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?
5 Answers
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 $\mathbb{CP}^{n1}$.

$\begingroup$ After thinking about it for a while, this looks good to me. I have to think about it some more before I'll be completely convinced, but so far it looks very nice. $\endgroup$ Aug 16, 2010 at 21:05

3$\begingroup$ According to the authors of uk.arxiv.org/abs/1012.3235 "no explicit triangulation of $CP^3$ was known so far". $\endgroup$ Dec 16, 2010 at 14:10
Here is an article on explicit triangulation on $CP^n$, http://arxiv.org/abs/1405.2568.
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 nondegenerate 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.
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.
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$cell. The $2k$skeleton is a $\mathbb{CP}^k$
EDIT: Sorry for the sloppiness!
Not every CWComplex is triangulable, but the following holds:
Every regular CWComplex (and $\mathbb{CP}^n$ is a regular complex $\oplus$) $X$ is triangulable.
This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example Cellular structures in topology (p.130) by Fritsch and Piccinini.
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?!

1$\begingroup$ Sorry, I guess I don't know much about triangulations. Why does a CWcomplex structure guarantee a simplicial complex structure? $\endgroup$ Apr 8, 2010 at 21:17

$\begingroup$ The characteristic map isn't 2to1, it collapses an entire dimension! That is to say, the big cell in $\mathbb{CP}^n$ is $2n$ dimensional, so its boundary should be $S^{2n1}$, but it is glued to $\mathbb{CP}^{n1}$, which has dimension $2n2$. (You might be thinking of $\mathbb{RP}^n$.) $\endgroup$ Aug 11, 2010 at 16:08

$\begingroup$ You're right, I was thinking of RP^n. Thanks for the correction $\endgroup$ Aug 11, 2010 at 16:30
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."

$\begingroup$ I've browsed through some of Sergeraert's work before, but I hadn't seen this paper. Unfortunately, it seems to deal with simplicial sets, not simplicial complexes, and it's not clear how to get from a simplicial set structure to a simplicial complex structure. Is it? (A preprint by Lutz (arxiv.org/abs/math/0506372) says that explicit triangulations, as simplicial complexes, of CP^n are not known for n>2.) $\endgroup$ Apr 13, 2010 at 21:43

$\begingroup$ Thanks, John. As Sergeraert says, he hasn't proved his triangulations actually are homeomorphic to $CP^n$! It's still a surprise that explicit triangulations are apparently not known, but on first sight, Tyler's idea looks sound to me. It shows the problem is harder than it looks. $\endgroup$ Apr 14, 2010 at 5:55

$\begingroup$ If anyone is still interested, a paper on triangulations of $CP^2$ by Bagchi and Datta appeared on the ArXiV today: uk.arxiv.org/abs/1004.3157 . $\endgroup$ Apr 20, 2010 at 12:48
$CP^{n1}$
, 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. $\endgroup$