# Gauss's lemma (Riemannian geometry)

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 T$_p$M 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
    $\exp _{p}:T_{p}M\supset B_{\epsilon }(0)\longrightarrow M,\quad vt\longmapsto \gamma _{p,v}(t),$
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 $t\in [0,1],vt\in B_{\epsilon }(0)\subset T_{p}M$ the geodesic $\gamma _{p,v}(t)$ is defined. 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:
    $T_{0}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl\exp _{p}(vt){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl\gamma _{p,v}(t){\Bigr )}{\Big \vert }_{t=0}=\gamma _{p,v}'(0)=v.$
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 $\gamma _{p,v}(1)=\exp _{p}(v)$ is well defined)! Then, radially, in all the directions permitted by the domain of definition of $\exp _{p}$, it remains an isometry.

## 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:
$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.$
* $\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =0$ :
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,
    ${\frac {\partial }{\partial s}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle ={\frac {\partial }{\partial s}}\left\langle v+sw_{N},v+sw_{N}\right\rangle =2\left\langle v,w_{N}\right\rangle =0.$