'Twisted duality' for Clifford Algebras
  Viewing the complex Clifford algebra $C(V)$ of a real inner product space $V$ as a superalgebra, we offer several proofs of the fact that if $W$ is a subspace of the complexification of $V$ then the supercommutant of the Clifford algebra $C(W)$ is precisely the Clifford algebra $C(W^{\perp})$. 
P. L. Robinson
math.RA