A geometric approach to the odd perfect number problem? 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 MO:
https://mathoverflow.net/questions/373475/a-geometric-approach-to-the-odd-perfect-number-problem
Here are some notes with more details of the claims I have written above.

We have for all $n$:
$$|\hat{\phi}(n)|^2 = |\sum_{d|n} \phi(d)|^2 = \left < \sum_{d|n} \phi(d),\sum_{d|n} \phi(d)\right >$$
$$= \sum_{d|n} |\phi(d)|^2 + 2 \sum_{d_1 < d_2,d_1,d_2|n} \left < \phi(d_1),\phi(d_2)\right >$$
$$= \tau(n) + 2 \sum_{d_1 < d_2} \frac{\gcd(d_1,d_2)^2}{d_1 d_2}$$
$$\ge \tau(n) + 2\sum_{d_1 < d_2} \frac{1}{d_1 d_2}$$
$$\ge \tau(n) + 2\sum_{d_1 < d_2} \frac{1}{n^2}$$
$$= \tau(n) + \frac{2}{n^2} \frac{\tau(n)(\tau(n)-1)}{2}$$
$$= \tau(n) + \frac{\tau(n)(\tau(n)-1)}{n^2}$$
hence:
$$|\hat{\phi}(n)| \ge \sqrt{\tau(n)(1+\frac{\tau(n)-1}{n^2})}\ge \sqrt{\tau(n)}$$
