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By Wolfgang Arendt, Annette Grabosch, Günther Greiner, Ulrich Groh, Heinrich P. Lotz, Ulrich Moustakas, Rainer Nagel, Frank Neubrander, Ulf Schlotterbeck
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Extra info for One-parameter Semigroups of Positive Operators
Example text
In this case for each integer k ≥ 1/s1 there exists an element in Zk ∈ B1/k (0) such that exp(Zk ) ∈ H and Zk ∈ / Lie(H). We write Zk = Xk +Yk with Xk ∈ Lie(H) and 0 = Yk ∈ K. Then ϕ(Zk ) = exp(Xk ) exp(Yk ) . Since exp(Xk ) ∈ H, we see that exp(Yk ) ∈ H. We also observe that Yk ≤ 1/k. Let εk = Yk . Then 0 < εk ≤ 1/k ≤ s1 . For each k there exists a positive integer mk such that s1 ≤ mk εk < 2s1 . Hence s1 ≤ mkYk < 2s1 . 21) Since the sequence mkYk is bounded, we can replace it with a subsequence that converges.
It follows that B([X,Y ]v, w) = −B(v, [X,Y ]w), and hence so(V, B) is a Lie subalgebra of gl(V ). Suppose V is finite-dimensional. Fix a basis {v1 , . . , vn } for V and let Γ be the n×n matrix with entries Γi j = B(vi , v j ). 2, we see that T ∈ so(V, B) if and only if its matrix A relative to this basis satisfies At Γ + Γ A = 0 . 8) can be written as At = −Γ AΓ −1 . In particular, this implies that tr(T ) = 0 for all T ∈ so(V, B). Orthogonal Lie Algebras Take V = Fn and the bilinear form B with matrix Γ = In relative to the standard basis for Fn .
It is a fundamental result in Lie theory that all homomorphisms from R to GL(n, R) are obtained in this way. 5. Let ϕ : R additive group R to GL(n, R). Then there exists a unique X ∈ Mn (R) such that ϕ(t) = exp(tX) for all t ∈ R. Proof. The uniqueness of X is immediate, since d exp(tX) dt t=0 =X . To prove the existence of X, let ε > 0 and set ϕε (t) = ϕ(εt). Then ϕε is also a continuous homomorphism of R into GL(n, R). 3 we can choose ε such that ϕε (t) ∈ exp Br (0) for |t| < 2, where r = (1/2) log 2.