Timeline for Integral positive definite quadratic forms and graphs
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 23, 2010 at 8:00 | comment | added | Roland Bacher | The quadratic form is the usual scalar product of $\mathbb R^4$. These five vectors define five roots contained in the root system $D_4$ so it is obvious that they are linearly dependent but this has no relation with the quadratic form which is always non-negative but not necessarily positive-definite. However I answered your question with respect to the framework of my answer which is related but not identical with the framework of your question. | |
| Sep 23, 2010 at 7:41 | comment | added | VA. | They are linearly dependent, so the quadratic form is not positive definite. This may work for the problem you investigated but not for my question. Indeed, I think that there is no positive definite form for this graph, for any way of putting solid and broken edges. | |
| Sep 22, 2010 at 11:55 | comment | added | Roland Bacher | Take the elements $(1,1,0,0)$, $(1,0,1,0)$, $(1,0,0,1)$, $(1,0,0,-1)$, $(0,1,1,0)$ of the root system $D_4$. | |
| Sep 20, 2010 at 12:52 | comment | added | VA. | Do you really get all graphs with ≤8 vertices, as you say? For example, consider the graph on 5 vertices with the edges 12,13,14,15,23,24,25. How do you get that? What is the root system? | |
| Sep 3, 2010 at 13:28 | comment | added | VA. | That is very interesting. This sort of implies that there is little hope for a nice classification. On the other hand, graphs for the $A_n$ and $D_n$ lattices do have a clear structure, and in the $E_6$, $E_7$, $E_8$ cases there are only finitely many graphs. | |
| Sep 3, 2010 at 9:01 | history | answered | Roland Bacher | CC BY-SA 2.5 |