Let $e\_d$ be the $d$-th standard-basis vector in the Hilbert space $H=l\_2(\mathbb{N})$.
Let $h(n) = J\_2(n)$ be the second Jordan totient function, defined by:


$$J\_2(n) = n^2 \prod\_{p|n}(1-1/p^2)$$
Define:


$$\phi(n) = \frac{1}{n} \sum\_{d|n}\sqrt{h(d)} e\_d.$$


Then we have:


$$ \left < \phi(a),\phi(b) \right > = \frac{\gcd(a,b)^2}{ab}=:k(a,b)$$


The vectors $\phi(a\_i)$ are linearly independent for each finite set $a\_1,\cdots,a\_n$ of natural numbers, [since (page 1,2 in the notes)](https://drive.google.com/file/d/1vPz4llzvv9dup3RcJxvCCgrE-RLXwu4h/view)


$$\det(G\_n) = \prod\_{i=1}^n \frac{h(a\_i)}{a\_i^2} $$
is not zero, where $G\_n$ denotes the Gram matrix.


We want to look at $a\_1,\cdots,a\_n = 1,\cdots,n$ and get:


$$d(n):=\det(G\_n) = \prod\_{i=1}^n \frac{h(i)}{i^2} = \prod\_{k=1}^n \prod\_{p|k} (1-1/p^2) = \prod\_{p \le n} (1-1/p^2)^{\operatorname{floor}(n/p)}$$


Supposing now, that there exist only finitely many primes $p\_1,\cdots,p\_r$ we get for $d(n)$:


$$d(n) = \prod\_{i=1}^r (1-1/p\_i^2)^{\operatorname{floor}(n/p\_i)}$$


Consider now the number $N = p\_1 \cdots p\_r$ then we have:


$$\operatorname{floor}(N/p\_i) = \operatorname{floor}((N+1)/p\_i)$$


hence also:


$$d(N) = d(N+1)$$


But this [should be empirically impossible](https://sagecell.sagemath.org/?z=eJwtjkEKhTAMRPeCd8gy0aqty483EQWhrQj-tES9v62aWc6bzFjnwVpk-pUFpBN3XsIQJVgc0TSmi3NPVeX3EAS5iwQ-CETYMrT93YFcG5qoLF5ll7MrC68Ojeq1TsD3PkX4RFZPZa1brZ72BAzvDLoBhYQm7g==&lang=sage&interacts=eJyLjgUAARUAuQ==), if the function $d(n)$ can be shown to be monotonically decreasing. Notice also that the volume $\operatorname{vol}(n) = \sqrt{d(n)}$ is the volume spaned by the vectors $\phi(k), k=1,\cdots,n$. Hence maybe a geometric inequality could be applied in this setting?


**Question:
Is it possible to show that $d$ is monotonically decreasing in $n$ and thereby giving a geometric proof of the infinitude of primes?**


**Edit**: With the help of @Mark the proof is now complete! I still do not understand the downvote yet. :-(



A geometric proof that there are infinitely many primes?
You have claimed twice that


$$
d(n)= \prod\_{k = 1}^n \prod\_{p\mid k}(1-p^{-2})
$$


If this is true, your desired claim immediately holds, namely $d(n+1) <d(n)$.
This is because the quotient


$$
\frac{d(n+1)}{d(n)} = \frac{\prod\_{k = 1}^{n+1}\prod\_{p\mid k}(1-p^{-2})}{\prod\_{k = 1}^n\prod\_{p\mid k}(1-p^{-2})} = \prod\_{p\mid (n+1)}(1-p^{-2}).
$$
Note that for any non-trivial divisor $p\mid (n+1)$, we have that $p>1$, and therefore $0 < p^{-2} < 1$, and $0<1-p^{-2} < 1$.
It therefore follows that the product of these numbers in $(0,1)$ is in $(0,1)$, i.e.


$$\frac{d(n+1)}{d(n)} < 1\implies d(n+1) < d(n),$$


as desired.


