Why is complex projective space triangulable?

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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?

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All smooth manifolds are triangulable. This is due to Whitehead. There's a nice write-up in Whitney's "Geometric Integration Theory". – Ryan Budney Apr 7 2010 at 22:00

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For a more direct proof, one might try using the fact that CP^n is homeomorphic to the n-fold symmetric product of S^2. Symmetric products don't take simplicial complexes to simplicial complexes, but the quotient of a subdivision of the n-fold product is itself a simplicial complex. – Tyler Lawson Apr 7 2010 at 22:09

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Deane, what sort of induction are you imagining? Given a simplicial structure on `$CP^{n-1}$`, 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 `$S^{2n-1} \to CP^{n-1}$`, 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 2010 at 23:18

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Oh, triangulating CP^n isn't an exercise in Munkres; rather, one of his exercises says something like, "Assume that CP^n can be triangulated (it can be). Then use the Lefschetz fixed point theorem to ..." -- the statement of the Lefschetz fixed point theorem requires that the space be triangulable. I was looking for justification for his parenthetical remark. – John Palmieri Apr 8 2010 at 21:14

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I'm still a bit surprised that there isn't a way to triangulate $CP^n$ more easily than an arbitrary smooth manifold. Using google, I found the following short paper by Cairns on triangulations of smooth manifolds: projecteuclid.org/… – Deane Yang Apr 19 2010 at 2:23
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3 Answers

I will present a triangulation of $\mathbb{CP}^{n-1}$. More specifically, I will give an explicit regular CW structure on $\mathbb{CP}^{n-1}$. 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 $\mathbb{CP}^{n-1}$. For $I$ a nonempty subset of `$\{ 1,2, \ldots, n \}$`, let $Z_I$ be the subset of $\mathbb{CP}^{n-1}$ 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 $Z_I \cong (S^1)^{|I|-1} \times D^{2(n-|I|)}$, where $D^k$ is the open $k$-disc. Also, $\overline{Z_I} = \bigcup_{J \supseteq I} Z_J \cong (S^1)^{|I|-1} \times \overline{D}^{2(n-|I|)}$ 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 $D^{2(n-|I|)}$. The result, if I am not confused, is a regular $CW$ decomposition of $\mathbb{CP}^{n-1}$.

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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. – John Palmieri Aug 16 2010 at 21:05

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According to the authors of uk.arxiv.org/abs/1012.3235 "no explicit triangulation of $CP^3$ was known so far". – Robin Chapman Dec 16 2010 at 14:10

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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 CW-Complex. (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 CW-Complex is triangulable, but the following holds:
Every regular CW-Complex (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 CW-structure of $\mathbb{CP}^n$ obtained by Schubert cells isn't regular (the characteristic map is 2-to-1) but I think there exists a regular CW-structure. But this might be harder to prove than I thought?!

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Sorry, I guess I don't know much about triangulations. Why does a CW-complex structure guarantee a simplicial complex structure? – John Palmieri Apr 8 2010 at 21:17
The characteristic map isn't 2-to-1, it collapses an entire dimension! That is to say, the big cell in $\mathbb{CP}^n$ is $2n$ dimensional, so its boundary should be $S^{2n-1}$, but it is glued to $\mathbb{CP}^{n-1}$, which has dimension $2n-2$. (You might be thinking of $\mathbb{RP}^n$.) – David Speyer Aug 11 2010 at 16:08
You're right, I was thinking of RP^n. Thanks for the correction – Spinorbundle 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://www-fourier.ujf-grenoble.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."

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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.)  – John Palmieri Apr 13 2010 at 21:43
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. – Robin Chapman Apr 14 2010 at 5:55
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 . – Robin Chapman Apr 20 2010 at 12:48
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