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5

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 ...


3

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.


2

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 ...


2

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 ...


2

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


2

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 ...


1

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.


1

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 ...


1

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 ...


1

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)= ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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, ...


1

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



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