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I am currently reading [$K$-Theory and $C^*$-Algebras: A Friendly Approach] by N. E. Wegge-Olsen. In p. 18 (section 1.3), there is a following sentence:

Allowing for bounded borel functions on $\operatorname{Spec}(a)$ and pointwise monotone convergence we end up with the weak (and strong) closure of $C^*(a,1)$, i.e. the von Neumann algebra $W^*(a,1)$. The spectral map is then no longer isometric but only norm decreasing.

Before this quote, the book defines an isometric homomorphism from functions in $C(\operatorname{Spec}(a))$ to $C^*(a,1)$, the smallest $C^*$-algebra containing $a, a^*,$ and $1$. $a$ is normal.

I know that the von Neumann algebra $W^*(a,1)$ is defined as the weak (and strong) closure of $C^*(a,1)$. I also know that the pointwise limit of continuous functions give Baire (category 1) functions, which is (subset of) Borel functions.

But I'm stuck: I think the main line reads as $L^{\infty} (\operatorname{Spec}(a))\simeq W^*(a,1)$, but I'm quite suspicious about it. Is it true? Of course the von Neumann algebra is isomorphic to the $L^\infty$-space, but I'm not sure it is $L^{\infty}(\operatorname{Spec}(a))$ here. (This maybe shows that this is true, but I'm not sure.) Furthermore, why does the main text contains the condition monotonicity? Is it crucial(if we consider w/o monotonicity then we get wrong result), or the monotonicity condition is just written here although the result w/ and w/o monotonicity is same?

haru
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First of all, Baire and Borel functions are the same here, since $\text{Spec}(a)$ is a compact subset of $\mathbb{C}$.

Iterated applications of monotone limits are enough to obtain all pointwise limits, whence all bounded Borel functions. Restricting to monotone convergence is not necessary - either iterated monotone convergence or iterated pointwise convergence will give you all bounded Borel functions. So the result is the same either way. (But it is easy to show that monotone limits exist for operators, so perhaps that’s why the author put it that way.)

$L^\infty(\text{Spec}(a)) \simeq W^\ast(a, 1)$ is true - under the assumption that your Hilbert space is separable, and that you can choose what measure to use on $\text{Spec}(a)$, as the correct measure to use may not be the Lebesgue measure - but it is not what the quote is claiming. Recall that $L^\infty(\text{Spec}(a))$ only makes sense when you have a measure on $\text{Spec}(a)$, and it identifies two functions if they are a.e. equal. Instead, the author simple claims there is a surjective $\ast$-homomirphism:

$$\mathcal{B}^\infty(\text{Spec}(a)) \to W^\ast(a, 1)$$

Where $\mathcal{B}^\infty(\text{Spec}(a))$ is the algebra of all bounded Borel functions on $\text{Spec}(a)$, where we do not have a measure specified, nor do we identify functions up to a.e. equivalence. This is indeed true, assuming the Hilbert space is separable (in that case you can actually choose a probability measure $\mu$ on $\text{Spec}(a)$ and the surjection is effectively identifying functions up to a.e. equivalence w.r.t. $\mu$). However, first remark: this may not be true if the Hilbert space is not separable. For example, the operator $a$ on the sequence space $\ell^2([0, 1])$ given by $ae_t = te_t$ for all $t \in [0, 1]$. $W^\ast(a, 1)$ consists of multiplication by all bounded functions on $[0, 1]$ along the diagonal, but the range of the map $\mathcal{B}^\infty(\text{Spec}(a)) \to W^\ast(a, 1)$ (known as the Borel functional calculus) only consists of multiplication by bounded Borel functions along the diagonal.

Second remark: even if the Hilbert space is separable, whence the map is surjective, it needs not be injective. For example, $a$ being multiplication by $t$ on $L^2([0, 1], m)$, where $m$ is the Lebesgue measure. Then for $f$ being the indicator function of a singleton (or any nonempty Lebesgue-null set, for that matter), $f(a) = 0$ despite $f \neq 0$. (In fact, one can show that, assuming the Hilbert space is separable, the Borel functional calculus map is injective iff $\text{Spec}(a)$ is countable and every point in the spectrum is an eigenvalue of $a$. In particular, even $\text{Spec}(a)$ being countable is not enough by itself. For example, if $a$ is a normal, compact, and injective operator on an infinite-dimensional Hilbert space, then $0 \in \text{Spec}(a)$. If $f$ is the indicator function of $\{0\}$, then $f \neq 0$ but $f(a) = 0$.)

David Gao
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  • Thank you for your nice and detailed answer. So basically, when the Hilbert space is unknown to be separable, the set of Borel functions are smaller than the von Neumann algebra - vN algebra is "very large". I might consider there is an assumption saying $\mathcal{H}$ is separable in the book. The last injectivity condition is also interesting. I have some additional questions: (1) for the "correct measure" of $\operatorname{Spec}(a)$, is Borel measure unsufficient? (2) are there nice sources of these facts? – haru Jun 07 '24 at 14:25
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    @haru Technically, what you need for both the Borel functional calculus map to be surjective as well as for $W^\ast(a, 1)$ to be isomorphic to $L^\infty(\text{Spec}(a))$ is that $W^\ast(a, 1)$ admits a separating vector, which can always be arranged if the Hilbert space is separable but may not exist otherwise. (I wouldn’t be surprised if the book assumes the Hilbert space is separable. That’s a very common assumption.) – David Gao Jun 07 '24 at 14:35
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    @haru When I said the “correct measure”, I meant a Borel measure - in fact, even a Radon probability measure on $\text{Spec}(a)$. But “Borel measure” or “Radon measure” is a class of measures, not just including the Lebesgue measure, so of course there are still measures that are correct to use and those that are not. (In fact, the measure $\mu$ is given by a separating unit vector $h$ of $W^\ast(a, 1)$. It is defined by $\mu(f) = \langle f(a)h, h \rangle$ for any continuous $f$ on $\text{Spec}(a)$.) – David Gao Jun 07 '24 at 14:39
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    @haru For references, most of these can be found in section 2.5 and section 4.4 of Murphy’s “$C^\ast$-algebras and operator theory”, as well sections 9, 10, 14, and 15 of Conway’s “A course in operator theory”. (The injectivity conditions are not really in there, but one can obtain those conditions using the tools contained in those references.) – David Gao Jun 07 '24 at 14:47