Gelfand transform and spectrum

Question

Let $\mathcal A$ commutative, unital Banach algebra and denote by $\mathcal M(\mathcal A)$ the space of multiplicative functionals on $\mathcal A$. The Gelfand transform is defined by

$$\Gamma: \mathcal A \rightarrow \mathcal M(\mathcal A), \quad A \mapsto \Gamma(A),$$

where $\Gamma(A)(m)=m(A),\forall m\in \mathcal M(\mathcal A)$.

Why is the following implication true?

$$\Gamma(A) = 0 \Rightarrow \sigma(A) = \{0\}$$

($\sigma (A)$ is the spectrum of $A\in \mathcal A$)

The set $\Omega(A)=\mathcal{M}(A)\setminus\left{0\right}$ is usually used instead of $\mathcal{M}(A)$, and the Gelfand transform takes values in $C(\Omega(A))$ (or $C(\mathcal{M}(A)))$! See G. J. Murphy, C$$-Algebras and Operator Theory*, Theorem 1.3.6. — Luiz Cordeiro, May 05 '13 at 18:08

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

In fact $A\in\mathcal{A}$ is invertible iff $m(A)\neq 0$ for all $m\in\Omega(\mathcal{A}):=\mathcal{M}(\mathcal{A})\setminus\{0\}$. As the consequence we have the chain equivalences $$ \begin{align} \lambda\in\mathbb{C}\setminus\sigma(A) &\Longleftrightarrow A-\lambda 1\in\mathrm{Inv}(\mathcal{A})\\ &\Longleftrightarrow \forall m\in \Omega(\mathcal{A})\quad m(A-\lambda 1)\neq 0\\ &\Longleftrightarrow \forall m\in \Omega(\mathcal{A})\quad\lambda\neq m(A)\\ &\Longleftrightarrow \lambda\in\mathbb{C}\setminus\{m(A): m\in \Omega(\mathcal{A})\} \end{align} $$ This means that $$ \sigma(A)=\{m(A):m\in \Omega(A)\}=\{\Gamma(A)(m):m\in \Omega(A)\} $$ For $A=0$ we obviously get $\sigma(A)=\{0\}$. For more details see this answer.

edited Apr 13 '17 at 12:21

Community

1

answered May 05 '13 at 18:45

Norbert

58,398

Thanks! Why is $m(A-\lambda I)\neq 0 \Leftrightarrow m(a)\neq \lambda$? Is every multiplicative functional linear? – madison54 May 06 '13 at 07:31
In Gelfand's theory they are assumed to be linear – Norbert May 06 '13 at 10:47

Gelfand transform and spectrum

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