Extensions: Spectrum

Question

Problem

Given a C*-algebra $\mathcal{A}_0$ and unital extensions $1\in\mathcal{A}$ and $1'\in\mathcal{A}'$.

Regard a common element: $$A_0\in\mathcal{A}_0:\quad A^{(\prime)}:=\iota^{(\prime)}(A_0)$$

Can it happen that its spectra really differ: $$\sigma(A)\cup\{0\}\neq\sigma(A')\cup\{0\}$$ (The scenario is inspired by possible noncanonical unital extensions.)

One might be tempted to conclude that they agree as: $$\langle\iota(\mathcal{A}_0)\cup\{1\}\rangle\cong\mathcal{A}_0\oplus\mathbb{C}\cong\langle\iota'(\mathcal{A}_0)\cup\{1'\}\rangle'$$ But there is a flaw as the example below illustrates.

Example

Given the matrix algebra $\mathcal{M}:=M(2;\mathbb{C})$.

Consider the strict subalgebra: $$\mathcal{M}':=\left\{\begin{pmatrix}a&0\\0&0\end{pmatrix}:a\in\mathbb{C}\right\}\subseteq\mathcal{M}$$

Then their units disagree: $$1=\begin{pmatrix}1&0\\0&1\end{pmatrix}\neq\begin{pmatrix}1&0\\0&0\end{pmatrix}=1'$$

Regard their common element: $$M:=\begin{pmatrix}1&0\\0&0\end{pmatrix}\in\mathcal{M}\cap\mathcal{M}'$$

Then it has at least distinct spectra: $$\sigma(M)=\{0,1\}\neq\{1\}=\sigma(M)'$$ (Still, leaving open the question wether they really can differ.)

Answer 1

Disclaimer

There was another answer giving the motivation for this one. Sadly, it got deleted.

Crucial Point

It boils down to the existence of an isomorphism: $\pi[A]=A'$ & $\pi[1]=1'$

This is ill-defined only for the case: $1\propto A$ & $1'!\propto A'$

Prework

Suppose proportionality holds: $1=\lambda A$

Then the original algebra was unital: $1_0=\lambda A_0\in\mathcal{A}_0$

Pathological Case

Assume proportionality is trivial too: $\lambda=0$

So one checks: $(A=0=1)$ & $\left(A'=0'\neq1'\right)$

Thus the spectra differ by zero: $$\sigma(A)=\varnothing\approx\{0\}=\sigma(A')$$

(Besides, the original algebra was pathological then!)

Proper Case

Assume proportionality is nontrivial only: $\lambda\neq0$

So one checks: $\left(A=\frac{1}{\lambda}1\right)$ & $(1\neq0)$ & $\left(A'=\frac{1}{\lambda}E'\right)$ & $\left(0'\neq E'\neq1'\right)$ & $\left(E^2=E=E^*\right)$

Thus the spectra differ by zero: $$\sigma(A)=\{\frac{1}{\lambda}\}\approx\{0,\frac{1}{\lambda}\}=\sigma(A')$$

(Besides, the original algebra was unital then!)

Standard Case

In any other case there exists an isomorphism: $\pi[1]=1'$

Thus the spectra agree by spectral permanence: $$\sigma(A)=\sigma(\pi[A])=\sigma(A')$$

(Besides, it didn't matter wether the algebra was pathologic or unital here!)

Key Observation

Observe that the spectra only can differ if the original algebra was unital: $A_0\propto1_0\in\mathcal{A}_0$

Thus one can unambigously define the spectrum by any unital extension!