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$\pmatrix{0 &0 &1 \\ 1&0&0\\0&1&0} $ You can solve this just by looking at it for one minute. See what happens to common elements... eg: $f((1,\textbf{1},0))=(0,1,\textbf{1}) $ $f((0,\textbf{1},1))=(1,0,\textbf{1}) $ and so on...


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Yes, when working with projective geometry using clifford algebra, one needs to take care that one's results do not depend on the metric of the base space. However, that doesn't mean one has to avoid the metric completely; rather, one just needs to use metrical products in combinations that are non-metrical. For example, consider a 2d projective geometry, ...


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Let's consider the cross-ratio in the coplex plane so that it looks like $$(A,B,C,D)=\frac{A-C}{A-D}\cdot\frac{B-D}{B-C}$$ Given trhee point $A,B,C$ in $\widehat{\mathbb C}=\mathbb C\cup\{\infty\}$ there is a unique moebius transformation $z\mapsto \frac{az+b}{cz+d}$ that sends $A\to 1, D\to 0, C\to\infty$ and this is exactly $$f(z)= ...


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You can map any three points on a line to any three other using a projective transformation. So there exists a projective transformation which fixes the two points you have but maps your supposed midpoint anywhere at all. Therefore there can be no definition of a midpoint which is invariant under projective transformations. If you fix one hyperplane as the ...


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Yes, you can treat the points on a line as the set of all linear combinations of the two defining points, the way you did. In my nomenclature, I'd usually use different vectors to describe the line itself. In $\mathbb P^2$ I'd use the cross product to obtain a vector normal to all the points. In $\mathbb P^3$ I'd use Plücker coordinates. But there are ...


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Just directly building upon bubba's answer, in Matrix form this can be computed by doing $$Y = \left(M^TM\right)^{-1}M^TW$$ where $M=\left\{\begin{matrix} |\\ U \;|\; V \\ | \end{matrix}\right\}$ is the matrix that contains the basis vectors and $Y=\left\{\begin{matrix} h\\ k \end{matrix}\right\}$ is the projection of W (your original vector) on said basis.


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We may assume that the sphere is given by $$S:\quad x^2+y^2+z^2=a^2, \qquad a>0\ ,$$ and the cylinder by $$C:\quad (x-m)^2 +y^2=b^2,\qquad m\geq0, \quad b>0\ .$$ The sphere intersects the $(x,y)$-plane in the circle $K_s:\ x^2+y^2=a^2$, and the cylinder in the circle $K_c: \ (x-m)^2+y^2=b^2$. The vertical projection of the curve $\gamma:=S\cap C$ is a ...


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Draw the circle passing through $B,C,D$ and the circle passing through $A,C,D$ (or the line for three collinear points). The argument of the cross-ratio $\arg(A,B,C,D)$ is the angle between the two circles* where they meet at $C$. To work this out, notice that the construction and the answer you get are invariant under Möbius maps, so you can make ...


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It is way easier to think when a plane section of a cylinder, that is an ellipse having ratio between the axis that depends on the angle between the cutting plane and the axis of the cylinder, is also a plane section for a sphere i.e. a circle. There is no inaccuracy in point $(2)$: in order to have a planar intersection, the center of the sphere has to lie ...


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If I take a photo of the circular cross section of the moon using my 35mm camera, then the resulting image on my 35mm film is certainly smaller than the moon itself, along any axis. Nevertheless, this transformation from plane in the world world to plane of the film would usually be considered a perspective transformation. So unless you have a more ...


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The projectivity defined by $f$ maps lines in $S$ to sets in $I$ (with $f$ at certain points perhaps being undefined); for most lines of $S$, the resulting image under $f$ is contained in a unique line of $I$. (Some lines in $S$, if $S$ happens to contain $f$, might map to single points in $I$). So we can consider it as a kind of map from lines to lines. ...


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The answer to the three questions is yes. If one wants to make it a little more precise, one could add the functor in play. This however isn't done in one line, which is why many people suppress it. The main ingredient of the proof is in fact Chow's theorem. Serre's GAGA is mainly concerned with the additional equivalence of coherent sheaves and cohomology ...


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This is a problem involving similar triangles. Can you see the related ratios: $$\frac{L-\sqrt{L^2-r^2}}{r}=\frac{2 L}{\rho }?$$


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Suppose a rational solution $(x,y,z)$. Let $d$ be a common denominator of $x,y,z$. Then $(dx)^n + (dy)^n= (dz)^n$ and we have an integral solution corresponding to it.


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Thank you everyone. It was "Geometry of Conics" by A. V. Akopyan (Author), A. A. Zaslavsky (Author).


