# Tag Info

1

This isn't rigorous but the only reason I am posting this is "...could help me understand these concepts intuitively...". So here goes. $\partial E$ would be a 'small change in E'. What elementary object is it, that if you keep adding side by side, will produce a surface? A line of course. It could be a curved one too, but a line nevertheless. And ...

0

Note: If you mention the level that you would like this question answered at (i.e. undergraduate differential geometry, beginning graduate level differential geometry, etc.) then the following response can be adjusted accordingly. A surface $S$ is said to be embedded in $\mathbb{R}^3$ if there is a one-to-one immersion $f : S \to \mathbb{R}^3R$ such that ...

2

Yes everything you write is correct. In particular your very last inequality follows from the implication for divisors on $S$ (or on any smooth variety for that matter): $$D\leq E\implies H^0(S, \mathcal O(D))\subset H^0(S, \mathcal O(E))$$ This implication is evident by interpreting $H^0(S, \mathcal O(D))$ as the vector space of rational functions ...

0

Well first of all these have total curvature $-4\pi$, not $4\pi$. The Gauss-Bonnet theorem tells us that the total curvature of a surface is equal to $2\pi$ times the Euler characteristic of the surface. Moreover we know that the Euler characteristic of a connected surface is always an integer that is at most 2. So in general we might want to ask: What are ...

0

If you know the normal vector $\vec N=(n_1,n_2,n_3)$ of a surface at a point $P=(x_0,y_0,z_0)$ then the plane tangent to the surface at $P$ is given by $n_1(x-x_0)+n_2(y-y_0)+n_3(z-z_0)=0$. This is just the equation of a plane given a point and it's normal vector (you should look into that). Any vector in this plane is tangent to the surface. This holds for ...

2

For a surface, the multiplicity of a singular point is the first invariant to look for. Suppose it is a double point. In that case, you want to look for the tangent cone. If the tangent cone is $xy$ then you have an $A_k$ singularity (basically of the form $x^2+y^2+z^k$). If the tangent cone has the form $x^2$. Then, the double point singularity has the form ...

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I think the simplest way to define tangent vectors is to not use curves at all. Just say that a vector $v$ is tangent to $M$ at point $p$ if $$\operatorname{dist}(p+tv,M)=o(t),\qquad t\to 0^+\tag{1}$$ This agrees with the usual definition at the non-boundary points. At the boundary points the above definition yields halfspace, as in wspin's comment. If you ...

-1

Try Eureqa. It's awesome. As their add says, it's a breakthrough technology that uncovers the intrinsic relationships hidden within complex data. It works very simply: after downloading the software, you can insert your data (as vectors or as a matrix) and describe the relationship you want to find (i.e. $z = f(x,y)$). Then, after selecting which functions ...

2

Each one of their horizontal cross-sections (with z being the vertical dimension, as in the $3$ figures you presented) is a superellipse (the case $d=4$ is called squircle), and are connected to the famous Gamma and Beta functions in terms of area. For rational values of the form $d=\dfrac1n$ , with $n\in\mathbb{N}$, they are linked to factorials and ...

2

Yes, the geometric classification of surfaces tells us that a simply connected Riemannian surface $S$ must be (up to diffeomorphism) the sphere $S^2$, the complex plane $\mathbf{C}$, or the hyperbolic plane $\mathbf{H}$. Given that $\mathbf{H}$ is the only one of these with negative curvature, $S$ must be the hyperbolic plane.

2

Yes, by Uniformization theorem

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Considering the shape of the $z(x,y)$ plot : It seems better to consider $z-1$ as a function of $(x-1)$ and of $y$ A polynomial function of $(x-1)$ might be convenient. The first trial is made with a second degree polynomial. Exponential functions of $y$ might be better than polynomial. The first trial is made with two exponentials. The coefficients ...

2

Real case The sphere is an unruled quadric, so the answer is no. Wikipedia distinguishes several cases, in particular three non-degenerate ones: The first case is the empty set. The second case generates the ellipsoid, the elliptic paraboloid or the hyperboloid of two sheets, depending on whether the chosen plane at infinity cuts the quadric in ...

3

The linear system $|D|$ is the set of all effective divisors linearly equivalent to $D$. Consider the map $H^0(S,\mathcal{O}_S(D))\to|D|$ where $f\mapsto\mbox{div}(f)+D$ (here I am identifying $H^0(S,\mathcal{O}_S(D))$ with the vector space of all rational functions $f$ such that $\mbox{div}(f)+D\geq0$). This map is surjective, since if $E\in |D|$, then ...

1

Let me explain why I would not expect this to be true. Suppose that $X$ is a $K3$ surface with the following properties: First we ask that the automorphism group $G=\operatorname{Aut}(X)$ is finite. (Remark: by a theorem of Sterk, this is equivalent to requiring that $X$ has fintely many elliptic pencils, though we don't use this.) Note that this gives ...

2

You have the associated form $$G=ydy\wedge dz+x^2ydz\wedge dx+y^2z dx\wedge dy$$ Then $$dG=({\rm div}\; F)dx\wedge dy\wedge dz=(x^2+y^2)dx\wedge dy\wedge dz$$ Now the flat part of the surface, $S_1$; can be parametrized by $(r\cos t,r\sin t,1)$ with $0\leqslant r \leqslant 1$ and $0\leqslant t\leqslant 2\pi$. The upper cap $S_2$ may be parametrized by ...

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Firstly, the isomorphism you state is more or less the definition of the fixed part of $L$. More precisely, I guess one could say that the fixed part of $L$ is the maximal divisor $F$ such that $H^0(X,F)$ is 1-dimensional and the multiplication map $$H^0(X,L-F) \otimes H^0(X,F) \rightarrow H^0(X,L)$$ is an isomorphism. To see what happens with higher ...

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By definition of the fixed part, we should have $h^0(D)=h^0(D')$. Let's show that implies the claim. First assume that $\Delta$ consists of a single curve $\Gamma$ which intersects $D'$. (It must be a (-2)-curve.) By the facts you mentioned, $h^0(D')=2+\frac12 (D')^2$, whereas $h^0(D) \geq 2+\frac 12 D^2$. But now  D^2 = (D' + \Gamma)^2 = (D')^2 + ...

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I had a similar question once, and at least part of your first question is addressed in this paper.

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