Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 6 / 1

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

The derived series of a Lie algebra isThe algebra is a solvable Lie algebra if ${\mathfrak{b}}^{(m)}=0$ for some $m$. Here a bracket of subspaces means the linear span of their Lie brackets.

The Lie theorem states that a nonzero finite-dimensional representation of a complex solvable Lie algebra has a common eigenvector; equivalently, its operators can be simultaneously upper triangularized. Applied to an Irreducible Lie algebra representation, the common eigenvector gives a nonzero invariant line, which must be the entire module. A scalar action $\rho (x)=\lambda (x)$ respects the Lie bracket exactly when $\lambda ([x,y])=0$. Thus, in the finite-dimensional setting,Every functional displayed gives a one-dimensional irreducible, and two such modules are isomorphic exactly when their functionals agree.

The finite-dimensional qualification is essential here, although the printed wording does not repeat it. For example, let $[x,y]=y$ and act on $\mathbb{C}[t]$ by $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. These operators satisfy $[x,y]f=yf$. A nonzero submodule stable under $x$ is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ forces $p(t)\mid p(t-1)$, so the two polynomials, having the same degree and leading coefficient, are equal. In characteristic zero a polynomial invariant under a nonzero shift is constant. The module is therefore an infinite-dimensional simple module for the two-dimensional affine Lie algebra. It rules out extending the one-dimensional classification to unrestricted representations.