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. 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

$$\det(G_n) = \prod_{i=1}^n \frac{h(a_i)}{a_i^2} $$ is not zero, where $G_n$ denotes the Gram matrix.

Define:

$$\hat{\phi}(n) := \sum_{d|n} \phi(d) = \frac{1}{n} \sum_{d|n} \sigma(\frac{n}{d})\sqrt{h(d)} e_d$$

Then we have:

$n$ is an odd perfect numbers, if and only if:

$$\left < \hat{\phi}(n),\phi(2) \right > = 1$$

By the triangle inequality we have:

$$|\hat{\phi}(n)| \le \tau(n)$$

where $\tau$ counts the number of divisors of $n$.

Geometric intuition: Since the vectors $\phi(d), d|n$ are almost orthogonal and have norm $1$, we should have by Pythagoras:

$$|\hat{\phi}(n)|^2 \approx \sum_{d|n} |\phi(d)|^2 = \tau(n)$$

**A more concrete claim, which I have not been able to prove yet is:
$$|\hat{\phi}(n)|^2 \ge \tau(n)$$
for all $n$?**

Let $\alpha$ be the angle between $\phi(2)$ and $\hat{\phi}(n)$, where $n$ is an OPN. Then, by Jordans inequality for the $\sin$-e we get after some algebraic manipulation (and using the last claim), the following upper and lower bound for $\tau(n)$ for the OPN $n$:

$$\frac{1}{\sqrt{1-\frac{4\alpha^2}{\pi^2}}} \le \tau(n) \le \frac{1}{1-\alpha^2}$$

However it seems that numerical experiments suggest, that the last inequality can hold only for $n=1$ or $n=$ a prime, which would contradict the OPN property.

My question is, if one can prove the **claim**.

Also asked on MSE, since it may not be research level: https://math.stackexchange.com/questions/3854989/a-geometric-approach-to-the-odd-perfect-number-problem

Here are some notes with more details of the claims I have written above.