Timeline for The Matrix-Tree Theorem without the matrix
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2011 at 4:17 | comment | added | Gjergji Zaimi | Smith's normal form shows it's advantages especially in q-enumeration. Why do you say it is not combinatorial? | |
| Oct 3, 2011 at 2:57 | comment | added | darij grinberg | Well, I prefer having signed quantities. I personally find nothing wrong with Sylvester-sieve direct proofs of the matrix-tree theorem, without the detour through the sandpile group. A proof using Smith's normal form cannot be considered combinatorial anyway. | |
| Oct 3, 2011 at 1:30 | comment | added | Gjergji Zaimi | @Darij: You are right, but I was only commenting on the Critical group/spanning trees bijection. I guess I'm saying that from a "combinatorics without linear algebra" perspective, the determinant of the Laplacian should be replaced by the cardinality of the critical group. | |
| Oct 3, 2011 at 1:17 | comment | added | darij grinberg | I am skeptical with regard to this proof: If you are using Smith's normal form, it only shows that the cardinality of the critical group is THE ABSOLUTE VALUE OF the determinant of the Laplacian. Now have fun proving that it is $\geq 0$ without linear algebra (such as properties of diagonally dominant matrices). | |
| Aug 22, 2011 at 10:39 | comment | added | Daniel Moskovich | Thanks! This looks extremely interesting, for me if not for the students! | |
| Aug 22, 2011 at 9:44 | history | answered | Gjergji Zaimi | CC BY-SA 3.0 |