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.

  • $\begingroup$ 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 $\endgroup$ – reuns May 19 '16 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.

Edit: see comments.

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: $$\sum_{i,j} V(g_i(X))g_j(Y)-g_i(\nabla_V(X))g_j(Y)+g_i(X)V(g_j(Y))-g_i(X)g_j(\nabla_V(Y))=V(g(X,Y)-(g(\nabla_VX,Y)+g(X,\nabla_VY))$$

  • $\begingroup$ how do you get your formula for the derivative of the norm from the definitions and $\nabla g = 0$ ? $\endgroup$ – reuns May 20 '16 at 6:58
  • $\begingroup$ is an equivalent definition: en.wikipedia.org/wiki/Metric_connection#Metric_compatibility $\endgroup$ – Spotty May 20 '16 at 7:05
  • $\begingroup$ you meant en.wikipedia.org/wiki/Metric_connection#Riemannian_connections $$\nabla_X \, g(Y,Z)=g(\nabla_X\, Y,Z)+g(Y,\nabla_X \, Z)$$ $\endgroup$ – reuns May 20 '16 at 7:09
  • $\begingroup$ yep with $X=\dot c$ $\endgroup$ – Spotty May 20 '16 at 7:14
  • $\begingroup$ I can't find how to write $\nabla_X (g(Y,Z))$ in general when the connection isn't Riemannian... (using the chain rule ?) $\endgroup$ – reuns May 20 '16 at 7:16

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