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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.


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A general Banach algebra does not come with a * operation. Did you mean to ask about general *-Banach algebras? –  tomasz Jul 1 '12 at 0:03
@tomasz I am actually asking about those banach algebras with involution but fail the norm identity for $C^*$. –  Hui Yu Jul 1 '12 at 0:19
Take the Laurent polynomials in one variable (=trigonometric polynomials) and give them the norm called "sum of absolute values of coefficients". Take the completion. The resulting algebra, with pointwise product, is isometrically isomorphic to $\ell^1({\bf Z})$ with convolution. –  user16299 Jul 1 '12 at 4:50
I recommend you edit your question to make it clear that you are asking about Banach $*$-algebras. There are many examples of Banach algebras (not *-algebras) which have no hope of being anything like a $C^*$-algebra, because of some kind of asymmetry. –  user16299 Jul 1 '12 at 4:52
(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.) –  user16299 Jul 1 '12 at 9:30
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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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Yes. They are both good examples. But the problem is that the multiplication on $\ell^1$ and the involution on the function algebra you mentioned are both strange to me. Maybe there are some more 'natural' examples? –  Hui Yu Jul 1 '12 at 0:22
@Hui: Then you could do worse than familiarize yourself with the strange multiplication and the strange involution... :) Maybe start with the latter and look at what the involution looks like if you develop $f$ as a power series around $0$. Then read about convolution in its various guises. –  t.b. Jul 1 '12 at 2:34
Slight correction/addition to Jonas's last comment: take the algebra of Schatten class operators $S_p(H)$ on any infinite-dimensional Hilbert space, for $1\leq p <\infty$. Then this is never a $C^*$-algebra. In fact, using some Banach space theory and a fairly deep facts about $C^*$-algebras, $S_p(H)$ cannot even be isomorphic as a Banach space to any $C^*$-algebra. –  user16299 Jul 1 '12 at 4:49
@Yemon: Thank you for the informative comment. One thing I don't see is, what part of your comment is a correction? (If something needs to be corrected, I'd like to know.) –  Jonas Meyer Jul 1 '12 at 4:52
@Jonas: I just meant that while some of us do believe $\ell^1({\bf Z})$ is more interesting than $M_n$ ;-) I don't think the reason you give is the compelling one. Indeed, $M_n$ with Hilbert-Schmidt norm has to be very far in Banach-Mazur distance from a $C^*$-algebra. –  user16299 Jul 1 '12 at 4:54
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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.

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