We a pretty neat binary operation: given two points, we draw a line through them, find the third intersection with E and reflect it through the x-axis. A manifold is simply connected if it has no ‘holes’: An equivalent formulation of being simply-connected is that each loop can be continuously tightened to a point. An elliptic curve E, is the set of solutions of an equation of the form y²=x³+Ax+B with the constraint that the discriminant ∆=-16(4A³ + 27B²) ≠0. The current version of the Hodge Conjecture says: every Hodge class is algebraic. − We can write the Birch and Swinnerton-Dyer conjecture very simply as. Ok maybe one string: the problems are somewhat hard. The first natural question to ask is what can we deduce about the structure of E(ℚ)?

Hopefully, you caught the weird terminology there. As you might imagine, a subvariety of X is just a subset that is also given by zero sets of polynomial equations, and hence also a variety. What the P vs NP problem asks is if in fact NP problems are distinct from P problems. Firstly, a manifold is an object or a generalisation of a space that is locally like Euclidean space. If the one loop is [A] and the second is [B], you’re allowed to make sums with these elements with rational number coefficients. An example of a problem that is thought to be NP (I say thought since it hinges on the verity of the hypothesis) is the (decision problem version of the) travelling salesman problem: Given a list of cities and the distance between each, can you construct a route that visits each city whose total length is less than a given distance? Official Problem Description — Pierre Deligne : http://www.claymath.org/millennium/Hodge_Conjecture/hodge.pdf\r\rIn the twentieth century mathematicians discovered powerful ways to investigate the shapes of complicated objects.

Aleph 0 … Then in 1961 Atiyah and Hirzebruch showed the integral version is false. The same is true if We’ll use the notation X for the space we’re working on.

H This makes a lot of stuff work like making sense of intersecting. Geometry can mean a lot of different things, but in this article, I’ll always use it to mean “algebraic geometry.” For our purposes, this means the study of shapes given by the zero sets of polynomials. A mass gap is essentially a stipulation that the mass of these particles have to be bounded below, so that you can’t find a particle that is arbitrarily light. is the subgroup of cohomology classes which are represented by harmonic forms of type z

Because we’re working over the complex numbers, “real” dimensions will always be even. This one is a doozy.

Another way of phrasing the Hodge conjecture involves the idea of an algebraic cycle.

When we can do that, we’ll call the loop “equivalent to 0” because we can deform it into a trivial thing. So for a Yang-Mills theory to be ‘good’ at describing reality must also exhibit this mass gap. Hodge came up with a neat and elegant idea to tell if a homology class was equivalent to an algebraic cycle and this is in essence the Hodge conjecture.

( n p + − {\displaystyle (p,q)} Now the rule book is in fact a field and the game is a Yang-Mills theory and swaping the colours locally is a gauge symmetry.

(

Because [Z] is a cohomology class, it has a Hodge decomposition. In topology, we want to classify all manifolds into classes where all manifolds within a certain class are all homeomorphic to each other.

As it turns out, it seems that if s is a non-trivial zero (i.e. …

Solving this problem is hard and very taxing to solve but a solution is easy to check — a solution is a list of cities to visit in order, and one can verify that it is a valid solution just by adding up the distances and comparing it to the given bound. n Taking wedge products of these harmonic representatives corresponds to the cup product in cohomology, so the cup product is compatible with the Hodge decomposition: Since X is a compact oriented manifold, X has a fundamental class. Hodge made an additional, stronger conjecture than the integral Hodge conjecture. So here everything hits the fan. be the inclusion map. More abstractly, the integral can be written as the cap product of the homology class of Z and the cohomology class represented by

It will enhance any encyclopedic page you visit with the magic of the WIKI 2 technology.

Unlike the Riemann Hypothesis that has overwhelming evidence to support it, the Hodge Conjecture seems like a work in progress that got stuck after a couple of refinements. A gauge group is a group of (possibly very bizzare) symmetries of a system, this gives rise to a conservation law and we can write a ‘rule book’ that is a field that defines how particles interact which is a Yang-Mills theory.

The Lefschetz theorem on (1,1)-classes also implies that if all Hodge classes are generated by the Hodge classes of divisors, then the Hodge conjecture is true: By the strong and weak Lefschetz theorem, the only non-trivial part of the Hodge conjecture for hypersurfaces is the degree m part (i.e., the middle cohomology) of a 2m-dimensional hypersurface {\displaystyle d\leq 5} Ok so what does this all mean? Then every Hodge class on X is a linear combination with rational coefficients of the cohomology classes of complex subvarietie… A very quick proof can be given using sheaf cohomology and the exponential exact sequence.

Here’s the key idea that will make the Hodge Conjecture interesting: from a topologist’s viewpoint, practically nothing is a (sub-)variety. That is, these are the cohomology classes represented by differential forms which, in some choice of local coordinates This technique turned out to be so useful that it got generalized in many different ways, eventually leading to powerful tools that enabled mathematicians to make great progress in cataloging the variety of objects they encountered in their investigations. Weil (1977) generalized this example by showing that whenever the variety has complex multiplication by an imaginary quadratic field, then Hdg2(X) is not generated by products of divisor classes. With this notation, the Hodge conjecture becomes: The assumption in the Hodge conjecture that X be algebraic (projective complex manifold) cannot be weakened. X

Imagine now however that I locally switch the colour of a certain square, and do so as much as I want throughout the board. In this case, we’ll suggestively write H₁(X, ℚ)=ℚ². The donut/torus is an elliptic curve given by a cubic equation. You could also do it yourself at any point in time. Two manifolds are homeomorphic if you can deform one into the other and back again continously. (

Given an elliptic curve E over ℚ, does the algebraic rank always coincide with the analytic rank? > d {\displaystyle p>k} Consequently, this integral is zero if The Hodge conjecture is one of the Clay Mathematics Institute's Millennium Prize Problems, with a prize of $1,000,000 for whoever can prove or disprove the Hodge conjecture. , One of the reasons for doing this is that cohomology has an interpretation using differential forms. Topological things can be crazy and weird and not smooth in the slightest. In some sense it was necessary to add pieces that did not have any geometric interpretation.

At the time I am writting this, only the Poincaré conjecture has been solved.

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