📂Geometry

Differentiation of Functions Defined on Differentiable Manifolds

Theorem¹

Let’s call $M_{1}^{n}, M_{2}^{m}$ and $m, n$, respectively, $m, n$-dimensional differentiable manifolds. Let’s say $\varphi : M_{1} \to M_{2}$ is a differentiable function. And for every point $p \in M_{1}$ and tangent vector $v \in T_{p}M$, choose a differentiable curve

$$\alpha : (-\epsilon, \epsilon) \to M_{1} \text{ with } \alpha (0) = p,\ \alpha^{\prime}(0)=v$$

Let’s set it as $\beta = \varphi \circ \alpha$. Then, the following mapping

$$d\varphi_{p} : T_{p}M_{1} \to T_{\varphi(p)}M_{2} \\[1em] d\varphi_{p}(v) = \beta^{\prime}(0)$$

is a linear transformation that is independent of the choice of $\alpha$.

Definition

The mapping $d\varphi_{p}$, defined as in the theorem above, is called the differential from $p$ to $\varphi$.

Description

It’s not the derivative, but the differential.

Since the tangent vector acts on functions defined on a differentiable manifold, the differential $d\varphi_{p}$ is a mapping from the function space to the function space. Moreover, it is clear from the conclusion of the proof that $d\phi_{p}$ is the Jacobian for the function $\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}$.

$$\text{differential} = \text{Jacobian}$$

Let’s recall the role of Jacobians. For example, suppose the integral of a function defined at $\mathbb{R}^{2}$ is given as follows.

$$\int \int f(x,y) dx dy$$

When changing coordinates to polar coordinates $(r,\theta)$, it is necessary to multiply by the determinant of the Jacobian $\displaystyle \begin{vmatrix}\dfrac{ \partial x}{ \partial r} & \dfrac{ \partial x}{ \partial \theta} \\ \textstyle{} \\ \dfrac{ \partial y}{ \partial r} & \dfrac{ \partial y}{ \partial \theta} \end{vmatrix}=r$.

$$\int \int f(x,y) dx dy = \int \int f(r, \theta) rdr d\theta$$

Therefore, the differential $d\phi_{p}$ of $\phi : M_{1} \to M_{2}$ can be thought of as implementing the coordinate transformation between the differentiable manifolds $M_{1}$ and $M_{2}$ through their coordinate systems $\mathbf{x}, \mathbf{y}$.

Since the Jacobian of the composition is equal to the product of the Jacobians, the following holds for $\phi : M \to N, \psi : N \to L$ and $p\in M$,

$$d(\psi \circ \phi)_{p} = d(\psi)_{\phi (p)} d(\phi)_{p}$$

Understanding the definition and significance of tangent vectors is crucial for comprehending the content of this document.

Proof

Let’s call the coordinate system at point $p \in M_{1}$ on $M_{1}$ as $\mathbf{x} : U \subset \mathbb{R}^{n} \to M_{1}$. Let’s denote the coordinates in $\mathbb{R}^{n}$ as $(r_{1}, \dots, r_{n}) \in \mathbb{R}^{n}$.

$$\mathbf{x}(r_{1}, \dots, r_{n}) = p \quad \text{and} \quad \mathbf{x}^{-1}(p) = \left( x_{1}(p), \dots, x_{n}(p) \right)$$

And let’s call the coordinate system at point $\phi (p) \in M_{2}$ on $M_{2}$ as $\mathbf{y} : V \subset \mathbb{R}^{m} \to M_{2}$. Let’s denote the coordinates in $\mathbb{R}^{m}$ as $(s_{1}, \dots, s_{m}) \in \mathbb{R}^{m}$.

$$\mathbf{y}(s_{1}, \dots, s_{m}) = \phi (p) \quad \text{and} \quad \mathbf{y}^{-1}(\phi (p)) = \Big( y_{1}(\phi (p)), \dots, y_{m}(\phi (p)) \Big)$$

