# Gelfand spectrum of a Banach algebra generated by a single element

Let $A$ be a Banach algebra generated by $a$ which means it is a closure of $span \{1, a, a^2, ... \}$. The problem is to show that its Gelfand spectrum $\Omega(A)$ is homeomorphic to the spectrum of $a$ which is $\sigma_A(a) = \{ \lambda \in \mathbb{C} : a - \lambda \cdot 1$ is not invertible$\}$.

So far I can prove that $\sigma_A(a) = \hat a(\Omega (A))$ where $\hat a: \Omega(A) \to \mathbb{C}$ is a Gelfand transform of $a$. It remains to show that $\hat a: \Omega(A) \to \sigma_A(a)$ is actually homeomorphism.

## 1 Answer

Consider the Gelfand transform of $a$: \begin{align*}\hat{a}:\Omega (A)&\to \sigma_A(a) \\ h &\mapsto \hat{a}(h)=h(a) \end{align*} The map is well defined and surjective, because for any element of a commutative unital Banach algebra we have $$\lambda \in \sigma_A(a)\iff \exists\;h\in \Omega(A):h(a-\lambda e)=0\iff \exists\;h\in \sigma_A(a):h(a)=\lambda$$ Moreover, since $A$ is generated by $a$, the map is also injective: indeed, since any $h\in \Omega(A)$ is a continuous algebraic homomorphism, the value of $h$ on $a$ uniquely determines $h$ on the closed subalgebra spanned by the element $a$, which by assumption is exactly $A$.

Finally, to show that it is a homeomorphism we only need to prove continuity, since $\Omega(A)$ is compact and $\sigma_A(a)$ is Hausdorff. But this is easy because $\Omega(A)$ has the weak*-topology, so if $\left\langle h_i\right\rangle_{i\in I}$ is a net that converges to $h$ in $\Omega(A)$ then $h_i(a)\to h(a)$ for all $a\in A$. (EDIT: This is also true on any commutative unital Banach algebra. It is the fact that the Gelfand transform of $a$ is also a bijection that allows us to conclude that it is in fact a homeomorphism).