From: Robert Israel <israel@math.ubc.ca>
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) = <u @ w,
A(v @ z)>
where < , > is the inner product and @ is the tensor product symbol
\otimes.
If A = B @ C then F(u,v,w,z) = <u,Bv> <w,Cz> 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) =
<w, C z>.  
Similarly there is a bounded operator B on H such that F(u,v,w_0,z_0) =
<u, Bv>.
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. <u @ w,A(v @ z)> = <u,Bv><w,Cz> = <u @ w, Bv @ Cz> 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