By the definition of the tangent vector, the tangent vector $\beta^{\prime}(0)$ at the point $\phi (p)$ on $M_{2}$ is as follows. For the differentiable function $g : M_{2} \to \mathbb{R}$ defined on $M_{2}$,

$$\begin{align*} \beta^{\prime}(0) g =&\ \dfrac{d}{dt}(g \circ \beta)(0) = \dfrac{d}{dt}(g \circ \mathbf{y} \circ \mathbf{y}^{-1} \circ \beta)(0) \\ =&\ \dfrac{d}{dt}\big( (g \circ \mathbf{y}) \circ (\mathbf{y}^{-1} \circ \beta)\big)(0) \\ =&\ \sum \limits_{j=1}^{m} \left.\dfrac{\partial (g\circ \mathbf{y})}{\partial s_{j}}\right|_{t=0} \dfrac{d (\mathbf{y}^{-1} \circ \beta)_{j}}{d t}(0) & \text{by } \href{https://freshrimpsushi.github.io/posts/chaine-rule-for-multivariable-vector-valued-funtion}{\text{chain rule}} \\ =&\ \sum \limits_{j=1}^{m} y_{j}^{\prime}(0) \left.\dfrac{\partial (g\circ \mathbf{y})}{\partial s_{j}}\right|_{t=0} \\ =&\ \sum_{j=1}^{m} y_{j}^{\prime}(0) \left.\dfrac{\partial g}{\partial y_{j}}\right|_{t=0} \end{align*}$$

The meaning of the operator $\left.\dfrac{\partial }{\partial y_{j}}\right|_{t=0}$ here is to differentiate by pulling the domain of the non-differentiable function $g$ defined on $M_{2}$ to $\mathbb{R}^{m}$ through composition with $\mathbf{y}$. Now, let’s find $y_{j}^{\prime}$.

$$y_{j}^{\prime} = \dfrac{d}{dt} (\mathbf{y}^{-1} \circ \beta)_{j}$$

As with calculating the tangent vector, let’s consider $\mathbf{y}^{-1} \circ \beta$ decomposed as follows:

$$\mathbf{y}^{-1} \circ \beta = \mathbf{y}^{-1} \circ \phi \circ \alpha = \mathbf{y}^{-1} \circ \phi \circ \mathbf{x} \circ \mathbf{x}^{-1} \circ \alpha$$

And consider the above equation as the composition of two functions, $\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}$ and $\mathbf{x}^{-1} \circ \alpha$.

• Part 1. $\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}$

  Since $\mathbf{y}^{-1} \circ \phi \circ \mathbf{x} : \mathbb{R}^{n} \to \mathbb{R}^{m}$,

  $$\begin{equation} \mathbf{y}^{-1}\left( \phi (\mathbf{x}(r_{1}, \dots, r_{n})) \right) = \left( y_{1}(\phi (\mathbf{x}(r_{1}, \dots, r_{n}))), \dots, y_{m}(\phi (\mathbf{x}(r_{1}, \dots, r_{n}))) \right) \end{equation}$$

  Simplified, this becomes

  $$\mathbf{y}^{-1} \circ \phi \circ \mathbf{x} (r_{1}, \dots, r_{n}) = \left( y_{1}, \dots, y_{m}\right)$$

  Here $y_{j}$ are technically functions concerning $\phi (\mathbf{x}(r_{1}, \dots, r_{n}))$ as in $(1)$, but for simplicity, let’s denote them as functions concerning $(r_{1}, \dots, r_{n})$.

  $$y_{j} = y_{j}(r_{1}, \dots, r_{n}),\quad 1\le j \le m$$

• Part 2. $\mathbf{x}^{-1} \circ \alpha$

  Since $\mathbf{x}^{-1} \circ \alpha : \mathbb{R} \to \mathbb{R}^{n}$,

  $$\mathbf{x}^{-1} (\alpha (t)) = \left( x_{i}(\alpha (t)), \dots, x_{n}(\alpha (t)) \right)$$

