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A while ago I heard that there was no known triangulation of the Grassmannian of 3-planes in 6-space.

To believe a statement like that, you have to be a little bit ungenerous about what you mean by "known." For instance if something is cut out by some polynomial equations and inequalities in R^n, I think there is an algorithm for writing down a simplicial complex homeomorphic to it. But it is not a very satisfying one, maybe even computationally infeasible in some sense.

Real projective space is easy to triangulate. Complex projective space is of the form (torus) x (simplex) divided by a simple equivalence relation on the boundary, so probably a triangulation can be cooked up from there. And there's a beautiful 9-vertex triangulation of CP^2 which is probably not of this form:

http://www.math.brown.edu/~banchoff/howison/newbanchoff/publications/pdfs/9Vertex.pdf

The real Grassmannian of 2-planes in n-space could probably be triangulated by refining the partition by oriented matroids. But for 3-planes and higher, this partition is too crazy.

What are some other triangulations of Grassmannians? Is the claim about 3-planes in 6-space accurate? Is there a reference for 2-planes in n-space?

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Grassmannians have Schubert cell decompositions which are more natural than triangulations in most situations where you need some kind of combinatorial decomposition of a Grassmannian. Is there a particular reason why you want a triangulation? The Schubert cells are PLCW complexes so you could in principle refine these in perhaps some canonical way to get a nice triangulation. –  Ryan Budney Mar 12 '13 at 19:09
    
I have nothing against Schubert cells, but the attaching maps are very complicated. I would be just as interested in hearing about a regular cell complex structure as a triangulation, but it's easy to get from one to the other so I asked about triangulations. If you know a triangulation that refines the Schubert stratification, so much the better! –  David Treumann Mar 12 '13 at 19:16

2 Answers 2

This is only a comment on your statement that a triangulation can "probably" be cooked up for complex projective space. Apparently, a triangulation for $\mathbb{CP}^3$ was only given in 2010 (by Palmieri?), and no triangulation is known for $\mathbb{CP}^n$ for $n\geq 4$. This is summarized in a talk of John Palmieri. It suggests that finding triangulations for other Grassmannians might be a difficult problem.

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The paper on triangulating $\mathbb{CP}^3$ is by Bagchi and Datta: uk.arxiv.org/abs/1012.3235. David Speyer's answer to mathoverflow.net/questions/20664/… seems to give an approach to triangulating $\mathbb{CP}^n$. The idea is to give a regular CW complex structure on complex projective space, and then to do barycentric subdivision. See also Spinorbundle's reply to the same question. Perhaps David's method would work for Grassmannians more generally. –  Benjamin Antieau Mar 12 '13 at 22:26
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I believe there is an easy way to triangulate complex projective spaces. Tyler Lawson already made the same suggestion in a comment at the MO thread menntioned in Antieau's comment. Namely, use that $\mathbb CP^n$ is the nth symmetric power of $\mathbb CP^1=S^2$. An (ordered) triangulation of $X$ yields an triangulation of $X^n$ invariant under the group action, and the second barycentric subdivision of this will yield a trangulation of the orbit space. But this does not seem to help with more general Grassmannians. –  Tom Goodwillie Mar 13 '13 at 15:24

I came across a weaker problem recently, namely finding a simplicial set (or complex) homotopy equivalent to $BO(n)$ was what I needed. The answer is in http://annals.math.princeton.edu/wp-content/uploads/annals-v158-n3-p04.pdf that states that certain "realizations of MacPhersonians" (they are simplicial complexes) are homotopy equivalent to Grassmannians $BO(n)$.

As for triangulation, I would bet on the approach via Schubert cell decomposition. In the case of $BO(n)$ it is not regular, but there is an analog in the case of "oriented Grassmanian" $BSO(n)$, a double cover of $BO(n)$ ($BSO(n)/Z_2=BO(n)$). In https://circle.ubc.ca/bitstream/handle/2429/21381/UBC_1979_A6_7%20J96.pdf?sequence=1 it is shown, that incidence nubers between cells of dimension $r$ and $r-1$ are $\pm 1$ or $0$. As far as I understand, they implicitly even prove that every face is regular. However, there is something still needed to prove that the cell structure is regular, if that is true. If that is true, then barycentric subdivision (normal subdivision) of the Schubert decomposition of $BSO(n)$ will be simplicial set homeomorphic to $BSO(n)$. As simplicial sets are closed under taking quotients, we get a simplicial set for $BO(n)$ as well. In the end, any simplicial set $X$ can be converted into a simplicial complex by taking certain double subdivision $B_{\ast}Sd(X)$ where $Sd$ is the normal subdivision (it takes into account degenerate simplices) and $B_*$ is its variant that ignores the degenerate simplices.

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This paper on realizations of MacPhersonians apparently has a fatal flaw, and was later retracted by Biss. –  Matthew Kahle Mar 18 '13 at 23:21
    
See also arxiv.org/abs/0709.1291 –  S. Carnahan Mar 19 '13 at 2:12
    
Thanks for your warning! Another popular-style reference is books.google.com/books?id=mXWr00HurmMC&pg=PA97 that I have read myself! (If I did not find the story boring then and remembered it, I would not get trapped now:)) –  Marek Krčál Mar 19 '13 at 11:24
    
At the moment I tend to believe that the Schubert cell decomposition of $BSO(n)$ is not regular (the attaching map can be 2:1 on the codimenison 2 "faces" already, I guess). –  Marek Krčál Mar 19 '13 at 11:28

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