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3

You're wrong about "trough-shaped." There are cylinders, cones, and so-called tangent developables, all of which are flat, ruled surfaces. As a hint, I'll suggest that you want to use the Mainardi-Codazzi equations.


2

Here is a proof of equivalence of 1, 3 and 4. Below, $S$ is a compact connected surface without boundary. Definition. A maximal cut in $S$ is a 1-dimensional submanifold $L\subset S$ such that $S\setminus L$ is connected and contains no nonseparating simple loops. The latter condition just means that $S\setminus L$ is the 2-dimensional sphere with ...


0

I don't have much time for elaborating now, but you definitely can do this yourself (feel free to add your own answer). Just observe that substitutions $x = r \cos \theta$ and $y = r \sin \theta$ to $z = a \left( x/y + y/x \right)$ turn this equation to $z = \tfrac{a}{\cos \theta \cdot \sin \theta}$ which clearly does not depend on $r$. This means that for ...


-1

I believe this surface can be described by two networks of lines. ( Shown in Red and Cyan). Make the substitution $nt$ for $u$ , and separately $nt$ for $t$ , so that the new parametrization might look like : $<t,nt,a(1/n + n/1)> $ and/or $<nt,t,a(1/n+ n)>$ these are the asymptotic lines. Then switch the sign of n to pick up the surface below ...


2

Rigidity persists a little below $C^2$, namely in the class $C^{1,\alpha}$ with $\alpha>2/3$. It breaks down for small Hölder exponents $\alpha$ in a rather strong way: any continuous embedding can be uniformly approximated by $C^{1,\alpha}$ isometric embeddings. (Key term: h-principle). This is a stronger form of Nash's $C^1$ isometric embedding ...


0

Write the equation for the plane using cylindrical coordinates: $y+2z=2\implies \rho\sin{\phi}+2z=2$. Solving for $z$ we have, $$z(\rho,\phi)=\frac{2-\rho\sin{\phi}}{2},~~\text{where}~0\leq\rho\leq1,~0\leq\phi\leq2\pi.$$ Then the surface parametrization is $$\vec{r}(\rho,\phi)=\langle\rho\cos{\phi},\rho\sin{\phi},\frac{2-\rho\sin{\phi}}{2}\rangle.$$


0

Intuitively, the first fundamental form tells you how to compute the distances along the paths within the surface (it is just a Riemannian metric of the surface thought as a standalone manifold, that is if we forget about the embedding/immersion). This explains why it is also called the intrinsic metric. The second fundamental form describes how "curved" ...


0

For calculate the finite total curvature of a complete minimal surface $\phi:M\longrightarrow\mathbb{R}^3$, by start from the topology of $\bar{M}$ and the geometry of ends. An $\mathbf{end}$ of a completle minimal surface of finite total curvature is the image $c_i=\phi(D_i-p_i)$ of a sufficiently small punctured disk $D_i-p_i$ on $\bar{M}$ with centered ...


2

Your second parametrization does not trace out the surface of a cylinder, but instead gives a disk of radius $R$ in the $z=2z_0$ plane, centered on the $z$-axis.


0

Intuition: a tangent plane intersects a convex surface (e.g. a paraboloid) in exactly one point. If that's true, and substituting one equation into the other, we seek the only solution $(x,y)$ which satisfies $x^2+x+(ay^2+y+1)=0$. Applying the formula for the roots of a quadratic equation (twice) and demanding a single root, I got the condition for ...


0

The surface integral is given by: $$\iint \left\| {\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y} \right\| dx\, dy$$ Now when you transfer to cylindrical coordinates, you must convert your derivatives and add the Jacobian, so you must convert: $${\partial \mathbf{r} \over \partial x}={\partial \mathbf{r} \over \partial ...


0

Basically you can construct the surface of genus $g$, by gluing a disk $\mathbb{D}^2$ to a wedge of $2g$ circles, $\bigvee\limits_{i=1}^{2g} \mathbb{S}_i^1$. First we think the border of $\mathbb{D}^2$ as $2g$-sided polygon, and we put labels in each side $a_1,\ b_1,\ a_1,\ldots,\ a_g,\ b_g$, and arrows as in the picture: Then we label each copy ...


0

Another way to find this is integrating on the lengths of the intersections of the surface of interest and the planes of the form $y=k$ for $-1\leq k \leq 1$. I haven't tried it in a while but here it goes. The sections look like parenthesis joined by flat top and bottom. By parameterizing one of the cylinders with $\theta$; the length of the part that looks ...


0

The surface area is $$S=2\int\int_D \sqrt{1+f_x^2+f_y^2}dxdy$$ where $z=f=\sqrt{1-y^2}$ and so $f_y=\frac{-y}{\sqrt{1-y^2}}$ so $$S=2\int\int_D \sqrt{1+\frac{y^2}{1-y^2}}dxdy=2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dxdy \\=2\int_{-1}^12dy=8 $$


1

If you are looking at a hypersurface which is defined by an equation of the form $F(x)=0$, then you are looking at the set of point along which $F$ does not change. The gradient of $F$ points into the direction in which $F$ changes most, so it is normal. In formulas: if $c=c(t)$ is a curve in $F=0$, then $F\circ c(t)=0$, hence also it's derivative: ...


2

The principle curvatures are extrinsic quantities, the depend on the embedding of the manifold into some exterior manifold, and describe how the embedded manifold curves in that space. The curvature tensor of a Riemannian (or Lorentzian) manifold is an intrinsic quantity which measures to which extend covariant derivatives commute.


