# Sum of certain integer numbers related to the elements of a finite abelian group and to its group of characters

We consider a finite abelian group $$G$$ and its group characters $$G^*$$. For each $$g\in G$$ and $$\chi\in G^*$$ we define $$0\leq r_g^\chi< o(g)-1$$ such that $$\chi(g):=e^{\frac{2\pi i}{o(g)}r_g^\chi}$$.

We fix two elements $$g,h\in G$$. I want to find a closed formula for the following sum:

$$\sum_{\chi \in G^*}r_g^\chi r_h^\chi$$

I found already the closed form for the following sum:

$$\sum_{\chi \in G^*}r_g^\chi=\frac{|G|}{2}(o(g)-1)$$

but I don't know how to get the other closed form.

A little remark can be the following

$$\frac{|G|}{2}(o(g)-1)\frac{|G|}{2}(o(h)-1)=(\sum_{\chi\in G^*}r_g^\chi )(\sum_{\eta\in G^* }r_h^\eta)=\sum_{\chi,\eta \in G^*}r_g^\chi r_h^\eta= (\sum_{\chi \in G^*}r_g^\chi r_h^\chi)+(\sum_{\chi\neq \eta \in G^*}r_g^\chi r_h^\eta)$$

and so

$$\sum_{\chi \in G^*}r_g^\chi r_h^\chi= \frac{|G|}{2}(o(g)-1)\frac{|G|}{2}(o(h)-1)- \sum_{\chi\neq \eta \in G^*}r_g^\chi r_h^\eta$$

Here is a partial answer. Let $$f(g,h)=\sum_{\mathbb\chi\in G^*}r_g^\chi r_h^\chi$$ be the required sum for $$g,h\in G.$$ I can express this in a similar form to your expression for $$\sum r^\chi_g,$$ but much more complicated, see equation (1). I'm not sure if there is a nice'' expression.

Let $$g$$ have order $$m$$ and $$h$$ have order $$n$$ and let $$|K|=d$$ where $$K=\langle g\rangle\cap\langle h\rangle.$$ Let $$H=\langle g,h\rangle,$$ so $$|H|=mn/d.$$ Since $$g^{m/d}$$ and $$h^{n/d}$$ both generate $$K,$$ there is unique $$u\in(\mathbb Z/d\mathbb Z)^\times$$ such that $$f^{n/d}=(g^{m/d})^u.$$

Let $$\zeta_m=e^{2\pi i/m}$$ and $$\zeta_n=e^{2\pi i/n}.$$ Then $$H^*=\{\lambda_{a,b}:au=b\text{ mod }d\},$$ where $$\lambda_{a,b}:H\mapsto\mathbb C^\times$$ is defined by $$\lambda_{a,b}(g)=\zeta_m^a$$ and $$\lambda_{a,b}(h)=\zeta_n^b.$$

To see this note every $$\lambda\in H^*$$ i.e. every homomorphism $$H\rightarrow\mathbb C$$ is of the form $$\lambda_{a,b}$$ for some $$a,b,$$ because $$g$$ and $$h$$ must be mapped to powers of $$\zeta_m$$ and $$\zeta_n$$ and the congruence condition ensures $$f^{-n/d}(g^{m/d})^u$$ maps to $$1.$$ Also $$a$$ and $$b$$ are unique mod $$m$$ and $$n$$ respectively. But the allowed $$a,b$$ are just those in the kernel of the map $$\mathbb Z/m\mathbb Z\oplus \mathbb Z/n\mathbb Z\rightarrow\mathbb Z/d\mathbb Z,$$ $$(a,b)\mapsto au-b,$$ so there are $$mn/d=|H|$$ such $$(a,b),$$ as required.

As $$r^{\lambda_{a,b}}_g=a$$ and $$r^{\lambda_{a,b}}_h=b,$$ and each $$\lambda_{a,b}$$ has $$|G:H|$$ extensions to $$G,$$ we get $$$$\tag{1} f(g,h)=|G:H|\sum_{a=0}^{m-1}\sum_{b=0,b=au\text{ mod }d}^{n-1}ab.$$$$

I will sketch some further evaluation of the RHS but it seems to be messy. Write $$m=dm_1,n=dn_1,$$ and $$a=rd+a_1,$$ $$b=sd+b_1$$ where $$0\le r $$0\le s $$0\le a_1,b_1 Then $$|G:H|^{-1}f(g,h)=\sum_{r=0}^{m_1-1}\sum_{s=0}^{n_1-1}\sum_{a_1=0}^{d-1}(rd+a_1)(sd+\overline{a_1u})$$ where $$\overline{a_1u}$$ means the remainder of $$a_1u$$ mod $$d.$$ The RHS expands to $$d^3\sum_{0}^{m_1-1} r\sum_0^{n_1-1}s+m_1d\sum_0^{d-1}a_1\sum_0^{n_1-1}s+n_1d\sum_0^{m_1-1}r\sum_{a_1=0}^{d-1}\overline{a_1u}+m_1n_1\sum_{a_1=0}^{d-1} a_1\overline{a_1u}.$$ The first three summands are straightforward (note the second factor in the third summand is just $$d(d-1)/2$$ because $$u$$ is prime to $$d$$) and I won't write them down. For the last term I find (verified by excel): $$\sum_{a_1=0}^{d-1} a_1\overline{a_1u}=d\left(\frac 12\sum_{j=1}^u\left(\left[\frac{jd}u\right]\left(\left[\frac{jd}u\right]-1\right)\right)-\frac{d-1}6(u(d-2)+3)\right).$$

In principle then we have an expression for the RHS of (1) which is polynomial in $$m,n,d,u$$ apart from the sum over $$j$$ above - I'm not sure if this can be improved or interpreted in a nice way.