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"::$mathrm\{exp\}\; :\; T\_pM\; o\; M$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 on $M$ the exponential map at $pin\; M$ by:$exp\_p:T\_pMsupset\; B\_\{epsilon\}(0)\; longrightarrow\; M,qquad\; vlongmapsto\; gamma(1,\; p,\; v),$where we have had to restrict the domain $T\_pM$ by definition of a ball $B\_epsilon(0)$ of radius $epsilon>0$ and centre $0$ to ensure that $exp\_p$ is well-defined, and where $gamma(1,p,v)$ is the point $qin\; M$ reached by following the unique geodesic $gamma$ passing through the point $pin\; M$ with tangent $frac\{v\}\{vert\; vvert\}in\; T\_pM$ for a distance $vert\; vvert$. It is easy to see that $exp\_p$ is a local diffeomorphism around $0in\; B\_epsilon(0)$. Let $alpha\; :\; I\; ightarrow\; T\_pM$ be a curve differentiable in $T\_pM$ such that $alpha(0):=0$ and $alpha\text{'}(0):=v$. Since $T\_pMcong\; mathbb\; R^n$, 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:

:$T\_0exp\_p(v)\; =\; frac\{mathrm\; d\}\{mathrm\; d\; t\}\; Bigl(exp\_pcircalpha(t)Bigr)Bigvert\_\{t=0\}\; =\; frac\{mathrm\; d\}\{mathrm\; d\; t\}\; Bigl(exp\_p(vt)Bigr)Bigvert\_\{t=0\}=frac\{mathrm\; d\}\{mathrm\; d\; t\}\; Bigl(gamma(1,p,vt)Bigr)Bigvert\_\{t=0\}=\; gamma\text{'}(t,p,v)Bigvert\_\{t=0\}=v.$

The fact that $exp\_p$ is a local diffeomorphism and that $T\_0exp\_p(v)=v$ for all $vin\; B\_epsilon(0)$ allows us to state that $exp\_p$ is a local isometry around $0$, i.e.

:$langle\; T\_0exp\_p(v),\; T\_0exp\_p(w)\; angle\_0\; =\; langle\; v,\; w\; angle\_pqquadforall\; v,win\; B\_epsilon(0).$

This means in particular that it is possible to identify the ball $B\_epsilon(0)subset\; T\_pM$ with a small neighbourhood around $pin\; M$. We can see that $exp\_p$ is a local isometry, but we would like it to be rather more than that. We assert that it is in fact possible to show that this map is a radial isometry !

The exponential map is a radial isometry

Let $pin\; M$. In what follows, we make the identification $T\_vT\_pMcong\; T\_pMcong\; mathbb\; R^n$.Gauss's Lemma states:

Let $v,win\; B\_epsilon(0)subset\; T\_vT\_pMcong\; T\_pM$ and $M\; i\; q:=exp\_p(v)$. Then, :$langle\; T\_vexp\_p(v),\; T\_vexp\_p(w)\; angle\_v\; =\; langle\; v,w\; angle\_q.$

For $pin\; M$, this lemma means that $exp\_p$ is a radial isometry in the following sense: let $vin\; B\_epsilon(0)$, i.e. such that $exp\_p$ is well defined. Moreover, 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.

Proof

Recall that

:$T\_vexp\_p\; :\; T\_pMcong\; T\_vT\_pMsupset\; T\_vB\_epsilon(0)longrightarrow\; T\_\{exp\_p(v)\}M.$

We proceed in three steps:
* "$T\_vexp\_p(v)=v$" : let us construct a curve $alpha\; :\; mathbb\; R\; supset\; I\; ightarrow\; T\_pM$ such that $alpha(0):=vin\; T\_pM$ and $alpha\text{'}(0):=vin\; T\_vT\_pMcong\; T\_pM$. Since $T\_vT\_pMcong\; T\_pMcong\; mathbb\; R^n$, we can put $alpha(t):=v(t+1)$. 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 $alpha(t)\; =\; vt$ (these are exactly the same curves, but shifted (###décalées###), because of the domain of definition $I$; however, the identification allows us to gather them (###ramener###) around $0$ !!!). Hence,

:$T\_vexp\_p(v)\; =\; frac\{mathrm\; d\}\{mathrm\; d\; t\}Bigl(exp\_pcircalpha(t)Bigr)Bigvert\_\{t=0\}=frac\{mathrm\; d\}\{mathrm\; d\; t\}gamma(t,p,v)Bigvert\_\{t=0\}\; =\; v.$

Now let us calculate the scalar product $langle\; T\_vexp\_p(v),\; T\_vexp\_p(w)\; angle$.

We separate $w$ into a component $w\_T$ tangent to $v$ and a component $w\_N$ normal to $v$. In particular, we put $w\_T:=alpha\; v$, $alphain\; mathbb\; R$.

The preceding step implies directly:

:$langle\; T\_vexp\_p(v),\; T\_vexp\_p(w)\; angle\; =\; langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_T)\; angle\; +\; langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle$

::$=alphalangle\; T\_vexp\_p(v),\; T\_vexp\_p(v)\; angle\; +\; langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle=langle\; v,\; w\_T\; angle\; +\; langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle.$

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

$langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle\; =\; langle\; v,\; w\_N\; angle\; =\; 0.$

* "$langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle\; =\; 0$" : Let us define the curve

:$alpha\; :\; ]\; -epsilon,\; epsilon\; [\; imes\; [0,1]\; longrightarrow\; T\_pM,qquad\; (s,t)\; longmapsto\; tcdot\; v(s),$with $v(0):=v$ and $v\text{'}(0):=w\_N$. We remark in passing that::$alpha(0,1)\; =\; v(0)\; =\; v,qquadfrac\{partial\; alpha\}\{partial\; t\}(0,t)\; =\; v(0)\; =\; v,qquadfrac\{partial\; alpha\}\{partial\; s\}(0,t)\; =\; tw\_N.$

Let us put:

:$f\; :\; ]\; -epsilon,\; epsilon\; [\; imes\; [0,1]\; longrightarrow\; M,qquad\; (s,t)longmapsto\; exp\_p(tcdot\; v(s)),$

and we calculate:

:$T\_vexp\_p(v)=T\_\{alpha(0,1)\}exp\_pleft(frac\{partial\; alpha\}\{partial\; t\}(0,1)\; ight)=frac\{partial\}\{partial\; t\}Bigl(exp\_pcircalpha(s,t)Bigr)Bigvert\_\{t=1,\; s=0\}=frac\{partial\; f\}\{partial\; t\}(0,1)$and:$T\_vexp\_p(w\_N)=T\_\{alpha(0,1)\}exp\_pleft(frac\{partial\; alpha\}\{partial\; s\}(0,1)\; ight)=frac\{partial\}\{partial\; s\}Bigl(exp\_pcircalpha(s,t)Bigr)Bigvert\_\{t=1,s=0\}=frac\{partial\; f\}\{partial\; s\}(0,1).$Hence:$langle\; T\_vexp\_p(v),\; T\_vexp\_p(w\_N)\; angle\; =\; langle\; frac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle(0,1).$We can now verify that this scalar product is actually independent of the variable $t$, and therefore that, for example:

:$langlefrac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle(0,1)\; =\; langlefrac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle(0,0)\; =\; 0,$because, according to what has been given above::$lim\_\{t\; ightarrow\; 0\}frac\{partial\; f\}\{partial\; s\}(t,0)\; =\; lim\_\{t\; ightarrow\; 0\}T\_\{tv\}exp\_p(tw\_N)\; =\; 0$being given that the differential is a linear map! This will therefore prove the lemma.
* We verify that "$frac\{partial\}\{partial\; t\}langle\; frac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle=0$" : this is a direct calculation. We first take account of the fact that the maps $tmapsto\; f(s,t)$ are geodesics, i.e. $frac\{D\}\{partial\; t\}frac\{partial\; f\}\{partial\; t\}=0$. Therefore,

:$frac\{partial\}\{partial\; t\}langle\; frac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle=langleunderbrace\{frac\{D\}\{partial\; t\}frac\{partial\; f\}\{partial\; t\_\{=0\},\; frac\{partial\; f\}\{partial\; s\}\; angle+langlefrac\{partial\; f\}\{partial\; t\},frac\{D\}\{partial\; t\}frac\{partial\; f\}\{partial\; s\}\; angle=langlefrac\{partial\; f\}\{partial\; t\},frac\{D\}\{partial\; s\}frac\{partial\; f\}\{partial\; t\}\; angle=frac\{partial\; \}\{partial\; s\}langle\; frac\{partial\; f\}\{partial\; t\},\; frac\{partial\; f\}\{partial\; t\}\; angle\; -\; langlefrac\{partial\; f\}\{partial\; t\},frac\{D\}\{partial\; s\}frac\{partial\; f\}\{partial\; t\}\; angle.$Hence, in particular,:$0=frac\{1\}\{2\}frac\{partial\; \}\{partial\; s\}langle\; frac\{partial\; f\}\{partial\; t\},\; frac\{partial\; f\}\{partial\; t\}\; angle=\; langlefrac\{partial\; f\}\{partial\; t\},frac\{D\}\{partial\; s\}frac\{partial\; f\}\{partial\; t\}\; angle=frac\{partial\}\{partial\; t\}langle\; frac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; s\}\; angle,$because, since the maps $tmapsto\; f(s,t)$ are geodesics, we have $langlefrac\{partial\; f\}\{partial\; t\},frac\{partial\; f\}\{partial\; t\}\; angle=mathrm\{cste\}$.

See also

* Riemannian geometry
* Metric tensor

References

* [http://www.amazon.fr/dp/0817634908]