Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In an introduction to the GNS construction, I'm told that the GNS construction is a generalization of the way that $L^{\infty} (X, \mu)$ has a representation on $L^2$ where $\mu$ is a measure on $X$. Can someone demonstrate that, with the appropriate notions of equivalence in place (that I hope you will specify), indeed the GNS construction applied to the unital C* algebra $L^\infty$ with respect to some choice positive linear functional, does furnish the Hilbert space $L^2$ and then the representation as stated? As a very brief reminder, a representation is a $*-homomorphism$ coming out of a unital C* algebra into $B(H)$ preserving the identity, and in the GNS construction, one starts out with a unital C* algebra and a positive linear functional on it, which is then necessarily continuous. (Although I always mean everything is continuous.) You take the induced sesquilinear form and quotient by its null space, and then take the Hilbert space completion of the resulting inner product space. The representation is, roughly speaking, by left multiplication. (Maybe that's reversed if you use the reverse convention for what it means to be a sesquilinear form.)

share|improve this question
    
Oh and just so that we all have the same notation, if you don't mind let's use the physicist convention where linearity of a sesquilinear form is in the second entry. –  Jeff Aug 13 '12 at 0:58
    
The positive linear functional is just integration. And I think the rest is bookkeeping. –  Qiaochu Yuan Aug 13 '12 at 1:11
    
Actually that was the only part of this problem where I had any inclination. I'm having a particular difficulty seeing why the abstractly obtained Hilbert space is anything like $L^2$ –  Jeff Aug 13 '12 at 1:21

1 Answer 1

up vote 3 down vote accepted

The positive linear functional is integration. Note that if you want the integral of the identity to be finite then $\mu$ needs to be a finite measure.

In that case, integration induces the usual $L^2$ inner product

$$\langle f, g \rangle = \int_X \overline{f(x)} g(x) \, d \mu$$

on $L^{\infty}(X, \mu)$. Null with respect to either $L^2$ or $L^{\infty}$ means equal to zero a.e. so this inner product is already positive-definite. By finiteness, $L^{\infty}$ contains the indicator function of every measurable subset of $X$, so the completion of $L^{\infty}$ with respect to the $L^2$ inner product is just $L^2(X, \mu)$ and $L^{\infty}$ acts on it by left multiplication as desired.

share|improve this answer
    
@ qiaochu yuan Oh actually this is not at all what I had in mind so it's a good thing you posted the solution. I agree that it seems to work for finite measures, although I'll think about it in greater detail soon. My main issue right now is that the document I'm reading does not say the measure must be finite. However, after looking at surrounding material, it seems my source is suggesting "integration" as the linear functional. Linear functionals must always be "finite." Can my question be answered for infinite measures, or was this perhaps an oversight of the author's? –  Jeff Aug 13 '12 at 15:39
    
Specifically, as long as we have any measure, then $L^\infty$ has a nice representation on $L^2$ but it appears it's just not obtained via this integration which is not defined. Is it obtained via some other positive linear functional? –  Jeff Aug 13 '12 at 15:42
    
@Jeff: I think some authors define positive linear functionals only on the positive elements and allow them to take values in $[0, \infty]$. In that case I guess the GNS construction is restricted to the subspace of elements with finite $L^2$ norm. The situation is not clear to me as far as conventions go; the Wikipedia article restricts to the $\sigma$-finite case. –  Qiaochu Yuan Aug 13 '12 at 15:55
    
When the measure is not finite, what you are doing is GNS for weight (and not a functional). –  Martin Argerami Aug 13 '12 at 21:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.