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## Summary

In Riemannian geometry, Gauss's lemma asserts that any sufficiently small sphere centered at a point in a Riemannian manifold is perpendicular to every geodesic through the point. More formally, let M be a Riemannian manifold, equipped with its Levi-Civita connection, and p a point of M. The exponential map is a mapping from the tangent space at p to M:

$\mathrm {exp} :T_{p}M\to M$ which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in TpM under the exponential map is perpendicular to all geodesics originating at p. The lemma allows the exponential map to be understood as a radial isometry, and is of fundamental importance in the study of geodesic convexity and normal coordinates.

## Introduction

We define the exponential map at $p\in M$  by

$\exp _{p}:T_{p}M\supset B_{\epsilon }(0)\longrightarrow M,\quad v\longmapsto \gamma _{p,v}(1),$

where $\gamma _{p,v}$  is the unique geodesic with $\gamma _{p,v}(0)=p$  and tangent $\gamma _{p,v}'(0)=v\in T_{p}M$  and $\epsilon$  is chosen small enough so that for every $v\in B_{\epsilon }(0)\subset T_{p}M$  the geodesic $\gamma _{p,v}$  is defined in 1. So, if $M$  is complete, then, by the Hopf–Rinow theorem, $\exp _{p}$  is defined on the whole tangent space.

Let $\alpha :I\rightarrow T_{p}M$  be a curve differentiable in $T_{p}M$  such that $\alpha (0):=0$  and $\alpha '(0):=v$ . Since $T_{p}M\cong \mathbb {R} ^{n}$ , it is clear that we can choose $\alpha (t):=vt$ . In this case, by the definition of the differential of the exponential in $0$  applied over $v$ , we obtain:

$T_{0}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}(vt){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\gamma (1,p,vt){\Bigr )}{\Big \vert }_{t=0}=\gamma '(t,p,v){\Big \vert }_{t=0}=v.$

So (with the right identification $T_{0}T_{p}M\cong T_{p}M$ ) the differential of $\exp _{p}$  is the identity. By the implicit function theorem, $\exp _{p}$  is a diffeomorphism on a neighborhood of $0\in T_{p}M$ . The Gauss Lemma now tells that $\exp _{p}$  is also a radial isometry.

## The exponential map is a radial isometry

Let $p\in M$ . In what follows, we make the identification $T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}$ .

Gauss's Lemma states: Let $v,w\in B_{\epsilon }(0)\subset T_{v}T_{p}M\cong T_{p}M$  and $M\ni q:=\exp _{p}(v)$ . Then, $\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle _{q}=\langle v,w\rangle _{p}.$

For $p\in M$ , this lemma means that $\exp _{p}$  is a radial isometry in the following sense: let $v\in B_{\epsilon }(0)$ , i.e. such that $\exp _{p}$  is well defined. And let $q:=\exp _{p}(v)\in M$ . Then the exponential $\exp _{p}$  remains an isometry in $q$ , and, more generally, all along the geodesic $\gamma$  (in so far as $\gamma (1,p,v)=\exp _{p}(v)$  is well defined)! Then, radially, in all the directions permitted by the domain of definition of $\exp _{p}$ , it remains an isometry.

The exponential map as a radial isometry

## Proof

Recall that

$T_{v}\exp _{p}\colon T_{p}M\cong T_{v}T_{p}M\supset T_{v}B_{\epsilon }(0)\longrightarrow T_{\exp _{p}(v)}M.$

We proceed in three steps:

• $T_{v}\exp _{p}(v)=v$  : let us construct a curve

$\alpha :\mathbb {R} \supset I\rightarrow T_{p}M$  such that $\alpha (0):=v\in T_{p}M$  and $\alpha '(0):=v\in T_{v}T_{p}M\cong T_{p}M$ . Since $T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}$ , we can put $\alpha (t):=v(t+1)$ . Therefore,

$T_{v}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}(tv){\Bigr )}{\Big \vert }_{t=1}=\Gamma (\gamma )_{p}^{\exp _{p}(v)}v=v,$

where $\Gamma$  is the parallel transport operator and $\gamma (t)=\exp _{p}(tv)$ . The last equality is true because $\gamma$  is a geodesic, therefore $\gamma '$  is parallel.

Now let us calculate the scalar product $\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle$ .

We separate $w$  into a component $w_{T}$  parallel to $v$  and a component $w_{N}$  normal to $v$ . In particular, we put $w_{T}:=av$ , $a\in \mathbb {R}$ .

The preceding step implies directly:

$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle =\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{T})\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle$
$=a\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(v)\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{T}\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle .$

We must therefore show that the second term is null, because, according to Gauss's Lemma, we must have:

$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.$

• $\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =0$  :

The curve chosen to prove lemma

Let us define the curve

$\alpha \colon [-\epsilon ,\epsilon ]\times [0,1]\longrightarrow T_{p}M,\qquad (s,t)\longmapsto tv+tsw_{N}.$

Note that

$\alpha (0,1)=v,\qquad {\frac {\partial \alpha }{\partial t}}(s,t)=v+sw_{N},\qquad {\frac {\partial \alpha }{\partial s}}(0,t)=tw_{N}.$

Let us put:

$f\colon [-\epsilon ,\epsilon ]\times [0,1]\longrightarrow M,\qquad (s,t)\longmapsto \exp _{p}(tv+tsw_{N}),$

and we calculate:

$T_{v}\exp _{p}(v)=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial t}}(0,1)\right)={\frac {\partial }{\partial t}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial t}}(0,1)$

and

$T_{v}\exp _{p}(w_{N})=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial s}}(0,1)\right)={\frac {\partial }{\partial s}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial s}}(0,1).$

Hence

$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle (0,1).$

We can now verify that this scalar product is actually independent of the variable $t$ , and therefore that, for example:

$\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle (0,1)=\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle (0,0)=0,$

because, according to what has been given above:

$\lim _{t\rightarrow 0}{\frac {\partial f}{\partial s}}(0,t)=\lim _{t\rightarrow 0}T_{tv}\exp _{p}(tw_{N})=0$

being given that the differential is a linear map. This will therefore prove the lemma.

• We verify that ${\frac {\partial }{\partial t}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle =0$ : this is a direct calculation. Since the maps $t\mapsto f(s,t)$  are geodesics,
${\frac {\partial }{\partial t}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle =\left\langle \underbrace {{\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}} _{=0},{\frac {\partial f}{\partial s}}\right\rangle +\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial t}}{\frac {\partial f}{\partial s}}\right\rangle =\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\right\rangle ={\frac {1}{2}}{\frac {\partial }{\partial s}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle .$

Since the maps $t\mapsto f(s,t)$  are geodesics, the function $t\mapsto \left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle$  is constant. Thus,

${\frac {\partial }{\partial s}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle ={\frac {\partial }{\partial s}}\left\langle v+sw_{N},v+sw_{N}\right\rangle =2\left\langle v,w_{N}\right\rangle =0.$