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The topological symmetry group of a canonically embedded complete graph in $$S^3$$. (English) Zbl 0891.05024
Let $$G$$ be a (finite and simple) graph with vertex set $$V(G)$$ and automorphism group $$\operatorname{Aut}(G)$$. If $$f:G\to \mathbb{S}^3$$ is an embedding of $$G$$ into the 3-sphere $$\mathbb{S}^3$$, then the topological symmetry group $$\text{TSG}(f)$$ of $$f$$ is defined by $\begin{split} \text{TSG} (f)=\{h\in \operatorname{Aut}(G) \mid\text{there is a homeomorphism }\phi: \mathbb{S}^3\to \mathbb{S}^3 \\ \text{with }\phi(f(G)) =f(G)\text{ such that } f\circ h= \phi\circ f|_{V(G)}\}. \end{split}$ The present paper faces the problem of determining $$\text{TSG} (f_n)$$, where $$f_n$$ is a particular embedding of the complete graph $$K_n$$ with $$n$$ vertices into $$\mathbb{S}^3$$, called canonical bud presentation, see T. Endo and T. Otsuki [Hokkaido Math. J. 23, No. 3, 383-398 (1994; Zbl 0814.57007)] and T. Otsuki [J. Comb. Theory, Ser. B 68, No. 1, 23-35, Art. No. 0054 (1996; Zbl 0858.05038)]. In particular, it is proved that, if the order $$n$$ is at least seven, then the topological symmetry group of $$f_n$$ is isomorphic to the dihedral group of order $$2n$$.
Note that, since the cases $$n\leq 5$$ and $$n=6$$ have already been solved by Y. Yoshimata [Topological symmetry group of standard spatial graph of $$K_5$$ (Japanese), Master Thesis, Tokyo Woman’s Christian Univ. (1992)] and by K. Kobayashi and C. Toba [Proc. TGRC-KOSEF 3, 153-171 (1993)] respectively, the previous result yields a complete answer to the considered problem.

##### MSC:
 05C10 Planar graphs; geometric and topological aspects of graph theory 57M15 Relations of low-dimensional topology with graph theory
##### Keywords:
topological symmetry group; embedding; complete graph
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