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## Hot answers tagged riemannian-geometry

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 ...

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(... 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 ... 3 Yes, for$\mathbb{S}^2$you can do it like that (but be careful to the poles!) Another possibility is to notice that for every two points$p,q$there is an isometry of the sphere sending$p$to$q$, so the curvature must be constant, say$K$, on the whole sphere. Then the Gauss-Bonnet theorem tells you that $$4\pi K = \int K dA = 4\pi.$$ Therefore,$K=1$. ... 2 Let$C_p$be the cut locus of$p$.$C_p$contains two type of points: (i) points$q$such that there exist at least two minimizing geodesics from$p$to$q$; (ii) points$q$that are conjugate to$p$. It can be proven that$C_p$is closed and also a null set (i.e. if$(h,U)$is a chart, then$h (U \cap C_p)$is a null set with respect to the usual Lebesgue ... 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

Because $\dfrac{\partial\phi}{\partial s}(0,a) = \dfrac{\partial\phi}{\partial s}(0,b) = 0$! As you yourself wrote, the endpoints stay fixed during the variation. Indeed, $\dfrac{\partial\phi}{\partial s}(s,a) = \dfrac{\partial\phi}{\partial s}(s,b) = 0$ for all $s$.

2

Consider, $\frac{d}{dt}Y(c(t)) = (\frac{d}{dt}Y_1(c(t)), \cdots , \frac{d}{dt}Y_n(c(t)))$. Moreover, each component function of $Y$ is a function of the coordinates $x_1, \dots , x_n$: $$Y_j = Y_j(x_1, \dots , x_n )$$ Also, $c'(0) = X(p)$ by construction ($c$ is the curve whose tangent vector produces the vector attached to $p$ by the vector field $X$). ...

2

Yes, a Riemannian metric is an algebraic object (or tensor field): That is, $g(X, Y)$ at a point $p$ depends only on the values $X_{p}$ and $Y_{p}$ of the vector fields $X$ and $Y$. Bilinearity over smooth functions is another way to say the same thing.

1

A slight extension of what you refer to as the "fundamental theorem of Riemannian geometry" says that mappsing linear connections on $TM$ which are compatible with a Riemannian metric $g$ to their torsion is an isomorphism between metric connections and smooth sections of the bundle $\Lambda^2T^*M\otimes TM$. So for any skew symmetric tensor field $T$, there ...

1

In polar coordinates $(r,\theta_1, \dots, \theta_n)$, the rays $t\mapsto (t,a_1,a_2, \dots, a_n)$ (where the $a_i$ are constant) are known to be geodesics, and the same applies to reparametrizations $t\mapsto (\phi(t),a_1,a_2, \dots, a_n)$ . The sketch of proof you replicated shows that, in polar coordinates, the length minimizer is of this form.

1

First, we recall the definition of the norm of a tangent vector $v\in T_p M$ $$\| v\|^ 2 = g_p(v,v)$$ in this case $$g_{(x,y)} = \frac{dx^2 + dy^2}{y^2}$$ so $$\| v\|^ 2 = \frac{dx(v)^2 + dy(v)^2}{y^2}$$ and by definition of $dx$ and $dy$ $$dx(v) = v_1$$ $$dy(v) = v_2$$ so $$\| v\|^ 2 = \frac{v_1^2 + v_2^2}{y^2}$$ Using this in $\dot \gamma$ we can express $... 1 From$g= 4{ d\rho ^2+ \rho ^2 d \theta ^2 \over (1-\rho^2)^2}$Let$R = \int _0^ \rho 2{1 \over (1-t^2)} dt= 2$artanh$(\rho)$, then$g=d R^2+\sinh^2 R d\theta ^2$, as${\tanh R/2\over (1-\tanh^2 R/2) }= \sinh R$. Now$g=dR^2+ \sinh^2R d\theta ^2$proves that$g$is rotationnaly invariant and that$R$is the distance to the origin, so these are polar ... 1 If the tangent vectors of$\gamma_1$and$\gamma_2$are equal at a common point then they cannot be distinct geodesics (that point and vector serve as initial data to a system of ODEs, which has a unique solution). Therefore their derivatives are different at$y$. If we assume$\gamma_1 \cup \gamma_3$is smooth, the tangent vector along this path has ... 1 Your notation are a bit confused. But I think there is a mistakes, it should be $$\nabla _{\dot \gamma (t)}Y_t=\left(\frac{\mathrm d a(\gamma (t))}{\mathrm d t}+\dot\gamma ^ia^j\Gamma_{ij}^k\right)\partial _k$$ Indeed, let$\gamma(t)= (\gamma ^1(t),...,\gamma ^n(t))$, and$X:=\dot\gamma =\dot\gamma ^i\partial _i$and$Y=a^j\partial _j\$. Now, \begin{align*} \...

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