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4

Following √Člie Cartan, you want to think of your differential system $dR - R\omega = 0$ on $M\times SO(n)$. This $\mathfrak{so}(n)$-valued $1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions $R\colon U\to SO(n)$.


3

Let $M$ and $N$ be complex manifolds. Because we have a local form (there is a neighborhood of a point isomorphic to the the complex unit disc) you can verify that $M \# \overline N$ has a natural complex structure - just line up the copies of $\Bbb C^n \setminus \{0\}$ you're gluing together so the almost complex structures match up. Then the Nijenhuis ...


2

In the equation $$ \int_{\partial \Omega} \alpha = \int_{\Omega} d\alpha $$ it is assumed that $\alpha$ is defined on $\Omega$. If $i\colon\partial\Omega\to\Omega$ is the inclusion, it induces a restriction map $i^*$ on differential forms in the opposite direction. Really the $\alpha$ on the left-hand side is $i^*\alpha$. We don't usually bother with ...


2

If $M$ is a closed codimension-1 submanifold of $\mathbb R^n$, smoothly embedded, then the first method is in fact completely general, in the sense that if you have such a manifold, you can compute the signed distance to $M$ and call this $f$; $0$ is then a regular value, and $f^{-1}(0) = M$. For things like the (open) Mobius band, however, that's not ...


1

1) The rank of $\pi\circ f$, not of $f$. Look at differentials. We may identify $\pi$ with its differential. The point is in showing that $\ker \pi\cap \mathrm{im}(d_pf)=0$. Once you have this you just apply the usual dimension formulas for kernel and image of a linear operator. Remember that you can use $\mathrm{rank} (\pi \circ d_pf)\le ...


1

Eric, even in Guillemin & Pollack you'll find a more general version. Look at Exercise 14 on p. 56. It removes the compactness hypothesis on $Z$. There are also interesting questions to ask along the lines of this: If $f\colon\Bbb R^n\to\Bbb R^n$ (replace with manifolds if you wish) is a local diffeomorphism at each point, what condition(s) are ...


1

For sake of completeness I am writing a solution based on the suggestion of Mike Miller: Theorem: There are no topology $\tau$ and a compatible (finite-dimensional) smooth structure $A$ on $\text{Diff(M)}$ satisfying: (1) The action of $\text{Diff(M)}$ on $M$ is continuous w.r.t $\tau$. (2) $\text{Diff(M)}$ is a Lie group w.r.t $(\tau,A)$. Proof: We ...


1

As Rob Arthan pointed out, to ask whether a quotient space $X$ is diffeomorphic to $S^2$ one has to define a differentiable structure on that space first. An easy way to define a differentiable structure is to pick a homeomorphism $\phi$ from $X$ onto some differentiable manifold $M$ (a sphere in your case), and declare it to be a diffeomorphism: that ...


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I reccomend Loring W. Tu And introduction to manifolds for your purposes, it has close to no prerequisities


1

As a commenters said, a single point is a $0$-dimensional manifold. This is consistent with the fact that $\mathbb{R}^0=\{0\}$, and also with the fact that the boundary of a $d$-dimensional manifold-with-boundary is a $(d-1)$-dimensional manifold. (The boundary of $[0,1]$ is $\{0,1\}$.) Since there is just one topology on a single-point space, it is always ...


1

The answer is obviously no, even if you relax $F$ to be continuous. Your requirement (1) asks for a sequence of continuous functions that approximate $H$ uniformly on $\mathbb{R}$. That is impossible since $H$ is not continuous.



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