- #1
TIMRENNER
- 1
- 0
Hello all,
I'm attempting to find in literature a method of determining from a Lie algebra's full root system in an arbitrary basis which roots are simple. It seems there are many books, articles, etc on getting all the roots from the simple roots but none that go the other way.
My task is to take a system of positive roots and write an algorithm which builds the Dynkin diagram. I can identify the group quite easily, but building the Dynkin diagram is difficult. The algorithm I'm currently trying picks the first root and then looks for ones that have the correct dot products one at a time, thus building the diagram. However, it sometimes stops at smaller groups, identifying a D5 (SO(10)) group as D4 (SO(8)) or an E7 as an E6.
Is there any way other than using the Dynkin diagrams to identify a root as simple, or should I just start building linear combinations?
By the way, the basis for the roots is not specified, so I can't use the generic root structures for the groups presented in the literature.
Thanks,
Tim
I'm attempting to find in literature a method of determining from a Lie algebra's full root system in an arbitrary basis which roots are simple. It seems there are many books, articles, etc on getting all the roots from the simple roots but none that go the other way.
My task is to take a system of positive roots and write an algorithm which builds the Dynkin diagram. I can identify the group quite easily, but building the Dynkin diagram is difficult. The algorithm I'm currently trying picks the first root and then looks for ones that have the correct dot products one at a time, thus building the diagram. However, it sometimes stops at smaller groups, identifying a D5 (SO(10)) group as D4 (SO(8)) or an E7 as an E6.
Is there any way other than using the Dynkin diagrams to identify a root as simple, or should I just start building linear combinations?
By the way, the basis for the roots is not specified, so I can't use the generic root structures for the groups presented in the literature.
Thanks,
Tim