$\mathfrak{sl}(3,F)$ is simple

Prove that $\mathfrak{sl}(3,F)$ is simple, unless $\operatorname{char}F=3$. [Use the standard basis $h_1,h_2,e_{ij}(i\neq j)$. If $I\ne 0$ is an ideal, then $I$ is the direct sum of eigenspaces for $\operatorname{ad}h_1$ or $\operatorname{ad}h_2$; compare the eigenvalues of $\operatorname{ad}h_1$, $\operatorname{ad}h_2$ acting on the $e_{ij}$.]

We have $h_1=e_{11}-e_{22}$ and $h_2=e_{22}-e_{33}$

So I found the following:

$\operatorname{ad}h_1(e_{12})=2e_{12}$, $\operatorname{ad}h_1(e_{13})=e_{13}$, $\operatorname{ad}h_1(e_{21})=-2e_{21}$, $\operatorname{ad}h_1(e_{23})=-e_{23}$, $\operatorname{ad}h_1(e_{31})=-e_{31}$, $\operatorname{ad}h_1(e_{32})=e_{32}$

And for $h_2$:

$\operatorname{ad}h_2(e_{12})=-e_{12},\operatorname{ad}h_2(e_{13})=e_{13}$, $\operatorname{ad}h_2(e_{21})=e_{21}$, $\operatorname{ad}h_{2}(e_{23})=2e_{23}$, $\operatorname{ad}h_{2}(e_{31})=-e_{31}$, $\operatorname{ad}h_{2}(e_{32})=-2e_{32}$

Both sets of eigenvalues are $2,1,1,-1,-1,-2$. I don't understand how to tie this into the ideal $I$.

Solution 1:

The idea might be the following: We show that

• If $I \neq 0$, then $I$ contains some (scalar multiple of) $e_{ij}$ ($i \neq j$),

and then, as per the relations in Patrick Da Silva's comment, must contain all of them, and hence (because also $h_i = [e_{i,i+1}, e_{i+1,i}]$), $I$ is the whole Lie algebra.

To show that, note first that if $h=a_1h_1 +a_2h_2 \in I$, then

$I \ni [h,e_{12}]= (2a_1-a_2) e_{12}$ and

$I \ni [h,e_{13}]= (a_1+a_2) e_{13}$

so we are done unless $2a_1-a_2 = a_1+a_2=0$ which forces either $a_1=a_2=0$, or $char(F) =3$ (where indeed $F \cdot (h_1-h_2) = F \cdot id$ is a non-trivial ideal, as noted in comments and the other answer.) So we can assume that

$(*) \qquad I\cap span(h_1, h_2) = 0$.

Now both $ad (h_1)$ and $ad (h_2)$ are diagonalisable endomorphisms on the whole Lie algebra, and because $I$ is an ideal, they restrict to that subspace, so their restrictions to $I$ are also diagonalisable. They also commute, which implies that they are simultaneously diagonalisable, i.e. there is a basis of $I$ all of whose elements are both in an $ad (h_1)$-eigenspace and in an $ad (h_2)$-eigenspace. Now according to your calculations, if $char(F) \neq 2,3$, each intersection of an $ad (h_1)$-eigenspace with any $ad (h_2)$-eigenspace is either

• $0$, or
• the (one-dimensional) span of one $e_{ij}$, or
• (the intersection of the respective $0$-spaces) $span(h_1, h_2)$.

Since by $(*)$ we had excluded the last case, we are done, unless $char(F)=2$. In this case, the only eigenvalues are $0$ and $1$, and one sees that the intersection of one of the $ad(h_1)$-eigenspaces with one of the $ad(h_2)$-eigenspaces is either contained in $span(h_1, h_2)$ -- which again we excluded by $(*)$ --, or of the form $span(e_{ij}, e_{ji})$. Now if $I$ contained anything in that span, say, $ae_{ij}+ be_{ji}$, then

$I \ni [e_{ij}, ae_{ij}+ be_{ji}]= bh$ with some $0 \neq h \in span(h_1, h_2)$, and since we assumed $(*)$ we have $b=0$ and hence $e_{ij} \in I$.

Solution 2:

You are almost done. Assume char $F\neq 3$, otherwise $\mathfrak{s}(3,F)$ would be a non-trivial ideal of $\mathfrak{sl}(3,F)$, implying $\mathfrak{sl}(3,F)$ is not simple. You have shown that $ad\; h_1$ has eight eigenvectors namely $h_1, h_2, e_{12}, e_{21}, e_{13}, e_{32}, e_{31}, e_{23}$. Note that these eight eigenvectors are linearly independent. Also note that dimension of $\mathfrak{sl}(3,F)$ is 8. So $ad\; h_1$ is an vector-space endomorphism. Now suppose that $I$ is an ideal of $\mathfrak{sl}(3,F)$. Therefore, $ad\; h_1(I) = [h_1,I]\subset I$, which means that $I$ is invariant under the map $ad\; h_1$. But then $I$ has to be span of some of eigenvectors of $ad\; h_1$. Easy calculation can reveal that if any of the eigenvector is there in $I$, then I being an ideal will force all eigenvectors to be there in $I$, that is $I = \mathfrak{sl}(3,F)$. Thus $\mathfrak{sl}(3,F)$ is simple.