Thm. f : M → N embedding implies f(M) is asubmanifold of N.Recall K is a submanifold of N if ∀q ∈ K ⊂ N ,∃gsmooth: Vopen⊂ N → Rn−m, q ∈ V such thatK ∩ V = g−1(0) and rank dpg = n − mProof. Since f : M → N embedding, f : M → N isa 1-1 immersion andf : M → f(M) is a homeomorphism where f (M ) isa subspace of NTake q ∈ f(M).Since f is 1:1, ∃!p ∈ M such that f (p) = q.f : M → N an immersion implies f has rank m ≤ n.Thus by the rank theorem,Defn. Suppose f : M → N is smooth.p ∈ M is a critical point and f(p) is a critical value ifrank dfp< n.If p ∈ M is not a criti cal point, then it is a regularpoint.If q ∈ N is not a critical value, then it is a regularvalue.Note: q ∈ N is a regular value iff f−1(q) = ∅ or∀p ∈ f−1(q), dfp= n.Thm 2.3.13: Let q be a regular value of f: M → N .Then either f−1(q) = ∅ or f−1(q) is an (m − n)-submanifold of M .Gl(n, R) is an n2manifold.A ∈ Gl(n, R) is orthogonal if AtA = I.The orthogonal gr oup =O(n) = {A ∈ Gl(n, R) | AtA = I}The special orthogonal group =SO(n) = {A ∈ O(n) | det(A) = 1}O(n), SO(n) are subgroups of Gl(n, R).O(n), SO(n) are closed in Gl(n, R).If A ∈ O(n), then det(A) = ±1SO(n) is open in 0(n).s : Gl(n, R) → Gl(n, R), s(A) = AtA is smooth.Let S = the set of symmetric matrices.Then S = is an manifold.s : Gl(n, R) → S, s(A) = AtA is smooth.s−1(I) =Claim: I is a regular value of s : Gl(n, R) → S,s(A) = AtA.That is, if A ∈ O(n), dAS has rankn(n+1)2.n2−n(n+1)2=n(n−1)2.Thus if I is a r egular value, O(n) is ann(n−1)2sub-manifold of Gl(n,
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