Timeline for Infinitesimal rigidity vs. local rigidity
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2012 at 2:17 | vote | accept | Earthliŋ | ||
| Jul 25, 2012 at 17:19 | comment | added | Misha | Yes, character schemes are universal, read my original answer. | |
| Jul 25, 2012 at 15:09 | comment | added | Earthliŋ | And this phenomenon (points with 1-dimensional tangent spaces) also occurs for Lie groups (or algebraic groups)? | |
| Jul 25, 2012 at 6:58 | comment | added | Misha | @s.barmeier: 1. Smoothness of character varieties to $SO(3,1)$ at discrete embeddings is proven in my book "Hyperbolic manifolds and discrete groups". It uses heavily 3-dimensional topology and fails in hyperbolic 4-space (an example is given by Johnson and Millson). 2. You have to know a bit of algebraic geometry. Consider the non-reduced scheme $x^2=0$ in the affine line. It consists of a single isolated point, but its Zariski tangent space is 1-dimensional. | |
| Jul 25, 2012 at 6:55 | history | edited | Misha | CC BY-SA 3.0 |
added 907 characters in body
|
| Jul 23, 2012 at 8:33 | comment | added | Earthliŋ | What is wrong with the following argument: If the identity representation $\phi$ is locally rigid, then $\phi\in\mathrm{Hom}(\Gamma,G)$ has an open neighbourhood $G\cdot\phi$. But that means that the subspace of the tangent space $T_\phi(\mathrm{Hom}(\Gamma,G))=Z^1(\Gamma,\mathfrak g)$ corresponding to the conjugation action (which is equal to $B^1(\Gamma,\mathfrak g)$) has maximal dimension, whence $H^1=Z^1/B^1=0$ and $\phi$ is infinitesimally rigid. | |
| Jul 23, 2012 at 8:12 | comment | added | Earthliŋ | Very helpful, as always. Thank you! Is there any geometric reason why local deformations of f.g. discrete subgroups in $\mathrm O(3,1)$ must be inifinitesimally rigid? Why is it just $\mathrm O(3,1)$? | |
| Jul 23, 2012 at 7:39 | history | answered | Misha | CC BY-SA 3.0 |