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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.


2

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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