📂Geometry

Definition of Parallel Vector Field along a Curve on Surface

Vector Field Along a Curve¹

Definition

Given a surface $M$ and a curve $\alpha : \left[ a, b \right] \to M$, let us consider a function $\mathbf{X}$ that maps each $t \in \left[ a,b \right]$ to a tangent vector at point $\alpha (t)$ on surface $M$. This function $\mathbf{X}$ is called a vector field along curve $\alpha$.

$$ \mathbf{X} : \left[ a, b \right] \to \mathbb{R}^{3} \\ \mathbf{X}(t) \in T_{\alpha (t)}M $$

Explanation

It’s important to note that the tangent vector mentioned in the definition is not the tangent vector of the curve $\alpha$, but rather the tangent vector at point $\alpha (t)$, which is an element of $T_{\alpha (t)}M$. Since the tangent vector at each point $\alpha (t)$ on the surface is not unique, the vector field along curve $\alpha$ is not unique either. There are infinitely many vectors in the tangent plane, hence there are infinitely many vector fields as well.

For a simple example, given a curve $\alpha (t)$ on surface $M$, the tangent vector field $\mathbf{T}(t)$ of $\alpha (t)$ becomes a vector field along $\alpha$. $\mathbf{S} = \mathbf{n} \times \mathbf{T}$ is also a vector field along $\alpha$.

$\mathbf{S}$ and $\mathbf{T}$ form a basis of the tangent space, so every vector field $\mathbf{X}$ along $\alpha$ can be expressed as a linear combination of these.

$$ \mathbf{X}(t) = A(t)\mathbf{T}(t) + B(t)\mathbf{S}(t)\quad \text{for some } A,B:[a,b]\to \mathbb{R} $$

Differentiable Vector Field

Definition

A vector field $\mathbf{X}(t)$ along $\alpha (t)$ is differentiable if the function $\mathbf{X} : \left[ a, b \right] \to \mathbb{R}^{3}$ is differentiable.

Explanation

Strictly speaking, it’s correct to say ‘$\mathbf{X}$ is differentiable’, but it’s also convenient to say ‘$\mathbf{X}(t)$ is differentiable’.

Parallel Vector Field

Definition

Given a differentiable vector field $\mathbf{X}(t)$ along $\alpha$, if $\dfrac{d \mathbf{X}}{dt}$ is perpendicular to surface $M$, then $\mathbf{X}(t)$ is defined to be parallel along $\alpha (t)$.

Explanation

As explained earlier, a vector field $\alpha$ can be chosen arbitrarily, but the condition ‘a differentiable vector field $\alpha$’ sets a restriction for defining the concept of ‘parallel lines’.

Being perpendicular to surface $M$ means that $\dfrac{d \mathbf{X}}{dt}$ should have no component in the tangent direction, and only in the normal direction. The definition might not make much sense at first glance, so let’s look at the following examples.

Example

In a 2D Plane

Consider a curve $\boldsymbol{\gamma}(t) = \left( a(t), b(t), 0 \right)$ on the $xy-$ plane. And let’s say $\mathbf{X}(t) = \left( A(t), B(t), 0 \right)$ is a vector field along $\boldsymbol{\gamma}$. Then,

$$ \dfrac{d \mathbf{X}}{dt} = \left( \dfrac{d A}{dt}, \dfrac{d B}{dt}, 0 \right) $$

For this vector to be perpendicular to the $xy-$ plane, the dot product with any vector $(x,y,0)$ must be $0$, leading to the following result.

$$ \dfrac{d A}{dt} = 0 = \dfrac{d B}{dt} $$

Therefore, $A(t), B(t)$ is a constant. If we illustrate this, it matches well with our intuitive understanding of ‘vectors that are parallel along curve $\boldsymbol{\gamma}$’.

On a Sphere

Let $M$ be the unit sphere. Let ${\color{6699CC}\boldsymbol{\gamma}(t)}$ be the equator. Consider a vector field ${\color{295F2E}\mathbf{X}_{\boldsymbol{\gamma}}(t) = (0, 0, 1)}$ along $\boldsymbol{\gamma}$. Then, since $\dfrac{d \mathbf{X}_{\boldsymbol{\gamma}}}{dt} = (0,0,0)$, it’s always perpendicular to $M$. Hence, $\mathbf{X}_{\boldsymbol{\gamma}}$ is a vector field parallel along $\boldsymbol{\gamma}$. As seen in the 2D example, a constant vector field naturally becomes a parallel vector field.

Now, consider a vector field ${\color{295F2E}\mathbf{X}_{\boldsymbol{\beta}} = \left( -\dfrac{\sqrt{2}}{2}\cos t, \dfrac{\sqrt{2}}{2}\sin t, \dfrac{\sqrt{2}}{2} \right)}$ along curves ${\color{6699CC}\boldsymbol{\beta}(t) = \left( \dfrac{\sqrt{2}}{2}\cos t, \dfrac{\sqrt{2}}{2}\sin t, \dfrac{\sqrt{2}}{2} \right)}$ and $\boldsymbol{\beta}$. Then, ${\color{f8c512}\dfrac{d\mathbf{X}_{\boldsymbol{\beta}}}{dt} = \left( \dfrac{\sqrt{2}}{2}\sin t, \dfrac{\sqrt{2}}{2}\cos t, 0 \right)}$, and if we plot this in 3D, it appears as follows.

Since ${\color{f8c512}\dfrac{d\mathbf{X}_{\boldsymbol{\beta}}}{dt}}$ is not perpendicular to the sphere, ${\color{295F2E}\mathbf{X}_{\boldsymbol{\beta}}}$ is not a vector field parallel along $\boldsymbol{\beta}$.