Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 102 (Lie Algebras and Their Representations)

For a finite-dimensional complex Lie algebra $\mathfrak{g}$, the following three conditions are equivalent descriptions of a semisimple Lie algebra.

First, its solvable radical is zero. An ideal of a Lie algebra is a linear subspace $I$ satisfying $[\mathfrak{g},I]\subseteq I$. A solvable Lie algebra is one whose derived series, defined by $I^{(0)}=I$ and $I^{(r+1)}=[I^{(r)},I^{(r)}]$, eventually becomes zero. The solvable radical is the largest solvable ideal of a Lie algebra; its vanishing is equivalent to the absence of nonzero solvable ideals.

Second, the Killing formis a nondegenerate bilinear form: if $\kappa (u,v)=0$ for every $v$, then $u=0$. Here $\mathrm{ad}u(v)=[u,v]$ is the Adjoint representation, and $\mathrm{tr}$ is the trace. This equivalence is the Cartan criterion for semisimplicity.

Third, $\mathfrak{g}$ is a direct sum of Lie algebras that are nonabelian simple Lie algebras. A simple Lie algebra is nonabelian and has no ideals of a Lie algebra other than zero and itself. Thus the three equivalent criteria are

The derived series gives a direct test, with no structural theorem needed. By bilinearity and antisymmetry of the Lie bracket, every bracket in $\mathfrak{h}$ is a multiple of $z$, and $z=[x,y]$ actually occurs. ThereforeConsequently this Heisenberg Lie algebra is a nonzero solvable Lie algebra. It is an ideal of a Lie algebra in itself, since $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{h}$. Its solvable radical is therefore all of $\mathfrak{h}$, contradicting the first criterion in part (a). Hence

A two-dimensional Lie algebra representation already provides a counterexample. In the basis $e_1,e_2$ of $V={\mathbb{C}}^2$, defineThe commutator $[\rho (x),\rho (y)]$ is zero, agreeing with $\rho (z)$; the other defining Lie brackets also map to zero. Thus $\rho$ is a Lie algebra homomorphism, although it is not a Faithful Lie algebra representation.

The line $L=\mathbb{C}{e}_1$ is an invariant subspace. Any complementary line has the form $\mathbb{C}(e_2+a{e}_1)$ for some $a\in \mathbb{C}$. Butso no complementary line is an invariant subspace. Equivalently, every invariant line is killed by $\rho (x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho (x)=0$, a contradiction. This representation is therefore not a semisimple representation. Thus

Use the coroot normalization ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, with the pairing understood via the root system and its invariant inner product. A dominant integral weight, called a dominant weight here, is a functional $\lambda \in {\mathfrak{t}}^{\ast }$ such thatEquivalently, $\lambda ={\sum }_i{n}_i{\omega }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, where the fundamental weights satisfy $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. If dominance and integrality are defined separately, dominance means the inequalities and membership in the weight lattice means integrality.

A highest-weight vector is a nonzero weight vector $v\in V_{\lambda }$, so $h\cdot v=\lambda (h)v$ for all $h\in \mathfrak{t}$, annihilated by every positive root space:The last condition is equivalently annihilation by the raising operators for the simple roots.

Fix a simple root ${\alpha }_i$ and choose generators $e_i,f_i,h_i$ of the sl2 subalgebra associated with a root, with $h_i={\alpha }_i^{\vee }$ andThe highest-weight vector satisfies $e_{i}v=0$ and $h_{i}v=m_{i}v$, where $m_i=⟨\lambda ,{\alpha }_i^{\vee }⟩$. The classification of finite-dimensional sl2 representations already implies that $m_i$ is a nonnegative integer. One can also see the necessary integrality directly from the lowering operator.

Indeed, $f_i^{k}v$, if nonzero, has $h_i$ eigenvalue $m_i-2k$. These distinct eigenvalues cannot occur indefinitely in a finite-dimensional space, so there is a largest $r\ge 0$ with $f_i^{r}v\ne 0$. The sl2 Lie algebra relations imply, by induction on $k$,For $k=r+1$ the left side is zero, hence $(r+1)(m_i-r)f_i^{r}v=0$. Thus $m_i=r\in {\mathbb{Z}}_{\ge 0}$. Since this holds for every simple root,

Choose the given short root $\alpha$ as a simple root, and let $\beta$ be the long simple root. This entails no loss of generality: the Weyl group is transitive on the short roots of the G2 root system. The relevant Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. The positive roots areUnder ${\mathfrak{m}}_{\alpha }\cong {\mathfrak{sl}}_2$, the root space for $m\alpha +n\beta$ has $h_{\alpha }$ eigenvalue $2m-3n$. The subalgebra ${\mathfrak{m}}_{\alpha }$ itself is the three-dimensional irreducible $V(2)$ in the Adjoint representation.

