Hot answers tagged projective-geometry
23
I have seen the impossible staircase used to give an intuitive picture of cohomology. One has a bunch of local pictures (the individual sides of the staircase) and one wants to patch them together into a global picture (the entire staircase), but there's an obstruction to doing this (the implied heights don't match up), and this obstruction is in some sense ...
21
Recently, the best freely available textbook on category-laden algebraic geometry seems to be:
Vakil - Foundations of Algebraic Geometry, Standford University.
The following reference is a great companion to the hard core of Vakil and/or Hartshorne:
Mumford/Oda - Algebraic Geometry II: Schemes and Sheaf Cohomology, (draft).
For deep classical ...
15
Monodromy doesn't literally work like a paradoxical staircase, but more like a completely possible spiral staircase that goes infinitely upwards and infinitely downwards. As illustration, see this graphic of the natural logarithm over multiple branches (cut off at top and bottom of course):
$\hskip 2.3 in$
(Source: Wikipedia. Bigger version.)
The ...
10
See the Algebraic Geometry site of Donu Arapura's from Purdue University, where you'll find links to:
A pre-introduction to algebraic geometry by pictures (html link)
Basic algebraic geometry (pdf): Described as "...sort of a 'prequel' to Hartshorne."
You might also want to check out J.S. Milne's site on Algebraic Geometry, where you'll find a list ...
9
In the case that you are willing to assume that the map is once differentiable, the answer is yes.
Sketch of proof:
There is a duality between great circles on $\mathbb{S}^2$ and lines in $\mathbb{R}^3$: you can identify a great circle on $\mathbb{S}^2$ with plane in $\mathbb{R}^3$ that contains the great circle, and then identify with the line that is ...
8
Your question is equivalent to the proposition that the linear system of hyperplane divisors of a smooth curve $V$ is complete. This is not true for $n\geq 3$, as the following example will show:
For any $n\geq 3$, consider $\sigma_n:\mathbb{P}^1\rightarrow \mathbb{P}^n$ defined by $[x,y]\mapsto[x^{n+1},x^ny,\cdots,\widehat{x^2y^{n-1}},xy^n,y^{n+1}]$, then ...
8
Intuitively and for my answer, I am working over an algebraically closed field $k$.
I personally think about invertible sheaves as sheaves of functions on a scheme, where $s\in\mathcal L(U)$ can be evaluated at a point $P\in U$ in the sense that $s(P)$ is the image of $s$ under the morphism $\mathcal L(U) \to \mathcal L_P \cong \mathcal O_{X,P} ...
7
Yes, there is a reason for $L$ to be trivial and here it is:
Let $0\neq s\in \Gamma(X,L)$ and $0\neq \sigma\in \Gamma(X,L^*)$ be two non zero sections.
Then $s\otimes \sigma\in \Gamma(X,L\otimes L^*)=\Gamma(X,\mathcal O)$ is a constant since $X$ is complete: $s\otimes \sigma =c\in k$ (the base field).
Now, since $s$ and $c$ are non-zero there is a ...
7
Yes, the Plücker relations are written down totally explicitly in terms of the polynomials you require on page 110, equation (3.4.10), of Jacobson's book Finite-Dimensional Algebras over Fields. The proof, attributed by the author to Faulkner (a student of his?), is completely down-to-earth: no identifications, no duality,...
Edit Since Martin doesn't have ...
7
Definition:
If $\mathcal I=\mathcal I_V\subset \mathcal O_{\mathbb P^n}$ is the ideal sheaf defining the curve $V$, we have the exact sequence of sheaves on $\mathbb P^n$: $$0\to \mathcal I\to \mathcal O_{\mathbb P^n} \to \mathcal O_V \to 0$$ which gives after twisting by $\mathcal O_{\mathbb P^n}(1)$: $$ 0\to \mathcal I(1)\to \mathcal O_{\mathbb P^n} ...
7
The elements of $\mathbb{P}^n(k)$ are equivalence classes of elements of $k^{n+1}-\{(0,\ldots,0)\}$, under the equivalent relation
$$(a_1,\ldots,a_{n+1}) \sim (b_1,\ldots,b_{n+1})\Leftrightarrow \text{there exists }\lambda\in k, \lambda\neq 0\text{ s.t. }b_i=\lambda a_i\text{ for }i=1,\ldots,n+1.$$
More conceptually: the elements of $\mathbb{P^1}(k)$ ...
7
The graph of a morphism $f: X \to Y$ is the pull-back under the product map
$f\times 1: X \times Y \to Y \times Y$ of the diagonal $\Delta(Y) \subset Y \times Y.$ Thus for the graph to be closed, what you need is the diagonal $\Delta(Y)$
to be closed in $Y \times Y$. This is true for all quasi-projective varieties,
and so in particular for projective ...
6
To expand on Asaf's comment, any ordinal number $\alpha$ can be made into a topological space by giving it with the topology generated by open intervals $(\beta,\gamma)$. This is called the order topology on $\alpha$. In this topology, the limit points are just the limit ordinals.
This space is compact if $\alpha$ is a successor ordinal ($\alpha = \beta+1$ ...
5
I presume you're working over the complex numbers, and in that case, as has already been mentioned, by Bezout's theorem, there are 12 intersection points, counted with their multiplicities.
But in this particular case, it is not that hard to find the actual points of intersection as follows. First we find the points at infinity, i.e., the points $[X, Y, 0] ...
5
In classical mechanics, the state of a particle on a smooth manifold $M$ is represented as a point of the cotangent bundle $T^{\ast}(M)$. In quantum mechanics, the state of a particle on a manifold $M$ is instead represented as a unit vector in the Hilbert space $H = L^2(M)$, except that two unit vectors differing in phase (that is, differing by ...
5
"Plücker coordinates", and "Grassmannian" are the right key words for what you are looking for. The object you want is the Grassmannian $G(2,4)$ (or $G(1,\mathbb P^3)$, depending on notations). You can realize this variety, the smallest Grassmannian which is not a projective space, as a quadric hypersurface in $\mathbb P^5$, via the "Plücker embedding".
4
David Huffman, though trained as an engineer, did lots of mathematical work. He is famous for the codes named after him (Huffman codes) that were a jumping off point for the mathematics of data compression.
http://www.huffmancoding.com/my-family/my-uncle/scientific-american
However, he also developed a coding system for taking drawings of polyhedra and ...
4
An alternative definition of the given blow-up is the following.
Take $\mathbb{C}^2\times\mathbb{CP}^1$ with coordinates $((z,w),[s:t])$ and take the submanifold $X=\{zt=ws\}$. We can restrict the projection $\pi:\mathbb{C}^2\times\mathbb{CP}^1\to\mathbb{C}^2$ to $X$, obtaining $\pi_X:X\to\mathbb{C}^2$.
We note that $\pi_1((z,w),[s:t])=(z,w)$, but if ...
4
The line through $(a_0 : ... : a_n)$ and $(b_0 : ... : b_n)$ is uniquely parameterized by $(a_0 X + b_0 Y : a_1 X + b_1 Y : ... a_n X + b_n Y)$. Note that this precisely describes a morphism $\mathbb{P}^1 \to \mathbb{P}^n$ which is an isomorphism onto its image. I am surprised this is not given somewhere in Fulton.
4
Regarding the third part of your question: The intersection of two lines in $\mathbb P^3$ is generically empty (if you write down two random lines, they will be skew), but sometimes the lines will be coplanar (i.e. lie in a common plane), and then they will meet in a point (as any two lines in $\mathbb P^2$ do).
If you know that a priori that the two lines ...
4
Are you asking why a projective complex variety has "enough" meromorphic functions to be Moisezon?
If so, the answer is that if $X$ is projective of dimension $d$, then the field of rational functions on $X$ (all of which are meromorphic --- and conversely, although we won't need this latter fact) is of transcendence dimension $d$,
i.e. contains (exactly) ...
4
Every point of $V=V(XT-YZ,\, Y^2-XZ,\,Z^2-YT)$ is in at least one of the four standard copies of $\mathbb A^3$affine spaces covering $\mathbb P^3$. So check them successively. I'll do the $T=1$ part.
When $T=1$, $V$ is given by $$x=yz, \: y^2=xz,\: z^2=y$$
But then a point $(x,y,z)=[x:y:z:1] \in V $ satisfying these equations is simply the image under ...
4
An alternative approach to this problem is to use the isomorphism from $\mathbb{P}^1 \times \mathbb{P}^1$ to the smooth quadric surface $S \subset \mathbb{P}^3$. In coordinates, the map $\mathbb{P}^1 \times \mathbb{P}^1 \rightarrow \mathbb{P}^3$ is given by
$$
(X_0;X_1), (Y_0;Y_1) \mapsto (X_0 Y_0; X_0 Y_1; X_1 Y_0; X_1 Y_1),
$$
with the image being $S = \{ ...
4
One way to see this is to look at the Picard groups: that of $\mathbb{P}^2$ is $\mathbb{Z}$ (this is true for any projective space over a field), while that of $\mathbb{P}^1$ is $\mathbb{Z} \times \mathbb{Z}$ (in fact, the Picard group of a projective bundle over a noetherian scheme $S$ is always $\mathrm{Pic} S \times \mathbb{Z}$). The case of interest here ...
3
It seems you can use the enormous symmetry of the Fano plane for this. You can pick a line ABC and a point off the line D and pick f(A) and f(B) arbitrarily, f(C) is the third point on the line AB, and pick f(D) arbitrarily different from the first three. Then if the lines through D are ADE, BDF, and CDG you can define f(E), f(F) and f(G) in the obvious ...
3
In the first sentence, the instructor makes the unreasonable request (which may be a command, in the way that classes are often organized) to perform a long, repetitive and largely useless computation. This is not something that people are likely to spend time doing in an online math site, and I am sorry to hear that this kind of time waste is being ...
3
Let's call your two "axis" vectors $U$ and $V$. The nice folks around here would probably call them "basis" vectors. And suppose you have some other vector $W$, lying in the plane of $U$ and $V$, that you want to decompose.
Essentially, you want to find two numbers $h$ and $k$ such that
$$W = hU + kV$$
The vector $hU$ will be the projection of $W$ onto the ...
3
The tangent space $T_p S$ can be thought as a hyperplane $H\subset \mathbb{P}^3$ and any line $l\subset S$ passing through $p$ is contained in this hyperplane.
On the other hand, as $S$ is smooth, it is irreducible, so the intersection $H\cap S$ is of dimension $1$ and has the same degree as $S$, namely $2$.
Each line in $H\cap S$ adds one to this degree so ...
3
Maybe picking $(a,b) \mapsto \text{span}(a,b,1)$ would be a better choice.
This map seem awfully arbitrary and unmotivated, and I find it easier to explain how there can be an isomorphism if I go the other way : start from $\Sigma$ and go to $\Pi'$ :
In order to see the points of $\Sigma$ as points on a normal plane, pick a 2-dimensional affine ...
3
The basic idea is that the "inverse map" is given by sending $\ell \in \mathbb{P}^2$ (identify the points in $\mathbb{P}^2$ with lines through $x$) to the point $y$ where $X \cap \ell = \{x, y\}$. (We are using that $X$ has degree 2, which means that it intersects a general line in 2 points.) It's pretty clear that this map is inverse to the one you ...
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