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I have 2 problems in Gelfand theory. I shall be thankful for any answers. 1)What is the gelfand spectrum of l^1(N)? A few of the elements are evaluations of functions(defined below) on closed unit disc. For an element a = {a(n)} of l^1, corresponding function is summation a(n)x^n over natural numbers. Are there any others?

2)Can we find a non unital Banach Algebra with a compact Gelfand spectrum? Sincerely, Madhuresh.

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The second question is very nice. Is it possible to construct a unit using compactness and holomorphic functional calculus? –  Martin Brandenburg Nov 22 '10 at 17:49
My recollection is that the answer to 2) should be "no" but I will need to either think some more on this, or look it up in Bonsall and Duncan –  Yemon Choi Nov 22 '10 at 18:56
[deleted earlier comment] –  Yemon Choi Nov 22 '10 at 19:03
Let me also just say that personally I find the level of Q1 a bit basic for MO; that's not to say the questioner is asking in bad faith, but Q1 is something I would set as an exercise in any course I taught which included the Gelfand representation for a CBA. (I guess the key point is that each character on a Banach algebras has norm $\leq 1$.) –  Yemon Choi Nov 22 '10 at 19:04

1 Answer 1

The Gel'fand spectrum of $\ell^1 \mathbb N$ is indeed the closed unit disc. After all, every functional to $\mathbb C$, must be given by sending the generator to some complex number. It is easy to see, that this works if and only if this complex number lies in the unit disc.

Your second question is rather nice. Anyhow, I think that there cannot be any commutative example (at least if it embeds into the algebra of continuous functions on the Gel'fand spectrum). As soon as there is the unit in the algebra of functions on the Gel'fand spectrum, then the Banach algebra contains at least an invertible element, and hence also the unit.

However, and now it is getting more interesting: there are non-unital Banach algebras whose universal $C^\star$-algebra has a unit. (Note that this is precisely the non-commutative analogue of your question in Gel'fand theory.) Indeed, consider a group $\Gamma$ with Kazhdan's property (T), its $\ell^1$-algebra and the augmentation ideal $\omega(\Gamma) \subset \ell^1 \Gamma$. It is well-known that $C^\star(\Gamma)$ splits of the unit as a direct summand. The remaining summand is the universal $C^*$-algebra of $\omega(\Gamma)$ and it is unital.

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Reps of $\omega(\Gamma)$ correspond to what in terms of reps of $\Gamma$? –  André Henriques Nov 22 '10 at 18:41
The cyclic $\star$-representations of $\omega(\Gamma)$ correspond to positive linear functionals on the $\star$-algebra $\omega(\Gamma)$. Those, by some tool analogous to the GNS-construction, can be used to construct an affine action of $\Gamma$ on a Hilbert space. Since $\Gamma$ has property $(T)$, this action has a fixed point and is thus just a unitary representation of $\Gamma$. Hence, any cyclic $\star$-representation of $\omega(\Gamma)$ is the restriction of a unitary representation of $\Gamma$. –  Andreas Thom Nov 22 '10 at 18:47
The special feature of representations of $\Gamma$ contstructed this way is that they have no fixed vectors. –  Andreas Thom Nov 22 '10 at 18:48
Regarding your last point: let $\Gamma$ be any group whose reduced $C&*$-algebra is simple (e.g. a non-abelian free group). Then I think the closure of $\omega(\Gamma)$ inside $C_r^*(\Gamma)$ would have to be a nonzero closed ideal, hence the whole algebra. Does this imply that the universal $C^*$-algebra of $\omega(\Gamma)$ will also have an identity? –  Yemon Choi Nov 22 '10 at 18:52
@Yemon: I considered $\mathbb N = \lbrace 0,1,2,\dots, \rbrace$, so that zero is in the spectrum and the algebra is unital. –  Andreas Thom Nov 22 '10 at 18:57

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