Timeline for Pfaffian representation of the Fermat quintic
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 12, 2018 at 0:06 | comment | added | Abdelmalek Abdesselam | @VictorProtsak: Nevermind my last comment. I see what you meant. Nice observation! | |
| Jan 11, 2018 at 17:31 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
deleted 112 characters in body
|
| Jan 11, 2018 at 17:12 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
added 723 characters in body
|
| Jan 11, 2018 at 15:25 | comment | added | Abdelmalek Abdesselam | @VictorProtsak: Could you explain a bit your statement about the four-cycle. I am a bit too tired this morning to see the connection with my condition: no other pair partition except $P_1,\ldots,P_n$. | |
| Jan 11, 2018 at 15:09 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
added 2893 characters in body
|
| Jan 11, 2018 at 7:13 | comment | added | Will Sawin | @VictorProtsak In the $n=3$, one can show that the Fermat cubic in 3 variables is the determinant of a $3 \times 3$ matrix of linear forms in $3$ variables, which gives it as the Pfaffian of a $6 \times 6$ matrix. However this method is based on cancellation rather than showing that all but the power terms don't appear. | |
| Jan 11, 2018 at 1:20 | comment | added | Abdelmalek Abdesselam | @VictorProtsak: Thank you for the clarification. I am not expecting a construction which works for any $n$. I believe if it works it should be special to $n=5$. | |
| Jan 11, 2018 at 1:06 | comment | added | Victor Protsak | @Abdelmalek: $n$ is the order of the Pfaffian, so that e.g. in your case, $n=5$. The fact that no such system of $n$ complete matchings of $[2n]$ exists for $n=3$ does not logically preclude its existence for $n=5$, but it does seem to make it less likely. | |
| Jan 11, 2018 at 0:34 | comment | added | Abdelmalek Abdesselam | @VictorProtsak: Thank you for your efforts. I did not quite get what $n$ is. | |
| Jan 11, 2018 at 0:10 | comment | added | Victor Protsak | By a somewhat laborious search, I verified that this approach does not work for $n=3$. The combinatorial condition is as follows: each $\{a,b\}$ occurs at most once and there are no "4-cycles" of the form $\{a,b\}, \{b,c\}, \{c,d\}, \{d,a\}$. | |
| Jan 10, 2018 at 23:12 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
deleted 12 characters in body
|
| Jan 10, 2018 at 22:12 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
added 230 characters in body
|
| Jan 10, 2018 at 22:06 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
added 88 characters in body
|
| Jan 10, 2018 at 22:05 | comment | added | Abdelmalek Abdesselam | @WillSawin: You are right. I went a bit too fast, so my argument as is does not work yet. | |
| Jan 10, 2018 at 21:49 | comment | added | Will Sawin | I see why the Fermat quintic terms appear, but why do the other terms vanish? For instance, it seems to me that the coefficient of $x_1^3 x_4 x_5$ in the Pfaffian of your matrix is nonzero, as the partition $\{ \{1,2\}, \{3,4\}, \{5,6\}, \{7,9\}, \{8,10\}\}$ contributes and no other partition does. | |
| Jan 10, 2018 at 21:34 | comment | added | Libli | This looks a rather interesting answer. I am not very familiar with this type of reasonning, but I am definitively very interested. Could you explain a bit more your sentence "A moment of thought..." | |
| Jan 10, 2018 at 21:21 | history | edited | Abdelmalek Abdesselam | CC BY-SA 3.0 |
added 573 characters in body
|
| Jan 10, 2018 at 21:01 | history | answered | Abdelmalek Abdesselam | CC BY-SA 3.0 |