# Gelfand transform explicity

Let $T$ be a bounded normal operator. Let $A$ be the algebra generated by $T$ and $T^*$. What is the explicit Gelfand transform $G:A\to C(\sigma(T))$?

My book says the image of $T$ is the identity but what about in general?

Thanks.

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What "in general"? For an abelian $C^*$-algebra? – Michael Feb 14 '14 at 0:51

The polynomial algebra $\mathbb{C}[T,T^*]$ is dense in $A$. A polynomial $p(T,T^*)$ is sent to the corresponding polynomial function $p(z,\overline{z})$ on the spectrum of $T$. The general description then comes from continuity. For example, when $||T|| < 1$, the element $\frac{1}{1-T} = \sum_{k=0}^{\infty} T^k$ is sent to $\frac{1}{1-z} = \sum_{k=0}^{\infty} z^k$.
Thanks a lot. How do we know $p(z,\overline{z})$ is in the spectrum. Also that this map is surjective? – user108605 Feb 14 '14 at 7:47
$p(z,\overline{z})$ is not in the spectrum - this doesn't make any sense. For the isomorphism $C^*(T) \cong C(\sigma(T))$ look at any introduction to operator algebras. It's all written down. – Martin Brandenburg Feb 14 '14 at 13:55