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It suddenly occurred to me almost every Banach algebra I know is actually a $C^{*}$ algebra. Several kinds of function algebras are definitely $C^{*}$ algebras. So is the matrix algebra. Although one gets a non-$C^*$ algebra by focusing on the upper triangular matrices, the norm still satisfies the $C^*$ identity.

The algebra of operators on a general banach space is not $C^*$, but at least for me this is a too abstract class that do not provide much intuition.

Thus I wonder whether someone has some good examples of banach algebras that fail the $C^*$ identity but are on the other hand elementary enough to provide intuition and direct computation, like the function algebras.

Thanks!

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    (For instance, the algebra of upper triangular $n\times n$ matrices has a non-zero 2-sided nilpotent ideal, which immediately stops it being isomorphic in any way to a $C^*$-algebra, as the only nilpotent element in a $C^*$-algebra of the form $x^*x$ is zero.)2012-07-01

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Adding the assumption that you're asking about Banach $*$-algebras with isometric involution, your question Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra? has the example $\ell^1(\mathbb Z)$.

Another is the algebra of bounded analytic functions on the unit disk with sup norm and involution $f^{*}(z)=\overline{f(\overline z)}$.


You mention matrices, and implicitly you seem to be assuming that $M_n$ is given the operator norm from acting as operators on $\mathbb C^n$ with the standard inner product, or equivalently $\|a\|=\sqrt{\text{the spectral radius of }a^*a}$, where $a^*$ is the conjugate transpose of $a$. But if you give $M_n$ another submultiplicative norm that makes the conjugate transpose norm-preserving, then you will not have a $C^*$-algebra. One example is the Frobenius norm, a.k.a. the Hilbert–Schmidt norm, $\|a\|=\sqrt{\mathrm{Trace}(a^*a)}$.

You say that the upper-triangular matrices satisfy the $C^*$-identity, but it is not clear what that means when they don't have an involution. If you are asking about $*$-algebras, then subalgebras of $*$-algebras that are not closed under the involution are out of the picture.

The first example above, $\ell^1(\mathbb Z)$, fits into a bigger picture of considering the Banach $*$-algebra $L^1(G)$ of a locally compact Hausdorff group $G$ with Haar measure, which sometimes arises in the study of group representations.

One thing that makes $\ell^1(\mathbb Z)$ more interesting than $M_n$ with the Frobenius norm as an example is that $\ell^1(\mathbb Z)$ is not even isomorphic as an algebra to any $C^*$-algebra.

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    @t.b. Thanks! Your advice is always helpful!2012-07-01
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I must say I find it odd (and a bit worrisome) that one can find textbooks whereby you learn the definition of a Banach algebra (and even a Banach *-algebra) without seeing examples that are not $C^*$-algebras. There is more to life than $B(H)$...

Anyway, some commutative examples which are naturally algebras of functions. In every case the involution is just conjugation of functions.

1) For $G$ a locally compact abelian group (think ${\mathbb Z}^k$ or ${\mathbb T}^k$ or ${\mathbb R}^k$) with dual group $\Gamma$, take $A(G) = \{ f\in C_0(G) \mid \widehat{f} \in \ell^1(\Gamma) \} $ the so-called Fourier algebra of $G$. (One can define $A(G)$ for arbitrary locally compact groups but the definition is more technical.)

2) Algebras of Lipschitz/H\"older functions. Take your favourite compact metric space $(X,d)$, take some $0<\alpha<1$, and define $ L_\alpha(f) = \sup_{x,y\in K; x\neq y} \frac{ \vert f(x)-f(y) \vert }{d(x,y)^\alpha} $ then take $ {\rm Lip}_\alpha(X,d) = \{ f: X\to {\mathbb C} \mid L_\alpha(f)<\infty \} $ equipped with the norm $\Vert f \Vert_\alpha := \Vert f\Vert_{\infty} + L_\alpha(f)$.

3) The algebra $C^k[0,1]$ of $k$-times continuously differentiable functions on $[0,1]$ (for $k\geq 1$), equipped with the natural norm built out of the sup-norms of the derivatives.

If you are willing to consider Banach algebras without involution then there are ${\rm many}^{\rm many}$ more examples.