Does Stinespring's dilation theorem hold if the algebra of interest is a topological $\ast$-algebra instead of the usual $C^{\ast}$-algebra?

I will now state the version of Stinespring's dilation theorem that I know, and am wondering what happens if I were to replace the $C^{\ast}$-algebra with a topological $\ast$-algebra.

Theorem. Let $\mathfrak{A}$ be a unital $C^{\ast}$-algebra, and let $\Phi : \mathfrak{A} \to \mathcal{B}(\mathcal{H})$ be a completely positive map. Then there exists a Hilbert space $\hat{\mathcal{H}}$, a unital $\ast$-homomorphism $\pi : \mathfrak{A} \to \mathcal{B}(\hat{\mathcal{H}})$ and a bounded operator $V : \mathcal{H} \to \hat{\mathcal{H}}$ with $||\Phi(1_{\mathfrak{A}})||=||V||^{2}$ such that \begin{equation} \Phi (a)= V^{*} \pi (a) V, \; \; \; \; \; a \in \mathfrak{A}. \end{equation}

Notation. Let $1_{\mathfrak{A}}$ denote the unit in $\mathfrak{A}.$

References to literature on the subject would be greatly appreciated as well.

  • $\begingroup$ I imagine that a sufficiently ugly $\ast$-algebra does not need to have any useful hilbert space representations at all and the theorem could fail for that reason. $\endgroup$ Jul 13, 2015 at 23:45
  • $\begingroup$ Banach star-algebras can look very very different from the Cstar case, as @JohannesHahn has alluded. For instance, consider the disc algebra with $f^*(z):=\overline{f(\overline{z})}$. (Palmer volume II is a good source for other examples if you have the patience to dig around in it.) $\endgroup$
    – Yemon Choi
    Jul 14, 2015 at 0:20
  • $\begingroup$ If, as I suspect, you are really thinking of topological star-algebra that somehow come from B(H) in some way, e.g. the algebra of operators affiliated to a ${\rm II}_1$-factor, or some kind of pro-Cstar algebra, then you should try to formulate your question in one of these precise settings, and make it clear what statement you actually hope to be true $\endgroup$
    – Yemon Choi
    Jul 14, 2015 at 0:21
  • $\begingroup$ Finally: why not try running through the proof of Stinepsring's theorem, applied to the star-algebras you are interested in, and seeing how much survives? $\endgroup$
    – Yemon Choi
    Jul 14, 2015 at 0:22
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    $\begingroup$ @Nik Weaver: (complete) positivity is in fact an algebraic notion. For *-algebras over $\mathbb{C}$ this has been investigated a lot by Schmuedgen in his unbounded operator algebra book, for more general *-algebras, please take a look at my answer. $\endgroup$ Jul 14, 2015 at 6:00

1 Answer 1


The surprising fact is that the GNSStinespringKasparov theory is in fact completely algebraic, at least to a very large extend: the following results have been obtained by a PhD of mine but are, unfortunately, not published (yet?).

So the setting is a *-algebra over a ring $C$ which is of the form $C = R(i)$ with $i^2 = -1$ and $R$ being an ordered ring. In this framework one can define positive functionals to be linear functionals $\omega\colon A \longrightarrow C$ satisfying $\omega(a^*a) \ge 0$. This gives also rise to the notion of a positive element by asking for $\omega(a) \ge 0$ for all positive $\omega$. One can then state what a positive linear map is (mapping positive elements to positive ones) and hence what $n$-positive and completely positive maps are. So far, the notion is entirely algebraic and, of course, reproduces the notions for a $C^*$-algebra you know.

Then one can proceed by constructiong pre-Hilbert modules over *-algebras. Without topology, we can of course not speak of completeness, but this "pre" is already interesting. The only catch is that one has to ask for completely positive inner products: for a Hilbert module over a $C^*$-algebra, this is a consequence of positivity, but in this algebraic setting, there seems to be no way to prove that (hint very welcome!).

From there on, one can proceed with many things like a strong Morita theory for *-algebras, etc. Among them is the algebraic part of the Stinespring construction: if you go through it, then we will see that most of it is algebraic and can be transfered to this setting.

The name of the PhD student is Florian Becher and you can find his thesis somewhere on the document server of Freiburg university.

On my homepage you find a lot of ref's on the algebraic *-representation theory of *-algebras, Morita theory and things, including a (not yet finished) lecture note on this.

  • $\begingroup$ The (draft?) monograph you mention looks interesting. How important are properties H and K to the theory you develop? $\endgroup$
    – Yemon Choi
    Jul 14, 2015 at 14:59
  • $\begingroup$ By the way, if you are looking for examples of Banach $*$-algebras that don't have property K, then I think $\ell^1({\bf F}_2)$ would do the job -- there are self-adjoint elements whose spectrum is not contained in the real line. Another example would be the disc algebra with $(\sum_{n\geq 0} a_nz^n)^* := \sum_{n\geq 0} \overline{a_n} z^n$, in which $z=z^*$, so that $1+zz^* = 1+z^2$ is not invertible. (Apologies if you know all of this!) $\endgroup$
    – Yemon Choi
    Jul 14, 2015 at 15:06
  • $\begingroup$ @YemonChoi. Thanks for your comments. And yes: H and K are sort of important. They are the abstracted version of what one needs in view of $C^*$-algebras. We (i.e. Henrique Bursztyn and I) used that a lot when discussing all sorts of Morita theory for -algebras. The upshot is somehow that the (strong) Morita theory of C-algebras is indeed algebraic, but the Picard group is not: there are subtle differences even for C*-alegbras. To get a reasonable theory for *-algebras, H and K are sort of essential (for technical reasons) The main point is that there are many examples beyond $C^*$... $\endgroup$ Jul 14, 2015 at 15:48
  • $\begingroup$ ... among them the star product algebras which are not even algebras over $\mathbb{C}$ but over the formal power series ring $\mathbb{C}[[\hbar]]$. This exmplains why we were interested to extend the notions to more general scalars. Thanks also for the $\ell^1$ example. This I didn't know. I always use the continuous functions on the sphere with -involution given by complex conjugation *and the antipodal map, as you indicated also in your above comment. This has really creepy features for many reasons :) $\endgroup$ Jul 14, 2015 at 15:49

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