A Hilbert Algebra?

January 8, 2012

Let $\delta:N \rightarrow H$ be a closable, real, densely defined derivation into an $N-N$ bimodule $H$ . Let $D$ denote the domain of $\delta$ . We suggest the following involutive algebra structure on the orthogonal (Hilbert space) direct sum $D \oplus H$ . Let $(x,\xi)(y,\eta):=(xy,x \eta+\xi y)$ denote the product, and $(x,\xi)^{\#}:=(x^{*},J\xi)$ the involution, where $J$ is assumed to be an antiunitary involution on $H$ that replaces the bimodule structure with its opposite: $J(x \xi y)=y^{*}J\xi x^{*}$ .

From Wikipedia, we have the following definition of Hilbert algebra:

A Hilbert algebra is an algebra $\mathcal{A}$ with involution $\#$ and an inner product $\langle,\rangle$ such that
1) $\langle a,b \rangle= \langle b^{\#},a^{\#} \rangle$ for $a, b \in \mathcal{A}$ ;
2) left multiplication by a fixed a in $\mathcal{A}$ is a bounded operator;
3) $\#$ is the adjoint, in other words $\langle xy,z\rangle = \langle y, x^{\#}z \rangle$ ;
4) the linear span of all products $xy$ is dense in $\mathcal{A}$ .

Let’s explicitly see where our algebra fails these axioms. The involution $J$ is antiunitary on $H$ , and the usual involution $x\Omega \mapsto x^{*}\Omega$ on $L^{2}(M)$ is antiunitary also, so (1) holds. For (2), we need to worry mostly about the “second entry”…perhaps we’ll need to restrict our attention to “bounded vectors”. Sample computation: $\|x \eta+\xi y \| \leq \|x\|\|\eta\|+\|\xi\|\|y\|$ …but the second term is troublesome because the norm on $y$ is the wrong sort. This should be the two-norm. Perhaps if we require $\xi$ to be a bounded vector, then these operators would be bounded, perhaps, and we’d have (2). The condition (4) is easy to satisfy, since $(1,0)$ is the unit element of the algebra we have that the set of products is the entire domain (we’ve taken no closures, yet).

It is easy to see that (3) is the weak point here, in the second coordinate. Let’s record the failure here, though:

$\langle (x,\xi)(y,\eta),(z,\zeta)\rangle=\langle (xy,x \eta+\xi y),(z,\zeta)\rangle$

but $\langle x\eta+\xi y,\zeta\rangle=\langle x\eta,\zeta\rangle+\langle \xi y,\zeta\rangle$

and this is $\langle \eta,x^{*}\zeta\rangle+\langle y, \xi^{*}\zeta\rangle$ . Now the first coordinates are orthogonal (we’re freewheeling here, now) and since $\xi^{*}=J\xi$ can we get this? Looking at this last bit, things don’t quite parse…and we’re left with the Radon-Nikodym business again…so the adjoints don’t work out. Unless, for some mysterious reason, we can get $\xi^{*}\zeta$ to be a vector in $L^{2}(M)$ ! Perhaps this is not so mysterious, as a bounded vector is viewed as a map from $L^{2}(M)$ into $H$ , and its adjoint a map the other way. The question is, what’s the domain of the adjoint? (Among the million other questions like…can we use the orthogonality like this?)