# Tag Info

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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$\...

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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 ...

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