 Solution

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 $W$.