The four root spaces along the root string $\beta ,\alpha +\beta ,2\alpha +\beta ,3\alpha +\beta$ have eigenvalues $-3,-1,1,3$. Consecutive spaces are connected by nonzero raising operators and lowering operators, so their sum is $V(3)$. The negative root string supplies another $V(3)$. The root spaces for $\pm (3\alpha +2\beta )$ have eigenvalue zero, and neither adding nor subtracting $\alpha$ gives a root; they are two copies of $V(0)$. Finally, the one-dimensional space $\ker \alpha \subset \mathfrak{t}$ commutes with ${\mathfrak{m}}_{\alpha }$, giving one more $V(0)$. We have accounted for all $14$ dimensions, and therefore the G2 adjoint branching to a short-root sl2 subalgebra isFor a nonzero root vector $x\in {\mathfrak{g}}_{\alpha }$, its Adjoint representation action is a nonzero scalar multiple of the raising operator on each summand. On each irreducible $V(n)$ this operator has a one-dimensional kernel, including $V(0)$. Since the Lie algebra centralizer is that kernel,

Suppose first that $x\ne 0$, and let $E$ be the real span of the root system, of dimension equal to the rank of a semisimple Lie algebra $\ell$. The images of the roots span $E/\mathbb{R}\alpha$, so select roots ${\beta }_1,\dots ,{\beta }_{\ell -1}$ whose images form a basis of this quotient.

For each $i$, take the highest endpoint ${\gamma }_i^{+}$ of the $\alpha$ root string through ${\beta }_i$, and the highest endpoint ${\gamma }_i^{-}$ of the root string through $-{\beta }_i$. By construction, ${\gamma }_i^{\pm }+\alpha$ is not a root. Moreover ${\gamma }_i^{\pm }\ne -\alpha$, since their images in $E/\mathbb{R}\alpha$ are nonzero. The root-space decomposition and its bracket rule therefore giveThese $2(\ell -1)$ roots are distinct: their quotient images are the two signs of a basis, which are distinct. Their one-dimensional root spaces consequently contribute $2(\ell -1)$ independent vectors to the Lie algebra centralizer.

In addition, the $\ell -1$ dimensional subspace $\ker \alpha \subset \mathfrak{t}$ commutes with $x$, because $[h,x]=\alpha (h)x$. The line $\mathbb{C}x\subset {\mathfrak{g}}_{\alpha }$ also commutes with $x$. These contributions are independent by the root-space decomposition; in particular none of the selected endpoint roots is $\alpha$. Thus the centralizer lower bound for a root vector isIf $x=0$, then $Z_{\mathfrak{g}}(x)=\mathfrak{g}$. The root system contains the $2\ell$ distinct roots $\pm {\alpha }_i$, so $\dim \mathfrak{g}=\ell +|\Phi |\ge 3\ell$, which also proves the desired inequality. The argument includes rank one, where the list of ${\beta }_i$ is empty.

Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are $H(t)\mapsto \pm t_i\pm t_j$, with the two signs independent. This is the D3 root system, with twelve roots.

For a direct check, the defining matrix identity gives the block form $X=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B^T=-B$ and $C^T=-C$. If $E_{ij}$ denotes a matrix unit, the root space for ${\epsilon }_i-{\epsilon }_j$ is spanned by $E_{ij}-E_{j+3,i+3}$; those for ${\epsilon }_i+{\epsilon }_j$ and $-{\epsilon }_i-{\epsilon }_j$ are spanned respectively by $E_{i,j+3}-E_{j,i+3}$ and $E_{i+3,j}-E_{j+3,i}$. Commuting these with $H(t)$ gives exactly the displayed maps.

Choose the root basisIn the usual invariant inner product, each simple root has squared length $2$, while $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$ and $({\alpha }_2,{\alpha }_3)=0$. The labeled Dynkin diagram is therefore the following chain, with a single bond between adjacent nodes:It is type $D_3$, equivalently the A3 root system. To check the root basis explicitly, the corresponding positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$; these and their negatives are all twelve roots.

Let $s_i=w_{{\alpha }_i}$. The root reflection formula is $s_i(\gamma )=\gamma -⟨\gamma ,{\alpha }_i^{\vee }⟩{\alpha }_i$. Applying the Cartan integers from part (b), the images of the chosen simple roots areIn particular, the two end-node Weyl reflections fix the opposite end node.

An automorphism of a root system outside the Weyl group is the sign changeIt preserves the set $\{\pm {\epsilon }_i\pm {\epsilon }_j\}$, fixes ${\alpha }_1$, and interchanges ${\alpha }_2$ and ${\alpha }_3$. It therefore realizes the nontrivial symmetry of the Dynkin diagram while preserving the chosen root basis.

To verify that $\tau \notin W$, note that $s_1$ swaps ${\epsilon }_1,{\epsilon }_2$, $s_2$ swaps ${\epsilon }_2,{\epsilon }_3$, and $s_3$ swaps ${\epsilon }_2,{\epsilon }_3$ while changing both their signs. Every product of these generators is a signed permutation with an even number of sign changes, whereas $\tau$ has one sign change. Thus it cannot be a Weyl group element. The parity here concerns sign changes, rather than the determinant of the permutation.

The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.

For a dominant integral weight $\lambda$, let $V(\lambda )$ be the finite-dimensional irreducible highest-weight representation with highest weight $\lambda$. Set $\rho =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha$, the half-sum of positive roots. With ${\alpha }^{\vee }$ the coroot of $\alpha$, the Weyl dimension formula isHere ${\Phi }^{+}$ is the positive system of a root system specified by the chosen root basis. The fundamental weights are the basis dual to the simple coroots: $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. Thus $\lambda ={\sum }_i{a}_i{\omega }_i$ with nonnegative integers $a_i$. The denominators are positive; equivalently one can use $(\lambda +\rho ,\alpha )/(\rho ,\alpha )$ in each factor, since the coroot normalization cancels.

Use the C2 root system realization in an orthonormal basis ${\epsilon }_1,{\epsilon }_2$ with simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$. This numbers the short root first, as required. The fundamental weights and half-sum of positive roots areso $\lambda +\rho =(a+b+2){\epsilon }_1+(b+1){\epsilon }_2$. For the four positive roots, the factors of the Weyl dimension formula areMultiplying yields the Weyl dimension formula for C2:As checks, $V({\omega }_1)$ has dimension $4$, $V({\omega }_2)$ dimension $5$, and $V(2{\omega }_1)$ dimension $10$.

In the defining representation of the symplectic Lie algebra ${\mathfrak{sp}}_4$, the weights are ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_1,-{\epsilon }_2$. Their highest weight, for the root basis of part (b), is thereforeThe flip on the tensor square commutes with the action of ${\mathfrak{sp}}_4$, giving $V\otimes V={\mathrm{Sym}}^{2}V\oplus {\Lambda }^{2}V$ with dimensions $10$ and $6$.

If $v_1$ is a highest-weight vector of weight ${\epsilon }_1$, then $v_1\otimes v_1$ is a highest-weight vector of weight $2{\omega }_1$ in the symmetric square. By Weyl complete reducibility theorem, this ensures an irreducible summand $V(2{\omega }_1)$ occurs. Its dimension is $10$ by part (b), exhausting the symmetric square.

Let $\omega$ denote the preserved symplectic form. The symplectic contraction of an exterior square is the nonzero equivariant mapwhere the target is a trivial Lie algebra representation. Thus its kernel has dimension $5$. Choose a weight vector $v_2$ of weight ${\epsilon }_2$ in a symplectic basis, with $\omega (v_1,v_2)=0$. The vector $v_1\wedge v_2$ belongs to this kernel and has weight ${\epsilon }_1+{\epsilon }_2={\omega }_2$. It is a highest-weight vector: none of ${\omega }_2+\alpha$, for $\alpha \in {\Phi }^{+}$, is a weight of ${\Lambda }^{2}V$, whose weights are $\pm {\epsilon }_1\pm {\epsilon }_2$ and zero. Hence the kernel contains $V({\omega }_2)$; its dimension $5$ exhausts the kernel. Weyl complete reducibility theorem supplies a complementary invariant line $V(0)$.

