# Gauss's lemma (Riemannian geometry)

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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"::$mathrm\left\{exp\right\} : T_pM o M$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:$exp_p:T_pMsupset B_\left\{epsilon\right\}\left(0\right) longrightarrow M,qquad vlongmapsto gamma\left(1, p, v\right),$where we have had to restrict the domain $T_pM$ by definition of a ball $B_epsilon\left(0\right)$ of radius $epsilon>0$ and centre $0$ to ensure that $exp_p$ is well-defined, and where $gamma\left(1,p,v\right)$ is the point $qin M$ reached by following the unique geodesic $gamma$ passing through the point $pin M$ with tangent $frac\left\{v\right\}\left\{vert vvert\right\}in T_pM$ for a distance $vert vvert$. It is easy to see that $exp_p$ is a local diffeomorphism around $0in B_epsilon\left(0\right)$. Let $alpha : I ightarrow T_pM$ be a curve differentiable in $T_pM$ such that $alpha\left(0\right):=0$ and $alpha\text{'}\left(0\right):=v$. Since $T_pMcong mathbb R^n$, it is clear that we can choose $alpha\left(t\right):=vt$. In this case, by the definition of the differential of the exponential in $0$ applied over $v$, we obtain:

:$T_0exp_p\left(v\right) = frac\left\{mathrm d\right\}\left\{mathrm d t\right\} Bigl\left(exp_pcircalpha\left(t\right)Bigr\right)Bigvert_\left\{t=0\right\} = frac\left\{mathrm d\right\}\left\{mathrm d t\right\} Bigl\left(exp_p\left(vt\right)Bigr\right)Bigvert_\left\{t=0\right\}=frac\left\{mathrm d\right\}\left\{mathrm d t\right\} Bigl\left(gamma\left(1,p,vt\right)Bigr\right)Bigvert_\left\{t=0\right\}= gamma\text{'}\left(t,p,v\right)Bigvert_\left\{t=0\right\}=v.$

The fact that $exp_p$ is a local diffeomorphism and that $T_0exp_p\left(v\right)=v$ for all $vin B_epsilon\left(0\right)$ allows us to state that $exp_p$ is a local isometry around $0$, i.e.

:$langle T_0exp_p\left(v\right), T_0exp_p\left(w\right) angle_0 = langle v, w angle_pqquadforall v,win B_epsilon\left(0\right).$

This means in particular that it is possible to identify the ball $B_epsilon\left(0\right)subset T_pM$ 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 $v,win B_epsilon\left(0\right)subset T_vT_pMcong T_pM$ and $M i q:=exp_p\left(v\right)$. Then, :$langle T_vexp_p\left(v\right), T_vexp_p\left(w\right) angle_v = langle v,w angle_q.$

For $pin M$, this lemma means that $exp_p$ is a radial isometry in the following sense: let $vin B_epsilon\left(0\right)$, i.e. such that $exp_p$ is well defined. Moreover, let $q:=exp_p\left(v\right)in M$. Then the exponential $exp_p$ remains an isometry in $q$, and, more generally, all along the geodesic $gamma$ (in so far as $gamma\left(1,p,v\right)=exp_p\left(v\right)$ 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

:$T_vexp_p : T_pMcong T_vT_pMsupset T_vB_epsilon\left(0\right)longrightarrow T_\left\{exp_p\left(v\right)\right\}M.$

We proceed in three steps:
* "$T_vexp_p\left(v\right)=v$" : let us construct a curve $alpha : mathbb R supset I ightarrow T_pM$ such that $alpha\left(0\right):=vin T_pM$ and $alpha\text{'}\left(0\right):=vin T_vT_pMcong T_pM$. Since $T_vT_pMcong T_pMcong mathbb R^n$, we can put $alpha\left(t\right):=v\left(t+1\right)$. 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 $alpha\left(t\right) = vt$ (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,

:$T_vexp_p\left(v\right) = frac\left\{mathrm d\right\}\left\{mathrm d t\right\}Bigl\left(exp_pcircalpha\left(t\right)Bigr\right)Bigvert_\left\{t=0\right\}=frac\left\{mathrm d\right\}\left\{mathrm d t\right\}gamma\left(t,p,v\right)Bigvert_\left\{t=0\right\} = v.$

Now let us calculate the scalar product $langle T_vexp_p\left(v\right), T_vexp_p\left(w\right) angle$.

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:

:$langle T_vexp_p\left(v\right), T_vexp_p\left(w\right) angle = langle T_vexp_p\left(v\right), T_vexp_p\left(w_T\right) angle + langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle$

::$=alphalangle T_vexp_p\left(v\right), T_vexp_p\left(v\right) angle + langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle=langle v, w_T angle + langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle.$

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

$langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle = langle v, w_N angle = 0.$

* "$langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle = 0$" : Let us define the curve

:$alpha : \right] -epsilon, epsilon \left[ imes \left[0,1\right] longrightarrow T_pM,qquad \left(s,t\right) longmapsto tcdot v\left(s\right),$with $v\left(0\right):=v$ and $v\text{'}\left(0\right):=w_N$. We remark in passing that::$alpha\left(0,1\right) = v\left(0\right) = v,qquadfrac\left\{partial alpha\right\}\left\{partial t\right\}\left(0,t\right) = v\left(0\right) = v,qquadfrac\left\{partial alpha\right\}\left\{partial s\right\}\left(0,t\right) = tw_N.$

Let us put:

:$f : \right] -epsilon, epsilon \left[ imes \left[0,1\right] longrightarrow M,qquad \left(s,t\right)longmapsto exp_p\left(tcdot v\left(s\right)\right),$

and we calculate:

:$T_vexp_p\left(v\right)=T_\left\{alpha\left(0,1\right)\right\}exp_pleft\left(frac\left\{partial alpha\right\}\left\{partial t\right\}\left(0,1\right) ight\right)=frac\left\{partial\right\}\left\{partial t\right\}Bigl\left(exp_pcircalpha\left(s,t\right)Bigr\right)Bigvert_\left\{t=1, s=0\right\}=frac\left\{partial f\right\}\left\{partial t\right\}\left(0,1\right)$and:$T_vexp_p\left(w_N\right)=T_\left\{alpha\left(0,1\right)\right\}exp_pleft\left(frac\left\{partial alpha\right\}\left\{partial s\right\}\left(0,1\right) ight\right)=frac\left\{partial\right\}\left\{partial s\right\}Bigl\left(exp_pcircalpha\left(s,t\right)Bigr\right)Bigvert_\left\{t=1,s=0\right\}=frac\left\{partial f\right\}\left\{partial s\right\}\left(0,1\right).$Hence:$langle T_vexp_p\left(v\right), T_vexp_p\left(w_N\right) angle = langle frac\left\{partial f\right\}\left\{partial t\right\},frac\left\{partial f\right\}\left\{partial s\right\} angle\left(0,1\right).$We can now verify that this scalar product is actually independent of the variable $t$, and therefore that, for example:

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

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

* Riemannian geometry
* Metric tensor

References

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

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