Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / A / 2

Choose a basis $(u,v)$ with $U=Fu$. The representation matrices areThe one-dimensional modules $U,V/U$ have characters $\lambda ,\mu$. If they coincide, these modules are isomorphic and the first alternative holds.

Otherwise choose $h$ with $\delta (h)\ne 0$, where $\delta =\lambda -\mu$. Replacing $v$ by $v-a(h)u/\delta (h)$ makes the upper-right entry of $\rho (h)$ zero. The representation identity on commutators now gives, for every $x\in H$,Iteration yields $a((\mathrm{ad}h{)}^{r}x)=\delta (h{)}^{r}a(x)$. Nilpotence of $H$ implies $(\mathrm{ad}h{)}^{r}x=0$ for sufficiently large $r$, by its lower central series. Since $\delta (h)\ne 0$, it follows that $a(x)=0$ for all $x$. The new line $W=Fv$ is therefore a submodule, andThe argument works over the stated arbitrary field, including positive characteristic.

The flag makes every composition factor one-dimensional, described by a character $\lambda :H\to F$. The Jordan–Hölder theorem for modules for finite-length modules says that the multiset of simple composition factors is independent of the chosen composition series.

Here is a construction valid over every $F$. For $h\in H$, every generalized eigenspace of $\rho (h)$ is an $H$-submodule. Indeed $\mathrm{ad}(h)$ is nilpotent on $H$, say of exponent $c$, andIf $v$ is killed by $(\rho (h)-aI{)}^{N_0}$, take $N\ge N_0+c$. Terms with $j\ge c$ vanish by adjoint nilpotence, and all other terms kill $v$. This proves invariance of the generalized eigenspace.

The flag also makes the characteristic polynomial of every $\rho (h)$ split over $F$. Whenever two composition characters differ, choose an $h$ on which they differ and split the module into the distinct generalized eigenspaces of $\rho (h)$. Repeat within each summand until all its composition factors have the same character. Every nontrivial split lowers dimensions, so the process terminates. Finally combine summands carrying the same character. Jordan–Hölder ensures that the resulting distinct blocks have distinct factor types, givingThis recursive method does not assume that a single element separates every character, which could fail over a finite field.

Writing matrices in $2\times 2$ blocks, the defining relation givesThe trace is automatically zero. The relation is linear, and if $as=-s{a}^T$, $bs=-s{b}^T$, thenThus $L$ is closed under the commutator and is a Lie subalgebra of ${\mathfrak{sl}}_4(F)$.

Its diagonal subalgebra isPut ${\epsilon }_1(D(a,b))=a$, ${\epsilon }_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition isThe four lines have respective characters ${\epsilon }_1-{\epsilon }_2$, $-{\epsilon }_1+{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $-{\epsilon }_1-{\epsilon }_2$. They are distinct and nonzero because $\mathrm{char}F\ne 2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the dimension obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.