# Tag Info

7

Most people are going about this from a projective geometry perspective. I will approach this from a differential geometry viewpoint instead, since I feel I can give a more intuitive understanding from this vantage point. Unfortunately I can't speak about $-\infty = \infty$ from this viewpoint without dealing with stereographic projection or ...

0

I will assume that when you say "projection", it is expressed as a vector of coordinates relative to the basis. It would not change the length, as you only consider orthogonal bases. By definition, orthogonal bases consist of vectors of length $1$, so the coordinates relative to all bases will have the same length.

1

If I take the leaning tower of Pisa, and hang a weight from the top to the ground, then I'm projecting it orthogonally into the plane of the ground. The point that that weight touches the ground is the same whether I choose to use a pair of basis vectors where one is a meter pointing north and the other is a meter pointing west or if I choose to use a pair ...

1

Yes, you are correct! The line defined by $y = mx$ in $\mathbb{R}^2$ embeds into the projective line defined by $y = mx$ in homogenous coordinates. The line defined by $y = mx + c$ embeds into the projective line defined by $y = m x + c z$ in homogenous coordinates. For $c\neq 0$, these two lines intersect when $y - mx = y-mx - cz = 0,$ that is $z = 0$ ...

1

Recall that $\mathbb{P}^2(\mathbb{C}) = (\mathbb{C}^3\setminus\{0\})/\sim$ where $(z_1, z_2, z_3) \sim (w_1, w_2, w_3)$ if there is $\lambda \in \mathbb{C}\setminus\{0\}$ such that $(z_1, z_2, z_3) = \lambda(w_1, w_2, w_3)$. Usually one writes an element of $\mathbb{P}^2(\mathbb{C})$ using homogeneous coordinates. That is, we write a point in ...

10

You lose the vector space structure and even the affine structure and even the additive group structure when you go from $\mathbb C$ to $\mathbb P^1_\mathbb C$. More concretely, given two points in $\mathbb P^1_\mathbb C$ it doesn't make sense to talk of their difference as a vector. The general context is that $\mathbb C$ has both algebraic and ...

3

Lost: the metric structure: distances, angles, areas. Conserved: incidences. See http://www.nct.anth.org.uk/basics.htm.

0

In projective three-space, a point $X$ described by a homogeneous vector $x$ lies on a plane $\pi$ described by a homogeneous coordinate vector $u$ if the dot product $\langle u,x\rangle$ is zero (or in the notation from your lecture notes, if $u^Tx=0$). So if you think of $X$ as the vector $x=(x_1,x_2,x_3,1)^T$ then that plane $u=(u_1,u_2,u_3,u_4)^T$ ...

0

This is a complete rewrite of this answer, following an approach which is very different from my first attempt. Homogeneous parametrization Suppose you have a point $x=(x_1,x_2,x_3)$ on a circle on the sphere. That means it belongs to the intersection of the sphere and a plane, or in formula, $\lVert x\rVert=1$ and $\langle x,n\rangle=d$. This $n$ is the ...

3

Assuming that you're either talking about great circles or circles of nonzero radius, then yes, the space of circles on a sphere is a manifold. To make sure I don't screw things up, I'm going to talk about the case of great circles on a real sphere, because that has all the features you care about. In this case, each great circle corresponds nicely to a ...

1

Observe that $\operatorname{GL}_3(K)$ acts transitively on $K^3\setminus\{0\}$ because for any nonzero $v\in K^3$, you can find an invertible linear map $\phi:K^3\to K^3$ that maps $(1,0,0)$ to $v$, say. Hence, the orbits in $K^3$ are $\{0\}$ and $K^3\setminus\{0\}$. In $\mathbb P^2_K$, there is no origin point, so the space itself is one big orbit. To ...

0

(1) If $T$ is not injective, then it simply induces a projective map from $P(V)\setminus P(\ker T)$, which is an open set in $P(V)$, to $P(W)$. The easiest examples are provided by the very projective maps, namely, the central projections from a projective space onto a smaller subspace. (2) The terminology is quite wild, however, the two names you mention ...

0

For part 1, you'll need $r+s$, $s+t$, and $r+t$ to all be less than $n$ for each pairwise intersection to be empty. Adding these equations, you see that $2(r+s+t)<3n$, and combining with your restriction $r+s+t\geq n$, we have $2n\leq 2(r+s+t)<3n$. Certainly solutions to this inequality exist- let $n=5, r=s=t=2$. So we can pick $X=(x_0,x_1,x_2)$, ...

2

Since an algebraically closed field is necessarily infinite, $\mathbb{A}^1$ and $\mathbb{P}^1$ have the same cardinality, and on both the Zariski topology is the cofinite topology. Therefore they are homeomorphic.

1

Here's one way to look at things: let $U$ be a matrix with columns $u_1 = X,u_2,u_3$ so that the columns of $U$ form an orthonormal basis of $\Bbb R^3$. Note that $UU^T = I$. We can write $$X^TX= U\pmatrix{1&0&0\\0&0&0\\0&0&0}U^T$$ To show this, try the block matrix-multiplication  \pmatrix{u_1&u_2&u_3} ...

0

Personally, I'm not aware of any connection of these theorems with conic sections other than the possibility of a conic section appearing in the diagram describing the configuration. The theorem itself would be in whatever geometry (real projective/Euclidean) that we are talking about. And even then, the only one I'm aware of at present is for Pascal's ...

1

If $\dim X>\dim Y$, taking $\mathcal{O}(1)$ be a very ample line bundle on $X$, then in the linear system $|\mathcal{O}(1)|$ there is a divisor dominates $Y$. This is because $|\mathcal{O}(1)|$ intersects the fiber of $f$ non-trivially. By this, we can cut down the dimension of $X$ and hence get an affirmative answer to your problem. On the other ...

0

The theorem of Desargues is true in the real projective plane, but not in every projective plane: there exist non-Desarguesian planes. I'm not perfectly sure, but I assume that Pascal's theorem should be not only a generalization of but also a consequence of Pappos' theorem, which in turn means you have a projective plane over some field. For Ceva's theorem ...

0

Your observation that we need $n=3$ for the second axiom to hold is correct (+1). Assuming $n=3$, consider the vectors with coordinates $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(1,1,1)$. Any three of them span all of $V$ (check this). Therefore no 2D subspace contains any three of them. Nor does any 2D subspace contain three 1D-subspaces spanned by those points. ...

3

There exist two disjoint irreducible closed subsets both containing more than one point in $\mathbb A^2_k$ but not in $\mathbb P^2_k$ (Bézout).

0

If you have a little more machinery from algebraic geometry you could notice that $\mathbb{P}^2$ is a complete variety while $\mathbb{A}^2$ is not. To explain a little: One says that a variety $X$ is complete if for all varieties $Z$ the projection map $X\times Z\rightarrow Z$ is closed, ie sends closed sets to closed sets. To see that $\mathbb{A}^2$ is ...

-2

Well, there are two things. Homeomorphism preserves compactness (can you see why? homeomorphisms being one-to-one preserves unions, inclusions and opennes). $\mathbb{P}^2$ is compact, while $\mathbb{A}^2$ is not.

1

If a vector $v$ represents a given point $p$, then any multiple $\lambda v$ of $v$ represents the same point (as long as $\lambda\neq0$). So you can think of a point in $\mathbb{RP}^2$ as a line through the origin in $\mathbb R^3$. The set of all points on a line in $\mathbb{RP}^2$ likewise corresponds to the union of many such lines, i.e. to a plane through ...

1

On the part about $uu^T$, your error is that $u$ needs to be considered a column vector, not a row vector. As for the part about $2P - I$, you seem to have generally the right idea, but you're not executing it properly. If you start with a vector $v$, its reflection $Rv$ is symmetric to $v$ with respect to the projection $Pv$. This means that the vector ...

0

Every invariant is a function of cross-ratio. See Olver, Classical invariant theory, Example 8.34, Aczel, Lectures on functional equations and their applications, Theorem on page 233.

1

It is from the definition. However you may be intereseting in that what it come from or why we called it by this name. In fact: Pick a point $(x,y)$ from the space $\Bbb R^2$, you will find that $x$ is just the projection of the point $(x,y)$, and $y$ is also!

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