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 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 $p\in M$ 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:

• $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 $T_vT_pM\cong T_pM\cong \mathbb R^n$, 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:

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

• $\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = 0$ :

The curve chosen to prove lemma

Let us define the curve

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

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.

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

$\frac{\partial}{\partial t}\left\langle \frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\right\rangle=\left\langle\underbrace{\frac{D}{\partial t}\frac{\partial f}{\partial t}}_{=0}, \frac{\partial f}{\partial s}\right\rangle + \left\langle\frac{\partial f}{\partial t},\frac{D}{\partial t}\frac{\partial f}{\partial s}\right\rangle = \left\langle\frac{\partial f}{\partial t},\frac{D}{\partial s}\frac{\partial f}{\partial t}\right\rangle=\frac12\frac{\partial }{\partial s}\left\langle \frac{\partial f}{\partial t}, \frac{\partial f}{\partial t}\right\rangle.$

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

• Riemannian geometry
• Metric tensor

References

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