F. Bimodules over von Neumann Algebras 483
the identity bimodule L^2 (M) as a space of square integrable closed operators
affiliated with M.
A densely defined operator Ton L^2 (M) is said to be affiliated with M if
u'T =Tu' (which includes the requirement that dom(T) = u' dom(T)) for
every unitary element u' in M'. We observe that if T is a closed operator
affiliated with M and T = ulTI is the polar decomposition of T, then u E
M and ITI is a positive self-adjoint operator affiliated with M. A closed
operator T which is affiliated with M is said to be square integrable if l E
dom(T). Although we won't need it, note that Tis square integrable if and
only if for the spectral decomposition ITI = f 000 tdE(t) of ITI, one has
r(T*T) := fo00
t^2 d(r o E)(t) < oo.Moreover, for square integrable T, one has r(T*T) = llTlll§. For every
e E L^2 (M), we define a densely defined linear operator L~ affiliated with M
by
L~: L^2 (M) :JM 3 x f-t ex E L^2 (M).
Proposition F.11. For every e E L^2 (M), the operator L~ is closable and
its closure Le is a square integrable operator affiliated with M. Moreover,
the map e c---t Le gives a bijective correspondence between L^2 (M) and the
space of closed square integrable operators affiliated with M.
Proof. Let J: L^2 ( M) --t L^2 ( M) be the conjugate linear isometry defined
by Jx = J? (cf. Section 6.1). Fore E L^2 (M), we have
(L~x, fj) = (Jfj, J(ex)) = (ly*, x* Je) = (xl, (Je)y) = (x, L~efi).This implies that (L~) is densely defined and L~e c (L~). It follows that
L~ and L~e are closable and their closures Le and LJe satisfy LJe c L~.
Since L~ is affiliated with M, so is the closure Le. Let Le = ulLel be the
polar decomposition. (We may assume that u E M is unitary.) Then,
we have L~ = ILelu = uLeu = Lueu· Since LJe C Lueu, we have
Je = ueu and L~ = LJe·
Let T be a closed square integrable operator affiliated with M. We
first show that T is also square integrable. Let T = ulTI be the polar
decomposition. We have that;;;; E dom(ITI) since ITI is a square integrable
operator affiliated with M. It follows that for every e E dom(T),
(Te, 1) =(!Tie,;;;;) = (e, !Tl;;;).This implies that l E dom(T). Let e = Tl and 'T/ = Tl. Since T and
T are affiliated with M, it is easy to see that L~ c T and Lg C T. This
implies that Le c T c LJw It is clear that e = JrJ, so T = Le. D