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

Path connected and connected.

  1. Feb 5, 2008 #1
    My assignment is to prove that the next groups: SO(n),U(n),SL(n,R) are path connected, and that the groups O(n),GL(n,R) are not connected.

    now for the first group I tried to do it with brute force, but with no success, i.e SO(n) are the nxn orthogonal matrices with determinant 1, so we need to find a function f:[0,1]->SO(n) s.t f(1)=A f(0)=B for every A,B in SO(n), so as always i tried to use this function: f(x)=xA+(1-x)B where x in [0,1] but I need to show that the multiplication of it with its transpose gives the identity matrix, any ideas here?

    also if you can help me with the other groups it would help very much.

    thanks in advance.
     
  2. jcsd
  3. Feb 5, 2008 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The continuous image of a connected set is connected. det is a continuous function.
     
  4. Feb 5, 2008 #3
    I don't see how this helps me, i need to show that they aren't connected, so you say that they are connected cause the determinant is a continuous function from R to those groups, perhaps the lecturer meant that they aren't path connected?

    still how do i see path connectedness for the other groups?
    thanks in advance.
     
  5. Feb 5, 2008 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I didn't say they are connected. :biggrin:

    Either reductio ad absurdum or contrapositive.
     
  6. Feb 5, 2008 #5
    then if we assume that O(n) which is the orthogonal matrices set and is connected, then because det is continuous then also the set {-1,1} is connected (cause the determinant of orthogonal matrix is +-1, but this set isn't connected, cause it equals: {-1}U{1} which is a seperation of {-1,1}.
    ok i think i got it.

    BTW, what with the other groups which i need to prove that they are path connected, any tips?


    I have another question (sorry for the flood of questions, but the lectrurer decided to shortcut a 10 weeks schedule to a 6 weeks schedule by adding three hours to a three hours class, all because of the damn of the lecturer's strike, so I'm having difficulty coping with my schedule, this thing can happen only in israel or other third world countries), the question is:
    find the: connectedness components, path components and equivalnce classes (by the equivalence relation: x~y in X iff there arent disjoint open sets A,B of X s.t x in A and y in B, and X=AUB.) of the next spaces in R^2 with the standard topology:
    A=(Kx[0,1]U{(0,0)}U{(0,1)}
    B=AU{(t,0)|t in [0,1]}
    C=(Kx[0,1])U(-Kx[-1,0])U([0,1]x(-K))U([-1,0]xK)

    well for A i think because A is locally connected then the equivalnce classes are the same as the connectedness components, and I think these sets are the ones constructing A, i.e: {(0,0)} , {(0,1)} and Kx[0,1] are the connectedness componenets and the path componenetes as well of the space A, am I way off here from the answer?

    also if you can help me with the other three sets.

    I feel burntout.
     
  7. Feb 7, 2008 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    SU(n) is a manifold, and thus is locally connected. Therefore, if SU(N) is connected, it is path-connected. SU(n-1) is a closed subgroup of SU(n), S^n is connected, and S^(2n-1) = SU(n)/SU(n-1). Consequently, an inductive argument can be used to show that SU(n) is connected.

    What is K?

    Are you using Munkres as a text? It doesn't do this question, but it considers similar examples? A looks like the deleted comb without its spine.
     
  8. Feb 7, 2008 #7
    K={1/n|n in N}
    -K={-1/n|n in N}

    yes, i'm using munkres, but those exercises are from the lecturer.
     
  9. Feb 7, 2008 #8

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Take any U in U(n) (n>1) and diagonalize it. The diagonal will have things of the form e^(ix) on it, and you can use these to form paths.

    As for the path-connectedness of SL(n,R) IIRC this can be done using polar decomposition. See B. Hall, Lie Groups, Lie Algebras, and Representations.
     
  10. Feb 8, 2008 #9
    well we havent yet even touched manifolds in the course, let alone Lie groups, so I guess there's a simpler way here.
     
  11. Feb 8, 2008 #10

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    It is simple - it just happens to be in a book on Lie groups.
     
  12. Apr 8, 2008 #11
    An element of O(n) may be thought of as an ordered orthonormal basis for R_n. If one thinks of rotating this basis around then one gets other bases with the same orientation (same determinant).

    So you need to convince yourself that

    1) rotation preserves determinant.

    2) Any rotation is connected to the identity map by a path of rotations.

    Intuitively this is clear.

    The determinant argument given to you is correct to show that O(n) is not connected.
    For instance, one could look at the inverse image of the interval (-2,0) in Gl(n,R) under the determinant. This is an open set in Gl(n,R) which is disjoint from SO(n). So O(n) is not connected in the subspace topology inherited from Gl(n,R).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?