Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 1 (Lie Algebras and Their Representations)

For the Adjoint representation of a Lie algebra, the operator associated with $x\in L$ isThe Jacobi identity gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, so this really is a Lie algebra representation. Define the lower central series of a Lie algebra by $L^1=L$ and $L^{r+1}=[L,L^r]$. A nilpotent Lie algebra is one for which $L^{c+1}=0$ for some integer $c$. If this holds, ${\mathrm{ad}}_x$ carries $L^r$ into $L^{r+1}$, and consequently $({\mathrm{ad}}_x{)}^{c}L=0$. This proves one implication without any condition on the field.

For the converse, we prove the linear form of the Engel theorem: if $A\subseteq \mathrm{End}(V)$ is a finite-dimensional Lie algebra consisting entirely of nilpotent linear maps, with $V\ne 0$ finite-dimensional, then there is a nonzero vector annihilated by all of $A$. This argument works over any field.

Induct on $\dim A$. The assertion is immediate when $A=0$. First observe that if $b^m=0$, then its commutator action on $\mathrm{End}(V)$ is nilpotent, becauseLet $B$ be a proper Lie subalgebra of $A$. The commutator action of $B$ on $A/B$ therefore consists of nilpotent linear maps. Its image has dimension at most $\dim B<\dim A$, so induction supplies a nonzero coset $a+B$ annihilated by $B$. Thus $[B,a]\subseteq B$, and the normalizer $N_A(B)=\{a\in A:[a,B]\subseteq B\}$ strictly contains $B$. This is the Engel normalizer lemma.

Choose $B$ maximal among proper Lie subalgebras. Its normalizer must be $A$, so $B$ is a Lie algebra ideal. Moreover $\dim (A/B)=1$: otherwise the inverse image of a one-dimensional subalgebra of $A/B$ would lie strictly between $B$ and $A$. By induction,Since $B$ is an ideal, $W$ is $A$-invariant: $b(av)=a(bv)+[b,a]v=0$. Take $x\in A\setminus B$. Its restriction to $W$ is nilpotent and hence has a nonzero kernel. A nonzero vector in that kernel is annihilated by $B$ and $x$, and therefore by all of $A$. This completes the induction.

Apply this result successively to $V$, then its quotient by a common annihilated line, and so on. We obtain a complete flag $0=V_0\subset V_1\subset \cdots \subset V_n=V$ with $A{V}_i\subseteq V_{i-1}$. Equivalently every element of $A$ is strictly upper triangular in one common basis; any product of $n$ such operators is zero.

Now take $V=L$ and $A=\mathrm{ad}(L)$. The hypothesis supplies exactly the nilpotence needed above. Every iterated bracket of length $n+1$, where $n=\dim L$, is an application of $n$ adjoint operators and vanishes. These brackets span $L^{n+1}$, giving $L^{n+1}=0$. The zero algebra is already nilpotent. Thus $L$ is nilpotent if and only if every ${\mathrm{ad}}_x$ is nilpotent. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

A finite-dimensional complex Lie algebra is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form is the symmetric bilinear formThe cyclic trace identity and ${\mathrm{ad}}_{[z,x]}=[{\mathrm{ad}}_z,{\mathrm{ad}}_x]$ giveThus the Killing form is an invariant bilinear form on a Lie algebra. In particular its radical $R=\{x:B_L(x,L)=0\}$ is an ideal.

We supply the trace argument needed for nondegeneracy rather than assuming the Cartan criterion for semisimplicity. The matrix form of the Cartan solvability criterion says: if $A\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ and $\mathrm{tr}(uv)=0$ for every $u\in [A,A]$, $v\in A$, then $A$ is solvable. To prove this direction, fix $u\in [A,A]$. Let $V={⨁}_{\lambda }{V}_{\lambda }$ be its generalized eigenspace decomposition, and define $s$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, the semisimple part of ${\mathrm{ad}}_u$ has eigenvalue $\lambda -\mu$, whereas ${\mathrm{ad}}_s$ acts by $\overline{\lambda }-\overline{\mu }=\overline{\lambda -\mu }$. Polynomial interpolation on the finitely many eigenvalues, with all derivatives through the sizes of the nilpotent blocks set to zero, therefore givesHere the same difference always has the same conjugate, so the interpolation is consistent; the prescribed zero derivatives remove every nilpotent block. Since ${\mathrm{ad}}_u(A)\subseteq [A,A]$, it follows that $[s,A]\subseteq [A,A]$.

Write $u={\sum }_j[a_j,b_j]$. By cyclicity and the trace hypothesis,On the other hand, $\mathrm{tr}(us)={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$. Thus all eigenvalues of $u$ vanish, and every element of $[A,A]$ is nilpotent. The Engel theorem makes $[A,A]$ nilpotent as a Lie algebra, hence solvable; $A/[A,A]$ is abelian, so $A$ is solvable. This is the Conjugate-spectrum proof of Cartan solvability.

