Parallel Transport

Christoffel Symbols (Gauss Equations)

Consider some surface \(\Sigma\) with (locally) parameterization \(\sigma\). Note that \(\sigma_u, \sigma_v, \mathbf N\) spans \(\mathbb R^3\), hence we can write the second order derivatives as

\[\begin{align*} \sigma_{uu} &= a_1\sigma_u + a_2\sigma_v + a_3\mathbf N\\ \sigma_{uv} &= b_1\sigma_u + b_2\sigma_v + b_3\mathbf N\\ \sigma_{vv} &= c_1\sigma_u + c_2\sigma_v + c_3\mathbf N\\ \end{align*}\]

Now, consider the second fundamental form

\[L = \sigma_{uu}\cdot N = a_1\sigma_u\cdot N + a_2\sigma_v \cdot N + a_3\mathbf N\cdot \mathbf N = 0 + 0 + a_3 = a_3\]

we have that

\[a_3 = L, b_3 = M, c_3 = N\]

Then, we take dot product with \(\sigma_u, \sigma_v\)

\[\begin{align*} \sigma_{uu}\cdot \sigma_u &= a_1 (\sigma_u\cdot\sigma_u) + a_2(\sigma_v\cdot \sigma_u) + a_3 (\mathbf N\cdot\sigma_u) = Ea_1 + Fa_2\\ \sigma_{uu}\cdot \sigma_v &= F a_1 + Ga_2\\ \sigma_{uv}\cdot\sigma_u &= Eb_1 + Fb_2\\ \sigma_{uv}\cdot\sigma_v &= Fb_1 + Gb_2\\ \sigma_{vv}\cdot\sigma_u &= Ec_1+ Gc_2\\ \sigma_{vv}\cdot\sigma_v &= Fc_1+Gc_2\\ \end{align*}\]

On the other hand, we have that

\[\begin{align*} &E_u = (\sigma_u\cdot\sigma_u)_u = 2 \sigma_{uu}\cdot\sigma_{u} &E_v = 2\sigma_{uv} \cdot\sigma_{u}\\ &F_u = \sigma_{uu}\cdot \sigma_{uv} &F_v = \sigma_{vv}\cdot\sigma_{uv}\\ &G_u = 2\sigma_{uv}\cdot \sigma_{v} &G_v = 2\sigma_{vv}\cdot \sigma_{v} \end{align*}\]

Combine the equations, we have 6 equations w.r.t. to the 6 unknowns

\[\begin{align*} E a_1 + F a_2 &= \frac{1}{2}E_u&F a_1 + G a_2 = F_u - \frac{1}{2}E_v\\ E b_1 + F b_2 &= \frac{1}{2}E_v&F b_1 + G b_2 = \frac{1}{2}G_u\\ E c_1 + F c_2 &= F_v - \frac{1}{2}G_u&F c_1 + G c_2 = \frac{1}{2}G_v \end{align*}\]

The solution is that

\[\begin{align*} \Gamma_{11}^1 = \frac{GE_u - 2FF_u+FE_v}{2(EG-F^2)}&&\Gamma_{11}^2 = \frac{2EF_u-EEv-FEu}{2(EG-F^2)}\\ \Gamma_{12}^1 = \frac{GE_v - FG_u}{2(EG-F^2)}&&\Gamma_{12}^2 = \frac{EG_u-FE_v}{2(EG-F^2)}\\ \Gamma_{22}^1 = \frac{2GF_v-GG_u-FG_v}{2(EG-F^2)}&&\Gamma_{22}^2 = \frac{EG_v-2FF_v+FG_u}{2(EG-F^2)} \end{align*}\]

so that the equations are called Gauss Equations

\[\begin{align*} \sigma_{uu} &= \Gamma_{11}^1\sigma_u + \Gamma_{11}^2\sigma_v + L\mathbf N\\ \sigma_{uv} &= \Gamma_{12}^1\sigma_u + \Gamma_{12}^2\sigma_v + M\mathbf N\\ \sigma_{vv} &= \Gamma_{22}^1\sigma_u + \Gamma_{22}^2\sigma_v + N\mathbf N\\ \end{align*}\]

and the \(\Gamma\)'s are Christoffel symbols.

