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 \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.

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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).

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 !

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.

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%2B1). 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 %2B \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 %2B \langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle=\langle v, w_T\rangle %2B \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.

Let us define the curve

\alpha \colon [-\epsilon, \epsilon]\times [0,1] \longrightarrow T_pM,\qquad (s,t) \longmapsto tv%2Btsw_N.

Note that

\alpha(0,1) = v,\qquad \frac{\partial \alpha}{\partial t}(s,t) = v%2Bsw_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%2Btsw_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 %2B \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%2Bsw_N,v%2Bsw_N\right\rangle =2\left\langle v,w_N\right\rangle=0.

See also

References