2
$\begingroup$

Is there a polynomial function $P: \mathbb{R}^3 \to \mathbb{R}$ with the following property?:

P does not have any critical value and for all $c \neq c'$, $f^{-1}(c)$ and $f^{-1}(c')$ are non isometric Riemannian manifolds(with the metric they inherit from the standard metric of $\mathbb{R}^3$)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
9
$\begingroup$

The polynomial $P(x,y,z)=x^2+y^2+\frac{z^3}{3}+z$ has no critical values. Any level set $L_c := P^{-1}(c)$ is a surface of revolution around the $z$-axis. Here's an image of the $L_c$ for $c=-30,-15,0,15,30$ (drawn in SageMath):

$L_c$ for $c=-30,-15,0,15,30$

Actually, any level set $L_c$ has a unique Killing vector field $X_c$ which vanishes just at one point $O_c$. Let $K(c)$ (resp. $LK(c)$) be the Gauss curvature of $L_c$ at $O_c$ (resp. the value of the Laplace-Beltrami operator applied to the Gauss curvature of $L_c$ at $O_c$). If $L_c$ and $L_{c'}$ are isometric then $K(c)=K(c')$ and $LK(c)=LK(c')$. Using software I got that $K(c)=K(c')$ iff $c=-c'$. Again using the computer, I got $LK(c) \neq LK(-c)$ for all $c$. Here are the formulae, where $u=(\sqrt{9 c^2+4}+3 c)/2$: $$K(c) = \frac{4}{(u^{2/3}-1+u^{-2/3})^2},$$ $$LK(c)=\frac{128 \sqrt[3]{u} \left(-1-\sqrt[3]{u}+u^{2/3}\right) \left(2u+3 c+9c^2u\right)}{\sqrt{9 c^2+4} \left(1-u^{2/3}+u^{4/3}\right)^4}.$$

Improve this answer
$\endgroup$
14
  • 1
    $\begingroup$ I added a picture; hope that's all right. $\endgroup$
    – j.c.
    Commented Jan 9, 2018 at 18:49
  • 1
    $\begingroup$ @j.c. : thank you for the picture and for improving the answer. $\endgroup$
    – Holonomia
    Commented Jan 10, 2018 at 15:17
  • $\begingroup$ @Holonomia Thank you very much for your great and interesting idea. I need times to understand its details. But just another question: On the opposite extreme : is there a polynomial $P(x,y,z)$ without critical value such that all level sets are isometric Riemannian manifold but $P$ is not in the form $z+Q(x,y)$, after a possible permutation in $x,y,z$ and scalar multiplication? $\endgroup$
    – Ali Taghavi
    Commented Jan 11, 2018 at 18:39
  • 1
    $\begingroup$ @Ali Taghavi: I think you mean that $P$ is not of the form $c \cdot H \circ g$ where $g$ is an isometry of $\mathbb{R}^3$, $c$ a constant and $H$ of the form $z + Q(x,y)$. But anyway I do not know if this happen. No idea about geometric Morse theory. $\endgroup$
    – Holonomia
    Commented Jan 11, 2018 at 21:15
  • 1
    $\begingroup$ @Ali Taghavi: The existence of the Killing vector field $X_c$ is obvious being $L_c$ a surface of revolution. The uniqueness of such Killing vector field (up to a multiple, of course) is a consequence of the fact that the Gauss curvature of the level sets $L_c$ is not constant i.e. the existence of another Killing vector field independent of $X_c$ would imply that $L_c$ has constant Gauss curvature. Finally, I do not understand you query about $p(z,w)=z^2 + w^2$. Are $z,w$ the complex coordinates of $\mathbb{C}$? who are $x, \hat{y}, \hat{w}$ ? $\endgroup$
    – Holonomia
    Commented Jan 14, 2018 at 14:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .