Samstag, 2. März 2024
Application of Ehrhart polynomials for counting of numbers and lattices for natural numbers.
This is a transcription of the MO question written here:
It is known that log(p) is independent as a vector over the rational numbers where p is a prime.
We have log(ab)=log(a)+log(b) and log(n)=∑p|nvp(n)log(p).
We define ψ(n):=∑p|nvp(n)ep where ep is the p-th standard basis vector of the Hilbert space of sequences.
for q=a/b∈Q>0 we define:
ψ(a/b):=ψ(a/gcd(a,b))−ψ(b/gcd(a,b)).
We have for n∈N:
‖ψ(n)‖2=∑p|nvp(n)2 is a natural number
Furthermore the function
K(a,b)=⟨ψ(a),ψ(b)⟩=∑p|gcd(a,b)vp(a)vp(b)
is a positive definite kernel on the natural numbers.
Let us define the ∞-dimensional lattice Γ as:
Γ:={ψ(q)|q∈Q>0,‖ψ(q)‖2≡0mod(2)}
and let us also define:
η(n):=(−1)‖ψ(n)‖2=∏p|n(−1)vp(n)2
For ψ(a),ψ(b)∈Γ we have:
‖ψ(ab)‖2=‖ψ(a)+ψ(b)‖2=‖ψ(a)‖2+‖ψ(b)‖2+2K(a,b)≡0+0+0≡0mod(2)
Therefore ψ(a)+ψ(b)∈Γ and ψ(1)∈Γ.
The lattice shares some properties with the Leech-lattice and the E8 lattice:
1. It is unimodular in the sense that: For a finite set of primes, the gram matrix (K(p,q))p,q is prime has det=1.
2. It is even: ∀x∈Γ:|x|2≡0mod(2)
3. For all nonzero vectors in the lattice the squared norm is at least 2.
My questions are these:
1) Probably an easy question:
η(mn)=η(m)η(n)∀m,n∈N
2) Does it satisfy a Riemann-Hypothesis type of random variable equality:
∀ϵ>0:limN→∞1N1/2+ϵN∑n=1η(n)=0
3) Is there a relationship to the Riemann Hypothesis and this lattice?
Here is some Sagemath code for experiments.
And here is some picture concerning the 2. question:
**Edit**:
Let λ(n)=(−1)Ω(n) be the Liouville function.
Then because vp(n)2≡vp(n)mod(2), we have indeed:
η(n)=λ(n)
This answers the first question since it is known that λ is multiplicative.
This also answers the second and third question.
Modified question: Is this lattice known in literature?
Thanks for your help.
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A maybe useful point of view, is to use the Ehrhart polynomial of a polytope, namely the simplex, to count the numbers as points in a simplex:
Let QN:= Polytope of ({ψ(p)|1≤p≤N,p prime })
Then the polytope generated by the primes is the simplex, and it is known ( "Computing the continuous discretely" Matthias Beck, Sinai Robins, Second edition, p.31 ff) that the Ehrhart polynomial is given by:
L(QN,t)=binomial(d+t,d)
where d=π(N) is the dimension of the simplex, and it is equal to the number of primes ≤N, which is denoted by π(N). Let ˆπ(n):={p prime:p|n} and let pn:=n-th prime.
Since L(QN,t) counts the points ψ(n)=(x1,⋯,xd) in the dilated polytope tQN and those points have coordinates ≥0 with x1+⋯+xd≤t, we conclude that, by observing that the sum of the coordinates correspond to Ω(n):=∑p|nvp(n):
L(QN,t)=|{ψ(n):Ω(n)≤t,ˆπ(n)⊂ˆπ(N!),1≤n≤ptπ(N)}|
The condition ˆπ(n)⊂ˆπ(N!) is for making sure we look at those numbers which have only prime divisors ≤N.
We also observe that:
L(QN,t)−L(QN,t−1)=|{n:1≤n≤ptd,Ω(n)=t,ˆπ(n)⊂ˆπ(N!)}|=|AN,t|
where I have defined:
AN,t={n:1≤n≤ptd,Ω(n)=t,ˆπ(n)⊂ˆπ(N!)}
We can now define and try to evaluate the following sum:
F(N,t):=t∑k=0∑n∈AN,kλ(n)
=t∑k=0(−1)k(L(QN,k)−L(QN,k−1))
=t∑k=0(−1)k(binomial(d+k,d)−binomial(d+k−1,d))
which according to Wolfram Alpha is equal to:
=((−1)tbinomial(d+t+1,d)F2,1(1,d+t+2;t+2;−1)+2−d−1)−2−d−1(2d+1(−1)tbinomial(d+t,d)F2,1(1,d+t+1;t+1;−1))
=((−1)tbinomial(π(N)+t+1,π(N))F2,1(1,π(N)+t+2;t+2;−1)+2−π(N)−1)−2−π(N)−1(2π(N)+1(−1)tbinomial(π(N)+t,π(N))F2,1(1,π(N)+t+1;t+1;−1))
where F2,1(a,b;c;z) is the hypergeometric function.
I must admit that the formula is not really "nice" but I think it should be useful, because using the RHS of the last equality with d could maybe be extended to other values for d,t other than the natural numbers.
I can also not see, how to relate the sum F(N,t) to the Liouville sum:
N∑n=1λ(n)
The naive idea is that in the limit N→∞,t→∞, the two sets:
N:=limN→∞{1,⋯,N} and limN,t→∞AN,t
should be intuitively equal.
There is also an equivalent formulation of the Riemann Hypothesis with the points on a different polytope, but while it uses the Ehrhardt polynomials, which in this case are very complicated at t=1 I can not see how to derive an analytic formula in this case.
Here you can find some sanity check in SageMath.
Let BN,t={n:Ω(n)≤t,ˆπ(n)⊂ˆπ(N!),1≤n≤ptπ(N)}, so that
binomial(d+t,d)=L(QN,t)=|BN,t|
The arguments above show that:
∑n∈BN,tλ(n)=F(N,t)=
=((−1)tbinomial(d+t+1,d)F2,1(1,d+t+2;t+2;−1)+2−d−1)−2−d−1(2d+1(−1)tbinomial(d+t,d)F2,1(1,d+t+1;t+1;−1))
hence dividing by the number of points under which the sum runs, we get:
1|BN,t|∑n∈BN,tλ(n)=1|BN,t|F(N,t)=
=(−1)tdt+1F2,1(1,d+t+2;t+2;−1)+(−1)tF2,1(1,d+t+2;t+2;−1)+12d+1binomial(d+t,d)+(−1)t+1F2,1(1,d+t+1;t+1;−1)
where d:=π(N).
Notice the similarity to the prime number theorem:
We have in both cases sets CN such that:
limN→∞1|CN|∑n∈CNλ(n)=0
The case CN:={1,⋯,N} corresponds by Landau equivalently to the prime number theorem.
We observe that:
limN→∞BN,t={n∈N|Ω(n)≤t}=:B∞,t
and that
limt→∞(limN→∞BN,t)=limt→∞B∞,t=limt→∞{n∈N|Ω(n)≤t}=N
Similarly:
limN→∞(limt→∞BN,t)=limN→∞BN,∞=limN→∞{n∈N|ˆπ(n)⊂ˆπ(N!)}=N
Hence we have:
limt→∞,N→∞BN,t=N
(Here we have used that there are infinitely many primes.)
Now let us put CN:=BN,π(N).
Then one empirical observation, which I was not able to prove directly, is:
limN→∞1|CN|∑n∈CNλ(n)=?0
which reminds a little bit on the prime number theorem where CN:={1,⋯,N}.
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