According to a well-known result in geometric topology, we have $(S^2)^n/Sym(n) = \mathbb{CP}^n$, where $Sym(n)$ acts on $(S^2)^n$ by coordinate permutation. We use this fact to explicitly construct a regular simplicial cell decomposition of $\mathbb{CP}^n$ for each $n > 1$. In more detail, we take the standard two triangle crystallisation $S^2_3$ of the 2-sphere $S^2$, in its $n$-fold Cartesian product. We then simplicially subdivide, and prove that naively taking the $Sym(n)$ 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 $n > 3$. This is a joint work with Jonathan Spreer, University of Sydney.

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Last updated: 15 Jul 2024