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It is true that $c=\lfloor \frac{h}{2} \rfloor$, here's a proof. Cutting along a circle does not change the Euler characteristic. So the connected surface-with-boundary that you get from $N_h$ by cutting along those circles, which I will denote $P_h$, has the same Euler characteristic as $N_h$: $$\chi(P_h) = \chi(N_h)$$ This Euler characteristic equals ...

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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 ...

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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 $f : (-\pi, \pi) \to ... 2 Why do you expect to get a relatively minimal elliptic fibration with singular fibres of type$I^*_0$? In my opinion, blowing up base-points and eliminating fixed components produces a fibration$X \longrightarrow \mathbb P^1$with$(-1)$-components in the two singular fibres. Blowing down these$(-1)$-curves produces the relatively minimal rational ... 2 If$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)$. Thus this fibre is a local complete intersection, ... 2 Corank$r$cannot be larger than$c$. Otherwise, consider a PL map$f$from$N_h$to the bouquet$B$of$r$circles which induces the epimorphism$\pi_1(N_h)\to F_r$. Next, take preimages of generic points in the edges of$B$under$f$. Due to genericity of the points, the preimages form a 2-sided 1-dimensional submanifold$L$in$N_h$. Some of the circles ... 2 Based on the wording of your question. No, you are not correct. As I stated in the comments, any point$(\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$: Here is your ... 2 It's just really a matter of notation. Suppose you have a function$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$at ... 2$\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$is a ... 2 Like you said, a property is intrinsic if it can be computed with knowledge of the first fundamental form (metric). I take it as meaning a property that a 'being' living on the surface could calculate. For example, Gaussian curvature is intrinsic. A being on a sphere, with only their metric, could see their world was 'curved' by finding a triangle whose ... 2 The degree$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$curves? 1 No need to solve for$\theta$. Notice that$z$is not an independent variable, thus not a good choice. Using$x = r \cos(\theta)$and$y = r \sin(\theta)$you obtain$z=r$. The length of the normal vector is $$\sqrt{ \left( \frac{\partial z}{\partial r} \right)^2 + \left( \frac{\partial z}{\partial \theta} \right)^2 + 1 } = \sqrt{2}.$$ The surface area is ... 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$1 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 $$(x,y,Z(u(x,y),v(x,y))) .$$ 1 A Mobius strip can be seen as$I^2 / \sim$where$(a,0) \sim (1-a, 1)$for$a \in I$. See picture. The boundary of the strip is$\{(0,i) \ \vert \ i \in I\} \cup \{(1,i) \ \vert \ i \in I\}$as seen from the picture. But recognize that$(0,1) \sim (1,0)$and$(0,0) \sim (1,1)$so this forms two lines that are really halves of a loop. So the boundary is ... 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 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 I would say, tasks properly, only when it is necessary to install the coordinate transformation Jacobian:$\displaystyle \iint_{s} f \, dS=\int_0^{2\pi}{} \int_0^2{\sqrt{4r^2+1}\,|J|\,dr d\theta}\displaystyle J=\begin{vmatrix} x'_r &x'_\theta \\ y'_r & y'_\theta \end{vmatrix}=\begin{vmatrix} \cos \theta & -r\sin \theta \\ \sin \theta & ...

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According to the picture (and the equation of the cone), the cone is above the XY plane (namely with positive z coordinate), so that the normal on its basis that points outward should be with a negative sign in the z coordinate. When using the divergence theorem you always need to choose the outward normal, so you first need to calculate a normal using the ...

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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 ...

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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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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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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 ...

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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 ...

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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 ...

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