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'':
:$\mathrm : T_pM \to M$
which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in ''T''_p''M'' 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
:$\exp_p: T_pM\supset B_(0) \longrightarrow M,\quad vt \longmapsto \gamma_(t),$
where $\gamma_$ is the unique geodesic with $\gamma_(0)=p$ and tangent $\gamma_'(0)=v \in T_pM$ and $\epsilon$ is chosen small enough so that for every $t \in$, 1 vt \in B_(0) \subset T_pM the geodesic $\gamma_(t)$ is defined. 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:

:$T_0\exp_p(v) = \frac \Bigl(\exp_p\circ\alpha(t)\Bigr)\Big\vert_ = \frac \Bigl(\exp_p(vt)\Bigr)\Big\vert_=\frac \Bigl(\gamma_(t)\Bigr)\Big\vert_= \gamma_'(0)=v.$
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 $\gamma_(1)=\exp_p(v)$ 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

:$T_v\exp_p \colon T_pM\cong T_vT_pM\supset T_vB_\epsilon(0)\longrightarrow T_M.$

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 $\alpha(t):=v(t+1)$.
Therefore,

$T_v\exp_p(v) = \frac\Bigl(\exp_p\circ\alpha(t)\Bigr)\Big\vert_=\frac\Bigl(\exp_p(tv)\Bigr)\Big\vert_=\Gamma(\gamma)_p^v=v,$

where $\Gamma$ is the parallel transport operator and $\gamma(t)=\exp_p(tv)$. The last equality is true because $\gamma$ is a geodesic, therefore $\gamma'$ is parallel.

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:

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

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

Let us define the curve

:
\alpha \colon \epsilon, \epsilontimes ,1\longrightarrow T_pM,\qquad (s,t) \longmapsto tv+tsw_N.

Note that
:$\alpha(0,1) = v,\qquad \frac(s,t) = v+sw_N, \qquad\frac(0,t) = tw_N.$

Let us put:

:
f \colon \epsilon, \epsilon times ,1\longrightarrow M,\qquad (s,t)\longmapsto \exp_p(tv+tsw_N),

and we calculate:

:$T_v\exp_p(v)=T_\exp_p\left(\frac(0,1)\right)=\frac\Bigl(\exp_p\circ\alpha(s,t)\Bigr)\Big\vert_=\frac(0,1)$
and
:$T_v\exp_p(w_N)=T_\exp_p\left(\frac(0,1)\right)=\frac\Bigl(\exp_p\circ\alpha(s,t)\Bigr)\Big\vert_=\frac(0,1).$
Hence
:$\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \left\langle \frac,\frac\right\rangle(0,1).$
We can now verify that this scalar product is actually independent of the variable $t$, and therefore that, for example:

:$\left\langle\frac,\frac\right\rangle(0,1) = \left\langle\frac,\frac\right\rangle(0,0) = 0,$
because, according to what has been given above:
:$\lim_\frac(0,t) = \lim_T_\exp_p(tw_N) = 0$
being given that the differential is a linear map. This will therefore prove the lemma.
* We verify that ''$\frac\left\langle \frac,\frac\right\rangle=0$'': this is a direct calculation. Since the maps $t\mapsto f(s,t)$ are geodesics,

:$\frac\left\langle \frac,\frac\right\rangle=\left\langle\underbrace_, \frac\right\rangle + \left\langle\frac,\frac\frac\right\rangle = \left\langle\frac,\frac\frac\right\rangle=\frac12\frac\left\langle \frac, \frac\right\rangle.$
Since the maps $t\mapsto f(s,t)$ are geodesics,
the function $t\mapsto\left\langle\frac,\frac\right\rangle$ is constant. Thus,
:$\frac\left\langle \frac, \frac\right\rangle =\frac\left\langle v+sw_N,v+sw_N\right\rangle =2\left\langle v,w_N\right\rangle=0.$

See also

* Riemannian geometry
* Metric tensor

References

*

{{Manifolds

Articles containing proofs
Lemmas
Riemannian geometry
Riemannian manifolds
Theorems in Riemannian geometry