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The Hodge Conjecture

  1. Feb 17, 2008 #1
    Can Anyone explain the hodge conjecture in simple terms?
     
  2. jcsd
  3. Feb 18, 2008 #2

    HallsofIvy

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  4. Feb 23, 2008 #3
    I think i found a way to simply explain the hodge conjecture to myself, comment on the accuracy please.

    We have a small smooth “space” (which at every neighborhood resembles Euclidean space but on a larger scale the “space” is different) that is described by a bunch of equations such that this space has even dimension. Then we take the basic “topological” information and break it into smaller geometric parts (labeled by pairs of numbers). The rational things inside the geometric parts are called “Hodge cycles”. Every smaller geometric part is a combination of geometric pieces called algebraic cycles.

    Basically we have a “pile.” We take a closer look at it to see that it is made up of lots of “chopped wood.” The “chopped wood” has “twigs” (Hodge cycles) inside it.

    The Hodge conjecture asserts that for the piles of chopped wood, the twigs are actually combinations of geometric pieces called atoms (algebraic cycles).
     
  5. Feb 23, 2008 #4

    mathwonk

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    in a smooth projective algebraic variety over the complex numbers, every al;gebraic subvariety determines a homology class. the hodge conjecture asks to characterize the converse, which homology classes do come from algebraic subvarieties?

    for curves it is trivial, since the only possible classes are zero dimensional, and all zero dimensional sets are algebraic.

    so the first case of interest is algebraic surfaces, which are 4 manifolds, and the relevant question is to describe which 2 dimensional homology classes come from algebraic curves living on the surface.

    looking at the dual cohomology classes, it turns out one can use the harmonic structure underlying the complex structure to filter the de rham cohomology, according to the differential forms representing these classes.

    i.e. every class has a harmonic representative, which is a wedge product of forms of form dz or dzbar. a form is called of type (1,1) if it has form dz^dwbar, i.e. it has one holomorphic and one antiholomorphic factor.

    the theorem of lefschetz is that every integral (1,1) form comes from a lineAR COMBINATION OF algebraic curves, i.e. is ALGEBRAIC.

    the hodge conjecture is an attempt to generalize this to higher dimensions.

    look on the clay mathematics institute website for an authoritative description of all the clay math institute problems, including i presume this one.

    about 40-50 years ago grothendieck observed the naive conjecture over Z was false for "trivial reasons" and it has been revised over Q, and generalized to the point i do not know myself exactly what it says anymore, but essentially it is an attempt to characterize those cohomology classes on algebraic varieties that "come from geometry".
     
    Last edited: Feb 23, 2008
  6. Feb 23, 2008 #5

    mathwonk

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  7. Feb 24, 2008 #6

    mathwonk

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    since not everyone is conversant with differential forms, and harmonic ones at that, let me try a more elementary discussion.

    topology is the most fundamental of all geometry, in that it is easier for two geometric objects to be topologically the same than to be equivalent in any other sense, whether differentiably, holomorphically, or algebraically.

    Thus a basic tool in studying the more rigid and restrictive geometries is to consider their topological structure, the most computable of which is homology.

    Just as one studies a vector space by looking at its subspaces, one studies a topological space by looking at nice subspaces, such as curves or surfaces lying on it. Since there are too many of these, one introduces an equivalence relation, the most geometric of which is homotopy, or deformation invariance.

    this equivalence relation is still notoriously difficult, and a more computable one is homology. two geomnetric figures inside a given space are homologous if together they form the boundary of a higher dimensional figure. Strictly speaking however this relation is also very hard, and is strictly called cobordism.

    An easier, but less geometric version of this is to let a boundary be a more algebraic object, a linear combination, with possibly negative coefficients, of geometric subobjects. Two figures are homologous if their formal algebraic difference is a boundary in this wider sense. e.g. the homology boundary of a triangle is the alternating sum of the three sides, where the sides have been given an ordering.

    Now after finding an invariant that one can compute, one wants to go backwards and assign some meaning to the results of the computation. Hence in an algebraic variety, defined by equations, the subvarieties, also defined by equations, can be separated into homology classes.

    It is not to be expected that every homology class arises from an algebraic subvariety, and a primary problem would be to determine which ones do. Then one could prove th existence of algebraic subvarieties with certain properties by making homology calculations.

    this is the hodge problem.
     
