Gauss's lemma (Riemannian geometry)

This article is about Gauss's lemma in Riemannian geometry. For other uses, see Gauss's lemma.

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 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 p\in M by

\exp_p: T_pM\supset B_{\epsilon}(0) \longrightarrow M,\quad v\longmapsto \gamma_{p,v}(1),

where \gamma_{p,v}\ is the unique geodesic with \gamma(0)=p and tangent \gamma_{p,v}'(0)=v \in T_pM and \epsilon_0 is chosen small enough so that for every v \in B_{\epsilon}(0) \subset T_pM the geodesic \gamma_{p,v} is defined in 1. So, if M is complete, then, by the Hopf–Rinow theorem, \exp_p is defined on the whole tangent space.

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.

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

The exponential map is a radial isometry

Let p\in M. In what follows, we make the identification 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_q = \langle v,w\rangle_p.

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. And 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

Proof

Recall that

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


We proceed in three steps:

\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 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 because of the domain of definition I; however, the identification allows us to gather them 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\ parallel to v\ and a component w_N\ normal to v\ . In particular, we put w_T:=a v\ , a \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
=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:

\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \langle v, w_N\rangle = 0.

The curve chosen to prove lemma

Let us define the curve

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

Note that

\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:

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

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 = \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 t\ , and therefore that, for example:

\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:

\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.

\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.

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

\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.

See also

References

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