Definition(Curviliear coordinates) Let $U$ be an open set in $\R^n$. A curvilinear coordinate on $U$ is a smooth function $f=(f_1,\cdots,f_n): U \to \R^n$, such that

• $f$ is a bijection between $U$ and its image $f(U)$
• and the inverse function $f^{-1}: f(U) \to U$ is also smooth.

Notation We reserve the notation $(x_1, \cdots, x_n)$ to be the standard Cartesian coordinate on $\R^n$. We use notation of a pair $(U, (u_1, \cdots, u_n))$, or $(U, (f_1, \cdots, f_n))$ for a coordinate on $U \subset \R^n$.

$$\gdef\b{\mathbf}$$ $$\gdef\d{\partial}$$

Consider the n-dimensional Euclidean space $\R^n$ with basis vectors $\b e_1, \cdots, \b e_n$. A vector $\b v = v^1 \b e_1 + \cdots + v^n \b e_n$ has two possible meansings

1. it can represent a location in the space $\R^n$. You cannot add two locations (can you add New York to San Francisco?)
2. it can represent a velocity vector (an arrow with direction and length).

In order to represent both the position and the velocity ¹⁾, we need consider the notion of a tangent vector on $\R^n$.

Definition (Tangent vector) A tangent vector on $U \subset \R^n$ is a pair $(\b a, \b v)$ representing the location and velocity of a particle, where $\b a \in U$ represent the position, and $\b v \in \R^n$ represent the velocity. We denote the set of tangent vectors over a point $p \in U$ as $T_p U$. It is an $n$-dimensional vector space.

Warning: Only tangent vectors standing over the same position can be added or subtracted.

Definition (Vector field) A vector field on $U \subset \R^n$ is an assignment of tangent vectors $\b v$ to each point $\b a \in U$, such that $\b v$ varies smoothly with respect to $\b a$.

Let $(U, (u_1, \cdots, u_n))$ be a coordinate system on $U$. Let $p \in U$ be a point, and choose an $i \in \{1, \cdots, n\}$. We will define a tangent vector $\frac{\d}{\d u_i} \vert_p$ at $p$ “physically” as follows. Consider the motion of a particle on $U$, describe by the following curve $\gamma: (-\epsilon, +\epsilon) \to U$, such that $\gamma(0) = p$, and for $t \in (-\epsilon, +\epsilon)$ $$ u_j(\gamma(t)) = \begin{cases} u_j(\gamma(0)) & j \neq i \cr u_j(\gamma(0))+t & j = i \end{cases} $$. Then, we define $\frac{\d}{\d u_i} \vert_p$ to be the velocity of the particle at the moment $t=0$.

As we vary $p \in U$, the tangent vectors $\frac{\d}{\d u_i} \vert_p$ forms a vector field, denoted as $\frac{\d}{\d u_i}$ or $\d_{u_i}$. This is called a coordinate vector field.

Without spelling out all the details, I will simply say that $\d_{u_i}$ generate a flow (or motion) of $U$, that moves each point on $U$ by keeping all the $u_j$ coordinates fixed and only increasing the $u_i$ coordinate “at unit speed”. Imagine $\d_x$ on $\R^3$ is moving everyone towards the positive $x$ axis.

Recall previously, at every point $p \in U$, we have the tangent vector space $T_p U$. We can consider the dual space there and we get $T^*_p U$. An element in $T_p^* U$ is called a cotangent vector.

Let $f$ be a smooth function on $U$. $p \in U$ a point. We will define $df(p) \in T_p^*U$ a cotangent vector at $p$. By definition, we need to specify for each tangent vector $v \in T_p U$, what is the value $df(p)(v)$. This is the directional derivative of $f$ at $p$ in the direction $v$: $$ df(p)(v) := \sum_{i=1}^n v^i \frac{\d f}{\d x_i}\vert_p $$

We can define $df(p)(v)$ without using coordinate. Let $\gamma: (-\epsilon, +\epsilon) \to U$ be a curve, such that $\gamma(0)=p$ and $\dot \gamma(0)=v$, then $$ df(p)(v) = \frac{d f(\gamma(t))}{d t} \vert_{t=0}.$$

Definition (Differential 1-form) A differential one-form is a assignment from $p \in U$ to elements in $T_p^*U$, that varies smoothly with $p$.

$df$ is a differential one-form. Since the coordinates $x_1, \cdots, x_n$ are also function on $U$, we also have $dx_1, \cdots, dx_n$ as differential one-forms.

Lemma Let $u_1, \cdots, u_n$ be a coordinate on $U$. Then for each point $p$, $du_1(p), \cdots, du_n(p)$ is a basis of the cotangent vectors $T_p^* U$.

Since $\{ d u_i(p) \}$ is a basis on $T_p^*U$, we can decompose the element $df(p)$, it turns out the decomposition is as following $$ df(p) = \frac{\d f}{\d u_1}(p) d u_1 + \cdots + \frac{ \d f}{\d u_n} (p) d u_n, $$ where the partial derivatives $$ \frac{\d f}{\d u_i}(p) = df(p) (\frac{\d }{\d u_i} ). $$ If one view $f$ as a function on the curvilinear coordinates $u_1, \cdots, u_n$, then these are indeed partial derivatives.