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Suppose that we have two Banach algebras A and B and subsets S_1 of A and S_2 such that there exists a "natural" bijection between S_1 and S_2. We are interested in A_1 and B_1 which are smallest closed subalgebras of A (B) generated by S_1 (S_2). In my particular case I would like to show that A_1 and A_2 are (isometrically) isomorphic. Problem is that in my situation I don't have complete description of neither A_1 or A_2. Do you have any suggestions on how to tackle problem of this form?

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I don't see how this can be answered without making the term "natural" more precise... For example, $S_1$ and $S_2$ could both be $\mathbb C[X]$, and so are "equal", but living inside $A=C[0,1]$ (with pointwise product) and $B=L^1[0,1]$ (with restricted convolution as the product). –  Matthew Daws May 23 '11 at 12:41
natural only means that there is a bijection f from S_1 to S_2 such that f should be a restriction to S_1 of desired isomorphism from A_1 to B_1 –  ivo May 23 '11 at 12:45
So do you mean to ask: $f:S_1\rightarrow S_2$ is a bijection. When does there exists an (isometric) isomorphism of Banach algebras $g:A_1 \rightarrow A_2$ such that $g(x)=f(x)$ for each $x\in S_1$??? –  Matthew Daws May 23 '11 at 12:51
So your defintion of "natural" is a posteriori, and not knowing your desired isomorphism yet, there are only two possibilities: Either, you absolutely have no clue how this isomorphism should look like. In this case yoiu can drop the term "natural". Or, you have a (vague) idea what this isomorphism should do to $S_1$, and thus you maybe hiding this valuable information... –  Abel Stolz May 23 '11 at 12:55
Another counter-example: $S_1$ and $S_2$ are the "point masses" in $\ell^p$ or $c_0$ (all with pointwise product) or in $\ell^1(\mathbb Z) \cong \ell^1(\mathbb N)$ with convolution product. There seems no hope in general; so you're going to have to tell us more background I think... –  Matthew Daws May 23 '11 at 19:46
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