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:

$\mathrm{e}\mathrm{x}\mathrm{p}:T_{p}M\to M$

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

${\exp }_p:T_{p}M\supset B_{ϵ}(0)⟶M,vt⟼{\gamma }_{p,v}(t),$

where ${\gamma }_{p,v}$ is the unique geodesic with ${\gamma }_{p,v}(0)=p$ and tangent ${\gamma }_{p^{\prime },v}(0)=v\in T_{p}M$ and $ϵ$ is chosen small enough so that for every $t\in [0,1],vt\in B_{ϵ}(0)\subset T_{p}M$ 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 $\alpha :I\to T_{p}M$ be a curve differentiable in $T_{p}M$ such that $\alpha (0):=0$ and ${\alpha }^{\prime }(0):=v$. Since $T_{p}M\cong {ℝ}^n$, 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:

$T_0{\exp }_p(v)=\frac{\mathrm{d}}{\mathrm{d}t}({\exp }_p\circ \alpha (t)){|}_{t=0}=\frac{\mathrm{d}}{\mathrm{d}t}({\exp }_p(vt)){|}_{t=0}=\frac{\mathrm{d}}{\mathrm{d}t}({\gamma }_{p,v}(t)){|}_{t=0}={\gamma }_{p^{\prime },v}(0)=v.$

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 $T_v{T}_{p}M\cong T_{p}M\cong {ℝ}^n$.

Gauss's Lemma states: Let $v,w\in B_{ϵ}(0)\subset T_v{T}_{p}M\cong T_{p}M$ and $M\ni q:={\exp }_p(v)$. Then, $⟨T_v{\exp }_p(v),T_v{\exp }_p(w){⟩}_q=⟨v,w{⟩}_p.$

For $p\in M$, this lemma means that ${\exp }_p$ is a radial isometry in the following sense: let $v\in B_{ϵ}(0)$, 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

$T_v{\exp }_p:T_{p}M\cong T_v{T}_{p}M\supset T_v{B}_{ϵ}(0)⟶T_{{\exp }_p(v)}M.$

We proceed in three steps:

• $T_v{\exp }_p(v)=v$ : let us construct a curve

$\alpha :ℝ\supset I\to T_{p}M$ such that $\alpha (0):=v\in T_{p}M$ and ${\alpha }^{\prime }(0):=v\in T_v{T}_{p}M\cong T_{p}M$. Since $T_v{T}_{p}M\cong T_{p}M\cong {ℝ}^n$, we can put $\alpha (t):=v(t+1)$. Therefore,

$T_v{\exp }_p(v)=\frac{\mathrm{d}}{\mathrm{d}t}({\exp }_p\circ \alpha (t)){|}_{t=0}=\frac{\mathrm{d}}{\mathrm{d}t}({\exp }_p(tv)){|}_{t=1}=\mathrm{\Gamma }(\gamma {)}_p^{{\exp }_p(v)}v=v,$

where $\mathrm{\Gamma }$ is the parallel transport operator and $\gamma (t)={\exp }_p(tv)$. The last equality is true because $\gamma$ is a geodesic, therefore ${\gamma }^{\prime }$ is parallel.

Now let us calculate the scalar product $⟨T_v{\exp }_p(v),T_v{\exp }_p(w)⟩$.

We separate $w$ into a component $w_T$ parallel to $v$ and a component $w_N$ normal to $v$. In particular, we put $w_T:=av$, $a\in ℝ$.

The preceding step implies directly:

$⟨T_v{\exp }_p(v),T_v{\exp }_p(w)⟩=⟨T_v{\exp }_p(v),T_v{\exp }_p(w_T)⟩+⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩$

$=a⟨T_v{\exp }_p(v),T_v{\exp }_p(v)⟩+⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩=⟨v,w_T⟩+⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩.$

We must therefore show that the second term is null, because, according to Gauss's Lemma, we must have:

$⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩=⟨v,w_N⟩=0.$

• $⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩=0$ :

The curve chosen to prove lemma

Let us define the curve

$\alpha :[-ϵ,ϵ]\times [0,1]⟶T_{p}M,(s,t)⟼tv+ts{w}_N.$

Note that

$\alpha (0,1)=v,\frac{\partial \alpha }{\partial t}(s,t)=v+s{w}_N,\frac{\partial \alpha }{\partial s}(0,t)=t{w}_N.$

Let us put:

$f:[-ϵ,ϵ]\times [0,1]⟶M,(s,t)⟼{\exp }_p(tv+ts{w}_N),$

and we calculate:

$T_v{\exp }_p(v)=T_{\alpha (0,1)}{\exp }_p(\frac{\partial \alpha }{\partial t}(0,1))=\frac{\partial }{\partial t}({\exp }_p\circ \alpha (s,t)){|}_{t=1,s=0}=\frac{\partial f}{\partial t}(0,1)$

and

$T_v{\exp }_p(w_N)=T_{\alpha (0,1)}{\exp }_p(\frac{\partial \alpha }{\partial s}(0,1))=\frac{\partial }{\partial s}({\exp }_p\circ \alpha (s,t)){|}_{t=1,s=0}=\frac{\partial f}{\partial s}(0,1).$

Hence

$⟨T_v{\exp }_p(v),T_v{\exp }_p(w_N)⟩=⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial s}⟩(0,1).$

We can now verify that this scalar product is actually independent of the variable $t$, and therefore that, for example:

$⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial s}⟩(0,1)=⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial s}⟩(0,0)=0,$

because, according to what has been given above:

$\underset{t\to 0}{\lim }\frac{\partial f}{\partial s}(0,t)=\underset{t\to 0}{\lim }{T}_{tv}{\exp }_p(t{w}_N)=0$

being given that the differential is a linear map. This will therefore prove the lemma.

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

$\frac{\partial }{\partial t}⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial s}⟩=⟨\underset{=0}{\underbrace{\frac{D}{\partial t}\frac{\partial f}{\partial t}}},\frac{\partial f}{\partial s}⟩+⟨\frac{\partial f}{\partial t},\frac{D}{\partial t}\frac{\partial f}{\partial s}⟩=⟨\frac{\partial f}{\partial t},\frac{D}{\partial s}\frac{\partial f}{\partial t}⟩=\frac{1}{2}\frac{\partial }{\partial s}⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}⟩.$

Since the maps $t\mapsto f(s,t)$ are geodesics, the function $t\mapsto ⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}⟩$ is constant. Thus,

$\frac{\partial }{\partial s}⟨\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}⟩=\frac{\partial }{\partial s}⟨v+s{w}_N,v+s{w}_N⟩=2⟨v,w_N⟩=0.$

See also

• Riemannian geometry
• Metric tensor

References

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