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.)


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 $\mathfrak{g}$.

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

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.