Theorem in manifold theory
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 :
e x p : T p M → M {\displaystyle \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 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 [ edit ] We define the exponential map at p ∈ M {\displaystyle p\in M} by
exp p : T p M ⊃ B ϵ ( 0 ) ⟶ M , v t ⟼ γ p , v ( t ) , {\displaystyle \exp _{p}:T_{p}M\supset B_{\epsilon }(0)\longrightarrow M,\quad vt\longmapsto \gamma _{p,v}(t),} where γ p , v {\displaystyle \gamma _{p,v}} is the unique geodesic with γ p , v ( 0 ) = p {\displaystyle \gamma _{p,v}(0)=p} and tangent γ p , v ′ ( 0 ) = v ∈ T p M {\displaystyle \gamma _{p,v}'(0)=v\in T_{p}M} and ϵ {\displaystyle \epsilon } is chosen small enough so that for every t ∈ [ 0 , 1 ] , v t ∈ B ϵ ( 0 ) ⊂ T p M {\displaystyle t\in [0,1],vt\in B_{\epsilon }(0)\subset T_{p}M} the geodesic γ p , v ( t ) {\displaystyle \gamma _{p,v}(t)} is defined. So, if M {\displaystyle M} is complete, then, by the Hopf–Rinow theorem , exp p {\displaystyle \exp _{p}} is defined on the whole tangent space.
Let α : I → T p M {\displaystyle \alpha :I\rightarrow T_{p}M} be a curve differentiable in T p M {\displaystyle T_{p}M} such that α ( 0 ) := 0 {\displaystyle \alpha (0):=0} and α ′ ( 0 ) := v {\displaystyle \alpha '(0):=v} . Since T p M ≅ R n {\displaystyle T_{p}M\cong \mathbb {R} ^{n}} , it is clear that we can choose α ( t ) := v t {\displaystyle \alpha (t):=vt} . In this case, by the definition of the differential of the exponential in 0 {\displaystyle 0} applied over v {\displaystyle v} , we obtain:
T 0 exp p ( v ) = d d t ( exp p ∘ α ( t ) ) | t = 0 = d d t ( exp p ( v t ) ) | t = 0 = d d t ( γ p , v ( t ) ) | t = 0 = γ p , v ′ ( 0 ) = v . {\displaystyle 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 ≅ T p M {\displaystyle T_{0}T_{p}M\cong T_{p}M} ) the differential of exp p {\displaystyle \exp _{p}} is the identity. By the implicit function theorem, exp p {\displaystyle \exp _{p}} is a diffeomorphism on a neighborhood of 0 ∈ T p M {\displaystyle 0\in T_{p}M} . The Gauss Lemma now tells that exp p {\displaystyle \exp _{p}} is also a radial isometry.
The exponential map is a radial isometry [ edit ] Let p ∈ M {\displaystyle p\in M} . In what follows, we make the identification T v T p M ≅ T p M ≅ R n {\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}} .
Gauss's Lemma states: Let v , w ∈ B ϵ ( 0 ) ⊂ T v T p M ≅ T p M {\displaystyle v,w\in B_{\epsilon }(0)\subset T_{v}T_{p}M\cong T_{p}M} and M ∋ q := exp p ( v ) {\displaystyle M\ni q:=\exp _{p}(v)} . Then, ⟨ T v exp p ( v ) , T v exp p ( w ) ⟩ q = ⟨ v , w ⟩ p . {\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle _{q}=\langle v,w\rangle _{p}.}
For p ∈ M {\displaystyle p\in M} , this lemma means that exp p {\displaystyle \exp _{p}} is a radial isometry in the following sense: let v ∈ B ϵ ( 0 ) {\displaystyle v\in B_{\epsilon }(0)} , i.e. such that exp p {\displaystyle \exp _{p}} is well defined. And let q := exp p ( v ) ∈ M {\displaystyle q:=\exp _{p}(v)\in M} . Then the exponential exp p {\displaystyle \exp _{p}} remains an isometry in q {\displaystyle q} , and, more generally, all along the geodesic γ {\displaystyle \gamma } (in so far as γ p , v ( 1 ) = exp p ( v ) {\displaystyle \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 {\displaystyle \exp _{p}} , it remains an isometry.
