Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 3 (Lie Algebras and Representation Theory)

Take a basis $x,y$ of the defining representation of the sl2 Lie algebra, with $Hx=x$, $Hy=-y$, $Ey=x$, $Ex=0$, $Fx=y$ and $Fy=0$. A monomial basis of the symmetric power ${\mathrm{Sym}}^{4}V$ isFor the exterior power $U={\bigwedge }^2{\mathrm{Sym}}^{4}V$, put $w_{ij}=v_i\wedge v_j$ for $i<j$. These ten elements form a basis, and their weights of a representation add:Thus the complete weight-space decomposition, with explicit bases, isThe weight diagram therefore has multiplicities $1,1,2,2,2,1,1$ at weights $6,4,2,0,-2,-4,-6$, respectively.

The induced raising operator and lowering operator act bySince $E{w}_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible sl2 Lie algebra module ${\mathrm{Sym}}^{6}V$. By complete reducibility of semisimple Lie algebra representations, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the weight diagram leaves weights $2,0,-2$ once each. HenceThese are the only two irreducible isomorphism types occurring as submodules.

For the three-dimensional summand, calculate $E{w}_{03}=3w_{02}$ and $E{w}_{12}=w_{02}$. Thereforeis a nonzero highest-weight vector of weight $2$. The classification of finite-dimensional sl2 representations shows that the submodule $W$ it generates is the irreducible module ${\mathrm{Sym}}^{2}V$, of dimension three.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.

Let $x_1,x_2,x_3$ be the coordinate basis of the defining representation $W$ of ${\mathfrak{sl}}_3$. Write $s_{ij}=x_i{x}_j$ for $1\le i\le j\le 3$. The six $s_{ij}$ form a basis of the symmetric power ${\mathrm{Sym}}^{2}W$, with weight $L_i+L_j$. A basis for the tensor product of Lie algebra representations $T$ consists of all 36 ordered tensorsThe order of the two factors matters; there is no quotient identifying their interchange.

Each weight of a representation has the form $k_1{L}_1+k_2{L}_2+k_3{L}_3$, where the nonnegative integers satisfy $k_1+k_2+k_3=4$. Its weight multiplicity is the number of ways to split this triple into two nonnegative triples, each summing to two. The possible patterns, including every permutation, areFor example, $(2,2,0)$ splits as $(2,0,0)+(0,2,0)$, the reverse split, or $(1,1,0)+(1,1,0)$, giving multiplicity three. For $(2,1,1)$, the four splits pair $(2,0,0)$ with $(0,1,1)$ and $(1,1,0)$ with $(1,0,1)$, each in both orders. The multiplicities sum to $3+12+9+12=36$, verifying the dimension.

The following weight diagram uses Dynkin labels $(p,q)=(k_1-k_2,k_2-k_3)$; the circle numbers give multiplicities and orange points mark the Closed dominant Weyl chamber.

Take the simple roots $L_1-L_2$ and $L_2-L_3$. A weight represented by $(k_1,k_2,k_3)$ is in the Closed dominant Weyl chamber exactly when $k_1\ge k_2\ge k_3$. Among nonnegative triples of total four, the possibilities are $(4,0,0),(3,1,0),(2,2,0),(2,1,1)$. Using $L_1+L_2+L_3=0$, their explicit weight space bases areEach listed tensor has the required summed weight, and the monomial basis from part (i) shows linear independence and exhausts all possible tensors of that weight. Thus these are all the requested bases.

Use the positive simple-root raising operators $E_{12}=x_1{\partial }_{x_2}$ and $E_{23}=x_2{\partial }_{x_3}$. On the weight space of $2L_1-L_3$, the first maps both $s_{11}\otimes s_{12}$ and $s_{12}\otimes s_{11}$ to $s_{11}\otimes s_{11}$, while the second annihilates both. Consequentlyis a nonzero highest-weight vector; the remaining positive-root operator $E_{13}=[E_{12},E_{23}]$ also annihilates it. Its Dynkin labels are $(2,1)$, and the generated submodule $Z$ is the irreducible highest-weight representation with that highest weight.

