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I am trying to solve exercises 17 and 18 on Marcus book (page 210, chapter 7). Let's look at the second. This exercise is essential in order to deduce the class number formula in Marcus approach.

18)Let $\chi$ be a primitive character mod $m\ge 3$ such that $\chi(-1)=-1$. Set: $u=\sum_{1\le k

a) Show that $u=2v-mv$

The claim to prove is $\sum_{1\le k

b) Suppose m is odd. Show that $u=4\chi(2)v-m\chi(2)w$ Hint: Replace odd values of $k$ by $m-k$, $k$ even.

I tried but I fail at some point. Thanks to part a), I get that the statement is equal to prove that $u=\chi(2)(u+2v)$. I obtain from one side $\sum_{1\le 1

c)Show that if $m$ is odd, then $u=mw/(\overline{\chi(2)}-2)$

d)Now suppose $m$ is even. Use exercise 17d at the link Two exercises on characters on Marcus (part 1) to prove that $u=-mw/2$.

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    Well, I really want to do it, so if you know how to do some points, you are welcome. – 2017-01-19
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    It is supposed to train for manipulating Dirichlet characters, but if you can't do it today it is not very important. Try the next chapter (and don't forget to look at [some examples](https://www.wolframalpha.com/input/?i=Table%5BDirichletCharacter%5B7,+2,+n%5D,+%7Bn,+10%7D%5D)). And writing "solved" for a) is a joke ! You are not helping us – 2017-01-19
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    I have to do it because this exercise is essential in order to deduce the class number formula in Marcus approach (he tells the reader to do it !). Also hints are welcome. Well I wrote "proved" just because it is quite easy from the definitions, but if someone is interested (maybe you?) I can give the details. – 2017-01-19
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    No, I mean I cannot solve your exercice without taking a paper and working 10min, because you didn't write what you found for a) – 2017-01-19

1 Answers 1

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Your calculation for a) is essentially correct, but it misses an argument why

$$\sum_{1 \leqslant t \leqslant m/2}\chi(t)(m-t) = \sum_{1 \leqslant t < m/2} \chi(t)(m-t).\tag{$\ast$}$$

The argument is simple of course: If $m$ is odd, then for an integer $t$ we have $t \leqslant m/2 \iff t < m/2$, and if $m$ is even, then $\gcd(m,m/2) > 1$, so $\chi(m/2) = 0$. Nevertheless, the argument must be made.

Also, you have mistyped what you want to prove. At the statement of a), it must be $u = 2v - mw$, not $u = 2v - mv$, and at the start of your calculation you wrote $\sum \chi(2k)$ instead of $\sum \chi(k)2k$.

Let's repeat the calculation laid out differently:

\begin{align} \sum_{1 \leqslant k < m} \chi(k)k &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 \leqslant k < m} \chi(k)k \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 < k < m} \chi(k)k \tag{$\ast$} \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant t < m/2} \chi(m-t)(m-t) \tag{$t = m-k$} \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant t < m/2} -\chi(t)(m-t) \tag{$\chi(m-t) = -\chi(t)$} \\ &= \sum_{1\leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant k < m/2} \chi(k)(k-m) \\ &= 2 \sum_{1 \leqslant k < m/2} \chi(k)k - m \sum_{1 \leqslant k < m/2} \chi(k). \end{align}

The calculation for part b) is similar. Following the hint (also strongly suggested by the appearance of $\chi(2)$ as a common factor), we start by splitting into sums for even $k$ and for odd $k$:

\begin{align} \sum_{1 \leqslant k < m} \chi(k)k &= \sum_{\substack{1\leqslant k < m \\ k \equiv 0 \pmod{2}}} \chi(k)k + \sum_{\substack{1\leqslant k < m \\ k \equiv 1 \pmod{2}}} \chi(k)k \\ &= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant s < m/2} \chi(2s-1)(2s-1) \\ &= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant r < m/2} \chi(m - 2r)(m-2r) \tag{$r = \frac{m+1}{2}-s$}\\ &= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant r < m/2} \chi(2r)(2r-m) \tag{$\chi(m-t) = -\chi(t)$} \\ &= 4 \sum_{1 \leqslant r < m/2} \chi(2r)r - m\sum_{1\leqslant r < m/2}\chi(2r) \\ &= 4\chi(2)\sum_{1 \leqslant r < m/2} \chi(r)r - m\chi(2)\sum_{1\leqslant r < m/2} \chi(r). \tag{$\chi(2r) = \chi(2)\chi(r)$} \end{align}

For part c), we use the results of a) and b) together with $\overline{\chi(2)}\chi(2) = 1$ for odd $m$:

\begin{align} \bigl(\overline{\chi(2)} - 2\bigr)u &= \overline{\chi(2)}u - 2u \\ &= \overline{\chi(2)}\bigl(\underbrace{4\chi(2)v-m\chi(2)w}_{b)}\bigr) - 2\bigl(\underbrace{2v-mw}_{a)}\bigr) \\ &= (4v-mw) - (4v-2mw) \\ &= mw, \end{align}

and dividing by $\overline{\chi(2)} - 2$ (which is nonzero) yields the result.

For part d), we again split the sum at $m/2$, and use $\chi(m/2+k) = -\chi(k)$ [17.d) gives that for odd $k$, and for even $k$ that follows from $\chi(k) = -\chi(m/2+k) = 0$ because $m/2$ is even by 17.a)]:

\begin{align} \sum_{1 \leqslant k < m} \chi(k)k &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 \leqslant r < m} \chi(r)r \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 < r < m} \chi(r)r \tag{$\ast$} \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant k < m/2} \chi(m/2 + k)\biggl(\frac{m}{2} + k\biggr) \\ &= \sum_{1 \leqslant k < m/2} \chi(k)k - \sum_{1 \leqslant k < m/2} \chi(k)\biggl(\frac{m}{2} + k\biggr) \tag{17.d)} \\ &= -\frac{m}{2}\sum_{1 \leqslant k < m/2} \chi(k). \end{align}

Note that since $\chi(2) = 0$ for even $m$, the results of c) and d) can both be written in the same way:

$$u = \frac{mw}{\overline{\chi(2)} - 2}.$$

The derivation however depends on the parity of $m$.

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    Oh thank you very much ! I was going to lose hope for this exercise. Anyway since you have been so precise with details (and I really appreciated it), maybe you should also say at point b), when you substitute $2s-1$ by $m-2r$, how the indexes in the summation change, because it is less obvious than point a) (I mean also $r$ is between 1-m/2). And to be super precise, at the end you use 17d which holds only for odd $k$, but in this case since $m$ is even, also $m/2$ is even and so $\chi(m/2+k)=\chi(k)=0$. Thank you again !! Just give me some time to clear my ideas, then I will mark the answer. – 2017-01-20
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    I've added these points. $2s-1 = m - 2r \iff r = \frac{m+1}{2} - s$. – 2017-01-20
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    What I meant is that if $s\ge 1$ then $r\le (m-1)/2$ which is equivalent to say that $r$r$ is an integer and $m$ is odd. Thank you again. – 2017-01-20
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    If $1 \leqslant s \leqslant \frac{m-1}{2}$, then $1 = \frac{m+1}{2} - \frac{m-1}{2} \leqslant \frac{m+1}{2} - s \leqslant \frac{m+1}{2} - 1 = \frac{m-1}{2}$. – 2017-01-20
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    To Daniel Fischer: please, see also this very related doubt: http://math.stackexchange.com/questions/2106173/dirichlet-series-in-terms-of-primitive-characters Thank you very much. – 2017-01-20