Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I might need some help on the following exercise :

Let $\mathbb{S}^{n} \subset \mathbb{R}^{n+1}$ be the unit $n$-sphere. For any $p \in \mathbb{S}^{n}$, we have $T_{p}\mathbb{S}^{n} = p^{\perp} = \lbrace v \in \mathbb{R}^{n+1}, \, p \cdot v = 0 \rbrace$. The object of this exercise is to compute the exponential map $\exp \, : \, T\mathbb{S}^{n} \, \rightarrow \, \mathbb{S}^{n}$.

  1. Explain why, if $(M,g)$ is a Riemannian manifold, $x \in M$ and $f \, : \, M \, \rightarrow \, M$ an isometry such that $\mathrm{D}_{x}f \cdot v = v$ for some $v \in T_{x}M$ then, for the geodesic $\gamma \, : \, [a,b] \, \rightarrow \, M$ with $\gamma(0) =x$ and $\gamma'(0)=v$, we have $f \circ \gamma = \gamma$.
  2. Show that if $(p,v) \in T\mathbb{S}^{n}$ with $v \neq 0$, the reflection $R \, : \, \mathbb{R}^{n+1} \, \rightarrow \, \mathbb{R}^{n+1}$ that fixes pointwise hte plane spanned by $p$ and $v$ and reverses all vectors perpendicular to $p$ and $v$ is an isometry of $\mathbb{S}^{n}$.
  3. Explain why this shows that the geodesic $\gamma \, : \, [a,b] \, \rightarrow \, \mathbb{S}^{n} \subset \mathbb{R}^{n+1}$ that satisfies $\gamma(0)=p$ and $\gamma'(0)=v$ must be of the form $ \gamma(t) = c(t) p + s(t) \frac{v}{\Vert v \Vert}$.
  4. Explain why we mush have $c(t) = \cos(t\Vert v \Vert)$ and $s(t) = \sin(t\Vert v \Vert)$.

Here is what I did :

  1. Let $t \in [a,b]$. By definition, $ (f \circ \gamma)'(t) = \mathrm{D}_{t}(f \circ \gamma) \cdot \frac{d}{dt}$ and by the chain rule, $(f \circ \gamma)'(t) = \mathrm{D}_{\gamma(t)} f \circ \mathrm{D}_{t} \gamma \cdot \frac{d}{dt} = \mathrm{D}_{\gamma(t)} f \cdot \gamma'(t)$. It follows that $(f \circ \gamma)'(0) = \mathrm{D}_{x} f \cdot v = v$. Furthermore, if $L(\gamma)$ denotes the length of the curve $\gamma$, let $\tilde{\gamma}(t) = (f \circ \gamma)(t)$ and we have : $$ L(\tilde{\gamma}) = \int_{a}^{b} \Vert \dot{\tilde{\gamma}}(t) \Vert_{\tilde{\gamma}(t)} \: dt$$ where $$ \begin{eqnarray*} \Vert \dot{\tilde{\gamma}}(t) \Vert_{\tilde{\gamma}(t)}^{2} & = & g_{\tilde{\gamma}(t)}\left( \dot{\tilde{\gamma}}(t) , \dot{\tilde{\gamma}}(t) \right) \\ & = & g_{(f \circ \gamma)(t)} \left( \mathrm{D}_{\gamma(t)} f \cdot \gamma'(t) , \mathrm{D}_{\gamma(t)} f \cdot \gamma'(t) \right) \\ & = & g_{\gamma(t)} \left( \gamma'(t) , \gamma'(t) \right) \\ & = & \Vert \dot{\gamma}(t) \Vert_{\gamma(t)} \end{eqnarray*} $$ because $f$ is an isometry. We get $L(\tilde{\gamma}) = L(\gamma)$. I think it suffices to prove that $\tilde{\gamma} = \gamma$ but I don't know which theorem it follows from.
  2. (I'm not so sure for this one) Let $W = \mathrm{Vect}(p,v) = \mathrm{Vect}(p,\frac{v}{\Vert v \Vert})$. Since $\mathbb{R}^{n+1} = W \oplus W^{\perp}$, I can define the reflection $R$ on $\mathbb{R}^{n+1}$ as follows : $R = \mathrm{Id}$ on $W$ and $R = -\mathrm{Id}$ on $W^{\perp}$. Let $x \in \mathbb{S}^{n}$. There exists $\alpha \in W$ and $\beta \in W^{\perp}$ such that $x = \alpha + \beta$. Then, $R(x) = \alpha - \beta$. If $\Vert \cdot \Vert$ denotes the usual euclidean norm in $\mathbb{R}^{n+1}$, it follows from Pythagore's theorem that $\Vert R(x) \Vert = \Vert x \Vert$. So, $R$ is an isometry of $\mathbb{S}^{n}$.
  3. I think we need to use the result from 2. but I don't know how to use that $R \circ \gamma = \gamma$.
  4. I can use $\Vert \gamma'(t) \Vert = \Vert v \Vert$ and $\Vert \gamma(t) \Vert = 1$ for all $t \in [a,b]$. Since $p \cdot v = 0$, we have $$ \Vert \gamma(t) \Vert^{2} = \vert c(t) \vert^{2} + \vert s(t) \vert^{2} = 1$$. So, we can say there exists some $\ell \in \mathbb{R}$ such that, for all $t \in [a,b]$, $c(t) = \cos(\ell t)$ and $s(t) = \sin(\ell t)$. From the condition $\Vert \gamma'(t) \Vert = \Vert v \Vert$, it follows that $\ell = \Vert v \Vert$.

Thank you for your help !

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted
  1. Since f is an isometry, $f\circ\gamma$ is also a geodesic. By the uniqueness of geodesic given the start point and velocity, we have $f\circ\gamma=\gamma$
  2. A reflection is an isometry of the ambient $R^{n+1}$, and hence is an isometry of $S^n$, since the inner product is induced.
  3. Based on 1 and 2, a geodesic is fixed under the reflection determined by p and v. So it is the intersection of the plane spanned by p, v and the sphere.
  4. is trivial.
share|improve this answer
    
Thanks a lot ! You're right, since the geodesic $\gamma$ is fixed under the reflection $R$, it lies in $W \cap \mathbb{S}^{n}$ (which is : $\forall t \in [a,b], \, \gamma(t) \in W$). This is why $\gamma$ is of the form $\gamma(t) = c(t) p + s(t) \frac{v}{\Vert v \Vert}$. –  jibounet Jul 30 '13 at 15:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.