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Suppose $\psi: GL_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ defined by sending $ g\mapsto gAg^{-1}. $

Then why is it that $d\psi:T_eGL_2(\mathbb{C})=M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ is defined to be $ C\mapsto [C,A]? $

It is sort of related to this link, but I am not sure if the same strategy as in that link will work here.

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The same strategy works (with $2$ replaced by $n$). The saving grace here is that $\text{GL}_n(\mathbb{C})$ naturally embeds into $\mathcal{M}_n(\mathbb{C})$ and its tangent space at any point can be canonically identified with $\mathcal{M}_n(\mathbb{C})$. So write $g = 1 + \epsilon C$ and compute the $\epsilon$ term of $gAg^{-1}$...

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    If you'd like to see a complete motivation of the definition of a Lie algebra using these nilpotent-style calculations, you might also be interested in http://qchu.wordpress.com/2011/02/26/the-quaternions-and-lie-algebras-i/ .2012-08-16