# Centre of Lie Algebra $sl_2(\mathbb{F})$

For $L=sl_2(\mathbb{F})$ i.e. matrices with trace zero, what is the centre i.e.
$Z(L)$= {$x\in L : [x,y]=[y,x] \ \forall\ y \in L$}.

I will have to find matrices $A \in L$ such that $AB=BA$ for all $B \in L$. But how to appoach this considering trace is zero here.

• A 2 by 2 traceless matrix has the form a b c -b. Fix such a matrix and impose conditions on a b and c to force it to commute with each of the trivial matrices 0 1 0 0, 0 0 1 0, and 1 0 0 -1. – Cass Feb 6 '15 at 4:26

## 2 Answers

The centre $Z$ of $L=\mathfrak{sl}_2(\mathbb{F})$ is an abelian ideal in $L$, different from $L$ itself. Hence $\dim(Z)\le 2$. Suppose that $\dim(Z)=1$ or $2$. Then $L/Z$ is $1$ or $2$-dimensional, hence solvable. It follows that $L$ is solvable, a contradiction. So the only possibility is that $Z=0$.

• I suspect this is quite more elaborate than what the OP is prepared to digest! – Mariano Suárez-Álvarez Jun 20 '15 at 18:46
• Yes, I admit that you are right. The elementary proof is already given in the comment. – Dietrich Burde Jun 20 '15 at 18:48
• Assuming the field has characteristic not equal to 2 right? – justanothermathstudent Jan 19 at 6:16
• @justanothermathstudent, when the characteristic is $2$, the identity matrix is also in the centre. – Zuriel Jun 6 at 0:37
• So in characteristic 2, there are Lie algebras of dimension 2 which are not solvable? Is that where the characteristic comes in in the above argument? – justanothermathstudent Jun 7 at 9:28

Be careful : when F has positive characteristic p and n an integral multiple of p, the identity matrix lies in sl(n,F), so in this case Z isn't trivial. ( To state it with a little more generality, we know that the center of a gl(n,F) contains the scalar matrices, so under the assumption above, sl(n,F) contains these matrices too )