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:
 * $$\mathrm{exp} : T_pM \to M$$

which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in TpM 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_pM\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_pM$$ and $$\epsilon$$ is chosen small enough so that for every $$ t \in [0, 1], vt \in B_{\epsilon}(0) \subset T_pM $$ the geodesic $$\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 $$\alpha : I\rightarrow T_pM$$ be a curve differentiable in $$T_pM$$ such that $$\alpha(0):=0$$ and $$\alpha'(0):=v$$. Since $$T_pM\cong \mathbb R^n$$, 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_pM$$) the differential of $$\exp_p$$ is the identity. By the implicit function theorem, $$\exp_p$$ is a diffeomorphism on a neighborhood of $$0 \in T_pM$$. The Gauss Lemma now tells that $$\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 $$T_vT_pM\cong T_pM\cong \mathbb R^n$$.

Gauss's Lemma states: Let $$v,w\in B_\epsilon(0)\subset T_vT_pM\cong T_pM$$ 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.



Proof
Recall that



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

We proceed in three steps: $$\alpha : \mathbb R \supset I \rightarrow T_pM$$ such that $$\alpha(0):=v\in T_pM$$ and $$\alpha'(0):=v\in T_vT_pM\cong T_pM$$. Since $$T_vT_pM\cong T_pM\cong \mathbb R^n$$, we can put $$\alpha(t):=v(t+1)$$. Therefore,
 * $$T_v\exp_p(v)=v$$ : let us construct a curve

$$ T_v\exp_p(v) = \frac{\mathrm d}{\mathrm d t}\Bigl(\exp_p\circ\alpha(t)\Bigr)\Big\vert_{t=0}=\frac{\mathrm d}{\mathrm d t}\Bigl(\exp_p(tv)\Bigr)\Big\vert_{t=1}=\Gamma(\gamma)_p^{\exp_p(v)}v=v, $$

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 $$w_T:=a v$$, $$a \in \mathbb R$$.

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:

$$\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = \langle v, w_N\rangle = 0.$$

Let us define the curve
 * $$\langle T_v\exp_p(v), T_v\exp_p(w_N)\rangle = 0$$ :



\alpha \colon [-\epsilon, \epsilon]\times [0,1] \longrightarrow T_pM,\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=\frac12\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. $$