Lie algebras course of the University of Oxford (C2.1a)

Source: Mathematics release, bundle Part C 2007 11.pdf, pages 405–407. The paper was sat on 14 June 2011, 2.30–4.00pm, and is shared by Mathematics and Mathematics and Computer Science Part C.

The department's answer-release policy excludes ordinary odd years. Bounded exact-paper checks found no matching advertised answers. All nine printed leaves are answered; the final unpartitioned question includes all of its requests.

Write the lower central series of a Lie algebra as ${\mathfrak{g}}^1=\mathfrak{g}$ and ${\mathfrak{g}}^{j+1}=[\mathfrak{g},{\mathfrak{g}}^j]$. Since the nilpotent Lie algebra is nonzero, there is a last nonzero term ${\mathfrak{g}}^c$, with ${\mathfrak{g}}^{c+1}=0$. Thus $[\mathfrak{g},{\mathfrak{g}}^c]=0$, soConsequently the centre is nonzero. This argument works over every field and includes an abelian algebra, for which $c=1$.

Fix $x$ with $x^N=0$ as a matrix. On the full matrix algebra define $L_x(y)=xy$ and $R_x(y)=yx$. These endomorphisms commute and each has $N$th power zero. The Adjoint representation of a Lie algebra is $\mathrm{ad}(x)=L_x-R_x$. The binomial theorem givesFor $m=2N-1$, each summand has either $j\ge N$ or $m-j\ge N$, and hence vanishes. Therefore $\mathrm{ad}(x)$ is nilpotent on the full matrix algebra and on its invariant subspace $\mathfrak{g}$.

Engel's theorem, in its adjoint form, states that a finite-dimensional Lie algebra is nilpotent if every $\mathrm{ad}(x)$ is a nilpotent endomorphism. Applying it to all $x\in \mathfrak{g}$ provesThis is distinct from assuming that an arbitrary representation of a nilpotent algebra must act by nilpotent matrices, which is false.

If $\mathfrak{g}$ is nilpotent, every Lie subalgebra $\mathfrak{a}$ is nilpotent, since ${\mathfrak{a}}^j\subseteq {\mathfrak{g}}^j$ in the respective lower central series of a Lie algebra. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a basis $(z,u)$; every bracket of these basis elements is zero. Thus $\mathfrak{a}$ is abelian.

Conversely, suppose $\mathfrak{g}$ is not nilpotent. By Engel's theorem, some $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, it has an eigenvalue $\lambda \ne 0$ with eigenvector $y\ne 0$. ThusThe vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional Lie subalgebra and it is nonabelian. This contradicts the hypothesis. Hence

Suppose $\mathrm{ad}(x)$ has a nonzero eigenvalue $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, soFor a diagonalizable endomorphism $T$, its eigenspace decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\mathrm{ad}(y)$.

Thus no $\mathrm{ad}(x)$ has a nonzero eigenvalue. Each is diagonalizable over the algebraically closed field, so each is the zero endomorphism. Hence $[x,z]=0$ for every $x,z\in \mathfrak{g}$, proving

By the classification of finite-dimensional sl2 representations over $\mathbb{C}$, every finite-dimensional module is a direct sum of simple modules $L_m$, indexed by integers $m\ge 0$. The module $L_m$ has dimension $m+1$ and a basis $v_0,\dots ,v_m$ withwith out-of-range vectors interpreted as zero. In particular $v_0$ is a highest weight vector of weight $m$. The weights are $m,m-2,\dots ,-m$, each with multiplicity one. The multiplicities of the $L_m$ in a finite module determine that module up to isomorphism.

In the simple module $L_m$, weight $0$ occurs once if $m$ is even and never if $m$ is odd. Weight $1$ occurs once if $m$ is odd and never if $m$ is even. Thus each simple summand contributes exactly one to $\dim V_0+\dim V_1$.

Complete reducibility of semisimple Lie algebra representations writes $V={⨁}_m{L}_m^{\oplus a_m}$. Taking the two weight spaces commutes with this finite direct sum, soThe right side counts irreducible summands with multiplicity, independently of a chosen decomposition.

Every $2\times 2$ matrix has the unique decompositionBoth summands define invariant subspaces for the Adjoint representation of a Lie algebra. Scalar matrices commute with everything, so $\mathbb{C}I$ is $L_0$. The traceless matrices form ${\mathfrak{sl}}_2$ with basis $e,h,f$ of weights $2,0,-2$. Their brackets satisfy $[e,f]=h$, $[h,e]=2e$, $[h,f]=-2f$; the raising and lowering operations connect all three weight lines. This is the simple module $L_2$. ThereforeAs a check, its zero-weight space has dimension two and its one-weight space is zero, agreeing with part (ii).

