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?
I will present a triangulation of $\mathbb{CP}^{n1}$. More specifically, I will give an explicit regular CW structure on $\mathbb{CP}^{n1}$. As spinorbundle says, the first barycentric subdivision of a regular CW complex is a simplicial complex homeomorphic to the original CW complex. Recall that to put a regular CW complex on space $X$ means to decompose $X$ into disjoint pieces $Y_i$ such that: (1) The closure of each $Y_i$ is a union of $Y$'s. (2) For each $i$, the pair $(\overline{Y_i}. Y_i)$ is homemorphic to $(\mbox{closed}\ d\mbox{ball}, \mbox{interior of that}\ d\mbox{ball})$ for some $d$. The barycentric subdivision of $X$ corresponding to this regular CW complex is the simplicial complex which has a vertex for each $Y_i$ and has a simplex $(i_0, i_1, \ldots, i_r)$ if and only if $\overline{Y_{i_0}} \subset \overline{Y_{i_1}} \subset \cdots \subset \overline{Y_{i_r}}$. Write $(t_1: t_2: \ldots: t_n)$ for the homogeneous coordinates on $\mathbb{CP}^{n1}$. For $I$ a nonempty subset of $\{ 1,2, \ldots, n \}$, let $Z_I$ be the subset of $\mathbb{CP}^{n1}$ where $t_i=t_{i'}$ for $i$ and $i' \in I$ and $t_i > t_j$ for $i \in I$ and $j \not \in I$. Note that $Z_I \cong (S^1)^{I1} \times D^{2(nI)}$, where $D^k$ is the open $k$disc. Also, $\overline{Z_I} = \bigcup_{J \supseteq I} Z_J \cong (S^1)^{I1} \times \overline{D}^{2(nI)}$ where $\overline{D}^k$ is the closed $k$disc. We now cut those torii into discs. For $i$ and $i'$ in $I$, cut $Z_I$ along $t_i=t_{i'}$ and $t_i =  t_{i'}$. So the combinatorial data indexing a face of this subdivision is a cyclic arrangement of the symbols $i$ and $i$, for $i \in I$, with $i$ and $i$ antipodal to each other. For example, let $I=\{ 1,2,3,4,5 \}$ and write $t_k=e^{i \theta_k}$ for $k \in I$. Then one of our faces corresponds to the situation that, cyclically, $$\theta_1 < \theta_2 = \theta_4 + \pi < \theta_3 = \theta_5 < \theta_1+ \pi < \theta_2 + \pi = \theta_4 < \theta_3 + \pi = \theta_5 + \pi < \theta_1.$$ This cell is clearly homeomorphic to $\{ (\alpha, \beta) : 0 < \alpha < \beta < \pi \}$. Similarly, each of these cells is an open ball, and each of their closures is a closed ball. We have put a CW structure on the torus. Cross this subdivision of the torus with the open disc $D^{2(nI)}$. The result, if I am not confused, is a regular $CW$ decomposition of $\mathbb{CP}^{n1}$. 


I think the comments answer the question, but to give you a reference: Milnor, Stasheff: Characteristic Classes, Chapter 6 They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CWComplex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion. EDIT: Sorry for the sloppiness! Edit 2: $\oplus$: Perhaps the next sloppiness: The CWstructure of $\mathbb{CP}^n$ obtained by Schubert cells isn't regular (the characteristic map is 2to1) but I think there exists a regular CWstructure. But this might be harder to prove than I thought?! 


An online search yielded a reference to Francis Sergeraert's paper, Triangulations of complex projective spaces, available at http://wwwfourier.ujfgrenoble.fr/~sergerar/Papers/ . But, to quote the author: "The Kenzo program is used to automatically produce triangulations of the complex projective spaces $P^nC$ as simplicial sets, more precisely of spaces having the right homotopy type. The homeomorphism question between the obtained objects and the projective spaces is open." 


$CP^{n1}$
, one might try to show that there's some triangulation of the next cell such that after attaching this cell we still have a simplicial complex. The attaching map is the quotient map$S^{2n1} \to CP^{n1}$
, whose fibers are copies of$S^1$
. The inverse image of a point under a simplicial map is always discrete, so this attaching map is definitely not a simplicial map, no matter what simplicial structures you use. So I think John's question is not so trivial. – Dan Ramras Apr 7 '10 at 23:18