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1 m dm+1 By the 0 (1 − s) dsm+1 (f (exp(sX))ds. 1 1 m ˜ m+1 f (exp(sX))ds. Let (m+1)! 0 (1 − s) X n of g and write X = Then i=1 xi Xi . first equation Rm (X) = {X1 , . . , Xn } be a basis ˜ m+1 = ( n xi X ˜ i )m+1 which is a finite sum of products of X ˜ i of X i=1 order m + 1 indexed by the various partitions of m + 1 into n parts with coefficients, the product of the corresponding xi ’s. Now each of these coefficients is ≤ ||X||n+1 . 1] m+1 1 m we get ||Rm (X)|| ≤ ||X|| (m+1)! cdn,m 0 (1 − s) ds.

In particular k is a subalgebra of gl(V ). When V has dimension n one also writes o(n, k) for k. Since in o(n, k), the trace of every element of o(n, k) is zero, we sometimes also write so(n, k). Similar arguments apply in the case of a vector space V = Cn over the complex field and a Hermitian form ·, · . Then we get the real Lie algebra of skew hermitian operators on V = (Cn , ·, · ) which we denote by u(n). We leave it to the reader to check that this is not a complex Lie algebra. 8. For positive integers n, p and q where p + q = n, consider the matrix A = diag(1, .

Let H be a closed subgroup of a Lie group G and h = {X ∈ g : exp(tX) ∈ H for all t ∈ R}. Then h is a Lie subalgebra of g. Proof. Clearly h is closed under scalar multiplication. 3 part 1, shows h is closed under addition, while part 2 shows h is closed under bracketing. 5. A closed subgroup H of a Lie group G is a Lie group with relative topology. Proof. It is enough to prove the theorem when H is a connected subgroup. 4. 3 there exists a Lie subgroup H ′ ⊂ G such that T1 H ′ = h. Let s be a complementary subspace for h in g such that g = h ⊕ s.

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