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Given a finite group $G$ we have that the $Z(\mathbb{C}G)$ (the center of the algebra group) has two basis the characteristic funtions on the conjugacy classes and the minimal idempotents for the convolution (that are the irreducible characters of $G$ "normalized").

And $Z(\mathbb{C}G)$ can be seen as an algebra for the pointwise product, or for the convolution product. Is there any natural/cannonical isomorphism between these algebras, or something done in this setting (besides having to choose a correspondence between irreducible characters and conjugacy classes and extending that map to $Z(\mathbb{C}G)$)

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    What do you mean "algebra for the pointwise product, or for the convolution product", exactly? I don't understand why any operation besides the one in $\mathbb C[G]$ comes into play. – 2017-02-22
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    If you see $\mathbb{C}[G]$ as the functions from $G$ to $\mathbb{C}$, so you can consider the pointwise product of functions $(f.g)(x)=f(x)g(x)$, and the convolution product $(f*g)(x)=\sum f(xy^{-1})g(x)$ – 2017-02-27
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    So you mean you want to compute $Z\mathbb C[G]$ with the ordinary operation on $\mathbb C[G]$, then switch to considering its multiplication as pointwise function multiplication? Odd. I would guess that this has more idempotents than the ordinary product, but I haven't worked through an example yet. I would start with $\mathbb C[S_3]$ to try to confirm or disprove this idea. – 2017-02-27
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    It has the same idempotents because in the convolution product the idempotents are in bijection with the irreducible characters and with pointwise they are the characteristic functions of the conjugacy classes. – 2017-02-27
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    I'm trying to figure out what correspondence that is. It makes sense to me to put the centrally primitive idempotents in correspondence with the (unique isoclass of) irreducible representation in its corner ring. However, there are many nonprimitive central idempotents that would be left out of this correspondence. So... what's the deal? – 2017-02-27

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