Computing the differential of the map $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$

Question

Let $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ be the map $$(B_1,B_2)\mapsto [B_1,B_2]$$ which takes two $2\times 2$ matrices to its Lie bracket.

Then why does $d\phi_{(B_1,B_2)}:M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ send $$(D_1,D_2)\mapsto [B_1,D_2]+[D_1,B_2]?$$

$$ $$ Taking $B_1=(g_{ij})$ and $B_2=(h_{ij})$, I do not think taking the partials of the Lie bracket $[B_1,B_2]=$ $$ \left[ \begin{array}{cc} g_{12} h_{21} - g_{21} h_{12} & -g_{12} h_{11} + g_{11} h_{12} - g_{22} h_{12} + g_{12} h_{22} \\ g_{21} h_{11} - g_{11} h_{21} + g_{22} h_{21} - g_{21} h_{22} & g_{21} h_{12} - g_{12} h_{21} \\ \end{array}\right] $$ is a clever way to figure out the map.

Answer 1

Taking differentials is all about looking at what happens to your map upon a very small perturbation. So compute the bracket $$[B_1 + \epsilon D_1, B_2 + \epsilon D_2]$$

and look at the coefficient of $\epsilon$.

Answer 2

This has actually nothing to do with Lie brackets nor Lie algebras!

Given finite-dimensional complex (or real) vector spaces $E,F,G$ and a bilinear map $f:E\times F\to G$, the differential of $f$ at $(a,b)\in E\times F$ is given by the formula (whose proof follows directly from the definition)
$$ df_{(a,b)} (x,y)=f(a,y)+f(x,b) $$

The only thing you have to check in your case is that the bracket is bilinear, which is obvious in the concrete case of matrices and in the abstract theory of Lie algebras is an axiom .