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Unformatted text preview: ations,
exp adx = exp (λx + ρ−x ) = exp λx exp ρ−x
But clearly, exp λx = λexp x and exp ρ−x = ρexp(−x) . So,
exp adx (y ) = (exp x)y (exp(−x)) = (exp x)y (exp(x)−1 )
8 MATH 223A NOTES 2011 LIE ALGEBRAS 2.3. Exercises. What about derivations in characteristic p?
(1) Show that δ p is a derivation.
(2) For the Lie algebra of an associative algebra over a ﬁeld of characteristic p show
that adp = adxp .
(3) (in any characteristic) Let ϕ : V × V → F be a nondegenerate skew-symmetric
bilinear pairing. Then the Heisenberg algebra of f is given by L = V ⊕ F with
[(v, a)(w, b)] = (0, f (v, w))....
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- Fall '11