1

The surface $X(t,\theta)$ is regular if $X_t,X_\theta$ (the partial derivatives wrt each variable) are linearly independent for all $(t,\theta)\in domain$. Compute $X_t\times X_\theta$. Showing that it is nowhere zero would imply regularity. Let $$f(t)=\pm (arctan{(\sqrt{1-\mathbb{e}^{2t}})}-\sqrt{1-\mathbb{e}^{2t}}).$$ Then $$f'(t)=\pm\frac{e^{2 t} ...


1

Confocal Quadric coordinates have been of real importance for the ellipsoid. A typical situation is given in the example, the umbilic points are as in the reference above. $$ \frac{x^2}{4} + \frac{y^2}{2} + \frac{z^2}{1} = 1 $$ $$ \frac{x^2}{4-\lambda} + \frac{y^2}{2-\lambda} + \frac{z^2}{1-\lambda} = 1 $$ $$ \frac{x^2}{4-\mu} + \frac{y^2}{2-\mu} + ...


1

Isothermal coordinates on surfaces were first introduced by Gauss. Korn and Lichtenstein proved that isothermal coordinates exist around any point on a two dimensional Riemannian manifold. On higher-dimensional Riemannian manifolds a necessary and sufficient condition for their local existence is the vanishing of the Weyl tensor and of the Cotton tensor. ...


0

As @AnonSubmitter85 says in comments, use hold between the plot commands to plot in the same figure.


0

this is a very interesting question. Your understanding is pretty accurate, though you do not technically need to represent the diagram in $\mathbb R^2$ to obtain a virtual knot. The embedding $K:\mathbb S^1 \to M\times I$ , where $M$ is a 2-manifold with genus g, and $I$ is the unit interval (the cartesian product yields a thickened surface) is enough to ...


0

$S^2$, which is the sphere,is a surface if we can find for each point $ x \in S^2$ a neigbourhood homeomorphic to $\mathbb{R^2}$. Now for $x$, a neighborhood would be $S^2-\{y\}$ where $y \neq x$. By the stereographic projection we know that this is homemorphic to $\mathbb{R^2}$.


1

Taking a bite from DoCarmo's book is the following theorem which is essentially the same as the one which you have written Theorem: Let $U \subset \mathbb{R}^3$ and $f:U \to \mathbb{R}$ be differentiable with $a$ as a regular value. Then $f^{-1}(a)$ is a regular surface in $\mathbb{R}^3$. Clearly $0$ is a regular value of your function here since all the ...


0

Given your parametrization, you should be able to show that the surface you get really is a rectangle with two of its opposing sides glued together after inversing their orientation. This is the definition of a Möbius band I would try to work with here. Let's say your parametrization $X$ has a rectangle $[a,b] \times [-c,c]$ for domain. Then showing that $X$ ...


3

The great dodecahedron and small rhombihexahedron may be viewed as closed, orientable surfaces whose genera are difficult to see geometrically. (Each has genus $4$.)


4

In general, two connected manifolds can have at most two different connected sums up to homeomorphism. The two possibilities are determined by whether the gluing map preserves or reverses orientation of the sphere (the circle, in the case you've drawn). But sometimes the two possibilities yield homeomorphic manifolds. More specifically, If either $M$ or ...


3

You can look around and record the directions in which you see yourself and at what distance. At least in surfaces of constant curvature, this information should be enough to compute the genus. This is precisely the idea behind COBE and other later experiments. Alternatively, you can count for each $\ell\geq0$ the number of simple closed geodesics of ...


0

The Dupin indicatrix for the pseudosphere is a hyperbola and the principal directions are the axes of the hyperbola (not the asymptotes), which are always orthogonal. Since the principal directions are the axes, they are necessarily orthogonal, as well. Therefore the "other" direction is uniquely determined.


4

When I was taught the classification theorem, my professor emphasized being able to compute the Euler characteristic, and his favorite examples that he set as exercises were Seifert surfaces. For example, you can try to compute the Euler characteristic of the Hopf link: http://en.wikipedia.org/wiki/File:Hopf_band_wikipedia.png. Or something weirder, like ...


2

The functions $2H = \operatorname{tr}(S)$ and $K = \det(S)$ are elementary symmetric polynomials in the eigenvalues of the shape operator $S$, hence "natural". Other prospective measures of non-flatness, such as $$ \kappa_{1}^{2} + \kappa_{2}^{2} = 4H^{2} - K, $$ can be expressed as polynomials in $H$ and $K$. Separately, $K$ turns out to be intrinsic, ...


1

Hint: The surface is determined uniquely up to transformation using its first and second forms If we get a parametrization of the unit sphere that gives these first and second fundamental forms, then these two surfaces coincides. Consider the parametrization of the unit sphere as $$r(u,v)=(\cos u\cos v,\cos u\sin v, \sin u)$$ Find the first and second ...


2

If $M$ is a connected closed (i.e. compact without boundary) 3-dimensional manifold, then $\pi_1(M)$ cannot be isomorphic to $\pi_1(S)$, where $S$ is an orientable surface of genus $\ge 2$. You can see this by first noting that $\pi_2(M)=0$ (otherwise, by the sphere theorem, $M$ is a nontrivial connected sum which will imply that $\pi_1(M)$ is a nontrivial ...


1

Let be the chart/parametrization near some point of $S$: $$h:V\subset\Bbb R^n\longrightarrow S\subset\Bbb R^{n+1}.$$ Define $$ H:V\times(0,\infty)\subset\Bbb R^{n+1}\longrightarrow S\times(0,\infty)\subset\Bbb R^{n+2} $$ by $$H(x,t)=(h(x),t).$$



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