Past exam of the mathematics course of the University of Cambridge / 2017 / iii

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

For the finite-dimensional Lie algebra representation specified in the PDF, a Lie-invariant bilinear form satisfiesThe dual Lie algebra representation has action $(x\varphi )(w)=-\varphi (xw)$, so the map $T_B:V\to V^{\ast }$ defined by $T_B(v)=B(v,-)$ is an intertwining operator. If $V$ is irreducible and $B\ne 0$, its kernel is zero. Since $V,V^{\ast }$ have equal finite dimension, $T_B$ is an isomorphism, and $B$ is a nondegenerate bilinear form.

For $B_1\ne 0$, the composition $T_{B_1}^{-1}{T}_{B_2}$ is a representation endomorphism. Over an algebraically closed field, the Schur lemma makes it scalar, provingTransposing $B$ gives another Lie-invariant bilinear form. For $B\ne 0$, write $B^T=\mu B$; transposing twice gives ${\mu }^2=1$. In field characteristic different from two, this means $\mu =1$ or $\mu =-1$. Thus the form is a symmetric bilinear form or an alternating bilinear form, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.

Use the classification of finite-dimensional sl2 representations. In the irreducible representation $V(m)$, choose $v_j=f^j{v}_0$, $0\le j\le m$, sowith vectors beyond the endpoints interpreted as zero. The invariant form on an irreducible sl2 module isIts anti-diagonal entries are nonzero, so it is nondegenerate. Lie-invariant bilinear form invariance under $h$ follows from the sum of the two weights. For $f$, the two potentially nonzero terms are $(-1{)}^{j+1}$ and $(-1{)}^j$. For $e$, their coefficients coincide when $j+k=m+1$, and their signs are opposite. These checks prove invariance under the generators of the sl2 Lie algebra.

Interchanging $j,k$ multiplies the form by $(-1{)}^m$, givingEquivalently it is a symmetric bilinear form in odd integer dimension and an alternating bilinear form in even number dimension. The Weyl complete reducibility theorem expresses any finite-dimensional representation as a direct sum of these irreducibles. Give each summand the displayed form and make different summands orthogonal. The resulting form is invariant and nondegenerate. On a reducible representation with both parities, this orthogonal sum need not itself be symmetric or alternating; the dichotomy in the preceding part required irreducibility.

The PDF's $J$ is the anti-diagonal identity matrix. Write $\bar{i}=2n+2-i$ and $o=n+1$. Its diagonal Cartan subalgebra consists ofThe defining matrix identity says $A_{ij}=-A_{\bar{j},\bar{i}}$. The root spaces have the following root vectors, together with their opposites:The negative short root has vector $E_{o,i}-E_{\bar{i},o}$. Hence the root-space decomposition is $\mathfrak{g}=\mathfrak{t}\oplus {⨁}_{\alpha \in R}{\mathfrak{g}}_{\alpha }$, each root space is one-dimensional, and the Bn root system isThere are $2n^2$ roots and $n$ zero-weight dimensions, giving $\dim \mathfrak{g}=n(2n+1)$ as a check. Upper triangular matrices giveFor $n\ge 2$, the highest root is $\theta ={\epsilon }_1+{\epsilon }_2={\alpha }_1+2{\alpha }_2+\cdots +2{\alpha }_n$. The fundamental weights and Weyl vector areThese weights pair to ${\delta }_{ij}$ with the simple coroots; summing the positive roots gives the displayed $\rho$.

The Dynkin diagram is the chain $1,\dots ,n$, with single edges except a double edge between $n-1$ and $n$, whose arrow points to the short root ${\alpha }_n$. In the Extended Dynkin diagram, ${\alpha }_0=-\theta$ attaches by a single edge to ${\alpha }_2$ when $n\ge 3$. For $n=2$, both long nodes ${\alpha }_0,{\alpha }_1$ have a double edge to the short node ${\alpha }_2$. The Bn Dynkin diagram and affine extension below distinguishes these small-rank cases.

For $n=1$, the system is $A_1$: $R^{+}=\{{\epsilon }_1\}$, $\theta ={\epsilon }_1$, ${\omega }_1=\rho ={\epsilon }_1/2$. Its ordinary diagram is one node; its affine diagram has two equal-length nodes with the usual affine $A_1$ double bond.

