Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 (Lie Algebras and Their Representations)

Write $e,h,f$ for the standard generators of the sl2 Lie algebra. For the representation $\varphi$, use the normalizationThis is the quadratic Casimir element, and by assumption it commutes with every $\varphi (x)$.

Schur lemma says that an endomorphism of a finite-dimensional irreducible complex representation which commutes with the representation is a scalar. Hence $\Omega =c{I}_V$ when $V$ is irreducible.

Let $v$ be a highest-weight vector of highest weight $m$, so $ev=0$ and $hv=mv$. Since $[e,f]=h$,ThereforeIt follows from scalarity thatThis is the Casimir eigenvalue for sl2 in the chosen normalization.

The one-dimensional quotient $V/W$ is trivial because every one-dimensional representation vanishes on the derived algebra $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$. Choose $v\in V$ mapping to $1\in V/W$. Thenis a $1$-cocycle:We show that it is a coboundary.

Decompose $W$ into generalized eigenspaces of its Casimir element $\Omega$. These are subrepresentations because $\Omega$ is central. On a generalized eigenspace with nonzero eigenvalue, $\Omega$ is invertible. If $(x_i)$ and $(x^i)$ are dual bases of ${\mathfrak{sl}}_2$ for the Killing form, putInvariance of the Killing form and the cocycle identity give the standard Casimir calculationThus on every nonzero generalized eigenspace, $c(x)=x({\Omega }^{-1}u)$.

On the zero generalized eigenspace, every irreducible composition factor has zero Casimir eigenvalue. By part i and the Classification of finite-dimensional sl2 representations, each such factor is trivial. In a basis adapted to a composition series, the image of ${\mathfrak{sl}}_2$ is therefore strictly upper triangular and hence solvable. Since ${\mathfrak{sl}}_2$ is simple and non-solvable, that image is zero. The cocycle then vanishes because it kills $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]$.

Combining the generalized eigenspaces gives $w\in W$ such that $c(x)=xw$ for every $x$. Hence $v-w$ is invariant, andis a decomposition into subrepresentations. This proves the codimension-one case of the Weyl complete reducibility theorem.

The relation $[h,e]=2e$ forces $e{w}_i$ to have weight $a+2(i+1)$, sofor scalars $c_i$. The relation $[e,f]=h$ becomesWith $c_0=b$, this recurrence has the unique solutionfor every $i\in \mathbb{Z}$. Direct substitution also verifies $[h,f]=-2f$ and $[h,e]=2e$, so these formulas define the unique required sl2 Lie algebra action. They form an Intermediate-series sl2 module.

Let $0\ne U\subseteq W$ be a subrepresentation and choosewith finite support. The $h$-eigenvalues $a+2r$ are pairwise distinct. By Lagrange interpolation, there is a polynomial $P$ which is one at one chosen eigenvalue appearing in $u$ and zero at all the others. Then $P(h)u$ is a nonzero scalar multiple of one basis vector $w_i$. Since $U$ is invariant under $h$, it contains $w_i$.

Ifthen the coefficient $c_j=b-ja-j(j+1)$ vanishes. Consequentlyis stable under $h$, $f$, and $e$: the only raising operation that could leave it is $e{w}_j$, and that is zero. Thus $W$ is reducible.

Conversely, let $U$ be a nonzero subrepresentation. By part ii it contains some $w_i$, and repeated application of $f$ gives every $w_r$ with $r\le i$. If every $c_r$ is nonzero, repeated application of $e$ also gives every $w_r$ with $r>i$, so $U=W$. Hence a proper nonzero subrepresentation exists exactly when some $c_j=0$, or

For a finite-dimensional Lie algebra representation $\varphi :\mathfrak{g}\to \mathfrak{gl}(V)$, the Trace form of a Lie algebra representation isThe Killing form is the trace form of the Adjoint representation of a Lie algebra:A bilinear form $B$ is $\mathfrak{g}$-invariant whenequivalently $B([x,y],z)=B(x,[y,z])$. For a trace form this follows from cyclicity of trace:

Use the nondegenerate restriction of the Killing form to $\mathfrak{t}$ to define $t_{\alpha }\in \mathfrak{t}$ byKilling-form invariance and the root-space decomposition show that ${\mathfrak{g}}_{\alpha }$ pairs nondegenerately with ${\mathfrak{g}}_{-\alpha }$ and orthogonally with every other root space. Choose nonzero $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $\kappa (e,f)\ne 0$. For $t\in \mathfrak{t}$,so

We need $\alpha (t_{\alpha })\ne 0$. If it were zero, the span of $e,f,t_{\alpha }$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By Lie theorem, its adjoint action on $\mathfrak{g}$ can be upper triangularized, so $\mathrm{ad}[e,f]$ is nilpotent. But $[e,f]\in \mathfrak{t}$, and elements of the Cartan subalgebra act semisimply; hence $\mathrm{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne 0$.

