# Tag Info

7

From the comments you included, I can see you are comparing the two geometries' synthetic axioms to see if one is a special case of the other. Of course, that is doomed to fail because the two lists contain mutually exclusive axioms about parallels (as you noticed.) The real idea is that hyperbolic, affine and Euclidean geometries can be modeled as subsets ...

5

Given a dominant morphism $f:X\to Y$ of varieties (or more generally of integral locally noetherian schemes) the degree of $f$ is the degree of the corresponding field extension $f^*:K(Y)\to K(X)$. For example the degree of $$f_n:\mathbb P^1_k\to \mathbb P^1_k: [x:y]\mapsto [x^n:y^n]$$ is the degree of the extension of fields $$f^*_n:k(t)\to k(t):t\mapsto ... 4 The so called Klein model for hyperbolic space gives a possible answer. Consider an ellipse in the projective plane, or equivalently a quadratic form on the 3-dimensional vector space with signature (+,+,-). The interior of the ellipse is a model for a hyperbolic plane where the lines are the intersections of usual projective lines with the ellipse. The ... 2 One way to motivate this is: in the projective plane, any two distinct lines meet at exactly one point. This is a general instance of homogeneity - any two pairs of (distinct) lines in the projective plane "look alike" in an appropriate (specifically, topological) sense. If you added two points at infinity for every family of parallel lines, then parallel ... 2 You better assume that the curve X is smooth: else the morphism \pi might not be defined everywhere and moreover ramification would not be a very clear concept. Your morphism \pi:X\to \mathbb P^1 is best seen as the projection of the curve X from the point O=(0:0:1) to the line at infinity z=0. Indeed that line at infinity is a copy of ... 2 Consider two hyperplanes [X],[Y] \subset \mathbb{R}P^n, defined by two linear hyperplanes X,Y \subset \mathbb{R}^{n+1}, such that S is contained in each of the two affine charts \mathbb{R}P^n - [X], \mathbb{R}P^n - [Y]. It follows that S \subset \mathbb{R}P^n - ([X] \cup [Y]). The subset \mathbb{R}^{n+1} - (X \cup Y) has four components, and ... 2 Take the cylinder x^2+y^2=a^2 (with z arbitrary), and make a plane passing through the x axis which tilts upward in such a way that the distance from the origin to where the plane cuts the vertical planes y=\pm a is the b of your major axis 2b. Now take two spheres of radius a which are to be bounded by the cylinder, one below and one above ... 2 A projective plane is defined just by the incidence relations, that is, by which points are on which lines. If we use an arc that is not the whole circle but that still connects all three points on the circular arc, it defines the same incidence relations as the usual diagram, and therefore defines the same projective plane. 1 Consider the first figure below where three points A,B,C are given on the same straight line. The task is to determine the point D so that that the cross ratio$$(AB;CD)=-1. \tag 1$$Let P=A, Q=B and R be an arbitrary point not on the staright line determined by AB. Also, let S be an arbitrary point inside PQR as shown in the second figure. ... 1 We want to show that$$FK\cdot FM = FC\cdot FB= FD\cdot FA.$$Since M is the midpoint of AD, it is equivalent to$$FK=\frac{2FD\cdot FA}{FD+FA},$$or$$DK=\frac{2FD\cdot FA}{FD+FA}-FD=\frac{FD(FA-FD)}{FD+FA}=\frac{FD\cdot AD}{FD+FA}.$$Thus we need to show that$$\frac{DK}{AD}=\frac{FD}{FA+FD},$$equivalently$$\frac{DK}{AK}=\frac{FD}{FA}. Finally, ...

1

$O(3)$ is the isometry group of $S^2$ with its usual round metric, and it acts linearly on $\mathbb{R}^3$ for geometric reasons. So the question is why $SU(2)$ acts on the Riemann sphere by isometries for this metric. The answer is that $SU(2)$ preserves the Fubini-Study metric on the Riemann sphere, and moreover the stabilizer of a point acts transitively ...

1

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 ...

1

The real reason is to avoid degenerate cases. Sometimes the properties are given as: Each two distinct points of P are incident with a unique line. Each line is incident with at least three points. Each two distinct lines of P meet in a unique point. There are four points of P with no three of them collinear. In other words, we want our ...

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