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Let ‎$‎‎A$ ‎be a unital ‎‎‎$‎‎C^*$-algebra.

‎‎ Assume that ‎$‎‎a\in A$ ‎is a ‎‎normal ‎and ‎invertible element ‎i.e ‎‎$‎‎aa^*=a^*a$ ‎and ‎‎$‎‎aa^{-1}=a^{-1}a=1$‎.‎

‎let $‎‎C^*({a}) $ be the ‎‎‎$‎‎C^*$-algebra generated by ‎$‎‎a$‎.

I know that ‎$‎‎C^*({a}) $ ‎is ‎the ‎closed ‎linear ‎span ‎of ‎‎$‎‎a^{m}a^{*{n}}$‎‎‎ such that $m,n\in N$.

‎ ‎ I want to know ‎$‎1 , a^{-1} \in ‎‎C^*({a}) ‎‎$‎‎ ‎

Q: Is it true?"$‎1 , a^{-1} \in ‎‎C^*({a}) ‎‎$‎‎"‎

How can I prove it? ‎‎

alex v
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2 Answers2

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The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform.

Edit: in view of Josse's comments, here's a clarification. Since $ a $ is invertible, $\sigma (a)\cap\{0\}=\varnothing $, so there exists a continuous function $f $ with $f (0)=0$ and $f (t)=1/t $ on $\sigma (a) $. By using Stone-Weierstrass on a closed disk, we can write $f $ as a uniform limit of polynomials (on $z $ and $\bar z $) with constant term zero. This implies that $a^{-1}\in C^*(a) $, and a fortiori $1\in C^*(a) $.

