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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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