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

We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence

Let $A$ be the two-dimensional affine Lie algebra with basis $u,v$ and bracket $[u,v]=v$, and putBoth factors have zero centre, so $Z(L)=0$. The line $\mathbb{C}v$ in the second factor is an ideal, making $L$ PSS. But its quotient onto the nonabelian simple algebra ${\mathfrak{sl}}_2$ is not soluble. A quotient of a soluble algebra would be soluble, so $L$ is not soluble.

If $L$ is abelian and nonzero, $L^{\prime }=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ writeThe coefficient $\lambda (a)$ is uniquely defined and linear. The Jacobi identity givesso $\lambda ([a,b])=0$. If $\lambda \ne 0$, then $L^{\prime }\subseteq \ker \lambda <L$.

If $\lambda =0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L^{\prime }\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both casesNo algebraic closure or characteristic restriction was used.

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 $B,C$.

A reduced crystallographic root system $\Phi$ is a finite spanning subset of a real Euclidean vector space, excluding zero, such that $\Phi \cap \mathbb{R}\alpha =\{\pm \alpha \}$, each reflection $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ preserves $\Phi$, and all Cartan integers $⟨\beta ,\alpha ⟩=2(\beta ,\alpha )/(\alpha ,\alpha )$ are integers. A base is a vector-space basis $\Delta \subset \Phi$ in which each root has integer coordinates either all nonnegative or all nonpositive. The Weyl group is generated by the root reflections.

For nonparallel roots put $p=⟨\beta ,\alpha ⟩$ and $q=⟨\alpha ,\beta ⟩$. Then $pq=4{\cos }^2\theta$ is an integer strictly below $4$ and at least zero. ThereforeIf $(\alpha ,\beta )>0$, the positive integers $p,q$ have product $1,2$ or $3$, so at least one is one. If $p=1$, reflection gives $\beta -\alpha \in \Phi$ and hence its negative. If $q=1$, reflection directly gives $\alpha -\beta \in \Phi$. This proves the positive-inner-product assertion.

To prove that an $\alpha$-string has no gaps, suppose consecutive occurring integer positions $a<b$ in $\beta +t\alpha$ had $b-a\ge 2$. If $(\beta +a\alpha ,\alpha )<0$, apply the previous result to $\beta +a\alpha$ and $-\alpha$ to obtain the missing root at $a+1$. Thus this inner product must be nonnegative. Similarly the absence of a root at $b-1$ forces $(\beta +b\alpha ,\alpha )\le 0$. But the latter inner product exceeds the former by $(b-a)(\alpha ,\alpha )>0$, impossible. Thus every intermediate integer position occurs.

Write the complete finite string as $\beta -r\alpha ,\dots ,\beta +s\alpha$. Reflection maps position $t$ to $-p-t$, so it interchanges the two endpoints. HenceHere $r,s$ are the maximal endpoint distances, not arbitrary shorter distances mentioned in the existence assertion.

Use the Cartan-matrix convention $C_{ij}=⟨{\alpha }_i,{\alpha }_j⟩$. The displayed matrix is realized bywith the standard orthonormal coordinates. Closure under the two simple reflections givesThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2$. Conversely the standard fact that every root is Weyl-conjugate to a simple root, or successive height reduction by simple reflections, shows these two reflection orbits exhaust the roots. This is the rank-two system $B_2$, also called $C_2$ after exchanging the length convention. With a transposed Cartan-matrix convention, interchange which simple-root label is short; the geometric system and Weyl group are unchanged.

The reflection in $e_2$ flips the second coordinate, while the reflection in $e_1-e_2$ swaps the two coordinates. These generate all eight signed permutation matrices in two coordinates. Thuswhere $D_8$ means the symmetry group of a square.