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A surface is a space whose points have each an open nghb homeomorphic to an open disc in $\mathbb R^2$. This is equivalent to have an open nghb homeomorphic to an open set in $\mathbb R^2$: the latter always contains an open disc! But the problem is that one cannot say $U=S\cap B$ is homeomorphic to an open disc, only that it contains some open set $W$ of ...

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Assume $a=1$ and use spherical coordinates. Compute $r(\phi,\theta)$ in order to obtain a parametric representation $$S:\quad(\phi,\theta)\mapsto{\bf r}(\phi,\theta)=\bigl(r(\phi,\theta)\cos\theta\cos\phi,r(\phi,\theta)\cos\theta\sin\phi,r(\phi,\theta)\sin\theta\bigr)\qquad\left(-{\pi\over4}\leq\phi\leq{\pi\over4}, \ ... 2 As I said in my comment, isometric or conformal type of a surface in R^3 is independent of a parameterization. This is an answer in the conformal setting (isometric setting is hopeless). You need: Uniformization Theorem: Every simply-connected Riemann surface is conformal either to the unit disk, or to the complex plane or to the 2-sphere. In view of ... 2 \newcommand{\Reals}{\mathbf{R}}In the usual sense of "depend",[*] no, the Gauss curvature, mean curvature, and shape operator of a (locally oriented) regular surface in \Reals^{3} do not depend on parametrization; that's what's meant by saying these are "geometric" data. :) Depending on your definition of the shape operator (e.g., O'Neill's: If U is a ... 0 Here's my first solution. It does not represent every possible monotonic surface over the domain x : [0,1], y : [0,1], but it's a good start. c, d, h, and k are tunable constants that range from -1 to 1. f(a, x) = (x - x * a) / (a - abs(x) * 2 * a +1) g(x, y) = (f(c, x) + f(d, y) - f(c, x) * k - f(d, y) * h) / (k + h - abs(f(c, x)) * 2 * k - abs(f(d, y)) ... 2 If y \in \mathbb P^1 is a (closed) point and V is an affine n.h. of y, then we may find a function a \in \mathcal O(U) which vanishes precisely at y. If we let U = f^{-1}(V), then U is an open set containing the fibre over y, and the fibre over y is cut out by f^* a \in \mathcal O(V). Thus this fibre is a local complete intersection, ... 2 The degree d curves in \mathbb P^2 form a linear system parameterized by \mathbb P^{(d+3)d/2}. If we embed \mathbb P^2 into \mathbb P^N via the d-uple embedding, these are the linear system of hyperplane sections. Now what you can say about the structure of singular degree d curves? 0 You are asking a very good question. The problem, however, that a proper answer to it lies beyond the class you are taking. You should consider reading, say, Spivak's Calculus on Manifolds or the book by Guillemin and Pollack Differential Topology. Here is the upshot: There is a concept of an (oriented) differential manifold with boundary, it is a bit ... 0 The standard metric is the only metric of constant sectional curvature 1 on S^2. More generally, for any positive integer n and any \lambda \in \mathbb{R}, there is a unique simply connected Riemannian manifold of dimension n and constant scalar curvature \lambda (which therefore looks like a scaling of either S^n, \mathbb{R}^n, or hyperbolic ... 0 The notion of a parameterization of a surface S in R^3 is notoriously sloppy. Here are some definitions and it is pretty much up to you (or your professor) which one you accept as valid. In all these definitions, smooth means _infinitely differentiable (one can require less differentiability). Local parameterization, which is what Mark McClure has in ... 0 One possible definition of a surface is that each point has a neighborhood N that "looks" two-dimensional. Rigorously, this means there is an open set U in R^2 and a smooth, one-to-one, and onto function f that maps U\mapsto N. If this is how you think of a surface, then every surface is, by definition, parametrizable. This certainly does not ... 1 If the parabola is to be rotated about its focus, then we must have parabolic sections of the form$$z = ax^2 - \frac{1}{4a}, \quad a \ne 0.$$Without loss of generality we may suppose a > 0, and suppose that for each unit increase in y, the parabola is rotated by some angle \theta. This yields the parametrization$$\begin{align*} x(u,v) &= u ...

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If the parabola has equation $y = x^{2} - a$, and is rotated about the origin at angular speed $k$ as the "horizontal" section moves along the $z$-axis, the resulting surface may be given the parametric description \begin{align*} x(u, v) &= u\cos(kv) - (u^{2} - a)\sin(kv), \\ y(u, v) &= u\sin(kv) + (u^{2} - a)\cos(kv), \\ z(u, v) &= v. ...

