Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 1

Use the complex inner product convention linear in its first argument. Give the exterior algebra its induced inner product, with an orthonormal wedge basis. Exterior multiplication is $e(v)\eta =v\wedge \eta$. Its adjoint is contraction:Applying this formula to $a\wedge \eta$ separates its first term from the remaining terms, giving $e(b{)}^{\ast }e(a)\eta =(a,b)\eta -e(a)e(b{)}^{\ast }\eta$. Hence the canonical anticommutation relations areFor irreducibility, choose an orthonormal basis $u_1,\dots ,u_d$. The occupancy projections are $N_j=e(u_j)e(u_j{)}^{\ast }$, and $P_0={\prod }_j(I-N_j)$ projects onto the vacuum $1\in {\Lambda }^{0}V$. For an increasing index set $J$, put $E_J=e(u_{j_1})\cdots e(u_{j_r})$. Then $E_I{P}_0{E}_J^{\ast }$ maps the wedge basis vector $u_J$ to $u_I$ and kills every other wedge basis vector. Thus these products are all matrix units and the generated star-algebra is $\mathrm{End}(\Lambda V)$. Any common invariant subspace for the creation and annihilation operators is therefore either zero or all of $W$.

The commutant is $A^{\prime }=\{T\in \mathrm{End}V:Ta=aT\text{ for every }a\in A\}$. Clearly $A\subseteq A^{\prime \prime }$. For the reverse inclusion, first note that $A\xi$ is an invariant subspace for the unital star-algebra $A$, and its orthogonal complement is invariant too. Its projection belongs to $A^{\prime }$. Thus if $B\in A^{\prime \prime }$, $B\xi \in A\xi$.

To make the same algebra element work for all vectors, amplify to $V^d$, where $d=\dim V$. The diagonal action $a^{(d)}$ has commutant consisting of block matrices whose entries belong to $A^{\prime }$. Therefore $B^{(d)}$ commutes with this amplified commutant. For a basis ${\xi }_1,\dots ,{\xi }_d$, the subspace $\{(a{\xi }_1,\dots ,a{\xi }_d):a\in A\}$ is invariant and reducing for the amplified algebra. Its projection lies in that commutant. The preceding argument gives some $a\in A$ with $(B{\xi }_1,\dots ,B{\xi }_d)=(a{\xi }_1,\dots ,a{\xi }_d)$. Since these vectors form a basis, $B=a$. HenceThis finite-dimensional double commutant theorem for star-algebras uses the star hypothesis to make the orbit subspaces reducing.

A strongly continuous unitary representation of the circle group decomposes as $H={\widehat{⨁}}_{m\in \mathbb{Z}}{H}_m$, with $U_z\xi =z^m\xi$ on $H_m$. Its positive energy unitary representation of the circle condition is that the occurring integers are bounded below. Equivalently $U_{e^{it}}=e^{itL}$ with a self-adjoint generator bounded below. An integer character can shift the lowest weight to zero.

Let $M=S^{\prime \prime }$. Because $S=S^{\ast }$, this is a Von Neumann algebra, and rotation invariance makes $U_z{M}^{\prime }{U}_z^{\ast }=M^{\prime }$. Joint irreducibility implies $M^{\prime }\cap \{U_z{\}}^{\prime }=\mathbb{C}I$: a nonscalar self-adjoint element of that joint commutant would have a nontrivial invariant spectral projection gives a reducing subspace.

For $T\in M^{\prime }$, take its strong Fourier coefficients $T_k={\int }_{\mathbb{T}}{z}^{-k}{U}_{z}T{U}_z^{\ast }dz$. They lie in $M^{\prime }$ and satisfy $U_z{T}_k{U}_z^{\ast }=z^k{T}_k$. Both $T_k^{\ast }{T}_k$ and $T_k{T}_k^{\ast }$ are rotation-invariant elements of $M^{\prime }$, hence scalar. If $T_k\ne 0$, both scalars equal $\Vert T_k{\Vert }^2$, so $T_k/\Vert T_k\Vert$ is a unitary. It shifts energy by $k$. For $k\ne 0$, applying it or its adjoint to a nonzero lowest-energy vector produces a vector below the lowest energy, a contradiction. Thus $T_k=0$ for every $k\ne 0$, while $T_0$ is scalar.

Fejér sum averaging of the strongly continuous orbit $U_{z}T{U}_z^{\ast }$ converges strongly to $T$. Its Fourier coefficients have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular every bounded operator commuting with $S$ commutes with every $U_z$. In fact the hypotheses force the stronger conclusion $S^{\prime }=\mathbb{C}I$.