0
$\begingroup$

What is an example of an algebraic (=Zariski closed) subset $C$ of $\mathbb{R}^2$ which is a topological retract of $\mathbb{R}^2$, but there is no algebraic retraction $P:\mathbb{R}^2 \to C$?

What is an example of such a situation in $\mathbb{C}^2$?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ $y^2=x^3-x+1$ should work. There's no algebraic retraction because as an elliptic curve, it's not unirational. $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 10:19
  • 2
    $\begingroup$ What about $xy=0$? $\endgroup$
    – Piotr Achinger
    Commented Mar 21, 2019 at 17:43
  • $\begingroup$ @PiotrAchinger in fact the motivation for this question was this curve xy=0. I was thinking whether it is a TE space. Then i realized it is a retract of the plane hence a TE space but I could not find an algebraic retraction. However i can not understand your comment $\endgroup$
    – Ali Taghavi
    Commented Mar 21, 2019 at 19:14
  • 1
    $\begingroup$ Yes $xy=0$ works, in the real or complex case, just using Zariski irreducibility (the image of the plane by a polynomial map should be Zariski-irreducible). $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 19:41
  • 1
    $\begingroup$ No, $xy=0$ is Zariski-connected. Use irreducibility instead. $\endgroup$
    – YCor
    Commented Mar 21, 2019 at 19:44

0

Reset to default

You must log in to answer this question.