Conjugate points in do Carmo's Riemannian Geometry

Question

Picture below is from the do Carmo's Riemannian Geometry.

First, what is the mean of "not identically zero" ? Does it mean that $J(t)\neq 0~~~\forall t\in [0,a]$ ? If so, it is contradictory with $J(0)=0=J(t_0)$. Therefore, I think it means that there is at least $\tilde t\in [0,a]$ such that $J(\tilde t) \ne 0$, yes or no ?

Second, why $J$ is non-zero if and only if $\omega \ne 0$ ? Seemly, it should be got from $J$ is a non-zero Jacobi field along $\gamma$. But it is also contradictory with $J(0)=J(t_0)=0$. So, how to understand it ?

PS: I feel my English is too poor to misunderstand the author, but I really don't know the mean of it.

Answer 1

1. For a function $f$ to be identically zero means that for every entry of that function, the result is zero, or in other words, $f$ is the zero function. Hence, $J$ is identically zero if $\forall t \in [0,a], J(t) = 0$. Its negation in english is "$J$ is not identically zero" and in mathematical language is "$\exists t \in [0,a], J(t) \neq 0$".

2. Recall that a Jacobi field $J$ along a geodesic $\gamma$ is a solution to the second order linear differential equation $J'' + R_{\gamma}J = 0$. A solution $J$ to this second order linear differential equation is uniquely determined by its initial data $J(0)$ and $J'(0)$. As $J_0(t)=0$ is solution to the Jacobi fields equation and satisfies $J_0(0) = J_0'(0) = 0$, it is the only solution with these initial data. It follows that if one requires $J(0)=0$, then $J$ is identically zero if and only if $J'(0) = 0$.