The exponential map as a radial isometry Recall that
T v exp p : T p M ≅ T v T p M ⊃ T v B ϵ ( 0 ) ⟶ T exp p ( v ) M . {\displaystyle 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 {\displaystyle T_{v}\exp _{p}(v)=v} : let us construct a curve α : R ⊃ I → T p M {\displaystyle \alpha :\mathbb {R} \supset I\rightarrow T_{p}M} such that α ( 0 ) := v ∈ T p M {\displaystyle \alpha (0):=v\in T_{p}M} and α ′ ( 0 ) := v ∈ T v T p M ≅ T p M {\displaystyle \alpha '(0):=v\in T_{v}T_{p}M\cong T_{p}M} . Since T v T p M ≅ T p M ≅ R n {\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}} , we can put α ( t ) := v ( t + 1 ) {\displaystyle \alpha (t):=v(t+1)} . Therefore,
T v exp p ( v ) = d d t ( exp p ∘ α ( t ) ) | t = 0 = d d t ( exp p ( t v ) ) | t = 1 = Γ ( γ ) p exp p ( v ) v = v , {\displaystyle 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 Γ {\displaystyle \Gamma } is the parallel transport operator and γ ( t ) = exp p ( t v ) {\displaystyle \gamma (t)=\exp _{p}(tv)} . The last equality is true because γ {\displaystyle \gamma } is a geodesic, therefore γ ′ {\displaystyle \gamma '} is parallel.
Now let us calculate the scalar product ⟨ T v exp p ( v ) , T v exp p ( w ) ⟩ {\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle } .
We separate w {\displaystyle w} into a component w T {\displaystyle w_{T}} parallel to v {\displaystyle v} and a component w N {\displaystyle w_{N}} normal to v {\displaystyle v} . In particular, we put w T := a v {\displaystyle w_{T}:=av} , a ∈ R {\displaystyle a\in \mathbb {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 ) ⟩ {\displaystyle \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 ⟨ 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 ) ⟩ . {\displaystyle =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:
⟨ T v exp p ( v ) , T v exp p ( w N ) ⟩ = ⟨ v , w N ⟩ = 0. {\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.}
⟨ T v exp p ( v ) , T v exp p ( w N ) ⟩ = 0 {\displaystyle \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
α : [ − ϵ , ϵ ] × [ 0 , 1 ] ⟶ T p M , ( s , t ) ⟼ t v + t s w N . {\displaystyle \alpha \colon [-\epsilon ,\epsilon ]\times [0,1]\longrightarrow T_{p}M,\qquad (s,t)\longmapsto tv+tsw_{N}.} Note that
α ( 0 , 1 ) = v , ∂ α ∂ t ( s , t ) = v + s w N , ∂ α ∂ s ( 0 , t ) = t w N . {\displaystyle \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 : [ − ϵ , ϵ ] × [ 0 , 1 ] ⟶ M , ( s , t ) ⟼ exp p ( t v + t s w N ) , {\displaystyle 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 α ( 0 , 1 ) exp p ( ∂ α ∂ t ( 0 , 1 ) ) = ∂ ∂ t ( exp p ∘ α ( s , t ) ) | t = 1 , s = 0 = ∂ f ∂ t ( 0 , 1 ) {\displaystyle 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 α ( 0 , 1 ) exp p ( ∂ α ∂ s ( 0 , 1 ) ) = ∂ ∂ s ( exp p ∘ α ( s , t ) ) | t = 1 , s = 0 = ∂ f ∂ s ( 0 , 1 ) . {\displaystyle 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
⟨ T v exp p ( v ) , T v exp p ( w N ) ⟩ = ⟨ ∂ f ∂ t , ∂ f ∂ s ⟩ ( 0 , 1 ) . {\displaystyle \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 {\displaystyle t} , and therefore that, for example:
⟨ ∂ f ∂ t , ∂ f ∂ s ⟩ ( 0 , 1 ) = ⟨ ∂ f ∂ t , ∂ f ∂ s ⟩ ( 0 , 0 ) = 0 , {\displaystyle \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 → 0 ∂ f ∂ s ( 0 , t ) = lim t → 0 T t v exp p ( t w N ) = 0 {\displaystyle \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 ∂ ∂ t ⟨ ∂ f ∂ t , ∂ f ∂ s ⟩ = 0 {\displaystyle {\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 ↦ f ( s , t ) {\displaystyle t\mapsto f(s,t)} are geodesics, ∂ ∂ t ⟨ ∂ f ∂ t , ∂ f ∂ s ⟩ = ⟨ D ∂ t ∂ f ∂ t ⏟ = 0 , ∂ f ∂ s ⟩ + ⟨ ∂ f ∂ t , D ∂ t ∂ f ∂ s ⟩ = ⟨ ∂ f ∂ t , D ∂ s ∂ f ∂ t ⟩ = 1 2 ∂ ∂ s ⟨ ∂ f ∂ t , ∂ f ∂ t ⟩ . {\displaystyle {\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 ↦ f ( s , t ) {\displaystyle t\mapsto f(s,t)} are geodesics, the function t ↦ ⟨ ∂ f ∂ t , ∂ f ∂ t ⟩ {\displaystyle t\mapsto \left\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\right\rangle } is constant. Thus,
∂ ∂ s ⟨ ∂ f ∂ t , ∂ f ∂ t ⟩ = ∂ ∂ s ⟨ v + s w N , v + s w N ⟩ = 2 ⟨ v , w N ⟩ = 0. {\displaystyle {\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 [ edit ] References [ edit ]
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