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:

$\mathrm{exp} : T_pM \to M$

which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in T_pM 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.

1 Introduction
2 The exponential map is a radial isometry
3 Proof
4 See also
5 References

[edit] Introduction

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

$\exp_p:T_pM\supset B_{\epsilon}(0) \longrightarrow M,\qquad v\longmapsto \gamma(1, p, v),$

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 $q\in M$ reached by following the unique geodesic $\gamma\$ passing through the point $p\in M$ with tangent $\frac{v}{\vert v\vert}\in T_pM$ for a distance $\vert v\vert\$ . It is easy to see that $\exp_p\$ is a local diffeomorphism around $0\in B_\epsilon(0)$ . Let $\alpha : I\rightarrow T_pM$ be a curve differentiable in $T_pM\$ such that $\alpha(0):=0\$ and $\alpha'(0):=v\$ . Since $T_pM\cong \mathbb R^n$ , 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:

$T_0\exp_p(v) = \frac{\mathrm d}{\mathrm d t} \Bigl(\exp_p\circ\alpha(t)\Bigr)\Big\vert_{t=0} = \frac{\mathrm d}{\mathrm d t} \Bigl(\exp_p(vt)\Bigr)\Big\vert_{t=0}=\frac{\mathrm d}{\mathrm d t} \Bigl(\gamma(1,p,vt)\Bigr)\Big\vert_{t=0}= \gamma'(t,p,v)\Big\vert_{t=0}=v.$

The fact that $\exp_p\$ is a local diffeomorphism and that $T_0\exp_p(v)=v\$ for all $v\in B_\epsilon(0)$ allows us to state that $\exp_p\$ is a local isometry around $0\$ , i.e.

$\langle T_0\exp_p(v), T_0\exp_p(w)\rangle_0 = \langle v, w\rangle_p\qquad\forall v,w\in B_\epsilon(0).$

The exponential map as a local isometry

This means in particular that it is possible to identify the ball $B_\epsilon(0)\subset T_pM$ with a small neighbourhood around $p\in M$ . 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 !

[edit] The exponential map is a radial isometry

Let $p\in M$ . In what follows, we make the identification Failed to parse (Cannot write to or create math output directory): T_vT_pM\cong T_pM\cong \mathbb R^n . Gauss's Lemma states:

Let $v,w\in B_\epsilon(0)\subset T_vT_pM\cong T_pM$ and $M\ni q:=\exp_p(v)$ . Then,

$\langle T_v\exp_p(v), T_v\exp_p(w)\rangle_v = \langle v,w\rangle_q.$

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. 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.

The exponential map as a radial isometry

[edit] Proof

Recall that

$T_v\exp_p : T_pM\cong T_vT_pM\supset T_vB_\epsilon(0)\longrightarrow T_{exp_p(v)}M.$

We proceed in three steps:

$T_v\exp_p(v)=v\$ : let us construct a curve $\alpha : \mathbb R \supset I \rightarrow T_pM$ such that $\alpha(0):=v\in T_pM$ and $\alpha'(0):=v\in T_vT_pM\cong T_pM$ . Since Failed to parse (Cannot write to or create math output directory): T_vT_pM\cong T_pM\cong \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,

$T_v\exp_p(v) = \frac{\mathrm d}{\mathrm d t}\Bigl(\exp_p\circ\alpha(t)\Bigr)\Big\vert_{t=0}=\frac{\mathrm d}{\mathrm d t}\gamma(t,p,v)\Big\vert_{t=0} = v.$

Now let us calculate the scalar product $\langle T_v\exp_p(v), T_v\exp_p(w)\rangle$ .

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\$ , $\alpha\in \mathbb R$ .

The preceding step implies directly:

$\langle T_v\exp_p(v), T_v\exp_p(w)\rangle = \langle T_v\exp_p(v), T_v\exp_p(w_T)\rangle + \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle$

$=\alpha\langle T_v\exp_p(v), T_v\exp_p(v)\rangle + \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle=\langle v, w_T\rangle + \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle.$

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

$\alpha : ]-\epsilon, \epsilon[\times [0,1] \longrightarrow T_pM,\qquad (s,t) \longmapsto t\cdot v(s),$

with $v(0):=v\$ and $v'(0):=w_N\$ . We remark in passing that:

$\alpha(0,1) = v(0) = v,\qquad\frac{\partial \alpha}{\partial t}(0,t) = v(0) = v,\qquad\frac{\partial \alpha}{\partial s}(0,t) = tw_N.$

Let us put:

$f : ]-\epsilon, \epsilon[\times [0,1] \longrightarrow M,\qquad (s,t)\longmapsto \exp_p(t\cdot v(s)),$

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

$\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\rangle(0,1).$

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

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

because, according to what has been given above:

$\lim_{t\rightarrow 0}\frac{\partial f}{\partial s}(t,0) = \lim_{t\rightarrow 0}T_{tv}\exp_p(tw_N) = 0$

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

We verify that $\frac{\partial}{\partial t}\langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\rangle=0$ : this is a direct calculation. We first take account of the fact that the maps $t\mapsto f(s,t)$ are geodesics, i.e. $\frac{D}{\partial t}\frac{\partial f}{\partial t}=0$ . Therefore,

$\frac{\partial}{\partial t}\langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\rangle=\langle\underbrace{\frac{D}{\partial t}\frac{\partial f}{\partial t}}_{=0}, \frac{\partial f}{\partial s}\rangle+\langle\frac{\partial f}{\partial t},\frac{D}{\partial t}\frac{\partial f}{\partial s}\rangle=\langle\frac{\partial f}{\partial t},\frac{D}{\partial s}\frac{\partial f}{\partial t}\rangle=\frac{\partial }{\partial s}\langle \frac{\partial f}{\partial t}, \frac{\partial f}{\partial t}\rangle - \langle\frac{\partial f}{\partial t},\frac{D}{\partial s}\frac{\partial f}{\partial t}\rangle.$

Hence, in particular,

$0=\frac{1}{2}\frac{\partial }{\partial s}\langle \frac{\partial f}{\partial t}, \frac{\partial f}{\partial t}\rangle= \langle\frac{\partial f}{\partial t},\frac{D}{\partial s}\frac{\partial f}{\partial t}\rangle=\frac{\partial}{\partial t}\langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\rangle,$

because, since the maps $t\mapsto f(s,t)$ are geodesics, we have $\langle\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}\rangle=\mathrm{cste}$ .