Why is complex projective space triangulable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:31:26Z http://mathoverflow.net/feeds/question/20664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable Why is complex projective space triangulable? John Palmieri 2010-04-07T21:31:05Z 2010-08-11T15:32:33Z <p>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?</p> http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable/20703#20703 Answer by Spinorbundle for Why is complex projective space triangulable? Spinorbundle 2010-04-08T09:20:00Z 2010-04-08T23:30:41Z <p>I think the comments answer the question, but to give you a reference:</p> <p>Milnor, Stasheff: <A href="http://books.google.de/books?id=5zQ9AFk1i4EC&amp;printsec=frontcover&amp;source=gbs_slider_thumb#v=onepage&amp;q&amp;f=false" rel="nofollow">Characteristic Classes</a>, Chapter 6</p> <p>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.<br> 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 $\mathbb{CP}^k$ </p> <p><strong>EDIT:</strong> Sorry for the sloppiness!<br> Not every CW-Complex is triangulable, but the following holds:<br> <strong>Every regular CW-Complex (<strike>and $\mathbb{CP}^n$ is a regular complex</strike> $\oplus$) $X$ is triangulable</strong>.<br> This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example <a href="http://books.google.de/books?id=J-2F641zx-MC&amp;pg=PA130&amp;lpg=PA130&amp;dq=CW-Complex+triangulable&amp;source=bl&amp;ots=857Bjqzpce&amp;sig=RZO51f71tCOvbnqiyIKfSAO_isQ&amp;hl=de&amp;ei=bla-S4-lOYOvOPqK1ZYE&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=2&amp;ved=0CA4Q6AEwATgK#v=onepage&amp;q=CW-Complex%20triangulable&amp;f=false" rel="nofollow">Cellular structures in topology</a> (p.130) by Fritsch and Piccinini.</p> <p><strong>Edit 2:</strong> $\oplus$: 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?!</p> http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable/21246#21246 Answer by Robin Chapman for Why is complex projective space triangulable? Robin Chapman 2010-04-13T18:57:37Z 2010-04-13T18:57:37Z <p>An online search yielded a reference to Francis Sergeraert's paper, Triangulations of complex projective spaces, available at <a href="http://www-fourier.ujf-grenoble.fr/~sergerar/Papers/" rel="nofollow">http://www-fourier.ujf-grenoble.fr/~sergerar/Papers/</a> . 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."</p> http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable/35241#35241 Answer by David Speyer for Why is complex projective space triangulable? David Speyer 2010-08-11T15:32:33Z 2010-08-11T15:32:33Z <p>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.</p> <hr> <p>Recall that to put a regular CW complex on space $X$ means to decompose $X$ into disjoint pieces $Y_i$ such that:</p> <p>(1) The closure of each $Y_i$ is a union of $Y$'s.</p> <p>(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 $d$.</p> <p>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 $\overline{Y_{i_0}} \subset \overline{Y_{i_1}} \subset \cdots \subset \overline{Y_{i_r}}$.</p> <hr> <p>Write $(t_1: t_2: \ldots: t_n)$ for the homogeneous coordinates on $\mathbb{CP}^{n-1}$. For $I$ a nonempty subset of <code>$\{ 1,2, \ldots, n \}$</code>, 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.</p> <p>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 <code>$I=\{ 1,2,3,4,5 \}$</code> and write $t_k=e^{i \theta_k}$ for $k \in I$. Then one of our faces corresponds to the situation that, cyclically, <code>$$\theta_1 &lt; \theta_2 = \theta_4 + \pi &lt; \theta_3 = \theta_5 &lt; \theta_1+ \pi &lt; \theta_2 + \pi = \theta_4 &lt; \theta_3 + \pi = \theta_5 + \pi &lt; \theta_1.$$</code> This cell is clearly homeomorphic to <code>$\{ (\alpha, \beta) : 0 &lt; \alpha &lt; \beta &lt; \pi \}$</code>. 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.</p> <p>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 $\mathbb{CP}^{n-1}$.</p>