2

Consider the algebra $C(S^1)$ of continuous functions $S^1 \to \mathbb C$ together with the $\|\cdot\|_\infty$ ($\sup$-norm). I am thinking that:

(?) The (sub-)algebra generated by $\rm{id}$ and $\overline{\cdot}$ (the complex conjugation) is the set of all polynomials.

(??) The closure of this subalgebra generated by $z$ and $\overline{z}$ is the entire algebra $C(S^1)$ (because of Stone-Weierstrass)?.

Is that accurate? If my thinking is wrong I'd greatly appreciate any corrections.

student
  • 1,617
  • More accurate: $\dots$ subalgebra generated by $1,z,\overline z$ $\dots$ – Yurii Savchuk Dec 20 '13 at 11:29
  • 2
    Note that $z\mapsto\frac{1}{z}$ ($=\overline{z}$ over $S^1$) cannot be approximated by polynomials. If it could, then $\int_{S^1}\frac{dz}{z}=0$, but it doens't. – Jonathan Y. Dec 20 '13 at 12:28
  • 1
    @JonathanY. Polynomials here means polynomials in $z$ and $\overline{z}$, for the circle, also trigonometric polynomials. – Daniel Fischer Dec 20 '13 at 13:15

2 Answers2

2

Regarding the first question, yes, it is the algebra of polynomials, if you consider polynomials in $z$ and $\overline{z}$. This is also the algebra of trigonometric polynomials, generated by $\cos nt,\, n \geqslant 0$ and $\sin nt,\, n\geqslant 1$.

By Weierstraß' theorem, this algebra is dense in $C(S^1)$.

Daniel Fischer
  • 206,697
  • But the algebra generated by $z$ and $\overline{z}$ does not contain $1$. To apply Stone-Weierstrass the algebra has to contain $1$, no? – student Dec 25 '13 at 15:41
  • On the unit circle, $1 = z\cdot \overline{z}$, so the algebra generated by $z$ and $\overline{z}$ contains all polynomials in $z$ and $\overline{z}$, including the constants. For other spaces than $S^1$, the generated algebra need not include $1$. – Daniel Fischer Dec 25 '13 at 16:25
1

By Stone-Weierstrass, the sub-$C^*$-algebra generated by $z\mapsto z$ and $z\mapsto\bar{z}$ (which is to say, that generated by $z\mapsto z$, as complex conjugation is simply the involution of this algebra), which naturally separates points, is indeed dense in $C(S^1,\mathbb{C})$.

However, the fact mentioned above ($\int_{S^1}\frac{dz}{z}=2\pi i$ and $\int_{S^1}p(z)dz=0$ for all $p\in\mathbb{C}[z]$, implying that uniform approximation is impossible) shows that it must be a proper super-set of the polynomials.

In fact, it's not difficult to verify that the $C^*$-algebra of rational functions (which naturally contains the $C^*$-algebra generated by $z\mapsto z$) is also contained in it.

Jonathan Y.
  • 4,222
  • Right, great thanks! I just had a new thought: isn't it true then that the polynomials are the algebra generated by the identity function together with the $1$ function? – student Dec 20 '13 at 13:04
  • Do you have any ideas about (??)? – student Dec 20 '13 at 13:11