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4

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 ...


1

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 ...



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