0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Cass Feb 6 '15 at 4:26
4
$\begingroup$

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 $Z=0$.

$\endgroup$
  • 2
    $\begingroup$ I suspect this is quite more elaborate than what the OP is prepared to digest! $\endgroup$ – Mariano Suárez-Álvarez Jun 20 '15 at 18:46
  • $\begingroup$ Yes, I admit that you are right. The elementary proof is already given in the comment. $\endgroup$ – Dietrich Burde Jun 20 '15 at 18:48
  • 1
    $\begingroup$ Assuming the field has characteristic not equal to 2 right? $\endgroup$ – justanothermathstudent Jan 19 at 6:16
  • $\begingroup$ @justanothermathstudent, when the characteristic is $2$, the identity matrix is also in the centre. $\endgroup$ – Zuriel Jun 6 at 0:37
  • $\begingroup$ 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? $\endgroup$ – justanothermathstudent Jun 7 at 9:28
0
$\begingroup$

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 )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.