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Prove that $\mathfrak{sl}(n,F)$ (matrices with trace zero) has center $0$, unless $\operatorname{char}F$ divides $n$, in which case the center is $\mathfrak{s}(n,F)$ (scalar multiples of the identity).

I don't understand why the center can be $0$. Aren't scalar multiples of the identity always in the center, because they commute when multiplied by any other matrix? ($AB=BA$ if $A$ is a scalar multiple of the identity.)

PJ Miller
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1 Answers1

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We can show the claim either by direct computation or by Lie algebra theory. In the first case let $E_{ij}$, $i\neq j$ and $H_i=E_{ii}-E_{i+1,i+1}$ for $i=1,\ldots n-1$ be a basis of $L={\frak{sl}} (n,k)$ and $A\in Z(L)$. Then $0=[A,E_{ij}]=AE_{ij}-E_{ij}A$ for all $i\neq j$ and $0=[A,H_i]=AH_i-H_iA$ for all $i$ imply that $A=0$, if $p\nmid n$.

On the other hand, we can use that $L$ is simple in characteristic zero. This means, that every ideal, in particular the center $Z(L)$, is zero or ${\frak{sl}}(n,F)$ itself. However $L=Z(L)$ would imply that $L$ is commutative, which means $n=1$ and $Z(L)=\frak{sl}(1)=0$. In the other case, $Z(L)=0$.
In characteristic $p$ this argument is more complicated, because for $p\mid n$ the center is in fact $1$-dimensional and $L$ then is not simple. However, $L/Z={\frak{psl}}(n,F)$ then is simple for $n>2$.

Dietrich Burde
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  • The last part is not correct. For example, if $n = 2$ and the field has characteristic $2$, then taking the quotient by the center gives a Lie algebra of dimension $2$, and such a Lie algebra cannot be simple. – Tobias Kildetoft Aug 13 '13 at 10:12
  • @Tobias Kildetoft: yes, $sl(2)$ for $p=2$ is the Heisenberg Lie algebra, this is always excluded -sorry. – Dietrich Burde Aug 13 '13 at 12:03