Martin Argerami
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  • Yes.I got.thanks alot – alex v Feb 01 '16 at 23:10
  • The functional calculus is obtained from the Gelfand transform of the $C^$-subalgebra $B \subseteq A$ generated by $1$ and $a$, so how do you know that $f(a)$ lies in $C^(a)$? The first part of the question is whether $1 \in C^*(a)$ holds, but you did not address this in your answer. – Josse van Dobben de Bruyn Aug 25 '17 at 17:16
  • If you solve this difficulty by considering the Gelfand representation of $C^(a)$, you run into a different problem: is $C^(a)$ unital? Is its character space once again homeomorphic to $\sigma(a)$, or do you get something else? (The spectrum of $a$ relative to $C^(a)$ need not be equal to the spectrum of $a$ in the ambient $C^$-algebra $A$.) And even if $C^*(a)$ is unital, its unit could in theory be different from the unit of $A$. – Josse van Dobben de Bruyn Aug 25 '17 at 17:19
  • I'm not sure where you are going. Since $t\mapsto 1/t$ is a uniform limit of polynomials, it is immediate that $a^{-1}\in C^(a)$, and so $C^(a)$ is unital. And that the spectrum of an element does not depend on the ambient C$^$-algebra is a basic fact; so the spectrum of $a$ relative to $C^(a)$ is indeed equal to the spectrum of $a$ in A$. – Martin Argerami Aug 25 '17 at 18:06
  • @MartinArgerami, you need $f : t \mapsto 1/t$ to be a uniform limit of polynomials with zero constant term (which it is, by the locally compact version of the Stone–Weierstrass theorem, since we have $0 \notin \sigma(a)$). But even then, how do you prove that $f(a)$ is the inverse of $a$? I'm sure this is basic knowledge to the expert in operator algebras, but it is a consequence of the present question. In other words, I think the question is more low level than you realise, and you need to be careful to avoid circular reasoning. (But maybe I missed something.) – Josse van Dobben de Bruyn Aug 25 '17 at 19:45
  • Oh, and regarding the spectrum being independent of the ambient $C^$-algebra: that's not quite the case. We always have $\sigma_B(b) \cup {0} = \sigma_A(b) \cup {0}$ whenever $B \subseteq A$ are $C^$-algebras. But it could happen that $0 \in \sigma_A(b) \setminus \sigma_B(b)$ holds. The present question shows that the other way around is impossible: if $0 \notin \sigma_A(b)$, then we automatically have $0 \notin \sigma_B(b)$. But I don't see a way to prove this without first solving the question posed by OP. (Again, maybe I'm missing something here.) – Josse van Dobben de Bruyn Aug 25 '17 at 19:51
  • Okay, I think I know what the confusion is about. I thought you suggested to look somehow at the Gelfand representation of the commutative $C^$-subalgebra $C^(a) \subseteq A$, and it was unclear to me what its character space might be. But you're using the functional calculus, so implicitly you're looking at the $C^*$-subalgebra generated by $1$ and $a$ instead. I didn't catch that, so I got confused and completely missed your point. In any case, I cleaned up my answer, and hopefully it corresponds with what you had in mind. :-) Sorry for being so verbose, and thank you for your time! – Josse van Dobben de Bruyn Aug 25 '17 at 21:21
  • Thanks to you for paying attention. I still think you make it unnecessarily complicated: in $A$, the spectrum of $a$ is disjoint with $0$, so $t\mapsto 1/t$ is continuous, and so $a^{-1}$ is a norm limit of polynomials; this makes $a^{-1}\in C^(a)$ (as I said, it is even in the Banach algebra generated by $A$), and a fortiori $1\in C^(a)$. – Martin Argerami Aug 25 '17 at 22:07
  • Hm, I think a difference of opinion persists. I respectfully disagree with your answer, for the following reason: unless you can convince me that $a^{-1}$ is a norm limit of polynomials with zero constant term, your proof only shows that $a^{-1}$ lies in the $C^*$ (or Banach) subalgebra generated by $1$ and $a$. After all, we're not assuming (sub)algebras to be unital, so the Banach subalgebra generated by $a$ is the closure of the space of all polynomials in $a$ with zero constant term (equivalently: the closed linear span of ${a,a^2,a^3,\ldots}$). – Josse van Dobben de Bruyn Aug 27 '17 at 01:50
  • But that's completely trivial. Take any continuous function $f$ with $f(0)=0$ and $f(t)=1/t$ for $t\in\sigma(a)$. Approximate it uniformly by polynomials in some (compact!) disk containing $\sigma(a)$ and $0$; if whatever method you use does not give you zero constant terms, you can replace them with zero because of $f(0)=0$. – Martin Argerami Aug 27 '17 at 03:13
  • Er, $1/t$ cannot be approximated by polynomials in general. If $\sigma(a)$ is the whole unit circle, then it cannot be. However, replace $a$ by $a^*a$. Then $\sigma(a)$ has no holes, so there is no problem. – Bruce Evans Aug 27 '17 at 12:53
  • @MartinArgerami Oh, yes, that works. Thanks for the clarification! It is now clear to me that your answer is complete and correct. :-) – Josse van Dobben de Bruyn Aug 27 '17 at 12:55
  • @BruceEvans, is that so? In any case, we can always approximate $1/t$ (uniformly on a compact set $\varnothing \subsetneq D \subseteq \mathbb{C} \setminus {0}$) by polynomials in $z$ and $\overline z$, which is sufficient since we're dealing with the $C^*$-subalgebra (rather than the Banach subalgebra) generated by $a$. – Josse van Dobben de Bruyn Aug 27 '17 at 13:01
  • @BruceEvans: here "polynomial" means "linear combination of powers of $z $ and $\bar z $", as required by Stone-Weierstrass and corresponding to the C $^$-algebra generated by an element. Using $a^a $ leads nowhere, as it remembers little to nothing about $a $ (in this question, $a^*a=1$). – Martin Argerami Aug 27 '17 at 13:09
  • @MartinArgerami I don't think we assume $a^a = 1$ in this question, but we don't exclude the possibility either. In any case, I agree that using $a^a$ leads nowhere, as $C^(a^a)$ does not typically contain $a^{-1}$ or even $a$. Indeed, a good example is when $a$ is any non-trivial unitary. – Josse van Dobben de Bruyn Aug 27 '17 at 13:15
  • Yes, I misread that, but it is as you say. – Martin Argerami Aug 27 '17 at 13:24
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    I want to emphasize that the Banach algebra case is very different. The analytic functional calculus cannot be applied in general because $1/z$ is not analytic on a large enough set. Using $a^a$ is just the standard trick of reducing to the self-adjoint case. Its inverse is more obviously a polynomial in $a^a$ (possibly with constant term. Multiply this by $a^a$ to get $1$ as a polynomial in $a$ and $a^a$ (now without constant term). Similarly for $a^{-1}$. – Bruce Evans Aug 27 '17 at 13:29
  • @BruceEvans, you're right, sorry, I just remembered I read this trick in Murphy's textbook (theorem 2.1.11), and it does work. You note that $a^a$ is also invertible, and show that its inverse lies in $C^(a^a) \subseteq C^(a)$. Then we have $(a^a)^{-1}a^a = 1$, hence $a^{-1} = (a^a)^{-1}a^ \in C^*(a)$. – Josse van Dobben de Bruyn Aug 27 '17 at 13:38
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Let $B \subseteq A$ denote the $C^*$-subalgebra generated by $1$ and $a$. Recall that the functional calculus gives us a unital $*$-homomorphism $\varphi : C(\sigma(a)) \to A$ with the following properties:

  • $\varphi(\iota) = a$, where $\iota : \sigma(a) \to \mathbb{C}$ denotes the inclusion $z \mapsto z$.
  • $\varphi$ is isometric;
  • The image of $\varphi$ is the $C^*$-subalgebra $B \subseteq A$ generated by $1$ and $a$.
  • Consequently, $\varphi$ restricts to a $*$-isomorphism $C(\sigma(a)) \to B$.

Now, since we have $0\notin \sigma(a)$, we see that $\iota \in C(\sigma(a))$ vanishes nowhere and separates points:

  • For all $x\in \sigma(a)$ we have $\iota(x) = x \neq 0$ (here we use that $0 \notin \sigma(a)$ holds);
  • For all $x,y\in\sigma(a)$ with $x \neq y$ we have $\iota(x) \neq \iota(y)$.

As such, it follows from the Stone–Weierstrass theorem (locally compact version) that $\iota$ generates $C(\sigma(a))$ as a $C^*$-algebra. Since $\varphi$ restricts to a $*$-isomorphism $C(\sigma(a)) \to B$ with $\iota \mapsto a$, it follows that $B$ is generated by $a$. In other words, we have $B = C^*(a)$. Now it is immediate that $1 \in C^*(a) = B$ holds. Furthermore, it is now easy to see that $a^{-1}\in C^*(a)$ holds, since $\iota$ is invertible in $C(\sigma(a))$.

Josse van Dobben de Bruyn
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