Английская Википедия:Gauss's lemma (Riemannian geometry)

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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:

<math>\mathrm{exp} : T_pM \to M</math>

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 <math>p\in M</math> by

<math>

\exp_p: T_pM\supset B_{\epsilon}(0) \longrightarrow M,\quad vt \longmapsto \gamma_{p,v}(t), </math> where <math>\gamma_{p,v}</math> is the unique geodesic with <math>\gamma_{p,v}(0)=p</math> and tangent <math>\gamma_{p,v}'(0)=v \in T_pM</math> and <math>\epsilon</math> is chosen small enough so that for every <math> t \in [0, 1], vt \in B_{\epsilon}(0) \subset T_pM </math> the geodesic <math>\gamma_{p,v}(t)</math> is defined. So, if <math>M</math> is complete, then, by the Hopf–Rinow theorem, <math> \exp_p</math> is defined on the whole tangent space.

Let <math>\alpha : I\rightarrow T_pM</math> be a curve differentiable in <math>T_pM</math> such that <math>\alpha(0):=0</math> and <math>\alpha'(0):=v</math>. Since <math>T_pM\cong \mathbb R^n</math>, it is clear that we can choose <math>\alpha(t):=vt</math>. In this case, by the definition of the differential of the exponential in <math>0</math> applied over <math>v</math>, we obtain:

<math>

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. </math> So (with the right identification <math>T_0 T_p M \cong T_pM</math>) the differential of <math>\exp_p</math> is the identity. By the implicit function theorem, <math>\exp_p</math> is a diffeomorphism on a neighborhood of <math>0 \in T_pM</math>. The Gauss Lemma now tells that <math>\exp_p</math> is also a radial isometry.

The exponential map is a radial isometry

Let <math>p\in M</math>. In what follows, we make the identification <math>T_vT_pM\cong T_pM\cong \mathbb R^n</math>.

Gauss's Lemma states: Let <math>v,w\in B_\epsilon(0)\subset T_vT_pM\cong T_pM</math> and <math>M\ni q:=\exp_p(v)</math>. Then, <math> \langle T_v\exp_p(v), T_v\exp_p(w)\rangle_q = \langle v,w\rangle_p. </math>

For <math>p\in M</math>, this lemma means that <math>\exp_p</math> is a radial isometry in the following sense: let <math>v\in B_\epsilon(0)</math>, i.e. such that <math>\exp_p</math> is well defined. And let <math>q:=\exp_p(v)\in M</math>. Then the exponential <math>\exp_p</math> remains an isometry in <math>q</math>, and, more generally, all along the geodesic <math>\gamma</math> (in so far as <math>\gamma_{p, v}(1)=\exp_p(v)</math> is well defined)! Then, radially, in all the directions permitted by the domain of definition of <math>\exp_p</math>, it remains an isometry.

Файл:Gauss lemma radial isometry.png

The exponential map as a radial isometry

Proof

Recall that

<math>

T_v\exp_p \colon T_pM\cong T_vT_pM\supset T_vB_\epsilon(0)\longrightarrow T_{\exp_p(v)}M. </math>

We proceed in three steps:

• <math>T_v\exp_p(v)=v</math> : let us construct a curve

<math>\alpha : \mathbb R \supset I \rightarrow T_pM</math> such that <math>\alpha(0):=v\in T_pM</math> and <math>\alpha'(0):=v\in T_vT_pM\cong T_pM</math>. Since <math>T_vT_pM\cong T_pM\cong \mathbb R^n</math>, we can put <math>\alpha(t):=v(t+1)</math>. Therefore,

<math> 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, </math>

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

Now let us calculate the scalar product <math>\langle T_v\exp_p(v), T_v\exp_p(w)\rangle</math>.

We separate <math>w</math> into a component <math>w_T</math> parallel to <math>v</math> and a component <math>w_N</math> normal to <math>v</math>. In particular, we put <math>w_T:=a v</math>, <math>a \in \mathbb R</math>.

The preceding step implies directly:

<math>

\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</math>

<math>=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.

</math>

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

<math>\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \langle v, w_N\rangle = 0.</math>

• <math>\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = 0</math> :

Файл:Gauss lemma proof.png

The curve chosen to prove lemma

Let us define the curve

<math>

\alpha \colon [-\epsilon, \epsilon]\times [0,1] \longrightarrow T_pM,\qquad (s,t) \longmapsto tv+tsw_N. </math> Note that

<math>

\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. </math>

Let us put:

<math>

f \colon [-\epsilon, \epsilon ]\times [0,1] \longrightarrow M,\qquad (s,t)\longmapsto \exp_p(tv+tsw_N), </math>

and we calculate:

<math>

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) </math> and

<math>

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). </math> Hence

<math>

\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). </math> We can now verify that this scalar product is actually independent of the variable <math>t</math>, and therefore that, for example:

<math>

\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, </math> because, according to what has been given above:

<math>

\lim_{t\rightarrow 0}\frac{\partial f}{\partial s}(0,t) = \lim_{t\rightarrow 0}T_{tv}\exp_p(tw_N) = 0 </math> being given that the differential is a linear map. This will therefore prove the lemma.

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

<math>

\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. </math> Since the maps <math>t\mapsto f(s,t)</math> are geodesics, the function <math>t\mapsto\left\langle\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}\right\rangle</math> is constant. Thus,

<math>

\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. </math>

See also

• Riemannian geometry
• Metric tensor

References

• Шаблон:Citation