Timeline for Orthogonal Complements of Root Lattices in E_8
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2011 at 14:06 | comment | added | Tony Pantev | The first time I encountered all this was in the excellent paper: Oguiso, Keiji; Shioda, Tetsuji The Mordell-Weil lattice of a rational elliptic surface. Comment. Math. Univ. St. Paul. 40 (1991), no. 1, 83–99. Among other things they show that both embeddings of all five exceptions do occur as narrow Mordell-Weil lattices of rational elliptic surfaces. | |
| Jun 21, 2011 at 14:06 | comment | added | Tony Pantev | The original paper is: Dynkin, E. B. Semisimple subalgebras of semisimple Lie algebras. (Russian) Mat. Sbornik N.S. 30(72), (1952), 349–462. There is also an English translation: E.B. Dynkin, Semisimple subalgebras of semisimple Lie algebras. AMS Translations, volume 6, (1957), 111-244. The part you need is in Table 11 in Chapter II. | |
| Jun 21, 2011 at 13:41 | vote | accept | user4192 | ||
| Jun 21, 2011 at 13:41 | comment | added | user4192 | Thanks so much, Tony. Could you give me the reference for the Dynkin paper? | |
| Jun 21, 2011 at 13:09 | comment | added | Tony Pantev | $A_{1}^{\oplus 4}$, ($D_{4}$, $A_{1}^{\oplus 4}$); \linebreak $A_{3}\oplus A_{1}^{\oplus 2}$, ($A_{3}$, $A_{1}^{\oplus 2}\oplus \langle 4\rangle$); \linebreak $A_{3}^{\oplus 2}$, ($A_{1}^{\oplus 2}$, $\langle 4 \rangle^{\oplus 2}$), \linebreak $A_{5}\oplus A_{1}$, ($A_{2}$, $A_{1}\oplus \langle 6 \rangle$); \linebreak $A_{7}$, ($A_{1}$, $\langle 8\rangle$). \linebreak So for the primitive embeddings you get a uniquely determined orthogonal complement. | |
| Jun 21, 2011 at 13:00 | comment | added | Tony Pantev | According to Dynkin's classification of root sublattices in $E_{8}$, there are only five root subblattices in $E_{8}$ (modulo the Weyl group action) that admit more than one embedding in $E_{8}$. Moreover each of the five admits exactly two non-equivalent embeddings. The five exceptions (and their possible complements) are: | |
| Jun 21, 2011 at 12:21 | comment | added | user4192 | Sorry, I should have said that PRIMITIVE embeddings. I think then that the discriminant groups of $L$ and $ L^\perp $ will be isomorphic, so at least I still have some hope. I've edited the question to reflect this. | |
| Jun 21, 2011 at 12:11 | history | answered | Tony Pantev | CC BY-SA 3.0 |