Definition 1: Let $$M$$ be a differentiable manifold with an affine connection $$\nabla$$. A vector field along a curve $$c:I\to V$$ is called parallel when $$\dfrac{DV}{dt}=0$$ for every $$t\in I$$.

Definition 2: Let $$M$$ be a differentiable manifold with an affine connection $$\nabla$$. Let $$c:I\to M$$ be a differentiable curve in $$M$$ and $$V_0\in T_{c(t_0)}M$$ with $$t_0\in I$$. Then there exists a unique parallel vector field $$V$$ along $$c$$, such that $$V(t_0)=V_0$$. $$V(t)$$ is called the parallel transport of $$V(t_0)$$ along $$c$$.

These definitions are from Do Carmo, Riemannian Geometry.

I don't think we can give an explicit formula for $$V(t)$$, which makes me wonder how to solve the many problems about the parallel transport. For example the first exercise is:

Let $$M$$ be a Riemannian manifold. Consider the mapping $$P=P_{c,t_0,t}:T_{c(t_0)}M\to T_{c(t)}M$$ defined by: $$P_{c,t_0,t}(v), v\in T_{c(t_0)}M$$, is the vector obtained by parallel transporting the vector $$v$$ along the curve $$c$$. Show that $$P$$ is an isometry and that, if $$M$$ is oriented, $$P$$ preserves the orientation.

So, basically my question is how do you use just the definition to prove $$P$$ is an isometry?

Thank you very much.

• the definition of $V_0 \in T_{\gamma(0)} M$ parallel transported along a curve is $\gamma$ is $\nabla_{\dot{\gamma}(t)} V(t) = 0$ for every $t$, with $V(t)$ being a function associating a vector of $T_{\gamma(t)} M$ to every point $\gamma(t)$ of the curve. from this definition, if $\nabla$ preserves the metric, then it is clear that $g(V(t),V(t)) = g(V_0,V_0)$, because of the definition https://en.wikipedia.org/wiki/Affine_connection#The_Levi-Civita_connection Commented May 19, 2016 at 23:17

Let $(M,g)$ a Riemannian manifold with $\nabla$ any $g$-compatible connection, then $\frac{g(V(t),V(t))}{dt}=g(\nabla_{\dot c} V(t),V(t))+g(V(t),\nabla_{\dot c}V(t))=0$ so $g(V(t),V(t))$ is constant so $P$ preserves lengths.
You have $0=\nabla_Vg$ which is the same as $0=\sum_{i,j} \nabla_V(g_i\otimes g_j)=\sum_{i,j}(\nabla_Vg_i\otimes g_j+g_i\otimes \nabla_Vg_j)$ where $g_i,j$ are 1-forms. Then we get: $$(\nabla_Vg)(X,Y)=\sum_{i.j}\nabla_Vg_i(X)g_j(Y)+g_i(X)\nabla_Vg_j(Y))$$ now by the rule of covariant derives in 1-forms we get $(\nabla_Vg_i)(X)=V(g_i(X))-g_i(\nabla_VX)$ by putting this in the previous formula we get: $$\sum_{i,j} V(g_i(X))g_j(Y)-g_i(\nabla_V(X))g_j(Y)+g_i(X)V(g_j(Y))-g_i(X)g_j(\nabla_V(Y))=V(g(X,Y)-(g(\nabla_VX,Y)+g(X,\nabla_VY))$$
• how do you get your formula for the derivative of the norm from the definitions and $\nabla g = 0$ ? Commented May 20, 2016 at 6:58
• you meant en.wikipedia.org/wiki/Metric_connection#Riemannian_connections $$\nabla_X \, g(Y,Z)=g(\nabla_X\, Y,Z)+g(Y,\nabla_X \, Z)$$ Commented May 20, 2016 at 7:09
• yep with $X=\dot c$ Commented May 20, 2016 at 7:14
• How can you assume that $\nabla$ is $g$-compatible from the start? Commented Jun 6, 2022 at 1:49