# Group isomorphism applied to sum of characters

Let $$G$$ be a finite abelian group and let $$\widehat{G}$$ be the group of group homomorphisms $$\chi:G\to\mathbb{C}^*$$ (i.e. the dual group of $$G$$, or the group of characters on $$G$$). Then there is an isomorphism $$\varphi:G\to \widehat{G}$$.

We have the following proposition: Given a fixed $$\chi\in\widehat{G}$$, we have $$\sum_{g\in G} \chi(g) = \begin{cases}|G| & \chi = 1_{\widehat{G}},\\0 & \text{otherwise.}\end{cases}$$ and given a fixed $$g\in G$$, we have $$\sum_{\chi\in \widehat{G}} \chi(g) = \begin{cases}|\widehat{G}| & g = 1_G,\\0 & \text{otherwise.}\end{cases}$$

I can prove both statements separately but all texts I have found about it state that due to the fact that $$G$$ and $$\widehat{G}$$ are isomorphic, then it is enough to prove just one of them and the other one is a consequence of the first. My question is: how is the isomorphism applied here?

• I think you need to understand the natural isomorphism between $\widehat{\widehat{G}}$ and $G$ to see the equivalence of these two statements. Commented Apr 13, 2023 at 19:25
• @DerekHolt care to elaborate? Isn't that isomorphism implied by that of $G$ and $\widehat{G}$ or is the fact that it's natural (not sure what that means) significant here? Commented Apr 13, 2023 at 19:37
• The isomorphism is defined by $g \leftrightarrow \hat{\hat{g}}$, where $\hat{\hat{g}}(\chi) = \chi(g)$. Commented Apr 13, 2023 at 19:41

To make your second set of equations look the first set, each $$g \in G$$ defines a character $${\rm ev}_g \colon \widehat{G} \to \mathbf C^\times$$ as "evaluate at $$g$$": $${\rm ev}_g(\chi) = \chi(g)$$ for all $$\chi$$ in $$\widehat{G}$$. That each $${\rm ev}_g$$ is a group homomorphism is due to the definition of multiplication in $$\widehat{G}$$ as pointwise products: $${\rm ev}_g(\chi\psi) = (\chi\psi)(g) = \chi(g)\psi(g) = {\rm ev}_g(\chi){\rm ev}_g(\psi).$$ Thus $${\rm ev}_g$$ is a character on $$\widehat{G}$$.
If $$g = 1$$, then $${\rm ev}_1$$ is the trivial character on $$\widehat{G}$$. Using the cyclic decomposition for finite abelian groups (or other methods), one can show for $$g \not= 1$$ in $$G$$ that there is $$\chi \in \widehat{G}$$ such that $$\chi(g) \not= 1$$. Writing that as $${\rm ev}_g(\chi) \not= 1$$, we see that if $$g \not= 1$$ in $$G$$, then $${\rm ev}_g$$ is a nontrivial character on $$\widehat{G}$$.
For each $$g \in G$$, let's evaluate $$\sum_{\chi \in \widehat{G}} \chi(g).$$ If $$g = 1$$ then each term in the sum is $$1$$, so the sum is $$|\widehat{G}|$$. We want to show if $$g \not= 1$$ that the sum is $$0$$. Rewrite the above sum as $$\sum_{\chi \in \widehat{G}} {\rm ev}_g(\chi),$$ which is the sum of a character $${\rm ev}_g$$ over the elements of a finite abelian group $$\widehat{G}$$. If $$g \not= 1$$ then $${\rm ev}_g$$ is a nontrivial character on $$\widehat{G}$$, so the above sum is $$0$$ by the general result that the sum of a nontrivial character over the elements of a finite abelian group is $$0$$. That is, in the equation $$\sum_{x \in A} \chi(x) = 0 \ {\rm if } \ \chi \not= {\mathbf 1}_A,$$ let $$A = \widehat{G}$$ and let the character on $$A$$ be $${\rm ev}_g$$ for $$g \in G$$ with $$g \not= 1$$. In this way, your second character sum formula can be regarded as a special case of your first character sum formula by using $$\widehat{G}$$ in place of $$G$$.
Here I am not explicitly relying on an isomorphism between $$G$$ and $$\widehat{G}$$, but there is a natural isomorphism $$G \to \widehat{\widehat{G}}$$ given by the mapping $$g \mapsto {\rm ev}_g$$, and this isomorphism is related to the argument above.