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The nearest point projection onto a closed subset $E\subset \mathbb R^n$ is single-valued if and only if $E$ is convex*. In this case, the Lipschitz constant is equal to $1$. If $E$ is not convex, there is at least one point $x\in \mathbb R^n$ for which $\min_{y\in E}\|x-y\|$ is attained at more than one point. We could try to discuss the continuity of ...


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No. The length of the semi-minor axis of the projected ellipse will be $r*cos(\theta)$, where $\theta$ is the angle of rotation from horizontal and $r$ is the radius of the circle.


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I. No, I haven't heard of this problem, and can give no other source, sorry. II. According to page 2, the Desargues theorem used in the document is the projective version in the wiki page. (The author of the interesting document employs of the theorem to prove that the point $C$ in the construction actually lies on the line $AB,$ so as to reduce the distance ...


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Your space $\mathbb{P}^n_{(a_0, \ldots, a_n)}$ is the quotient of $\mathbb{A}^{n+1} - 0$ (not an affine space, since its global functions are teh same as $\mathbb{A}^{n+1}$) by a $\mathbb{G}_m$ action with weights $a_1, \ldots, a_n$. Say $\mathbb{A}^{n+1} = k[x_0, \ldots, x_n]$. The $\mathbb{G}_m$ stable open affines are still given by $x_i \ne 0$. So you ...


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The mathematical reason for this is like this: suppose you look at an infinitely large plane from a point 1 meter above that plane and you have two infinitely long parallel lines drawn on it. Further, suppose the distance between lines is also 1 meter. Then, when you look at these lines right under your point of view the light has to travel 1 meter from each ...


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In $\mathbb R^4,$ take the ordinary 3-sphere $$ x^2 + y^2 + z^2 + w^2 = 1. $$ For anything on this, centrally project from $(0,0,0,0)$ to the hyperplane $$ w = -1. $$ The points that are pushed out to infinity are the equator of the 3-sphere, which is a 2-sphere, but with antipodal points identified, so it is an RP2


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Think of a family of all lines passing through a point in $\mathbb{R}^3$. Now start moving this point to infinity. In the limit the family turns into a family of parallel lines. By definition, the plane at infinity is the set of these families, it's a projective plane, not an ordinary Euclidean plane. Different families have different directions so it ...


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Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$. Now consider two ...


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Let's start with $\Bbb RP^2$ since the rest are analogous. Description 1: The points of $\Bbb RP^2$ are the 1-dimensional subspaces of $\Bbb R^3$. A line in $\Bbb RP^2$ is a set of such points such that the 1-d subspaces all lie in a single 2-d subpace of $\Bbb R^3$. A point is on a line if the 1-d subspace for the point is contained in the 2-d subspace for ...


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I agree with every word written in rschwieb's answer, and would like to add another equivalent definition, which can be useful in some topological and geometric contexts. Let $n$ be any positive integer (the case $n=0$ is not so interesting topologically, and thus omitted here), and let $\mathbb{Z}/2$ act on $S^n$ by the antipodal map. The projective space ...


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Claim: $AB$ intersects $CD$ if and only if the underlying vectors of $A,B,C,D$ are linearly dependent. (in fact, if $AB \neq CD$, they span a space of dimension $3$)


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General approach I'm more used to writing down the Plücker description of a line as a single six-dimensional vector of Plücker coordinates, as opposed to Plücker matrices, but that's mostly notation. Anyway, one possible approach to decide whether two lines are parallel is the following: intersect one line with the line at infinity, then use that point of ...


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A plane in $\mathbb{R}^3$ is implicitly defined by $$ F(x,y,z) = ax+by+cz = \delta. $$ The points $(x,y,z)$ that verify this equation all lie in the plane perpendicular to the vector $(a,b,c)$. The gradient of $F$ is just $\nabla F(x,y,z) = (a,b,c)$.


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If I misinterpreted something, let me know: the presentation above is a little foreign to me. It seems to me that we're discussing the conic $C=\{q\mid q^tAq=0\}$, and that the pole-polar correlation in question is the one where $q\mapsto (q^tA)^\perp$. That is, the right hand side is a plane in $\Bbb R^3$ that determines a projective line in $\Bbb R P^2$. ...



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