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A revision:

According to the comment of Nate Eldredge, in order to avoid the triviality, we revise the property $P$.

Assume that $A$ is a commutative unital Banach algebra. Its maximal ideal space is denoted by $\Delta_{A}$. For $a\in A$, the Gelfand transform of $a$ is denoted by $\hat{a}$. An element $a\in A$ is called an injective (constant) element if $\hat{a}$ is an injective(constant) function from $\Delta_{A}$ to $\mathbb{C}$. Consider the following property "$P$" for a commutative unital Banach algebra:

$A$ satisfies $P$ if $A$ has at least one injective element and the limit of every sequence of injective elements is either injective or constant.

The main motivation for this property (and this post) is that the algebra of all bounded holomorphic functions on a connected open subset of the plane, satisfies this property

Some questions:

1.What is an example of a commutative $C^{*}$ algebra, different from $\mathbb{C}$, with this property?

2.Does The disc algebra satisfies $P$?(We know that a limit of a sequence of injective holomorphic maps defined on an open connected set is either injective or constant. So I think that we should be carefull about the boundary points of the disc ). If the answer to this question is negative, is it true to say that every infinite dimensional banach algebra with this property is isomorphic to $Hol(U)$, the algebra of bounded holomorphic functions on an open subset of the plane?

3.Is there any relation between "existence of injective element in $A$" and the real or topological stable rank of $A$?

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    $\begingroup$ Property P could be satisfied vacuously. Consider the C* algebra $C(S^n)$ for n at least 2. There is no continuous injection from $S^n$ to C (by compactness it would be an embedding which is absurd) so there are no injective elements. $\endgroup$ – Nate Eldredge May 15 '14 at 8:07
  • $\begingroup$ @NateEldredge thank you for your comment. it seems there are obstructions which are related to dimension(that is real or topological stable rank of the banach algebra.for example: is it true to say for real rank at least 3, there is no an injective element in a commutative unital banach algebra(not necessarily C* algebra)? $\endgroup$ – Ali Taghavi May 15 '14 at 8:42
  • $\begingroup$ @NateEldredge So $C(S^{n}),\;\; n\geq 2$, as you said satisfies property $P$. because there is no an injective element. $C[0,1]$ or $C(S^{1}$ does not satisfy. (consider figure 8 space as alimit of topological circle $\endgroup$ – Ali Taghavi May 15 '14 at 8:51
  • $\begingroup$ $\mathbb C^2$ has your revised property P. Maybe it would be helpful to discuss why you are interested in such properties? $\endgroup$ – Nate Eldredge May 15 '14 at 14:24
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    $\begingroup$ Yes, I agree that $\mathbb{C}$ and $\mathbb{C}^2$ are the only commutative $C^*$ algebras with property P. I was going to sketch a proof but the details are tedious so I'll let you do it. At least for unital commutative $C^*$ algebras $C(K)$, any injective element of $C(K)$ is an embedding of $K$ into $\mathbb{C}$ as I mentioned, so $C(K)$ has an injective element iff $K$ is (homeomorphic to) a compact subset of $\mathbb{C}$. $\endgroup$ – Nate Eldredge May 15 '14 at 15:05
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$A(D)$ does not satisfy P. I will actually work on the upper semi-disk, but of course one could map things back to $D$. Consider $f_{\epsilon}(z)=1-z^2-i\epsilon z$. The other point with the same image as $z$ is $w=-z-i \epsilon$, which is never in the (closed) semi-disk if $z$ is, so all $f_{\epsilon}$ are injective and they converge to $f(z)=1-z^2$, which is not.

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  • $\begingroup$ thank you for your answer. Are we sure that $f_{n}$ converges? $\endgroup$ – Ali Taghavi May 16 '14 at 0:33
  • $\begingroup$ it is an equicontinuous familly so it has a subsequence converging? $\endgroup$ – Ali Taghavi May 16 '14 at 0:39
  • $\begingroup$ Ali's comments refer to an earlier, less explicit version of my answer. $\endgroup$ – Christian Remling May 16 '14 at 4:16

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