Apply this to the Killing radical $R$. For $x,y\in R$, the adjoint actions preserve $R$ and act as zero on $L/R$, because $R$ is an ideal. Computing traces in a basis adapted to $R$ givesThe matrix Lie algebra ${\mathrm{ad}}_R(R)$ consequently satisfies the trace criterion and is solvable. Its kernel is $Z(R)$, an abelian ideal; a central extension of a solvable algebra is solvable. Thus $R$ is solvable. This proves the reusable assertion that the Killing radical is a solvable ideal. Semisimplicity forces $R=0$, and hence $B_L$ is nondegenerate.

A Cartan subalgebra is a nilpotent subalgebra $H$ which is self-normalizing:For a complex semisimple algebra this is equivalently a maximal toral subalgebra. The nilpotent, self-normalizing definition permits a proof of the restricted nondegeneracy without first assuming the toral characterization.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the adjoint action of $H$:For completeness, the stability underlying this decomposition follows directly from nilpotence of $H$. For fixed $h$, every ${\mathrm{ad}}_h$ on $H$ is nilpotent. If $T=\rho (h)$ and $S=\rho (y)$ in a finite-dimensional representation, then $({\mathrm{ad}}_T{)}^{r}S=0$ for sufficiently large $r$. The identityshows that $S$ preserves each generalized eigenspace of $T$. Starting with a basis of $H$, refine these primary decompositions successively; all summands remain $H$-invariant. Each resulting summand has only one eigenvalue for each basis element. The Lie theorem triangularizes the action on that summand, so those eigenvalues extend to a single linear character $\lambda$ on all of $H$. This proves the displayed decomposition and nilpotence of all shifted operators there.

Since $H$ is nilpotent, $H\subseteq L^0$. The space $L^0$ is a subalgebra: repeated use of the derivation rule for ${\mathrm{ad}}_h$ shows that the bracket of two generalized zero-eigenvectors is another such vector. If $L^0/H\ne 0$, the adjoint action of $H$ on this quotient consists entirely of nilpotent maps. The Engel theorem supplies a nonzero coset $x+H$ with $[H,x]\subseteq H$, contradicting self-normalization. Therefore the zero generalized weight space of a Cartan subalgebra is exactly $H$.

Finally, $B_L(H,L^{\lambda })=0$ when $\lambda \ne 0$. Choose $h\in H$ with $\lambda (h)\ne 0$ and put $T={\mathrm{ad}}_h$. On $H$, a power $T^m$ vanishes. On $L^{\lambda }$, $T$ is invertible. For $u\in H$ and $v\in L^{\lambda }$, write $v=T^{m}w$; invariance givesIf $u\in H$ is orthogonal to $H$, it is now orthogonal to every summand of $L$, so nondegeneracy of $B_L$ implies $u=0$. The restriction $B_L{|}_{H\times H}$ is nondegenerate. This establishes the nondegeneracy of the Killing form on a Cartan subalgebra for every Cartan subalgebra, without requiring a chosen root basis.

A Lie algebra representation $V$ is completely reducible when it is a direct sum of irreducible submodules. In finite dimension this is equivalent to saying that every submodule $W\subseteq V$ has an invariant complementary subspace. Indeed, one may split off a minimal nonzero submodule and induct on dimension; conversely, if $V$ is a finite sum of simple submodules, choose a maximal sum $U$ of these with $U\cap W=0$. If $W+U\ne V$, some simple summand $S$ is not contained in $W+U$. Then $S\cap (W+U)=0$ by simplicity, so $U+S$ is a larger sum disjoint from $W$, a contradiction. Thus $V=W\oplus U$.

We prove the required splitting by Haar averaging produces invariant complements. The structural input is the compact real form of a complex semisimple Lie algebra: there is a real semisimple algebra $\mathfrak{k}$ with $L=\mathfrak{k}\oplus i\mathfrak{k}$, whose simply connected integrating Lie group $K$ is compact. This structural fact is independent of the complete-reducibility assertion. One standard construction uses root vectors normalized so that $[e_{\alpha },e_{-\alpha }]=h_{\alpha }$, with real structure constants compatible with the involution $e_{\alpha }\mapsto -e_{-\alpha }$ and $h_{\alpha }\mapsto -h_{\alpha }$. The real spanis closed under brackets, complexifies to $L$, and has negative-definite Killing form. The compactness criterion for real semisimple Lie algebras then gives a compact simply connected $K$. Thus this route uses root structure and compact integration, not the theorem being proved.

Restrict $\rho :L\to {\mathrm{End}}_{\mathbb{C}}(V)$ to $\mathfrak{k}$. The integration of a Lie-algebra representation gives a representation $R:K\to {\mathrm{GL}}_{\mathbb{C}}(V)$, since $K$ is simply connected. Choose any positive-definite Hermitian inner product $⟨,{⟩}_0$ and average using normalized Haar measure:The integral exists by compactness. It is positive-definite because $R(g)v\ne 0$ for $v\ne 0$, and it is $K$-invariant by translation invariance of Haar measure. Equivalently, every $\rho (a)$ with $a\in \mathfrak{k}$ is skew-Hermitian for this inner product.

