📂Geometry

Directional Derivatives in Differential Geometry

Definition¹

Let $\mathbf{X} \in T_{p}M$ be a tangent vector, and let $\alpha (t)$ be a curve on the surface $M$. Then, we have $\alpha : (-\epsilon, \epsilon) \to M$ and it satisfies $\alpha (0) = p$. In other words, $\mathbf{X} = \dfrac{d \alpha}{d t} (0)$. Now, let’s say the function $f$ is a differentiable function defined in some neighborhood of point $p \in M$ on the surface $M$. Then, the directional derivative $\mathbf{X}f$ of $f$ in the direction of $\mathbf{X}$ is defined as follows.

$$\mathbf{X} f := \dfrac{d}{dt_{}} (f \circ \alpha) (0)$$

Explanation

The reason we use such notation is that whenever there is a fixed tangent vector $\mathbf{X}$, a unique $\mathbf{X}f$ is determined every time $f$ is given, considering the tangent vector as an operator. Therefore, in differential geometry, we think of “tangent vector = function = differentiation”. In other words, the tangent vector is treated as a functional.

$$\mathbf{X} : \mathcal{D} \to \mathbb{R}, \quad \text{where } \mathcal{D} \text{ is set of all differentiable functions near } p$$

The following notations are commonly used. When denoting the tangent vector as $\mathbf{X}, \mathbf{v}$,

$$\mathbf{X}f,\quad \mathbf{X}_{p}f,\quad \nabla_{\mathbf{v}}f, \quad \mathbf{v}_{p}f, \quad \mathbf{v}_{p}\left[ f \right], \quad \alpha^{\prime}(0)f$$

From the theorem below, such directional derivatives can be represented through the given coordinate chart mapping $\mathbf{x}$.

Theorem

Let $\mathbf{x} : U \subset \mathbb{R}^{2} \to M$ be a coordinate chart mapping, and let $p=\mathbf{x}(0,0) \in M$. Let $(u^{1}, u^{2})$ be the coordinates of $U$. The tangent vector $\mathbf{X} \in T_{p}M$ is expressed as $\mathbf{X} = X^{1}\mathbf{x}_{1} + X^{2}\mathbf{x}_{2}$. Then, the directional derivative of $f$ is as follows.

$$\mathbf{X}f = \sum \limits_{i=1}^{2} X^{i} \dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0)$$

Especially, if the curve $\alpha$ is $\mathbf{X} = \alpha^{\prime}(0)$ and satisfies $\alpha (0) = p$, then $\mathbf{X}f$ does not depend on the choice of any curve.

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In vector analysis,

$$\nabla _{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = \dfrac{\partial f}{\partial x_{1}} u_{1} + \dfrac{\partial f}{\partial x_{2}} u_{2} + \dots + \dfrac{\partial f}{\partial x_{n}} u_{n}$$

the formula is as shown.

Proof

Since $\alpha$ is a curve on the coordinate chart $\mathbf{x}$, let’s denote it as $\alpha (t) = \mathbf{x}(\alpha^{1}(t), \alpha^{2}(t))$. Then, the following holds.

$$(f \circ \alpha) (t) = ( f \circ \mathbf{x} ) (\alpha^{1}(t), \alpha^{2}(t))$$

Consider the above formula as a composition of the function $f \circ \mathbf{x} : \mathbb{R}^{2} \to \mathbb{R}$ and the function that maps as $t\mapsto (\alpha^{1}(t), \alpha^{2}(t))$. Differentiating this, according to the chain rule, results in the following.

$$\begin{equation} \dfrac{d}{dt}(f \circ \alpha)(t) = \dfrac{d}{dt} ( f \circ \mathbf{x} ) (\alpha^{1}(t), \alpha^{2}(t)) = \sum \limits_{i} \dfrac{\partial ( f \circ \mathbf{x} ) }{\partial u^{i}} \dfrac{d \alpha^{i}}{dt} \end{equation}$$

Note that the variable in $\dfrac{\partial ( f \circ \mathbf{x} ) }{\partial u^{i}}$ is $(u^{1}, u^{2})$, and $\dfrac{\partial ( f \circ \mathbf{x} ) }{\partial u^{i}} (u^{1}, u^{2})$ is simply shorthand notation. Similarly, $\dfrac{d \alpha^{i}}{dt}(t)$ is also shortened by omitting variables.