Let $U$ be a four-dimensional symplectic vector space. The symplectic contraction of an exterior square splits ${\Lambda }^{2}U$ into its five-dimensional primitive exterior square $W$ and the invariant line spanned by the inverse symplectic bivector $\Omega$. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$. Exterior multiplication defines a symmetric bilinear form on ${\Lambda }^{2}U$ by $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$; it is symmetric because both degrees are two. In a symplectic basis, $\Omega \wedge \Omega =2\mathrm{vol}\ne 0$, and $W={\Omega }^{\perp }$, so the restriction to $W$ is nondegenerate.

The symplectic Lie algebra ${\mathfrak{sp}}_4$ preserves this form, giving a homomorphism ${\mathfrak{sp}}_4\to \mathfrak{so}(W)$. It is injective: an element acting trivially on $W$ also acts trivially on $\Omega$, hence on all of ${\Lambda }^{2}U$. In a four-vector basis, the identities $A{e}_i\wedge e_j+e_i\wedge A{e}_j=0$ first force every off-diagonal coefficient of $A$ to vanish, then force $A_{ii}+A_{jj}=0$ for every pair. Over $\mathbb{C}$ these equalities imply $A=0$.

Both Lie algebras have dimension ten: $\dim {\mathfrak{sp}}_4=4\cdot 5/2$ and $\dim {\mathfrak{so}}_5=5\cdot 4/2$. The injective map is therefore an isomorphism, proving the exceptional isomorphism between sp4 and so5:

Identify the diagonal Cartan subalgebra with coordinates $(h_1,\dots ,h_n)$. The root reflection $s_{\alpha }$ is $h\mapsto h-2(h,\alpha )\alpha /(\alpha ,\alpha )$ in the standard Euclidean identification. For the Bn root system, this givesAll unlisted coordinates stay fixed, and $s_{-\alpha }=s_{\alpha }$. These transformations are coordinate permutations and independent sign changes. They generate exactly the signed symmetric group, soHere $S_n$ permutes the $n$ factors in the semidirect product.

The PDF uses $V={\mathbb{C}}^{2n+1}$ for ${\mathfrak{so}}_{2n+1}$; the TeX incorrectly turns this into a power of two. The crystal of the defining odd-orthogonal representation is the chain with weights ${\epsilon }_1,\dots ,{\epsilon }_n,0,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Label its vertices $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$. Its directed edge colors areIn particular the short-root string $n\to 0\to \bar{n}$ has two edges of color $n$.

For a full drawing of the tensor product of crystals in any rank, take every ordered pair of chain vertices $(a,b)$. Let ${\epsilon }_i(b)$ count incoming edges of color $i$ along its string, and let ${\phi }_i(a)$ count outgoing ones. We use the conventionomitting an edge when the indicated operator vanishes. This crystal tensor-product rule specifies every vertex and every edge; the raising rule uses $\ge$ for its first-factor branch. The picture shows the chain and all 25 tensor vertices for $n=2$, with connected components distinguished by node color.

For $n\ge 2$, the three highest-weight vectors of the tensor crystal are $1\otimes 1$, $1\otimes 2$ and $1\otimes \bar{1}$. They have weights $2{\epsilon }_1$, ${\epsilon }_1+{\epsilon }_2$ and zero. There are no others: a highest tensor vertex must have highest first factor, and the string inequalities then restrict its second factor to these three vertices. ThereforeThe summands are the traceless symmetric square, the exterior square (the Adjoint representation), and the invariant scalar line. Their dimensions are $(2n+1)(n+1)-1$, $n(2n+1)$ and one, summing to $(2n+1{)}^2$. For $n\ge 3$ the first two highest weights are $2{\omega }_1,{\omega }_2$; for $n=2$ they are $2{\omega }_1,2{\omega }_2$. For $n=1$, the three tensor highest vertices are $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, giving $L(4{\omega }_1)\oplus L(2{\omega }_1)\oplus L(0)$ of dimensions $5,3,1$.

Choose a positive root system $R^{+}$ for a complex semisimple Lie algebra, with Weyl group $W$ and Weyl vector $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote its finite-dimensional irreducible representation, and use formal symbols with $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. Its formal character of a weight module is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The Weyl character formula saysHere $\det (w)=(-1{)}^{\ell (w)}$, the sign of the product of simple root reflections. The Weyl denominator formula isThe quotient is initially interpreted in the fraction field of the group algebra of the weight lattice; the theorem makes it a finite character.