SetRescale $f$ so that $[e_{\alpha },f_{\alpha }]=h_{\alpha }$. Since $e_{\alpha }$ and $f_{\alpha }$ lie in the $\alpha$ and $-\alpha$ root spaces,Thus ${\mathfrak{m}}_{\alpha }=⟨e_{\alpha },h_{\alpha },f_{\alpha }⟩$ is the sl2 subalgebra associated with a root.

Bilinearity and alternatingness ofare immediate. For three elements, the $\mathfrak{g}$ component of the Jacobi sum vanishes by the Jacobi identity in $\mathfrak{g}$. In the $V$ component, the coefficient of a vector such as $u$ isbecause the action is a Lie algebra representation; the other terms cancel cyclically in the same way. Hence the bracket satisfies Jacobi and defines the semidirect product of a Lie algebra and a module $\mathfrak{g}⋉V$.

Let $\mathfrak{g}={\mathfrak{sl}}_2(\mathbb{C})$ and let $V={\mathbb{C}}^2$ be its defining irreducible representation. SetSince $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ and $\mathfrak{g}V=V$,If $(x,v)$ is central, commuting with every $(0,w)$ gives $xw=0$ for all $w$, so faithfulness of the defining representation gives $x=0$. Commuting with every $(y,0)$ then gives $yv=0$ for all $y$; irreducibility and nontriviality give $v=0$. Thus $Z(\mathfrak{h})=0$.

The nonzero abelian subspace $V$ is a proper ideal of a Lie algebra, so $\mathfrak{h}$ is not simple. It is not a direct product of simple Lie algebras either, because such a product is semisimple and has no nonzero solvable ideal, whereas $V$ is one.

Relative to $\mathfrak{g}\oplus V$, the adjoint action has block formMultiplying two such block-triangular matrices and taking the trace givesthe Killing form of a semidirect product with a module. It does not depend on $v$ or $w$, so $V$ lies in its radical. Therefore $K$ can be nondegenerate only if $V=0$. In that case $K=\kappa$, which is nondegenerate exactly when $\mathfrak{g}$ is semisimple by the Cartan criterion for semisimplicity. Thus

A Weyl chamber is a connected component ofA root basis $\Delta \subseteq \Phi$ is a vector-space basis of $E$ such that every root is an integer combination of elements of $\Delta$ with all nonzero coefficients of one sign.

To construct one, choose a regular vector $\gamma \in E$, meaning $(\gamma ,\alpha )\ne 0$ for every root. DeclareThe positive roots in ${\Phi }_{\gamma }^{+}$ which cannot be written as sums of two positive roots form a root basis ${\Delta }_{\gamma }$. Vectors $\gamma$ in the same Weyl chamber give the same basis.

Write a positive nonsimple root asIf $(\alpha ,\beta )\le 0$ for every simple root with $n_{\beta }>0$, thenwhich is impossible. Hence $(\alpha ,\beta )>0$ for some simple $\beta$. The root-string property then gives $\alpha -\beta \in \Phi$, and its simple-root coefficients remain nonnegative. This is the simple-root subtraction lemma.

Induct on the height $\sum n_{\beta }$. Applying the induction hypothesis to $\alpha -\beta$ and appending $\beta$ writesso that every partial sum is a root.

Finally let $\alpha$ be simple and let $\gamma \ne \alpha$ be positive. In the simple-root expansion ofall coefficients except possibly that of $\alpha$ are unchanged, and at least one of those unchanged coefficients is positive. Since a root has coefficients all of one sign, the image cannot be negative. Thus $s_{\alpha }$ permutes ${\Phi }^{+}\setminus \{\alpha \}$, as stated by action of a simple reflection on positive roots.

The diagonal Cartan subalgebra consists ofLet ${\epsilon }_i\in {\mathfrak{t}}^{\ast }$ read off $t_i$. The roots arethe D3 root system.

One root basis isAll roots have the same length. The nonzero inner products are $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$, so the labeled Dynkin diagram isIt is the three-node $A_3$ diagram.

The simple reflections act on the root basis byby the Weyl reflection formula $s_{\alpha }(\beta )=\beta -⟨\beta ,{\alpha }^{\vee }⟩\alpha$.

The linear mappreserves $\Phi$ and exchanges ${\alpha }_2$ with ${\alpha }_3$ while fixing ${\alpha }_1$. It is not in the $D_3$ Weyl group, whose signed permutations change an even number of signs. Thus it is an outer automorphism of the root system.

The preceding Dynkin diagram identifies the root system $D_3$ with $A_3$. The classification of finite-dimensional complex Simple Lie algebras by connected Dynkin diagrams therefore givesThis is the Isomorphism between so6 and sl4. Both algebras have dimension $15$, consistently with the classification.

