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Since it preserves orientation, your homeomorphism is going to be a rotation of the simplex. Take a narrow annulus $A$ surrounding $\Delta$ with one boundary component $\partial_0A$ equal to the boundary of the simplex $\Delta$. Now we want to extend $g$ from $\partial_0A$ to $A$ in such a way that it restricts to the identity on $\partial _1A$. To do this ...

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Here's an idea which may or may not help in your particular problem - but if it does it should be reasonably accurate and efficient. Find the convex hull of your surface (there are algorithms for that for various inputs). Your book should come to rest on a facet. If there are no facets (or are too few, or they are too small) then you're in an unstable ...

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If you only know the $x$ value this is not enough to find the $y$ value or $z$ value. If you know the $x$ value AND the $y$ value, you could find the $z$ value. For example $x=2$, $y=1$, you may consider $x=2$ to be a plane $p$ parallel to the $yz$-coordinate plane, and consider $y=1$ to be a plane $q$ parallel to the $xz$-coordinate plane, the intersection ...

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Choose a half-plane that represents the rotation of the graph by $\phi$ from the $(+y)z$-halfplane. The horizontal direction on this half-plane is given by the unit vector $(\sin \phi, \cos \phi, 0)$. Call this the "s" axis. The intersection of this half-plane with the surface looks exactly like the same in all of the half-planes. In particular, like it does ...

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This link in wikipedia explains anything you want about finding envelopes. God bless wiki! :) The solution to this problem is a piece of parabola. The procedure is thoroughly explained in the link I mentioned.

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Here is a sketch (let me know if you want to see more details on any of these items) of a direct proof which does not require hard classification results which appear in the links, it inly requires differential topology which you can find the the book of Guillemin and Pollack: Each nonempty connected closed subsurface $S\subset M= int(S_h\times [-1,1])$ ...

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More or less.There is a formula which you can try to apply in some cases. I'll state here lemma $5.1$ in O'Neill's Elementary Differential Geometry, page $236$: Let $Z$ be a nonvanishing vector field on $M$. If $V$ and $W$ are tangent vector fields such that $V \times W = Z$ then: $$K =\frac{Z \cdot \nabla_VZ \times \nabla_WZ}{\|Z\|^4}\qquad H = ... 2 Yes, this is true. We may as well suppose f is a homeomorphism that already preserves one of the fibers. Now cut the homeomorphism open along that fiber. Then both manifolds are now homeomorphic to \Sigma \times [0,1], and we may identify your homeomorphism with a homeomorphism \Sigma \times [0,1] to itself; assume it preserves \Sigma \times \{0\} or ... 1 The curve C is unique. As you say, of course if B is a curve linearly equivalent to C then B^2=C^2. However, in this case by Riemann-Roch you find h^0(\Bbb{F}_n,\mathcal{O}( C ))=1 and therefore the linear system |C| (of curves equivalent to C) actually contains C only. As for your second question maybe it will be clearer if you write ... 2 Take an equilateral triangle with sides of length 1, and then use each vertex as the center of a circular arc passing through the other two vertices: This is at least a contender, with area 3\cdot\frac{\pi}{6}-2\cdot\frac{\sqrt3}{4}, less than a circle of diameter 1 or a quarter-circle of radius 1, which are other convex shapes meeting the ... 1 If you write your vectors in the basis of eigenvectors f_1,\ldots,f_k of C, then if u=\sum t_jf_j we have$$ u^TCu=\sum t_j^2\lambda_j, $$where \lambda_1,\ldots,\lambda_k are the eigenvalues of C (counting multiplicities). So, in each direction, you are stretching the unit circle by \sum t_j^2\lambda_j a convex combination of the eigenvalues. In ... 2 Eular characteristic of a non-orientable surface is 2-g where g denotes the genus of the surface. So if we consider g=3 i.r if we consider a hexagon and identify its edges with the relation a^2b^2c^2 , the resultant surface we will get as our required surface. 2 The Euler characteristic of a connected sum of surfaces S_1 and S_2 is given by$$\chi(S_1 \# S_2) = \chi(S_1) + \chi (S_2) -2.$$So the connected sum of a torus with the projective plane will have Euler characteristic -1. 2 The curve is actually unique, not unique as linear equivalence class. Consider blow up one point in \mathbb{P}^2, then it is \mathbb{F}_1 1 \newcommand{\Proj}{\mathbf{P}}One approach is to note that two-dimensional integral homology of \Proj^{1} \times \Proj^{1} is generated by the classes of the base B = \Proj^{1} \times \{\text{pt}\} and the fibre F = \{\text{pt}\} \times \Proj^{1}, and$$ [B] \cdot [B] = [F] \cdot [F] = 0,\qquad [B] \cdot [F] = 1. $$If a and b are integers and ... 1 Stokes theorem is basically relating the flux through a surface with a closed path around the surface. Intuitively it says that a vector field, the total flux of the surface or flow through the closed path, must be equal to the dot product of the vector field along the path. \iint \nabla \times \mathbf {\vec F}\cdot \mathbf n dA = \iint \nabla \times ... 0 Let think of a rational map \phi: S\dashrightarrow X, and \phi is not defined at an isolated point p\in X. What happened such that \phi can not be extend to p? there must be 2 points near p such that their images are far away to each other. Otherwise by continuity, the map can be extended. So take a neighborhood of p, the image of this ball(removing ... 3 No, a section cannot be tangent to a fibre at any point. \pi \circ \sigma = \operatorname{id} implies that at any point p = \sigma(q) in the image \Gamma of the section, (d_p \pi)_{|T_p \Gamma} must be an isomorphism onto T_q C. On the other hand, (d_p \pi)_{|T_p F} =0 because \pi is constant along F. So if \Gamma and F share a tangent ... 4 There is a little bit of ambiguity in your question: are you asking this for any \eta that satisfies the conclusion of the theorem, or specifically for the one that is constructed in the proof? The first interpretation is silly: for any S \dashrightarrow X, given one (S',\eta,f) satisfying the conclusion, we could just do another 'unnecessary' ... 3 a) No, it is not true that \mathcal{O}_X(C)=f^*\mathcal{O}_Y(C'). For example take X=\mathbb P^2_{x:y:z}, Y=\mathbb P^2_{u:v:w}, C=V(x) and let$$f(x:y:z)=(u:v:w)=(x^2:yz:z^2) so that $C'= V(u)$. Then $\mathcal O_X(C)=\mathcal O_{\mathbb P^2}(1)$, whereas $f^*(\mathcal O_Y(C'))=f^*(\mathcal O_{\mathbb P^2}(1))=\mathcal O_{\mathbb P^2}(2)$ b) The ...

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The basic idea is that: When you remove a disk in a surface, you basically introduced a new boundary into the surface word. For instance, when you remove a disc from the torus $aba^{-1}b^{-1}$, you get a surface with symbol $aba^{-1}b^{-1}c$. (Of course, the process is more complicated. You need to add the new symbol at the location where it is not ...

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The formula you see is the result of projecting $ds$ onto the plane. When we calculate any integral we are taking an infinite sum of small pieces. In this case, $ds$ represents an infinitesimal piece of surface area. Imagine a surface in $3D$ space and a little square on the surface. This little square is $ds$ and it will have a unit normal vector pointing ...

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1) They may or may not contain $E$. 2) Union is a set theoretic notion. $C+E$ is a scheme theoretic notion.Locally, they are defined by the product of the equations defining $C$ and $E$. 3) No, the ideal sheaf is in fact $\mathcal{O}_E(-C)$. 4) The degree does not make sense for a possibly non-irreducible curve. You only have a multi degree, restricting to ...

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