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Let $p,q$ be two relatively prime positive integers.In euclidean 3-space we take a regular p-gonal region $P$ with centre of gravity origin.$a_{0},a_{1},....,a_{p-1}$ are the vertices of the regular polygon.$b_{0}=(0,0,1)$ , $b_{1}=(0,0,-1)$.We join each point of the polygonal region with these two point by straight lines to obtain a solid double pyramid.We identify the triangle $a_{i}a_{i+1}b_{0}$ with $a_{i+q}a_{i+q+1}b_{1}$ in such a way that $a_{i}$ is identified to $a_{i+q}$,$a_{i+1}$ is identified to $a_{i+q+1}$ and $b_{0}$ is identified to $b_{1}$ ...How can I show that the resulting identification space is homeomorphic to the lens space $L(p,q)$ ..

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I have always thought that in fact this is precisely why lens spaces are called like that... – Mariano Suárez-Alvarez Jun 27 '11 at 17:35
It's not clear what your question is. If it is along the lines of "How do I prove...?" then it is useful to the community to know what you have tried and what exactly is confusing you. In general, stating questions in the form "Prove this" or "Show that" come across as rude to this community. – wckronholm Jun 27 '11 at 17:35
@wckronholm ofcourse I mean "How do I prove?"..Sorry if that sounds rude.. – godely Jun 27 '11 at 17:38
@Mariano Suárez-Alvarez well I am taking the lens space to be the orbit space S_3/Z_p...where the action is what generally found in textbooks... – godely Jun 27 '11 at 17:42
@godely: I know. But if you draw that double pyramid (and make a draw base polygon so that it is not too tall...) you will see the famous lense. – Mariano Suárez-Alvarez Jun 27 '11 at 17:46
up vote 3 down vote accepted

The sphere can be identified with $$S^3=\{(z,w)\in\mathbb C^2:|z|^2+|w|^2=1\},$$ and the action of a generator of $\mathbb Z_p$ is then given by $$(z,w)\mapsto(\lambda z,\lambda^q w)$$ with $\lambda$ a primitive $p$th root of unity.

The orbit of each point of $S^3$ has a point $(z,w)$ such that the argument of $z$ is in $[0,2\pi/p]$. Consider the subset $L$ of $S^3$ of such points. If you look at it correctly, you will see it is a (curved) bipiramid, whose central vertical axis is the curve of points of the form $(z,0)$ with $z$ of modulus $1$. Moreover, it is easy to see that the lens space can be obtained from $L$ by doing identifications along its boundary, which is the set of points of the form $(z,w)$ in $S^3$ with $z$ of argument either $0$ or $2\pi/p$.

If you work out exactly what identifications are induced by the action of the group, you will find your description.

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oh!!thanks a lot... – godely Jun 27 '11 at 18:10
This is verbatim an exercise in Armstrong Basic Topology. I'm having trouble making this intuitive argument rigorous. In particular this sentence "If you look at it correctly, you will see it is a (curved) bipiramid, whose central vertical axis is the curve of points of the form (z,0) with z of modulus 1." I don't even know what a "bipiramid" is. If anybody can fill in some of the details that would be awesome. – Gregory Grant May 2 '15 at 13:42
Have you tried googling 《bipyramid》? – Mariano Suárez-Alvarez May 2 '15 at 18:42
Ok that's the shape we need but I can't visualize it in the 4th dimension as you suggest. I don't know what "look at it correctly" means. Is there no rigorous proof of this anywhere? – Gregory Grant May 3 '15 at 21:10
Nice, I found a copy on amazon for $24 shipped. – Gregory Grant May 3 '15 at 21:30

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