A geometric proof that there are infinitely many primes?

Question

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)

$$\det(G_n) = \prod_{i=1}^n \frac{h(a_i)}{a_i^2} $$ 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, 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. :-(

Answer 1

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.