# Tag Info

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 ...

6

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 ...

3

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

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 ...

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).

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.

2

Let $F,U,x,y$ be as in your attempt. For every $\theta$ the projection map $\operatorname{proj}_\theta$ is continuous; hence, $\operatorname{proj}_\theta (U)$ is connected. A connected subset of a line with two distinct points has positive length. For all $\theta\in [0,\pi)$ except one (the one parallel to the line segment $xy$), ...

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 ...

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} ...

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

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 ...

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

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

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 ...

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 ...

1

There is another common model where the points of the projective plane consist of the points of the Euclidean plane and equivalence classes of lines for the equivalence relation "is parallel to". The mental image is that the latter types of points are to be thought of as the "point at infinity" that the class of parallel lines intersects at. (and the lines ...

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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