📂Geometry

Parameterized Surfaces

Definition¹

• For an open set $U \subset \mathbb{R}^{2}$, let’s say the connected set $A\subset \mathbb{R}^{2}$ satisfies the following:

  $$ U \subset A \subset \overline{U} \quad \text{and} \quad \partial A \text{ is piecewise differentiable} $$

  $s : A \to M$ is called a parameterized surface within the manifold $M$.

• A vector field $V$ along $s$, maps each $q \in A$ to $V(q) \in T_{s(q)}M$ and is a differentiable function in the following sense:

  If $f$ is a differentiable function over $M$, then the function $q \in A \mapsto V(q)f \in \mathbb{R}$ is also differentiable.

Description

Let’s call $(u, v)$ the Cartesian coordinates over $\mathbb{R}^{2}$. For a fixed $v_{0}$, the function $u \mapsto s(u,v_{0})$ is a curve over $M$. Considering the standard basis of the tangent space over $\mathbb{R}^{2}$, it leads to $\left\{ \frac{\partial }{\partial u}, \frac{\partial }{\partial v} \right\}$. Simplify the function value $ds(\frac{\partial }{\partial u})$ of derivative $ds$ of $s$ as

$$ \dfrac{\partial s}{\partial u} $$

Then $\dfrac{\partial s}{\partial u}$ is a vector field following curve $u \mapsto s(u,v_{0})$. Similarly, $\dfrac{\partial s}{\partial v}$ is defined as a vector field along $s$. In differential geometry, similar concepts are considered for coordinate patch maps.

Consider a vector field $V$ along $s$. Now, we will define the covariant derivatives $\dfrac{D V}{\partial u}$, $\dfrac{D V}{\partial v}$ of this $V$. Think about the reduction mapping of $V$ over the curve $u \mapsto s(u, v_{0})$. Then $\dfrac{DV}{du}(u,v_{0})$ can be defined as the covariant derivative of this reduction mapping. As this can be considered for all $v_{0}$, $\dfrac{DV}{\partial u}(u, v)$ is defined for all $(u,v) \in A$. $\dfrac{D V}{\partial v}$ is defined in the same manner.

Symmetry

If $M$ is a manifold with a symmetric connection, and $s : A \to M$ is a parameterized surface, then the following holds:

$$ \dfrac{D}{\partial v}\dfrac{\partial s}{\partial u} = \dfrac{D}{\partial u}\dfrac{\partial s}{\partial v} $$

Proof

Let $\mathbf{x} : U \subset \mathbb{R}^{n} \to M$ be a coordinate system including a neighborhood of point $s(A)$. In other words, the coordinate system is chosen such that $s(A) \subset \mathbf{x}(U)$. Then $\mathbf{x}^{-1} \circ s(u,v)$ is a point of $\mathbb{R}^{n}$, and it’s represented as follows:

$$ \mathbf{x}^{-1} \circ s (u,v) = \left( s^{1}(u,v), \dots, s^{n}(u,v) \right) $$

Also, $s : A \to M$ and the derivative of $s$ is $ds : T_{(u,v)}A \to T_{s(u,v)}M$. The coordinates of $A$ are $(u, v)$, and the coordinates of $M$ are $\mathbf{x}^{-1} = \left( s^{1}, \dots, s^{n} \right)$, therefore,

$$ ds = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\ \vdots \\ \dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v} \end{bmatrix} $$

The basis of $T_{(u,v)}A$ is, the coordinate vector of $\left\{ \dfrac{\partial }{\partial u}, \dfrac{\partial }{\partial v} \right\}$ is $\begin{bmatrix} 1 \\ 0\end{bmatrix}$. Thus,

$$ \dfrac{\partial s}{\partial u} = ds(\dfrac{\partial }{\partial u}) = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\ \vdots & \vdots \\ \dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v} \end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} \\ \vdots \\ \dfrac{\partial s^{n}}{\partial u} \end{bmatrix} = \sum_{i} \dfrac{\partial s^{i}}{\partial u}\dfrac{\partial }{\partial x_{i}} $$

Now, calculating $\dfrac{D }{\partial v}\left( \dfrac{\partial s}{\partial u} \right)$, by the properties of covariant derivatives,

$$ \begin{align*} \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) &= \dfrac{D }{\partial v} \left( \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{\partial }{\partial x_{j}} \right) \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{D }{\partial v}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{{\partial s}/{\partial u}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{\frac{\partial s^{i}}{\partial u}\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\nabla_{\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial v \partial u}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \\ \end{align*} $$

Similarly, $\dfrac{D }{\partial u}\left( \dfrac{\partial s}{\partial v} \right)$ is

$$ \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial u \partial v}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial v} \frac{\partial s^{i}}{\partial v}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} $$

At this point, as these are differentiable functions where $s^{k} : \mathbb{R}^{2} \to \mathbb{R}$, $\dfrac{\partial ^{2}}{\partial u \partial v} = \dfrac{\partial ^{2}}{\partial v \partial u}$ holds. Therefore,

$$ \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) $$

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