Hot answers tagged

9

This follows from Gauss-Bonnet Theorem: If $f$ is the Gaussian curvature of a compact surface $S$ without boundary, then $$\int_S f=2\pi\chi(S)$$ where $\chi(S)$ is the Euler characteristics of $S$. In particular, if $S$ is $T^2$ the torus, we have $\chi(S)=\chi(T^2)=0$. Therefore, it is impossible for $f>0$ everywhere. BTW, for higher dimensional ...


9

Since the product rule tells us $0 = \partial( g g^{-1} ) = (\partial g) g^{-1} + g (\partial g^{-1})$, we have a formula for the derivative of the inverse metric: $$ \partial_l g^{ij} = -g^{ia} g^{jb} \partial_l g_{ab}.$$ Substituting this in to your expression we get $$ -g^{ia} g^{jb} \partial_l g_{ab} \partial_k g_{ij}.$$ If we swap the dummy indices $...


7

By the Gauss–Bonnet theorem, the Euler characteristic of $T^2$ is given by $$\chi(T^2) = \frac{1}{2\pi} \int_{T^2} K dA$$ where $K$ is the curvature and $dA$ is the element area of $T^2$. If $K$ were everywhere positive, then this would be a positive number, for the same reason that the integral of a positive function is positive; but $\chi(T^2) = \chi(...


6

No, $f$ needn't be globally injective. A counterexample is $$f:\mathbb C\to \mathbb C:z\mapsto \int_0^ze^{t^2}dt$$ Why is that entire map $f$ surjective? Because by Picard's theorem it could at most skip one value $b\in \mathbb C$ i.e. $f(\mathbb C)=\mathbb C\setminus \{b\}$. Of course that potential $b$ is nonzero since $f(0)=0$. But since $f(-z)=-f(z)...


5

There are topological obstructions to a vector bundle admitting a flat connection: most simply, by Chern-Weil theory the real Pontryagin classes of such a bundle must all vanish. So, for example, any closed $4$-manifold with nonzero signature, such as $\mathbb{CP}^2$, does not admit a flat connection. Also by Chern-Weil theory, or by the Chern-Gauss-Bonnet ...


5

Ok, so lets start by identifying what group we're actually working with. We're acting with the action of a fractional linear transformation right? So we are looking at subgroups of $PSL(2, \mathbb{C})$. Ok, so with the restrictions given we know that we have to be in the matrix group generated by $$ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} $$ ...


5

It does appear to be symmetric, though the proof I came up with requires the introduction of a covariant derivative operator. There may be another proof out there that doesn't require quite so much heavy machinery. Let $\nabla_k$ be a torsion-free derivative operator defined such that $\nabla_k g_{ij} = 0$. By the general properties of derivative ...


4

Yes. The box $B$ is homeomorphic to the $2$-sphere $S^2$, so you can choose a smooth structure on $S^2$ and a homeomorphism $f: B \to S^2$ and use that to transfer all your charts: a chart $\varphi : S^2 \supseteq V \to U \subseteq \mathbb{R}$ becomes $\varphi \circ f$. Transferring a smooth structure will result in a smooth structure, since if a transition ...


4

How to obtain the first definition from the second Let $E$ be the bundle $\mathcal{J}^{r,s}$ and $F$ be the bundle $\Omega^1 \times \mathcal{J}^{r,s}$. The precise definition of $\mathcal{J}^{r,s}$ doesn't matter that much here, but for concreteness let's say it is the set of rank $(r,s)$ tensor fields over $M$ on which the metric inner product extends as $\...


3

It seems that you forgot that $\theta$ depends on $y^0$ when you computed $\dfrac{\partial x^1}{\partial y^0}$. The correct answer should be $$\frac{\partial x^1}{\partial y^0} = -\omega(y^1\sin\theta+y^2\cos\theta),$$ and then $$\frac{\partial^2 x^1}{\partial y^1\partial y^0} = -\omega\sin\theta,$$ as you desired.


3

The exponential map is a bijection implies that $G$ is contractible since the Lie algebra is a vector space, Let ${\cal G}$ be the Lie algebra of $G$, write ${\cal G}=V\oplus R$ where $R$ is the radical and $V$ is semi-simple, consider the subgroup $V_G$ of $G$ associated to $V$, the restriction of $exp$ to the Lie algebra of $V_G$ is injective if and only ...


3

Your first question is trivial as written: just take $\phi = 0$, $u_1 = v$, $u_2=u_3=0$. Perhaps you meant for the $u_i$ to be independent of $v$? In this case it's almost true, but you need to allow arbitrary linear combinations of the $u_i$: There are three curl-free, divergence-free vector fields $u_i$ on $\mathbb T^3$ such that for every curl-free ...


3

In order to answer this question, to start with it helps to write the parameterizations of $U_x$, $U_y$, $U_z$ as functions, where $U_x$, $U_y$, $U_z$ are the ranges of the functions, and the domains are open subsets of $\mathbb{R}^2$ (in this case, all of $\mathbb{R}^2$). So for example your parameterization of the set $U_z$ is in the form $$L \in U_z: (...


2

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$Your understanding is perfectly correct, and although it may be overly-careful for everyday use, it's a good idea to be certain what symbols mean! :) (If you can find a library copy, Jan J. Koenderink's Solid Shape is a highly worthwhile if somewhat idiosyncratic read, ...


