The paraboloid $z = x^2 + y^2$ has no conjugate points at $p = (0,0,0)$ Can someone check if I am correct? This is exercise 2 in Do Carmo (Surface and curves) in chapter 5.5.
Let $S = \{ z = x^2 + y^2 \}$ with geodesic $\gamma(0) = p = (0,0,0)$ then clearly $T_pS = \{ (x,y,0) \}$. 
Now since the paraboloid has positive Gaussian curvature, we cannot invoke the theorem involving empty conjugate locus. So I thought we have to go to Proposition5 to show that the exponential map $\exp_p:T_pS \to S$ is regular. That is it has no critical points (and hence no conjugate points relative to $p$). But I can't think of how to do it without knowing what the map should be.
So for the surface $\sigma (x,y) = (x,y,x^2 + y^2)$, the velocity of the curve on the surface is $\gamma' = (x',y',0)$. Now it is also equivalent to show that the Jacobi field along this geodesic is in fact $0$. 
By Corollory 4 (of proposition 4), Jacobi fields along $\gamma$ with $J(0) = J(\ell) = 0$ must be orthogonal to the velocity, that is $(J(s),\gamma'(s)) = 0$.
Hence by inspection, one could take $J(s) = (-y',x',-)$. Here is where I am not sure if I can do this. Since $\gamma$ is a geodesic, $\gamma'' = (x'',y'',z'') = (0,0,0)$. Thus $J'(s) = (-y'',x'',-) = (0,0,0)$. Now I want to say something like since $J = 0$ on the end points and it has vanishing derivative that $J = 0$. But there is a lot of shaky arguments I have written. 
EDIT: It just occured to me that in fact the last component of $J(s)$ is $0$. But whether $J(s)=(-y',x',0)$ is debatable since we can insert a smooth function $h(s)$ such that $J(s) = -h(s)(-y',x',0)$. 
 A: Let $S$ be the paraboloid $S = \{(x,y,z) : z = x^2 + y^2\}$, and let $\gamma : [0, a] \to S$ be a geodesic starting at $p = (0,0,0)$. Let $J$ be a Jacobi field along $\gamma$ with $J(0) = 0$. By the definition of a Jacobi field given in do Carmo's Curves and Surfaces on page 363, we can find a smooth variation $f : (-\varepsilon, \varepsilon) \times [0, a] \to S$ of $\gamma$ such that

*

*$J$ is the variational field of $f$, i.e. $J(t) = \frac{\partial f}{\partial s}(0, t)$, and

*for each $s \in (-\varepsilon, \varepsilon)$, the variational curve $f_s(t) = f(s, t)$ is a geodesic starting from $p$.

Any geodesic of the paraboloid $S$ starting at $p = (0,0,0)$ must be a meridian, which is proven in Example 5 on page 258 of do Carmo. Therefore
$$
f_s(t) = (t \cos \theta(s), t \sin \theta(s), t^2)
$$
(up to reparametrization in $t$) for some angle function $\theta(s)$. We show that $\theta : (-\varepsilon, \varepsilon) \to \mathbb{R}$, defined by the above equation, is differentiable at $0$. If $\phi(u,v) = (v \cos u, v \sin u, v^2)$ is the parametrization of $S$ described in the aforementioned example, then for sufficiently small $s$, we can write the function $\theta$ as
$$
\theta(s) = (\pi_2 \circ \phi^{-1})(f(s, a)).
$$
As a composition of differentiable functions (for sufficiently small $s$, so that $f(s,a)$ is in a neighbourhood of $\gamma(a)$ in which $\phi$ is a diffeomorphism), this must be differentiable at $s = 0$. (In fact, it is differentiable for all sufficiently small $s$.)
Now, since $J$ is the variational field, we see that if $J(a) = 0$, then
$$
(0,0,0) = J(a) = \frac{\partial f}{\partial s}(0, a) = (-a \theta'(0) \sin \theta(0), a \theta'(0) \cos \theta(0), 0).
$$
Since $a \neq 0$, and since $\sin \theta(0)$ and $\cos \theta(0)$ are not simultaneously zero, we have that $\theta'(0) = 0$. Then, for any $t$,
$$
J(t) = \frac{\partial f}{\partial s}(0, t) = (-t \theta'(0) \sin \theta(0), t \theta'(0) \cos \theta(0), 0) = (0, 0, 0).
$$
But then $J \equiv 0$. Therefore no non-trivial Jacobi field along $\gamma$ can vanish at more than one point, which proves that there are no conjugate points to $p$ along $\gamma$.
