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 $L(p,q)$ ..
Alternative Construction of Lens Space
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0There is a third definition, in terms of gluing solid tori (ie copies of $S^1 \times D^2$). – 2011-06-27
1 Answers
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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1I'll try to understand from that book, if I'm still stuck I may appeal here for more help. I appreciate the guidance. – 2015-05-03