Let $W$ be an $L$-submodule. It is invariant under $\mathfrak{k}$, hence under the exponentials generating the connected group $K$. For $v\in W^{\perp }$, $w\in W$ and $g\in K$,Consequently the orthogonal complement $W^{\perp }$ is $K$-invariant, and differentiation makes it $\mathfrak{k}$-invariant. It is a complex vector subspace, so it is also invariant under $i\mathfrak{k}$ and hence under $L$. Therefore $V=W\oplus W^{\perp }$ as $L$-modules.

Induction on $\dim V$ now expresses $V$ as a finite direct sum of irreducible submodules. Every finite-dimensional representation of a finite-dimensional complex semisimple Lie algebra is completely reducible. The compact real form need not be the real form originally used to present $L$; in particular the averaging argument does not require the given presentation to be unitary.

Here root systems are finite, reduced and crystallographic, as for complex semisimple Lie algebras. For a base of a root system $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, the Dynkin diagram has one vertex for each simple root. Two distinct vertices have $a_{ij}{a}_{ji}$ bonds, with the arrow on a multiple bond pointing toward the shorter root. The classification of finite crystallographic Dynkin diagrams consists of the following connected diagrams; the integers describing arms count vertices away from the unique branching vertex.

This is classification up to diagram isomorphism and overall rescaling of the root system. It includes the conventional identifications $D_3\cong A_3$ and $B_1\cong A_1$ when those low-rank notations are used. The weighting here encodes root lengths and bonds; it is not the separate theory of vertex labels attached to nilpotent orbits. Dropping the crystallographic or reduced hypothesis changes the classification.

Precisely, a base of a root system is a vector-space basis $\Delta \subset \Phi$ such that each root has a unique expansion $\beta ={\sum }_i{n}_i{\alpha }_i$, with integer coefficients all nonnegative or all nonpositive. This defines the positive and negative roots. We use the column-coroot convention for the Cartan matrix:Some conventions transpose this matrix; specifying which index labels the coroot resolves that difference. The Weyl group is the subgroup of orthogonal transformations generated by the Weyl reflections

For the C3 root system, take an orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{R}}^3$ andThere are six long roots of squared length $4$ and twelve short roots of squared length $2$. A base is ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. The Cartan matrix convention for C3 above givesThus the double bond connects ${\alpha }_2$ to ${\alpha }_3$, with its arrow directed toward ${\alpha }_2$.

An explicit complex Lie algebra realizing this root system is the symplectic Lie algebraIt is closed under commutators, since the two identities $X^{T}J=-JX$, $Y^{T}J=-JY$ imply $[X,Y{]}^{T}J=-J[X,Y]$. Equivalently,so its dimension is $9+6+6=21$. A Cartan subalgebra isWith $E_{ab}$ denoting a matrix unit, the root spaces have generators $E_{ij}-E_{3+j,3+i}$ for $e_i-e_j$ ($i\ne j$), $E_{i,3+j}+E_{j,3+i}$ for $e_i+e_j$ ($i<j$), and $E_{i,3+i}$ for $2e_i$; the analogous lower-left generators give the negative roots. These are all eighteen root spaces, each one-dimensional, and together with $H$ they exhaust the algebra.

For an explicit semisimplicity check, let $I\ne 0$ be an ideal. Simultaneous diagonalization of $\mathrm{ad}H$ makes $I$ a sum of its intersections with $H$ and these root spaces. If it contains a nonzero $h\in H$, some root has $\alpha (h)\ne 0$, and $[h,L_{\alpha }]$ supplies a root vector in $I$. Thus $I$ always contains a root vector. Bracket it with its opposite root vector to obtain a nonzero coroot $h_{\alpha }\in I$. Whenever $(\alpha ,\beta )\ne 0$, bracketing $h_{\alpha }$ with either $L_{\beta }$ or $L_{-\beta }$ puts both root spaces into $I$. The graph of the roots with edges for nonzero inner products is connected: $2e_i$ connects to $e_i+e_j$, which connects to $2e_j$, and every short root connects to an axis root. Repeating the argument yields all root spaces and then all of $H$ from their brackets. Hence $I=L$: the constructed algebra is simple, and therefore semisimple.

The reflections in $e_1-e_2$ and $e_2-e_3$ exchange adjacent coordinates, while reflection in $2e_3$ changes the sign of the third coordinate. They generate every signed permutation. Conversely reflection in any displayed root is a signed permutation. ThusThe reflecting hyperplanes are $x_i=0$ and $x_i=\pm x_j$. A fundamental open Weyl chamber is $x_1>x_2>x_3>0$. Every regular point has three nonzero coordinates of distinct absolute values; its signs and the ordering of those absolute values specify exactly one of the $2^3\cdot 3!=48$ chambers. The closure of this chamber is generated by the rays through $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$. This is the Weyl chamber geometry of C3.

The left panel retains the actual long and short root lengths. The right panel radially projects a hemisphere of the unit sphere, draws every visible reflecting great-circle arc, and highlights the chamber $x_1>x_2>x_3>0$. Each spherical triangle is the section of a three-dimensional chamber by the sphere; opposite triangles on the rear hemisphere supply the remaining chambers. Roots of a root system are normal to the chamber walls, rather than generally lying along their edges.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincaré-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.