Download Pythagoreans quadruples on the future light cone

Pythagoreans quadruples on the future light cone
Arkadiusz Jadczyk and Jerzy Kocik
February 18, 2013
Draft, Preliminary version
1
The group SL(2, C)
The approach taken here is based on relativistic interpretation of our constructions (cf. also [1, 2]), although it is rather easy to extract pure number
theoretic issues.
The group SL(2, C) acts naturally on the two–sphere S 2 thought of as a projective future light cone in Minkowski space. This action can be also thought
of as coming from the natural action of SL(2, C) as the group of linear transformations on C2 - the complex space of spinors for the restricted Lorentz
↑
group SO+
, and the derived fractional linear transformations (Möbius transformations) on C :
az + b
a b
: z 7→
A=
.
(1)
c d
cz + d
These transformations should be undersood as acting on the one-point compactification C̃ = C ∪ ∞ of C - the Riemann sphere.
1.1
Classification of Möbius transformations
Möbius transformations are classified according to the value of the trace of
implementing matrices [4, 5]. A matrix A in SL(2, C) is called
i) loxodromic if tr(A) 6∈ [−2, 2],
ii) elliptic if tr(A) ∈ (−2, 2)
iii) parabolic if tr(A) = 2 or tr(A) = −2.
1
Loxodromic transformations with real trace are called hyperbolic. Hyperbolic transformations represent relativistic boosts. Elliptic transformations
represent three-dimensional rotations. Parabolic transformations are distinguished by the fact that they have just one fixed point on the Riemann
sphere. All other transformations (except of the identity) have exactly two
fixed points.
1.2
Special subgroup of parabolic transformations
Matrices
A(x, y) =
1 x + iy
, x, y ∈ R,
0
1
(2)
form a special abelian subgroup of parabolic transformations. We have
A(x, y)A(x0 , y 0 ) = A(x + x0 , y + y 0 ), A(x, y)−1 = A(−x, −y).
(3)
The action of A(x, y) on C is a pure complex translation:
A(x, y)z = z + x + iy.
(4)
All A(x, y) share the same unique fixed point in C - the infinity Each
matrix A(x, y) has the following explicit polar decomposition A(x, y) =
U (x, y)T (x, y), with U, T ∈ SL(2, C), where U is unitary and T is positive:
1
2
x + iy
U (x, y) = p
,
(5)
2
4 + x2 + y 2 −(x − iy)
T (x, y) = p
1.3
2
x + iy
.
x − iy 2 + x2 + y 2
1
4 + x2 + y 2
(6)
↑
SL(2, C) to SO+
double covering homomorphism
For Minkowski space-time indices we use coordinates xµ , (µ = 1..4), with
signature (−1, −1, −1, 1). The natural 2 : 1 homomorphism is explicitly
2
given by:
Λ11 = <(bc̄ + ad¯),
Λ1 = <(ac̄ + bd¯),
4
2
Λ3
Λ32
Λ34
Λ42
Λ44
= =(ac̄ − bd¯),
= =(cd¯ − ab̄),
Λ12 = =((bc̄ − ad¯), Λ13 = <(ac̄ − bd¯),
Λ21 = =(bc̄ + ad¯), Λ22 = <(ad¯ − bc̄),
Λ2 = =(ac̄ + bd¯), Λ3 = <(ab̄ − cd¯),
4
3
Λ3=
1
(7)
(8)
(9)
(aā − bb̄ − cc̄ + dd¯)/2,
= (aā + bb̄ − cc̄ − dd¯)/2), Λ41 = <(ab̄ + cd¯),
= −=(ab̄ + cd¯), Λ4 = (aā − bb̄ + cc̄ − dd¯)/2),
(10)
= (aā + bb̄ + cc̄ + dd¯)/2.
(13)
3
(11)
(12)
<(z) and =(z) denote respectively the real and the imaginary part of a
complex number z.
1.4
↑
Action of SO+
on S 2 .
Given a point n = (n1 , n2 , n3 ), (n1 )2 + (n2 )2 + (x3 )2 = 1 in S 2 ⊂ R3
one builds the four-vector x = (n1 , n2 , n3 , 1) on the future light cone in
Minkowski space. Acting with a Lorentz matrix Λ on x we get another
vector x0 = Λx, on the future light cone
0
2
2
4
4
(x 1 )2 + (x0 )2 + (x0 )2 = (x0 )2 , x0 > 0.
(14)
We then rescale x0 so as to obtain the fourth coordinate equal 1. Explicitly:
P3
i j
i
j=1 Λ j n + Λ 4
i
0i
(Λn) = n = P3
.
(15)
4 j
4
j=1 Λ j n + Λ 4
1.5
Lorentz subgroup
The image of the subgroup A(x, y) under the homomorphism (13) is given
by the following formula:


