Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 1

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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