Chapter 19: Problem 8

Show that the group of all Lie algebra automorphisms of a Le algebra \(A\) form a Lie subgroup of \(\operatorname{Aut}(A) \subseteq G L(\mathcal{A})\). (b) A lincar operator \(D: \mathcal{A} \rightarrow \mathcal{A}\) is called a deruation on \(A\) ?f \(D[X, Y]=[D X, Y]+[X, D Y]\). Prove that the set of all derivations of \(\mathcal{A}\) form a Lie algebra, \(\partial(\mathcal{A})\), which is the Lie algebra of Aut( \(\mathcal{A}\) ).

Short Answer

Expert verified

The group of all Lie algebra automorphisms of \( A \) forms a Lie subgroup of \( Aut(A) \). The set of all derivations of \( \mathcal{A} \) forms a Lie algebra, denoted as \( \partial(\mathcal{A}) \), and is the Lie algebra of \( Aut(\mathcal{A}) \).

Step by step solution

Show Lie algebra automorphisms form a Lie subgroup

In order to prove that the group of all Lie algebra automorphisms form a Lie subgroup of Aut(A), we must show two things: closure under group operation and inversion. The group operation for Aut(A) is function composition, which is associative. If \( \phi, \psi \in Aut(\mathcal{A}) \), then \( \phi \circ \psi \) is an automorphism of \( A \), which shows closure under group composition.Inversion in Aut(A) is function inversion. If \( \phi \in Aut(\mathcal{A}) \), then \( \phi^{-1} \) exists and is an automorphism of \( A \), which shows closure under inversion.

Define the derivation on A

A linear operator \( D: \mathcal{A} \rightarrow \mathcal{A} \) is called a derivation of \( A \) if it follows the Leibniz rule: \( D[X, Y] = [DX, Y] + [X, DY] \).

Prove derivations form a Lie algebra

In order to show that the set of all derivations of \( \mathcal{A} \) form a Lie algebra, we must show two things: closure under addition and closure under Lie bracket.For closure under addition: if \( D_1, D_2 \) are derivations, then \( D_1 + D_2 \) is a derivation since \( (D_1 + D_2)[X, Y] = D_1[X, Y] + D_2[X, Y] = [D_1X, Y] + [X, D_1Y] + [D_2X, Y] + [X, D_2Y] = [(D_1 + D_2)X, Y] + [X, (D_1 + D_2)Y] \)For closure under Lie bracket: if \( D_1, D_2 \) are derivations, then \( [D_1, D_2] \) is a derivation since \( [D_1, D_2][X, Y] = D_1[D_2[X, Y]] - D_2[D_1[X, Y]] = D_1[[D_2X, Y]] + D_1[[X, D_2Y]] - D_2[[D_1X, Y]] -D_2[[X, D_1Y]] \).

Show that derivations form the Lie algebra of Aut(A)

Finally, we prove that the Lie algebra of \(Aut(\mathcal{A})\) is the set of all derivations of \( \mathcal{A} \). Given \( \phi_t \in Aut(\mathcal{A}) \) is a smooth curve, we define \( D = \frac{d \phi_t}{dt}\Big|_{t=0} \). By applying the equation from previous step, we can find that \( D[X, Y] = [DX, Y] + [X, DY] \), which is the definition of a derivation. This implies that \( \phi \) is really a derivation of \( A \), hence \( \partial(\mathcal{A}) \) is a Lie algebra of \( Aut(\mathcal{A}) \).