Covariant Derivative

Consider a particle's trajectory on some surface \(\Sigma\) as \(\gamma:(a, b) \rightarrow\Sigma\) and its velocity is the tangent vector \(\mathbf v = \gamma'\) and acceleration \(\mathbf a = \mathbf v' = \gamma''\). Suppose that \(\Sigma\) is a plane, then \(\gamma\) can be considered as a 2D curve, and both \(\mathbf v\) and \(\mathbf a\) resides on the plane.

However, \(\Sigma\) is not always a plane. For each \(t, \mathbf v(t) \in T_{\gamma(t)}\Sigma\) being a vector field, which is (most likely) constantly changing. Since each tangent plane is spanned by \(\sigma_u(t), \sigma_v(t)\), we can write

\[\mathbf v(t) = a(t)\sigma_u(t) + b(t)\sigma_v(t)\]

Then, acceleration \(\mathbf a= \mathbf v'\) is a 3D vector, hence can be spanned by \(\{\sigma_u(t), \sigma_v(t), \mathbf N(t)\}\). Then, consider the particle's perspective, since it moves along the surface, from its perspective, it treats its movement as on a plane, and only perceive the change on this plane (think of car's velocity, while earth is actually a sphere). Therefore, we can interested in the perceived rate of change of \(\mathbf v\) as the tangential component of \(\mathbf v'\), a.k.a. the orthogonal projection of \(\mathbf v'\) onto \(T_{\gamma(t)}\Sigma\), as

\[\nabla_\gamma \mathbf v = \mathbf v' - (\mathbf v' \cdot \mathbf N)\mathbf N\]

Note that orthogonal projection is unchanged of the sign of \(\mathbf N\).

Therefore, we define \(\nabla_\gamma \mathbf v: (a,b)\rightarrow T_{\gamma(t)}\Sigma\) as the covariant derivative of \(\mathbf v\) along \(\gamma\). If \(\nabla_\gamma\mathbf v = 0\) for all \(t\), then \(\mathbf v\) is parallel along \(\gamma\).

Theorem 1

\(\mathbf v\) is parallel along \(\gamma\) IFF \(\mathbf v'\) is perpendicular to the tangent plane of \(\Sigma\) for all \(t\).

proof. This is obvious, since \(\nabla_\gamma\mathbf v\) is the projection of \(\mathbf v'\) onto \(T_p\Sigma\), if \(\mathbf v'\) is perpendicular to \(T_p\Sigma\), then it is parallel to \(\mathbf N\), and its projection onto \(T_p\Sigma\) is always 0.

Theorem 2

\(\mathbf v\) is parallel along \(\gamma\) IFF

\[a' + (\Gamma_{11}^1 u' +\Gamma_{12}^1v')a + (\Gamma_{12}^1u' + \Gamma_{22}^1v')b = 0\]

\[b' + (\Gamma_{11}^2 u' + \Gamma_{12}^2v')a + (\Gamma_{12}^2u' + \Gamma_{22}^2v')b = 0\]

For \(\mathbf v(t) = a(t)\sigma_u (t) + b(t)\sigma_v(t), a,b\) are smooth scalar functions.

proof. We have that \(\mathbf v'\)

\[\begin{align*} \mathbf v' &= a' \sigma_u + a(\sigma_{uu}u'+\sigma_{uv}v') + b'\sigma_v + b(\sigma_{uv}u' + \sigma_{vv}v')\\ &= a'\sigma_u + b'\sigma_v + au'\sigma_{uu} + (av'+bu')\sigma_{uv} + bv'\sigma_{vv} \end{align*}\]

where each of \(\sigma_{uu}, \sigma_{uv}, \sigma_{vv}\) can be written into Gauss equations, and only involves \(\sigma_u, \sigma_v, \mathbf N\).
Then, since \(\mathbf v\) is parallel along \(\gamma\), we must have that \(\mathbf v' = \lambda N\) and all coefficient of \(\sigma_{u}, \sigma_{v}\) must be \(0\), a.k.a the equations in the statement must be zero.