Continuing similarly, by applying the chain rule to differentiate $\alpha (t) = \mathbf{x}(\alpha^{1}(t), \alpha^{2}(t))$, we obtain the following.

$$\begin{align*} \alpha^{\prime} (t) =&\ \dfrac{d \mathbf{x}}{d u^{1}} \dfrac{d \alpha^{1}}{d t}(t) + \dfrac{d \mathbf{x}}{d u^{2}} \dfrac{ d \alpha^{2}}{dt}(t) \\ =&\ \mathbf{x}_{1}(\alpha^{1}(t), \alpha^{2}(t)) (\alpha^{1})^{\prime}(t) + \mathbf{x}_{2}(\alpha^{1}(t), \alpha^{2}(t)) (\alpha^{2})^{\prime}(t) \\ =&\ (\alpha^{1})^{\prime}(t) \mathbf{x}_{1} + (\alpha^{2})^{\prime}(t) \mathbf{x}_{2} \end{align*}$$

By substituting $t=0$, we obtain the following.

$$\alpha^{\prime}(0) = \mathbf{X} = X^{1}\mathbf{x}_{1} + X^{2}\mathbf{x}_{2}$$

Therefore, $X^{i} = \dfrac{d\alpha^{i}}{dt} (0)$ holds. Moreover, when $t=0$, since $\alpha (0) = p = \mathbf{x}(0,0)$ then by substituting $t=0$ into $(1)$d, we obtain the following.

$$\begin{align*} \mathbf{X} f = \dfrac{d}{dt_{}} (f \circ \alpha) (0) =&\ \sum \limits_{i} \dfrac{\partial ( f \circ \mathbf{x} ) }{\partial u^{i}}(0,0) \dfrac{d \alpha^{i}}{dt}(0) \\ =&\ \sum \limits_{i} X^{i}\dfrac{\partial ( f \circ \mathbf{x} ) }{\partial u^{i}}(0,0) \end{align*}$$

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Corollary

Let’s say $\mathbf{X}, \mathbf{Y} \in T_{p}M$. Suppose $f, g$ is a differentiable function in the neighborhood of $p \in M$. Let’s denote $r \in \mathbb{R}^{3}$. Then, the following formula holds.

$$\left( r\mathbf{X} + \mathbf{Y} \right) f = r\mathbf{X}f + \mathbf{Y} f \\ \mathbf{X}(rf+g) = r\mathbf{X}f + \mathbf{X}g \\ \mathbf{X}_{p}(fg) = \mathbf{X}_{p}(f) g(p) + f(p)\mathbf{X}_{p}(g)$$

The third formula implies differentiation of products $(fg)^{\prime} = f^{\prime}g +fg^{\prime}$.

Proof

This can be easily shown by definition.

$$\begin{align*} \left( r\mathbf{X} + \mathbf{Y} \right) f =&\ \sum \limits_{i=1}^{2} (rX^{i} + Y^{i}) \dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0) \\ =&\ r\sum \limits_{i=1}^{2} X^{i}\dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0) + \sum \limits_{i=1}^{2}Y^{i}\dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0) \\ =&\ r\mathbf{X}f + \mathbf{Y} f \end{align*}$$

Since $fg(p)= f(p)g(p)$, the following holds.

$$(fg)\circ \mathbf{x} (u^{1}, u^{2}) = f(\mathbf{x}(u^{1}, u^{2})) g(\mathbf{x}(u^{1}, u^{2})) = f\circ \mathbf{x}(u^{1}, u^{2}) g\circ \mathbf{x}(u^{1}, u^{2})$$

Therefore, by the product differentiation rule, we obtain the following.

$$\begin{align*} \mathbf{X}_{p}(fg) =&\ \sum \limits_{i=1}^{2} X^{i} \dfrac{\partial ((fg)\circ \mathbf{x})}{\partial u^{i}} (0,0) \\ =&\ \sum \limits_{i=1}^{2} X^{i} \dfrac{\partial ((f\circ \mathbf{x}) (g \circ \mathbf{x}) )}{\partial u^{i}} (0,0) \\ =&\ \sum \limits_{i=1}^{2} X^{i} \left[ \dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0) (g \circ \mathbf{x}) (0,0) + (f \circ \mathbf{x}) (0,0)\dfrac{\partial (g\circ \mathbf{x})}{\partial u^{i}} (0,0) \right] \\ =&\ \sum \limits_{i=1}^{2} X^{i} \dfrac{\partial (f\circ \mathbf{x})}{\partial u^{i}} (0,0) g(p) + f(p) \sum \limits_{i=1}^{2} X^{i} \dfrac{\partial (g\circ \mathbf{x})}{\partial u^{i}} (0,0) \\ =&\ \mathbf{X}_{p}(f) g(p) + f(p) \mathbf{X}_{p}(g) \end{align*}$$

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See Also

• Directional Derivatives in Analysis
• Tangent Vectors in Riemannian Geometry