Simplicial cell decompositions of $\mathbb{CP}^{\hspace{.3mm}n}$
Abstract
According to a well-known result in geometric topology, we have $\left (\mathbb{S}^2 \right)^{n}\!\!/\operatorname{Sym}(n) = \mathbb{CP}^{n}$, where $\operatorname{Sym}(n)$ acts on $\left (\mathbb{S}^2 \right)^{n}$ by coordinate permutation. We use this fact to explicitly construct a regular simplicial cell decomposition of $\mathbb{CP}^{n}$ for each $n \geq 2$. In more detail, we take the standard two triangle crystallisation $S^2_3$ of the $2$-sphere $\mathbb{S}^2$, in its $n$-fold Cartesian product. We then simplicially subdivide, and prove that naively taking the $\operatorname{Sym}(n)$ quotient yields a simplicial cell decomposition of $\mathbb{CP}^n$. Taking the first derived subdivision of this cell complex produces a triangulation of $\mathbb{CP}^n$. To the best of our knowledge, this is the first explicit description of triangulations of $\mathbb{CP}^n$ for $n \geq 4$.
- Publication:
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arXiv e-prints
- Pub Date:
- January 2024
- DOI:
- arXiv:
- arXiv:2401.09891
- Bibcode:
- 2024arXiv240109891D
- Keywords:
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- Mathematics - Combinatorics;
- Mathematics - Geometric Topology;
- 57Q15;
- 57Q05;
- 05C15;
- 06A06
- E-Print:
- 14 pages, 1 figure, 16 pages of appendix