The resulting tensor-square decomposition of the defining sp4 representation is

The formal character of a weight module is $\mathrm{ch}V={\sum }_{\mu }(\dim V_{\mu })e^{\mu }$, where $e^{\mu }$ are formal basis symbols in the group algebra of the weight lattice, satisfying $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. For a finite-dimensional irreducible highest-weight representation, the highest weight has weight multiplicity one. Another standard result is that weight multiplicities are invariant under the Weyl group: $\dim V_{w\mu }=\dim V_{\mu }$.

A minuscule representation has all its weights in a single Weyl group orbit. Since the highest weight $\lambda$ is present, that orbit is $W\lambda$, and the invariance just stated shows every point of it is present with multiplicity one. Consequently the formal character of a minuscule representation isThe sum is over distinct weights, rather than over all elements of $W$: summing $e^{w\lambda }$ over $w\in W$ would count each weight $|{\mathrm{Stab}}_W(\lambda )|$ times.

Use the coroot ${\alpha }^{\vee }$ in the pairing. The original PDF retains this mark, which is lost in the local TeX transcription. Fix any positive root $\alpha$, and put $n=⟨\lambda ,{\alpha }^{\vee }⟩$. For a dominant integral weight, $n$ is a nonnegative integer; positive coroots are nonnegative integral combinations of simple coroots.

Let $v_{\lambda }$ be a highest-weight vector, and restrict to the sl2 subalgebra associated with a root $\alpha$. It is killed by its raising operator $e_{\alpha }$ and has $h_{\alpha }$ eigenvalue $n$. If $n\ge 1$, its lowering operator produces a nonzero vector: otherwise $[e_{\alpha },f_{\alpha }]v_{\lambda }=n{v}_{\lambda }$ would vanish. Therefore $\lambda -\alpha$ is a weight.

Choose a Weyl group invariant positive-definite inner product on the real weight space. All weights of a minuscule representation have the same norm, since they belong to $W\lambda$. ButIf $n\ge 2$, this is negative, a contradiction. Thus for every positive root,When $n=1$, the lower weight is precisely the Weyl reflection $s_{\alpha }\lambda =\lambda -\alpha$, consistent with the orbit condition.

We first prove the useful lemma that dominant root-lattice highest weights have zero weight. Write $Q=\mathbb{Z}\Phi$ for the root lattice, and let a dominant integral weight $\lambda \in Q$ be the highest weight of a finite-dimensional irreducible representation.

First, $\lambda$ is a nonnegative integral combination of simple roots. Indeed, write $\lambda ={\lambda }_{+}-{\lambda }_{-}$, separating its positive and negative coefficients in the root basis. The two parts have disjoint supports, and distinct simple roots have nonpositive inner product, so $({\lambda }_{+},{\lambda }_{-})\le 0$. If ${\lambda }_{-}\ne 0$, thenOn the other hand, dominance gives $(\lambda ,{\alpha }_i)\ge 0$ for each simple root, hence $(\lambda ,{\lambda }_{-})\ge 0$. This contradiction proves ${\lambda }_{-}=0$.

Now suppose a nonzero weight $\mu ={\sum }_i{m}_i{\alpha }_i$ has all $m_i\in {\mathbb{Z}}_{\ge 0}$. The identityshows that some $i$ satisfies $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. We use the standard sl2 Lie algebra fact that its lowering operator is injective on any positive $h$ eigenspace in a finite-dimensional representation. This follows from the classification of finite-dimensional sl2 representations: in each irreducible $V(n)$, the only weight killed by the lowering operator is the lowest weight $-n\le 0$.

Consequently a nonzero vector of weight $\mu$ lowers to a nonzero vector of weight $\mu -{\alpha }_i$. Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at $\lambda$, repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.

For a minuscule representation, every weight belongs to $W\lambda$, so the zero weight just obtained lies in this orbit. Every Weyl group element is invertible, and $w\lambda =0$ forces $\lambda =0$. This proves that minuscule weights in the root lattice are zero. Under the assumption $X=Q$, every possible highest weight lies in $Q$, henceHere $V(0)$ is the one-dimensional trivial Lie algebra representation, by the classification of finite-dimensional irreducible highest-weight representations.