TITLE: Why is complex projective space triangulable?
QUESTION [21 upvotes]: In an exercise in his algebraic topology book, Munkres asserts that $\mathbf{C}P^n$ is triangulable (i.e., there is a simplicial complex $K$ and a homeomorphism $|K| \rightarrow \mathbf{C}P^n$).  Can anyone provide a reference or a proof?

REPLY [3 votes]: Although I am more than a decade late to the discussion, here is an alternate approach to an explicit triangulation of $\mathbb{C}P^n$.
(1) A simplicial set $X$ is a natural generalization of a simplicial complex with locally ordered vertices.  $X$ has a realization $|X|$ which is a CW complex, with a cell for each non-degenerate simplex $\sigma \in X$.  However, $X$ may or may not be regular as a CW complex.
(2) If $X$ is a simplicial set, then so is its $n$th symmetric power $X^n/S_n$, and miraculously $|X^n/S_n| \cong |X|^n/S_n$ (since geometric realization preserves finite limits).  In particular we can let $|X| \cong S^2 \cong \mathbb{C}P^1$ by collapsing the boundary of a triangle to a point.   Then $|X^n/S_n| \cong \mathbb{C}P^n$ is a canonical simplicial set model of $\mathbb{C}P^n$.
(3) Finally, there is a canonical way to subdivide a simplicial set to make a simplicial complex; see arXiv:2001.04339. It is a modified second barycentric subdivision and is not all that efficient of a triangulation, but it works.
(4) You can make the construction a little more efficient by letting $|X| \cong S^2$ be the boundary of a tetrahedron.  Then I believe that $|X^n/S_n|$ is a regular cellulation of $\mathbb{C}P^n$ with simplices, so that its first barycentric subdivision is already an honest simplicial complex.