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 {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 v\longmapsto \gamma _{p,v}(1),

where $\gamma _{p,v}\$ is the unique geodesic with $\gamma (0)=p$ and tangent $\gamma _{p,v}'(0)=v\in T_{p}M$ and $\epsilon _{0}$ is chosen small enough so that for every $v\in B_{\epsilon }(0)\subset T_{p}M$ the geodesic $\gamma _{p,v}$ is defined in 1. 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 (1,p,vt){\Bigr )}{\Big \vert }_{t=0}=\gamma '(t,p,v){\Big \vert }_{t=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 (1,p,v)=\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_{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 $\alpha (t):=v(t+1)$ . We find that, thanks to the identification we have made, and since we are only taking equivalence classes of curves, it is possible to choose $\alpha (t)=vt\$ (these are exactly the same curves, but shifted because of the domain of definition $I$ ; however, the identification allows us to gather them around $0$ . Hence,

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}}\gamma (t,p,v){\Big \vert }_{t=0}=v.

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:

$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.$

$\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =0$ :

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.

References

do Carmo, Manfredo (1992), Riemannian geometry, Basel, Boston, Berlin: Birkhäuser, ISBN 978-0-8176-3490-2

Introduction

The exponential map is a radial isometry

Proof

See also

References