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O(3) a 3-manifold in R^9

  1. May 7, 2007 #1
    Hi folks,

    How would I go about showing that O(3) (the set of all orthogonal 3x3 matrices) is a compact 3-manifold (without boundary) in R^9?
     
  2. jcsd
  3. May 8, 2007 #2

    matt grime

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    Which part is troubling you? That is in R^9, is a manifold, or is compact? The first and the last two are straightfoward - it is a closed, bounded subset of R^9 (3x3 matrices form a 9-dim real space), and the inclusion makes it a manifold naturally.
     
  4. May 8, 2007 #3
    I understand that it is in R^9. "Intuitively", it makes sense that it's a 3-manifold, but I'm not entirely clear on the details of the proof. Also, how do you show it is closed and bounded? I guess the fact that we are dealing w/ matrices rather than real numbers is troubling me here...

    Thanks for your help :smile:
     
  5. May 8, 2007 #4

    matt grime

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    M(3)=R^9, is, in coordinates {m_ij : 1<=i,j<=3}. O(3) is given by equations in the m_ij. E.G. M is in O(3) if and only if MM^t=I, giving equations the m_ij must satisfy. What are the equations that define O(3)? These make it clear that the set is closed, and bounded. O(3) is just cut out by some (nonsingular) equations, so it's a manifold (even a variety).
     
    Last edited: May 8, 2007
  6. May 8, 2007 #5

    mathwonk

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    if its a variety its a manifold, since its a group, hence homogeneous.
     
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