# Tag Info

## Hot answers tagged projective-geometry

3

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.

3

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 ...

2

As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane $\Sigma$ is orthogonally projected onto another plane $\Pi$ then the area is multiplied by $|\cos\phi|$, where $\phi$ is the angle between the planes, or equivalently: the angle between the ...

1

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 ...

1

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?

1

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 ...

1

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 ...

1

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. ...

1

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 ...

1

The synthetic viewpoint is more general! This is a benefit since stating things abstractly in terms of primitive notions is probably easier to grasp rather than delving immediately into coordinates. Compare: Two lines intersect in at most one point The system $\{ax+by+c=0;dx+ey+f=0\}$ has at most one solution. The first comes with less baggage than the ...

1

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible