I'm reading DoCarmo's book, Riemannian Geometry and i dont understand a step in the proof of this lemma.
Namely the last one in the ss. I don't get the fact that inner product is zero. And also i don't know how to use Gauss's lemma in that regard. Can some one fill in the details for me please?
On
By shifting 0 to a $w\in T_pM$ in Proposition 2.9, $d(\exp_p)_w$ is an identity between $T_wT_pM\to T_pM$. Let $w:=u(0,q,v)$. We now use a tilde to denote a point in $T_wT_pM$. For example, $v=d(\exp_p)_w(\tilde v)$. Then it is easy to see from the Gauss Lemma, we have \begin{align*} \langle\frac{\partial u}{\partial t}(0,q,v),\tilde{w}\rangle=\langle v,w\rangle=0, \end{align*} where the first equality comes from the definition of differentiate and the second equality comes from the tangent condition. Since $u(0,q,v)\in T_pM$ and $\partial u/\partial t(0,q,v)\in T_wT_pM$, I think the inner product will be confusing if we don't use the identity as an accessory.
I‘ll ignore $q$ and $v$ in the following, so that $\gamma$ is a unit speed geodesic and $u(t)=exp_p^{-1}(\gamma(t)).$ Furthermore let $\sigma(t)=exp_p(tu(1)).$
Then by the Gauss Lemma \begin{equation} \langle u‘(t),u(0)\rangle=\langle (dexp_p)_{u(0)}(u‘(0)),\sigma‘(1)\rangle=\langle \gamma‘(0),\sigma‘(1)\rangle. \end{equation}
The fact that $\gamma‘(0)$ is tangent to the geodesic sphere means that this expression is $0$ because geodesics which pass through $p$ are orthogonal to spheres centered at $p$ and because $\sigma$ is such a geodesic. This in turn follows again by the Gauss Lemma as follows:
Let $w\in T_q S(r,p),$ where $q=exp_p(v)$ and $\sigma(t)=exp_p(tv)$ is the geodesic connecting $p$ and $q.$ Then there exists a curve $v(s)$ in $T_pM$ with constant length, $v(0)=v$ and $(dexp_p)_{v}(v‘(0))=w.$ Then the Gauss Lemma inpliess \begin{equation} \langle w, \sigma‘(1)\rangle =\langle (dexp_p)_{v}(v‘(0)), (dexp_p)_{v}(v)\rangle =\langle v‘(0),v(0)\rangle=0, \end{equation} where the lase equality follows because $||v(s)||$ is constant.