Relation between the spectrum of a Banach algebra and the spectrum of its unitization.

Question

Let $A$ be any Banach algebra. We can always consider its unitization $A_{1}:=A\oplus\mathbb{C}$, even if $A$ is already unital. The multiplication on $A_{1}$ is defined as $$(a,\lambda)(b,\mu):=(ab+\lambda a+\mu b,\lambda\mu).$$ One can easily prove that a character $\phi$ on a unital algebra $A$ satisfies $\phi(1)=1$. The spectrum $\Omega(A)$ of $A$ is defined as the set of all non-zero homomorphisms $A\to\mathbb{C}$. We endow $\Omega(A)$ with the topology of pointwise convergence.

Note that $\hat{0}\colon A_{1}\to\mathbb{C}$, $(a,\lambda)\mapsto\lambda$ is a character on $A_{1}$.

I am trying to prove that $\Omega(A)$ is homeomorphic to $\Omega(A_{1})\setminus\{\hat{0}\}$. Also, I want to prove this directly from the definitions, i.e. without using (local) compactness or Hausdorffness.

Here is what I tried:

For $\phi\in\Omega(A)$ I define $\hat{\phi}\colon A_{1}\to\mathbb{C}$ by $\hat{\phi}(a,\lambda):=\phi(a)+\lambda$. Now I tried to prove that $\phi\mapsto\hat{\phi}$ is a homeomorphism.

Injectivity: Assume that $\hat{\phi}=\hat{\psi}$. Then $$\phi(a)=\hat{\phi}(a,0)=\hat{\psi}(a,0)=\psi(a)$$ for all $a\in A$, thus $\phi=\psi$.

Surjectivity: Let $\alpha\in\Omega(A)\setminus\{\hat{0}\}$ be given. Define $\phi\colon A\to\mathbb{C}$ by $\phi(a):=\alpha(a,0)$. Then $\phi\neq0$, otherwise $\hat{\phi}=\hat{0}$. Also, it is easy to see that $\phi$ is a homomorphism, thus $\phi\in\Omega(A)$. Now observe that $$\hat{\phi}(a,\lambda)=\phi(a)+\lambda=\alpha(a,0)+\lambda\alpha(0,1)=\alpha(a,\lambda).$$

Continuity: Let $(\phi_{i})$ be a net in $\Omega(A)$ that converges to a character $\phi\in\Omega(A)$. Since $\phi_{i}(a)\to\phi(a)$ for all $a\in A$, it is easy to see that $\hat{\phi}_{i}(a,\lambda)\to\hat{\phi}(a,\lambda)$ for all $(a,\lambda)\in A_{1}$, so $\hat{\phi}_{i}\to\hat{\phi}$ in $\Omega(A)\setminus\{\hat{0}\}$.

But I dont know how to conclude that $\phi\mapsto\hat{\phi}$ is a homeomorphism, i.e. that its inverse is also continuous. Any help would be greatly appreciated!

Answer 1

The inverse is given by $$\Omega(A_1) \setminus\{\hat{0\}}\to \Omega(A): \chi \mapsto \chi\vert_A$$

Clearly this is weak$^*$-continuous, because if $\chi_i\to \chi$ pointwise on $A_1$, then also on $A$ since $A$ embeds in $A_1$.

Answer 2

When you proved surjectivity, you already showed how to construct an explicit inverse to your map (i.e., the inverse is restriction to $A$). This map is obviously continuous.