$\mathfrak {sl}_{3}\mathbb{C}$ is the Lie algebra of $3\times3$ matrices with complex entries and trace $0$ and Lie bracket $[X,Y] = XY-YX \hspace{3mm} \forall \hspace{3mm} X,Y\in \mathfrak {sl}_{3}\mathbb{C}$. The centre of $\mathfrak {sl}_{3}\mathbb{C}$ is the set of all $Z\in \mathfrak {sl}_{3}\mathbb{C}$ s.t. $[Z,X]=0 \hspace{2mm} \forall \hspace{2mm} X \in \mathfrak {sl}_{3}\mathbb{C}$.

I want to show that this centre is $\left\{\mathbf{0}\right\}$. Is there a way to show this without ending up with many equations to solve? I was thinking of representing $X$ in terms of the basis for $ \mathfrak{sl}_{3}\mathbb{C}$ and then it would be enough to show that $[Z,K]=0$ implies $Z=\mathbf{0}$ for every basis element $K$ of $ \mathfrak{sl}_{3}\mathbb{C}.$ But I'm not sure if this is true and it also means solving many equations.


  • $\begingroup$ ok thanks. I got my ordering a bit confused in that sentence thanks for pointing that out. $\endgroup$
    – Sam Forbes
    Feb 10, 2013 at 12:07
  • $\begingroup$ You're welcome! The order is important, if only because you loose points with the wrong one ;-) $\endgroup$ Feb 10, 2013 at 12:12
  • $\begingroup$ Oh, no, I am very sorry, I stated it wrong myself. The correct statement is "$Z$ is in the centre if and only if $[Z, K] = 0$ for all $K$ in a basis." Sorry. $\endgroup$ Feb 10, 2013 at 12:26
  • $\begingroup$ Checking $Z$ against the natural basis of $E_{ij},1\le i,j\le 3, i\neq j$ and $E_{11}-E_{22}, E_{22}-E_{33}$ is good enough here. Courage! Here $E_{ij}$ is the matrix with $1$ at position $(i,j)$ and zeros elsewhere. $\endgroup$ Feb 10, 2013 at 13:08
  • $\begingroup$ @Andreas Sorry for late reply. How do you show that if $Z$ is in the centre then $[Z,K]=0$ for all $K$ in the basis. I know by linearisation that $[Z,X]=\sum_{i=1}^{n} a_{i}[Z,K_{i}]$ where $K_{1},K_{2},\dots, K_{n}$ are basis elements for an arbitrary element $X$ in the Lie algebra.But how do you know that if $\sum_{i=1}^{n} a_{i}[Z,K_{i}]=0$ implies $[Z,K_{i}]=0$ for all $i=1,2, \dots, n$ ? Thanks $\endgroup$
    – Sam Forbes
    Feb 10, 2013 at 17:52

1 Answer 1


As some people already said in the comments, it is necessary for intuition that you work on the basis $(E_{ij})$ (the matrices with $1$ in one entry and $0$ in all the others). But there are other means of proving that the center $\mathfrak z(\mathfrak{sl}_n\mathbb C)$ of $\mathfrak{sl}_n\mathbb C$ is null as, for example, using the Schur's Lemma.

For a better understanding of which hypothesis are needed, let us generalize the discussion to a finite dimensional vector space $V$ under a algebraically closed field $\mathbb K$ of characteristic $0$ instead of $\mathbb C^n$.

We say that a subset $\Gamma\subset\mathfrak{gl}(V)$ is irreducible if for all proper subspace W (i.e. $\neq 0$ or $V$) of $V$ there are $v\in W$ and $X\in\Gamma$ such that $Xv\notin W$.

One can state a version of the Schur's Lemma as:

Let $V$ (under the hypothesis above) and $\Gamma$ a irreducible subset of $\mathfrak{gl}(V)$. If $Z$ commutes with every element of $\Gamma$ then $Z=\lambda I$ (where $I$ is the identity function) for some $\lambda\in\mathbb K$.

So, one can check that $\mathfrak{sl}(V)$ is a irreducible set of $\mathfrak{gl}(V)$ (you can do it chosing a basis for $V$). Hence, if $Z\in\mathfrak z(\mathfrak{sl}(V))$ then $$ZX-XZ=[Z,X]=0,$$ for all $X\in\mathfrak{sl}(V)$, and, by the Schur's Lemma, $Z=\lambda I$, for some $\lambda\in\mathbb K$. Thus, since $\operatorname{tr}Z=0$ for all $Z\in\mathfrak z(\mathfrak{sl}(V))$, we must have that $\mathfrak z(\mathfrak{sl}(V))=0$.

Let us prove the above version of Schur Lemma.

Suppose $\lambda\in\mathbb K$ is an eigenvalue (possibly $0$) of $Z$. Define $$T:=Z-\lambda I.$$ Since $Z$ and $I$ commute with every element of $\Gamma$ then so does $T$.

The image $\operatorname{Im}(T)$ of $T$ is invariant by $\Gamma$ (i.e. there are no $v\in\operatorname{Im}(T)$ and $X\in\Gamma$ such that $Xv\notin\operatorname{Im}(T)$). In fact, for every $v\in V$, $$X(Tv)=T(Xv)\in\operatorname{Im}(T).$$ So, since $\Gamma\subset\mathfrak{gl}(V)$ is irreducible, we must have that $\operatorname{Im}(T)=V$ or $0$. Also, $\operatorname{Im}(T)$ can not be $V$ because $T=Z-\lambda I$ has non-null kernel and $\dim V<\infty$. Therefore, $$Z-\lambda I = T = 0.$$


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