Put $F_1=E_{21}=x_2{\partial }_{x_1}$ and $F_2=E_{32}=x_3{\partial }_{x_2}$, acting on both tensor factors. Direct calculation givesDefine the linearly independent tensorsThey both have weight $L_1$, andTherefore $A,B\in Z_{L_1}$. To prove that they exhaust this weight space, its weight differs from the highest weight by $(L_1-L_2)+(L_2-L_3)$. The Poincaré-Birkhoff-Witt theorem expresses all descendants at that difference using $F_2{F}_{1}v$ and $F_1{F}_{2}v$: the third possible operator $E_{31}=[F_2,F_1]$ is their difference. Hence the space has dimension at most two, and the two independent vectors already found give

In the finite-dimensional setting, let $R(a,b)$ denote the irreducible ${\mathfrak{sl}}_3$ highest-weight representation of highest weight $a{L}_1-b{L}_3$, where $a,b$ are nonnegative integers. Thus $(a,b)$ are its Dynkin labels. The target $\lambda =L_1-2L_3$ has labels $(1,2)$. Write the highest-weight difference in simple roots asThe dominant weight multiplicity formula for sl3 gives zero unless $r,s$ are nonnegative integers, and otherwise givesFor completeness, this multiplicity is an actual count. In the sl3 interlacing character formula, use top row $(a+b,b,0)$ and bottom entry $s+3$; the target diagonal exponents are $(s+3,s+2,s)$. The middle row $(P,Q)$ obeys $P+Q=2s+5$ andEach allowable integral $P$ gives one Gelfand–Tsetlin basis vector. Subtracting the lower bound from the upper bound gives the minimum of $a,s,r,r+3,b,s+2$, which reduces to $\min (a,b,r,s)$. This proves the displayed multiplicity formula in the case at hand.

Since both summands have a nonzero weight space and their dimensions at $\lambda$ sum to three, their multiplicities must be one and two. For multiplicity one, the minimum is zero. The cases $a=0$ and $b=0$ give $(0,4+3j)$ and $(5+3j,0)$; the remaining cases $r=0$ or $s=0$ give $(1,2)$ and $(3,1)$. Hence the full family isFor multiplicity two, the minimum is one. The cases $a=1$ and $b=1$ give $(1,5+3j)$ and $(6+3j,1)$; the remaining cases $r=1$ or $s=1$ give $(2,3)$ and $(4,2)$. ThusThe minimum conditions exhaust all cases and each listed module has the stated multiplicity. All candidate unordered pairs are $\boxed{\{U,V\}\text{ with }U\in {\mathcal{C}}_1,V\in {\mathcal{C}}_2}$. There are infinitely many at this stage; the question imposes no upper bound on their highest weights.

Let $\nu =-L_3$, whose Dynkin labels are $(0,1)$. Since $\lambda -\nu =L_1-L_3$ is the sum of the two simple roots, its boundary distances in the dominant weight multiplicity formula for sl3 are $r+1,s+1$. ConsequentlyFor a module in ${\mathcal{C}}_1$, this multiplicity is one in the two families having $a=0$ or $b=0$, and two for $R(1,2)$ and $R(3,1)$. For a module in ${\mathcal{C}}_2$, it is two in the families having $a=1$ or $b=1$, and three for $R(2,3)$ and $R(4,2)$. In each case, increasing $r,s$ by one increases the minimum only if neither $a$ nor $b$ already limits it.

To obtain total multiplicity five at $\nu$, the two summands must therefore contribute two and three. The only surviving pairs, up to interchange, areTheir multiplicities at $\lambda$ remain one and two, so all four pairs satisfy both conditions.

Work with a complex semisimple Lie algebra and its root system $R$ in the real span $E$ of the roots of a root system. The Killing form induces a positive definite inner product on this real root space, after the usual identification with the dual of the real span of the coroots. For each root $\alpha$, its Weyl reflection isThe Weyl group is $\boxed{W=⟨s_{\alpha }:\alpha \in R⟩}$. The structural facts being used are that these reflections preserve the finite root system, that the roots span $E$, and that the reflections in the simple roots already generate $W$. Thus $W$ is finite: it acts faithfully as permutations of $R$. Equivalently, it is the normalizer of the Cartan subalgebra modulo its centralizer in the associated adjoint group.

Choose a positive system $R^{+}$ and its simple roots $\Delta =\{{\alpha }_1,\dots ,{\alpha }_{\ell }\}$. The corresponding fundamental Weyl chamber is the open coneIts Closed dominant Weyl chamber replaces all strict inequalities by weak inequalities. The facts used here are that every positive root is a nonnegative integral combination of the simple roots, and that the Weyl chambers are the connected components left after removing the root-orthogonal hyperplanes. Hence the displayed inequalities describe one such component. The Weyl group permutes these components simply transitively, and every orbit in $E$ meets the Closed dominant Weyl chamber in exactly one point. A point on a wall may have a nontrivial stabilizer, which is why uniqueness of the chamber point does not always mean uniqueness of the group element carrying it there.

A highest-weight vector is a nonzero weight vector $v$ such that ${\mathfrak{g}}_{\alpha }v=0$ for every positive root $\alpha$. Its weight of a representation is a highest weight. For an irreducible finite-dimensional Lie algebra representation, this is the unique highest weight of a representation, and $v$ generates the module; a reducible module can have several highest weights of different summands.

Here is an existence argument for every nonzero finite-dimensional module $V$. For each simple root, its root operators and coroot form an sl2 subalgebra associated with a root. The classification of finite-dimensional sl2 representations shows that this coroot acts diagonalizably with integral eigenvalues. These commuting coroots span the Cartan subalgebra, so simultaneous diagonalization gives a finite weight-space decompositionChoose a real linear functional $\ell$ on the weight lattice span that is positive on every simple root, hence on every positive root, and choose a weight $\lambda$ maximizing $\ell$ among the finitely many weights of $V$. If $v\in V_{\lambda }$ and $X_{\alpha }\in {\mathfrak{g}}_{\alpha }$, thenso $X_{\alpha }v$ belongs to $V_{\lambda +\alpha }$ if nonzero. For positive $\alpha$, that would contradict maximality because $\ell (\lambda +\alpha )>\ell (\lambda )$. Thus every nonzero $v\in V_{\lambda }$ is a highest-weight vector. Every nonzero finite-dimensional module therefore has a highest weight. The zero module has no nonzero vector or weight, so it is the vacuous exception to the literal unrestricted wording.

Let $v$ be a highest-weight vector of weight $\lambda$. For a simple root ${\alpha }_i$, choose its sl2 subalgebra associated with a root, with operators $E_i,F_i,H_i$ satisfying $[E_i,F_i]=H_i$, $[H_i,F_i]=-2F_i$, and $H_i={\alpha }_i^{\vee }$. We have $E_{i}v=0$ and $H_{i}v=\lambda (H_i)v$.

Since $F_i$ decreases the weight by ${\alpha }_i$ and there are only finitely many weights of a representation, let $n\ge 0$ be maximal with $F_i^{n}v\ne 0$. Induction using the two bracket relations gives the sl2 highest-weight lowering formulaApplying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^{n}v$ on the right forces $\lambda (H_i)=n$. This holds for every simple root, soThus the highest weight is a dominant integral weight, in particular an element of the Closed dominant Weyl chamber.

Use the ${\mathfrak{sl}}_3$ weight of a representation convention from Question 1: $L_1+L_2+L_3=0$, with simple roots $L_1-L_2$ and $L_2-L_3$. This identification is necessary: the general preceding setup does not by itself specify which root system the symbols $L_i$ refer to.

