What are some triangulations of Grassmannians? 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?
 A: 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.
A: 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.
