16
$\begingroup$

tl;dr: Is there such a thing as a W*-completion of a C*-algebra, and if so, where can I read about it?

I'm wondering about the relationship between (abstract) C*-algebras and W*-algebras. On the one hand, every W*-algebra is a C*-algebra. On the other hand, it seems to me that it should be possible to complete any C*-algebra to a W*-algebra. (Categorially, this would be a reflection.) In the case of commutative algebras, I even think that I know how this works: every commutative C*-algebra is the algebra of continuous functions on some compact Hausdorff space, and we extend this to the W*-algebra of essentially bounded Borel-measurable functions on the space (considered up to equality almost everywhere). [Warning: this is determined in the comments to be wrong.]

So, is this correct? Does it work for noncommutative algebras as well? Is there a good algebro-analytic (without passing through topology) description of this? Is there a good reference, especially online?

Also, in the commutative case, it seems that every state (positive normal linear functional) on a C*-algebra extends uniquely to its W*-completion, so they have the same space of states. Is this correct? Does it extend to the noncommutative case?

Another question is how this relates to concrete algebras (those given as algebras of operators on some Hilbert space). One way to complete a C*-algebra would be to pick a concrete representation and take its weak closure (or double commutant). But I expect that this will depend on the representation chosen. (And my analysis is bad enough that I can't check this for even the commutative case.)

I'd appreciate any help even for the main question, never mind this stuff about states and representations!

Improve this question
$\endgroup$
28
  • 2
    $\begingroup$ Take the second dual (sorry for terseness, am in a bit of a rush) -- this is called the universal or enveloping W*-algebra of your original B*-algebra. (Yes, I believe in the distinction between B* and C*.) Note that in the commutative case this does not give what you guessed, I would have to think a bit more about what exactly goes wrong. $\endgroup$
    – Yemon Choi
    Commented Aug 1, 2011 at 19:39
  • 2
    $\begingroup$ I continued searching M.O after asking, and part of the answer is in question 60328, What kind of completion is this?. * The standard term for the W*-completion of a C*-algebra is "enveloping von Neumann algebra"; it may be constructed as the double dual. * In the commutative case, viewed as topology, the term is "hyperstonean cover"; viewed as analysis, I can read about it in Conway (which I own). So I'm left with asking for confirmation and references to enveloping W*-algebras. $\endgroup$
    – Toby Bartels
    Commented Aug 1, 2011 at 19:45
  • 3
    $\begingroup$ @Toby: if $X$ is, say, the unit interval with usual topology, and $L^\infty(X)$ is the bounded Lebesgue measurable functions on it (modulo almost everywhere caveats, yadda yadda) then "evaluation at $x$" is a (pure) state on $C(X)$ but isn't even well-defined on $L^\infty(X)$. The "nice" states on $L^\infty(X)$, i.e. those respecting the weak-star topology that it has as a $W*$-algebra, turn out to be given by "integration against $f$" for $f$ a positive integrable Lebesgue measurable function. $\endgroup$
    – Yemon Choi
    Commented Aug 6, 2011 at 3:38
  • 6
    $\begingroup$ I will risk bumping this question to reduce the net wrongness of statements on the internet. $L^\infty([0,1],\mu)$ is not the enveloping algebra of $C([0,1])$, but a proper quotient thereof. The predual of $C([0,1])^{**}$ is $C([0,1])^*$. Now, $\delta$-measures of points in $[0,1]$ form a norm-discrete subset of continuum cardinality in $C([0,1])^*$, so $C([0,1])^{**}$ does not have separable predual. But the predual of $L^\infty([0,1],\mu)$ is $L^1([0,1],\mu)$, which is norm-separable. $\endgroup$
    – Robert Furber
    Commented Oct 23, 2018 at 22:00
  • 5
    $\begingroup$ In some sense that can be made precise, $C(X)^{**}$ is a space that puts together $L^\infty(X,\mu)$ for all Borel measures $\mu$ on $[0,1]$. In a compatible way (i.e. $L^\infty(X,\mu)$ is always a quotient of $C(X)^{**}$). $\endgroup$
    – Robert Furber
    Commented Oct 23, 2018 at 22:01

1 Answer 1

Reset to default
13
$\begingroup$

The universal enveloping W* algebra of a C* algebra is discussed in detail in chapter III.2 of volume 1 of Takesaki's work on "Theory of operator algebra". It is universal in the sense that any map to an W* algebra factors through it (modulo some mumbling about topologies), and as mentioned above is given by the double dual of the C* algebra. This is called the Sherman-Takeda theorem, and was announced by Sherman in 1950 and proved by Takeda in 1954.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, I can find that book at my university library, so I'll check it out! $\endgroup$
    – Toby Bartels
    Commented Aug 3, 2011 at 16:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .