# Tag Info

## New answers tagged projective-geometry

1

Okay, assume the characteristic is $0$ (over complex numbers). Let's stare at your last exact sequence: $$0 \rightarrow H^0(S,O_S(H-C))\rightarrow H^0(S,O_S(H)) \rightarrow H^0(C,O_C(H)) \rightarrow H^1(S,O_S(H-C)) \rightarrow ....$$ In the surface $S$, $C=H|_S$ as divisor classes. So we know your first term has dimension 1 ($S$ is smooth and simply ...

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 ...

0

Why did you feel in 4., that $\mathcal{O}_L$ is flat over $\mathcal{O}_{\mathbb{P}^3}$? This is false. For, 3., yes the tors vanish since tor localizes for coherent sheaves. For the general case, the best path is to use the standard resolution for $L_1$ say. You have, $0\to \mathcal{O}_{\mathbb{P}^3}(-2)\to \mathcal{O}_{\mathbb{P}^3}(-1)^2\to ... 2 Take the cylinder$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 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 ... 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. ... 0 it is much easier to find the projection of$(-3, 2, 3)^\top$onto$(1,1,-2)^\top$first and then subtract that from$(-3,2,3)^\top.(-3,2,3)^\top = k(1,1,-2)^\top + \epsilon$so that$\epsilon$is orthogonal to$(1,1,-2)^\top.$taking the dot product gives you$-3 + 2 -6=k(1+1+4) \to k = -\frac 76and \begin{align}\epsilon &= (-3,2,3)^\top +\frac ... 0 First, the 9+4+36 in the square root is wrong, you should have used 1+1+4 instead if you look at the formula more carefully. Secondly, the formula is \Large{{a \cdot b}\over\bigl|a\bigr|^2} and you forgot a square in the bottom. Thirdly, the formula gives you the projection to the normal vector and you need to subtract this from a to get the ... 0 A normal vector of the plane is (1,1,-2)^t, it has length \sqrt{6}. Thus the part of v in direction of the above as unit normal vector n is: n\cdot v=(1/\sqrt{6})(1,1,-2)^t\cdot(-3,2,3)^t = -7/\sqrt{6} $$We subtract this from v to have the rest within the plane:$$ (-3,2,3)^t-(-7/\sqrt{6})(1/\sqrt{6})(1,1,-2)^t = (-3,2,3)^t + (7/6,7/6,-7/3)^t = ... 0 I have found an answer myself, I'll post it for readers from the future. It is key to calculate the intersection products (inS$) of the three curves. The easy part is $$C_{02} \cdot C_{13} = 0, C_{02} \cdot C_{v2} = 1, C_{13} \cdot C_{v2} = 0$$ (by simply counting points). For calculating the self-intersection product of a curve$C$in$S$, we will use ... 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 $$f^*_n:k(t)\to k(t):t\mapsto ... 0 Assuming your idea about \angle KBA = \angle MCD is correct, will \angle AKB = \angle MCB? 1 We want to show that$$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, ... 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 ...

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