The crystallographic axiom makesintegers. If $\theta$ is the angle between the roots, thenThis is a nonnegative integer. Since $\beta \ne \pm \alpha$, the roots are not parallel, so ${\cos }^2\theta <1$. Thereforewhich is the root-system finiteness lemma.

Choose simple roots $\alpha ,\beta$. Their inner product is nonpositive, so their angle lies in $[\pi /2,\pi )$. Part i leaves four possibilities for $mn$:

The root strings generated by the two simple reflections produce exactly the roots in those four standard systems. Hence these are all possibilities, proving the classification of rank-two root systems.

An irreducible root system is simply laced when every root has the same length, equivalently when its Dynkin diagram has no multiple edge.

If all roots have the same length, the two Cartan integers for $\alpha$ and $\beta$ are equal. The root-system finiteness lemma then makes their product either zero or one, sofor $\alpha \ne \pm \beta$.

Conversely, when all such Cartan integers lie in $\{0,\pm 1\}$, any two nonorthogonal roots have Cartan integers of absolute value one in both directions. Their squared lengths are therefore equal. Irreducibility makes the graph joining nonorthogonal roots connected, so all roots have the same length. Thus the system is simply laced.

On $P=\mathrm{span}\{\alpha ,\beta \}$, each $s_{\alpha }$ is reflection in the line ${\alpha }^{\perp }\cap P$. The product of two plane reflections is a rotation. If $\theta$ is the oriented angle from $\beta$ to $\alpha$, thenrotates through $2\theta$ modulo $2\pi$.

If this rotation has order $m$, conjugation by either reflection inverts it. Henceis a dihedral group, with rotational subgroup $⟨r⟩=⟨s_{\alpha }{s}_{\beta }⟩$.

For the simple-root angles from part ii, the rotation orders and Weyl groups areThese are the rank-two Weyl groups.

Choose a Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$. For $\lambda \in {\mathfrak{t}}^{\ast }$, let ${\mathbb{C}}_{\lambda }$ be the one-dimensional $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isIts universal property of a Verma module says that any vector of weight $\lambda$ annihilated by ${\mathfrak{n}}^{+}$ receives the canonical highest-weight vector under one unique module homomorphism from $M(\lambda )$.

The sum of the proper submodules of $M(\lambda )$ is its unique maximal proper submodule, because no proper submodule contains the highest-weight vector. Its quotient $V(\lambda )$ is therefore the unique irreducible quotient of a Verma module, and hence the unique irreducible highest-weight module of weight $\lambda$.

The module $V(\lambda )$ is finite-dimensional exactly when $\lambda$ is a dominant integral weight:for every simple root ${\alpha }_i$.

For ${\mathfrak{sl}}_2$, the Verma module has basisby the Poincare-Birkhoff-Witt theorem. If $hv=\lambda v$, thenFor $\lambda =-d$ with $d>0$, the coefficient isfor every $r\ge 1$. Thus no $f^{r}v$ with $r>0$ is a singular vector.

Any nonzero submodule contains a weight vector $f^{r}v$ because the $h$-weights are distinct. Applying $e^r$ gives a nonzero multiple of $v$, after which applying powers of $f$ generates all of $M(-d)$. HenceThis is also the negative-highest-weight case of Reducibility of an sl2 Verma module.

For a dominant integral weight $\lambda$, the Weyl character formula iswhere $W$ is the Weyl group, $\ell$ is Coxeter length, and $\rho$ is the half-sum of positive roots. The Weyl denominator formula is

Set $\lambda =k\rho$. Apply the denominator identity after replacing every formal exponential $e^{\mu }$ by $e^{(k+1)\mu }$:Dividing this by the ordinary denominator gives

Realize the B2 root system in ${\mathbb{R}}^2$ asChoose the short simple root and long simple rootThenas follows from $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. The roots form a square from the long roots with the four short roots on the coordinate axes; the double edge in the Dynkin diagram points toward ${\alpha }_1$.

The positive roots areFor $\lambda =a{\omega }_1+b{\omega }_2$, substituting their coroot pairings in the Weyl dimension formula givesThis is the Weyl dimension formula for B2.

The weights of the defining five-dimensional representation areIts highest weight is ${\epsilon }_1={\omega }_2$, so irreducibility identifies it as

If $v_{+}$ is a highest-weight vector, then $v_{+}\otimes v_{+}$ is a highest-weight vector of weight $2{\omega }_2$ in $V\otimes V$. The subrepresentation it generates is therefore $V(2{\omega }_2)$. Equivalently, it is the Traceless symmetric square of the defining so5 representation; the invariant quadratic form supplies the complementary trivial line in $S^{2}V$.

Putting $a=0$ and $b=2$ into the formula from part i gives