Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Hamiltonian systems, symplectic flows, classical integrable systems
10
votes
Accepted
Why are the following varieties symplectomorphic?
In order to obtain an explicit description of the diffeomorphism, one can use the following argument.
Take $B=\mathbb{C}^{n-1}$, with coordinates $t_1, \ldots, t_{n-1}$, and consider the complex spac …
2
votes
Accepted
Computing the coefficients of the polynomial $\dim H^0(X,L^k)$ in non-smooth case
Sometimes one can at least compute $\chi(\mathcal{O}_X(X, \, L^k))$.
The formula that one expects will be of the form "ordinary Riemann-Roch formula plus correction terms depending on the singularit …
41
votes
Accepted
What is a Lagrangian submanifold intuitively?
Lagrangian submanifolds arise naturally in Hamiltonian Mechanics, because of the classical Arnold-Liouville theorem. Let me state it here:
Theorem (Arnold-Liouville). Let $(M, \omega, H)$ be an inte …
4
votes
Calculating the decomposition of a vector bundle over rational curve
Since you know the explicit equation of the conic, you can compute everything by using Macaulay2.
The following script should be clear:
i1 : k=ZZ/32003;
i2 : ringP1=k[x, y];
i3 : ringP4=k[z1, z2, …
8
votes
Accepted
Hyperbolicity for algebraic varieties and relation to curves on them
Take a very general hypersurface $j \colon X \hookrightarrow \mathbb{P}^n$ of sufficiently high degree, $3 \leq n \leq 4$. Then $X$ is Kobayashi hyperbolic (Kobayashi actually conjectured that this is …
3
votes
Accepted
Are Gromov-Witten invariants birational invariants?
On a complex projective variety, Gromov-Witten invariants can be interpreted as virtual counts of curves, so they are biregular invariants.
However, they are not birational invariant in general. The …
7
votes
Is the mirror of a hyperkaehler manifold always a hyperkaehler manifold?
I think Verbisky proves a refined form of the Mirror Conjecture for hyperkaehler manifolds, not the conjecture in the strict form. This is explained at page 3 of the paper that you link.
In fact, th …