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I suppose given a Lie Group (G) and its corresponding Lie Algebra (g) every element in its dual defines a Maurer-Cartan form on the whole Lie Group?

Let $\omega \in g^*$ be a Maurer-Cartan form and let $X$ and $Y$ be two elements of $g$ then in what sense are $\omega(X)$ and $\omega(Y)$ "constant functions" on G ? (such that we can write $X\omega(Y)=Y\omega(X) =0$)

Assuming the above one can immediately write the Maurer-Cartan equation, $d\omega(X,Y)= -\frac{1}{2}\omega([X,Y])$

Thinking of $\omega$ as Lie Algebra valued 1-form on the Lie Group and using the fact from linear algebra that $V^* \otimes W = Hom(V,W)$ one can write them as $\omega = \sum _i \omega_i \otimes B_i$ where $\omega_i$ are ordinary 1-forms on G and B_i are a basis on g. (should there be some arbitrary coefficients in front of every term in the above sum?)

Say $c^i_{jk}$ are the structure constants of the Lie Algebra then I do not understand how the Maurer-Cartan equations can be recast as,

$$d\omega_i = -\frac{1}{2}\sum_{j,k} c^i_{jk} \omega_j \wedge \omega_k$$

which apparently can be further recast as the equation,

$$d\omega = -\frac{1}{2} [\omega,\omega]$$

I would be happy if someone can explain how the above two forms of the Maurer-Cartan equation can be obtained knowing the first form which is more familiar form to me.

Also finding the structure constants of a Lie Algebra is not so hard for at least the common ones. Knowing that one fully "knows" the Maurer-Cartan Equation. Now is there any sense in which one can "solve" this to find out the Maurer-Cartan forms? (I would guess a basis might be obtainable)

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I did not understand what the Maurer-Cartan form was until I worked it out for a matrix group where it is always just the one for GL(n) restricted to the group. After thar the abstract version is a lot easier to deal with. –  Deane Yang Aug 3 '10 at 22:57
    
This was one of the things I was looking for. To understand if there is any sense in which once can compute an explicit solution to the Maurer-Cartan equation. I haven't seen anything like that even hinted towards in any book. If you could enlighten about that. –  Anirbit Aug 5 '10 at 15:20

2 Answers 2

Note: As explained below, there is a clash of nomenclature between what Morita calls a Maurer--Cartan form and what Cartan introduced (which is described in the wikipedia page, say).

First of all there are two Maurer-Cartan forms: left-invariant and right-invariant. They are one-forms with values in the Lie algebra. If we identify the Lie algebra (=left-invariant vector fields) with the tangent space at the identity, then the left-invariant MC form $\omega$ is such that acting on a vector field $\xi$ on $G$ gives for all $g \in G$, $$ \omega(\xi)_g = (L_g)_*^{-1} \xi_g, $$ where $L_g$ means left multiplication by $g\in G$. There is a also a right-invariant one-form defined similarly but using right multiplication.

Now suppose that $\xi$ is a left-invariant vector field on $G$. This means that $$\xi_g = (L_g)_* \xi_e,$$ where $\xi_e$ is the value of $\xi$ at the identity $e\in G$. In that case, $$\omega(\xi)_g = (L_g)_*^{-1} (L_g)_* \xi_e = \xi_e,$$ which is constant, since it does not depend on $g$.

Now, as you point out, if $X$ and $Y$ are left-invariant vector fields, then it is immediate that $\omega$ satisfies the structure equation: $$d\omega(X,Y) = -\omega([X,Y]).$$

Now choose a basis $(e_i)$ for the Lie algebra, so that we can write $\omega = \sum_i \omega^i e_i$, where the $\omega^i$ are one-forms on $G$. Notice that $\omega(e_i)=e_i$, whence $\omega^j(e_i) = \delta^j_i$.

Applying the structure equation to $X=e_i$ and $Y=e_j$ you see that, on the one hand, $$d\omega(e_i,e_j)=-\omega([e_i,e_j]) = - [e_i,e_j] = - f_{ij}{}^k e_k,$$ whence $$d\omega^k(e_i,e_j) = f_{ij}{}^k.$$ But this is precisely the result of applying $$-\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j$$ on $e_i$ and $e_j$, hence the identity $$d\omega^k = -\tfrac12 \sum_{i,j} f_{ij}{}^k \omega^i \wedge\omega^j.$$

To write down explicitly the Maurer-Cartan forms, it is not hard. You have to compute the derivative of $L_g$ in your chosen coordinates. It is particularly easy if the group $G$ is a matrix group, in which case you have $\omega_g = g^{-1}dg$ and again you have to compute this in your favourite coordinates for $G$.


Added

I just realised that I forgot to answer the bit about the second form of the structure equation. That equation is usually confusing at first because the notation hides the fact that $[\omega,\omega]$ also involves the wedge product of one-forms. By definition, $[\omega,\omega]$ is the Lie-algebra valued 2-form on $G$ whose value on vector fields $X,Y$ is given by $$[\omega,\omega](X,Y) = [\omega(X),\omega(Y)] - [\omega(Y),\omega(X)] = 2 [\omega(X),\omega(Y)].$$ If you now take $X=e_i$ and $Y=e_j$, left-invariant vector fields, you see that $$-\tfrac12 [\omega,\omega](e_i,e_j) = -[e_i,e_j] = -\sum_k f_{ij}{}^k e_k,$$ agreeing again with $d\omega(e_i,e_j)$.


Further addition

This is in response to one of Anirbit's comments. In Morita's book Geometry of Differential Forms, he calls any left-invariant form on $G$ a Maurer--Cartan form. I don't think that this is standard. For me, as my answer above, the Maurer--Cartan form is Lie algebra valued. The two notions of Maurer--Cartan forms can of course be reconciled. Choose a basis $(e_i)$ for $\mathfrak{g}$ and a canonical dual basis $e^i$ for $\mathfrak{g}^*$. Let $\omega^i$ be the left-invariant one-form which agrees with $e^i$ at the identity. Then $\omega = \sum_i \omega^i e_i$ is what I have been calling the (left-invariant) Maurer--Cartan form.

While I'm at it, let me explain the nature of my factors of $2$, since that seems also to be in dispute. For me the wedge product is defined as follows $\alpha \wedge \beta := \alpha \otimes \beta - \beta \otimes \alpha$, without a factor of $\frac12$.

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Hmm... perhaps "particularly easy" is not an accurate assessment of the situation, since this can be quite a lengthy calculation. In exponential coordinates there are a couple of tricks one could use, but it is usually a pain, especially when the Lie algebra has a large semisimple part. For nilpotent and solvable Lie algebras it's not so bad. –  José Figueroa-O'Farrill Aug 3 '10 at 18:40
    
Thanks for this beautiful explicit explanation. There is one point that looks a bit unfamiliar to me. Could you clarify as to which calculation are you alluding to in your comment as a "lengthy calculation" ? Do you mean the proof of the formula that $(L_g)_*^{−1}=g^{−1}dg$? I am new to this equation. –  Anirbit Aug 5 '10 at 15:25
    
Thanks -- the lengthy calculation I referred to is simply the one one needs to do in order to write down explicitly what the MC forms look like in your chosen coordinate system. Once you parametrise a group, in the sense that group elements are given in terms of some local coordinates by $g(x,y,...,z)$, say, then $gˆ{-1}dg$, which is the pullback to the coordinate space of the MC forms, can be quite a complicated object. –  José Figueroa-O'Farrill Aug 5 '10 at 16:11
    
I strongly recommend just working in $GL(n)$, where you have natural global co-ordinates. You get the same solutions as you would if you restrict to a subgroup. There is no point in using local co-ordinates for the subgroup itself. That is, as José says, messy, complicated, and painful. –  Deane Yang Aug 5 '10 at 19:30
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I also recommend unwinding all the abstract notation (push-forwards, pull-backs, differential forms, etc.) and figuring out what everything is on $GL(n)$ in terms of plain old matrices. I found this exercise to be extremely helpful when I struggled with this stuff. –  Deane Yang Aug 5 '10 at 19:33

The Maurer-Cartan form for matrix groups is very well explained in Santalo's book about Integral Geometry.

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