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Hurkyl

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No, it won't work for non-C* algebras.

It can't be isometric for anything else, because C(M) is always a C* algebra.

For example, consider the set of integrable functions f:R->C with convolution as the 'multiplication' operator.

This is a Banach algebra with the L^{1} norm, [itex]\Vert f\Vert = \int |f(x)|\,dx[/itex]. However, ||a* a|| <= ||a||^{2}, and equality doesn't always hold, so it isn't a C*-algebra.

It can't be isometric for anything else, because C(M) is always a C* algebra.

For example, consider the set of integrable functions f:R->C with convolution as the 'multiplication' operator.

This is a Banach algebra with the L

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Also, in my example, you can see that the Gelfand transform is a Fourier transform, so C(M) is isomorphic to C(R) and the original algebra maps to those elements of C(R) which can be written as a Fourier transform of an integrable function.

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I think the way you phrased your question, you were only talking about unitial alegebras (otherwise you should replace C(M) by the continuous functions vanishing at infinity).

In that case, you could let A be the alegebra of functions f:Z->C such that [itex]\sum_n |f(n)| < \infty[/itex], again with convolution as the multiplication operator.

Using fourier series, C(M) is (isomorphic to) the set of continuous functions f:R->C of period 1 or, equivalently, C(S^{1}). However, the Gelfand transform maps A only to those elements whose fourier series is absolutely convergent.

Interesting to note that in both my examples, C(M) is isomorphic to the C*-algebra generated by the Banach alegebra. Wouldn't be surprised if that is always the case.

In that case, you could let A be the alegebra of functions f:Z->C such that [itex]\sum_n |f(n)| < \infty[/itex], again with convolution as the multiplication operator.

Using fourier series, C(M) is (isomorphic to) the set of continuous functions f:R->C of period 1 or, equivalently, C(S

Interesting to note that in both my examples, C(M) is isomorphic to the C*-algebra generated by the Banach alegebra. Wouldn't be surprised if that is always the case.

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If B is semisimple, then its Gelfand transform f(B) is dense in C(M) - so if Gelfand transform isn't "onto", f(B) isn't closed in C(M)?

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I'm not sure if this is standard terminology. You could try taking the completion w.r.t. the largest continuous C*-norm, which should give a closed subspace of C(M), and maybe is isometric to C(M).Thanks. What do you mean by "C* algebra generated by a Banach algebra"?

It isn't closed in my examples above. It is closed (and complete) under the original Banach norm, but not the C*-norm on C(M).If B is semisimple, then its Gelfand transform f(B) is dense in C(M) - so if Gelfand transform isn't "onto", f(B) isn't closed in C(M)?

Do you need the semisimple condition?

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I need semisimplicity only to assure that the translation is injective.

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That's what I meant by the C* algebra generated by A.

If this C* algebra is B, then the map A -> C(M) extends uniquely to a continuous homomorphism B->C(M), which I was suggesting gave an isometry.

However, you've already stated that semisimple => map is injective, and that the map is dense, so it seems that you know quite a bit already.