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I am reading HX Lin's book, named "An introduction to the classification of amenable C*-algebras", I can not understand a corollary of Gelfand theorem(Corollary 1.3.6): If a is a normal element in a unital C*-algebra A, then there is an isometric *-isomorphism from $C^{\ast}(a)$ to $C_{0}(sp(a)\setminus\{0\})$, which sends $a$ to the identity function on $sp(a)$.

Here, $C^{\ast}(a)$ denotes the smallest C*-subalgebra of A containing $a$ and $C_{0}(X)$ denotes all the continuous functions vanishing at infinity on locally compact Hausdorff space $X$. And $sp(a)$ denotes the spectrum of $a$.

My question are:

  1. How to explain the $C_{0}(sp(a)\setminus\{0\})$, I do not understand why deduct the zero point of $sp(a)$.

  2. Does $C^{\ast}(a)$ contains the unit of A? In my view, I take $f(t)=1/t$ on $sp(a)\setminus \{0\}$, then there exists an element $b$ in $C^{\ast}(a)$, corresponding to $f(t)$, hence b is the inverse element of $a$. This implies $1=a b\in C^{\ast}(a)$.

Stephen Curry
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  • I think question 2 gives the answer to question 1: $C^(a)$ does not* contain $1$. So all Gelfand transforms of elements of $C^(a)$ (viewed as functions on $sp(a)$) vanish at $0$ if $0\in sp(a)$, which explains the $C_0(sp(a)\setminus{ 0}$. As for $f(t)=1/t$, it is not in $C_0(sp(a)\setminus{ a})$; so your argument cannot show that $a$ is in invertible (and this is fine because $a$ is not* invertible if $0\in sp(a)$!). – Etienne Oct 17 '13 at 18:40
  • @Etinenne, Why $f(t)=1/t$ is not in $C_{0}(sp(a)\setminus{0})$? Is it not continuous on $sp(a)\setminus{0}$? – Stephen Curry Oct 18 '13 at 03:22
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    @Belle-tiantian: $C_0(\sigma(a)\setminus{0})$ is by definition the set of continuous functions $f$ from $\sigma(a)\setminus{0}$ to $\mathbb C$ such that for all $\varepsilon>0$, the set ${z\in \sigma(a)\setminus{0}: |f(z)|\geq \varepsilon}$ is compact. If $a$ is not invertible and $0$ is not an isolated point of the spectrum, then $1/t$ does not satisfy this definition. If $0$ is an isolated point, then you can talk about $1/t$ in the $C_0$ algebra, but it may be misleading because it gives only an inverse within the subalgebra. – Jonas Meyer Oct 18 '13 at 03:58
  • @Jonas Meyer, Why 1/t does not satisfy this definition? – Stephen Curry Oct 18 '13 at 17:51
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    @Belle-tiantian: It does not satisfy the definition if $0$ is a limit point of $\sigma(a)$, because for example ${z:|1/z|\geq 1}$ would not be closed as a subset of $\sigma(a)$, hence it would not be compact. $0$ would be a limit point not in the set. One necessary condition for a function to be in $C_0$ is boundedness, and $1/t$ is bounded on $\sigma(a)\setminus{0}$ if and only if $0$ is an isolated point of the spectrum. – Jonas Meyer Oct 18 '13 at 20:17

1 Answers1

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In this context $C^*(a)$ is defined to be the (not-necessarily-unital) C*-algebra generated by $a$. If $a$ is invertible, then $C^*(a)=C^*(1,a)$, $\sigma(a)=\sigma(a)\setminus\{0\}$, and $C_0(\sigma(a))=C(\sigma(a))$.

If $a$ is not invertible, then $0\in\sigma(a)$, and $C_0(\sigma(a)\setminus\{0\})$ can be identified with the ideal $M_0=\{f\in C(\sigma(a)): f(0)=0\}\subset C(\sigma(a))$. If $\Gamma:C^*(1,a)\to C(\sigma(a))$ is the Gelfand isomorphism, then $\Gamma(a)\in M_0$, from which it follows that $\Gamma(C^*(a))\subseteq M_0$. Since $\Gamma$ is an isomorphism and both $C^*(a)$ and $M_0$ are maximal ideals in the respective algebras $C^*(1,a)$ and $C(\sigma(a))$, this implies that $\Gamma(C^*(a))=M_0\cong C_0(\sigma(a)\setminus\{0\})$.