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You could try solving the system $$\begin{cases} x=X(u,v) \\y=Y(u,v) \end{cases}$$ for the functions $u=u(x,y),v=v(x,y)$. After that you can reparametrize as $$(x,y,Z(u(x,y),v(x,y))) .$$

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All of my geometric intuition for "immersed" versus "embedded" surface is contained in my geometric intuition for "immersions" versus "embeddings". In particular, as many people have pointed out, immersions need not be injective. But, of course, even injective immersions need not be embeddings. As a very simple example, consider the map $f : (-\pi, \pi) \to ... 0 Two comments: 1. You don't need to consider$P\mathbb{R}^n$. Locally, it is the same as$\mathbb{S}^{n-1}$. 2. Actually, there is no need to diagonalize$A$in my previous answer. 0 Suppose the eigenvalues of$A$are$\lambda_1\leq\dots\leq\lambda_n$, and$e_1,\dots,e_n$are corresponding orthonormal eigen vectors. Any$x\in\mathbb{S}^{n-1}$can be written as$x=\sum x_ie_i$. With frame {$e_i$},$A$is diagonalized, and$xAx^t=\sum\lambda_ix_i^2$. Let$x\in\mathbb{S}^{n-1}$, and$v$be any vector tangent to$\mathbb{S}^{n-1}$. Take ... 0 The answer to my question is yes, and is given by Theorem 5-2 in Spivak Calculus on Manifolds combined with Theorem 5.8 in Lee Introduction to Smooth Manifolds. See also here. 2 It's just really a matter of notation. Suppose you have a function$f$defined on a parametrized curve$\gamma\colon I\to \Omega$. What do you mean when you write$\dfrac d{dt} f$? Of course, you mean$\dfrac d{dt} (f\circ\gamma)$, and, by the chain rule, this is the directional derivative$D_{\gamma'} f$, i.e., the directional derivative of$f$at ... 0 The answer was almost given by amomin, let us just formulate it precisely. Take$B_0\subset B$, homeomorphic to the Möbius Band, that you have seen. Glueing a disk to$M\setminus B_0$yields a boundaryless surface$M'$. Then,$B_0$is homeomorphic to$\mathbb{R}P^2$minus a disc (see$\Bbb RP^2$as the union of a Möbius band and a ... 1 For convenience I'll write$N = f(\Omega)$and$N' = f'(\Omega')$. It's evident that at any point along$c$, say$c(t)$, the inner product on$T_{c(t)}N$and$T_{c(t)}N'$agree, since both are the restriction of the ambient inner product on$\mathbb{R}^3$. I'm not sure what it would mean for the Christoffel symbols to "agree", since there doesn't seem to ... 1 Let's look at what do Carmo actually writes for the definition of a regular surface: A subset$S \subset \mathbb{R}^3$is a regular surface if, for each$p \in S$, there exists a neighborhood$V \subseteq \mathbb{R}^3$and a map${\bf x} : U \to V \cap S$of an open set$U \subseteq \mathbb{R}^2$onto$V \cap S \subset \mathbb{R}^3$such that: ... 1 You have to be careful : glueing bounderies of 2 manifolds (with boundary) does not define a smooth manifold. Take for example$M_1 =(-\infty,0]$and$M_2 = [0,+\infty)$, then there are a lot of way to glue$M_1$and$M_2$at$0$(consider an homeomorphism$\mathbb{R} \rightarrow \mathbb{R}$which is diffeomorphic on$(-\infty,0]$and$[0,+\infty)$, but not ... 1 To answer your question directly: the connected sum is a smooth surface, but with a different atlas than the one you're using. Formally speaking, connected sums are defined using adjunction spaces: you don't reuse the same charts, but instead you introduce an abstract disjoint union space on which you perform identifications. Your perspective seems to be ... 2 Like you said, a property is intrinsic if it can be computed with knowledge of the first fundamental form (metric). I take it as meaning a property that a 'being' living on the surface could calculate. For example, Gaussian curvature is intrinsic. A being on a sphere, with only their metric, could see their world was 'curved' by finding a triangle whose ... 4 What you're looking for is known as a Riemannian metric. One way of phrasing it is as follows. (Lee talks about these later in a slightly different way of phrasing later in his book.) A Riemannian metric on a smooth manifold$M$is a choice of inner product$g_p$on each tangent space$T_pM$that varies smoothly, in the sense that given any two smooth ... 0 I got this from this Source it is just so easy to understand and well explained, i had to post it, i hope it will one day help someone like it did for me. The best way to tell is probably to draw a picture. I'm going to assume you're talking about normal vectors to surfaces in 3-space. A similar discussion holds for curves in the plane. Firstly, there ... 2 Based on the wording of your question. No, you are not correct. As I stated in the comments, any point$(\cos \theta, \sin \theta, \cdot )$lies strictly on the unit circle on the$xy$-plane. What you want is the plane region entrapped in the cylinder$x^2 + y^2 \le 1$. Here is a plot of both surfaces for$x \ge 0$, and$y \ge 0$: Here is your ... 1 The equation$y=x$is not the equation of a line, it is the equation of a plane. The intersection of the surface$z=4-y^2$with the plane surface$y=x$is the set of points $$\vec{r} = (x,x,4-x^2).$$ This is the parametrization of the curve. The tangent vector is found by differentiating each coordinate, and we obtain $$\frac{d\vec{r}}{dx} = (1,1,-2x).$$ The ... 1 Surface area = Area of (top+bottom)+Curved surface area Area of (top+bottom)=$2\pi r^2$Curved surface area:- Area of this rectangle is$=l\cdot b$, but here,$l$is$2\pi r$and$b$is$h$. So the area becomes ,$2\pi rh$Total area$=2\pi r^2+2\pi rh$0 If you make a cylinder out of cardboard, you can do it with three pieces. Two are circles, and one is a rectangle. The width of the rectangle is the height of the cylinder, and the length of the rectangle is the circumference of either of the cylinder's bases. The two circles have the same radius as the cylinder. If the radius is$r\$ and the height of the ...

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