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A Maurer–Cartan forms for a matrix group?

  1. Sep 7, 2016 #1
    I am very confused about that in some literature the Maurer Cartan forms for a matrix group is written as

    ##{\omega _g} = {g^{ - 1}}dg##

    what is ##dg## here? can anyone give an example explicitly?
    My best guess is
    ##

    dg = \left( {\begin{array}{*{20}{c}}
    {d{x^{11}}}& \ldots &{d{x^{1m}}}\\
    \vdots & \ddots & \vdots \\
    {d{x^{m1}}}& \cdots &{d{x^{mm}}}
    \end{array}} \right)% MathType!End!2!1!

    ##
    and if ## V \in {T_e}G##, I can find

    ##{\left. {{X_V}} \right|_g} = {L_{g * }}V = {\left. {{{(gV)}^{kj}}\frac{\partial }{{\partial {x^{kj}}}}} \right|_g}##

    in this way i seem to be able to pullback ##{\left. {{X_V}} \right|_g}##

    ##\begin{array}{l}
    {g^{ - 1}}dg({X_V}) = {({g^{ - 1}})^{ik}}d{x^{kj}}\left( {{{(gV)}^{mn}}\frac{\partial }{{\partial {x^{mn}}}}} \right)\\
    = {({g^{ - 1}})^{im}}{(gV)^{mn}} = {({g^{ - 1}}gV)^{in}} = {V^{in}}
    \end{array}##

    am I right ?
     
    Last edited: Sep 7, 2016
  2. jcsd
  3. Sep 8, 2016 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    The Lie group is a smooth manifold, which means that the elements of the group are at a one-to-one correspondence with points on the manifold. ##g : M \to G## is a map from the manifold into the group. ##g^{-1}##, however, does not mean the inverse of this map, but rather it means the group inverse of the group element in the image of ##g##.

    Take ##SU(2) \simeq S^3## for a concrete example. ##g : S^3 \to SU(2)## gives you the corresponding group element living at each point of the 3-sphere. Choose some convenient coordinates ##x^i## on the 3-sphere, and now ##dg## can be given in coordinates:

    $$dg = \partial_i g \, dx^i$$
     
  4. Sep 8, 2016 #3
    Thanks very much for the help, but I am still not clear what does ##{\partial _i}g## mean here. Can you write it more explicitly if ##g## have coordinates ##\{ {x^i}(g)\} ##?
     
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