Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 102 (Lie Algebras and Their Representations)

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

The distinct affine algebraic groupsboth have the Special linear Lie algebra ${\mathfrak{sl}}_2(\mathbb{C})$. Quotienting a Lie group by a discrete central subgroup does not change its tangent Lie algebra.

The finite-dimensional irreducible rational representations of $S{L}_2(\mathbb{C})$ areof dimension $n+1$. The central element $-I$ acts on $V_n$ by $(-1{)}^n$. Therefore precisely the even-indexed representationsdescend to irreducible representations of $PG{L}_2(\mathbb{C})$. This is the Descent of an irreducible SL2 representation to PGL2.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

Use the standard basisof ${\mathfrak{sl}}_2$. For the invariant trace form $B(x,y)=\mathrm{tr}(xy)$, the dual basis is $(f,e,h/2)$, soOn a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=nv$ and $fev=0$,Centrality and Schur lemma make this the eigenvalue on the whole module. Thus $\lambda =n(n+2)/2$, the Casimir eigenvalue for sl2. If the form is instead the Killing form, which is four times the trace form on ${\mathfrak{sl}}_2$, the corresponding Casimir and eigenvalue are divided by four.

No. Let $\mathfrak{g}={\mathfrak{sl}}_2$, let $\mathfrak{s}=\mathbb{C}e$ be the one-dimensional subalgebra generated by the standard raising operator, and let $L={\mathbb{C}}^2$ be the irreducible defining representation. On restriction to $\mathfrak{s}$, the element $e$ acts by a nonzero Nilpotent Jordan block. A direct sum of irreducible representations of the one-dimensional abelian Lie algebra would make $e$ diagonalizable, so this restriction is not completely reducible. The Complete reducibility of semisimple Lie algebra representations applies when the restricting algebra is semisimple, which $\mathfrak{s}$ is not.

The Cartan solvability criterion states that a finite-dimensional complex Lie algebra $\mathfrak{g}$ is solvable exactly whenwhere $\kappa$ is the Killing form. In the matrix form of the criterion, a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ is solvable iffor all $x\in [\mathfrak{g},\mathfrak{g}]$ and $y\in \mathfrak{g}$.

A torus $\mathfrak{t}$ in a Lie algebra is an abelian subalgebra whose elements act semisimply in the Adjoint representation. Its weight-space decomposition is

Invariance of the nondegenerate Trace form of a Lie algebra representation givesNondegeneracy therefore forces $\beta =-\alpha$. The standard sl2 subalgebra associated with a root argument gives one-dimensional opposite root spaces with vectors $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ and $h_{\alpha }\in \mathfrak{t}$ satisfyingTheir span is a copy of ${\mathfrak{sl}}_2$.

SetThen $\dim \mathfrak{h}=\dim \mathfrak{t}-1$, $\mathfrak{t}=\mathbb{C}{h}_{\alpha }\oplus \mathfrak{h}$, and every element of $\mathfrak{h}$ commutes with $e$, $f$, and $\mathfrak{t}$. Hence $\mathfrak{h}$ is abelian andas a direct sum of commuting Lie algebras. This is the two-root decomposition with a nondegenerate trace form.

Let $r=\dim \mathfrak{h}$ and choose a dual basis of weights ${\epsilon }_1,\dots ,{\epsilon }_r\in {\mathfrak{h}}^{\ast }$. TakeThe ${\mathfrak{sl}}_2$ factor acts in its defining representation on ${\mathbb{C}}^2$ and trivially on the other summands; $\mathfrak{h}$ acts trivially on ${\mathbb{C}}^2$ and by the displayed characters on the one-dimensional summands. The resulting trace form is the nondegenerate trace form on ${\mathfrak{sl}}_2$, ison $\mathfrak{h}$, and has zero cross terms. It is therefore nondegenerate on ${\mathfrak{sl}}_2\oplus \mathfrak{h}$.

Write an element of the diagonal torus asand let ${\epsilon }_i$ extract $t_i$. The root-space decomposition iswith one-dimensional root spaces. Thus the $D_n$ root system is

The upper-triangular choice givesA compatible simple system isThe highest root and Weyl vector areThe fundamental weights are

Using the paper's letters, the root lattice and weight lattice are respectivelyTheir quotient is

The Dynkin diagram is the $D_n$ diagram: a chain ${\alpha }_1-\cdots -{\alpha }_{n-2}$ whose last node is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. The Extended Dynkin diagram adds ${\alpha }_0=-\theta$ joined to ${\alpha }_2$. For $D_4$, the central node ${\alpha }_2$ consequently has the four leaves ${\alpha }_0,{\alpha }_1,{\alpha }_3,{\alpha }_4$.

Let $V={\mathbb{C}}^4$. The Special linear Lie algebra $\mathfrak{sl}(V)$ acts on ${\Lambda }^{2}V$. The wedge productis a nondegenerate symmetric bilinear form, and the action preserves it because $\mathfrak{sl}(V)$ acts trivially on ${\Lambda }^{4}V$. This gives an injective homomorphismBoth Lie algebras have dimension $15$, so the map is an isomorphism. This realizes the Isomorphism between so6 and sl4.

For every root, the Weyl reflection isFor $\alpha ={\epsilon }_i-{\epsilon }_j$, it swaps the $i$th and $j$th coordinates. For $\alpha ={\epsilon }_i+{\epsilon }_j$, it sendsand fixes all other coordinates. The Weyl group of $D_n$ is therefore the group of signed permutations with an even number of sign changes,

For a dominant integral weight $\lambda$, the Weyl character formula isHere $W$ is the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the Weyl vector, and $\mathrm{ch}$ the formal character of a weight module. Taking the limit at the identity gives the Weyl dimension formula

Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$. The G2 root system has positive rootsThe two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights areThe second is the highest root, so the irreducible module $L({\omega }_2)$ is the Adjoint representation of a Lie algebra.

The seven weights of $L({\omega }_1)$ are zero and the six short roots, each with multiplicity one. Its crystal, with arrows denoting the lowering operators ${\tilde{f}}_i$, is the chain

Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ gives

For type $B_n$, the spinor representation has highest weightand its weights are the $2^n$ sign vectorsEach weight has multiplicity one. Along the simple root ${\alpha }_j={\epsilon }_j-{\epsilon }_{j+1}$, the Kashiwara operator ${\tilde{e}}_j$ can raise a weight exactly when $(i_j,i_{j+1})=(-1,+1)$, when it replaces that pair by $(+1,-1)$. For the short root ${\alpha }_n={\epsilon }_n$, ${\tilde{e}}_n$ replaces a final $-1$ by $+1$. This proves the stated crystal by the root-string property of a crystal.

For $n=3$, the complete list of raising edges isThe tensor product of crystals has four highest-weight connected components, of highest weightsConsequently, for the eight-dimensional spin representation $S$ of ${\mathfrak{so}}_7$,with dimensionsEquivalently these summands are ${\Lambda }^k({\mathbb{C}}^7)$ for $0\le k\le 3$.