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Let $K$ be a compact subset of $\Bbb C$. Let $\mathcal P(K)$ be the closed algebra generated by the complex polynomials on $K$.

What is the character space $\Phi_{\mathcal P(K)}$ of $\mathcal P(K)$?

It is commonly known that $\Phi_{C(K)}=\{ \delta_z|z\in K\}$ where $\delta_z$ is the pointwise evaluation at $z$. Since $\mathcal P(K)\subset C(K)$ we have $\Phi_{C(K)}\subset\Phi_{\mathcal P(K)}$. It appears that the converse does not hold when $\Bbb C\backslash K$ is not connected. What are the other characters in $\Phi_{\mathcal P(K)}$?

This is related to a question I asked earlier on MathOverflow. The answer I got motivated me to ask this question here.

This is another similar question I posted yesterday on math.stackexchange with no answer so far. I modified some parts to make the question more precise.

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    $\begingroup$ The character space of $P(K)$ is the polynomial hull of $K$, which in turn equals the union of $K$ with all the bounded components of ${\bf C}\setminus K$. This should also be in Allan's book; it should also be in the book of Bonsall and Duncan, but I don't have my copy at hand right now. $\endgroup$
    – Yemon Choi
    Commented Jan 8, 2017 at 1:47
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    $\begingroup$ If you are relatively new to spectral theory/Gelfand theory for commutative Banach algebras, then a good example to think about is when $K={\bf T}$ is the unit circle sitting inside ${\bf C}$. Then $P({\bf T})$ consists of the closed subalgebra of $C({\bf T})$ generated by $1$ and the function $z$; this is (isomorphic to) the disc algebra, and interior points of the disc will give you characters on this Banach algebra that do not come from points of ${\bf T}$. $\endgroup$
    – Yemon Choi
    Commented Jan 8, 2017 at 1:49
  • $\begingroup$ Incidentally, I have just found the theorem that you mentioned in the previous question. I'll finish reading it tomorrow since it's pretty late at night now. I really appreciate all your answers so far and yes, I am relatively new to this. $\endgroup$
    – BigbearZzz
    Commented Jan 8, 2017 at 1:53
  • $\begingroup$ @BigbearZzz: By "complex polynomial on $K$" do you mean the restriction to $K$ of some polynomial function defined on $\mathbb C$? $\endgroup$
    – Alex M.
    Commented Jan 25, 2018 at 19:18
  • $\begingroup$ @AlexM. Yes, just any polynomial $p(z)$ with takes value in $\Bbb C$, restricted to K. $\endgroup$
    – BigbearZzz
    Commented Jan 25, 2018 at 21:20

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