1 0
−x
x
0 1

−y
y
.
L(x, y) = 
(16)
1
2
2
x y 1 −x2 − y 2 + 2

2
2 x +y
1
1
2
2
2
2
x y
2 −x − y
2 x +y +2
↑
Notice that, being in SO+
, the matrices L(x, y) preserve the sign of the
fourth coordinate. They all share the unique fixed point on the sphere S 2 :
the North Pole (0, 0, 1).
3
Let us choose x = 2, y = 0, so that our matrix is in SL(2, R). The the
corresponding Lorentz matrix L(2, 0) has integer coefficients:


1 0 −2 2
0 1 0 0

(17)
L(2, 0) = 
2 0 −1 2 .
2 0 −2 3
Its Inverse is L(−2, 0) :

1
0
L(−2, 0) = 
−2
−2
1.6

0 2 −2
1 0
0
.
0 −1 2 
0 −2 3
(18)
The 6 generators
The pair of matrices L(2, 0), L(−2, 0) distinguishes a direction in space the North-South axis on the heavenly sphere. For this reason we introduce
the 3-element group of transpositions of the axes). These are all orthogonal
matrices that do not affect the fourth coordinate. Explicitly, let L[i), (i =
1..3) be the following set of Lorentz matrices in one row simplified matrix
notation:
L(1) = [1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1]
L(2) = [0, 1, 0, 0; 0, 0, 1, 0; 1, 0, 0, 0; 0, 0, 0, 1]
L(3) = [0, 0, 1, 0; 1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 0, 1]
Using these transformations we rotate (L(i)hL(i)−1 )the original h1 = L(2, 0), h2 =
L(−2, 0) matrices to obtain 6 Lorentz matrices h(i), with integer coefficients
h(1) = [−1, −2, 0, 2; 2, 1, 0, −2; 0, 0, 1, 0; −2, −2, 0, 3]
h(2) = [1, 0, −2, 2; 0, 1, 0, 0; 2, 0, −1, 2; 2, 0, −2, 3]
h(3) = [1, 0, 0, 0; 0, −1, −2, 2; 0, 2, 1, −2; 0, −2, −2, 3]
h(4) = [1, 0, 0, 0; 0, −1, 2, 2; 0, −2, 1, 2; 0, −2, 2, 3]
h(5) = [1, 0, 2, −2; 0, 1, 0, 0; −2, 0, −1, 2; −2, 0, −2, 3]
h(6) = [−1, 2, 0, 2; −2, 1, 0, 2; 0, 0, 1, 0; −2, 2, 0, 3]
4
Our matrices, if we consider only those acting on (1,3,4) coordinates have
a similar structure to Hall’s matrices [3], except for the fact that all our
matrices are of determinant +1. The set contains, with each matrix h(i), its
inverse matrix h(j), where we have ordered our six matrices in such a way
that h(j) = h(i)−1 for i + j = 7.
1.7
Generalized Pythagorean quadruples
Definition 1. A quadruple a, b, c, d of integers, not all zero, d >= 1, with
the property
a2 + b2 + c2 = d2
(19)
is called a generalized Pythagorean quadruple (GPQ). If a, b, c, d are coprime, the quadruple is called primitive.
We allow all integers here, including negative and two zero. Therefore
the standard Pythagorean quadruples and also triples (when one of the four
numbers is zero) belong to this set. We introduce the following generating
set of primitive degenerate quadruples:
q1 = (1, 0, 0, 1)
q2 = (0, 1, 0, 1)
q3 = (0, 0, 1, 1)
Theorem 1. The set of 6 matrices h generates, from the generating set
(q1 , q2 , q3 ), all primitive generalized Pythagorean quadruples.
Proof. We first show that acting with the matrices h(i) on a primitive
GPQ produces again a primitive GPQ. Due to the symmetry it is enough to
consider h(1). Other cases will differ only by permutation and sign changes
that bear no influence on the final result. We have
h(1) : (a, b, c, d) 7→ (−a − 2b + 2d, 2a + b − 2d, c, −2a − 2b + 3d).
(20)
Suppose the resulting GPQ has a common divisor k, that is: −a − 2b + 2d =
km, 2a + b − 2d = kn, c = kp, −2a − 2b + 3d = kq, all integers. These four
equations for a, b, c, d solve to a = −k(m − 2n − 2q), b = k(−2m + n + 2q),
c = kp, d = k(−2m + 2n + 3q). Thus k would be then a common divisor for
a, b, c, d.
Since our generating set consist of primitive GPQ’s, we can restrict ourselves
to primitive ones in the following. Let (a, b, c, d) be a primitive generalized
GPQ. If d = 1, then two of a, b, c must be zero, the third one must be ±1.
5
We have h(5)q1 = (−1, 0, 0, 1), h(1)q2 = (0, −1, 0, 1), h(3)q3 = (0, 0, −1, 1).
Therefore it will be enough to prove that starting with any GPQ with d > 1,
we can always reach one of the three elements of the generating set q1 , q2 , q3 .
To prove that it will be enough to show that unless two elements are zero
and the third is greater or equal to 1 (what implies d = 1), we can always
find a transformation in our set that decreases the value of d.
Suppose, to the contrary, that (a, b, c, d) is a primitive GP Q such that for
all h(i), the resulting quadruple (a0 , b0 , c0 , d0 ) has d0 ≥ d. For any Lorentz
transformation L we have
d0 = Λ41 a + Λ42 b + Λ43 c + Λ44 d.
(21)
Denote w(i)µ = h(i)4µ . Since h(i)44 = 3 for all i, the inequality d0 ≥ d
translates to
w(i)1 a + w(i)2 b + w(i)3 c + 2d ≥ 0.
(22)
√
For instance,
d > 0 and d = a2 + b2 + c2 ,
√ for i = 1, we get d ≥ a + b. Since
2
2
2
2
we have a + b + c ≥ a + b, therefore a + b2 + c2 ≥ (a + b)2 , or c2 ≥ 2ab.
On the other hand h(6) leads to d ≥ a − b, which, by the same reasoning,
implies c2 ≥ −2ab. In this way we obtain the following system of inequalities:
a2 ≥ 2bc,
b2 ≥ 2ac,
c2 ≥ 2ab,
and similarly with minuses on the RHS. Suppose first that (a, b, c) are all
positive. Then, by multiplying the LHS’s and RHS’s, we get a2 b2 c2 ≥
8a2 b2 c2 . With a, b, c being positive integers it is possible only if one of them
is zero. Suppose a = 0. Then, from a2 ≥ 2bc, we deduce that b or c must
be zero. Suppose b = 0. Then, from a2 + b2 + c2 = d2 , we get c = d. If
d > 1, then the quadruple (0, 0, d, d) is not a primitive one, contrary to our
assumption.
If some of the (a, b, c) are negative, we can choose from our set of inequalities
some with minuses on the RHS’s, and arrive to the same conclusion.
1.8
Results of Robert Spira
For the sake of convenience we remind here the results obtained by Robert
Spira in his 1962 paper “The Diophantine equation x2 + y 2 + z 2 = m2 [6].
However, we will change his original notation as follows: his (x, y, z, m) we
translate to ours (y, x, z, t), his (t, u, v, w) we translate into ours (q, n, m, p).
Withe these changes his theorems 1 and 2 read as follows:
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Theorem 2 (Spira). Let (x, y, z, t) be a primitive Pythagorean quadruple
with x, y, z, t > 0, x, y even, z odd. Then there exist integers m, n, p, q such
that
x = 2(nq + mp)
y = 2(np − mq),
z = m 2 + n2 − p 2 − q 2
2
2
2
t = m +n +p +q
(23)
2
The parameters n, m, p, q can be chosen to satisfy the following conditions
(where gcd denotes “greatest common divisor”)
i) n, p ≥ 1, m, q ≥ 0, m + q ≥ 1,
ii) mn > mq, m2 + n2 > p2 + q 2 ,
iii) m + n + p + q ≡ 1 ( mod 2),
iv) gcd(m2 + n2 , p2 + t2 , nq + mp) = 1,
v) If q = 0 then n ≤ m, if m = 0 then p ≤ q.
With the conditions above satisfied, the representation of (x, y, z, t) in terms
of (m, n, p, q) is unique.
Remark 1. The representation (23) remains valid if y = 0, in which case
we have a Pythagorean triple. One should put then n = 0, q = 0. See e.g [7,
p. 38].
1.9
Interpretation of Spira’s representation in terms of two–
component spinors
Let C2 be the two-dimensional complex vector space. Its elements are pairs
of complex numbers ab , sometimes called also “qubits”.
Given a vector
a
ā
v = ( b ) , its complex conjugate vector is v̄ = b̄ , while its hermitian
conjugate v † = ( ā, v̄ ) . We can take the product of v and v† to obtain
