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Let $\mathbf{n}$ be a unit vector in $\mathbb{R}^3$ be normal to the plane $P$. The projection $p$ of a vector $v\in\mathbb{R}^3$ onto the plane $P$ is given by $p(v)=v-(v\cdot\mathbf{n})\mathbf{n}$. The projection of the unit cube onto $P$ along the vector $\mathbf{n}$ is a hexagon whose interior has preimage which intersects the boundary of the unit cube ...


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The original work here would be Beweis des Desarguesschen Satzes aus dem Pascalschen by Hessenberg (1905), which is the reason why this is called Hessenberg's theorem. My lecture notes suggest the following: Given $AB\Vert A'B'$ and $BC\Vert B'C'$ you want to show $AC\Vert A'C'$. This is a Euclidean formulation of Desargues' theorem, and I'll also use ...


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This is hard to make rigorous, but if assume by contradiction that all the space inside $C$ is taken, then there are couples and triples of ellipses arbitrarily close. Assume that we have $3$ ellipses very close each other, like in figure: then they define a region $D$ with a concave boundary with the property that any entering ellipse cannot have a minor ...


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This reference clarified everything for me. The grid $j$'th gridline is given by $$\vec{x} \cdot \vec{\epsilon}_j + \gamma_j = N,$$ for some integer $N$. Where the grid directions are given by: $$\epsilon_j = \left(\cos\left(\frac{2\pi j}{5}\right), \ \sin\left(\frac{2\pi j}{5}\right) \right) ,$$ for j = (0, 1, 2, 3, 4) and the shift of the each of the ...


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Let me answer by showing how a homogenous polynomial of degree $n$ in the projective line has $n$ solutions. A typical such polynomial is $$a_nx^n+a_{n-1}x^{n-1}y+\cdots +a_0y^n$$ let us assume that $a_{m}\neq 0$ is the first nonzero coefficient then the equation factors $$a_m y^{n-m}(x-\beta_1 y) \cdots (x-\beta_m y)$$ now for $y=1$ we have the finite ...


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A point $p$ is incident with a set $S$ iff $p\in S$. Given how we define the image of $S$ under a map, how could a map fail to preserve incidence?


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Every Möbius transformation is an automorphism of the extended plane. So $Tz\in \mathbb{R}\cup\{\infty\}$ if and only if $z\in T^{-1}(\mathbb{R}\cup\{\infty\})$. Since Möbius transformations map circles (in the extended plane) to circles, that means $z$ lies on a circle passing through $z_1,z_2,z_3$. But through any three points in the extended plane, there ...


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The way I read it, the book does not say that $AB\cap A'B'=AD\cap A'D'$. What it does say is this: But these points are projected for $X$ into $P,Q,R$, on $\Sigma$, hence $P,Q,R$ are collinear. Do you have some point $P'=AD\cap A'D'$ and $X,P,P'$ are collinear, so if you project $P'$ into $\Sigma$ with $X$ as the center of projection, then you end up ...


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No, this is not true. Take the ellipse with semiaxes $a$ and $a^{-1}$ where $a$ is very large. It looks like a long thin needle. The area is $\pi$, independent of $a$. But the integral $\int e(p)\,dp$ tends to $\infty$ as $a\to \infty$. Indeed, $e(p)\ge a$ for all $p$ (consider the distances from $0$ to the pointy vertices of ellipse), and the perimeter ...


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Suppose we think of a projection as a continuous mapping from $\mathbb{R}^4$ to $\mathbb{R}^3$. There are many kinds of projections used in computer graphics, but the computations generally represent continuous functions, sometimes affine or even linear. In any case continuous linear mappings (which is a bit redundant; all finite dimensional linear maps ...


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In this post I worked out, as an exercise, the answer for smooth projective complex hypersurfaces, not using de Rham cohomology. After applying the Lefschetz hyperplane theorem, the basic tool is characteristic classes.


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There is a general algorithm for computation of (co)homology groups of real-algebraic subsets in $\mathbf{R}^n$. Being a hypersurface does not particularly help in this computation. The algorithm goes back to Tarski's work (on elimination of quantifiers). The entire book Algorithms in Real Algebraic Geometry is pretty much all about such computations. ...


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It appears my comment was enough to answer the question, so I'll promote it to an answer: Projective space doesn't have an origin. To elaborate a little - the origin of a vector space is special because it is the additive identity under the group structure. But once you projectivise the vector space, the group structure disappears - and there is no ...


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I will sketch the argument, since writing all the details would be too tedious, and would certainly include typos... An important idea about blowups of points in the plane is the following: given two curves passing through a point $P$, the blowup $\operatorname{Bl}_P(\mathbb P^2)$ separates the curves (i.e. their strict transforms do not intersect) iff they ...


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Edit: This is just a worked out example of a blow-up. A solution was posted by Andrew here. I will work out the blow up for the function $f(x,y) = y^2 - x^3 - x^2$. Note that $\nabla f(0,0) = 0$, so there is a singularity at $(0,0)$. If you graph the function (over $\mathbb{R}$), you will see that it crosses over itself at $(0,0)$. Consider $f(x,xy) = ...



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