English Spanish online dictionary Term Bank, translate words and terms with different pronunciation options. Products of it with itself give candidates for counterexamples to the Hodge conjecture which may be of interest. We also study the Kuga-Satake. The Hodge conjecture asserts that, for particularly nice types of spaces called projective algebraic varieties, the pieces called Hodge cycles are actually rational.
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Mordell proved Mordell’s theorem: With this notation, the Hodge conjecture becomes:.
Term Bank – conjetura de hodge – Spanish English Dictionary
The Millennium prize problems. Weil generalized this example by showing that whenever the variety has complex multiplication by an imaginary quadratic fieldthen Hdg 2 X is not generated by products of divisor classes.
If the rank connjetura an elliptic curve is 0, then the curve has only a finite number of rational points. An algebraic cycle on X is a formal combination of subvarieties of X ; that is, it is something of the form:. In Zucker showed that it is possible to construct a counterexample to the Hodge conjecture as complex tori with analytic rational cohomology of type ppwhich is not projective algebraic.
Hodge made an additional, stronger conjecture than the integral Hodge conjecture. Around 0, we can choose local coordinates z 1However, for large primes it is computationally intensive. Views Read Edit View history.
conjetura de hodge
From Wikipedia, the free encyclopedia. Say that a cohomology class on X is of co-level c coniveau c if it is the pushforward of a cohomology class on a c -codimensional subvariety of X. In fact, it predates the conjecture and provided some of Hodge’s motivation. Voisin proved that the Chern classes of coherent sheaves give strictly more Hodge classes than the Chern classes of vector bundles and that the Chern classes of coherent sheaves are insufficient to generate all the Hodge classes.
By the strong and weak Lefschetz theoremthe only non-trivial part of the Hodge conjecture for hypersurfaces is the degree m part i. We define the cohomology class of an algebraic cycle to be the sum of the cohomology classes of its components. Because [ Z ] is a cohomology class, it has a Hodge decomposition.
The cohomology class of a divisor turns out to equal to its first Chern class. They show that the rational Hodge conjecture is equivalent to an integral Hodge conjecture for this modified motivic cohomology.
It was formulated by the Scottish mathematician William Vallance Douglas Hodge as a result of a work in between and to enrich the description of de Rham cohomology to include extra structure that is present in the case of complex algebraic varieties.
Loosely speaking, the Hodge conjecture asks:. A projective complex manifold is a complex manifold which can be embedded in complex projective space.
In other projects Wikiquote. The rank of certain elliptic curves can be calculated using numerical methods but in the current state of knowledge it is unknown if these methods handle all curves.
See Hodge theory for more details.
Analytic class number formula Riemann—von Mangoldt formula Weil conjectures. This is also false. In other projects Wikiquote. Let Z be a complex submanifold of X of connjetura kand let i: Since X is a compact oriented manifold, X has a fundamental class. Such a class is necessarily a Hodge class.
Such a cohomology class is called algebraic. However, Griffiths transversality theorem shows that this approach cannot prove the Hodge conjecture for higher codimensional ds.
Hodge conjecture – Wikipedia
conjjetura The Lefschetz theorem hodgr 1,1 -classes also implies that if all Hodge classes are generated by the Hodge classes of divisors, then the Hodge conjecture is true:.
That is, these are the cohomology classes represented by differential forms which, in some choice of local coordinates z 1Let X be a compact complex manifold of complex dimension n. Retrieved from ” https: A very quick proof can be given using sheaf cohomology and the exponential exact sequence.
Then the topological cohomology of X does not change, but the Hodge decomposition does change. Connetura the number of rational points on a curve is infinite then some point in a finite basis must have infinite order.
Finding the points on an elliptic curve modulo a given prime p is conceptually straightforward, as there are only a finite number of possibilities to check.