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:

$\mathrm {exp} :T_{p}M\to M$

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

$\exp _{p}:T_{p}M\supset B_{\epsilon }(0)\longrightarrow M,\quad vt\longmapsto \gamma _{p,v}(t),$

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

Let $\alpha :I\rightarrow T_{p}M$ be a curve differentiable in $T_{p}M$ such that $\alpha (0):=0$ and $\alpha '(0):=v$. Since $T_{p}M\cong \mathbb {R} ^{n}$, 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:

$T_{0}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}(vt){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\gamma _{p,v}(t){\Bigr )}{\Big \vert }_{t=0}=\gamma _{p,v}'(0)=v.$

So (with the right identification $T_{0}T_{p}M\cong T_{p}M$) the differential of $\exp _{p}$ is the identity. By the implicit function theorem, $\exp _{p}$ is a diffeomorphism on a neighborhood of $0\in T_{p}M$. 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 $T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}$.

Gauss's Lemma states: Let $v,w\in B_{\epsilon }(0)\subset T_{v}T_{p}M\cong T_{p}M$ 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 $\gamma _{p,v}(1)=\exp _{p}(v)$ is well defined)! Then, radially, in all the directions permitted by the domain of definition of $\exp _{p}$, it remains an isometry.

The exponential map as a radial isometry

Proof

Recall that

$T_{v}\exp _{p}\colon T_{p}M\cong T_{v}T_{p}M\supset T_{v}B_{\epsilon }(0)\longrightarrow T_{\exp _{p}(v)}M.$

We proceed in three steps:

• $T_{v}\exp _{p}(v)=v$ : let us construct a curve

$\alpha :\mathbb {R} \supset I\rightarrow T_{p}M$ such that $\alpha (0):=v\in T_{p}M$ and $\alpha '(0):=v\in T_{v}T_{p}M\cong T_{p}M$. Since $T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}$, we can put $\alpha (t):=v(t+1)$. Therefore,

$T_{v}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}(tv){\Bigr )}{\Big \vert }_{t=1}=\Gamma (\gamma )_{p}^{\exp _{p}(v)}v=v,$

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 $w_{T}:=av$, $a\in \mathbb {R}$.

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

$\alpha \colon [-\epsilon ,\epsilon ]\times [0,1]\longrightarrow T_{p}M,\qquad (s,t)\longmapsto tv+tsw_{N}.$

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,

${\frac {\partial }{\partial t}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\right\rangle =\left\langle \underbrace {{\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}} _{=0},{\frac {\partial f}{\partial s}}\right\rangle +\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial t}}{\frac {\partial f}{\partial s}}\right\rangle =\left\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\right\rangle ={\frac {1}{2}}{\frac {\partial }{\partial s}}\left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle .$

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