The Weyl group is $S_3$, acting by permutations of the $L_i$. Let $s_1=(12)$ and $s_2=(23)$, with composition read from right to left. ThenHenceFor direct verification, the diagonal coefficients of $\mu$ are $(1,0,-2)$, so its two Dynkin labels are $1-0=1$ and $0-(-2)=2$, both strictly positive. Thus $\mu$ is in the open Weyl chamber. The original three coefficients $(-2,1,0)$ are distinct, so its stabilizer is trivial and this permutation is unique.

Preserving the symmetric bilinear form infinitesimally means $Z^{T}M+MZ=0$. In blocks of sizes $2,2,1$, write $Z$ as a general block matrix and multiply by the given $M$. The condition says that $MZ$ is skew-symmetric. It forces the lower-right entry to be zero, the middle diagonal block to be the negative transpose of the upper-left block, and the two off-diagonal two-by-two blocks to be skew-symmetric. Thus the general element of the Special orthogonal Lie algebra iswhere $A$ is an arbitrary two-by-two matrix and $p,q$ are arbitrary two-component columns. More explicitly, using ten independent complex parameters,Conversely, substituting this matrix into $Z^{T}M+MZ$ gives zero, so there are no further restrictions. In particular $\dim {\mathfrak{so}}_5=4+1+1+2+2=10$.

The diagonal Cartan subalgebra isPut ${\epsilon }_1=L_1$ and ${\epsilon }_2=L_2$; then $L_3=-{\epsilon }_1$, $L_4=-{\epsilon }_2$ and $L_5=0$ on $\mathfrak{h}$. For the matrix unit $E_{ij}$, $[H,E_{ij}]=(L_i-L_j)(H)E_{ij}$. Applying this to the general matrix in part (i) gives the following one-dimensional root spaces and chosen root vectors:Each row satisfies $[H,X_{\alpha }]=\alpha (H)X_{\alpha }$ and belongs to the Special orthogonal Lie algebra. The two diagonal parameters and eight root vectors span the ten-parameter matrix from part (i), giving the explicit root-space decompositionThis is the B2 root system, with short roots on the coordinate axes and long roots on the diagonals.

For the defining representation, the standard coordinate vectors $e_1,e_2,e_3,e_4,e_5$ are weight vectors of weights ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_1,-{\epsilon }_2,0$, respectively. Thus the weight diagram has four axial points and the origin, each of multiplicity one.

For the Adjoint representation, the eight nonzero weights of a representation are the roots from part (ii), each of multiplicity one, with the displayed $X_{\alpha }$ as basis. The zero weight space is the Cartan subalgebra, with basisThe Adjoint representation diagram therefore has the four axial points, four diagonal points, and the origin of multiplicity two. The multiplicities sum to ten, as they must.

For a finite-dimensional Lie algebra, the Killing form is the symmetric bilinear formIn the root-space decomposition of this semisimple Lie algebra, $\mathrm{ad}H$ acts as zero on the abelian Cartan subalgebra and as multiplication by $\alpha (H)$ on ${\mathfrak{g}}_{\alpha }$. For $H,K\in \mathfrak{h}$, their product thus acts as multiplication by $\alpha (H)\alpha (K)$ on each root space. Every root space here is one-dimensional. Taking the trace over the two-dimensional Cartan subalgebra and the eight root spaces provesFor a general weight decomposition the trace would include the corresponding weight multiplicities; the root multiplicities are one in the semisimple setting here.

The PDF's specified five-by-five diagonal matrix is $H=H_1$, so ${\epsilon }_1(H)=1$ and ${\epsilon }_2(H)=0$. In the root sum for the Killing form, the two short roots $\pm {\epsilon }_1$ contribute $1+1$, the two short roots $\pm {\epsilon }_2$ contribute zero, and the four long roots $\pm {\epsilon }_1\pm {\epsilon }_2$ each contribute one. ThereforeThis computation uses the original PDF matrix; the converted TeX incorrectly gives its dimensions.

In the crystallographic convention for this course, an admissible Coxeter graph comes from linearly independent unit vectors $v_1,\dots ,v_r$ in a Euclidean space, such that for $i\ne j$,The vertices $i,j$ have no connection when $n_{ij}=0$, and a single, double or triple connection when $n_{ij}=1,2,3$. Equivalently, the symmetric Gram matrixmust be positive-definite. This equivalence follows because a positive definite Gram matrix can be realized by linearly independent vectors, while such vectors necessarily have a positive definite Gram matrix. In the edge-labelled Coxeter graph convention, these connections have labels $3,4,6$, since the corresponding inner products are $-\cos (\pi /3)$, $-\cos (\pi /4)$ and $-\cos (\pi /6)$.

Suppose the underlying graph has a cycle through $k\ge 3$ distinct vertices, and let $v_1,\dots ,v_k$ be their unit vectors from admissibility. Adjacent vertices have $(v_i,v_j)\le -1/2$, and every other pair has nonpositive inner product. ThereforeBut linear independence makes the sum nonzero, whose squared norm in a Euclidean space must be positive. This contradiction proves that the underlying graph of an admissible Coxeter graph has no cycles. Double and triple lines record edge multiplicities; they are not cycles of length two. The argument also tolerates chords, which only make the bound smaller.

In the finite-dimensional complex semisimple Lie algebra setting implicit in this root-system construction, the given Cartan subalgebra and root spaces determine a reduced crystallographic root system $R$. The ordering divides $R$ into positive roots and their negatives. The simple roots are the positive roots that cannot be written as a sum of two positive roots; a standard root-system fact is that they form a basis and that every positive root is a nonnegative integral combination of them.

Index the simple roots as ${\alpha }_1,\dots ,{\alpha }_{\ell }$ and form the Cartan matrixFor $i\ne j$, these entries are nonpositive integers and $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Construct one vertex per simple root and join distinct vertices by $A_{ij}{A}_{ji}$ lines. If root lengths differ, direct the arrow toward the shorter simple root. This labelled and oriented graph is the Dynkin diagram. Its unoriented Coxeter graph is admissible: normalizing the simple roots to unit length gives $4(v_i,v_j{)}^2=A_{ij}{A}_{ji}$. A general Lie algebra without the semisimple root-system hypotheses need not produce such a diagram.

Let ${\alpha }_1,{\alpha }_2$ be distinct simple roots. Their difference is not a root, since its simple-root coefficients have opposite signs. The root string through ${\alpha }_2$ in the direction ${\alpha }_1$ therefore starts at ${\alpha }_2$ and has the formHere we use the root-string theorem: if the full string runs from $\beta -p\alpha$ through $\beta +q\alpha$, then $p-q=⟨\beta ,{\alpha }^{\vee }⟩$; in this case $p=0$. Thus ${\alpha }_1+{\alpha }_2$ is a root exactly when $A_{12}<0$, equivalently when the corresponding Dynkin diagram nodes are connected.

The root-space bracket facts are $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, with a zero space when the sum is not a root, and $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]={\mathfrak{g}}_{\alpha +\beta }\ne 0$ when the sum is a root and $\beta \ne -\alpha$. Applying these stated structural facts proves

If the two names denote the same simple root, its one-dimensional root space has zero bracket with itself, and a Dynkin diagram has no self-edge. Thus the equivalence also holds in that case.

If the three simple roots are distinct and every pair of root spaces had nonzero Lie bracket, part (iv) would put all three connections in the Dynkin diagram. Their underlying Coxeter graph would contain a triangle, contradicting part (ii). Hence at least one pair satisfies $\boxed{[{\mathfrak{g}}_{{\alpha }_i},{\mathfrak{g}}_{{\alpha }_j}]=0}$. If two of the listed roots coincide, their common root space is one-dimensional, so its bracket with itself is already zero. Thus the conclusion also covers that case.

