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Extra info for The Classical Groups [Lecture notes]

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If α is an isomorphism, it is easy to check that if U is a subspace of V and α(U ) = S ⊂ W , then α∗ (S ◦ ) = U ◦ . Although there is not a natural isomorphism between V and V ∗ , if we dualize again and consider V ∗∗ = (V ∗ )∗ , then in fact there is a natural map S : V → V ∗∗ given by S(v)(φ) = φ(v). It is immediate that S is linear, and moreover it is injective. Hence if V is finite dimensional, S is an isomorphism. e. a hyperplane in V ∗ . Similarly, given a point p ∈ P(V ∗ ), its annihilator in V ∗∗ ∼ = V is a hyperplane in V , thus we see that the set of hyperplanes in V can naturally be identified with a projective space.

When we remove the poles, the range of values for θ and φ are 0 ≤ θ < 2π, 0 < φ < π, and the antipodal map corresponds to the map (θ, φ) → (θ + π, π − φ). (where the first component must be read modulo 2π). But then we may identify the space of lines in R2 with pairs (θ, φ) ∈ [0, π] × (0, π) where we identify (0, φ) with (π, π − φ). Drawing a picture of a square with the appropriate identifications, we immediately see that this space is a Mobius band. THE CLASSICAL GROUPS 33 7. T HE GENERAL LINEAR GROUP Recall that given a vector space V over a field k the group GL(V ) is the set of invertible linear transformations of V .

Our goal, of course, is to show that in fact this group is all of Sp(V ). 3. The group T acts transitively on V − {0}. Indeed T acts transitively on the set of hyperbolic pairs. Proof. Suppose that v, w ∈ V . If B(v, w) = 0, then set u = w − v and a = B(v, w)−1 so that τu,a (v) = v + aB(v, u)u = v + B(v, w)−1 B(v, w)(w − v) = w. If B(v, w) = 0, then we may find φ ∈ V ∗ such that φ(v) and φ(w) = 0. Then setting −1 u = RB (φ) we have B(v, u) = 0 and B(v, w) = 0. Then by what has already been established, there are symplectic transvections τ1 , τ2 taking v to u and w to u respectively.