Thm. f : M → N embedding implies f(M) is a submanifold of N. Recall K is a submanifold of N if ∀q ∈ K ⊂ N, ∃g^{smooth} : V^{open} ⊂ N → R^{n−m}, q ∈ V such that K ∩ V = g^{−1}(0) and rank d_{p}g = n − m Proof. Since f : M → N embedding, f : M → N is a 1-1 immersion and f : M → f(M) is a homeomorphism where f(M) is a subspace of N Take 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, 2 Defn. Suppose f : M → N is smooth. p ∈ M is a critical point and f(p) is a critical value if rank df_{p} < n. If p ∈ M is not a critical point, then it is a regular point. If q ∈ N is not a critical value, then it is a regular value. Note: q ∈ N is a regular value iff f^{−1}(q) = ∅ or ∀p ∈ f^{−1}(q), df_{p} = 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 n^{2} manifold. A ∈ Gl(n, R) is orthogonal if A^{t}A = I. The orthogonal group = O(n) = {A ∈ Gl(n, R) | A^{t}A = 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) = ±1 SO(n) is open in 0(n). s : Gl(n, R) → Gl(n, R), s(A) = A^{t}A is smooth. Let S = the set of symmetric matrices. Then S = is an manifold.

s : Gl(n, R) → S, s(A) = A^{t}A is smooth. s^{−1}(I) = Claim: I is a regular value of s : Gl(n, R) → S, s(A) = A^{t}A. That is, if A ∈ O(n), d_{A}S has rank \frac{n(n+1)}{2}. n^{2} − \frac{n(n+1)}{2} = \frac{n(n−1)}{2}. Thus if I is a regular value, O(n) is an \frac{n(n−1)}{2} submanifold of Gl(n, R).