Timeline for Developable 3-manifolds in $\mathbb{R}^4$
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2021 at 1:47 | comment | added | Geoffrey Sangston | I tried to understand why $M$ is flat. Hopefully the following helps someone. $T_{q(t) + v}M = \mathrm{span}\{q'(t), w : w \cdot p(t) = w \cdot p'(t) = 0\}.$ Since $w \cdot p(t) = 0$ and $q'(t) \cdot p(t) = g p'(t) \cdot p(t) = 0$, and $|p(t)| = 1$, the curve $p(t)$ parameterizes the image of the Gauss map. Hence the differentials $DU_{q(t) + v}$ of the Gauss map $U : M \to S^n, U(q(t) + v) := p(t)$ are $0$ (when restricted to $T_{q(t) + v}M$). I.e. the second fundamental form of $M$ is trivial. So the Gauss Equation implies the Riemann curvature tensor of $M$ is trivial, so $M$ is flat. | |
| Jul 30, 2011 at 16:20 | comment | added | Deane Yang | Thanks, Robert. I completely forgot that ruled does not imply flat. | |
| Jul 30, 2011 at 15:42 | history | answered | Robert Bryant | CC BY-SA 3.0 |