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If by “cross ratio equation” you mean the equation $$\operatorname{CR}(A,B;C,D)=\frac{[A,C][B,D]}{[A,D][B,C]}$$ or something similar, then usually $A,B,C,D$ are assumed to be collinear points in the underlying projective space. According to this post of mine, such a point in the projective space (e.g. on the hyperboloid model for hyperbolic geometry) ...


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Two opposite sides of the quadrilateral, if prolonged, meet at a vanishing point. So two of your three perspective points are easy to find and the third one should be arbitrary. But it is not true that every quadrilateral can be the face of a cube, see for instance the image below. If $ABCD$ is your quadrlateral, $E$ and $F$ are two vanishing points, while ...


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Can I simply take: $$\begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} h_{11} & h_{12} & h_{13} \\ h_{21} & h_{22} & h_{23} \\ h_{31} & h_{32} & h_{33} \end{bmatrix}. \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ Almost. For a single point you can indeed use that equation. But if you have more ...


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Here's a way of doing what I think you want to do. First, find two perpendicular unit-vectors that are parallel to the plane (i.e. perpendicular to the normal vector). You might want to use Gram-Schmidt here. Call the vectors $u_1$ and $u_2$. These are the "axes" along the plane you're projecting onto. Any point on the plane can be written in the form ...


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The intersection of two varieties is given by the sum of their ideals. So you're looking at the ideal generated by the union of the generators of $V_V$ and $V_S$. Immediately we see that both $x_1^2-x_0x_3$ and $x_0x_3-x_1x_2$ are generators of $I=I_{S \cap V}$. But this implies that $x_1^2-x_1x_2=x_1(x_1-x_2)$ is a generator of $I$. Thus $I$ is reducible. ...


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As far as I understand Affine transformation preserves parallelism and ratios of lengths. Consider a transformation which maps distinguished line from one space to the distinguished line of other space. Now consider two parallel lines in a world plane which meet at a point Q on the distinguished line in one space. There will definitely be a corresponding ...


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Sorry to resurrect my old question, but for the sake of completeness, I will provide a complete solution. Consider the tangent space at the origin of our divisor $D$. We have$$k[D] = k[a, b, c, d]/(a, b, ac - bd) \cong k[c, d].$$This Zariski cotangent space has dimension $2$ here (generated by $c$, $d$). Now, let us consider a tangent space of $X \cap ...


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This is false: let $C = \mathbb{P}^1_{\mathbb{C}}$ and let $L = \mathcal{O}(-1)$ and $L' = \mathcal{O}(1)$. Then, $$2 = \underbrace{h^0(C,L)}_{=0} + \underbrace{h^0(C,L')}_{=2} \neq h^0(C, L \otimes L') = 1.$$ It might be more reasonable to expect the dimensions to multiply i.e. have $$ h^0(C,L) \cdot h^0(C,L') = h^0(C,L\otimes L'), $$ because there is a ...


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Here is a construction that is not "tricky" or perhaps elegant but is straightforward. In your desired formula, let's replace the segment length $RS$ with the equivalent $(PS-PR)$, so we have only one segment length that is dependent on point $R$. Then we get $$ \frac{AC \cdot BD} {AB \cdot CD} = \frac{PR \cdot QS} {PQ \cdot (PS-PR)} $$ The only unknown ...


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Mariano in the comments makes a very good suggestion to look at Veblen and Young. But the following axiomatic description is the heart of it: A projective space $\mathbb{P}$ is a set of objects of two types. An object of the first type is called a point; one of the second type is called a line. A point and a line may or may not be incident, and the ...


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There is a bijection between the hyperplanes and the projective points. But defining in terms of hyperplanes is strange (as mentioned in another answer, it is the dual; where did you find this definition, by the way?). Projective geometry is sometimes defined, not in terms of the one-dimenional spaces, but in terms of ALL the subspaces of $V$. This gives ...


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$\phi$ is degree one if and only if it is an isomorphism from an open subset of $X$ to an open subset of $Y$ (i.e. it is birational). So it is generically one-to-one. But birational maps can be far from injective. As counterexamples, consider blowups of varieties along closed subvarieties. These are always birational but never injective. What you can say ...


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There is a bijection between the set of lines and hyperplanes in any vector space, but it is not canonical (it depends on a choice). Namely, choose a non-degenerate bilinear form (this you can do for instance by choosing a basis and taking the standard non-degenerate bilinear form with respect to it). Then the orthogonal complement of a line is a ...


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The line through $T$ and the midpoint of $\overline{AB}$ passes through the center of the ellipse. (In fact, it's one of the axes, which will be important later.) Moreover, if the tangents at $B$ and $C$ meet at, say, $U$, then the line through $U$ and the midpoint of $\overline{BC}$ also passes through the center. With two lines pinpointing its location, we ...


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My solution, for now: We take the line $l$ through $BT\cap AC,AT\cap BC$, $D=l\cap AB$, then $DC$ is tangent to the ellipse at $C$; Let $E=AT\cap DC$ and $\Gamma$ be the circle tangent to $AT,BT$ at $A,B$; Let $F=CT\cap\Gamma$ (one of the two intersections); Let $G$ be the intersection between $AT$ and the tangent to $\Gamma$ at $F$; Let $H$ be one ...



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