By Jerry R. Shipman

**REA’s Algebra and Trigonometry challenge Solver **

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**Extra info for Algebra & Trigonometry Problem Solver**

**Sample text**

22 §3. Real forms and involutive automorphisms Proposition 4. Any w ∈ W is of the form w = ϕ|t(R), where ϕ ∈ Int g has the following properties: ϕ(t) = t, ϕτ = τ ϕ. Proof. 17), any w ∈ W is a product of reﬂections ri = rαi for some simple roots αi ∈ Π. 12), ri = ϕi |t(R), where ϕi = ϕαi = exp(ad π2 (ei − fi )). Thus, we may choose ϕ as a product of ϕi , and it is suﬃcient to verify that ϕi τ = τ ϕi . Since τ ∈ Aut gR , we have τ ϕi τ −1 = exp(ad π2 τ (ei − fi )) = ϕi . Theorem 1. The real form gτ is compact.

Proof. The assertions (i)–(iii) are trivial. Let us prove (iv). Let G and H be connected Lie groups with tangent Lie algebras g and h such that there exists a homomorphism F : G → H satisfying de F = f (it always exist, if G is simply connected). 2), Int g = Ad G and Int h = Ad H. , F αg = αF (g) F . Diﬀerentiating this relation, we get f ϕ = (Ad F (g))f , and so we may set ϕ = Ad F (g). In the case when ϕ = exp(ad x) = Ad(exp x), x ∈ g, we have ϕ = Ad F (exp x) = exp(ad f (x)). To prove (v), we note that, by (iii), idg ↑ ψ means ψ|f (g) = id.

Then f (gσ ) ⊂ hσ if and only if f σ = σ f . In fact, if the latter relation holds, then we get immediately f (x) = σ (f (x)) for any x ∈ gσ . Conversely, suppose that f (gσ ) ⊂ hσ . Then f σ and σ f coincide on the real form gσ . Since both mappings are antilinear, they should coincide on the entire vector space g. Generally, we say that a mapping a : g → g extends by f to a mapping a : h → h, whenever f a = a f . In this case, we write a ↑f a . The subscript f may be omitted if it is clear which homomorphism is considered.