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how many injective homomorphism between two lie algebra $sl_2 $and $sp_6$ up to conjugate by$Sp_6$ ?

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Please be more specific about the meaning of "how many". Presumably you want to take into account the adjoint group action on the bigger Lie algebra. Since there are long and short roots there, that will largely shape the answer. –  Jim Humphreys Aug 26 '10 at 13:36
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I assume you are working over an algebraically closed field of characteristic 0? –  Theo Johnson-Freyd Aug 26 '10 at 15:07
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Suppose first the field $F$ is algebraically closed. Then every automorphism of $sl_2$ is conjugation by an element of $SL_2(F)$. Therefore the question is the same as asking: Up to conjugation under $Sp_6(F)$, how many copies of $sl_2$ are there in $sp_6$? Over an algebraically closed field of characteristic zero, this question was solved for semisimple Lie algebras by Dynkin in his early 1950s paper "Semisimple subalgebras of semisimple Lie algebras". –  Skip Aug 26 '10 at 15:11
    
Just of curiosity, and because it's a friendly thing to do, would you elaborate as to why you are actually interested in this? I have been interested in this sort of question in the past in two separate contexts: possible topological twistings of quantum field theories and classification of simple W-algebras. –  José Figueroa-O'Farrill Aug 26 '10 at 15:56
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TOM, I don't think that you've got the right embeddings. The first three are clearly conjugate, as are the second set of three. In the reference in my answer you can see the defining vectors for each of the 7 inequivalent embeddings: the first three cases are the ones in your comment above. –  José Figueroa-O'Farrill Aug 26 '10 at 17:26
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2 Answers

up vote 5 down vote accepted

As a follow-up to Jim's answer (which came in as I was typing an inferior answer), let me add that the 7 possible embeddings are given in the $C_3$ entry of Table VI in the paper: Classification of semisimple subalgebras of simple Lie algebras by Lorente and Gruber. It's of course based on Dynkin, but they work out the details up to rank 6.


Added

The defining vectors for the 7 embeddings are given by: (1,0,0), (1,1,0), (1,1,1), (2,2,0), (3,1,0), (3,1,1) and (5,3,1). Recall that the embedding with defining vector $(a,b,c)$ is one for which the Cartan generator $H$ of the $\mathfrak{sl}(2)$ subalgebra is given by $H = a H_1 + b H_2 + c H_3$, where $(H_i)$ is an orthonormal basis of a Cartan subalgebra of $C_3$ containing $H$.

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May I ask one more ? is it that beside A-->diag(A,A,A),all others factor through $SP_4$ ,ie.$SL_2$−−>$SP_4$−−>$SP_6$ –  TOM Aug 27 '10 at 10:01
    
TOM: I don't think so. The table in the paper I mentioned also lists the minimal embedding subalgebras, and, for instance, the last embedding with defining vector (5,3,1) has $C_3$ as minimal embedding subalgebra, whereas the embedding you mention has defining vector (1,1,1). –  José Figueroa-O'Farrill Aug 27 '10 at 10:34
    
(cont'd) Naively I would say that (1,1,1), (3,1,1) and (5,3,1) do not factor through $\mathfrak{sp}_4$. –  José Figueroa-O'Farrill Aug 27 '10 at 10:34
    
Thank you for your great help. –  TOM Aug 27 '10 at 12:18
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To follow up Skip's comment, assuming the underlying field is algebraically closed of characteristic 0, the number of possible embeddings (= injective homomorphisms) of $\mathfrak{s}\mathfrak{l}_2$ into $\mathfrak{s}\mathfrak{p}_6$ up to conjugacy by the adjoint group should be 7. This is the number of nonzero nilpotent conjugacy classes in the simple Lie algebra $C_3$, which in turn are in natural bijection (using Jacobson-Morozov theory) with the $\mathfrak{s}\mathfrak{l}_2$-triples. A good source for the basic theory is Section 11 of Chapter 8 in the Bourbaki treatise Groupes et algebres de Lie (published in English by Springer), supplemented by data on hilpotent orbits in books like those by Collingwood-McGovern and Carter. As Skip points out, general questions of this sort were studied systematically by Dynkin and later refined or generalized by others.

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