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In the context of $C^*$-algebras , why is the $C^*$-identity a "natural" one to choose ? ($||a^* a||=||a||^2$).

Some books try to motivate this by noting that bounded operators on a Hilbert space have the above property. Is this an optimal choice in some sense ? ($||a^*||=||a||$ could, for example, be another candidate for defining $C^*$-algebras).

From what I gather, historically $B^*$-algebras were defined first. ($||a^* a||=||a|| \cdot ||a^*||$). In that case, the same question may be directed towards the $B^*$-identity for $B^*$-algebras.

The fact that this identity is strongly used throughout and I haven't found a natural explanation for the choice, slightly dims the awesomeness of the GNS representation theorem for $C^*$-algebras for me.

Thanks.

nsoum
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Recall that in a unital $C^\ast$-algebra $A$, you have the spectral radius formula, i.e., for $a \in A$ normal, $$ \|a\| = \rho(a) := \sup\{|\lambda| \mid \text{$a-\lambda 1$ is not invertible in $A$}\}. $$ So, the $C^\ast$-identity $\|a\|^2 = \|a^\ast a\|$ gives you the norm of an arbitrary element $a$ in terms of the norm of self-adjoint element $a^\ast a$, which can be computed in terms of the spectral radius formula: $$ \|a\| = \sqrt{\|a^\ast a\|} = \sqrt{\rho(a^\ast a)} = \sqrt{\sup\{|\lambda| \mid \text{$a^\ast a-\lambda 1$ is not invertible in $A$}\}}. $$ Hence, phrasing things in terms of the $C^\ast$-identity at least emphasizes, if not makes manifest, the deep fact that the norm of a unital $C^\ast$-algebra $A$ is entirely determined by the (purely!) algebraic properties of the unital $\ast$-algebra $A$, or in other words, a unital $\ast$-algebra $A$ admits at most one norm making it into a pre-$C^\ast$-algebra.

Branimir Ćaćić
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  • In the book, Fundamentals of the Theory of Operator Algebras by Kadison, Ringrose the inequality $||ab|| \le ||a|| \cdot ||b||$ for Banach algebras is motivated by continuity of the left/right multiplication operator. I was wondering if my question is amenable to such an answer. $||a^* a||=||a^|| \cdot ||a||$ is the equality case of the Banach norm inequality for $a^$ and $a$. Would that mean something in that sense? – nsoum Jul 31 '13 at 05:43
  • I don't really know what you're asking. I mean, sure, $|a^\ast| = |a|$ is precisely the statement that the $\ast$-operation is isometric, but neither the $B^\ast$ nor $C^\ast$-algebras admit ready interpretations in terms of continuity of something, as far as I can gather, though a $C^\ast$-algebraist might well correct me on this point. I do still think that the main motivation for the $C^\ast$-identity is the facilitation of the spectral radius formula. Also, note that the resulting uniqueness of the norm very much distinguishes $C^\ast$-algebras amongst general Banach $\ast$-algebras. – Branimir Ćaćić Jul 31 '13 at 07:56
  • But seriously, the $B^\ast$- or $C^\ast$-algebra identity—they're equivalent, so your choice is purely a matter of taste—are precisely what make the theory of $C^\ast$-algebras so much more tractable than the theory of Banach algebras, precisely analogously to how having the norm induced by a positive-definite inner product is what makes the theory of Hilbert spaces so much more tractable than the theory of Banach spaces. Indeed, it's precisely the inner product on a Hilbert space $H$ that induces the $C^\ast$-structure on $B(H)$... – Branimir Ćaćić Jul 31 '13 at 08:10
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    I mean, if you want, you can view the Gel'fand–Naimark representation theorem as no more or no less than the statement that the $B^\ast$/$C^\ast$-identity is a necessary and sufficient condition for a Banach $\ast$-algebra to be isometrically $\ast$-isomorphic to a closed $\ast$-subalgebra of $B(H)$ for some Hilbert space $H$. – Branimir Ćaćić Jul 31 '13 at 08:14