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Spectral permanence for unital c star subalgebras

Statement

Assume AC. If $B$ is a unital C*-subalgebra of a unital C*-algebra $A$ with the same identity, then $σ_{B} (b) = σ_{A} (b)$ for every $b \in B$ .

Facts & Assumptions

[A1]

A unital C*-algebra is a nonzero complex Banach algebra with a unit of norm one and an involution satisfying the C*-identity; a unital C*-subalgebra with the same identity is a closed unital $*$ -subalgebra containing that unit (C star algebra, Unital Banach algebra).

[A2]

For every element $b$ invertible in $A$ the adjoint $b^{*}$ is invertible with $(b^{*})^{- 1} = (b^{- 1})^{*}$ , and $b^{*} b$ is positive; positivity of an element means $x = y^{*} y$ for some element $y$ (Self-adjoint positive unitary and normal elements for positivity, C star algebra for the involution rules).

[A3]

If $B \subseteq A$ is a unital subalgebra with the same unit and $b \in B$ , then $σ_{A} (b) \subseteq σ_{B} (b)$ , since invertibility in $B$ implies invertibility in $A$ (Spectrum and resolvent set in a Banach algebra).

[A4]

For a nonzero unital commutative complex C*-algebra $C$ the Gelfand transform $Γ : C \to C (Δ (C))$ is an isometric unital $*$ -isomorphism onto $C (Δ (C))$ , and $Δ (C)$ is a nonempty compact Hausdorff space (Commutative Gelfand Naimark, Maximal ideal space is compact Hausdorff, Character and maximal ideal space).

[A5]

A unital self-adjoint complex function algebra that separates the points of a nonempty compact Hausdorff space is uniformly dense in all continuous complex functions (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

[A6]

Characters of a unital commutative complex C*-algebra satisfy $χ (c^{*}) = \overline{χ (c)}$ (Characters on a unital commutative C star algebra preserve star).

[A7]

AC is the global hypothesis of the Gelfand-theoretic suppliers (The Axiom of Choice).

Proof

technique · direct

Given: A unital C*-algebra $A$ , a unital C*-subalgebra $B \subseteq A$ with $1_{B} = 1_{A}$ , and $b \in B$ .

1.1

Since invertibility in $B$ implies invertibility in $A$ , one has $σ_{A} (b) \subseteq σ_{B} (b)$ ; it remains to prove the reverse inclusion, that is, that $b$ invertible in $A$ is invertible in $B$ with inverse in $B$ .

1.2

If $b$ is invertible in $A$ , then $x : = b^{*} b \in B$ is positive and invertible in $A$ with $x^{- 1} = b^{- 1} (b^{- 1})^{*}$ ; and if $x$ positive and invertible in $A$ has $x^{- 1} \in B$ , then $(b^{*} b)^{- 1} \in B$ and $b^{- 1} = (b^{*} b)^{- 1} b^{*} \in B$ .

A1A2algebra

1.3

It suffices to treat a self-adjoint $x \in B$ invertible in $A$ , because the positive element $b^{*} b$ in the preceding reduction is self-adjoint. For such $x$ , let $C : = C_{A}^{*} (1, x, x^{- 1})$ be the closed unital star-subalgebra of $A$ generated by $x$ and $x^{- 1}$ , and let $D : = C_{A}^{*} (1, x) \subseteq B$ . Both are commutative, because $x = x^{*}$ and $(x^{- 1})^{*} = x^{- 1}$ , and the Gelfand transform is an isometric unital star-isomorphism $Γ : C \to C (Δ (C))$ on the nonempty compact Hausdorff character space.

A1A2A4A7algebra

2.1

The function $\hat{x}$ separates the points of $Δ (C)$ . Indeed, if characters $h, h^{'}$ have $h (x) = h^{'} (x)$ , then $h (x^{- 1}) = h (x)^{- 1} = h^{'} (x)^{- 1} = h^{'} (x^{- 1})$ ; they therefore agree on the unital star-algebra generated by $x, x^{- 1}$ and, by continuity, on its closure $C$ , so $h = h^{'}$ . Moreover $\hat{x}$ is real-valued because $x = x^{*}$ and characters preserve the involution.

step 1.3A6algebra

3.1

Under $Γ$ , the algebra $D$ is the closed unital self-adjoint function algebra generated by $\hat{x}$ . It separates points by step 2.1, so Stone--Weierstrass gives $Γ (D) = C (Δ (C)) = Γ (C)$ . Injectivity of $Γ$ yields $D = C$ , hence $x^{- 1} \in D \subseteq B$ .

step 2.1A4A5

4.1

Therefore every $b \in B$ invertible in $A$ has $(b^{*} b)^{- 1} \in B$ by step 3.1 and then $b^{- 1} = (b^{*} b)^{- 1} b^{*} \in B$ by step 1.2. Thus $σ_{B} (b) \subseteq σ_{A} (b)$ ; with step 1.1 this gives $σ_{B} (b) = σ_{A} (b)$ .

step 1.1step 1.2step 3.1∎

Depends on

Used by

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Dana P. Williams, Lecture Notes on the Spectral Theorem, Theorem 4.8, pp.13–15 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)