# The differential of $\psi: GL_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ which sends $g\mapsto gAg^{-1}$

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 $gAg^{-1}$...
You will notice that this computation uses almost nothing about either $\text{GL}_n(\mathbb{C})$ or $\mathcal{M}_n(\mathbb{C})$ (just the ring axioms). Actually it works (on a formal level) for an arbitrary ring $R$ and the "differential" of the conjugation map $A \mapsto gAg^{-1}$ where $g$ is a unit in $R$ if we interpret differentials appropriately by adjoining a nilpotent $\epsilon$. This is basically the motivation behind considering $R$ to be a Lie ring with commutator; morally it is the "Lie algebra" of its unit group. – Qiaochu Yuan Aug 16 '12 at 6:50
Thanks again Qiaochu! I tried a number of methods before posting this question and I just realized that one can write $(I+\epsilon C)^{-1}$ as $I -\epsilon C + O(\epsilon^2)$. And thanks for the additional thoughts! Definitely worth thinking about. – math-visitor Aug 16 '12 at 6:55