# derivation of Jacobi field equation (from do Carmo's riemannian geometry)

I have been reading about Jacobi fields in do Carmo's book "Riemannian geometry", and I have a little question.

Let $$(M,g)$$ be a riemannian manifold and let $$p\in M$$ and $$v\in T_p M$$ such that $$\exp_p v$$ is defined.

It begins with a parametrized surface $$f(t,s)=\exp_{p}tv(s)\qquad 0\leq t\leq 1,\, -\varepsilon\leq s\leq\varepsilon$$ where $$v(s)$$ is a curve in $$T_p M$$ such that $$v(0)=v$$ and $$v'(0)=w\in T_v(T_p M)$$.
By the Gauss lemma we have $$(d\exp_p)_v w=\frac{\partial f}{\partial s}(1,0)$$ So far so good. Then it goes on by saying

"It is convenient to extend our objective slightly and study the field $$(d\exp_p)_{tv}(tw)=\frac{\partial f}{\partial s}(t,0)$$ along the geodesic $$\gamma(t)=\exp_p(tv)$$."

Since $$\gamma$$ is a geodesic, we have $$\nabla_{\dot\gamma}\dot\gamma=0$$. But in the book they claim that we also have $$\nabla_{\dot\gamma(t)}\frac{\partial f}{\partial t}(t,s)=0$$ for all $$(t,s)$$. Why is that?
I see why this holds for $$s=0$$, since $$v(0)=v$$.
But why would that be for $$s\neq 0$$?

Any help would be very much appreciated!

• It is not true that $\dot{\gamma}(t)=v$ for all $t$. In fact, the above phrase does not even make any sense, as $v$ is a tangent vector at $p$ whereas $\gamma(t)\ne p$ for $t\ne0$. – Amitai Yuval Mar 16 at 14:03
• On the other hand, letting $\gamma_s(t):=f(t,s)$, the path $\gamma_s$ is a geodesic for any $s$. By construction, we have $\dot{\gamma}_s(t)=\partial f/\partial t$, and the claim follows. – Amitai Yuval Mar 16 at 14:07
• @AmitaiYuval thank you, i have edited my question. So we have $\nabla_{\dot\gamma}=\nabla_{\dot\gamma_0}$. Why would that imply $\nabla_{\dot\gamma_0}\dot\gamma_s=0$ for all $s$? – Pink Panther Mar 16 at 14:30
• No, this is not what do Carmo says. The direction of derivative, $\dot{\gamma}(t)$ is taken with respect to the varying value of $s$. Otherwise it makes no sense. – Amitai Yuval Mar 16 at 15:12
• @AmitaiYuval I think I got it now: In the statement (this is a quote from the book) "In fact, since $\gamma$ is a geodesic, we have for all $(t,s), \frac{D}{dt}\frac{\partial f}{\partial t}=0$", the 'since' seems kinda redundant then, right? Because it seems clear that $f(t,s)=\gamma_s(t)$ is a geodesic for fixed $s$. Also the $\frac{D}{dt}\dot\gamma_s$ then becomes $\nabla_{\dot\gamma_s}\dot\gamma_s$, right? – Pink Panther Mar 17 at 0:26