Timeline for Question about immersability of a 3d and 2d Riemannian metrics
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2013 at 0:32 | vote | accept | user61566 | ||
| Oct 10, 2013 at 0:32 | vote | accept | user61566 | ||
| Oct 10, 2013 at 0:32 | |||||
| S Oct 9, 2013 at 23:31 | comment | added | user41101 | ... $z=\vec b$ and (iii). So, I thought that (ii) is a linearized condition for solving for the isometric immersion of G, and that for small $h$ it is also sufficient for it is existence. But evidently, it is not. Thanks again. | |
| S Oct 9, 2013 at 23:31 | comment | added | user41101 | ... $Q=\left[\begin{array}{ccc}\partial_1y & \partial_2 y& \vec b\end{array}\right]\in R^{3\times 3}$, then: $$\det Q>0 \quad \mbox{ and } \quad Q^TQ=G$$ Of course one easily can compute $\vec b$ explicitly, but it is not important. The point is that the condition (ii) is equivalent to: (iii) $(\nabla y)^T (\nabla \vec b)$ is skew-symmetric (at each point $x'=(x_1, x_2)\in\Omega$). Now, attempting finding an isometric immersion of $G$, one writes: $$u(x',x_3)= y(x')+x_3\vec z(x') +\mbox{higher order terms}.$$ Then, the leading order terms in $(\nabla u)^T(\nabla u) - G$ vanish exactly when ... | |
| S Oct 9, 2013 at 23:31 | comment | added | user41101 | Yes, this answers my question. Thank you very much. In fact, I have in the meantime proved that condition (ii) is equivalent to the vanishing of the following three Riemann curvatures of the 3d metric $G$: $$R^3_{221} = R^3_{112} = R_{1212} = 0.$$ Of course, this does not imply flatness of $G$, as your example shows. Let me nonetheless explain how I came up with condition (ii). Let $y$ be an isometric immersion of $G_{2\times 2}$ into $R^3$. Define now the vector field $\vec b:\Omega\to R^3$ with the property that it we write the matrix field with three columns: ... | |
| Oct 9, 2013 at 15:58 | history | answered | Anton Petrunin | CC BY-SA 3.0 |