Brett Nortj

Hodge Conjecture solutions.

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\operatorname {Hdg}^{k}(X)=H^{{2k}}(X,{\mathbf  {Q}})\cap H^{{k,k}}(X).

This statement is where we could use my own new type of maths I am trying to bring into secondary education through my company "Hydra." I have spent a lot of time planning new approached to maths, at secondary and tertiary levels, so, please bear with me and observe where the strengths are of my new system? This is just an example of how it can be applied, of course.

[9H] * [6K] * [2X] * [1Q] = [108A]. This is because the powers can be applied to the symbols and then taken as '[X1]'. This leads to a simple set of symbols to multiply.

Then, we take the [108A] divided into the symbols on the left that we recognize, being [H], [X] and [K], coming to [HXK] = [108].

This leaves the [DG] being equal to 'the left over bits.' This means we can say [HKX] = [108] and [DG] = [X], so, that means the sum on the left equals [HK2X] = [108].

Or, did I make a mistake somewhere... I am in a heck of a hurry!

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[X1]

perhaps leading to [X2], [X3], [X4] etc.

 

Is this any relation to the person with the Chinese avatar, who is spreading havoc in a Physics forum website about this nonsense?

 

Quote
I think I've let this go on long enough.

Thread closed.

-Dan

 

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1 hour ago, studiot said:

[X1]

perhaps leading to [X2], [X3], [X4] etc.

Is this any relation to the person with the Chinese avatar, who is spreading havoc in a Physics forum website about this nonsense?

I am not sure I know what you mean, but I like your take on this.

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Posted (edited)
On 8/3/2018 at 4:13 AM, Brett Nortj said:

I am not sure I know what you mean, but I like your take on this.

Without looking it up, can you state the Hodge conjecture and explain in your own words what it means?

Heck I'll make it easier. Go ahead and look it up and then explain to us in your own worlds what it means! 

Edited by wtf

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