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Can someone give me an example of a Banach Algebra which does not have an isometric representation in a Hilbert Space ?

(if possible, can you add a proof or a reference ? )

Thank you very much !


Note added by YC: this question has also been asked on MSE where someone has given a much better, elementary example to the OP's question. (However, my example also works for the question of topologically isomorphic representations, not just the isometric ones.)

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The Wiener algebra $A(\mathbb{T})$ is not an operator algebra. –  user33709 Jun 18 '13 at 16:21
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I might be mistaken, but I think if you take any Banach $^*$ algebra that does not have a $C^*$-norm (eg $$L^1(G)$ for a locally compact group $G$) then you cannot represent the $\textit{algebra}$ on a Hilbert space. Note for the example I gave above you can represent the Banach space $L^1(G)$ on a Hilbert space, but as far as I know it will not be a representation of the algebra. –  Owen Sizemore Jun 18 '13 at 16:23
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Thank you Owen. To be honest, im just an undergraduate student trying to learn something on my free time, so im sorry if my questions are easy at an absurd level. Anyway, where can you give me at least some references regarding the algebras on groups ? Thank you. –  jpp Jun 18 '13 at 16:41
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@jpp: Conway's book "A course in Functional Analysis" is a fairly standard text for a grad course in Functional Analysis. It has the basics on of what I was saying. In particular, it shows that the usual norm on $L^1(G)$ is not a $C^*$-norm and thus cannot be the norm on a subalgebra of B($\mathcal{H})$. –  Owen Sizemore Jun 18 '13 at 16:49
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@jpp: if you're posting a question both here and on math.stackexchange, please tell this in the post, and include a link. This will prevent duplication of effort by people. (Shouldn't this be in the faq or something?) –  Jan Jitse Venselaar Jun 18 '13 at 20:45

1 Answer 1

up vote 6 down vote accepted

Any Banach algebra which is not Arens regular cannot be embedded as a closed subalgebra of B(H), even if you allow for isomorphic embeddings that have closed range yet are not isometric.

If you are only interested in Banach $\ast$-algebras and isometric $\ast$-homomorphic embeddings, then it is easier to find examples, as Owen Sizemore has indicated.


[Not directly relevant, but perhaps of background interest to the OP] By the way, although the question asks about isometric embeddings, there are interesting and slightly unexpected examples of Banach algebras $A$ for which there is an injective homomorphism $A\to B(\ell^2)$ that has closed range, showing that non-selfadjoint operator algebra theory has to be a lot wilder than the self-adjoint case. Examples include: $\ell^p$ for $1\leq p <\infty$ with pointwise product; and the algebras $C^k([0,1]^m)$ of $k$-times continuously differentiable functions on the $m$-cube. You can even get radical commutative Banach algebras embedded into $B(\ell^2)$ in this way, see

MR0410386 (53 #14136) P. G. Dixon, Radical Q-algebras. Glasgow Math. J. 17 (1976), no. 2, 119--126.

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Thank you very much, Yemon ! –  jpp Jun 18 '13 at 19:03

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