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Suppose that $A$ is a commutative, unital $C^*$-algebra. Then it is isomorphic to $C(X)$ for some compact Hausdorff topological space $X$. $X$ can be identified as the space of all unital homomorphisms $A \to \mathbb{C}$ with the topology of pointwise convergence. Any such homomorphism automatically preserves involution and is of norm one. We call $X$ to be the spectrum of $A$. However, if we drop the assumption of being commutative it may happen that there are no such homomorphisms so we are forced to redefine the spectrum. One posible aproach is via pure states (which, in commutative case reduce to the classical spectrum) but a priori they hev nothing to do with algebra structure on $A$. The most common definition of the spectrum is via representations: we collect all unital $*$ representations $\pi:A \to B(H_{\pi})$ which are irreducible and then divide by the equivalence relation $\sim$ where $\pi \sim \sigma$ if there is a unitary operator such that $U \pi=\sigma U$. One can show that indeed, for commutative $C^*$ algebra this coincides with the classical spectrum but may be very strange space for noncommutative algebras. My first impression was the following: we divide by the equivalence relation because we can form two different but isomorphic Hilbert spaces such as $\ell^2$ or $L^2[0,1]$ and consider the representations which acts "in the same way" meaning that $\pi(a)$ and $\sigma(a)$ are unitarily equivalent and we wish to treat these two representations as being the same. Moreover, I though that the fact that spectrum is badly behaved is due to the fact that it consists from differnt "pieces" (we can collect irreducible representations of various dimensions). It appears that this is not true: bad behaviour of the spectrum is due to the equivalence relation. One can choose, for each $n \in \mathbb{N} \cup \{\infty\}$ a standard Hilbert space $H_n$ of dimension $n$ and consider the set $Rep_n(A)$ and $Rep(A)=\bigcup_n Rep_n(A)$. With the assumption of $A$ being separable it can be shown that $Rep(A)$ is a Polish space and the suitable metric dictates the topology of pointwise, strong operator topology convergence (each $f_{a,\xi}:\pi \mapsto \pi(a)\xi$ should be continuos). Moreover, the set $Irr(A)$ of all irreducible representations is $G_{\delta}$ in $Rep(A)$ and therefore is also Polish space (see the book of Arveson). So my qestion is the following: what is the purpose for trading nice Polish space for badly behaved (almost never Hausdorff) space? Why spectrum is defined as the quotient space?

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    $\begingroup$ The spectrum is made for the representation theory, not representation theory for the spectrum, and from the standpoint of the representation theory of $C^\ast$-algebras, unitarily equivalent representations really are equal in every meaningful way. In other areas of mathematics, moduli spaces can be extraordinarily difficult to handle, but this doesn't make them any less meaningful or interesting. $\endgroup$ – Branimir Ćaćić Mar 12 '14 at 17:07
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    $\begingroup$ In any event, the fact that any reasonable notion of spectrum can act very strangely for noncommutative $C^\ast$-algebras should rather be seen as an indication that "noncommutative topology" contains fundamentally new phenomena that necessitate new tools. $\endgroup$ – Branimir Ćaćić Mar 12 '14 at 17:12
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I think that the thing to do is to consider an elementary example such as the C*-algebra $A$ of continuous functions from the unit interval $[0,1]$ to the $n\times n$-matrices or to the compact operators on a Hilbert space. Then the spectrum is isomorphic to $[0,1]$ and one can see why it deserves that name and why it is more useful than the space $Rep(A)$; for example because of its direct connection with the ideal structure of $A$. For a slightly non-Hausdorff example, look at the C*-algebra $A$ of continuous function from $[0,1]$ to the $2\times 2$-matrices which are diagonal-valued at the point $1$. Then the spectrum of $A$ has a double-point at $1$ but is still very useful. Furthermore, when the spectrum of $A$ is non-Hausdorff, one can consider its complete regularisation, which turns out to be the spectrum of the centre of $A$.

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