Friday, May 11, 2007
Algebraic residues and tautological spaces
One of my most powerful simple ideas was to use identities against a hard math problem, as it is such a simple idea—once you consider it—to use identities and subtracting out an equation to be analyzed so that you use the algebraic residue.
For instance, consider the maybe seemingly trivial identity:
x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2
where since residues are what's important you can move to
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
and now manipulate the equations in various ways, where because it is an identity, x, y, z, and v can be ANY numbers that you choose. So like you can say the ring is algebraic integers, and any algebraic integer will be ok, for x, y, z or v, because you're just manipulating an identity.
I call
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
a tautological space, as it is a space in that you have variables, and in this case 4 variables so I call it a 4-dimensional space, and it's tautological in that it's always true, as you're just using an identity, where I figured out a way to use tautologies in mathematics to do detailed analysis.
For instance, consider x^3 + y^3 = z^3, and you can subtract that out of your tautological space, do some algebra and find that
(v^3+1)z^6 - 3x^2y^2(vz^2) - 2x^3y^3 = (a_1 z^2 + b_1 xy)(a_2 z^2 + b_2 xy)(a_3 z^2 + b_3 xy)
where the a's are roots of
a^3 -3va^2 + v^3+1 = 0
and for the b's you have b_1*b_2*b_3 = -2.
Lot of complexity there that just exploded out at you, right? But what I did was just go from
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
to
x^2 + y^2 = -vz^2 (mod x^2 + y^2 + vz^2)
and cubed, and I took x^3 + y^3 = z^3 and squared and subtracted, to get the algebraic residue
(v^3+1)z^6 - 3x^2y^2(vz^2) - 2x^3y^3 = 0 (mod x^2 + y^2 + vz^2)
so it's not really so complicated but it is powerful, as remember, I subtracted from an identity, so in analyzing the residue I'm actually analyzing x^3 + y^3 = z^3.
But crucially v is a free variable, so I can make it whatever I wish, which is the handle given to me by this analysis technique. Now you can proceed to prove that x, y and z cannot be integers in a straightforward way that is easily extended to p odd prime.
And that is a quick introduction to tautological spaces and algebraic residues as I extended concepts began by Gauss.
Algebraic residues are the natural next step from his research where it just took a hundred years or so for the progression but that is the way mathematics actually works, as new techniques cannot be discovered until humanity is ready for them.
For instance, consider the maybe seemingly trivial identity:
x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2
where since residues are what's important you can move to
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
and now manipulate the equations in various ways, where because it is an identity, x, y, z, and v can be ANY numbers that you choose. So like you can say the ring is algebraic integers, and any algebraic integer will be ok, for x, y, z or v, because you're just manipulating an identity.
I call
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
a tautological space, as it is a space in that you have variables, and in this case 4 variables so I call it a 4-dimensional space, and it's tautological in that it's always true, as you're just using an identity, where I figured out a way to use tautologies in mathematics to do detailed analysis.
For instance, consider x^3 + y^3 = z^3, and you can subtract that out of your tautological space, do some algebra and find that
(v^3+1)z^6 - 3x^2y^2(vz^2) - 2x^3y^3 = (a_1 z^2 + b_1 xy)(a_2 z^2 + b_2 xy)(a_3 z^2 + b_3 xy)
where the a's are roots of
a^3 -3va^2 + v^3+1 = 0
and for the b's you have b_1*b_2*b_3 = -2.
Lot of complexity there that just exploded out at you, right? But what I did was just go from
x^2 + y^2 + vz^2 = 0 (mod x^2 + y^2 + vz^2)
to
x^2 + y^2 = -vz^2 (mod x^2 + y^2 + vz^2)
and cubed, and I took x^3 + y^3 = z^3 and squared and subtracted, to get the algebraic residue
(v^3+1)z^6 - 3x^2y^2(vz^2) - 2x^3y^3 = 0 (mod x^2 + y^2 + vz^2)
so it's not really so complicated but it is powerful, as remember, I subtracted from an identity, so in analyzing the residue I'm actually analyzing x^3 + y^3 = z^3.
But crucially v is a free variable, so I can make it whatever I wish, which is the handle given to me by this analysis technique. Now you can proceed to prove that x, y and z cannot be integers in a straightforward way that is easily extended to p odd prime.
And that is a quick introduction to tautological spaces and algebraic residues as I extended concepts began by Gauss.
Algebraic residues are the natural next step from his research where it just took a hundred years or so for the progression but that is the way mathematics actually works, as new techniques cannot be discovered until humanity is ready for them.