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By Miller G.A.

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Consider the bilinear form B(u, v) on W defined by u ∧ v = B(u, v)ω. (a) Show that B is nondegenerate. (b) Show that B is skew-symmetric if n is odd and symmetric if n is even. (c) Determine the signature of B when n is even and F = R, (Notation of the previous exercise) Let V = F4 with basis {e1 , e2 , e3 , e4 } and let ω = e1 ∧ e2 ∧ e3 ∧ e4 . Define ϕ(g)(u ∧ v) = gu ∧ gv for g ∈ SL(4, F) and u, v ∈ F4 . / SO( 2 F4 , B) is a group homomorphism with kernel Show that ϕ : SL(4, F) {±I}. ) (Notation of the previous exercise) Let ψ be the restriction of ϕ to Sp(2, F).

16. The differential of the inclusion map satisfies (dιG )I (T (G)I ) = {(XA )I : A ∈ Lie(G)} . 3 Closed Subgroups of GL(n, R) 33 Proof. 12). We define the tangent vector vA ∈ T (G)I by vA f = d f exp(tA) dt t=0 for f ∈ C∞ (G) . By definition of the differential of a smooth map, we then have (dιG )I (vA ) f = (XA )I . This shows that (dιG )I (T (G)I ) ⊃ {(XA )I : A ∈ Lie(G)} . 26) are the same. Define the left translation operator L(y) on C∞ (G) by L(y) f (g) = f (y−1 g) for y ∈ G and f ∈ C∞ (G).

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. If we can show that ϕε (t) = exp(tX) for some X ∈ Mn (R) and all t ∈ R, then ϕ(t) = exp (t/ε)X . Thus it suffices to treat the case ε = 1.

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