Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra

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Theorem

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra with identity element ${\mathbf 1}_A$.

Let $B \subseteq A$ be a unital $\text C^\ast$-subalgebra of $A$.

Let $\sigma_A$ and $\sigma_B$ be the spectrum in $A$ and $B$ respectively.

Let $x \in B$.


Then we have:

$\map {\sigma_A} x = \map {\sigma_B} x$


Proof

First take $x$ to be Hermitian.

Let $B' \subseteq B$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x}$.

By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B'$ is commutative.

From Spectrum of Hermitian Element in Unital C*-Algebra is Real, we have $\map {\sigma_{B'} } x \subseteq \R$.

From Spectrum of Element in Unital Subalgebra, we therefore have $\map {\sigma_B} x \subseteq \map {\sigma_{B'} } x \subseteq \R$.

From Interior of Set of Real Numbers in Complex Numbers is Empty: Corollary, we have $\map {\sigma_B} x^\circ = \O$ where $\map {\sigma_B} x^\circ$ denotes the interior of $\map {\sigma_B} x$.

From the definition of boundary, we have that $\partial \map {\sigma_B} x = \map {\sigma_B} x^- \setminus \map {\sigma_B} x^\circ$.

Since $\map {\sigma_B} x$ is closed from Spectrum of Element of Banach Algebra is Closed and $\map {\sigma_B} x^\circ = \O$, we then have that $\map {\sigma_B} x = \partial \map {\sigma_B} x$.

Then, from Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra we have:

$\map {\sigma_B} x = \partial \map {\sigma_B} x \subseteq \partial \map {\sigma_A} x$

From Set is Closed iff it Contains its Boundary, we have $\partial \map {\sigma_A} x \subseteq \map {\sigma_A} x$.

So $\map {\sigma_B} x \subseteq \map {\sigma_A} x$.

From Spectrum of Element in Unital Subalgebra we have $\map {\sigma_A} x \subseteq \map {\sigma_B} x$.

Hence we conclude $\map {\sigma_A} x = \map {\sigma_B} x$ in this case.

In particular, $0 \in \map {\sigma_A} x$ if and only if $0 \in \map {\sigma_B} x$.

Hence if $x$ is Hermitian, it is invertible in $B$ if and only if it is invertible in $A$.


Let $b \in B$.

Suppose that $b$ is invertible in $A$.

We show that it is invertible in $B$.

Take $a \in A$ such that $a b = b a = {\mathbf 1}_A$.

From Inverse of Star of Element in Unital *-Algebra, we have $b^\ast a^\ast = a^\ast b^\ast = {\mathbf 1}_A$.

We therefore have:

\(\ds b b^\ast a^\ast a\) \(=\) \(\ds b \paren {b^\ast a^\ast} a\)
\(\ds \) \(=\) \(\ds b {\mathbf 1}_A a\)
\(\ds \) \(=\) \(\ds b a\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A\)

and:

\(\ds a^\ast a b b^\ast\) \(=\) \(\ds a^\ast \paren {a b} b^\ast\)
\(\ds \) \(=\) \(\ds a^\ast {\mathbf 1}_A b^\ast\)
\(\ds \) \(=\) \(\ds a^\ast b^\ast\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A\)

Hence $b b^\ast$ is invertible in $A$.

From Product of Element in *-Star Algebra with its Star is Hermitian, $b b^\ast$ is Hermitian.

Hence from what we have already shown, $b b^\ast$ is invertible in $B$.

So there exists $c \in B$ such that $b b^\ast c = {\mathbf 1}_A$.

Then:

\(\ds a\) \(=\) \(\ds a b b^\ast c\)
\(\ds \) \(=\) \(\ds \paren {a b} b^\ast c\)
\(\ds \) \(=\) \(\ds {\mathbf 1}_A b^\ast c\)
\(\ds \) \(=\) \(\ds b^\ast c\)

Since $b^\ast \in B$ and $c \in B$, we have $a \in B$.

So $b$ is invertible in $B$.

Clearly conversely if $b$ is invertible in $B$, then it is invertible in $A$.

Hence we have shown that if $b \in B$, then $b$ is invertible in $B$ if and only if it is invertible in $A$.

In particular, if $x \in B$ then ${\mathbf 1}_A - x$ is invertible in $B$ if and only if it is invertible in $A$.

This is precisely the statement $\map {\sigma_A} x = \map {\sigma_B} x$.

$\blacksquare$


Sources

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