Talk:Maurer-Cartan form

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

It's Ludwig Maurer. Charles Matthews 06:06, 30 September 2005 (UTC)

[edit] If G is embedded n GL(n)

We know that $\omega:T_gG\rightarrow T_eG$ . Quoting from the article, "If G is embedded in GL(n), then $ω = g - 1 d g$ ."

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

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

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

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Talk:Maurer-Cartan form

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[edit] If G is embedded n GL(n)

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