Let $r(x)=d_g(x,p), p\in (M,g)$ denote the radial distance function on a Riemannian manifold $(M,g).$ I'm studying the Hessian comparison theorem from "Lee: Introduction to Riemannian manifolds", and I see on P. 321 that:
So if I think of the above point ($q$)-wise, this gives that for constant section curvature space with constant sectional curvature $c,$ the Hessian $Hess(r)_{q}:T_qM\to T_qM, Hess(r)(q)=\frac{s_c'(r(q)=d(p,q))}{s_c(r(q)=d(p,q))}\pi_r(q).$ So at any point $q$ in a normal neighborhood of $p,Hess(r)_q,$ as a bilinear form on $T_qM$(or linear operator on $T_qM$), is proportional to the bilinear form or linear operator on $T_qM$ that projects every vector $v\in T_qM$ to its component orthogonal to the unit vector $\partial_r, i.e. \pi_r (v):= v- g_q(v,\partial_r)\partial_r.$ But this means that $\pi_r$ doesn't have full rank, and since Hessian is pointwise proportional to $\pi_r, Hess(r)$ also doesn't have full rank (perhaps its rank is exactly equal to $dim(M)-1?$ I wanted to check if the my conslusion about the rank definitness of the Hessian is correct, and if yes, is $rank(Hess_r(q):T_qM\to T_qM)$ exactly $dim(M)-1?$