Theorem 3

Let \(\gamma: (a,b)\rightarrow \Sigma\) and some \(\mathbf v_0\) be a tangent vector of \(\Sigma\) at \(\gamma(t_0)\), then \(\exists !\mathbf v\) s.t. \(\mathbf v\) is parallel along \(\gamma\) and \(\mathbf v(t_0) = \mathbf v_0\).

proof. Note that the statement is equivalent to say that there exists unique scalar function \(a,b\) s.t.

\[a(t_0)\sigma_u(t_0) + b(t_0) \sigma_v(t_0) = \mathbf v_0\]

\[a' + (\Gamma_{11}^1 u' +\Gamma_{12}^1v')a + (\Gamma_{12}^1u' + \Gamma_{22}^1v')b = 0\]

\[b' + (\Gamma_{11}^2 u' + \Gamma_{12}^2v')a + (\Gamma_{12}^2u' + \Gamma_{22}^2v')b = 0\]

Then, this is a ODE

\[a' = f(a(t), b(t), t) = -((\Gamma_{11}^1 u' +\Gamma_{12}^1v')a + (\Gamma_{12}^1u' + \Gamma_{22}^1v')b)\]

\[b' = g(a(t), b(t), t) = -((\Gamma_{11}^2 u' + \Gamma_{12}^2v')a + (\Gamma_{12}^2u' + \Gamma_{22}^2v')b)\]

given the initial conditions, it is proven to have a unique solution.

Example: circle on sphere

Let \(\gamma\) be a circle of latitute \(\theta_0 \in (-\pi/2, \pi)\) on the standard unit sphere. \(\gamma(t) = \sigma(\theta_0, t)\), and the first fundamental form is \(E=1, F=0, G=\cos^2\theta\).
The Christoffel symbosls are then

\[\Gamma_{12}^1 = -\tan\theta, \Gamma_{22}^1 =\sin\theta\cos\theta\]

others are all 0.

Therefore, we have that

\[a' = -(\cos\theta_0\sin\theta_0)b, b' = \tan\theta_0 a\]

solves to be

\[a = A\cos(t\sin\theta_0) + B\sin(t\sin\theta_0)\]

\[b = A\frac{\sin(t\sin\theta_0)}{\cos\theta_0} - B\frac{\cos(t\sin\theta_0)}{\sin\theta_0}\]

Consider the case when \(t = 0\), then \(a = 0, b = 1\), which gives \(A = 0, B = -\sin\theta_0\) so that

\[\mathbf v(t) = -\sin\theta_0 \sin(t\sin\theta_0)\sigma_\theta + \cos(t\sin\theta_0)\sigma_\phi\]

Thus the tangent vector of \(\gamma\) is parallel along \(\gamma\) IFF \(\gamma\) is a great circle.

Example: Triangle on the Sphere

Suppose that a triangle \(T\) on the unit sphere whose sides are arcs of great circles has vertices \(p, q, r\). Let \(v_0\) be a non-zero tangent vector to the arc \(pq\) through \(p, q\) at \(p\). Show that, if we parallel transport \(v_0\) along \(pq\), then \(qr\) and then \(rp\), the result is to rotate \(v_0\) through an angle \(2\pi - A(T)\).

proof. Since the sides of the triangle are arcs of great circle. The parallel transporting of the tirangle sides is always parallel along the arc. Suppose that the internal angles are \(a,b,c\), then at each point of \(p, q, r\), the angle is \(\pi -a,\pi-b, \pi-c\). Therefore, the total angle made is

\[\pi-a+\pi-b+\pi -c = 2\pi - (a+b+c - \pi) = 2\pi - A(T)\]