A rather elementary question for the differential geometers. Let $M$ be a Riemannian manifold, let $X\colon M \to TM$ be a vector field, and let $$\phi_t = \exp(tX)$$ be the "geodesic flow" of the vector field $X$, that is $\phi_t(p) = \exp_p(t X(p))$ for $p \in M$. I cannot seem to find anything about this notion of flow, does it perhaps have a name I am unaware of?

I am interested in the directional derivative along the flow, that is $$\frac{d}{dt} (f \circ \phi_t)(p) = \nabla_{\dot{\phi}_t(p)} f(\phi_t(p))$$ for fixed $p$ and varying $t$. I would hope that, because $\dot{\phi}_t(p)$ is the parallel transport of $X(p)$ along the geodesic $\gamma$ with $\gamma(0) = p$, $\dot{\gamma}(0) = X(p)$, there might be an expression involving only $X$ and the map $\phi_t$.


I think the question might be whether one can use the fact that $\phi_t$ is actually a map on $M$. The directional derivative is e.g. in terms of the exterior derivative$$\nabla_{\dot{\phi}_t(p)} f(\phi_t(p)) = df_{\phi_t(p)}(\dot{\phi}_t(p)).$$ If, instead of the parallel transport $\dot{\phi}_t(p)$, we had the pushforward $d\phi_t X$ of $X$ by $\phi_t$, then this would simply be the pullback $$df_{\phi_t(p)}(d\phi_t X) = d(\phi_t^*f)_p(X) = X(\phi_t^*f)_p.$$ So I guess this boils down to a question about the difference between the pushforward of the exponential map and parallel transport, which should be a property of the (smoothness?) of the vector field $X$. Is there a way to make this idea more precise?

  • $\begingroup$ If you understand the tangent space as set of point derivations, then the directional derivative is the vector field itself. $\endgroup$ – Zero Aug 16 at 4:08
  • $\begingroup$ Since you’re fixing $p$, doesn’t your expression depend only on $X(p)$? I don’t see any alternative to saying it’s the directional derivative in the direction of the parallel translate of $X$. $\endgroup$ – Deane Yang Aug 16 at 14:12
  • $\begingroup$ @DeaneYang Well, you are right of course that there is nothing deep about the problem. I am merely curious if there is a straightforward way to bring the computation back to $t = 0$, if you will, and to understand when that can or cannot be done. My particular reason is that I have $X$ and $f \circ \phi_t$ readily available, but cannot easily evaluate the parallel transport. $\endgroup$ – user144437 Aug 16 at 14:18
  • $\begingroup$ In that case there’s no need to define $X$ as a vector field. You’re looking simply at the geodesic starting at a point $p$ with initial velocity X(p). Since the parallel transport depends on the geometry of the space along the geodesic and not just at $p$, I don’t see how to write the directional derivative using only data at $p$. $\endgroup$ – Deane Yang Aug 16 at 15:51
  • $\begingroup$ @DeaneYang I agree, and perhaps it is naive to expect that anything can be said at all. But I am still thinking that something must be known about the derivative from the structure of the maps: Looking at e.g. this somewhat related question I take it that the difference between $\dot{\phi}_t(p)$ and $d\phi_t X$ is second order in $tX$ -- that would be a start. $\endgroup$ – user144437 Aug 16 at 16:49

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