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Lie groups as riemann manifolds

  1. Jan 3, 2008 #1
    What Lie groups are also Riemann manifolds?

  2. jcsd
  3. Jan 4, 2008 #2
    Well any smooth manifold admits a Riemannian structure. For a lie group one could give it a metric by specifying the value at the identity component and extend to other tangent spaces by pushing forward along the left multiplication map, that is, by defining the metric to be left invariant.
  4. Jan 4, 2008 #3
    Thanks DeadWolfe, that helped.

    Follow-up question 1: giving a smooth manifold are there many Riemann structures we can assign or just one? I assume the later.

    Question 2: sometimes I read the rotation groups SO(n) are diffeomorphic to the n-sphere with antipodal points identified, sometimes that they are groups of isometries that preserve the metric on an n-sphere. How does that go together?

    thank you
  5. Jan 4, 2008 #4
    If you mean in the sense that the inherit connection can be given as a metric: This is only possible for abelian Lie groups and will turn such a group into a Euclidean space.
  6. Jan 5, 2008 #5
    1: There are many possible metrics that can be assigned to a smooth manifold. Although a sphere and an ellipsoid are equivalent as smooth manifolds, the geodesics are very different. The problem of calculating geodesics of an ellipsoid is quite interesting, it turns out that the geodesics are integrable due to "hidden symmetries".

    2: The definition of SO(3) is the set of 3x3 matrices (actually a representation) preserving the distance to the origin and not reversing orientation. These are the ortogonal matrices with unit determinant. But a 3x3 matrix can also be thought of as a point in [itex]R^9[/itex]. The SO(3)-matrices form a 3-dimensional submanifold in this space and this manifold is [itex]P^3[/itex].
  7. Jan 5, 2008 #6
    Thanks OrderofThings!

    Answer 1 makes perfect sense to me.

    But for 2 I have to further ask: when SO(n) can be understood as projective n-space or the n-sphere why are you saying only abelian Lie groups can be given a metric? SO(n) is non-abelian.
  8. Jan 6, 2008 #7
    The smooth manifold P³ can be assigned different connections. One possibility is an isotropic constant-curvature torsion-free connection. This is a metric connection and makes P³ look locally as a 3-sphere. Another possibility is an isotropic no-curvature connection with torsion. This is the one that makes P³ into a Lie group. Since the connection has torsion it cannot be stated as a metric. The torsion manifests itself as non-commutativity of the Lie group.

    Only abelian Lie groups have connections with no curvature and no torsion. Having no torsion they can be stated as metric connections (but quite dull such).
  9. Jan 6, 2008 #8
    Cool, that's some great information!

    Many thanks again, OrderOfThings.
  10. Jan 6, 2008 #9
    One more thing, though.

    While google searching, I saw the term 'bi-invariant Riemann metric on Lie groups' quite often. What does it mean? Is it different from the 'ordinary' Riemann metric', because it seems it applies to all compact, semi-simple Lie groups?
  11. Jan 7, 2008 #10
    Still love to know what 'bi-invariant Riemann metric on Lie groups' means and how it corresponds to what OrderOfThings wrote so far.

    Any input is welcome.
  12. Jan 8, 2008 #11
    An invariant metric on a Lie group is a scalar product on the tangent spaces that is preserved when one moves around in the group (i.e. parallel transport using the inherit connection). Such a metric is very easy to construct: Choose a scalar product at the identity and define the scalar product at any other point to be the result of transporting the vectors to the identity and calculating the scalar product there.

    The transport of tangent vectors can be done using either left or right multiplication and consequently the metric will be either left- or right-invariant. For compact groups it doesn't matter and the metric will be bi-invariant (also called two-sided invariant).
  13. Jan 8, 2008 #12
    You the man, OrderOfThings. All what you wrote here went straight in my notebook.

    I honestly hope you get paid for your knowledge.

    Now a complete different things, but which is not worth a own thread.

    John Baez defines http://math.ucr.edu/home/baez/octonions/node13.html" [Broken] the classical Lie groups. But why xx*=1 (conjugate), shouldn't it be the transpose?
    Last edited by a moderator: May 3, 2017
  14. Jan 8, 2008 #13
    The joy of others is my award. o:)

    Yes, that is conjugate transpose. For real matrices it reduces to ordinary transpose.
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