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Recently I came across this mathoverflow question, in which the number of homeomorphism classes of 3-dimensional lens spaces $L(p, q)$ is computed as a function of $p$. Using the OEIS, I found a connection to so-called generalized Petersen graphs: For $m\geq 3$ and $1\leq k <m$ relatively prime to $m$, we denote by $P(m, k)$ the graph on vertices $v_1,\dots, v_m, w_1, \dots, w_m$ and with edges $\{v_i, w_i\}, \{v_i, v_{i+1}\}$ and $\{w_i, w_{i+k}\}$ for $1\leq i \leq m$, where the indices are taken modulo $m$.

As was shown by Steimle and Staton, the graphs $P(m, k)$ and $P(m, l)$ are isomorphic if and only if $k\equiv\pm l \text{ mod } m$ or $kl\equiv\pm 1 \text{ mod } m$. The classic result about the homeomorphism classes of 3-dimensional lens spaces states that this condition on $l$ and $k$ is fulfilled if and only if the lens spaces $L(m, k)$ and $L(m, l)$ are homeomorphic. Hence we reach the suggestive conclusion $$P(m, k) \cong P(m, l) \Leftrightarrow L(m, k) \cong L(m, l).$$

On the one hand, I find that the proof for the classification of generalized Petersen graphs given in the paper lacks a conceptual approach. On the other hand the definition of a generalized Petersen graph really captures my intuition on how lens spaces are 'built'. This motivates my

Question: Is there a 'direct' structural correspondence between lens spaces and generalized Petersen graphs which allows to obtain one of the classification results from the other?

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    $\begingroup$ Can $P(m, k)$ be realized as the 1-skeleton of a distinguished CW structure on $S^3$ covering a distinguished CW structure on $L(m, k)$? $\endgroup$ Commented Dec 9, 2014 at 3:46
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    $\begingroup$ @Qiaochu: it looks like it. I don't know what the definition of a Peterson graph is, in general, but the one that has a Wikipedia page looks much like the lift of the standard 1-skeleton for $L_{5,2}$ to the 3-sphere. $\endgroup$ Commented Dec 9, 2014 at 4:03
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    $\begingroup$ @RyanBudney for more on generalized Petersen graphs (including a clearer definition; I think the OP wrote $\{w_i,v_{i+k}\}$ when he intended $\{w_i,w_{i+k}\}$) see en.wikipedia.org/wiki/Generalized_Petersen_graph $\endgroup$ Commented Dec 9, 2014 at 5:56
  • $\begingroup$ @DavidEppstein : Definition seems correct else both wickipedia and Wolfram are wrong. $\endgroup$ Commented Dec 10, 2014 at 0:47
  • $\begingroup$ @Jerome: As David remarked, I first had a typo in the definition. It should now be correct. $\endgroup$ Commented Dec 10, 2014 at 0:49

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As Qiaochu Yuan suggested, there is a CW structure on $S^3$ with $P(m,k)$ as its 1-skeleton, invariant under the ${\mathbb Z}_m$ action defining the lens space $L(m,k)$. The quotient CW structure on $L(m,k)$ subdivides the standard one that has a single cell in each dimension 0, 1, 2, 3. To construct this subdivision, start with the usual picture of $L(m,k)$ as a lens-shaped ball with its upper and lower faces identified via a $k/m$ "rotation". Put $m$ vertices $v_1,\cdots,v_m$ equally spaced around the rim of the lens, numbered consecutively (mod $m$). Factoring out the ${\mathbb Z}_m$ action identifies all the vertices $v_i$ and all the edges $[v_i,v_{i+1}]$, and also identifies the top and bottom faces of the lens to a single 2-cell, so this gives the standard CW structure on $L(m,k)$. To subdivide it, put a new vertex $w_0$ at the center of the lower face of the lens and a new vertex $w_k$ at the center of the upper face. Join these two vertices by an edge $[w_0,w_k]$ running straight through the lens, then join $w_0$ to $v_0$ and $w_k$ to $v_k$ by geodesic arcs in the lower and upper faces of the lens. The loop $v_0,v_1,\cdots,v_k,w_k,w_0,v_0$ then spans a 2-cell in the interior of the lens, whose complement in the lens is a 3-cell. This gives a CW structure on $L(m,k)$ with two 0-cells, three 1-cells, two 2-cells, and one 3-cell. Lifting to the cover $S^3$ gives a CW structure with $m$ times as many cells in each dimension. The preimage of the edge $[w_0,w_k]$ is a circle with vertices $w_0,w_k,w_{2k},w_{3k}\cdots$ in that order, so these give edges $[w_i,w_{i+k}]$ for each $i$. There are also edges $[v_i,w_i]$ for each $i$, so the 1-skeleton is the graph $P(m,k)$.

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  • $\begingroup$ Thank you very much for your answer! Do you think that there is a way to construct $L(m, k)$ out of the abstract graph $P(m, k)$, i.e. without having distinguished vertices $v_1, \dots, v_n$ or knowing the value of $k$? This is the connection I had in mind when asking this question and one that could make it possible to derive one classification result from the other. $\endgroup$ Commented Dec 9, 2014 at 22:25

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