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Aug 27, 2010 at 7:49 vote accept TOM
Aug 26, 2010 at 18:49 vote accept TOM
Aug 27, 2010 at 7:49
Aug 26, 2010 at 17:26 comment added José Figueroa-O'Farrill 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.
Aug 26, 2010 at 17:01 history edited TOM
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Aug 26, 2010 at 17:01 comment added TOM because I want to see how upper half plane$H_1$ can be holomorphic ibemdding to siegel space $H_3$ as totally geodesic ,see Ichiro Statke's paper"Holomorphic embedding of symmetric domains into a seigel space". And what is the corresponding map of lie groups $SL_2$ to $SP_6$ , if it is arise from map of lie groups? is it just the following type? A-->diag(A,1,1);A-->diag(1,A,1);A-->diag(1,1,A);A-->diag(A,A,1); A-->diag(A,1,A);A-->diag(1,A,A);A-->diag(A,A,A) – TOM 27 secs ago
Aug 26, 2010 at 16:58 comment added José Figueroa-O'Farrill Thanks! It's satisfying to know about yet another context in which Dynkin's results are useful.
Aug 26, 2010 at 16:30 vote accept TOM
Aug 26, 2010 at 17:05
Aug 26, 2010 at 15:56 comment added José Figueroa-O'Farrill 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.
Aug 26, 2010 at 15:53 answer added José Figueroa-O'Farrill timeline score: 5
Aug 26, 2010 at 15:48 answer added Jim Humphreys timeline score: 6
Aug 26, 2010 at 15:11 comment added Skip 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".
Aug 26, 2010 at 15:07 comment added Theo Johnson-Freyd I assume you are working over an algebraically closed field of characteristic 0?
Aug 26, 2010 at 13:42 history edited TOM CC BY-SA 2.5
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Aug 26, 2010 at 13:36 comment added Jim Humphreys 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.
Aug 26, 2010 at 13:31 history asked TOM CC BY-SA 2.5