Timeline for Computation of the pfaffian of a particular matrix
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2021 at 3:56 | answer | added | SiOn | timeline score: 3 | |
| Oct 9, 2021 at 16:10 | answer | added | Hjalmar Rosengren | timeline score: 2 | |
| Oct 8, 2021 at 11:35 | comment | added | Hjalmar Rosengren | Ordering the exponents decreasingly corresponds to some total order on the perfect matchings. I guess that neighbors in this total order are always related by a transposition. E.g. when n=6 the top coefficient comes from (12)(34)(56) and the second highest from (14)(23)(56) so they are related by flipping 2 and 3. But that is just my guess, it would need a formal proof. I hope someone else can give you a better answer. | |
| Oct 8, 2021 at 11:14 | comment | added | SiOn | Thanks a lot for your comments. Do you see quickly why are the intermediate coefficients alternating? | |
| Oct 8, 2021 at 10:30 | comment | added | Hjalmar Rosengren | I think this should follow rather directly from the definition of a pfaffian as a sum over perfect matchings. If you expand the pfaffian, the condition on $s_i$ means that all resulting exponents of $\alpha$ are distinct, so it is a polynomial in $\alpha$ with coefficients $\pm 1$. The top term comes from the matching $(12),(34),(56),\dots$ which has no crossings and hence coefficient $1$. The lowest term comes from the matching $(1m), (2,m-1),\dots$ (where $n=2m$) which also has coefficient $1$. Of course, it remains to explain why all the intermediate coefficients are alternating. | |
| Oct 8, 2021 at 6:36 | history | asked | SiOn | CC BY-SA 4.0 |