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I know that spectrum of an element $x$ of a unital C*-algebra is nonempty. I like to find an example of a non unital C*-algebra that has an element with empty spectrum, if it exists.

Motivation

I saw a theorem about "spectrum of an element x of a unital C*-algebra is nonempty" so I wanted to know if the unital assumption is essential.

niki
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    How are you defining spectrum of $x$ in a non-unital $C^{\star}$ alebra? A $C^{\star}$ algebra $A$ be enlarged to a unital $C^{\star}$ algebra $\tilde{A}$ so that $A$ is an ideal in $\tilde{A}$ and $\tilde{A}/A$ is isomorphic to $\mathbb{C}$, which may settle your question once you define spectrum for $x$. – Disintegrating By Parts Oct 04 '14 at 19:32
  • @ T.A.E.: Thanks for your comment. I saw a theorem about "spectrum of an element x of a unital C*-algebra is nonempty" so I got surprised. – niki Oct 05 '14 at 06:28
  • I didn't get your message, but happened to check back. I think there's a problem if the '@' symbol is followed by a space. The usual definition of spectrum is as the set of scalars $\lambda$ for which $x-\lambda 1$ is not invertible, which means you need a unit $1$. There are probably other ways, but that's why I needed to ask. Without a unit, nothing in the algebra can be invertible because there's no unit. So, technically, I think $0$ is in the spectrum of everything in a non-unital algebra. – Disintegrating By Parts Oct 05 '14 at 15:23
  • @T.A.E. : Thanks, I got it – niki Oct 05 '14 at 15:26
  • @T.A.E.: for an arbitrary commutative Banach algebra, you would define spectrum of the algebra as the set of homomorphisms into the complex numbers or equivalently, the set of its maximal ideals. If my memory serves right, for a unital algebra generated by a single element, it turns out those ideals are exactly those which are generated some nonunit $x-\lambda\cdot 1$. I think I recall that in non-commutative context, a maximal left ideal (of a $C^*$-algebra?) is actually also a right ideal, so this should translate directly to non-commutative context. – tomasz Oct 05 '14 at 19:09
  • @tomasz : Using your definition, do you always get $\lambda=0$ in the spectrum for all $x$ in a $C^{\star}$ algebra without unit? – Disintegrating By Parts Oct 05 '14 at 19:39
  • @T.A.E.: Good point. Of course, we would want to consider only nontrivial homomorphisms. – tomasz Oct 05 '14 at 20:01
  • @tomasz : The reason I was asking is that when you embed the $C^{\star}$ algebra in one with a unit, then all of the original elements are non-invertible. So I think $0$ ends up in the spectrum for all of the original elements, provided the original algebra did not contain a unit. But something seems strange about that to me. Oh, I see what you mean by non-trivial now: the one that is 0 on the original algebra. Is that what you're thinking? – Disintegrating By Parts Oct 05 '14 at 20:07
  • @T.A.E.: Yeah, a nontrivial homomorphism is one which is not identically zero. Because we're talking about complex algebras, it follows that it is onto the complex numbers. It's not tied to invertibility of anything, as far as I can tell. – tomasz Oct 05 '14 at 20:28

1 Answers1

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Spectrum

For Banach algebras the spectrum is always nonempty.

(This is due to the Neumann series.)

For algebras the spectrum may be empty.

(Consider the algebra of polynomials.)

Extensions

For Banach algebras the spectrum depends on extensions.

(Consider the Banach algebra of the bilateral shift.)

For C*-algebras the spectrum is independent of extensions.

(This is due to the spectral permanence.)

Reference

An investigation was done in: Extensions: Spectrum

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freishahiri
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