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Jun 21, 2017 at 0:06 vote accept D_S
Jun 18, 2017 at 13:01 comment added Mikhail Borovoi (2) $X_\alpha$ is a nilpotent matrix, so in char. 0 $\exp X_\alpha$ is defined. (3) You have $H_\alpha$, then you choose $X_\alpha$ and get $X_{-\alpha}$ such that $[X_\alpha,X_{-\alpha}]=H_\alpha$. This triple $(X_\alpha, H_\alpha,X_{-\alpha})$ defines an embedding ${\rm Lie\,SL}(2)\to {\rm Lie}\,G$, which is the differential of some homomorphism ${\rm SL}(2) \to G$. This homomorphism is what you need.
Jun 17, 2017 at 16:49 history edited D_S
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Jun 17, 2017 at 15:56 answer added Jim Humphreys timeline score: 1
Jun 17, 2017 at 3:12 comment added LSpice (1) $H_\alpha$ as an element of $\mathfrak t$ is $\mathrm dH_\alpha(1)$ in your sense. (2) Yes, precisely; it uses an identification of the root subspace with the root subgroup. This is canonical only up to conjugation by $T$. (3) This is the homomorphism coming from the theory of Jacobson–Morosov triples. It too is canonical only up to conjugation by something or other.
Jun 17, 2017 at 2:36 history asked D_S CC BY-SA 3.0