{\partial }{\partial x}_{i}$ and $Y = b_{j}\dfrac{\partial }{\partial x}_{j}$,

$$\begin{align*} X(fY) &= a_{i}\dfrac{\partial }{\partial x_{i}}\left( fb_{j}\dfrac{\partial }{\partial x_{j}} \right) \\ &= a_{i}\dfrac{\partial }{\partial x_{i}}\left( fb_{j}\dfrac{\partial }{\partial x_{j}} \right) \\ &= a_{i}\dfrac{\partial f}{\partial x_{i}} b_{j}\dfrac{\partial }{\partial x_{j}} + a_{i}f\dfrac{\partial b_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + a_{i}f b_{j}\dfrac{\partial^{2} }{\partial x_{i}\partial x_{j}}\\ &= a_{i}\dfrac{\partial f}{\partial x_{i}} b_{j}\dfrac{\partial }{\partial x_{j}} + f\left( a_{i}\dfrac{\partial b_{j}}{\partial x_{i}}\dfrac{\partial }{\partial x_{j}} + a_{i} b_{j}\dfrac{\partial^{2} }{\partial x_{i}\partial x_{j}} \right)\\ &= X(f)Y + fXY \end{align*}$$

See Also

• Vector field $X : M \to TM$
• Set of differentiable functions on a differentiable manifold $\mathcal{D}(M)$