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9

[2016-07-25]: Section Differential Geometry added. Although OP narrowed down the post, there are still many more important historical facts which should be addressed to adequately answer the question, than I can give in this answer. Nevertheless here are some aspects, which might be interesting. At least we will see, OP is right when he thinks that many ...


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We provide a careful proof of Theorem 5.1 from Bott & Tu. The main body of the proof is taken from this MO post. Let $M^n$ be a smooth manifold. An open cover $\{U_\alpha\}$ is good if each nonempty finite intersection $U_{\alpha_1}\cap\cdots\cap U_{\alpha_k}$ is diffeomorphic to $\Bbb R^n$. The goal here is to prove the following Theorem 1. Every ...


5

Let $x\in M$, let $\gamma : [0, t] \to M$ be $\gamma(s) = \exp_x (sX)$ for some $X \in T_xM$, with $\|X\| =1$. Then $\ell (\gamma) = t$. Thus $$ \int_0^t \omega_{ \gamma(s)} (\gamma'(s)) ds = t$$ for all $t$. Differentiating gives $$1 = \omega_{\gamma(t)} (\gamma'(t))$$ for all $t$. Putting $t=0$ gives $1 = \omega_x (X)$. But $X \in T_pM$ is ...


5

If you identify $\mathbb{R}^{2n+2}$ with $\mathbb{C}^{n+1}$ then not every two-dimensional real plane in $\mathbb{R}^{2n+2}$ is a one-dimensional complex plane in $\mathbb{C}^{n+1}$. The identification between $\mathbb{R}^{2n+2}$ and $\mathbb{C}^{n+1}$ endows $\mathbb{R}^{2n+2}$ with a complex structure $J \colon \mathbb{R}^{2n+2} \rightarrow \mathbb{R}^{2n+...


4

The range of $c$ is $\{(x,x^{3/2}): x\ge 0\}.$ Let's call the range $c^*.$ Suppose $\gamma (t) =(f(t),g(t))$ is a $C^1$ map of $\mathbb R$ onto $c^*.$ Then $f$ maps $\mathbb R$ surjectively onto $[0,\infty).$ Hence $f(t_0) = 0$ for some $t_0.$ This implies $f$ has a minimum at $t_0,$ which tells us $f'(t_0) = 0.$ Now $g(t)=f(t)^{3/2}$ for all $t.$ Thus $g'(...


4

This depends on how the manifold is defined. In general a metric is a symmetric bilinear form on the tangent space of a manifold $M$, that is you have, in each point $p$ of $M$, a scalar product which maps pairs of tangent vectors to real numbers: $$X_p,Y_p\mapsto g_M(p)(X_p, Y_p) \in \mathbb{R}$$ For a Riemannian metric you require this to be a scalar ...


3

To show that it is infinite, one need only to calculate the expression locally around the simple pole. Locally on an open ball (identified with the unit ball $B$ in $\mathbb C$) we have $$\omega = \left(\frac{1}{z} + g(z)\right) dz,$$ where $g$ is holomorphic (thus bounded). Since $$\int_B \left(\frac 1z\right) \left(\frac 1{\bar z}\right) dz \wedge d\...


3

This is not true. Note that in the above drawing your map never reaches $(1,0)$.


2

Jacob, you probably won't like this, but the easiest way I see to do the calculation rigorously is to use differential forms. (One reference that's somewhat accessible is my textbook Multivariable Mathematics ..., but you can find plenty of others.) If $S$ is a closed (oriented) surface in $\Bbb R^3$ not containing the origin and $S^2$ is the unit sphere, ...


2

According to section 4 of this paper: Stephanie B. Alexander, I. David Berg, and Richard L. Bishop, Cut loci, minimizers, and wavefronts in Riemannian manifolds with boundary, Michigan Math Journal, Volume 40, Issue 2 (1993), 229-237 for a connected complete Riemannian manifold $(M,g)$ without boundary the cut-locus $C(p)$ of a point $p\in M$ equals the ...


2

Use the fact that $B^{n+1}\setminus\{p,q\}\cong S^n\times(0,1)$, and define your map $f\colon S^n\times(0,1)\to S^n$ to just be projection onto the first factor.


2

This is similar to the answer from anomaly, but I hope to make it easier to understand by moving the essential point out of a little subscript and into the text. The first equation in the question, $A=\iint dx'\,dy'$ is fine when the integral is over $A$, as you said. But the next bit, $\dots=\iint G\,dx\,dy$ requires that the integral be over a different ...


2

You could have $\vec n\cdot\vec d = \cos(\theta(s)-\pi/2)$. It depends which direction the curve is bending. In this case, you would have a negative sign.


2

Unfortunately, you are not free to morph the two halves in the very restricted fashion described in your question. You simply do not have that much control over the three given sets. Those three sets can each be much wilder and off kilter than you are imagining. If your proof were correct, then the given plane $P$ that you found, passing through the centers ...


2

Because $\mathbb{H}$ has many symmetries doing it for one geodesic is enough. You can use an appropriate Mobius transformation, $g$, to realize any geodesic $\gamma$ of $\mathbb{H}$ as $\gamma(t)=g\cdot ie^{-t}$. Edit: I saw in your question you didn't want to use explicit geodesics, so you could also do something like this (to see why the denominator ...


1

"Pullback" and "pushforward" are often used when you have a map $f:X \to Y$ and do things with structures built on $X$ and $Y$. The pullback of category theory can be viewed as the specific case where you're pulling "maps with codomain $Y$ back to "maps with codomain $X$". In fact, in a Cartesian category $\mathcal{C}$, the pullback can be used to define a ...


1

The key issue was that $\dot{a}\left(t\right)^{2}+\dot{b}\left(t\right)^{2}=1$ because the curve was parametrized for lenght of arc. Then we had $$ \frac{\partial\varphi}{\partial t}= \left(\dot{a}\left(t\right)\cos\theta,\,\dot{a}\left(t\right)\sin\theta,\,\dot{b}\left(t\right)\right),$$ $$\frac{\partial\varphi}{\partial\theta}= \left(-a\left(t\right)\...


1

In differential geometry there are lots of curvatures. In your case normal curvature is what you are looking at. Compound Gauss curvature surfaces have either positive or negative sign. The sign allows you to divide either type (elliptic/hyperbolic points) into 4 quadrants, the dividing lines having of max/min or extreme curvatures. Cyclic variation of ...


1

A bit silly, since as mentioned, you can prove it by hand, but: covers of complete manifolds are complete, and compact manifolds are automatically complete. There are a great deal of hyperbolic structures on $\Sigma_g$, $g>1$.


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It's only in cartesian (rectangular) coordinates that $A = \iint_\Omega{\rm d}x\,{\rm d}y$ denotes the area of the region $\Omega$. When we change to a different coordinate-system then the Jacobian describes how the area of a small element of size ${\rm d}x'\,{\rm d}y'$ in the new coordinates is related to the corresponding area of size ${\rm d}x\,{\rm d}y$ ...


1

Let $X$ be a disjoint union of countably many copies of $S^1$, which we can embed in $\mathbb{R}^2$ as a series of disjoint small circles of radius $r_n \downarrow 0$. Let $f_n$ denote the map given by $z \to z$ on the first $n$ copies of $S^1$ and $z \to z^2$ on all subsequent ones. Then $|f_n-\operatorname{id}| < Cr_n$ for some constant $C$, but no $f_n$...


1

The author is not claiming that such a diffeomorphism always exists. For instance, if $K = M = S^1$ or $M = S^1\times S^1, K = S^1 \times \{0\}$, then it is clear that no neighborhood of $K$ can be homeomorphic to an open ball.


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One way to prove this would be to construct a diffeomorphism between $\operatorname{PSL}_2(\mathbb{R})$ and the unit tangent bundle $\operatorname{UT}(\mathbb{H})$ of the upper-half plane $\mathbb{H} = \mathbb{R} \times \mathbb{R}_{>0} = \{ z \in \mathbb{C} \mid \Im{z} > 0 \}$ with its standard hyperbolic metric. As $\mathbb{H}$ is contractible, $\...


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Hmm. I haven't read through the whole comment discussion but I'm finding it hard to understand the question. A function is a function. By itself it has nothing to do with coordinates. Its on the level of sets: you just have domain, codomain and function. So choosing coordinates can't change a function; the coordinates are extraneous. It's like defining an ...


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You can use that $T_{(t,0)}\Theta$ is linear. Compute seperately $T_{(t,0)}\Theta(s,0)=\frac{d}{du}|_0 \Theta(t+us, 0)=\frac{d}{du}|_0c(t+us)=s\dot c(t)$ and $T_{(t,0)}\Theta(0,Y)=\frac{d}{du}|_0\Theta(t, uY)=\frac{d}{du}|_0\gamma_{(c(t), Pt(c,t)Y)}(u)= Pt(c,t)Y$.


1

May be here you are doing some confusion. In the first case you are allowed to define a global 1-form in $\textit{local}$ coordinates just because you are in a lucky case in which you can cover your manifold by a single coordinate chart. In the second case the form $\beta$ is defined on $\mathbb{R}^{2n+2}$, and in order to get a 1-form over $S^{2n+1}$ (...


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You may try Reyer Sjamaar's note "Manifolds and Differential Forms": http://www.math.cornell.edu/~sjamaar/manifolds/manifold.pdf



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