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I can only provide an interpretation of the discreteness for the spectrum of a compact group in terms of discrete and continouos projection valued measures. Note every measure can be uniquely decomposed in a atomic measure and a measure without atoms.

Perhaps, one should also mention that the dual of a locally compact group has also been regarded long time just as measure space. The decomposition of the group von Neumann algebra $L(G)$ of a compact group $G$ decomposes then into a direct sum of unitary irreducible representation, which are finite dimensional $$L(G) = \bigoplus_{\pi \; irred.} M_{dim (\pi)} ( End_{G} (\pi)).$$ This decomposition is in my opinion the analogue of a noncommutative functional calculus, with the projection valued measures appearing discretely. So the spectrum of compact groups is discrete in this sense. So maybe, you're looking for the decomposition of the von Neumann group algebra of a locally compact group? For general noncompact groups, there will (necessary?) appear direct integrals. Wether these direct integrals are over compact spaces, if $G$ is discrete, I can not answer!

Factors are classified by type $1-3$. Usually people talk say that the The classification is just feasible for type $1$ groups, which are nice to describe, meaning that every factor appearing in the decomposition is type $1$. 1$, are easier to handle than the other types. Compact groups and abelian are hence of type $1$. Being type $1$ is also equivalent to certain regularity conditions on the measure space structure of the irreducible dual.

But the irreducible unitary representation of a discrete groups can be really hard to describe, e.g. a discrete group is only of type I if it contains a normal abelian subgroup of finite index. I'll give an example: take the free group $F_n$ in $n$ generators, then the group vNa $L(F_n)$ is a factor, so in this case your "integral" is over one point. It is not known wether $L(F_n) \cong L(F_m)$ for any $n \neq m$.

Addenum: I also have another idea for a topological picture. The Kernel-Hull topology on the primitive ideal space of $L(G)$ would be discrete for $G$ being compact.

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I can only provide an interpretation of the discreteness for the spectrum of a compact group in terms of discrete and continouos projection valued measures. Note every measure can be uniquely decomposed in a atomic measure and a measure without atoms.

Perhaps, one should also mention that the dual of a locally compact group has also been regarded long time just as measure space. The decomposition of the group von Neumann algebra $L(G)$ of a compact group $G$ decomposes then into a direct sum of unitary irreducible representation, which are finite dimensional $$L(G) = \bigoplus_{\pi \; irred.} M_{dim (\pi)} ( End_{G} (\pi)).$$ This decomposition is in my opinion the analogue of a noncommutative functional calculus, with the projection valued measures appearing discretely. So the spectrum of compact groups is discrete in this sense. So maybe, you're looking for the decomposition of the von Neumann group algebra of a locally compact group? For general noncompact groups, there will (necessary?) appear direct integrals. Wether these direct integrals are over compact spaces, if $G$ is discrete, I can not answer!

Factors are classified by type $1-3$. Usually people talk say that the classification is just feasible for type $1$ groups, which are nice to describe, meaning that every factor appearing in the decomposition is type $1$. Compact groups and abelian are hence of type $1$. Being type $1$ is also equivalent to certain regularity conditions on the measure space structure of the irreducible dual.

But the irreducible unitary representation of a discrete groups can be really hard to describe, e.g. a discrete group is only of type I if it contains a normal abelian subgroup of finite index. I'll give an example: take the free group $F_n$ in $n$ generators, then the group vNa $L(F_n)$ is a factor, so in this case your "integral" is over one point. It is not known wether $L(F_n) \cong L(F_m)$ for any $n \neq m$.

Addenum: I also have another idea for a topological picture. The Kernel-Hull topology on the primitive ideal space of $L(G)$ would be discrete for $G$ being compact.

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I can only help with the provide an interpretation of the discreteness for the spectrum of a locally compact group in terms of discrete and continouos projection valued measures. Note every measure can be uniquely decomposed in a atomic measure and a measure without atoms.

Perhaps, one should also mention that the dual of a locally compact group has also been regarded long time just as measure space. The decomposition of the group von Neumann algebra $L(G)$ of a compact group $G$ decomposes then into a direct sum of unitary irreducible representation, which are finite dimensional $$L(G) = \bigoplus_{\pi \; irred.} M_{dim (\pi)} ( End_{G} (\pi)).$$ This decomposition is in my opinion the analogue of a noncommutative functional calculus, with the projection valued measures appearing discretely. So the spectrum of compact groups is discrete in this sense. So maybe, you're looking for the decomposition of the von Neumann group algebra of a locally compact group? For general noncompact groups, there will (necessary?) appear direct integrals. Wether these direct integrals are over compact spaces, if $G$ is discrete, I can not answer!

Factors are classified by type $1-3$. Usually people talk say that the classification is just feasible for type $1$ groups, which are nice to describe, meaning that every factor appearing in the decomposition is type $1$. Compact groups and abelian are hence of type $1$. Being type $1$ is also equivalent to certain regularity conditions on the measure space structure of the irreducible dual.

But the irreducible unitary representation of a discrete groups can be really hard to describe, e.g. a discrete group is only of type I if it contains a normal abelian subgroup of finite index. I'll give an example: take the free group $F_n$ in $n$ generators, then the group vNa $L(F_n)$ is a factor, so in this case your "integral" is over one point. It is not known wether $L(F_n) \cong L(F_m)$ for any $n \neq m$.

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