The Jordan–Chevalley decomposition is characterized byOver $\mathbb{C}$, a semisimple linear operator is a diagonalizable one. These conditions determine the two parts uniquely. Concretely, on the generalized eigenspace for an eigenvalue $\lambda$, the semisimple part is $\lambda I$ and the nilpotent part is $X-\lambda I$; both are polynomials in $X$. To see uniqueness, any commuting candidate parts preserve each generalized eigenspace; diagonalize the proposed semisimple part there, and nilpotence then forces its eigenvalue to be the sole eigenvalue $\lambda$ of $X$ on that space.

For the displayed matrix, $\det (tI-X)=(t-1{)}^2$ andSince $I$ is semisimple and commutes with $N$, the required parts are

We prove closure using complete reducibility of semisimple Lie algebra representations, rather than assuming that a Lie subalgebra is closed under polynomials in its elements. Write the matrix Jordan–Chevalley decomposition as $X=S+N$. On $\mathrm{End}(V)$, the commutator operator $\mathrm{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_{\lambda }$, then its eigenvalue on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ is $\mu -\lambda$. The operator $\mathrm{ad}N$ is nilpotent: if $N^r=0$, thenThe two operators commute because $[S,N]=0$. Uniqueness of the Jordan–Chevalley decomposition therefore says that $\mathrm{ad}S$ is the semisimple part of $\mathrm{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak{g}$ is invariant under $\mathrm{ad}X$, it follows thatNow regard $\mathrm{End}(V)$ as a Lie algebra representation of $\mathfrak{g}$ under commutators. By complete reducibility of semisimple Lie algebra representations, there is an invariant complement $M$ with $\mathrm{End}(V)=\mathfrak{g}\oplus M$. Write $S=Y+C$, with $Y\in \mathfrak{g}$ and $C\in M$. For $A\in \mathfrak{g}$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in \mathfrak{g}$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole Lie algebra.

Decompose $V={⨁}_j{V}_j$ into irreducible $\mathfrak{g}$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the Schur lemma, $C{|}_{V_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every element is a sum of Lie brackets and has trace zero on every finite-dimensional representation. Consequently $\mathrm{tr}(X{|}_{V_j})=\mathrm{tr}(Y{|}_{V_j})=0$. Also $N{|}_{V_j}$ is nilpotent, so it has trace zero; henceThus every $c_j$ is zero, $C=0$, and $S=Y\in \mathfrak{g}$. Finally $N=X-S\in \mathfrak{g}$. We have proved

A complex semisimple Lie algebra has zero centre, so its Adjoint representation is faithful. Its absolute Jordan decomposition is the unique expression $x=x_s+x_n$ with $[x_s,x_n]=0$, $\mathrm{ad}{x}_s$ diagonalizable and $\mathrm{ad}{x}_n$ nilpotent. Existence follows by applying part (ii) to the adjoint embedding, and uniqueness follows from uniqueness of the matrix Jordan–Chevalley decomposition. The representation-compatibility theorem states that for every finite-dimensional Lie algebra representation $\rho$, $\rho (x_s)$ and $\rho (x_n)$ are respectively the semisimple and nilpotent parts of $\rho (x)$. Thus these parts are intrinsic to the Lie algebra, independent of a chosen matrix embedding, and their semisimplicity and nilpotence are preserved in every finite-dimensional representation.

Use the same nonsemisimple matrix from part (i) and the one-dimensional Lie subalgebraIt is a Lie subalgebra because $[aX,bX]=0$ for all scalars $a,b$. Its Jordan–Chevalley decomposition has $X_s=I_2$ and $X_n=X-I_2\ne 0$. If $I_2=cX$, comparing off-diagonal entries forces $c=0$, which contradicts the diagonal entries. If $X_n=cX$, taking traces gives $0=2c$, again forcing $c=0$ although $X_n\ne 0$. ThusThe semisimplicity assumption on the Lie algebra in part (ii) is therefore essential.