By Hall B.C.

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**Example text**

EX ≤ e X . It is not true in general that eX+Y = eX eY , although by 4) it is true if X and Y commute. This is a crucial point, which we will consider in detail later. ) Proof. Point 1) is obvious. Points 2) and 3) are special cases of point 4). To verify point 4), we simply multiply power series term by term. ) Thus eX eY = I +X+ X2 + ··· 2! I +Y + Y2 +··· 2! 2. 5) eX eY = ∞ m m=0 k=0 ∞ 1 X k Y m−k = k! (m − k)! m=0 m! m k=0 m! X k Y m−k . (m − k)! Now because (and only because) X and Y commute, m n (X + Y ) = k=0 m!

The only non-trivial point is the Jacobi identity. The only way to prove this is to write everything out and see, and this is best left to the reader. Note that each triple bracket generates four terms, for a total of twelve. Each of the six orderings of {X, Y, Z} occurs twice, once with a plus sign and once with a minus sign. 29. A subalgebra of a real or complex Lie algebra g is a subspace h of g such that [H1, H2] ∈ h for all H1, H2 ∈ h. If g is a complex Lie algebra, and h is a real subspace of g which is closed under brackets, then h is said to be a real subalgebra of g.

Proof. 11 in Br¨ ocker and tom Dieck. In view of what we have proved about the matrix logarithm, we know this result for the case of GL(n; C). To prove the general case, we consider a matrix Lie group G < GL(n; C), with Lie algebra g. 24. Suppose gn are elements of G, and that gn → I. Let Yn = log gn, which is defined for all sufficiently large n. Suppose Yn / Yn → Y ∈ gl (n; C). Then Y ∈ g. Proof. To show that Y ∈ g, we must show that exp tY ∈ G for all t ∈ R. As n → ∞, (t/ Yn ) Yn → tY . Note that since gn → I, Yn → 0, and so Yn → 0.