  Here again, for simplicity, each $x_{i}$ is denoted as if they are functions concerning $t$.

  $$\mathbf{x} \circ \alpha (t) = ( x_{i}(t), \dots, x_{n}(t) )$$

Now, $(\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}) \circ (\mathbf{x}^{-1} \circ \alpha)$ is the composition of the function that is $\mathbb{R} \to \mathbb{R}^{n}$ and the function that is $\mathbb{R}^{n} \to \mathbb{R}^{m}$, so by the chain rule, we obtain the following.

$$\begin{align*} \dfrac{d}{dt} (\mathbf{y}^{-1} \circ \beta)(0) =&\ \dfrac{d}{dt}(\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}) \circ (\mathbf{x}^{-1} \circ \alpha)(0) \\ =&\ \begin{bmatrix} \sum \limits_{i=1}^{n} \left.\dfrac{\partial (\mathbf{y}^{-1} \circ \phi \circ \mathbf{x})_{1}}{\partial x_{i}}\right|_{t=0} \dfrac{d (\mathbf{x}^{-1} \circ \alpha)_{i}}{d t}(0) \\ \vdots \\ \sum \limits_{i=1}^{n} \left.\dfrac{\partial (\mathbf{y}^{-1} \circ \phi \circ \mathbf{x})_{m}}{\partial x_{i}}\right|_{t=0} \dfrac{d (\mathbf{x}^{-1} \circ \alpha)_{i}}{d t}(0) \end{bmatrix} \\ =&\ \begin{bmatrix} \sum \limits_{i=1}^{n} \left.\dfrac{\partial y_{1}}{\partial x_{i}}\right|_{t=0} \dfrac{d x_{i}}{d t}(0) \\ \vdots \\ \sum \limits_{i=1}^{n} \left.\dfrac{\partial y_{m}}{\partial x_{i}}\right|_{t=0} \dfrac{d x_{i}}{d t}(0) \end{bmatrix} \\ =&\ \begin{bmatrix} \sum \limits_{i=1}^{n} \dfrac{\partial y_{1}}{\partial x_{i}} x_{i}^{\prime}(0) \\ \vdots \\ \sum \limits_{i=1}^{n} \dfrac{\partial y_{m}}{\partial x_{i}} x_{i}^{\prime}(0) \end{bmatrix} \end{align*}$$

Therefore,

$$y_{j}^{\prime}(0) = \dfrac{d}{dt} (\mathbf{y}^{-1} \circ \beta)_{j}(0) = \sum \limits_{i=1}^{n} \dfrac{\partial y_{j}}{\partial x_{i}} x_{i}^{\prime}(0),\quad 1\le j \le m$$

Thus, if we express $\beta^{\prime}(0)$ as a coordinate vector against the basis $\left\{ \left.\dfrac{\partial }{\partial y_{j}}\right|_{t=0} \right\}$, it is as follows.

$$\beta^{\prime}(0) = \sum_{j=1}^{m} y_{j}^{\prime}(0) \left.\dfrac{\partial}{\partial y_{j}}\right|_{t=0} = \begin{bmatrix} y_{1}^{\prime}(0) \\ \vdots \\ y_{m}^{\prime}(0) \end{bmatrix} = \begin{bmatrix} \sum \limits_{i=1}^{n} \dfrac{\partial y_{1}}{\partial x_{i}} x_{i}^{\prime}(0)\\ \vdots \\ \sum \limits_{i=1}^{n} \dfrac{\partial y_{m}}{\partial x_{i}} x_{i}^{\prime}(0) \end{bmatrix}$$

Therefore, it can be seen that $\beta^{\prime}(0)$ does not depend on $\alpha$.