    Last edited: Feb 24, 2008
  8. Feb 24, 2008 #7

    mathwonk

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    to continue, homology classes are in some sense more difficult to calculate with than their duals, cohomology classes. one big advantage of cohomology is it has a natural multiplicative structure dual to the intersection of homology classes, but which is easier to define and compute with than is intersection theory.

    a differential form is something we integrate over a geometric subspace. in good cases, e.g. when cauchy's theorem holds, integration is invariant over all homologous subvarieties. I.e. one can often calculate an integral by deforming the surface of integration. this is justified by stokes theorem.

    recalling that stokes theorem says that integrating one thing over a subspace can also be done by ointegrating soemthing else over the boundary of that subspace, suggests th other thing is in some sense a "coboundary" of the original integrand.

    since differential forms spit out numbers when faced with a subspace, and are constant under homology of the subspace, they are dual to homology classes and hence represent cohomology classes. then the wedge product provides a quick easy formal version of a multiplication dual to intersection.

    now that we are looking at differential forms, things made up of symbols like dz, dw,....we are beginning to see that the complex structure may be brought to bear to distinguish different forms from each other.

    indeed one can also integrate dzbar and dwbar, and their products, dz^dwbar,....


    then we can distinguish differential forms according to how many holomorphic coordinates they have, dz,.., and how many antiholomorphic ones, dzbar,....


    it turns out that on an algebraic surface, a homology class coems from an algebric curve on the surface if it is dual to a form represented by one dz and one dwbar, i.e. if it has type (1,1).

    the corresponding conjecture in higher dimensions, still open, concerns the role of forms of type (p,p).
     
  9. Feb 24, 2008 #8

    mathwonk

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    very briefly, there is a theory of cohomology where the coefficients are not just integers but locally defined functions, like O = holomorphic functions, O* = never zero ones, Ω^p = holomorphic p forms, etc...

    By hodge and dolbeault theory some of these give a direct sum decomposition of the usual topological cohomology, e.g. H^2(C) with complex coefficients, =

    H^2(O) + H^1(Ω^1) + H^0(Ω^2).

    then the exponential sequence 0-->Z-->O-->O*-->0 defined by f-->e^(2pif), gives an exact sequence of cohomologies containing the piece

    H^1(O*)-->H^2(Z)-->H^2(O)-->0.

    where the right hand map is the composition of the inclusion H^2(Z)-->H^2(C) with the projection

    H^2(C)-->H^2(O). By the exactness of the sequence, the group H^1(O*)maps onto the kernel of this map.

    but H^1(O*) is the group of line bundles, which is equivalent to the group of linear equivalence classes of formal sums of curves on the surface, so this sequence says that classes of curves are precisely the integral cohomology classes in H^2(C) that have no (2,0) part, (hence also no (0,2) part).

    I.e. integral cohomology classes of type (1,1) in H^2(C) are exactly the ones represented by curves on the surface.

    this solves the hodge problem for surfaces.
     
    Last edited: Feb 24, 2008
  10. Feb 24, 2008 #9

    mathwonk

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    so the first interesting open problem is understanding the classes in H^4 of a threefold that represent classes dual to curves on the threefold.

    thus one takes certain candidate classes in H^4 and tries to find curves representing them.

    a typical case is a principally polarized abelian variety of dimension 3, which is a usually jacobian of a curve of genus 3, and tries to find other curves inside the abelian variety.
     
  11. Feb 24, 2008 #10

    mathwonk

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    kurt your version is so picturesque i hesitate to comment on its accuracy. it is not really wrong, but not too precise either, i.e. it is too imprecise to be inaccurate.

    the hodge conjecture is not exactly chopped liver.

    by the way thank you for asking a question which is a good bit more interesting then the average.
     
    Last edited: Feb 24, 2008
  12. Feb 24, 2008 #11

    mathwonk

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    by the way i have concealed the harmonic type of the forms in the latest post by using dolbeault's isomorphism H^p,q = H^q(Ω^p).
     
  13. Feb 26, 2008 #12

    mathwonk

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    was this intelligible to someone interested in it, or have we confirmed the original suggestion that a simple explanation of hodge's conjecture is unfeasible?
     
  14. Mar 1, 2008 #13

    mathwonk

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    in spite of lack of responses i take some heart from the 344 hits to this thread. i deduce i am not speaking only to myself.
     
  15. Mar 2, 2008 #14
    I got lost on "principally polarized abelian variety", as I don't yet know a thing about varieties, sheaves, etc., but I suspect the rest of your post is understandable to anyone who has studied basic differential geometry/algebraic topology. Interesting stuff!
     
  16. Mar 2, 2008 #15

    CRGreathouse

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    You lost me as well, but for as long as I was able to follow it was interesting. I agree with hallsofivy that there's really no easy way to explain it.
     
  17. Mar 2, 2008 #16

    mathwonk

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    an abelian variety is a product of circles witha complex structure and admitting a projective embedding.

    i.e. a quotient of a complex space C^n, by a lattice, i.e. the Z linear span of an R basis of the vector space, which has a projective embedding.

    such a thing of dimension n is a p.p.a.v. iff it has a complex subspace of codimension one, whose self intersection number is n!
    the simplest example is the complex numbers modded out by a lattice, i.e. a complex torus.
     
    Last edited: Mar 2, 2008
  18. Mar 1, 2011 #17
    Answering directly -- with a delay of nearly 3 years -- you were NOT providing an explanation to an empty room. A very concise and step-by-step approach towards basic understanding, with hints along the path for excursions on related background; nicely done.

    My thanks as well to the original questioner.

    W.
     
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