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A triangulation of $\CC P^3$ as symmetric cube of $S^2$

Published 15 Dec 2010 in math.AT and math.CO | (1012.3235v1)

Abstract: The symmetric group $S_3$ acts on $S2 \times S2 \times S2$ by coordinate permutation, and the quotient space $(S2 \times S2 \times S2)/S_3$ is homeomorphic to the complex projective space $\CC P3$. In this paper, we construct an 124-vertex simplicial subdivision $(S2 \times S2 \times S2)_{124}$ of the 64-vertex standard cellulation $S2_4 \times S2_4 \times S2_4$ of $S2 \times S2 \times S2$, such that the $S_3$-action on this cellulation naturally extends to an action on $(S2 \times S2 \times S2)_{124}$. Further, the $S_3$-action on $(S2 \times S2 \times S2)_{124}$ is "good", so that the quotient simplicial complex $(S2 \times S2 \times S2)_{124}/S_3$ is a 30-vertex triangulation $\CC P3_{30}$ of $\CC P3$. In other words, we construct a simplicial realization $(S2 \times S2 \times S2)_{124} \to \CC P3_{30}$ of the branched covering $S2 \times S2 \times S2 \to \CC P3$. Finally, we apply the BISTELLAR program of Lutz on $\CC P3_{30}$, resulting in an 18-vertex 2-neighbourly triangulation $\CC P3_{18}$ of $\CC P3$. The automorphism group of $\CC P3_{18}$ is trivial. It may be recalled that, by a result of Arnoux and Marin, any triangulation of $\CC P3$ requires at least 17 vertices. So, $\CC P3_{18}$ is close to vertex-minimal, if not actually vertex-minimal. Moreover, no explicit triangulation of $\CC P3$ was known so far.

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