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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.)

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    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$2012-08-13

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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.

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    When the measure is not finite, what you are doing is GNS for weight (and not a functional).2012-08-13