Timeline for Is polynomial convexity a topological invariant?
Current License: CC BY-SA 3.0
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| May 24, 2016 at 14:07 | comment | added | Oleg Eroshkin | @AliTaghavi Of course not. Any linear (over $\mathbb{C}) transformation preserves polynomial convexity. Moreover, a polynomial map, obviously, also has this property. I don't know, if there are not polynomial maps with the same property. | |
| May 24, 2016 at 13:23 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 23, 2016 at 18:47 | comment | added | Ali Taghavi | Yes, Not only your example about circle but also your general statment is correct: every compact set in $\mathbb{R}^{n}$ is polynomialy convex in $\mathbb{R}^{n}+i\mathbb{R}^{n}$. So the next question: Assume that $\phi$ is a homeomorphism of $\mathbb{C}^{2}$ such that For every compact polynomialy convex set $K$, $\phi(K)$ is polynomialy convex. Does this imply that $\phi $ is a scalar map $\lambda Id$? | |
| May 23, 2016 at 17:26 | comment | added | Francesco Polizzi | Good. So you are saying that my example is correct, right? | |
| May 23, 2016 at 17:22 | comment | added | Ali Taghavi | and for $a\in \mathbb{R}^{n} \setminus K$, as you said ,the Weierestrass theorem, works. | |
| May 23, 2016 at 17:18 | comment | added | Ali Taghavi | @Ferancesco Thank you for this interesting point: If $K \subset \mathbb{R}^{n}$ is compact then its convex hull is a compact convex subset of $\mathbb{R}^{n}$ so its convex hull is polynomially convex. Then for every point $a\in \mathbb{C}^{n} \setminus \mathbb{R}^{n}$ there is a polynomial $f$ such that $|f(a)|> sup |f|$ on $K$. | |
| May 23, 2016 at 13:29 | comment | added | Francesco Polizzi | I'm not an expert. However, it seems to me that by Stone-Weierstrass approximation any compact subset $K \subset \mathbb{R}^n$ should give a polynomially convex subset of $\mathbb{C}^n$, where $\mathbb{C}^n = \mathbb{R}^n + i \mathbb{R}^n$. So the usual circle $S^1 \subset \mathbb{R}^2$ should provide the desired example in $\mathbb{C}^2$. | |
| May 23, 2016 at 13:21 | comment | added | Ali Taghavi | @Ferancesco By topological circle I mean a subset of $\mathbb{C}^{2}$ which is homeomorphic to $\mathbb{T}^{1}=\{z : |z|=1 \}$. | |
| May 23, 2016 at 13:13 | comment | added | Francesco Polizzi | By "circle" you mean $\{z \, \colon \, |z| \leq 1\}$? If so, yes: any compact, convex set of $\mathbb{C}^N$ is topologically convex. | |
| May 23, 2016 at 13:08 | comment | added | Ali Taghavi | Is there a topological circle in $\mathbb{C}^{2}$ which is polynomial convex? | |
| May 21, 2016 at 13:18 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 21, 2016 at 9:10 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 21, 2016 at 8:56 | comment | added | Ali Taghavi | Thank you very much for your complete answer. I thank you too for your help on revision of my post. | |
| May 21, 2016 at 8:54 | vote | accept | Ali Taghavi | ||
| May 21, 2016 at 7:45 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 21, 2016 at 7:32 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 21, 2016 at 7:27 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| May 21, 2016 at 7:20 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |