Timeline for How far can the analogy between a Cayley graph and a symmetric space be pushed?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2010 at 5:51 | comment | added | HenrikRüping | isn't there now still a problem. A group element g might be expressed in different ways as a product of the generators and the map might give different results depending on which choice you make, e.g. s1s2=s3, but s^1^-1s_2^-1 \neq s3^-1 | |
| Mar 25, 2010 at 13:52 | comment | added | Guntram | Thank you for the correction. What I meant was the symmetry which is locally given by $x \mapsto x^{-1}$; globally it will have the description as I have now given above. | |
| Mar 25, 2010 at 13:50 | history | edited | Guntram | CC BY-SA 2.5 |
Clarified definition of i
|
| Mar 25, 2010 at 13:22 | answer | added | Matthew Stover | timeline score: 7 | |
| Mar 25, 2010 at 12:56 | comment | added | HenrikRüping | I don't think, that the inversion is a symmetry of the Cayley- graph. Two vertices are adjacent in the Cayley-graph, if they differ by a (say) right multiplication with an element in S. After inverting this turns into left multiplication. For example in the free group inversion is not even a quasi-isometry: a^nb and a^n have distance 1, but their inverses have distance 2n+1. So this question makes only sense, if one considers a modified version of the Cayley-graph, where left and right multiplication is allowed. | |
| Mar 25, 2010 at 7:36 | history | asked | Guntram | CC BY-SA 2.5 |