# Sum of divisors and unitary divisors as the eigenvalue and the spectral norm of some addition matrix?

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:

$$(1,\cdots,1)$$

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?

Related questions:

https://math.stackexchange.com/questions/3800389/ring-of-divisors-of-a-natural-number-and-the-sum-of-divisors-as-an-eigenvalue-an

Boolean ring of unitary divisors / Structure of unitary divisors?

https://math.stackexchange.com/questions/3799607/does-this-characteristic-polynomial-factor-into-linear-factors-over-the-integers/3799759

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.

• Faleminderit Gjergj. – user6671 Aug 24 '20 at 3:57