2

I don't know exactly, how general you want to situation to be. Under weak assumptions, which basically say that the image of $\phi$ stays away from $\gamma(0)$, I think this is easy to get as follows: The derivative $\phi'$ defines a vector field along the curve $\phi$. Choose open subsets $U\supset V\supset \phi((-\epsilon,\epsilon))$ such that $\gamma(0)\...


2

$\newcommand{\Reals}{\mathbf{R}}$"Yes." Here's a sketch. Let $U$ be a coordinate neighborhood about a point $p$ and let $(E_{i})_{i=1}^{n}$ an orthonormal frame in $U$. The tangent bundle $TU$ is trivialized by the mapping $E:U \times \Reals^{n} \to TU$ defined by $$ E(p, x) = \sum_{i} x^{i} E_{i}(p). $$ The smooth mapping $\Phi:U \times \Reals^{n} \to U \...


2

As Mariano mentioned not even homeomorphism is enough. Consider this concise case: Take two exotic $\mathbb{R}^4$ (see Wikipedia's exotic $\mathbb{R}^4$ article) as $P$ and $N$, i.e. $P$ and $N$ are homeomorphic to $\mathbb{R}^4$ but $P$ is not diffeomorphic to $N$. Take $M = \mathbb{R}^n$ with $n>0$, then $M\times N$ and $M\times P$ are homeorphic to $\...


2

An abstract manifold of dimension $k$ is defined by a family of charts (local coordinate systems). In particular, such a family of charts exists for every $k$-dimensional submanifold of $\mathbb{R}^n$. On the other hand, manifolds may be constructed independent of a realization in a Euclidean space, e.g., by surgery. There are 2-dimensional manifolds such ...


2

Yes you're right; if $M$ is presented to you as being embedded in $\mathbb R^n$, the default atlas would be some covering of $M$ with open sets in $\mathbb R^m$ (standard topology) and the identity map on $\mathbb R^m$, where $m=\dim(M)$.


2

Yes, since the projection $p_H:TM\rightarrow {\cal H}$ is a differentiable map. You can show this by considering a trivialization $(U_i)$ of $TM$, the restriction of $TM$ to $U_i$ is parallelizable, you can find vector fields $(e_1,...,e_m,e_{m+1},...,e_n)$ which does not vanish on $U_i$ such that $(e_1,...,e_m)$ generates ${\cal H}$, (they can be determined ...


2

Yes, this is correct and your argument is correct. (One can also think about this in terms of linear systems. If $D$ is a divisor of degree $0$, then $h^0(D)\le 1$, with equality holding precisely when $D$ is the trivial divisor.)


2

There is a really cool example that I love, given in a paper by Arsigny, et al. which shows that the symmetric positive definite matrices can be given a Lie group structure, and this is one such example where the exponential map is a bijection from the Lie group to its Lie algebra. Let $SPD(n)$ be the manifold of symmetric positive definite matrices. This ...


2

$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Yes, $T\Cpx\Proj^{n}$ is globally generated. Note that it doesn't matter whether or not all global holomorphic vector fields on $\Cpx\Proj^{n}$ are of the stated form; what matters is: At some point $p$ (and hence every point, by homogeneity) there exist vector fields of the stated form ...


1

First of all, to comment on Hoot's comment, spacetime is (unless you assume there is no gravity, but then you are really doing special relativity and the whole manifold aspect becomes a bit obsolete as you have global coordinates available) definitively not $\mathbb{R}^4$. It is a 4-dimensional connected (and usually taken to be) non-compact smooth manifold ...


1

A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$). Notice ...


1

No. Consider the curve $$ \gamma(t) = (s \cos t, s \sin t), 0 \le t \le \frac{L}{2\pi s} $$ where $s < r$. its curvature is $\frac{1}{s}$, which is clearly unbounded. If you don't like that the path intersects itself, just make $s$ a very slowly increasing function of $t$ with mean $S$. Then the curvature will be approximately $\frac{1}{S}$.


1

Indeed, these conditions together with the assumption that $f^{-1}$ is continuous are enough, see https://en.wikipedia.org/wiki/Embedding The first conditions guarantee that $f$ is a local diffeomorphism, but without the latter condition that $f^{-1}$ is continuous, i.e. that $f$ is a homeomorphism onto its image, you can have global problems. The typical ...


1

Depends on what context did you study (old school, coordinate-based) differential geometry. Sometimes in physics courses, you have something called tensor calculus, which often just deals with doing tensor calculus in general coordinates in euclidean space. In that case, if you have a coordinate system $(u^1,...,u^n)$, and coordinate basis vectors $\mathbf{...


1

Set $u = \sqrt{x^2 + y^2} - R$. Then the level surface is given by $u^2 + z^2 = r^2$ which is the equation of a circle of radius $r$ in $(u,z)$. Hence, we can find $\varphi$ such that $u = r \cos(\phi)$ and $z = r \sin \phi$. Then $$ u + R = \sqrt{x^2 + y^2} \implies (u+R)^2 = x^2 + y^2 $$ which is the equation of a circle of radius $u + R$ in $(x,y)$ ...


1

I have at least managed to construct a left adjoint to the associated bundle functor $$P [-]: \mathsf{Space}_G \to \mathsf{Bund}(M),$$ Given by the "fibre product" functor $$ P \times_M (-): \mathsf{Bund}(M) \to \mathsf{Space}_G,$$ where the $G$ action is the obvious one $(u_x, v_x).g = (u_x.g, v_x)$. The counit is given by the $G$-equivariant maps $$\...



Only top voted, non community-wiki answers of a minimum length are eligible