Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 (Lie Algebras and Their Representations)

For this Heisenberg Lie algebra,because $c$ belongs to the center. Thus its Lower central series of a Lie algebra is $\mathfrak{g},\mathbb{C}c,0$, so $\mathfrak{g}$ is a two-step Nilpotent Lie algebra.

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Let $A,B,C$ represent $a,b,c$. Since $c$ is central, $C$ commutes with $A$ and $B$. Over the complex number field $\mathbb{C}$, $C$ has an eigenvalue $\gamma$, and its corresponding eigenspace is invariant under all three operators. The irreducibility of $V$ therefore makes this eigenspace all of $V$, so $C=\gamma I$. Taking the trace ofgives $(\dim V)\gamma =0$ by the cyclic property of the trace; hence $C=0$.

The remaining operators $A$ and $B$ commute. Two commuting operators on a nonzero finite-dimensional complex vector space have a common eigenvector, whose span is invariant. Irreducibility therefore forces $\dim V=1$. Conversely, every pair $(\alpha ,\beta )\in {\mathbb{C}}^2$ defines a one-dimensional irreducible representation byThese are all the finite-dimensional irreducible representations.

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For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

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Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

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Choose the Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$ determined by the positive roots. Regard the one-dimensional space ${\mathbb{C}}_{\lambda }$ as a $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isThe Poincare-Birkhoff-Witt theorem identifies it as a vector space with $U({\mathfrak{n}}^{-})$ acting on a highest-weight vector $v_{\lambda }$.

For each positive root $\alpha$, arbitrary powers of a negative-root vector contribute the geometric series $1+e^{-\alpha }+e^{-2\alpha }+\cdots$. Consequently the formal character of a weight module isThis product is interpreted in the completion of the group algebra in the negative-root direction; the PBW basis proves that every coefficient is the correct finite weight space dimension.

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The weights of $M_{\lambda }$ lie below $\lambda$ in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule $V$ is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight $\mu$ occurring in $V$. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero $v\in V_{\mu }$. Thus $v$ is a singular vector.

The Casimir element is central and acts throughout $M_{\lambda }$ byThe same element acts on the highest-weight vector $v$ of weight $\mu$ byBoth are the action of one operator on the same module, so the scalars agree and

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For the sl2 Lie algebra, let $v_{\lambda }$ be the highest-weight vector. The Poincare-Birkhoff-Witt theorem gives the basis $(f^r{v}_{\lambda }{)}_{r\ge 0}$, and the defining Lie brackets implyA positive-degree basis vector is singular exactly when $r=\lambda +1$ is a positive integer. Therefore $M_{\lambda }$ is irreducible when $\lambda \notin {\mathbb{Z}}_{\ge 0}$.

If $\lambda =m\in {\mathbb{Z}}_{\ge 0}$, the vector $f^{m+1}{v}_{\lambda }$ has weight $-m-2$ and generates a submodule isomorphic to $M_{-m-2}$. The latter is irreducible because $-m-2\notin {\mathbb{Z}}_{\ge 0}$. Every nonzero submodule contains a singular vector by the preceding part, and the displayed coefficient shows that this is the only possible proper singular vector. Henceis the unique proper nonzero submodule, as summarized by the Reducibility of an sl2 Verma module.

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After ordering a symplectic basis in two blocks, writeMatrices in the Symplectic Lie algebra have block formThe root-space decomposition isFor example, these one-dimensional spaces are spanned respectively byThus this is the Cn root systemThe upper-triangular choice givesIts simple roots, highest root, fundamental weights, and half-sum of positive roots are

Using the notation requested in the paper, the root lattice $P$ and weight lattice $Q$ areso $Q/P\cong \mathbb{Z}/2\mathbb{Z}$. This reverses the common notation in which the root lattice is called $Q$ and the weight lattice is called $P$.

Since a multiple-edge arrow in a Dynkin diagram points toward the shorter root, the finite and extended diagrams areand

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The Weyl reflection formula $s_{\alpha }(x)=x-⟨x,{\alpha }^{\vee }⟩\alpha$ becomes especially concrete on $x=(x_1,\dots ,x_n)$:with all unlisted coordinates fixed. The Weyl group is therefore the group of signed permutations

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Withthe equation $AJ+J{A}^T=0$ reduces to $p+s=0$. Henceas matrix Lie algebras, with the same commutator bracket.

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Let $R^{+}$ be the positive roots, $W$ the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the half-sum of positive roots, and ${\alpha }^{\vee }$ a coroot. For a dominant integral highest weight $\lambda$, the Weyl character formula isTaking the value at the identity gives the Weyl dimension formula

For the q-character convention relevant to the Principal sl2 subalgebra, set $h_{\mathrm{pr}}=2{\rho }^{\vee }$, so ${\alpha }_i(h_{\mathrm{pr}})=2$ for every simple root, and defineThe q-character formula is the principal specialization of the Weyl character formula:

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Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$, with $|{\alpha }_2{|}^2=3|{\alpha }_1{|}^2$ and angle $150^{\circ }$. The six positive roots of the G2 root system areand their negatives complete the two concentric hexagons of short and long roots. The fundamental weights areso ${\omega }_1$ is itself a short root and ${\omega }_2$ is the highest root.

The seven-dimensional representation $L_{{\omega }_1}$ has weight seteach with weight multiplicity one. Their positive heights are $1,2,3$, so the q-character of a highest-weight representation isThis is one ${\mathfrak{sl}}_2$ weight string, hence

The representation $L_{{\omega }_2}$ is the fourteen-dimensional Adjoint representation. Its nonzero weights are the twelve roots, and its zero-weight space is the two-dimensional Cartan subalgebra. The positive root heights are $1,1,2,3,4,5$, soSplitting this into ordinary ${\mathfrak{sl}}_2$ strings gives

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Use the B2 root system conventionThus $V=L_{{\omega }_2}$ is the five-dimensional vector representation of the Special orthogonal Lie algebra ${\mathfrak{so}}_5$. Label its weight verticesThe crystal basis is the colored chainbecause each Kashiwara operator ${\tilde{f}}_i$ subtracts ${\alpha }_i$.

For the tensor product of crystals, write $XY$ for $X\otimes Y$. The complete colored-arrow graph is compactly specified byIts three connected highest-weight components start at $AA$, $AB$, and $AE$. Their vertex sets areTheir highest weights and dimensions identify the ten-vertex component with the exterior square and the other two with the symmetric square. Thereforeof dimensions $10$ and $14+1$, respectively.

The module $L=L_{{\omega }_1}$ is the four-dimensional spin representation. Its weights are $(\pm {\epsilon }_1\pm {\epsilon }_2)/2$, and its crystal is

Every weight of $V$ lies in the root lattice, so every weight of every tensor power $V^{\otimes n}$ also lies in that lattice. But ${\omega }_1=({\epsilon }_1+{\epsilon }_2)/2$ represents the nonzero coset in the quotient of the weight lattice by the root lattice. Consequently no irreducible constituent of $V^{\otimes n}$ can have highest weight ${\omega }_1$, and $L$ never occurs.

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