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**Example text**

So that |eu − 1| ≤ e|u| − 1 < 1. Thus log(exp u) makes sense for all such u. Since log(exp u) = u for real u with |u| < log 2, it follows by analyticity that log(exp u) = u for all complex numbers with |u| < log 2. 6. 6) log A = ∞ (−1)m+1 m=1 (A − I)m m is defined and continuous on the set of all n×n complex matrices A with A − I < 1, and log A is real if A is real. For all A with A − I < 1, elog A = A. For all X with X < log 2, eX − 1 < 1 and log eX = X. Proof. 6) converges absolutely whenever A − I < 1.

M , 6. PROPERTIES OF THE LIE ALGEBRA d dt 41 d Now for point 3. 4) that dt etX = X. It follows that t=0 etX Y = XY , and hence by the product rule (Exercise 1) t=0 d dt etX Y e−tX = (XY )e0 + (e0 Y )(−X) t=0 = XY − Y X. 15, etX Y e−tX is in g for all t. Since we have (by points 1 and 2) established that g is a real vector space, it follows that the derivative of any smooth curve lying in g must be again in g. Thus XY − Y X is in g. 17. Given two n × n matrices A and B, the bracket (or commutator) of A and B is defined to be simply [A, B] = AB − BA.

Note that gn ∈ G and gn → I. Since the unit ball in D is compact, we can choose a subsequence of {Yn } (still called {Yn}) so that Yn/ Yn converges to some Y ∈ D, with Y = 1. But then by the Lemma, Y ∈ g! This is a contradiction, because D is the orthogonal complement of g. So for every neighborhood U of zero in g, exp (U ) contains a neighborhood of the identity in G. If we make U small enough, then the exponential will be one-toone on U . ) Let log denote the inverse map, defined on exp U . Since U is compact, and exp is one-to-one and continuous on U , log will be continuous.