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Jan 16, 2018 at 8:41 comment added Ali Taghavi Let we have a smooth ( vector or fiber) bundle. Can we equipe the total space to a Riemannian metric such that all fibers are isometric.(In the opposite direction we may require that all fibers all mutually non isometric). Existence of obstructions for such metric is an indication of high strongly non triviality of the bundle? What type of "characteristics" would appear in this way?
Jan 16, 2018 at 8:37 comment added Ali Taghavi @Holonomia Thanks again for your answer. I get the sketch of strategy of your answer but I confess thatI have to effort to understand its details. But this post is a motivation to consider a "Bundle" analogy of this question as follows:
Jan 15, 2018 at 17:54 vote accept Ali Taghavi
Jan 15, 2018 at 7:31 comment added Ali Taghavi @Holonomia You are write. The correct is the following $(z,w) \in \mathbb{C}^2$ correspond to $(x_1,y_1,x_2,y_2) \in \mathbb{R}^4$ where $z=x_1+iy_1,\; w=x_2+iy_2$. The metric of $\mathbb{R}^4$ is read as $dx_1^2+dy_1^2 +dx_2^2+dy_2^2$.
Jan 14, 2018 at 17:42 comment added Holonomia @Ali Taghavi: still do not understand. You wrote $z,w$ are complex numbers, but then you also wrote $z + i w$ which means that $z,w$ are reals.
Jan 14, 2018 at 17:28 comment added Ali Taghavi x,y^,w^ they were typos. They are usual x,y,z,w. The metric of $\mathbb{R}^4$ is denoted by $dx^2+dy ^2 +dz^2 + dw ^2$ where we identify $(x,y,z,w)$ with $(x+iy, z+iw)$.
Jan 14, 2018 at 17:20 comment added Ali Taghavi @Holonomia thank you for your comment. Yes, z and w are complex numbers. Th
Jan 14, 2018 at 14:39 comment added Holonomia @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 $x, \hat{y}, \hat{w}$ ?
Jan 12, 2018 at 19:07 comment added Ali Taghavi Are all (cylinder) level sets of $p(z,w)= z^2 + w^2$ mutually isometric? (We consider them as 2,dimensional submanifolds of $\mathbb{R}^4 \simeq \mathbb{C}^2$ with the standard metric $dx^2+dŷ2+dz^2+dŵ2$.?
Jan 12, 2018 at 18:56 comment added Ali Taghavi Can I ask you to expand your answer?In particular how you find the unique Killing vector field? Moreover can the same argument as in your answer be applied to the following question:
Jan 11, 2018 at 21:15 comment added Holonomia @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.
Jan 11, 2018 at 18:49 comment added Ali Taghavi Moreover, is there some thing as geometric version of Morse theory?
Jan 11, 2018 at 18:39 comment added Ali Taghavi @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?
Jan 10, 2018 at 15:17 comment added Holonomia @j.c. : thank you for the picture and for improving the answer.
Jan 9, 2018 at 18:49 comment added j.c. I added a picture; hope that's all right.
Jan 9, 2018 at 18:47 history edited j.c. CC BY-SA 3.0
add picture, minor grammar
Jan 9, 2018 at 18:28 history edited user44143 CC BY-SA 3.0
deleted 271 characters in body; deleted 7 characters in body; deleted 30 characters in body
Jan 9, 2018 at 17:40 history answered Holonomia CC BY-SA 3.0