the hermitian 2 × 2 matrix denoted, using Dirac’s bra and ket notation as
|v >< v| :
a
aā ab̄
ā, b̄ =
|vi hv| =
.
(24)
b
bā bb̄
The matrix |vi hv| is automatically hermitian, and it projects onto the subspace spanned by v and is always singular. Its determinant is automatically
7
zero.
Hermitian 2 × 2 matrices have simple interpretation in terms of coordinates
(x, y, z, t) of Minkowski space given by the following well known one–to–one
correspondence:
1 t + z x + iy
(x, y, z, t) 7→
.
(25)
2 x − iy t − z
The factor 12 is convenient, so that the coordinate t comes as the trace of
the associated matrix. Its determinant is (t2 − x2 − y 2 − z 2 )/4. Therefore
the matrix is singular if and only if the point (x, y, z, t) is on the light cone.
Let us take a spinor v written explicitely in terms of real numbers m, n, p, q,
a = m + in, b = p + iq. Then the associated matrix |vi hv| reads as follows:
m2 + n2
mp + nq + i(np − mq)
.
(26)
|vi hv| =
mp + nq − i(np − mq)
p2 + q 2
In terms of associated Minkowski coordinates (x, y, z, t) we obtain:
x = 2(mp + nq),
y = 2(np − mq),
z = m 2 + n2 − p 2 − q 2 ,
t = m2 + n2 + p2 + q 2 ,
that is exactly the same formulas as in Spiras’s parametrization (23), provided m, n, p, q are integers.
1.10
Analysis of the six transformations h(i) in terms of integer qubit coordinates m, n, p, q.
Let (x, y, z, t) be a primitive Pythagorean quadruple with all x, y, z, t > 0,
with x, y even. We can use then the representation (23) with n, p ≥ 1,
m, q ≥ 0. Acting (x, y, z, t) with any of the six transformations h(i) we get
another triple (x0 , y 0 , z 0 , t0 ). Let us calculate the difference (t0 − t) for each of
the six transformations. Notice that t, t0 are always odd, therefore (t0 − t) is
always even - for this reason we express the results in terms of ∆t given by
∆t = (t0 − t)/2.
(27)
. The results are as follows: (this part is tottaly wrong, based on my error!!!)
1. For i = 1,
∆t = m2 +2n2 −2mp−2np+2p2 +2mq−2nq+2q 2 = 2((m−p)2 +(n−p)2 +(n−q)2 +q(2m+q)).
(28)
8
It follows that ∆t > 0. (∆t = 0 would imply q = 0, n = 0, p = 0, m = 0,
which is excluded.)
2. For i = 2,
∆t = n2 + 2mp + 3p2 + 2nq + 3q 2 .
(29)
It follows that ∆t is always strictly positive: ∆t > 0.
3. For i = 3,
∆t = n2 − 2np + 3p2 + 2mq + 3q 2 = (n − p)2 + 2p2 + q(2m + 3q). (30)
Therefore we always have ∆t > 0. |Deltat = 0 would imply q = 0, p =
0, n = 0, which is impossible with our assumptions.
4. For i = 4,
∆t = 2m2 +3n2 −2np+p2 +2mq+q 2 = 2m2 +2n2 +(n−p)2 +q(2m+q).
(31)
Again, we have that always ∆t > 0.
5. For i = 5,
δt = n2 − 2mp + 3p2 − 2nq + 3q 2 = (n − q)2 + 3p2 + 2q 2 − 2mp. (32)
Here ∆t can be positive (e.g. (m, n, p, q) = (255, 101, 150, 5), corresponding to (x, y, z, t) = (77510, 27750, 52701, 97751), negative (e.g.
(m, n, p, q) = (8, 3, 3, 1), corresponding to (x, y, z, t) = (54, 2, 63, 83),
or zero (e.g. (m, n, p, q) = (5, 2, 3, 1), corresponding to (x, y, z, t) =
(34, 2, 19, 39).
References
[1] To be added
[2] To be added
[3] A. Hall, Genealogy of Pythagorean triads, Mathematical Gazette, LIV,
No. 390 (1970), 377–379.
[4] Ford, Lester R. (1929), Automorphic Functions McGraw-Hill, Inc.
[5] Ahlfors, Lars V. (1979), Complex Analysis McGraw-Hill, Inc.
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[6] Spira, Robert (1962), The American Mathematical Monthly , Vol. 69,
No. 5 (1962),360-365
[7] Sierpiński, Waclaw (1988), Elementary Theory of Numbers , North–
Holland,PWN
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