# Complete reducibility of sl(3,F) as an sl(2,F)-module

I was reading the Weyl's theorem on the complete reducibility of a finite dimensional representation of semi-simple Lie algerba and wanted to apply the theorem to the following problem which was unsuccessful. I hope someone can make it clear to me.

Let $M=\mathbf{sl}(3,F)$ be the Lie algebra of trace zero $3 \times 3$ matrices over characteristic zero, algebraically closed field $F.$

$M$ contains a copy of $L=\mathbf{sl}(2,F)$ in its upper left-hand $2 \times 2$ position. Now, write $M$ as direct sum of irreducible $L$-submodules ($M$ viewed as $L$-module via adjoint representation.)

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One way to know which are the simple modules appearing in the decomposition of $M=\mathfrak{sl}_3$ as a $\mathfrak{g}=\mathfrak{sl}_2$ module is to look at the eigenvalues of the action of $H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ on $M$.

An easy computation, which I hope I did correctly, shows that they are $$2,\quad 1,\quad1,\quad0,\quad0,\quad-1,\quad-1,\quad-2.$$ We can collect this information in a polynomial in the variable $q$, summing $q^\lambda$ for all eigenvalues $\lambda$, to obtained the character $\chi_M$ of $M$: in this case we get $$\chi_M=q^2+2q+2+2q^{-1}+q^{-2}.$$

Now, the character is additive with respect to direct sums, and in fact uniquely identifies the isomorphism class of modules. Moreover, for each $\ell\geq1$, the character of the unique simple $\mathfrak{g}$-module $V_\ell$ of dimension $\ell$ is $$\chi_{V_\ell}=\frac{q^\ell-q^{-\ell}}{q-q^{-1}}.$$

Now, an easy computation shows that $$\chi_M=\chi_{V_3}+2\chi_{V_2}+\chi_{V_1}.$$ The above remarks imply that $$M\cong V_3\oplus V_2\oplus V_2\oplus V_1.$$

Now of course you may want to do this explicitly, actually finding the submodules---people want all sort of wierd things! In this case it is not difficult to find them by looking.

• The $\mathfrak{g}$-module $V_3$, the unique simple module of dimension $3$, is of course isomorphic to $\mathfrak g$ with its adjoint action. It is immediate that inside out module $M$ there is a copy of $\mathfrak{g}$.

• Next, the subspace of matrices of the form $\begin{pmatrix}0&0&*\\0&0&*\\0&0&0\end{pmatrix}$ is easily seen to be a $\mathfrak{g}$-module isomorphic to $V_2$ (which is the "taulotogical" module), just as the subspace of matrices of the form $\begin{pmatrix}0&0&0\\0&0&0\\*&*&0\end{pmatrix}$.

• Finally, the matrix $\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$ spans a subspace of matrices which commute with $\mathfrak{g}$.

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You should try to decompose $\mathfrak{sl}_n$ as a $\mathfrak{sl}_2$-module! –  Mariano Suárez-Alvarez Oct 11 '11 at 0:41
Great! I really appreciate it. –  Ehsan M. Kermani Oct 11 '11 at 1:05