Lang's theorem: Difference between revisions

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Revision as of 08:17, 19 June 2014

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== Proof of Lang's theorem ==

Define:

:<math>f_a: G \to G, \quad f_a(x) = x^{-1}a\sigma(x).</math>

Then we have: (identifying the tangent space at ''a'' with the tangent space at the identity element)

:<math>(d f_a)_e = d(h \circ (x \mapsto (x^{-1}, a, \sigma(x))))_e = dh_{(e, a, e)} \circ (-1, 0, d\sigma_e) = -1 + d \sigma_e</math>&nbsp;