Meanwhile, the following is true for $\alpha^{\prime}(0) = v$.

$$v = \alpha^{\prime}(0) = \sum \limits_{i=1}^{n}x_{i}^{\prime}(0) \left.\dfrac{\partial }{\partial x_{i}}\right|_{t=0} = \begin{bmatrix} x_{1}^{\prime}(0) \\ \vdots \\ x_{n}^{\prime}(0) \end{bmatrix}$$

Thus, organizing $\beta^{\prime}(0) = d\phi_{p}(v)$ results in the following.

$$\begin{align*} \beta^{\prime}(0) =&\ \begin{bmatrix} \sum \limits_{i=1}^{n} \dfrac{\partial y_{1}}{\partial x_{i}} x_{i}^{\prime}(0)\\ \vdots \\ \sum \limits_{i=1}^{n} \dfrac{\partial y_{m}}{\partial x_{i}} x_{i}^{\prime}(0) \end{bmatrix} \\ =&\ \begin{bmatrix} \dfrac{\partial y_{1}}{\partial x_{1}} & \dfrac{\partial y_{1}}{\partial x_{2}} & \dots & \dfrac{\partial y_{1}}{\partial x_{n}} \\[1em] \dfrac{\partial y_{2}}{\partial x_{1}} & \dfrac{\partial y_{2}}{\partial x_{2}} & \dots & \dfrac{\partial y_{2}}{\partial x_{n}} \\[1ex] \vdots & \vdots & \ddots & \vdots \\[1ex] \dfrac{\partial y_{m}}{\partial x_{1}} & \dfrac{\partial y_{m}}{\partial x_{2}} & \dots & \dfrac{\partial y_{m}}{\partial x_{n}} \end{bmatrix} \begin{bmatrix} x_{1}^{\prime}(0) \\ \vdots \\ x_{n}^{\prime}(0) \end{bmatrix} \\ =&\ \begin{bmatrix} \dfrac{\partial y_{1}}{\partial x_{1}} & \dfrac{\partial y_{1}}{\partial x_{2}} & \dots & \dfrac{\partial y_{1}}{\partial x_{n}} \\[1em] \dfrac{\partial y_{2}}{\partial x_{1}} & \dfrac{\partial y_{2}}{\partial x_{2}} & \dots & \dfrac{\partial y_{2}}{\partial x_{n}} \\[1ex] \vdots & \vdots & \ddots & \vdots \\[1ex] \dfrac{\partial y_{m}}{\partial x_{1}} & \dfrac{\partial y_{m}}{\partial x_{2}} & \dots & \dfrac{\partial y_{m}}{\partial x_{n}} \end{bmatrix} v \end{align*}$$

Therefore, $d_{p}\phi$ is a linear transformation represented by the following matrix.

$$d_{p}\phi = \begin{bmatrix} \dfrac{\partial y_{1}}{\partial x_{1}} & \dfrac{\partial y_{1}}{\partial x_{2}} & \dots & \dfrac{\partial y_{1}}{\partial x_{n}} \\[1em] \dfrac{\partial y_{2}}{\partial x_{1}} & \dfrac{\partial y_{2}}{\partial x_{2}} & \dots & \dfrac{\partial y_{2}}{\partial x_{n}} \\[1ex] \vdots & \vdots & \ddots & \vdots \\[1ex] \dfrac{\partial y_{m}}{\partial x_{1}} & \dfrac{\partial y_{m}}{\partial x_{2}} & \dots & \dfrac{\partial y_{m}}{\partial x_{n}} \end{bmatrix}$$

This is also the Jacobian of the function $\mathbf{y}^{-1} \circ \phi \circ \mathbf{x}$.

$$d\phi_{p} = \text{Jacobian of } \mathbf{y}^{-1} \circ \phi \circ \mathbf{x} = \dfrac{\partial (y_{1}, \dots, y_{m})}{\partial (x_{1}, \dots, x_{n})}$$

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