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](https://drive.google.com/file/d/1vPz4llzvv9dup3RcJxvCCgrE-RLXwu4h/view?usp=sharing) with more details of the claims I have written above.



A geometric approach to the odd perfect number problem?
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:


$$|\hat{\phi}(n)| \ge \sqrt{\tau(n)(1+\frac{\tau(n)-1}{n^2})}\ge \sqrt{\tau(n)}$$


