# Does parallel transport change the subspace?

Let $$M$$ be a Riemannian manifold and $$N$$ be an immersed submanifold. Take $$\gamma$$ a geodesic starting at and perpendicularly to $$N$$. Let $$X(t)$$ be a vector field along $$\gamma$$ such that $$X(0) \in T_{\gamma(0)}N$$. Assume that $$X(t)$$ is not necessarily parallel along $$\gamma$$. Take an instant $$t_0$$ and consider $$X(t_0)$$. Denote by $$E_X(t)$$ the parallel transport of $$X(t_0)$$ along $$\gamma$$. Does $$E_X(0) \in T_{\gamma(0)}N$$?

• I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) \in T_{\gamma(0)}N$? Is that right? Cheers! – Robert Lewis Feb 2 at 20:12
• @RobertLewis, yes, I will rephrase. – L.F. Cavenaghi Feb 2 at 20:12

Consider the plane $$\{(x, y, 0)\mid x,y \in \Bbb R\}\subset\mathbb{R}^3$$, the line $$t\mapsto (0, 0, t)$$ and the vector field $$t\mapsto(1, 1, t)$$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.

Negative answer. Counter-example: $$(M,g) = (\Bbb R^3, \langle\cdot, \cdot \rangle)$$, $$N = \Bbb R^2\times \{0\}$$, $$\gamma(t) = (0,0,t)$$ and $$X(t) = (0,0,\sin t)$$. Then $$X(0) = 0 \in T_{\gamma(0)}N$$, but the parallel transport of $$X(\pi/2) = (0,0,1)$$ from $$\gamma(\pi/2) = (0,0,\pi/2)$$ to $$\gamma(0) =0$$ via $$\gamma$$ is $$(0,0,1) \not\in T_{\gamma(0)}N$$. The values $$X(t)$$ for $$t\neq t_0$$ should not influence the result of transporting $$X(t_0)$$, since a priori one transports vectors and not fields.

• I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it. – L.F. Cavenaghi Feb 3 at 3:36

We of course understand that $$\dot \gamma(0) \ne 0$$, lest $$\gamma(t)$$ be a rather trivial geodesic.

We observe that, since $$\dot \gamma(0)$$ is "perpendicular" to $$N$$, the condition

$$\langle \dot \gamma(0), X(0) \rangle = 0 \tag 0$$

is necessary, but not sufficient, for

$$X(0) \in T_{\gamma(0)}N, \tag 1$$

for (0) says that $$X(0)$$ has no component along $$\dot \gamma(0)$$; in the event that $$N$$ is a hypersurface in $$M$$, that is,

$$\dim N = \dim M - 1, \tag 2$$

(0) is in fact also sufficient, for then $$\dot \gamma(0)$$ and $$T_{\gamma(0)}N$$ span $$T_{\gamma}(0)N$$; in fact, (2) implies

$$T_{\gamma(0)}M = \langle \dot \gamma(0) \rangle \oplus T_{\gamma(0)}N, \tag 3$$

where $$\langle \dot \gamma(0) \rangle$$ is the one-dimensional subspace of $$T_{\gamma(0)}M$$ generated by $$\dot \gamma(0)$$.

Now consider any vector field $$X(t)$$ along $$\gamma(t)$$; we have

$$\nabla_{\dot \gamma} \langle \dot \gamma, X \rangle = \langle \nabla_{\dot \gamma} \dot \gamma, X \rangle + \langle \dot \gamma, \nabla_{\dot \gamma} X \rangle; \tag 4$$

if $$X(t)$$ is parallel along $$\gamma(t)$$, then

$$\nabla_{\dot \gamma}X = 0; \tag 5$$

thus (4) becomes

$$\nabla_{\dot \gamma} \langle \dot \gamma, X \rangle = \langle \nabla_{\dot \gamma} \dot \gamma, X \rangle; \tag 6$$

now the general formulation of the geodesic equation as I understand it is

$$\nabla_{\dot \gamma} \dot \gamma = f(t) \dot \gamma; \tag 7$$

here $$f(t)$$ is determined by the choice of the running parameter $$t$$ of $$\gamma(t)$$; if we substitute this into (6), we obtain

$$\nabla_{\dot \gamma} \langle \dot \gamma, X \rangle = f(t) \langle \dot \gamma, X \rangle; \tag 8$$

the solution to this linear, first order differential equation for $$\langle \dot \gamma, X \rangle$$ with $$\langle \dot \gamma(t_0), X(t_0) \rangle = \langle \dot \gamma, X \rangle(t_0)$$ is

$$\langle \dot \gamma, X \rangle(t) = \exp \left (\displaystyle \int_{t_0}^t f(s) \; ds \right ) \langle \dot \gamma, X \rangle (t_0); \tag 9$$

since

$$\forall t \in \Bbb R, \;\exp \left (\displaystyle \int_{t_0}^t f(s) \; ds \right ) \ne 0, \tag{10}$$

(9) shows that

$$\langle \dot \gamma(t), X(t) \rangle = \langle \dot \gamma, X \rangle(t) = 0 \Longleftrightarrow \langle \dot \gamma, X \rangle (t_0) = \langle \dot \gamma(t_0), X(t_0) \rangle = 0; \tag{11}$$

of course, if $$\gamma(t)$$ is affinely parametrized then

$$f(t) = 0, \tag{12}$$

and we in fact have

$$\langle \dot \gamma, X \rangle(t) = \langle \dot \gamma, X \rangle (t_0), \; \forall t \in \Bbb R; \tag{13}$$

now taking $$t_0 = 0$$ yields

$$\langle \dot \gamma(t), X(t) \rangle = 0 \Longleftrightarrow \langle \dot \gamma(0), X(0) \rangle = 0; \tag{14}$$

if we now choose $$X(t)$$ such that

$$\langle \dot \gamma(t), X(t) \rangle \ne 0, \tag{15}$$

(14) together with (0) shows the parallel transport of $$X(t)$$ along $$\gamma(t)$$ cannot satisfy

$$X(0) \in T_{\gamma(0)} N, \tag{15}$$

since it has a component in the direction $$\dot \gamma(0)$$.