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Take $G := \{Id,h,\ldots,h^{q-1}\}$ the group of isometries of $S^3$ seen as subset of $\mathbb{C}^2$, where each $$h(z_1,z_2) := (\exp({\frac{2\pi i}{q})z_1},\exp({\frac{2\pi ir}{q})z_2}),$$ $(z_1,z_2) \in S^3, \gcd{(q,r) = 1}, q,r\in \mathbb{Z}.$

We can show that $S^3$ is the covering space of $S^3\big/G$ and with the covering metric, these spaces are locally isometric. In particular, the geodesics of $S^3\big/G$ are closed since they are the projections of the great circles in $S^3.$ However, it seems that the geodesics on the quotient does not have the same length. Can anyone show me that this is true?

Here we are considering $S^3$ with the induced metric.

Thanks.

L.F. Cavenaghi
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    It seems you have a particular case of a lens space.

    I think that you mean "lengths of simple closed geodesics", otherwise the question makes little sense. That being the case, this is a comment because I have not done the computations, but I think that if you take $q=3, r=2$, by projecting the geodesics of the first "coordinate" and the geodesics of the second "coordinate" you will fall on different identifications on "different moments" and have a difference of length.

     – Aloizio Macedo Oct 23 '16 at 19:31
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    Also, related: http://mathoverflow.net/questions/103396/geodesics-in-lens-spaces – Aloizio Macedo Oct 23 '16 at 19:32
  • thank you @AloizioMacedo, Indeed, I am talking about simple closed geodesics, that are the projections of the great circles on $S^3.$ – L.F. Cavenaghi Oct 23 '16 at 19:34
  • can you explain how do you get the intuition of taking $q=3$ and $r=2$? – L.F. Cavenaghi Oct 23 '16 at 19:40
  • Take $\Bbb{R}P^3$ (which is a particular case). When you pass to the quotient from $S^3$, you collapse antipodal points. Since geodesics on $\Bbb{R}P^3$ come from the great circles, the simple closed geodesics below will be given by the projections of "half" great circles. But the way $\Bbb{R}P^3$ identifies points forces the only way to "close" a geodesic which is not closed on $S^3$ to be through identification of antipodal points (half-circles), making them all have same length. The idea is to force another identification on different distances on $S^3$. – Aloizio Macedo Oct 23 '16 at 19:49
  • @Aloizio Macedo In your first comment on the case where $q=3, \ r=2$ two geodesics have projections which are closed geodesic of same length ? – HK Lee Oct 24 '16 at 17:08

1 Answers1

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Fix two pt $\overline{p},\ \overline{q}\in X$ where $$\pi: S^3\rightarrow X:= S^3/G,\ \pi (p)=\overline{p}$$

If $\overline{p},\ \overline{q}$ are close then there is unique geodesic $\overline{c}$ joining them in $X$ If $c$ is a geodesic which is a lifting of $ \overline{c}$ then it is also $unique$ In further $c$ can be uniquely extended into a great circle $C$

Since $\pi$ is locally isometric so $\pi (C)$ is a closed smooth geodesic That is any geodesic in $X$ is a closed smooth geodesic If $G=\mathbb{Z}_3$, and $$ g\cdot (Z_1,Z_2)=(gZ_1,gZ_2),\ g\in G$$ then $\{(e^{it},0) | t\in [0,2\pi] \}$ is projected into a closed smooth geodesic of length $\frac{2\pi}{3}$

Consider great circle which is union of $$ (\cos\ t,\sin\ t),\ (-\sin\ t,\cos\ t),\ (-\cos\ t,-\sin\ t),\ (\sin\ t,-\cos\ t),\ t\in [0,\frac{\pi}{2}]$$ This its projection is a closed smooth geodesic whose length $2\pi$ in $X$

HK Lee
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  • Can't we simply write the great circle as, e.g. $(e^{it},0)$, $t \in [0,2\pi]$? I'm asking because I am currently working on Exercise 8.4(b) of do Carmo. – New day rising Jun 29 '17 at 06:16
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    (1) Consider a intersection between a plane $\mathbb{R}^2\oplus {0}$ and $S^3$ It is a great circle $C$, which is invariant under $G$-action Hence its quotient has length $2\pi/3$ (2) Note that $G$-action can be extended into $S^1$-action, which is a Hopf-fibration. Consider a geodesic which is perpendicular to a fiber. For example $c(t)=(\cos\ t,\sin\ t)\in S^3,\ t\in [0,2\pi]$ That is local portion of $c$ can not be shrinked under $G$-action so that $c$ in quotient space has a length $2\pi$. – HK Lee Jun 29 '17 at 06:38