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The Hodge Conjecture

The Hodge Conjecture

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Every harmonic differential form (of a cetain type) on a non-singular projective algebraic variety is a rational combination of cohomology classes of algebraic cycles.

In this thread discuss concepts you believe useful in proving or disproving this, your inclinations as to it's truth, progress you have made and potential proofs.
I will try to answer any questions non-mathmaticians have.

Currently I am in the process of trying a new attempt at disproving the conjecture and have made not a small amount of progress. The recent proof of Salmon's Theorem has helped me a great deal. Has anyone else found it useful?

EDIT: Sorry, I forgot the 'popular version' of the Hodge Conjecture which may be of use to the less good mathematicians among us who wish to follow along:

Let X be a projective algebraic manifold and p a positive integer. Also, let H[2p](X,Q)alg union H[2p](X,Q) be the subsapce of algebraic cocycles, i.e., the Q-vector space generated by the fundamental classes of algebraic subvarieties of codimension p in X. The Hodge conjecture assert that one can "compute" the subspace H[2p](X,Q)alg using Hodge theory, specifically H[2p](X,Q)alg = H[p,p](X) intersection H[2p](X,Q).

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Originally posted by XanthosNZ
Every harmonic differential form (of a cetain type) on a non-singular projective algebraic variety is a rational combination of cohomology classes of algebraic cycles.

In this thread discuss concepts you believe useful in proving or disproving this, your inclinations as to it's truth, progress you have made and potential proofs.
I will try to a ...[text shortened]... lg using Hodge theory, specifically H[2p](X,Q)alg = H[p,p](X) intersection H[2p](X,Q).[/i]
Wow, is that the recipe for "irrational" cookies?

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Originally posted by XanthosNZ
blah blah blah


Hahaha - as if anyone here gives a monkeys. 😕


Well you may get one or two interested I guess.

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Originally posted by Dr Strangelove
Originally posted by XanthosNZ
[b]blah blah blah



Hahaha - as if anyone here gives a monkeys. 😕


Well you may get one or two interested I guess.[/b]
Not from me. Never wanted to learn Russian. 😛

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Originally posted by XanthosNZ
... Currently I am in the process of trying a new attempt at disproving the conjecture ...
it does not surprise me that you took this approach 😉

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Originally posted by flexmore
it does not surprise me that you took this approach 😉
Who was that masked man anyway? Tonto? Look he even caries a badge on a stick! 😉

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I am willing answer any questions that don't directly discuss my approach to the Proof.

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Originally posted by XanthosNZ
I am willing answer any questions that don't directly discuss my approach to the Proof.
what does it mean?

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Originally posted by flexmore
what does it mean?
Get back to me after you've completed at least a BSc majoring in Math and I'll explain it.

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Originally posted by XanthosNZ
Get back to me after you've completed at least a BSc majoring in Math and I'll explain it.
ok .. i am back 😉

(ps you promised to answer questions by nonmathematicians.)

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Originally posted by XanthosNZ
...at least a BSc majoring in Math...
.....which, given your present progress may be "a bridge too far" for you, in my opinion.

skeeter

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... i had a racist teacher in primary school, will this help? hmmm... i think it was my spelling class.

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Originally posted by skeeter
.....which, given your present progress may be "a bridge too far" for you, in my opinion.

skeeter
Go read some poetry and leave the important stuff to the rest of us. Perhaps you could make some tea?

EDIT: Flex, I haven't the time nor the energy to devote to explaining complex and extremely abstract mathmatical concepts to anyone. The whole thing took me months to understand and even now I learn more everyday. May I suggest Five Golden Rules: Great Theories of the 20th Century - And Why They Matter as a good starting place.

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Originally posted by XanthosNZ
...your inclinations as to it's truth...
Is it truth?

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Originally posted by ivangrice
Is it truth?
That's the question.