# Gauss's lemma (Riemannian geometry)

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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:

${\displaystyle \mathrm {exp} :T_{p}M\to M}$

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 ${\displaystyle p\in M}$ by

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

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

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

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

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

## The exponential map is a radial isometry

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

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

The exponential map as a radial isometry

## Proof

Recall that

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

We proceed in three steps:

${\displaystyle \alpha :\mathbb {R} \supset I\rightarrow T_{p}M}$ such that ${\displaystyle \alpha (0):=v\in T_{p}M}$ and ${\displaystyle \alpha '(0):=v\in T_{v}T_{p}M\cong T_{p}M}$. Since ${\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}}$, we can put ${\displaystyle \alpha (t):=v(t+1)}$. We find that, thanks to the identification we have made, and since we are only taking equivalence classes of curves, it is possible to choose ${\displaystyle \alpha (t)=vt\ }$ (these are exactly the same curves, but shifted because of the domain of definition ${\displaystyle I}$; however, the identification allows us to gather them around ${\displaystyle 0}$. Hence,

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

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

The preceding step implies directly:

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

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

The curve chosen to prove lemma

Let us define the curve

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

Note that

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

Let us put:

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

and we calculate:

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

and

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

Hence

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

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

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

because, according to what has been given above:

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

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

${\displaystyle {\frac {\partial }{\partial t}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle =\left\langle \underbrace {{\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}} _{=0},{\frac {\partial f}{\partial s}}\right\rangle +\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial t}}{\frac {\partial f}{\partial s}}\right\rangle =\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\right\rangle ={\frac {1}{2}}{\frac {\partial }{\partial s}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle .}$

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

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

## References

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