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This seems like a very standard notation in analytic number theory, and I see it a lot. But I was confused with it and I would greatly appreciate any clarification.

When one writes sum of the shape $$ \sum_{q \leq Q} \ \sum_{\chi (mod \ q)} ' $$ where $\sum_{\chi (mod \ q)} '$ is the sum over the primitive characters, I am wondering how the principal characters being taken into account here.

My questions regarding this: Is the convention to take the principal character as the primitive character mod 1 (so it appears only when $q=1$ but for no other $q$?), or are the principal characters mod q are primitive character mod $q$ so they actually appear for all $q$? or is the sum simply an empty set when $Q <2$?

Thank you very much.

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By definition, a Dirichlet character is primitive if it is not induced by a Dirichlet character of smaller modulus. In particular, the trivial Dirichlet character modulo $1$ (i.e. the constant function $\mathbb{Z}\to\{1\}$) is primitive.

The above definition is convenient in the sense that every Dirichlet character (including the principal Dirichlet characters) is induced from a unique primitive Dirichlet character. This fact generalizes to automorphic forms, e.g. every Hecke eigenform on the upper half-plane comes from a unique primitive Hecke eigenform, but only when the level $1$ forms are regarded primitive.

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  • $\begingroup$ Is this the definition used in zero density estimates like in link.springer.com/article/10.1007/BF01403187 ? I was also confused about this, because in Davenport (first page of chapter 5), he only defines primitive characters for nonprincipal characters. But I suppose he is ignoring the modulo 1 case? $\endgroup$
    – Johnny T.
    Commented Oct 2, 2020 at 17:29
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    $\begingroup$ @TakeshiGouda: I think that Gallagher regards the constant function $\mathbb{Z}\to\{1\}$ a primitive character. This is apparent from his display (5), where he imposes the lower bound $1<q$. Also, Gallagher's remark below (2) makes it clear that he includes $\chi=1$ in (2). $\endgroup$
    – GH from MO
    Commented Oct 2, 2020 at 17:57
  • $\begingroup$ Good point! Thank you very much! $\endgroup$
    – Johnny T.
    Commented Oct 3, 2020 at 6:58

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