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The Gelfand-Naimark theorem says that if $A$ is a commutative unital $C^*$-algebra, then $C(Spec(A))=A$, where $Spec(A)$ is the set of all characters on $A$.

Does the theorem fail for commutative unital Banach algebras?

To be clear, is there a counter-example to see that $C^*$ part is indeed important?

Guest
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  • To be precise, the commutative Gelfand-Naimark theorem does not assert that $A=C(\mathrm{Spec}(A))$, but only that these $C^\ast$-algebras are $\ast$-isomorphic. Of course this notion makes sense only for $\ast$-algebras, so the first question would be in which sense you expect $A$ and $C(\mathrm{Spec}(A))$ to be isomorphic. Of course the answer show that even in a fairly weak sense (say isomorphism of Banach spaces) they are not isomorphic in general.# – MaoWao Jul 01 '21 at 11:46
  • @MaoWao Yes, that makes sense! – Guest Jul 01 '21 at 12:41

2 Answers2

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Hint.

Let $A=\ell^1(\mathbb Z)$ and for $z\in \mathbb T$, define $\phi_z: \ell^1(\mathbb Z) \to\mathbb C$ as $\phi_z(f)=\sum_n f(z)z^{-n}$ for all $f\in \ell^1(\mathbb Z)$.

Show that $\operatorname{Spec}(A)=\{\phi_z:z\in\mathbb T\}$ and then $C(\operatorname{Spec}(A)) = C(\mathbb T)$; the latter is not isomorphic to $A$.

Sahiba Arora
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  • Why is $C(\mathbb T)$ not isomorphic to $l^1(\mathbb Z)$? – Guest Jul 01 '21 at 10:57
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    Because $\ell^1$ is weakly sequentially complete while $C(\mathbb T)$ is not. – Sahiba Arora Jul 01 '21 at 11:11
  • In fact we don't need that the two are not isomorphic, just that the canonical map is not an isomorphism. Which it seems to me is just saying that there exists a continuous function with $\sum|\hat f(n)|=\infty$. – David C. Ullrich Jul 01 '21 at 12:18
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Since $C(X)$ is always a $C^*$-algebra (where $X$ is a compact Hausdorff space), it suffices to give an example of a unital Banach algebra which is not a $C^*$-algebra. See for instance Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra? and also Non-$C^{*}$ Banach algebras?.

Martin Brandenburg
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