Let $n$ be a natural number and $D_n$ be the set of divisors. We can make this set to a ring by observing that each divisor $d$ has

$$0 \le v_p(d) \le v_p(n)$$

Hence we can add two divisors $d,e$ by setting:

$$d \oplus e := \prod_{p | n} p^{v_p(d)+v_p(e) \mod (v_p(n)+1)}$$

and similarily we can multiply them by setting: $$d \otimes e := \prod_{p | n} p^{v_p(d) \cdot v_p(e) \mod (v_p(n)+1)}$$

Then, if $n = p_1^{a_1} \cdots p_r^{a_r}$, this ring will be isomorphic to the ring

$$\mathbb{Z}/(a_1+1) \times \cdots \times \mathbb{Z}/(a_r+1)$$

If $n$ is squarefree, than this reduces to :

$$d\oplus e = \frac{de}{\gcd(d,e)^2}$$

$$d\otimes e = \gcd(d,e)$$

and the ring of divisors is a boolean ring as defined here Boolean ring of unitary divisors / Structure of unitary divisors? and here https://math.stackexchange.com/questions/3799607/does-this-characteristic-polynomial-factor-into-linear-factors-over-the-integers/3799759

If we consider the addition table ($\oplus$) of this ring as a matrix, than it is clear that the sum of divisors $\sigma(n)$ is an eigenvalue to the eigenvector:


Here is as an example the addition ($\oplus$) table for $n=12$:

$$\left(\begin{array}{rrrrrr} 1 & 2 & 3 & 4 & 6 & 12 \\ 2 & 4 & 6 & 1 & 12 & 3 \\ 3 & 6 & 1 & 12 & 2 & 4 \\ 4 & 1 & 12 & 2 & 3 & 6 \\ 6 & 12 & 2 & 3 & 4 & 1 \\ 12 & 3 & 4 & 6 & 1 & 2 \end{array}\right) $$

I have checked numerically ($n=1,\cdots,60$) that

$$\sigma(n) = |A_n^k|_2^{1/k}, \forall k \ge 1$$

where $A_n$ is the addition matrix of this ring.

  1. Is there a proof for this last equality (where $|.|_2$ denotes the spectral norm)? (This question is proved here: https://math.stackexchange.com/questions/3800389/ring-of-divisors-of-a-natural-number-and-the-sum-of-divisors-as-an-eigenvalue-an )

Similarliy we can make the set $U_n$ of unitary divisors into a boolean ring by setting:

$$a\oplus b = \frac{ab}{\gcd(a,b)^2}$$

$$a \otimes b = \gcd(a,b)$$

I have checked numerically similarliy to the above ($\sigma^*(n) = $ sum of unitary divisors):

$$\sigma^*(n) = |B_n^k|_2^{1/k}, \forall k \ge 1$$

where $B_n$ is the addition matrix of $U_n$.

  1. Can this be proven?

To each eigenvalue $\lambda$ with eigenvector $v_{\lambda}$ of $B_n$ we can associate a "stabilizer group" $V_{\lambda} \le U_n$:

$$V_{\lambda} = \{u \in U_n| \left < (u\oplus u_1,\cdots,u \oplus u_r)^T ,v_{\lambda}\right >=\lambda \}$$

Then it seems that:

$$\lambda = \sum_{v \in V_{\lambda}} v - \sum_{u \in V_{\lambda}^C} u$$

  1. Is $V_n$ a subgroup of $U_n$?

  2. Is $\lambda$ equal to the right hand side of the last equality?

Thanks for your help!

Related questions:


Boolean ring of unitary divisors / Structure of unitary divisors?



In both cases you are really only using the additive structure of your rings, so this is really a question about abelian groups.

Assuming $n = p_1^{a_1} \cdots p_r^{a_r}$, when studying $A_n$ we are working with the abelian group $$G=\mathbb{Z}/(a_1+1)\mathbb Z \times \cdots \times \mathbb{Z}/(a_r+1)\mathbb Z.$$ We can think of the elements of $G$ as tuples $s=(s_1,\dots,s_r)$ where $s_i\in \mathbb Z/(a_i+1)\mathbb Z$. Here $A_n$ coincides with the group matrix $(x_{s+t})_{s,t\in G}$ where $x_{(s_1,s_2,\dots,s_r)}$ is set equal to the unique positive divisor of $n$ that satisfies $\nu_{p_i}(x)=s_i\pmod{a_i+1}$ for all $1\le i\le r$.

Now, $A_n$ is a symmetric matrix so we are really just trying to prove that the spectral radius is $\sigma(n)=\sum_{s\in G}x_s$. The good news is that we can say way more: we can write down all eigenvalues of this matrix.

Let's define vectors $\mathbf v(\chi)$ indexed by irreducible characters of $G$, to be given by $\mathbf v(\chi)_{s}=\chi(s)$. Then we can check that the following holds $$A_n\mathbf v(\chi)=\left(\sum_{s\in G}\chi(s)x_s\right)\mathbf v(\bar{\chi}).\tag{*}$$ If we let $\lambda_{\chi}=\sum_{s\in G}\chi(s)x_s$, then the eigenvalues of $A_n$ are either equal to $\lambda_{\chi}$ for some $\chi$ that is equal to it's own conjugate, or equal to $\pm \sqrt{\lambda_{\chi}\lambda_{\bar{\chi}}}$ for some $\chi$ that is not equal to it's own conjugate.

To prove this fact notice that $(*)$ tells us that when $\chi =\bar{\chi}$ we have $\mathbf v(\chi)$ as an eigenvector with eigenvalue $\lambda_{\chi}$, and when $\chi\neq \bar{\chi}$ we see that $A_n$ acts as $$\begin{pmatrix}0 & \lambda_{\chi}\\ \lambda_{\bar{\chi}} & 0\end{pmatrix}$$ on the span of $\{\mathbf v(\chi),\mathbf v(\bar{\chi})\}$.

Finally it remains to notice that since all $\chi(s)$ are roots of unity, the largest eigenvalue is $\sum_{s\in G}x_s$ corresponding to the trivial character.

For $B_n$ you can repeat a similar argument but for the group $$G=(\mathbb Z/2\mathbb Z)^r$$ with group matrix $(x_{s+t})_{s,t\in G}$ and $x_{(s_1,\dots,s_r)}$ chosen to be the unique positive divisor of $n$ that satisfies $$\nu_{p_i}(x)=\begin{cases} 0 & \text{if } s_i=0\\ a_i & \text{otherwise}, \end{cases}.$$ This also gives positive answers to your last two questions because every character of $G$ in this case is $\pm 1$ valued, and the elements where the value is $+1$ form a subgroup.

If you want to read more about such type of results you can look up K. Konrad's notes The origin of representation theory which focus on the related group matrix $(x_{s-t})_{s,t\in G}$, and how understanding its determinant/spectrum for (first abelian and later general) groups began the study of representation theory.

| cite | improve this answer | |
  • 1
    $\begingroup$ Faleminderit Gjergj. $\endgroup$ – user6671 Aug 24 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy