# Talk:Maurer–Cartan form

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## Who is Maurer?

Who is Maurer in the Maurer-Cartan form? --romanm (talk) 21:22, 19 August 2005 (UTC)

## If G is embedded n GL(n)

We know that ${\displaystyle \omega :T_{g}G\rightarrow T_{e}G}$. Quoting from the article, "If G is embedded in GL(n), then ${\displaystyle \omega =g^{-1}dg}$."

This definition confused me for a while. As one would correctly assume, the first ${\displaystyle g}$ is really ${\displaystyle L_{g}}$, left multiplication by ${\displaystyle g}$. However, the ${\displaystyle g}$ in ${\displaystyle dg}$ is not ${\displaystyle L_{g}}$, but rather a (local) function ${\displaystyle R^{k}\rightarrow R^{n^{2}}}$, where ${\displaystyle k}$ is the dimension of ${\displaystyle G}$. Thus ${\displaystyle dg}$ is essentially the identity map ${\displaystyle T_{g}G\rightarrow T_{g}G}$, since ${\displaystyle g}$ in this case takes any point in ${\displaystyle G}$ (viewed in ${\displaystyle R^{k}}$) to itself (now viewed in ${\displaystyle R^{n^{2}}}$).

If we were to interpret (incorrectly, as I had) the second ${\displaystyle g}$ also as ${\displaystyle L_{g}}$, then ${\displaystyle dg}$ would denote a map ${\displaystyle dg:T_{h}G\rightarrow T_{gh}G}$, in which case the composite ${\displaystyle g^{-1}dg}$, evaluated at the point ${\displaystyle g}$, would be a map from ${\displaystyle T_{g}G\rightarrow T_{g}G\neq T_{e}G}$ (unless ${\displaystyle g=e}$).

You can regard it as a formal identity in Rn x n, so that g = (xij) and dg = (dxij). This is useful for concrete calculations. More formally, g-1 is ${\displaystyle (L_{g^{-1}})_{*}}$, and dg is the identity map of the tangent space. Silly rabbit 23:37, 16 June 2006 (UTC)

## A Simpler Characterization of the Cartan-Maurer Form

A much simpler way of describing (and understanding) the Cartan-Maurer form should be incorporated into the article. The group quotient (${\displaystyle Q_{g}(h)=g^{-1}h}$) extends to a quotient operation on the tangent spaces through its differential map ${\displaystyle g^{-1}v={Q_{g}}_{*}(v)}$. This is the algebraic generalization of the Cartan-Maurer Form; which is the special case of this operation restricted to tangent vectors ${\displaystyle v\in T_{g}(G)}$.

This should also address the issue raised by the previous comment. If the product operation ${\displaystyle L_{g}(h)=gh}$ is similarly extended to a tangent space operation by ${\displaystyle gv={L_{g}}_{*}(v)}$, then an invariant field ${\displaystyle X}$ is characterized by ${\displaystyle X(g)=gX(e)}$, and the application of the Cartan-Maurer form to it by

${\displaystyle g^{-1}X(g)=g^{-1}(gX(e))=X(e)}$.

These characterizations apply independently of any question of an embedding into GL(n), though they reduce to the corresponding matrix operations in GL(n), when an embedding exists. —Preceding unsigned comment added by 4.159.174.19 (talkcontribs)

I agree that the statement in the article is awkward, and I suppose I assume some responsibility for it. I will see what I can do to make it more palatable. Silly rabbit 14:52, 19 June 2007 (UTC)
I think I know why I had introduced the Maurer-Cartan form in this strange fashion. At the time, I was working on a circle of articles dealing with integrability conditions and Cartan connections. From this point of view, it was desirable to have a version of the MC form which imitated the definition of a Cartan connection by using the right action rather than the left action. I suppose I never came around to finishing off my revisions here, and the article now needs some rather significant organizational changes. Silly rabbit 15:20, 19 June 2007 (UTC)

## Properties

By the definition of the differential, if ${\displaystyle X}$ and ${\displaystyle Y}$ are arbitrary vector fields then

${\displaystyle d\omega (X,Y)=X(\omega (Y))-Y(\omega (X))-\omega ([X,Y])}$.

What is the sense of ${\displaystyle X(\omega (Y))}$ here? If ${\displaystyle \omega }$ were a real-valued 1-form then it would be OK, because ${\displaystyle \omega (Y)}$ would be a real-valued function and X acts on real-valued functions. But ${\displaystyle \omega (Y)}$ is now a ${\displaystyle {\mathfrak {g}}}$-valued function. How acts X on it? 89.135.19.155 (talk) 21:09, 11 August 2008 (UTC)

I have added an explanation. The bottom line is that you can still calculate the Lie derivative of a function with values in a fixed vector space. Hopefully this addresses your question satisfactorily. siℓℓy rabbit (talk) 21:29, 11 August 2008 (UTC)
Yes, thank you (really, it was trivial). 89.135.19.155 (talk) 22:02, 11 August 2008 (UTC)

## 1/2 is missing?

--刻意(Kèyì) 00:02, 28 November 2012 (UTC)

Do you mean that there is 1/2 in the formula ${\displaystyle d\omega +{\frac {1}{2}}[\omega ,\omega ]=0}$ but not in ${\displaystyle d\omega (X,Y)+[\omega (X),\omega (Y)]=0?}$ There is no contradiction: by the definition of bracket of Lie algebra-valued forms, ${\displaystyle {\frac {1}{2}}[\omega ,\omega ](X,Y)=[\omega (X),\omega (Y)].}$ (I elaborated on that definition a little and added a link to it in the Properties section). Jaan Vajakas (talk) 02:46, 10 December 2012 (UTC)

These equations are OK. I suppose I mean the equation
${\displaystyle d\theta ^{i}(E_{j},E_{k})=-\theta ^{i}([E_{j},E_{k}])=-\sum _{r}c_{jk}^{r}\theta ^{i}(E_{r}),}$

by duality yields

${\displaystyle d\theta ^{i}=-\sum _{jk}c_{jk}^{i}\theta ^{j}\wedge \theta ^{k}}$ (2),

the equation (2) is missing a constant 1/2 on the right. But this probably depends on the definition of wedge product of Differential forms.--刻意(Kèyì) 12:08, 10 December 2012 (UTC)

You are right. If we do not require j < k then there should be a coefficient 1/2 by the usual definition of wedge product in differential geometry. Now I fixed it. Jaan Vajakas (talk) 13:01, 10 December 2012 (UTC)
Hah hah. See also Talk:Connection form#Factors of two This is an endemic problem that one must be careful about. 67.198.37.16 (talk) 00:07, 24 October 2016 (UTC)

## Examples needed

This stackexchange post discusses a good example of the Mauer-Cartan form:

https://math.stackexchange.com/questions/1102383/how-does-maurer-cartan-form-work?rq=1 — Preceding unsigned comment added by 161.98.8.1 (talk) 20:59, 12 August 2017 (UTC)