Why is complex projective space triangulable? - MathOverflow

Why is complex projective space triangulable?

2010-04-07T21:31:05Z

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?

Answer by Spinorbundle for Why is complex projective space triangulable?

2010-04-08T09:20:00Z

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?!

Answer by Robin Chapman for Why is complex projective space triangulable?

2010-04-13T18:57:37Z

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."

Answer by David Speyer for Why is complex projective space triangulable?

2010-08-11T15:32:33Z

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