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

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.

-

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 $\epsilon$.

-
Thanks Qiaochu! I will give that a try. Definitely better than working with coordinates... – math-visitor Aug 16 '12 at 6:02
I know that we "ignore" higher order perturbations $O(\epsilon^2)$, but why are we ignoring $[B_1,B_2]$ that appears in the expansion as well? – math-visitor Aug 16 '12 at 6:09
@math-visitor: because it's a constant. We only care about how the expression changes as $\epsilon$ changes. – Qiaochu Yuan Aug 16 '12 at 6:11
Another way to say that is we ignore $[B_1,B_2]$ because that's the constant part of our estimate, which we'd subtract out before computing the limit of a difference quotient in a more elementary computation. The differential's just the linear portion-if we wanted the quadratic part of the estimate, we'd take $[D_1,D_2]$ and forget above both the constant $\textit{and}$ the linear part. – Kevin Carlson Aug 16 '12 at 9:29
Congratulations on your graduation Qiaochu! Just one more question: how can one see that the dimension of the cokernel of $d\phi_{(B_1, B_2)}$ equal to 2? It is equal to 2 right? – math-visitor Aug 18 '12 at 2:25

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 .

-
Thank you for the generalization Georges! This is a very good remark and an answer especially if others want to do similar computations for other bilinear maps. One question though: for the above example, shouldn't the dimension of the image of $d\phi_{(B_1, B_2)}$ equal 2? How can one see that? – math-visitor Aug 18 '12 at 18:34
Dear math-visitor, no, the dimension of the image is not always 2: what makes you believe that? For example for $B_1=B_2=0$ the differential is zero and so its image has dimension zero. – Georges Elencwajg Aug 18 '12 at 22:10
Very good point. Thank you Georges. – math-visitor Aug 18 '12 at 23:16
You are welcome, math-visitor. – Georges Elencwajg Aug 18 '12 at 23:22