If $(M,g)$ is a riemannian manifold, and $N$ a submanifold in $M$ with $\sigma$ a geodesic normal to $P$ at $p=\sigma(0)$ under the hypotheses:
-$H(\sigma'(0))=g\left(\sigma'(0)),\vec{H_p}\right)>0$
-Ric($\sigma',\sigma'$)$\geq 0$
-$\sigma$ defined in at least in $[0,1/H(\sigma'(0))]$
I know how to prove that $\sigma$ admits a conjugate point ($P-$conjugate since $\sigma(0)$ is normal to $P$). My question is how can I prove the same with $H(\sigma'(0))\geq0$, $\sigma$ defined now in $[0,\infty)$ and Ric($\sigma',\sigma'$)$\geq K> 0$. Maybe I can modifify the following proof for the previous result, using also Myers theorem:
Let $e_1,\dots,e_m$ be a orthonormal basis of $T_pP$, and $E_1,\dots,E_m$ its parallel transport.With $f(t)=1-kt$ with $k=1/H(\sigma'(0))$ and $V=fE_i$ the expresion for the index form tell us $$ I(V_i,V_i)=k-\int_o^{1/k}f^2g(R(\sigma',E_i)E_i,\sigma')dt-g(\sigma'(0),II(e_i,e_i)), $$ where $II$ is the second fundamental form and we have used that $\int_0^{1/k}g(V',V')dt=k^2/k=k$. Adding the expresions we have $$ \sum I(V_i,V_i)=mk-\int_0^{1/k}f^2 \text{Ric}(\sigma',\sigma')dt-g(\sigma'(0),m\vec{H}_p)=-\int_0^{1/k}f^2 \text{Ric}(\sigma',\sigma')dt\leq 0 $$
So the index form is not positive definite and thefore we have a conjugate ($P-$conjugate) point in $(0,1/k]$.
Can someone give me a hint?