Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 102 / 1

A simple Lie algebra is a nonabelian Lie algebra whose only Lie algebra ideals are zero and the whole algebra. Assume $n\ge 2$; ${\mathfrak{sl}}_1=0$ is not simple. We prove the assertion by the matrix-unit extraction lemma for special linear ideals, avoiding any classification theorem.

Let $I$ be a nonzero ideal in the special linear Lie algebra and $0\ne A\in I$. If an off-diagonal entry $A_{ji}$ is nonzero, two commutators with the matrix unit $E_{ij}$ giveso $E_{ij}\in I$. If $A$ is diagonal and nonscalar, choose unequal diagonal entries; $[E_{ij},A]=(A_{jj}-A_{ii})E_{ij}$ again supplies a matrix unit. A nonzero scalar matrix cannot have zero trace over $\mathbb{C}$, so these cases cover every nonzero $A$.

Once $E_{ij}\in I$, its bracket with $E_{ji}$ gives $H=E_{ii}-E_{jj}$, and $[H,E_{ji}]=-2E_{ji}$ puts the reverse unit in $I$. Bracketing with the other off-diagonal matrix units generates every off-diagonal unit: for a third index $k$, $[E_{ij},E_{jk}]=E_{ik}$ and $[E_{ki},E_{ij}]=E_{kj}$; using the reverse unit fills the remaining row and column, then $[E_{ki},E_{i\ell }]=E_{k\ell }$ fills the others. Opposite pairs generate every trace-zero diagonal matrix. These vectors span ${\mathfrak{sl}}_n$, so

In field characteristic $p$, the identity matrix has trace $p=0$, and hence belongs to ${\mathfrak{sl}}_p$. Its span is a nonzero proper central ideal, so ${\mathfrak{sl}}_p$ is not a simple Lie algebra. In fact its center of a Lie algebra is precisely $k{I}_p$: commuting with every off-diagonal matrix unit forces a matrix to be scalar.

For $p>2$, the matrix-unit extraction lemma for special linear ideals still works because two is invertible. Any ideal containing a nonscalar matrix contains every off-diagonal matrix unit and every trace-zero diagonal matrix, hence equals ${\mathfrak{sl}}_p$. Therefore an ideal of the quotient Lie algebra pulls back either to the center or to the whole algebra. The quotient is nonabelian, since $[E_{12},E_{23}]=E_{13}$ remains nonzero modulo the center. ConsequentlyNo separation of diagonal root spaces is needed, so this proof also handles $p=3$ without assuming their weights are distinct.

For $p=2$, every trace-zero two-by-two matrix has the form $aI+b{E}_{12}+c{E}_{21}$. The only nonzero bracket between these basis vectors is $[E_{12},E_{21}]=I$, which is central. Thus