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We can use the result proved in this question to build one. There it is shown that $f : (-\frac{\pi}{2}, \frac{\pi}{2}) \to \mathbb{R}$ defined by $$f(x) = \tan(x)$$ is a diffeomorphism. To obtain the function you are looking for, use four copies of that one: $f : (-\frac{\pi}{2}, \frac{\pi}{2})^4 \to \mathbb{R}^4$ $$f(x_1, x_2, x_3, x_4) = ... 3 It follows from a much more general fact: Whenever a Lie group G acts smoothly on a smooth manifold M, its orbits are immersed smooth manifolds. This can be derived from the following observations: For each p\in M, the isotropy group G_p = \{g\in G: g\centerdot p=p\} is a closed subgroup of G. The quotient space G/G_p has a unique smooth ... 3 It is not true that every topological 4-manifold admits a Kirby diagram. In fact, a handle-decomposition of a 4-manifold and a Kirby diagram for that 4-manifold are equivalent notions. The Kirby diagram is exactly the attaching data for the handles. By a theorem of Morse, every smooth n-manifold admits a handle-decomposition so in particular every ... 2 This is a consequence of Sard's Theorem. Since \dim\Bbb R^{n-1}<\dim S^n, a map F:\Bbb R^{n-1}\to S^{n} has critical values everywhere and hence its image F(\Bbb R^{n-1}) has measure zero in S^n. This is not true in general for continuous functions, for example because of the existence of space filling curves \Bbb R\to S^2. 2 Yes, this is correct. Another way of proving this fact is to pick an ordered basis \{v_1, \dots, v_n\} for T_eG and define vector fields V_1, \dots, V_n by V_i(g) = dL_g(v_i) where L_g : G \to G is given by v \mapsto gv. As dL_g is an isomorphism ((dL_g)^{-1} = dL_{g^{-1}}), \{V_i(g) \mid i =1, \dots, n\} is a basis of T_gG for every g ... 2 For putting a manifold structure on the quotient, it's irrelevant whether G_x is normal. The quotient of a Lie group modulo a closed subgroup always has a unique smooth manifold structure such that the group action is smooth -- this is the Quotient Manifold Theorem (see Theorem 21.10 in my Introduction to Smooth Manifolds, 2nd ed.). Normality of the ... 2 Hint The elements of the Lie algebra \mathfrak{g} \cong T_{\Bbb I} G are the tangent vectors at the identity element \Bbb I \in G to curves in G through that point. 1 Every non-zero holomorphic function has a holomorphic square root on any simply connected open set. Pick any open \Omega with a function f:\Omega\to\mathbb C which does not has a square root and consider the covering of \Omega by the discs it contains. 1 Let P(x_0, \dots, x_n) = (P_0(x_0, \dots, x_n), \dots, P_k(x_0, \dots, x_n)) where P_0, \dots, P_k : \mathbb{R}^{n+1} \to \mathbb{R} are smooth homogeneous polynomials of degree d. If you had done this, I think, based on what you had already done, you would have been able to complete the proof. Just in case, I have included the proof below. Let ... 1 Let f(x) = \exp (-1/(x(1-x)), x\in (0,1), f(x) = 0 elsewhere. Then f\in C^\infty(\mathbb {R}), and f(1/2) = e^{-4} is a critical value. Let q_1,q_2, \dots  be the rationals. Then$$F(x)=\sum_{n=-\infty}^{\infty}q_nf(x-n)$$is C^\infty on \mathbb {R}, and the critical values of F include q_1e^{-4},q_2e^{-4},\dots , which comprise a dense ... 1 My friend helped me solve this problem. His solution uses more analysis than I would like, but here it is. Choose a chart for M on an open set U identifying x with the origin, and then choose V \subset U so that f(V) \subseteq U. Now we may view f as a diffeomorphism in a neighborhood of the origin in \mathbb{R}^n that has 0 as a non-isolated ... 1 "Why would one define B(X,Y)=tr(ad(X)ad(Y)) ?" You are right, for matrix algebras I would define a bilinear form more simply, namely just by C(X,Y)=tr(X)tr(Y). This is very natural, because a trace form for linear operators is the easiest thing you can imagine. If we do not have linear operators X,Y, then we can enforce this, by using the adjoint ... 1 I am not sure If I understand the definition correctly, but if I do, then take a look at the maps$$ f : (0,1) \rightarrow [0,1], x \mapsto x$$and$$ g : [0,1] \to (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$These are embeddings and composing these we get$$ f \circ g : [0,1] \rightarrow [0,1], x \mapsto \frac{1}{3} x + \frac{1}{3}$$and$$ g \circ f ...
Let $\{e_1, \dots ,e_n,f_1,\dots, f_n \}$ be a basis for $T_xM$ with $\omega (e_i,f_i)=1$ and all other pairs vanish. Such a basis exists by the standard diagonalization theorem for symplectic matrices. Then $\omega ^n(e_1,f_1, \dots , e_n,f_n)=\prod_i \omega(e_i,f_i)=1$, so $\omega^n$ is non-zero at $x$. Since this is true for each $x$, $\omega^n$ is a ...