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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(...


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 ...


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

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

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

Here's an easy derivation of the surface area of a torus (without using metric tensor or the Gramian determinant). All we have to do is compute the following integral: $$2\pi\int_{0}^{2\pi}\left(R+r\cos\theta\right)\cdot rd\theta$$ Which can be explained through this diagram: Since $\displaystyle \int(R+r\cos\theta)\cdot rd\theta=r\left(r\sin(\theta)+R\...


1

Separated, let $x,y\in Y$ if $p(x)\neq p(y)$, since $X$ is separated, there exists open subsets $p(x)\in U_x, p(y)\in U_y$ of $X$ such that $U_x\cap U_y$ is empty, $p^{-1}(U_x)\cap p^{-1}(U_y)$ is also empty and $x=\in p^{-1}(U_x), y\in p^{-1}(U_y)$. If $p(x)=p(y)$, since $p$ is a covering, there exists an open subset $p(x)\in U$ such that $p^{-1}(U)=\...


1

You are correct. Because each point in a manifold has a neighborhood homeomorphic to some Euclidean space, any manifold is locally contractible, which implies that is it both locally path connected and locally simply connected. Therefore if we restrict our attention to connected manifolds (which we usually do), we see that all manifolds admit universal ...


1

Your Gramian determinant is incorrect. To retain your definition of volume element $\phi'$ must be defined as the transpose of Jacobian or else the volume element should be $\sqrt{\det(\phi'\phi'^\top)}$. $$ \phi'=\begin{bmatrix} -\sin\alpha\cos\beta&-\cos\alpha\sin\beta\\ \cos\alpha\cos\beta&-\sin\alpha\sin\beta\\ 0&\sin2\beta \end{bmatrix}\\ \...


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

Note you have a sum of squares equal a constant. Hence you should use trigonometric functions. $$ \left(\sqrt{x^2+y^2}-R\right)^2+z^2=r^2 $$ So we set $z=r\cos(\theta),\sqrt{x^2+y^2}-R=r\sin(\phi)$. Squaring the second relation we get $$ x^2+y^2=(R+r\sin(\phi))^2 $$ So we may introduce another angle an set $x=(R+r\sin(\phi))\cos(\theta),y=(R+r\sin(\phi))\sin(...


1

We can describe the surface of the torus using parameters $(\phi,\alpha)$ by the position vector $\vec r(\phi,\alpha)$ $$\begin{align} \vec r(\phi,\alpha)&=\hat \rho(\phi) R+(\hat \rho(\phi) r\cos(\alpha)+\hat z r\sin(\alpha))\\\\ &=\hat \rho(\phi)\,(R+r\cos(\alpha))+\hat z r\sin(\alpha) \end{align}$$ where $\hat \rho(\phi)=\hat x \cos(\phi)+\hat ...



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