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Is the property of being polynomially convex a topological invariant?

In other words, let $M$ and $N$ be two homeomorphic, compact subsets of $n$-dimensional complex Euclidean space, and assume in addition that $M$ is polynomially convex.

Is $N$ necessarily polynomial convex?

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The answer is no.

In fact, Kallin has shown in [Kal64] that the union of three disjoint closed balls is polynomially convex, but the union of three disjoint closed polydisks needs not to be polynomially convex.

Actually, it turns out that polynomial convexity is not even preserved by (real) linear transformations. For instance, Khudaĭberganov and Kytmanov have given in [KK84] an example of three closed disjoint ellipsoids the union of which is not polynomially convex.

References

[Kal64] E. Kallin: Polynomial convexity: The three spheres problem, Proc. Conf. Complex Analysis (Minneapolis, 1964), Springer, Berlin, pp. 301–304, doi:10.1007/978-3-642-48016-4_26.

[KK84] G. Khudaĭberganov and A. M. Kytmanov: An example of a nonpolynomially convex compact set consisting of three non-intersecting ellipsoids, Sibirsk. Mat. Zh. 25 (5) (1984), 196-198 (in Russian).

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    $\begingroup$ Thank you very much for your complete answer. I thank you too for your help on revision of my post. $\endgroup$
    – Ali Taghavi
    Commented May 21, 2016 at 8:56
  • $\begingroup$ Is there a topological circle in $\mathbb{C}^{2}$ which is polynomial convex? $\endgroup$
    – Ali Taghavi
    Commented May 23, 2016 at 13:08
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    $\begingroup$ By "circle" you mean $\{z \, \colon \, |z| \leq 1\}$? If so, yes: any compact, convex set of $\mathbb{C}^N$ is topologically convex. $\endgroup$
    – Francesco Polizzi
    Commented May 23, 2016 at 13:13
  • $\begingroup$ @Ferancesco By topological circle I mean a subset of $\mathbb{C}^{2}$ which is homeomorphic to $\mathbb{T}^{1}=\{z : |z|=1 \}$. $\endgroup$
    – Ali Taghavi
    Commented May 23, 2016 at 13:21
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    $\begingroup$ 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 $\mathbb{C}^2$. $\endgroup$
    – Francesco Polizzi
    Commented May 23, 2016 at 13:29

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