Take an integer $k\ge 0$, so $k\rho$ is a dominant integral weight. Apply the dilation $e^{\mu }\mapsto e^{(k+1)\mu }$ to the Weyl denominator formula:This is exactly the numerator of the Weyl character formula for $L(k\rho )$. Divide by its denominator and use a finite geometric series in each root direction:This character of a Weyl-vector multiple also gives $\dim L(k\rho )=(k+1{)}^{|R^{+}|}$ by evaluating every formal exponential at one. This evaluation is safe in the final polynomial expression, without dividing by a zero denominator.

Number the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, normalized by $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=6$ and $({\alpha }_1,{\alpha }_2)=-3$. Its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2,3{\alpha }_1+{\alpha }_2,3{\alpha }_1+2{\alpha }_2$, and the negatives complete the twelve-root system. The short roots form one hexagon, and the long roots form a rotated hexagon whose radius is $\sqrt{3}$ times larger. The picture also supplies the crystal requested in the next part.

The fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$, ${\omega }_2=3{\alpha }_1+2{\alpha }_2$, and $\rho ={\omega }_1+{\omega }_2$. The six positive coroots, in coordinates relative to ${\alpha }_1^{\vee },{\alpha }_2^{\vee }$, are $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ therefore gives the G2 dimension polynomialThe denominator is the product $1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5$ of the corresponding pairings with $\rho$. As checks, the weights ${\omega }_1,{\omega }_2,2{\omega }_1$ give dimensions $7,14,27$ respectively.

The G2 dimension polynomial gives dimension seven at the short-root fundamental weight, soThe crystal of the seven-dimensional G2 representation is a chain whose weights, in order, areIts successive arrow colors are $1,2,1,1,2,1$, as drawn above. Each arrow subtracts its indicated simple root; in particular the central three vertices form a length-two string of color one. The six nonzero weights are exactly the six short roots, each with multiplicity one, and the zero weight also has multiplicity one.

Choose the chain basis $v_0,\dots ,v_6$ from the preceding part, normalized by $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. The associated sl2 Lie algebra strings give $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$ and $e_2{v}_2=v_1$; other raising actions used below vanish because their proposed weights do not occur.

In the exterior square, $v_0\wedge v_1$ is a highest-weight vector of weight $2{\omega }_1-{\alpha }_1={\omega }_2$: $e_1$ produces $v_0\wedge v_0=0$, and $e_2$ kills both factors. Alsois a nonzero highest-weight vector of weight ${\omega }_1$. Its two $e_1$ terms cancel, and its $e_2$ image is zero. The Weyl complete reducibility theorem supplies irreducible summands with those weights; their dimensions $14+7=21$ exhaust $\dim {\Lambda }^{2}V$. ThusFor the symmetric square, $v_0\otimes v_0$ is a highest-weight vector of weight $2{\omega }_1$, giving a 27-dimensional summand. The longest Weyl-group element of $G_2$ acts as minus the identity, so the highest weight of a dual representation is $-w_0{\omega }_1={\omega }_1$: $V$ is self-dual. A nonzero intertwiner $V\to V^{\ast }$ gives a nondegenerate Lie-invariant bilinear form. It is symmetric, since an alternating bilinear form cannot be nondegenerate in odd dimension. Its inverse is a nonzero invariant vector in $S^{2}V$, supplying the scalar summand. Since $27+1=28=\dim S^{2}V$,These are the exterior and symmetric squares of the seven-dimensional G2 representation; all highest-weight and dimension claims are specified rather than relying on an unstated cross-product identity.

Work over the complex numbers. The chain $S_0\subset S_1\subset \cdots \subset S_n$ has multiplicity-free restriction: each irreducible representation restricts to a direct sum of pairwise inequivalent irreducible representations. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted Olshanskii centralizer lemma makes $C_{\mathbb{C}{S}_r}(\mathbb{C}{S}_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative matrix algebra factor. Thus no identification of the branching graph with Young diagrams is being assumed here.

Successively restricting an irreducible representation $V$ gives one-dimensional spaces $V_T$ indexed by paths of irreducible representations from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a Gelfand–Tsetlin basis. Define the Gelfand–Tsetlin algebra as the subalgebra of $\mathbb{C}{S}_n$ acting diagonally in all these bases. The Artin–Wedderburn theorem identifiesUnder this identification the Gelfand–Tsetlin algebra is the direct sum of the full diagonal matrix algebras in the indicated bases, so it is commutative.

To see that it is actually available inside the group algebra, let $e_{\alpha }^{(r)}$ be the central primitive idempotent selecting the irreducible representation $\alpha$ of $S_r$. For a path $T=({\alpha }_0,\dots ,{\alpha }_n)$ the productis the projection onto $V_T$ in its final irreducible representation and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal matrix units, and therefore span the proposed algebra.

If an element of the group algebra commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the Gelfand–Tsetlin algebra. Thuswhich proves that it is a maximal commutative subalgebra.

The projections $P_T$ in the Gelfand–Tsetlin algebra satisfyConsequently the map sending a diagonal element to its scalar on each Gelfand–Tsetlin basis vector is a ring isomorphismEach factor is a simple module over this algebra, and the algebra is their direct sum as a module over itself. This proves that $G{Z}_n$ is a semisimple algebra. The argument uses its explicit diagonal structure; being a subalgebra of a semisimple algebra by itself would not suffice.

The Young–Jucys–Murphy elements arewith products of permutations composed from right to left. Fix $i<j$. In $[X_i,X_j]$, all pairs of disjoint transpositions commute. The only terms left, for each $a<i$, areIndeed $(ai)(ij)=(aj)(ai)$ and $(ij)(ai)=(ai)(aj)$. Summing directly provesThus all the Young–Jucys–Murphy elements commute.

The primitive idempotent products used above show that the Gelfand–Tsetlin algebra is generated by the centers at all levels: those products give every diagonal matrix unit, whereas each center acts diagonally. Hence

For the second generation assertion use the permitted Olshanskii centralizer lemma, in the precise formIn particular $Z_r\subseteq \mathrm{alg}(Z_{r-1},X_r)$. Induction puts every $Z_r$ in $\mathrm{alg}(1,X_1,\dots ,X_r)$. Conversely, if $A_r$ is the sum of all transpositions in $S_r$, then $A_r\in Z_r$ and $X_r=A_r-A_{r-1}\in G{Z}_n$. ThereforeHere generation is unital; the zero generator $X_1$ causes no problem.

The deletion projection for a symmetric group removes $n$ from the cycle containing it. A cycle $(a_1\cdots a_{r}n)$ becomes $(a_1\cdots a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine permutation of $\{1,\dots ,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma ,\theta \in H$ and $k<n$, if $\pi (\theta (k))\ne n$, deletion gives $\sigma \pi (\theta (k))$. If $\pi (\theta (k))=n$, it gives $(\sigma \pi \theta )(n)=\sigma \pi (n)$. These are exactly the two cases in $\sigma {\pi }_n\theta (k)$. Thus

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual centralizer of a subalgebra. Let $A_{n-1}$ be the sum of all transpositions in $H$. Direct expansion in the group algebra givesThus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies ${\Pi }_n(Y)=(n-1)1$. For $n\ge 3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\mathrm{span}\{1,X_n\}$ is supported only on the identity and transpositions. Therefore $Y$ is an explicit counterexample to the asserted equality.

One correct description of the entire preimage is obtained by putting $C=C_{\mathbb{C}{S}_n}(\mathbb{C}H)$. Equivariance gives ${\Pi }_n(C)\subseteq Z(\mathbb{C}H)$, and ${\Pi }_n$ is the identity on that center. It follows thatA useful corrected two-dimensional statement restricts the support to the identity and the transpositions $(in)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\mathrm{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.

Use the Young–Jucys–Murphy elements in the group algebra $\mathbb{C}{S}_n$. Their joint spectrum isEquivalently one can use the regular representation, which contains every irreducible representation. The commuting elements are self-adjoint in a unitary representation, so there is a simultaneous eigenbasis.

A standard Young tableau has entries increasing along each row and down each column. If entry $r$ occupies cell $(i_r,j_r)$, its Content of a Young-diagram cell is $c_T(r)=j_r-i_r$. Define the set of content vectors of standard Young tableaux byThe shape here is a partition of an integer; it is not itself the vector of eigenvalues. For $n=1$ both sets consist of $(0)$.

The first Young–Jucys–Murphy element is the empty sum $X_1=0$. Applying it to the nonzero simultaneous eigenvector defining a point of the joint spectrum gives $0=a_{1}v$. HenceFor the remaining spectral arguments use the following local facts, without assuming the restriction branching rule for a symmetric group: consecutive eigenvalues are distinct; a difference $a_{r+1}-a_r=\pm 1$ makes $s_r=(rr+1)$ act on the corresponding line as $\pm 1$; every other adjacent interchange gives another point of the joint spectrum in the same irreducible representation. These are the local spectral rules for Young–Jucys–Murphy elements. They follow from $s_r{X}_r{s}_r+s_r=X_{r+1}$ and the one- or two-dimensional analysis of the two neighboring eigenspaces. They are the appropriate local results used below.