English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Tools annual

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Tools monthly

Global: Zendy Plus monthly

Presents the access of premium content as premium feature

Premium Content

Global: Zendy Plus annual

Global: Zendy Free Plan

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 coordinatepermutation. We use this fact to explicitly construct a regular simplicial celldecomposition of $\mathbb{CP}^{n}$ for each $n \geq 2$. In more detail, we takethe standard two triangle crystallisation $S^2_3$ of the $2$-sphere$\mathbb{S}^2$, in its $n$-fold Cartesian product. We then simpliciallysubdivide, and prove that naively taking the $\operatorname{Sym}(n)$ quotientyields a simplicial cell decomposition of $\mathbb{CP}^n$. Taking the firstderived subdivision of this cell complex produces a triangulation of$\mathbb{CP}^n$. To the best of our knowledge, this is the first explicitdescription of triangulations of $\mathbb{CP}^n$ for $n \geq 4$.

Simplicial cell decompositions of $\mathbb{CP}^{\hspace{.3mm}n}$