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Note that $GL(2,\mathbb{C})$ acts transitively on $\mathbb{C}P^1$. Take a diffeomorphism $\phi$. Let $\phi ( \infty ) = \alpha$. Take $g\in GL(2,\mathbb{C})$ with $g(\alpha) = \infty$ ( for instance $g(z) = \frac{1}{z-\alpha}$). Consider the diffeomorphism $\psi = g \circ \phi$. We have $\psi( \infty) = \infty$, $\psi(\mathbb{C}) \subset \mathbb{C}$ and ...


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It is known as Monge's theorem. On the wikipedia page you can also find different proofs.


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Let $m_1$ and $m_2$ be two sides of the original triangle that are not parallel to the image plane, meeting at $A$. Consider the plane through $A$ parallel to the image plane and choose two lines $k_1$ and $k_2$ in it, which meet in a right angle at $A$, such that neither $k_1$ nor $k_2$ is parallel to the plane of the original triangle. Let $P_1$ be the ...


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If $X,Y\subset \mathbb P^n_k$ are irreducible subvarieties (smoothness is irrelevant), then for every irreducible component $Z$ of $ X\cap Y$ we have $$\operatorname {codim} Z\leq\operatorname {codim} X+\operatorname {codim}Y $$ Moreover $$\operatorname {codim} X+\operatorname {codim}Y \leq n \implies X\cap Y\neq \emptyset$$ The fact ...


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I got my answer here: http://en.wikipedia.org/wiki/M%C3%B6bius_transformation I'd like to post it to share. Consider a projectivity $g$ that maps $u,v$ to $0,\infty$: $$g(\lambda)=\frac{\lambda-u}{\lambda-v}$$ Then the projectivity $g\circ\omega\circ g^{-1}$ maps $0,\infty$ to $0,\infty$. So it is a dilation\rotaion: $$g\circ\omega\circ ...


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Let $ f_1, f_2$ and $f_3$ be involutions and $f\colon \mathbb{R}\mathrm{P}^1 \to \mathbb{R}\mathrm{P}^1 $ be a projective transformation. Now let $A, B, C$ and $D$ be four points in $\mathbb{R}\mathrm{P}^1$. Weobtain the following construction: $$ P \overset{f_1}{\to} Q \overset{f_2}{\to} R \overset{f_3}{\to} S \\ Q \to P \to S \to R \\ R \to S \to P \to Q ...


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Doran and Lasenby's book Geometric algebra for physicists does an OK job of presenting those three topics. Normally the way physicists write does not make me very happy, but their exposition for these particular topics was very helpful to me.


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There is a natural map $\varphi_n : \mathbb{CP}^n \to \mathbb{CP}^{n+1}$ given by $\varphi_n([z_0, \dots, z_n]) = [z_0, \dots, z_n, 0]$. This defines a direct system and therefore has a direct limit denoted $\mathbb{CP}^{\infty}$. More concretely, an element of $\mathbb{CP}^{\infty}$ can be written as $[z_0, z_1, z_2, \dots]$ where each component $z_i$ is ...


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You are trying to solve $\min \{ f(x) | x \in C \}$, where $f(x) = {1\over 2} \|u-x\|^2$ and $C = \bar{B}(y, \epsilon)$. Since $C$ is compact, we know there is a solution $\hat{x}$. Since the Euclidean norm is strictly convex, we see that the solution is unique. Since $f$ and $C$ are convex, we have that $\hat{x}$ solves the problem iff $D f(\hat{x}) ...


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Scroll to the bottom of Clifford Algebra: A Visual Introduction, right under Clifford the Big Red Algebra and you will find two links: The story continues with Geometric Algebra: Projective Geometry. The final chapter is Geometric Algebra: Conformal Geometry. If you like this intuitive graphical presentation of Clifford Algebra you might also be ...


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You know that $$ \begin{align} P = \alpha A + \beta B + \gamma C &= \beta(B-A) + \gamma(C-A) + A\\ &= \alpha (A-B) + \gamma(C-B) + B \\ &= \alpha (A-C) + \beta(B-C) + C. \end{align} $$ We can classify the outside of the triangle in six regions: 1) $0\leq\beta$, $0\leq\gamma$, and $1\leq\beta+\gamma$, in which case the point is closest to $BC$. ...


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In projective coordinates, $[x:y:z]$ denotes the same point as $[tx:ty:tz]$ (if $t\ne 0$), hence indeed $[0:1:0]$ and $[0:2:0]$ denote the same point. It is convenient but not mandatry to take the "simplest" representative, which would be $[0:1:0]$. While $[1:1:0]$, $[2:3:0]$, $[1:0:0]$ etc. are also infinte points they are different points; they all lie on ...



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