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I'm reading DoCarmo's book, Riemannian Geometry and i dont understand a step in the proof of this lemma.

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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?

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  • $\begingroup$ This what I've done so far, $p,q\in W$ so $q=\text{exp}_p(w)$ thus $u(0,q,v)=\text{exp}_p^{-1}(\gamma(0,q,v))=\text{exp}_p^{-1}(q)=w$ From the fact that $\gamma$ si tangent to the $S_r(p)$ we get that $<v,w>=0.$ I don't know how to use the Gauss lemma or how to calculate that derivative.. $\endgroup$ – Hurjui Ionut Sep 7 '18 at 22:01
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    $\begingroup$ The Gauss lemma says that geodesics emanating from $p$ are orthogonal to the (tangent space of the) geodesic sphere $S_r(p)$. $\gamma$ is a geodesic passing through $q$ at time $t=0$ with tangent vector $v\in T_qS_r(p)$. Thus, going back by $\exp_p^{-1}$, the position vector $q$ is orthogonal to the tangent vector to the preimage of $S_r(p)$ at $q$. $\endgroup$ – Ted Shifrin Sep 7 '18 at 22:54
  • $\begingroup$ ok, got this, but i still don't get the fact that $<\frac{\partial u}{\partial t}(0,q,v), u(0,q,v)>=0$ :( $\endgroup$ – Hurjui Ionut Sep 8 '18 at 9:51
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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.

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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.

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