On a monomial $X^a{Y}^b$, direct differentiation givesFor example $(EF-FE)(X^a{Y}^b)=[a(b+1)-b(a+1)]X^a{Y}^b=(a-b)X^a{Y}^b$. The same calculation with $H$ givesThese are exactly the brackets of $h,e,f$, so the linear map $\rho$ respects every bracket and is a Lie algebra representation.

For $m\ge 0$, let $W_m$ be the space of homogeneous polynomials of total degree $m$. Each operator preserves total degree. Every polynomial is a finite sum of homogeneous polynomials, soIn $W_m$, the monomials $v_j=X^{m-j}{Y}^j$ have distinct weights $m-2j$ and exactly the actions described in part (i). To see simplicity directly, a nonzero invariant subspace is invariant under $H$ and hence contains one of its weight components: polynomial interpolation in $H$ projects onto each of its distinct eigenspaces. Starting with a nonzero multiple of $v_j$, repeated application of $E$ reaches $v_0$, and repeated application of $F$ then reaches all $v_j$. All coefficients used are nonzero in characteristic zero. Therefore $W_m$ is simple and isomorphic to $L_m$.

ThusThis is an algebraic direct sum, not an infinite product or completion. Its simple submodules are precisely the $W_m$: they are pairwise nonisomorphic and each occurs once, so projection onto homogeneous degrees and Schur's lemma exclude diagonal copies between distinct degrees. Complete reducibility here follows from the explicit decomposition, rather than applying a finite-dimensional theorem directly to the infinite-dimensional $W$.

Equip the finite-dimensional real vector space $V$ with a positive-definite inner product. A reduced crystallographic root system is a finite set $R\subset V\setminus \{0\}$ spanning $V$ such that, for every $\alpha \in R$,where $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ is the orthogonal reflection in ${\alpha }^{\perp }$. The Weyl group is $W(R)=⟨s_{\alpha }:\alpha \in R⟩$. It permutes the finite set $R$. Its action is faithful because $R$ spans $V$: an element fixing every root fixes every vector. Hence it embeds in the finite symmetric group of $R$ and is finite.

A Cartan subalgebra is a nilpotent self-normalizing Lie subalgebra. For a finite-dimensional complex semisimple Lie algebra it is equivalently a maximal toral subalgebra: a maximal abelian subalgebra whose elements have diagonalizable adjoint actions. The commuting semisimple operators $\mathrm{ad}(h)$ give the root-space decompositionThe nonzero weights $\alpha$ are the roots. Their real span inside the complex dual ${\mathfrak{h}}^{\ast }$, with the positive-definite real inner product induced by the Killing form, is the Euclidean space in which they form a reduced crystallographic root system. In particular one uses this real span, rather than treating all of ${\mathfrak{h}}^{\ast }$ as a real vector space of twice the rank.

For the specified Special orthogonal Lie algebra, put $\bar{i}=7-i$. The matrix relation is $x_{ij}=-x_{\bar{j},\bar{i}}$. Thus its diagonal Cartan subalgebra consists ofLet ${\epsilon }_i(H(t))=t_i$. The matrix units satisfy $[H,E_{ij}]=(d_i-d_j)E_{ij}$, where $d=(t_1,t_2,t_3,-t_3,-t_2,-t_1)$. A basis of root vectors is supplied, for $1\le i<j\le 3$, byThese are twelve independent off-diagonal matrices. Together with the three diagonal generators, they account for the full dimension $6\cdot 5/2=15$, so no further roots occur. ConsequentlyChoose the simple rootsThe six positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$, and every other root is their negative. In the standard orthonormal coordinates all roots have squared length $2$, and their Cartan matrix isIts Dynkin diagram is the connected chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$. A decomposition into mutually orthogonal nonempty root systems would separate this connected set of simple roots, which is impossible. Thus the root system is irreducible. The chain is also the diagram $A_3$, which is the root system of ${\mathfrak{sl}}_4(\mathbb{C})$. The classification of complex semisimple Lie algebras by their root systems yields the Lie algebra isomorphismAn explicit explanation of this coincidence uses the exterior-power Lie algebra representation of ${\mathfrak{sl}}_4$ on ${\bigwedge }^2{\mathbb{C}}^4$, a six-dimensional space. After choosing a volume form, $u\wedge v$ defines a nondegenerate symmetric bilinear form on that exterior square. The induced action preserves it because a traceless endomorphism acts trivially on the top exterior power. This gives a nonzero homomorphism ${\mathfrak{sl}}_4\to {\mathfrak{so}}_6$. Its kernel is an ideal in the simple algebra ${\mathfrak{sl}}_4$, so it is zero; equal dimensions $15$ then make it an isomorphism. Over $\mathbb{C}$ all nondegenerate symmetric forms in dimension six are equivalent, including the form with matrix $J$.