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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 the 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 !

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1 Answer 1

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  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.
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  • $\begingroup$ 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 $\gamma(t) = c(t) p + s(t) \frac{v}{\Vert v \Vert}$. $\endgroup$
    – pitchounet
    Jul 30, 2013 at 15:26
  • $\begingroup$ In the first step, why does the isometry acting on the geodesic give a geodesic with the same start point? $\endgroup$
    – Bizwhiz
    Apr 29, 2020 at 12:28

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