# A Maurer–Cartan forms for a matrix group?

1. Sep 7, 2016

### lichen1983312

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. Sep 8, 2016

### Ben Niehoff

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:

$$dg = \partial_i g \, dx^i$$

3. Sep 8, 2016

### lichen1983312

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 $\{ {x^i}(g)\}$?