Note that $t\mapsto 1/t$ is not always an element of $C_0(\sigma(a)\setminus\{0\})$. It is if $a$ is invertible, or if $0$ is an isolated point in $\sigma(a)$. The latter is the tricky part, because in that case $C^*(a)$ is actually a unital $C^*$-algebra, but its unit is not the unit of $A$. Since $\sigma(a)\setminus \{0\}$ is compact in that case, you would have $C_0(\sigma(a)\setminus\{0\})=C(\sigma(a)\setminus\{0\})$, and the constant function $1$ would be in the space as the unit. Hence $\Gamma^{-1}(1)\in C^*(a)$ is a unit for $C^*(a)$ distinct from $1\in A$.

To hopefully clear up this confusion you can go back to $M_0$, where there is no function $t\mapsto 1/t$, but rather the function that sends $t$ to $1/t$ if $t\neq 0$, and sends $0$ to $0$. This is not the inverse of $\Gamma(a)$ in $C(\sigma(a))$, hence the inverse Gelfand map does not send it to $a^{-1}$ (which doesn't exist) in $C^*(1,a)$.

In general, if $p=p^*=p^2$, then $pAp$ is a C*-subalgebra of $A$ with unit $p$, but unless $p$ is the unit of $A$, it is not a unital subalgebra in the usual sense. In the case where $a$ is not invertible and $0$ is isolated in its spectrum, and the function $f:\sigma(a)\to\mathbb C$ is defined by $f(0)=0$, $f(t)=1$ if $t\neq 0$, then $p=\Gamma^{-1}(f)$ is a unit for $C^*(a)$, but $p$ is not the unit of $A$.

Jonas Meyer
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  • I am really sorry. I can not understand your answer completely. Could you recommend me a legible book which include this knowledge? – Stephen Curry Oct 18 '13 at 17:49
  • @Belle-tiantian: I have no specific recommendations at the moment, but if you are more specific about what is not understood I or someone else may try to help make it clearer. – Jonas Meyer Oct 18 '13 at 20:14
  • In the first line of your proof, you said that $a$ is invertible implies $C^{\ast}(a)=C^{\ast}(a,1)$. Why does this equation hold? I suppose, although $a\in C^{\ast}(a)$, $a^{-1}$ may not contain in $C^{\ast}(a)$. So you can not ensure $1\in C^{\ast}(a).$ Is there anything wrong in my view point? – Stephen Curry Oct 19 '13 at 02:36
  • @Belle-tiantian: You are right to be questioning that, as it would be a problem in some analogous situation, but it is not a problem for C-algebras. One way to see it in this case is as follows. $\Gamma:C^(1,a)\cong C(\sigma(a))$, $\Gamma(a)= f$ with $f(z)=z$. Note that $C^(a)$ is an ideal in $C^(1,a)$. Suppose that $C^(a)\neq C^(1,a)$. Then it is a maximal ideal, and hence $\Gamma(C^*(a))$ is a maximal ideal in $C(\sigma(a))$, call it $M$. Thus there exists $z_0\in\sigma(a)$ such that $g(z_0)=0$ for all $g\in M$. In particular this applies to $f$, so $z_0=0$. Thus $a$ isn't invertible. – Jonas Meyer Oct 19 '13 at 23:50
  • @Belle-tiantian: Another question asked whether a nonunital C-subalgebra can contain an invertible element (so this is more general and does not assume normality). The answer is still no, and the answer there uses spectral permanence to show this, hence this gives another way to see that $C^(a)=C^*(1,a)$ if $a$ is invertible (even if $a$ is not normal). – Jonas Meyer Oct 21 '13 at 23:41
  • Thanks for your patient and detailed answer, Jonas Meyer. But I suppose I need more time to understand your answer. Maybe I am poor in General topology. – Stephen Curry Oct 22 '13 at 15:02