Paul Bird's Scientific Research

Hyper-determinant With the Levi-Civita symbol we can define a determinant: ε^{a1,a2,a3,...} = 0 if any 2 of the indices the same. (1 otherwise) (Also, antisymmetric) Determinant of a 2-tensor is: det_2(A) = ε^{a}ε^{b}A^{a1,b1}A^{a2,b2}A^{a3,b3} .... We can define f^{a1,a2,a3,...} = 0 if any 3 of the indices the same. Determinant of a 3-tensor is: det_3(A) = f^{a}f^{b}f^{c}A^{a1,b1,c1}A^{a2,b2,c2}A^{a3,b3,c3} .... In the case where it is a 2x2x2 symmetric 3-tensor this should give the discriminant of a cubic. Let g_n{a1,a2,a3...} =0 if any no n of the indices are the same. Determinant of a n-tensor is: det_n(A) = g_n^{a}g_n^{b}g_n^{c}...A^{a1,b1,c1,...}A^{a2,b2,c2,...}A^{a3,b3,c3,...} .... Properties Assuming all components of A are independent (not true for symmetric tensors!). det_1( A ) = 0 by definition. d/dAin d/dAim det_2( A ) = 0 d/dAin d/dAim d/Aio det_3( A ) = 0 Inverses We can generalise the definition of an inverse for a 2-tensor: ...

Converting Inequalities to Equalities All strict and unstrict inequalities can be written as existence theorems on equalities: x!=0  <==>  ∃y{xy+1 =0} x>0   <==>  ∃y{xy 2 -1=0}    x<0   <==>  ∃y{xy 2 +1=0} x>=0  <==>  ∃y{x-y 2  =0}   x<=0  <==>  ∃ y{x+y 2  =0} Note that the first is the same as  ∃y{y=1/x}  which is true as long as we don't let x be infinite! Two equalities can be combined to a single equality like this: x=0 ∧ y=0  <==>  x 2 +y 2 = 0 A polynomial in a high degree can be written as a system in a lower degree: ax^6+bx^5+cx^4+dx^3+ex 2 +f = 0 <==>  y-x 2 =0 ∧ z-y 2 =0 ∧ azy+bzx+cz+dxy+ey+f = 0 <==> (y-x^2) 2 +(z-y^2) 2 +(azy+bzx+cz+dxy+ey+f) 2 = 0 Hence it turns out that any system of polynomials equalities and inequalities and can be written as a quartic in multiple variables.

Tarski-Seidenberg theorem States that a set of equalities and inequalities f(x1,x2,x3,....xn)>=0 can be reduced down to a set of equalities and inequalities f(x1,x2,x3,....xn-1)>=0 with one less variable. Thus all equations can be reduced to a set of equalities and inequalities on their coefficients. See also: cylindrical algebraic decomposition See:  Elimination These algorithms aren't very simple.

When does a quartic have a real root? This is a method for deciding if a polynomial has any real roots (or equivalently when it has no real roots). Start with a quartic polynomial. Ax 4 +4Bx 3 +6Cx 2 +4Dx+E = 0 or Q nmop X n X m X o X p = 0 The (det_4( Q )) discriminant is: also D=A^3.E-4.A^2.B.D-9.A^2.C^2+24.A.B^2.C-12.B^4 D' = (D + P^2) = d^2/dE^2  det_4(Q) also P = AC-B^2  = discriminant(  d^2/d x ^2 p( x ))   It has NO real roots when det_4( Q ) >0  and D >0 and P >0 or det_4( Q ) =0 and D =0 and P >0 Question How to get these polynomials as determinants of matrices? Or just find a simple algorithm. Guess : Are these the resultants of f (x) with it's derivatives? f '''''(x). How are D and P generalized? Extensions How to extend this to multi-variable polynomials? (This would solve kissing number problem). Obviously, we could have the coefficients depend on values of y; a(y), b(y), c(y), etc. Then we would ha...

Sphere Packing in Distance Geometry In distance geometry the only condition we have is that any (n+1)-simplex made from (n+2) points is zero in n dimensions. Let R_nm be the squared distance between point n and point m. The kissing number problem is then stated as: R 0n = 1 R nm > 1 Then we ask is there a solution in terms of R such that the above holds and also the Cayley-Menger determinants are all zero? In 2 dimensions with 7 circles this gives us R 0n = 1 R nm > 1 VolumeOfTetrahedron ( R_ab,R_ac,R_ad,R_bc,R_bd,R_cd) = 0 Which gives cubic equalities for the R. We could set: R nm = 1+(S nm )^2 This gives us a set of 6th degree polynomials in S. Then we should be able to use the resultant to see if they all have a simultaneous solution! But the resultant does not distinguish between real and imaginary solutions. :( e.g. ( x ^2+1)( x -3)=0 ( x ^2+1)( x -4)=0 Share an imaginary solution! (Only possible if share common factor?) What we need...

