Gauss's lemma (Riemannian geometry)

Short description: Theorem in manifold theory

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:

[math]\displaystyle{ \mathrm{exp} : T_pM \to M }[/math]

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.

Introduction

We define the exponential map at [math]\displaystyle{ p\in M }[/math] by

[math]\displaystyle{ \exp_p: T_pM\supset B_{\epsilon}(0) \longrightarrow M,\quad vt \longmapsto \gamma_{p,v}(t), }[/math]

where [math]\displaystyle{ \gamma_{p,v} }[/math] is the unique geodesic with [math]\displaystyle{ \gamma_{p,v}(0)=p }[/math] and tangent [math]\displaystyle{ \gamma_{p,v}'(0)=v \in T_pM }[/math] and [math]\displaystyle{ \epsilon }[/math] is chosen small enough so that for every [math]\displaystyle{ t \in [0, 1], vt \in B_{\epsilon}(0) \subset T_pM }[/math] the geodesic [math]\displaystyle{ \gamma_{p,v}(t) }[/math] is defined. So, if [math]\displaystyle{ M }[/math] is complete, then, by the Hopf–Rinow theorem, [math]\displaystyle{ \exp_p }[/math] is defined on the whole tangent space.

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

[math]\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_{p, v}(t)\Bigr)\Big\vert_{t=0}= \gamma_{p, v}'(0)=v. }[/math]

So (with the right identification [math]\displaystyle{ T_0 T_p M \cong T_pM }[/math]) the differential of [math]\displaystyle{ \exp_p }[/math] is the identity. By the implicit function theorem, [math]\displaystyle{ \exp_p }[/math] is a diffeomorphism on a neighborhood of [math]\displaystyle{ 0 \in T_pM }[/math]. The Gauss Lemma now tells that [math]\displaystyle{ \exp_p }[/math] is also a radial isometry.

The exponential map is a radial isometry

Let [math]\displaystyle{ p\in M }[/math]. In what follows, we make the identification [math]\displaystyle{ T_vT_pM\cong T_pM\cong \mathbb R^n }[/math].

Gauss's Lemma states: Let [math]\displaystyle{ v,w\in B_\epsilon(0)\subset T_vT_pM\cong T_pM }[/math] and [math]\displaystyle{ M\ni q:=\exp_p(v) }[/math]. Then, [math]\displaystyle{ \langle T_v\exp_p(v), T_v\exp_p(w)\rangle_q = \langle v,w\rangle_p. }[/math]

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

The exponential map as a radial isometry

Proof

Recall that

[math]\displaystyle{ T_v\exp_p \colon T_pM\cong T_vT_pM\supset T_vB_\epsilon(0)\longrightarrow T_{\exp_p(v)}M. }[/math]

We proceed in three steps:

• [math]\displaystyle{ T_v\exp_p(v)=v }[/math] : let us construct a curve

[math]\displaystyle{ \alpha : \mathbb R \supset I \rightarrow T_pM }[/math] such that [math]\displaystyle{ \alpha(0):=v\in T_pM }[/math] and [math]\displaystyle{ \alpha'(0):=v\in T_vT_pM\cong T_pM }[/math]. Since [math]\displaystyle{ T_vT_pM\cong T_pM\cong \mathbb R^n }[/math], we can put [math]\displaystyle{ \alpha(t):=v(t+1) }[/math]. Therefore,

[math]\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}\Bigl(\exp_p(tv)\Bigr)\Big\vert_{t=1}=\Gamma(\gamma)_p^{\exp_p(v)}v=v, }[/math]

where [math]\displaystyle{ \Gamma }[/math] is the parallel transport operator and [math]\displaystyle{ \gamma(t)=\exp_p(tv) }[/math]. The last equality is true because [math]\displaystyle{ \gamma }[/math] is a geodesic, therefore [math]\displaystyle{ \gamma' }[/math] is parallel.

Now let us calculate the scalar product [math]\displaystyle{ \langle T_v\exp_p(v), T_v\exp_p(w)\rangle }[/math].

We separate [math]\displaystyle{ w }[/math] into a component [math]\displaystyle{ w_T }[/math] parallel to [math]\displaystyle{ v }[/math] and a component [math]\displaystyle{ w_N }[/math] normal to [math]\displaystyle{ v }[/math]. In particular, we put [math]\displaystyle{ w_T:=a v }[/math], [math]\displaystyle{ a \in \mathbb R }[/math].

The preceding step implies directly:

[math]\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 }[/math]

[math]\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. }[/math]

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

[math]\displaystyle{ \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \langle v, w_N\rangle = 0. }[/math]

• [math]\displaystyle{ \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = 0 }[/math] :

The curve chosen to prove lemma

Let us define the curve

[math]\displaystyle{ \alpha \colon [-\epsilon, \epsilon]\times [0,1] \longrightarrow T_pM,\qquad (s,t) \longmapsto tv+tsw_N. }[/math]

Note that

[math]\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. }[/math]

Let us put:

[math]\displaystyle{ f \colon [-\epsilon, \epsilon ]\times [0,1] \longrightarrow M,\qquad (s,t)\longmapsto \exp_p(tv+tsw_N), }[/math]

and we calculate:

[math]\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) }[/math]

and

[math]\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). }[/math]

Hence

[math]\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). }[/math]

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

[math]\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, }[/math]

because, according to what has been given above:

[math]\displaystyle{ \lim_{t\rightarrow 0}\frac{\partial f}{\partial s}(0,t) = \lim_{t\rightarrow 0}T_{tv}\exp_p(tw_N) = 0 }[/math]

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

• We verify that [math]\displaystyle{ \frac{\partial}{\partial t}\left\langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\right\rangle=0 }[/math]: this is a direct calculation. Since the maps [math]\displaystyle{ t\mapsto f(s,t) }[/math] are geodesics,

[math]\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=\frac12\frac{\partial }{\partial s}\left\langle \frac{\partial f}{\partial t}, \frac{\partial f}{\partial t}\right\rangle. }[/math]

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

[math]\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. }[/math]

See also

• Riemannian geometry
• Metric tensor

References

• do Carmo, Manfredo (1992), Riemannian geometry, Basel, Boston, Berlin: Birkhäuser, ISBN 978-0-8176-3490-2