Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
11  
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 '10 at 22:00
11  
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 '10 at 22:09
3  
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 '10 at 23:18
2  
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 '10 at 21:14
2  
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 '10 at 2:23
show 6 more comments

4 Answers 4

up vote 6 down vote accepted

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 $\mathbb{CP}^{n-1}$.

share|improve this answer
    
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 '10 at 21:05
3  
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 '10 at 14:10
add comment

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?!

share|improve this answer
1  
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 '10 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 '10 at 16:08
    
You're right, I was thinking of RP^n. Thanks for the correction –  Spinorbundle Aug 11 '10 at 16:30
add comment

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."

share|improve this answer
    
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 '10 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 '10 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 '10 at 12:48
add comment

Here is an article on explicit triangulation on $CP^n$, http://arxiv.org/abs/1405.2568.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.