General Systems Polynomial Equations We want to find a solution to the general problem of systems of polynomial equations. (There is a simple algorithm to solve these problems but maybe a neater way can be found?) Firstly all systems of polynomial equations can be written in the following form: ∪ i=1..n {X T A i X = 0} with X a vector, A a matrix with X 0 =1. The solutions should all be written in terms of determinants { A i +..} . Or using the antisymmetric tensor. det(Q) = {Q} One equation - n variables For one equation a solution exists if:  {Q}  > 0 This can be written in terms of traces. One variable: 2 {Q}  = [Q] 2 -[Q 2 ] Two variables: 6{ Q}  = [Q] 3 -3[Q][Q 2 ]+2[Q 3 ] etc. The general formula from traces comes from it's expression as a determinant. Q Q Q... ε ε Two Equations Two equations - One(n?) Variable(s) A solution exists (common root) if (each has a solution as above): ([Q] 2 -[Q 2 ]) ([R] 2 -[R 2 ])  - ([Q][R]...

  Sphere Packing   The sphere packing problem or kissing number problem can be stated as a set of quadratic inequalities: {A i nm x n x m  + C i   ≥ 0 } Can we find a general algorithm to solve these equations? (Don't see why not!) (Actually need for pair of equations in N variables.) One Equation - One variable  ✓ (1) A x 2 +C ≥ 0 has  no  solution if: C < 0 & A <= 0 Two Equation - One variable  ✓ (1) A x 2 +c ≥ 0 (2) D x 2 +f ≥ 0 has no solution if: A<0 & c>0 & D>0 & f<0 &  (D c -A f )>0 or vice versa  N Equations - One variable (SOLVED!)  ✓ (1) A x 2 +C ≥ 0 (2) D x 2 +F ≥ 0 (3) G x 2 +I ≥ 0 has NO solution if (and only if) one pair does not have a solution or any single equation has no solution. ( If all pairs have solution then all MUST have a solution ). <-- This only works in one variable!  One Equation - M variables  ✓ (1)  x T A...

Solving two polynomials in two variables I had an algorithm for solving two polynomials in 2 variables by converting it into an equation in 1 variable. I'll see if I can remember it. Let us start with 2 polynomials. (solution: x=2 y=3). We try to eliminate x. (1) x 2 +4xy+2y 2 +3x+4y - 64 = 0 (2) 3x 2  -xy + y 2 +5x-y -22 = 0 First eliminate x 2 (3) =- (2) +3 (1) :  13xy +5y 2  +4x +13y  -170 = 0 or (13y+4)x + (5y 2 +13y  -170) = 0 Next use this equation to eliminate x 2  again from (1) . (1) .(13y+4) :  (13y+4)x 2 + (52y 2 +55y+12)x + (13y+4)(2y 2 +4y - 64) = 0 - (3) .x:             (13y+4)x 2  + (5y 2 +13y  -170)x = 0 =(4) : (47y 2 +42y+182)x + (13y+4)(2y 2 +4y - 64) = 0 Now use these two linear equations to eliminate x: (13y+4) (4) -  (47y 2 +42y+182) (3)  :   (13y+4)(13y+4)(2y 2 +4y - 64) - (47y 2 +42y+182)(5y 2 +13y  -170) =0 or  (y-3) ...

Existence of Solutions to Systems of Polynomial Inequalities I was wondering whether there is a general method to deciding if a system of polynomial inequalities has a solution. If there was such an algorithm this could be used to solve questions such as the sphere packing question. (The kissing number). Simply find all the inequalities for each sphere and determine if a solution can be found. To skip to the end it turns out that all systems of polynomial inequalities can be put in the following form: {A nm x n x m +B n x n +C  ≥ 0} or using the notation X ={x1,x2,x3,...,1} {Q nm X n X m   ≥ 0} Thus all conditions are built up from matrices, A , vectors B and constants C. We must deal with all the special cases where various components are zero. For the sphere packing problem we can set all the { B =0} which makes it much simpler. Simplest Case The simplest case is a single linear equation A x +B ≥ 0 This has no solution when A=0 & B<0 Star...

N=8 Supergravity coupled to E 8 xE 8 N=4 Super Yang-Mills unified in 19 Dimensions? Pure N=8 supergravity can be derived from a compactification of an 11 dimensional action. The massless(?) sector of Heterotic String Theory contains 256 + 248x2x16 = 2 13 particles. This suggests that it could be the compactification of a higher dimension action in 19 dimensions . It should be related to a 16 dimensional lattice. Only up to 11D has group structure . fermion = (D-3)x2 (D-3)/2 graviton = (D-2)(D-1)/2-1 antisymmetric n-tensor (D-2)(D-3)(D-4)..(D-n-1)/n! 11D: graviton ( 44 ) + antisymmetric-3-tensor( 84 ) + fermion ( 128 )  = 2 8 components. (Root vectors of E8 240 roots, 16 less) 19D: graviton ( 152 ) +  232  x vector( 17 )+ fermion ( 4096 ) = 2 13 components. (136, 680, 2380, 6188) (Root vectors of 16D lattice...? E 8 xE 8 ? D 16 ? (too small) Coexter? ~ 8160 roots? 32 less? 4320 densest in 16 dimensions?) 27D:  196608 components. ...