Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / A (Lie Algebras)

Define $L_1=L$ and $L_{n+1}=[L,L_n]$, where brackets of subspaces mean their linear span. Each term is an ideal. Inductively, if $L_n$ is an ideal, the Jacobi identity givesso $L_{n+1}$ is an ideal as well. Also $L_{n+1}\subseteq L_n$. Nilpotence means that this descending series eventually reaches zero; class $c$ means $L_{c+1}=0$ with $L_c\ne 0$ for nonzero $L$.

The idealizer here is the normalizer ${\mathrm{Id}}_L(A)=\{x\in L:[x,A]\subseteq A\}$. Since $A$ is a subalgebra it contains its own brackets, so $A\subseteq {\mathrm{Id}}_L(A)$.

Because $L_1⊈A$ and the lower central series terminates, choose $j$ with $L_j⊈A$ but $L_{j+1}\subseteq A$. Take $x\in L_j\setminus A$. Then $[x,A]\subseteq [L_j,L]=L_{j+1}\subseteq A$. Thus $x$ belongs to the idealizer and not to $A$, proving $\boxed{A<{\mathrm{Id}}_L(A)}$.

Start with $I_0=I$ and set $I_{k+1}=[L,I_k]$. Every term lies in $I$, and $I_k\subseteq L_{k+1}$, so the sequence eventually vanishes. Choose its last nonzero term $I_j$. Then $[L,I_j]=0$, hence $0\ne I_j\subseteq I\cap Z(L)$. Therefore $\boxed{I\cap Z(L)\ne \{0\}}$.

Induct on $i$, proving the claim for all $j$ simultaneously. The case $i=1$ is the definition $[L,L_j]=L_{j+1}$. If it holds for $i$, Jacobi givesBoth uses of the induction hypothesis have first index $i$, so the induction is valid. Thus $\boxed{[L_i,L_j]\subseteq L_{i+j}}$ for all positive indices.

Let $M=L_r$ and write $M_k$ for its own lower central series. Part (c) proves inductively that $M_k\subseteq L_{rk}$: it holds at $k=1$, and $[M,M_k]\subseteq [L_r,L_{rk}]\subseteq L_{r(k+1)}$. If $L$ has class less than $rs$, then $L_{rs}=0$, so $M_s=0$. Consequently $\boxed{\mathrm{class}(L_r)<s}$, with the zero algebra understood to have class zero.

The Killing form is ${\kappa }_L(x,y)=\mathrm{tr}({\mathrm{ad}}_{L}x{\mathrm{ad}}_{L}y)$. Symmetry follows from $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Since $\mathrm{ad}[z,x]=[\mathrm{ad}z,\mathrm{ad}x]$, cyclicity of trace gives the invariance identityIf $x\in I^{\perp }$ and $y\in I$, then $[z,y]\in I$, so the right side is zero. Hence $[z,x]\in I^{\perp }$ for every $z$, proving that $\boxed{I^{\perp }\text{ is an ideal}}$.

For $x\in I$, use a basis of $I$ followed by a complementary basis of $L$. Because $[x,L]\subseteq I$, the matrix of ${\mathrm{ad}}_{L}x$ has blocks $\left(\begin{array}{cc}{\mathrm{ad}}_{I}x& B_x\\ 0& 0\end{array}\right)$. Multiplying two such matrices showsThe restriction statement has both arguments in $I$; it does not identify the two forms for arbitrary elements of $L$ outside $I$.

Now use the ordered basis in the question for a central ideal $J$. The images of the $u_j$ in the quotient have adjoint matrix $A$, so the $u$-coordinate block of their images is $A$; arbitrary $J$-components form $B$. Each $v_j$ is central, so its image under every adjoint map is zero. Thus the matrix is exactly $\left(\begin{array}{cc}A& 0\\ B& 0\end{array}\right)$. Products have top-left block $A_x{A}_y$ and zero bottom-right block. Taking traces givesFinally, a nonzero finite-dimensional nilpotent Lie algebra has nonzero centre by (1)(b). Induct on its dimension: the Killing form descends to the quotient by its centre, which is again nilpotent of smaller dimension, and is therefore zero by induction. The zero algebra starts the induction. Apply this to the nilpotent ideal $I$ and then use the restriction identity to obtain $\boxed{{\kappa }_L(x,y)=0\text{ for }x,y\in I}$. This avoids the false general claim that the product of two nilpotent linear maps must be nilpotent.

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Use the convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. For a finite reduced crystallographic root system, diagonal entries are two and off-diagonal entries belong to $\{0,-1,-2,-3\}$, with $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Transposing this convention merely transposes the Cartan matrix.

The $\alpha$-string through $\beta$ has the consecutive roots $\beta -p\alpha ,\dots ,\beta +q\alpha$, where $p,q\ge 0$ and $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. For distinct simple roots ${\alpha }_i,{\alpha }_j$, the negative step is absent, so $p=0$ and $q=-A_{ij}$.

The matrix determines each simple reflection on the simple-root basis:It therefore determines their generated Weyl group as linear transformations. Every root is in the Weyl orbit of a simple root. For a positive nonsimple root $\beta =\sum c_i{\alpha }_i$, some $(\beta ,{\alpha }_i)$ is positive because $\sum c_i(\beta ,{\alpha }_i)=(\beta ,\beta )>0$. Its reflection subtracts a positive integer multiple of ${\alpha }_i$ and lowers height. It stays positive unless the root is simple, since the other simple-root coefficients are unchanged and at least one is positive. Induction on height reaches a simple root; negative roots follow by sign. Hence $\Phi =W\Delta$, which is completely reconstructed by those reflection matrices. This determines the system up to root-system isomorphism, rather than fixing an absolute length scale; irreducible components may be rescaled.

For positive roots for a chain Cartan matrix, symmetry and adjacency imply all simple roots have equal length; normalize their squared lengths to two. Adjacent inner products are minus one and all others zero. Since every root is Weyl-conjugate to a simple root, every root also has squared length two. Write a positive root as $\beta ={\sum }_{i=1}^l{c}_i{\omega }_i$ with nonnegative integers $c_i$, and set $c_0=c_{l+1}=0$. ThenThere must be exactly two nonzero integer differences, one $+1$ and one $-1$. Nonnegativity and the two zero endpoint values imply that the coefficients are one on one consecutive interval and zero elsewhere.

Conversely a single simple root is a root. If ${\omega }_i+\cdots +{\omega }_j$ is a root and $j<l$, its pairing with ${\omega }_{j+1}^{\vee }$ is minus one. Reflecting in ${\omega }_{j+1}$ therefore adds that simple root, making the next longer interval a root. Induction constructs every interval. Thus