- \(s(x,y) \ge 0 \forall x,y \in X\)
- \(s(x,y) = s(y,x) \forall x,y \in X\)
- \(s(x,y) \le s(x,x) \forall x,x \in X\)
- \(s(x,y) = s(x,x) \iff x=y\)
Edit: I think that \(4.\) can be proved by considering that if \(K(x,y)=1\) and assuming there exists a prime \(p\) which divides \(x\) but not \(y\), leads to a contradiction.
Second edit: The abc conjecture (in the version above) would follow, if one could prove that the function \[K(a,b) := \frac{2(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \] is positive definite over the natural numbers.
Proof: If \(K\) is a positive definite function over the natural numbers, then by the Moore-Aronszajn theorem , there exists a Hilbert space \(H\) and a feature mapping \(\phi: \mathbb{N} \rightarrow H\) such that: \[K(a,b) = \left < \phi(a),\phi(b) \right >_{H}\] But by Cauchy-Schwarz inequality we get: \[K(a,b) = |K(a,b)| = |\left < \phi(a), \phi(b) \right >_{H}|\] \[\le |\phi(a)||\phi(b)| = \sqrt{K(a,a)K(b,b)} = \sqrt{1 \cdot 1} = 1\] But then we get by division of \(2\) and application of the last inequality: \[\frac{(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \le \frac{1}{2} < 1 \] hence it would follow that: \[\forall a,b \in \mathbb{N}: \frac{a+b}{\gcd(a,b)} < \operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2 \] Modified question: Is there a way to speed up the computations to try and test if the Gram matrix for \(n=1,\cdots,10000 \) is positive definite? Here is my code so far, but it is rather slow:
# inductive cholesky decomposition
def rad(n):
return prod(prime_divisors(n))
def kk(a,b):
return (2*(a+b)*a**1*b**1/gcd(a,b))*1/rad(a*b*(a+b)/gcd(a,b)**3)**2
def notPositiveDefinite(a,b):
return rad(a*b/gcd(a,b))
#kk = notPositiveDefinite
def grammat(kk,n):
return matrix([[kk(a,b) for a in range(1,n+1)] for b in range(1,n+1)],ring=QQ)
PHI = dict([])
def phi(n,useN=False):
if n in PHI.keys():
return PHI[n]
if n==1:
if not useN:
x = vector([sqrt(kk(1,1))])
else:
x = vector([sqrt(kk(1,1)+0.0)])
PHI[n] = x
return x
else:
w = CC(n-1)**(-1)*vv(n-1)
lw = list(w)
if not useN:
lw.append(sqrt(kk(n,n)-w.norm()**2))
else:
lw.append(sqrt(kk(n,n)-w.norm().n()**2).n())
x = vector(lw)
PHI[n]= x
return x
def vv(n):
return vector([kk(i,n+1) for i in range(1,n+1)])
CCDict = dict([])
def CC(n):
if n in CCDict.keys():
return matrix(CCDict[n])
mm = []
for i in range(1,n+1):
vi = list(phi(i))
if len(vi) < n:
vi.extend((n-i)*[0])
mm.append(vi)
M= matrix(mm)
CCDict[n] = mm
return M
for n in range(1,200):
print(n,all(x in RR for x in phi(n,useN=True)))
#print(grammat(kk,n).rational_form())
```
Comment on code: It is based on inductive Cholesky decomposition as described below and keeps track in a dictionary if the \(\phi(n)\) or \(C_n\) have already been computed. It uses float approximation for greater speed and in the last step we check if each component of the vector \(\phi(n)\) is a real number. If it is for all \(1 \le k \le n\) then we conclude that the Gram matrix of \(n\) is positive definite. If it is not, then at least the Gram matrix for \(n\) is not positive definite:
Inductive Cholesky decomposition to find an embedding \(\phi(n)\) for each number \(n\):
Equation 1): \[\phi(n+1) = \left(\begin{array}{r} {C_{n}^{-1} v_n} \\ \sqrt{k(n+1,n+1)-{\left| {C_{n}^{-1} v_n} \right|}^{2} } \end{array}\right)\] Equation 2): \[v_n = \left(\begin{array}{r} k(1,n+1) \\ k(2,n+1) \\ \cdots\\ k(n,n+1)\\ \end{array}\right)\] Equation 3): \[C_n = \left(\begin{array}{r} \phi(1)^T \\ \phi(2)^T \\ \cdots\\ \phi(n)^T\\ \end{array}\right)\] Equation 4): \[\phi(1) := \sqrt{k(1,1)} e_1\] If we assume that the function \(k\) is positive definite, then this inductive method, should give us the vectors \(\phi(a),\phi(b)\) such that under the normal inner product we have: \[k(a,b) = \left < \phi(a), \phi(b) \right > \]. Edit (11.03.2024): This conjecture about the positive semidefiniteness of \(K\) is wrong as the determinant of \(G_{192}\) is < \(0\)
