Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Grassmanian manifold's topology

  1. Oct 11, 2014 #1
    Let n <= m and G:=Gr(n,m) be the (real) Grassmanian manifold. I understand the topology of the simplest case, that of projective space, and am wondering if there is a way to interpret the topology of the G to similar to projective space, with the according generalizations needed.

    If V^n is an element (vector subspace) of G, then how can I picture a neighborhood of it? The book I have is crap, has no proofs and the author (some Harvard guy) is basically writing for people who know this sh!t already, so in my opinion it defeats the purpose. But back to the question, can anyone help me out here in understand the topology of the Grassmanian, or point me to a good reasonable source where I can read this in?

    Thanks in advance!
    CM
     
  2. jcsd
  3. Oct 11, 2014 #2
    Yeah, it's exactly like projective space. It's a manifold, so one picture of a neighborhood is just R^n, but I don't think that's quite what you want. A point in the Grassmannian is a plane and a neighborhood of it is just all the "nearby" planes, so all the planes you get by tweaking the position of your original plane by a little bit.

    There's some good stuff about Grassmannian manifolds in here:

    http://www.math.cornell.edu/~hatcher/VBKT/VBpage.html

    That's probably your best bet.

    Baez has some really interesting stuff about projective spaces in his "This Week's Finds" blog. I was lazy, so you might have to track down more stuff to understand this well enough, but here's one example:

    http://math.ucr.edu/home/baez/week184.html

    Back when I was studying this stuff, I had a sort of vision for how all the puzzle pieces here should fit together, but I was never quite able to sit down and work it all out completely.
     
  4. Oct 13, 2014 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    One can easily write down a Grassmannian manifold as a coset (of Lie groups). Generically, one has

    $$G(k,n)= \frac{O(n)}{O(k) \times O(n-k)}.$$
    You can obtain this as follows: the Grassmannian ##G(k,n)## is the set of ##k##-planes through the origin in ##n##-dimensional space. A given ##k##-plane can be rotated into any other ##k##-plane, so we start with the rotation group in ##n## dimensions and mod out by the rotations that leave a ##k##-plane invariant. The manifold obtained has a single point for each ##k##-plane. We must use ##O## groups rather than ##SO## groups because the ##k##-planes are un-oriented.

    In the simplest case, you should be able to convince yourself that

    $$G(1,n) = \frac{O(n)}{O(1) \times O(n-1)} = \frac{SO(n)}{\mathbb{Z}_2 \times SO(n-1)} = S^{n-1} / \mathbb{Z}_2 = \mathbb{RP}^{n-1}$$
    for the projective spaces.

    From the coset expression you should be able to compute things like homology groups.
     
  5. Oct 14, 2014 #4

    lavinia

    User Avatar
    Science Advisor

    A classic description is in terms of Shubert cells. Milnor's Characteristic Classes has a chapter.

    Schubert cells form a cell decomposition of the Grassman manifolds. That is: they describe the Grassman as a CW complex, a space constructed from n cells with identifications along their boundaries.

    Here is the Shubert cell description of the Grassman of 2 planes in 3 space.

    The xy-plane projects to a single point. Call this point, P.

    Now consider those planes that contain the x-axis.These project to a 1 cell whose end points are both glued to the point, P. (The planes containing the x-axis have 1 degree of freedom and are parameterized by the unit positive half circle in the yz-plane.) Thus they project to a closed loop with one point attached to the projection of the xy-plane. Call this loop, L

    Now consider the planes that intersect the xy-plane in a line. These project to a 2 cell whose boundary circle is attached to the loop,L.

    In summary, the Grassman of 2 planes in 3 space is homeomorphic to a CW complex that has one cell in dimensions 0 1 and 2.
     
    Last edited: Oct 14, 2014
  6. Oct 14, 2014 #5
    Hatcher covers that stuff, too, in the book that I linked to.
     
  7. Oct 16, 2014 #6
    Thanks a lot guys, I especially appreciate the different responses/viewpoints. I'm going to give them all a try and see which fits my mental conception of the manifold best.

    Thanks again!
    CM
     
  8. Nov 4, 2014 #7

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    Think of your k plane in coordinate n space, and assume it projects isomorphically onto the 1st k coordinate axes (it does project onto some choice of k coordinate axes). then look which vectors of your k plane map onto the standard basis vectors. They form the k rows of a k byn matrix in echelon form. so the remaining entries, which determine the k plane are arbitrary entries in the residual k by (n-k) places at the right end of the matrix. to vary you k plane, just vary those entries, thus a nbhd of your k plane looks like k times (n-k) space. In particular the grassmannian of k planes in n space, has dimension k(n-k).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook