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8

this is in between. The second is the definition of the bracket, and requires very little. The first is the definition of a torsion free connection. This also does not require a metric, certainly not the positive definite kind. In physics, they often use connections that have torsion, meaning the equation is false for some useful connections. The ...

5

There are two scales at play here. Let $M$ be a closed Riemannian manifold and let $\Delta$ be the Laplace-Beltrami operator on it. There is an orthonormal eigenbasis $\{\phi_k\}_{k=1}^\infty$ of $L^2(M)$ and an increasing sequence of eigenvalues $\lambda_k\geq0$ so that $\Delta\phi_k=-\lambda_k\phi_k$. The piece of an article you linked to discusses the ...

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) = ... 4 You should have a look at Berger's monumental A Panoramic View of Riemannian Geometry (specifically section 3.4 p. 131-142 and section 4.6, p. 216-218). Two striking theorems are the following: Every Riemannian metric g on S^2 with positive curvature is isometrically embeddable in \mathbb R^3, in a unique way. [Alexandrov, Weyl, Nirenberg, ... 4 The factor a makes the length of each circle centered at the origin equal to 2\pi a r, where r is the radius. Case a<1: not enough circumference To realize this case in practice, cut off a part of the circle, specifically (1-a) part of it (in angle terms, 2\pi (1-a)). This is well explained in wikiHow: Note that this surface is ... 4 This is a mistake -- I should have written that a deck transformation is a diffeomorphism satisfying \renewcommand\phi{\varphi}\pi\circ\varphi = \pi. I've added this to my online correction list. (It's amazing that nobody has noticed this in the 18 years the book has been in print!) It's actually true that just assuming \varphi is a smooth map ... 3 Yes, by using a density. See John Lee's book Introduction to Smooth Manifolds Chapter 16 for a good introduction to the subject. 3 @John Ma's answer is a good one, but I'd also like to point out that parallel vector fields are even rarer that Killing vector fields. For example, on a round sphere, there are plenty of Killing vector fields but no nontrivial parallel fields. Basically, the existence of a parallel vector field is equivalent to the condition that the metric splits locally ... 3 This is too strong a condition. \nabla V = 0 would imply that V is a Killing vector field, thus the local one parameter subgroup of diffeomorphisms are isometries. As isometry preserves curvature, one can construct examples so that such a local vector field can't be found. For example, one can take a surface so that it's Gauss curvature attains a strict ... 3 Here's a way to construct a lot of non-flat examples. Let (\widehat M,\widehat g) be any Riemannian manifold whatsoever, and define a metric g on M = \mathbb R^+\times \widehat M by$$ g = dt^2 + t^2 \hat g, $$where t is the standard coordinate on \mathbb R^+. Then for each \lambda>0, the map \phi_\lambda\colon M\to M given by ... 3 You seem to have used the wrong transformation rule for \frac{\partial}{\partial x^i}. In TM it is not true that$$\frac{\partial}{\partial\tilde x^i}=\frac{\partial x^j}{\partial\tilde x^i}\frac{\partial}{\partial x^j}\in\mathfrak{X}(TM)\tag{Wrong!}.$$This is how \frac{\partial}{\partial x^i} transforms as a vector field over M, but not as a vector ... 3 The best complete, concise, clear and rapid overview of the essentials of Riemannian geometry that I know is found in the little book Morse Theory, by John Milnor. As far as delivering a more detailed review of this book: well, I've already said Milnor; need I say more? Note Added Wednesday 8 July 2015 9:33 PM PST: Despite my previous invocation of the ... 3 Always, since p\not\in C_p(M). 3 Here you have an example in the Riemannian case. Take M=\mathbb{R}^3 as smooth manifold. Then M is diffeomorphic to the the universal covering of the Lie group E(2) of rigid motions of the Euclidean 2-space. So we can consider M with two different non isomorphic Lie group structure. Namely, the above one and the obvious one as 3-dimensional ... 2 Frankly, I don't understand this type of arguments, they look to me as if taken from a 19th century treatise. I think that the stupid but straightforward approach is best here, and not really difficult. Assuming that the geodesic equation is \ddot x ^i (t) + \sum \limits _{j, k}\Gamma ^i _{jk} \dot x ^j (t) \dot x ^k (t) = 0 and that \Gamma ^i _{jk} = ... 2 We call Minkowski spacetime and denote with \Bbb{M} the smooth manifold \Bbb{R}^{4} endowed with the constant, symmetric, non-degenerate, covariant 2-tensor field \eta of coefficients$$ (\eta_{\mu\nu})=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} $$in the standard coordinate ... 2 Here's one way to make it precise. Theorem. Suppose (M_1,g_1) and (M_2,g_2) are Riemannian manifolds, \Delta_1,\Delta_2 are their respective Laplace operators, and \phi\colon M_1\to M_2 is a diffeomorphism. Then \phi is an isometry if and only if for every f\in C^\infty(M_2) we have \Delta_1(f\circ\phi) = (\Delta_2 f)\circ \phi. (Of course, ... 2 @JamesS.Cook is right -- that equation is meant to be interpreted pointwise. It's confusing, because in that book I was using the same notation (\varphi_*X) for the global pushforward of a vector field and the pointwise pushforward (or differential) of \varphi acting on a vector at a point. I've added a correction to my online correction list, clarifying ... 2 Perhaps the issue is to show that \nabla_Y X is a tensor. That is, you need:$$ \langle \phi Z, \nabla_Y X \rangle = \phi \langle Z, \nabla_Y X \rangle $$for any scalar field \phi. Or to put it another way,  \langle Z, \nabla_Y X \rangle depends only on the pointwise value of Z, and not on any of its derivatives. Then you will also need to show ... 2 For the first, you might want to look at a tech report I wrote several years back: http://cs.brown.edu/~jfh/papers/Hughes-DGO-2003/paper.pdf For the second, there's a quite general paper (of which why tech report is a distillation in the 3D case, with a correction): Peter Dombrowski. Krümmungsgrößen gleichungsdefinierter untermannigfaltigkeiten ... 2 Yes. For each p\in M, the set U_p = M\smallsetminus C_p is a neighborhood of p, and these neighborhoods cover M. Every open cover of a manifold has a countable subcover. [Note that the question in your title is different from the one you asked in the text, and has a different answer -- it's certainly possible to find a countable collection of cut ... 2 Here is an example showing that both things are not equivalent. Take M = S^1 the unit circle with its Riemannian metric (induced by \mathbb{R}^2) and let N=\mathbb{R} with its stantard metric. Then \dim (M) = \dim (N) = 1 and M,N are locally isometric because of the arclength parameter. But there are no map F : M \to N satisfying$$g(v,w) = ...

2

I thought the following was all true and perfectly standard just for topological covering maps $\pi$: If we say a deck transformation is just a continuous $\phi$ with $\pi\circ\phi=\pi$ then the deck transformations form a group, and in particular each one is bijective. (Right? Say $\phi(p)=q$. You can certainly find a $\psi$ defined only in some ...

2

It seems natural to require that the operators $d$, $\nabla$, $\text{div}$, $\Delta$, etc., are complex linear and to work with a Hermitian metric $g$ on vector fields that's conjugate linear, say, in its second entry, as you describe. Then we have Hermitian $L^2$ inner products on the spaces of complex-valued functions:  \langle f, h \rangle_{L^2, ...

2

no, take $N=4,$ one figure is a "square" with four equal edges and four equal angles at the vertices. The other figure is a rhombus, same four edges but hte vertex angles in two pairs. For example, we could just glue two equilateral triangles together along one edge. Isometries preserve angles too.

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Yes, there are non-complete manifolds for which this is true. Any convex proper open subset of $\mathbb R^n$ with its Euclidean metric, for example. The condition that there exists a minimizing geodesic between any two points in $M$ is called geodesic convexity.

2

For Riemann surfaces, the formulas are very simple. If $z=x+iy$ is a local holomorphic coordinate, the Kähler form can be written $\omega = \tfrac i 2\, h^2\, dz\wedge d{\bar z}$ for some smooth positive function $h$. The Riemannian metric is then $g= h^2\, dz\,d\bar z$ (where juxtaposition denotes the symmetric product), and the Gaussian curvature is $K = - ... 2 Here are some rough hints on how to analyze the convergence to get you started. Since the integrand is continuous on$(d,\infty)$it suffices to check the integral close to the singularity of the integrand at$x=d$and in the tail that goes off to$x=\infty$. For$r\gg 1$we have$\cosh(r) \sim \frac{e^{r}}{2}$and if$r\gg d$then$\cosh(r) \gg ...

2

Regarding item 1, in the context of 9.1 the function $\gamma : I \to M$ is given. Thus, the endpoint $s$ of $I$ determines the point $p = \gamma(s)$ and the vector $v = \partial I/dt |_{t=s} \in TM|_p$ which determines $\epsilon$. Regarding item 2, let $J$ be the union of all subintervals of $R$ that contain $I$ such that $\gamma$ extends to a geodesic ...

2

Proposition 8.3 does not seem to be very conveniently formulated for the purpose you need it for here. This is one way to see the required conclusion: You can take a coordinate chart around the point $p$ so that it contains the ball $B(p,2r)$ and choose $q\in B(p,r)$. (Use either the Riemannian distance or the Euclidean distance in a chart; it doesn't ...

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