From: Robert Israel Subject: Re: Tensor product of operators Date: Fri, 23 Apr 1999 17:26:49 -0700 Newsgroups: sci.math.research,sci.math Stefan Boller wrote: > Supose you have two Hilbert spaces H and K, resp., and a (linear, i.g. > unbounded) operator A on the Hilbert tensor product of these two > spaces. Now I am looking for criteria (neccessary or sufficient) for > this operator being a tensor product of two operators B and C on H and > K, resp. Suppose for simplicity that A is bounded. Consider F(u,v,w,z) = where < , > is the inner product and @ is the tensor product symbol \otimes. If A = B @ C then F(u,v,w,z) = so F(u,v,w,z) F(u',v',w',z') = F(u,v,w',z') F(u',v',w,z) for all u,v,u',v' in H and w,z,w',z' in K. Conversely, suppose this equality holds. We may assume F is not identically 0. Fix some u_0,v_0,w_0,z_0 so that some F(u_0,v_0,w_0,z_0) = 1, and consider the bounded sesquilinear form Q(w,z) -> F(u_0,v_0,w,z). By the Riesz lemma there is a bounded operator C on K such that Q(w,z) = . Similarly there is a bounded operator B on H such that F(u,v,w_0,z_0) = . Then F(u,v,w,z) = F(u,v,w,z) F(u_0,v_0,w_0,z_0) = F(u,v,w_0,z_0) F(u_0,v_0,w,z) i.e. = = for all u,v,w,z. Since the linear span of products u @ v is dense in H @ K, we conclude that A(v@z) = Bv @ Cz, i.e. A = B @ C. If A is an unbounded operator, things could well be more complicated, because the domain of a densely-defined unbounded operator on H @ K might not even contain any vectors of the form u @ v. But suppose there are linear subspaces V and Z of H and K such that V @ Z (the algebraic tensor product this time) is in the domain of A. Then F(u,v,w,z) is defined for u in H, v in V, w in K and z in Z, and our condition is F(u,v,w,z) F(u',v',w',z') = F(u,v,w',z') F(u',v',w,z) for u,v,w,z,u',v',w',z' in the appropriate spaces. The conclusion is that there are linear operators B: V -> H and C: Z -> K such that A(v@z)= Bv @ Cz for v in V and z in Z. -- Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2