After the GNS representation for $C^{\*}$-algebras is presented in Thirring's book [Quantum mathematical physics](https://doi.org/10.1007/978-3-662-05008-8), the author states the following theorem.



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> **The Spectral Theorem:** For any given Hermitian (self-adjoint) element $a$ of a $C^{\*}$-algebra $A$, every representation of $A$ is equivalent to a representation $\mathscr{H} = \bigoplus\_{i}\mathscr{H}\_{i}$, for which $\mathscr{H}\_{i} = L^{2}(\sigma(a),d\mu\_{i})$ and $\pi(a)\vert\_{\mathscr{H}\_{i}}: \varphi(\alpha) \mapsto \alpha \varphi(\alpha)$. In this representation, $a$ acts as a multiplication operator.
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I want to understand this theorem, but I did not follow Thirring's arguments which led to its proof. The argument uses the GNS construction: $A$ is a $C^{\*}$-algebra with unit and $\omega$ a state, there exists a representation $\pi\_{\omega}: A \to \mathscr{B}(\mathscr{H})$, where $\mathscr{H}$ is just the completion of $A/J$, $J$ being the left ideal defined by the set of $a \in A$ such that $\omega(a^{\*}a) = 0$. In his notation, $\pi\_{\omega}(a): b \mapsto ab$.



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> By the axiom of choice, we can choose $b\_{i} \in \mathscr{H}\_{i} \equiv $ the completion of the sets of linear combinations of $a^{n}b\_{i}$, $n=0,1,\dotsc$ spans all of $\mathscr{H}$. Each $\mathscr{H}\_{i}$ provides a representation of the (Abelian) $C^{\*}$-algebra generated by $a$ and has $b\_{i}$ as a cyclic vector.
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I really don't follow the argument. $\mathscr{H}\_{i}$ spans $\mathscr{H}$ in which sense? Is it always possible to find $b\_{i}$ such that $\mathscr{H}\_{i}$ spans $\mathscr{H}$? Is the index $i$ countable, uncountable? $\mathscr{H}\_{i}$ is a Hilbert space, so does a representation between $A$ and $\mathscr{H}\_{i}$ always exist?


Can someone help with these arguments? Maybe giving more details at each step? Or maybe providing a reference in which the theorem is proved more carefully?



GNS Representation — A theorem from Thirring’s book
The C${}^\*$-algebra $A$ is a red herring here. All the result is really saying is that if $T$ is a self-adjoint operator on a Hilbert space $H$ then we can find a family of measures $\mu\_i$ on $\sigma(T)$ and an isomorphism $H \cong \bigoplus L^2(\sigma(T), d\mu\_i)$ which takes $T$ to the operator of multiplication by $x$ on each $L^2(\sigma(T), d\mu\_i)$. Not my favorite version of the spectral theorem on aesthetic grounds, but it is easy to use.


To prove this, let $v$ be any nonzero vector in $H$ and let $A\_0$ be the C${}^\*$-algebra generated by $T$. Then $A\_0$ is abelian and by Gelfand's theorem it is isomorphic to $C(\sigma(T))$. Thus the map $\omega: S \mapsto \langle Sv, v\rangle$ is a positive linear functional on $A\_0$ of norm at most $\|v\|^2$, so it is given by $\omega(f(T)) = \int\_{\sigma(T)} f\, d\mu$ for some positive measure $\mu$ on $\sigma(T)$. Now $A\_0v = \{f(T)v: f \in C(\sigma(T))\}$ is a (not necessarily closed) subspace of $H$ and one can check that the map $f(T)v \mapsto f$ is an isometry from $\overline{A\_0v}$ onto $L^2(\sigma(T), d\mu)$. Also the restriction of $T$ to $\overline{A\_0v}$ is taken to the operator of multiplication by $x$ on $L^2(\sigma(T), d\mu)$.


Now $\overline{A\_0v}$ is invariant for $T$, and since $T$ is self-adjoint so is its orthocomplement $H \ominus \overline{A\_0v}$. So we can now choose another nonzero vector $w$ in the orthocomplement of $\overline{A\_0v}$, and keep going until there is nothing left in the orthocomplement. This argument can be made rigorous using Zorn's lemma, or just a simple transfinite induction.


If $H$ is nonseparable this is going to require uncountably many summands, since each summand is separable. If $H$ is separable then obviously there will only be countably many summands.


If I recall correctly, this version of the spectral theorem is given a simple proof, probably essentially the same as the one I just gave, in Reed and Simon vol. 1.


