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

Tangent Space to Unitary Group

  1. Apr 16, 2010 #1
    This may seem like an easy question, but my differential geometry is a little rusty. I'm trying to find the tangent space to the Lie group [itex] U(n) [/itex]; that is, for an arbitrary [itex] X \in U(n) [/itex] I'm trying to find an expression for [itex] T_X U(n) [/itex].

    I can't quite remember how to do this. I've been playing around with the idea that if we define a function
    [tex] F(X) = X^\dagger X - I [/tex]
    where [itex] \dagger [/itex] is the conjugate transpose, then [itex] U(n) = F^{-1}(0) [/itex]. I think [itex] F: M_n(\mathbb C) \to Sp(n) [/itex] where the domain is nxn matrices and the codomain is the symplectic group, though I don't think this is too important.

    So if [itex] U(n) = F^{-1}(0) [/itex], can I compute the tangent space as
    [tex] T_X U(n) = DF^{-1}(0)[X] = \left\{ Z \in M_n(\mathbb C) : X^\dagger Z + Z^\dagger X = 0 \right\} [/tex]

    This feels like it would be a suitable candidate, since if X is the identity matrix, this reduces to
    [tex] T_I U(n) = \left\{ Z \in M_n(\mathbb C) : Z + Z^\dagger = 0 \right\} = \fraktur{u}(n) [/itex]
    the corresponding Lie Algebra of traceless skew-Hermitian matrices.
  2. jcsd
  3. Aug 29, 2014 #2
    Does somebody know if this is correct?
  4. Aug 30, 2014 #3


    User Avatar
    Science Advisor
    Gold Member

    Usually the way it's usually derived is to first find the tangent space at the identity, which is the important one anyways since that is the Lie algebra. From that, you can use either the left or right action of the group to find a 1-1 isomorphism between the Lie algebra and all other tangent spaces of the manifold.

    The easiest way to find the tangent space at the identity is to look for paths through it, and then taking the derivatives of these paths to find the tangent vectors.

    For example, if A(t) is a particular path in manifold (i.e. it is a unitary matrix for all t) and A(0)=I, then the tangent space is the set of all matrices A'(0) such that the above is the case.

    In particular we know that ##A(t)A^{\dagger}(t)=I##. Taking the derivative of both sides at t=0 gives us the condition that ##A'(0)A^{\dagger}(0)+A(0)A'^{\dagger}(0)=0##. Since we know that ##A(0)=I## and therefore ##A^{\dagger}(0)=I## we conclude that the tangent at the identity of a unitary group is the set of all matrices B such that ##B+B^{\dagger}=0## which is exactly the set of all anti-Hermitian matrices (which we know to be the tangent space of the unitary group).

    I'm not sure if this is what you wanted. But I can't follow your derivation as I am unfamiliar with such formal mathematical notation. Sorry.

    EDIT: I just realized this thread is 4 years old...so...
    Last edited: Aug 30, 2014
  5. Aug 30, 2014 #4
    The thread is old, But I'm interested :)

    Thanks for your post Matterwave, now I understand how one obtain that the lie algebra of the unitary group are antihermitian matrix.

    Waht about the tanget space for general A?

    Repeting your procedure the answer seems to agree with the one in the original post.
  6. Aug 30, 2014 #5


    User Avatar
    Science Advisor
    Gold Member

    Well, we obtained the tangent space. I probably should have been clearer in my terminology. To get the Lie algebra requires us to to further show that matrix commutation is the correct Lie bracket (e.g. that it satisfies the Jacobi identity among others).

    For a general matrix group, you can do this same exact procedure, with the Lie algebra being specified by the matrix commutator (which I won't prove here). For example, using this procedure, you will also find that the tangent space to the orthogonal matrices are the anti-symmetric matrices.

    If A is a more abstract Lie group, e.g. an extraordinary Lie group, then I'm not sure how one would apply this construction. But of course abstractly speaking, if A(t) is any path through the identity such that A(0)=e (e=identity element), then A'(0) will be a member of the Lie algebra. But without the useful matrix commutator acting as your Lie bracket, you will have to start working with isomorphisms induced by the right action (or the left action) and taking the actual Lie bracket on the manifold as the Lie bracket...
  7. Aug 30, 2014 #6

    Sorry for being annoying, But what about the tanget space of a general at a A that belongs to U(n)?
  8. Aug 30, 2014 #7


    User Avatar
    Science Advisor
    Gold Member

    I'm not sure what your question is, we have proved earlier that the tangent space to U(n) is the set of all n-dimensional Anti-hermitian matrices.
  9. Aug 30, 2014 #8
    I'm seeing the prove for the identity. You showed the equation


    and then at the identity


    Do we get the same result for other ##A≠I##?
  10. Aug 30, 2014 #9


    User Avatar
    Science Advisor
    Gold Member

    A=I only at t=0. A(t) is not generally equal to the identity.

    Also, I proved the equality ##A'(0)+A'^{\dagger}(0)=0##. Notice the primes on them. These are the derivatives, and not the A's themselves.

    Although I'm still not understanding your question...
  11. Aug 30, 2014 #10
    Sorry, I'm not native speaker.

    What I want to know is if tanget space to an arbitrary ##A## that belongs to SU(n) (not just at the identity) is still the set of antihermitian matrices?

    I'm inclined to say yes, but I want to be sure.
  12. Aug 30, 2014 #11


    User Avatar
    Science Advisor
    Gold Member

    Oh, in that case you will get something like this:

    $$A(P)A'^\dagger (P)+A^\dagger(P)A'(P)=0$$

    Now, this condition actually depends on what ##A(P)## is, and I am not seeing any way of reducing this to a form easily recognizable as showing A' is an anti-hermitian matrix.

    However, the tangent spaces on a particular manifold are certainly isomorphic to each other, so at least the tangent space at P must be isomorphic to the set of all anti-hermitian matrices of dimension n.
  13. Aug 30, 2014 #12


    I think that if I write $$A(t)=Exp[tH]$$, with $$A(P)=Exp[PH]$$ then I can get $$H=-H^\dagger$$
  14. Aug 31, 2014 #13


    User Avatar
    Science Advisor
    Gold Member

    If we care only about the connected component of the identity (which are those points which can be reached by the exponential map) then we can have ##A(t)=\exp(tH)## indeed. However, recall that ##H## is an element of the tangent space at the identity. This means that ##H=-H^\dagger## as we already showed in my previous posts. The element we care about now is ##A'(P)## which given this map is now simply ##A'(P)=H\exp(tH)|_P##. Now ##\exp(tH)|_P## is some unitary matrix (it is in fact ##A(P)##), so then ##A'(P)## is the original member of the Lie algebra (at the identity) rotated by some unitary matrix ##A(P)##:


    I'm sure that this is related to the induced isomorphism by the right action of the group, but I'm not seeing a concrete say to develop that notion at the moment.
  15. Sep 1, 2014 #14
    I don't know if that's whay you meant by
    But the exponential map is not in general surjective form the Lie algebra of the group to the identity component of the group.
  16. Sep 1, 2014 #15


    User Avatar
    Science Advisor
    Gold Member

    It is surjective for the Unitary groups since they are compact.
  17. Sep 18, 2014 #16
    It seems that is the case. So far the Only explicit reference I have found is in Wulf Rossman "lie Groups: And introduction through Linear groups", pag 46. Not sure if it is true for more general Lie groups.

    Thanks for your help.
  18. Sep 30, 2014 #17
    You should use the level set theorem.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook