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: $u=\sum_{1\le k<m}\chi(k)k, v=\sum_{1\le k<m/2}\chi(k)k, w=\sum_{1\le k<m/2}\chi(k)$

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

The claim to prove is $\sum_{1\le k<m}\chi(k)k=\sum_{1\le k<m/2}\chi(2k)-\sum_{1\le k<m/2}\chi(k)m$. Now one side gives me $\sum_{1\le k<m/2}\chi(k)k+\sum_{m/2\le k<m}\chi(k)k$. The other side gives me $\sum_{1\le k<m/2}\chi(k)(2k-m)=-\sum_{1\le k<m/2}\chi(k)(m-k)+\sum_{1\le k<m/2}\chi(k)k$. So it remains to prove that $\sum_{m/2\le k<m}\chi(k)k=-\sum_{1\le k<m/2}\chi(k)(m-k)$. But this is true because setting $k=m-t$ you obtain $\sum_{m/2\le k<m}\chi(k)k=\sum_{1\le t\le m/2}\chi(m-t)(m-t)=-\sum_{1\le t\le m/2}\chi(t)(m-t)$.

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 $\sum_{1\le 1<m, k even}\chi(k)k-\sum_{1\le k<m,k even}\chi(k)(m-k)$. From the other side $\sum_{1\le k<m}\chi(2k)k+\sum_{1\le k<m/2}\chi(2k)2k$. So it remains to prove that $\sum_{1\le k<m}\chi(2k)k=\sum_{1\le k<m, k even}\chi(k)(k-m)$ which seems false to me. Any help??

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 $u=-mw/2$.

  • $\begingroup$ Well, I really want to do it, so if you know how to do some points, you are welcome. $\endgroup$ – Richard Jan 19 '17 at 21:38
  • $\begingroup$ 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). And writing "solved" for a) is a joke ! You are not helping us $\endgroup$ – reuns Jan 19 '17 at 21:40
  • $\begingroup$ 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. $\endgroup$ – Richard Jan 19 '17 at 21:44
  • $\begingroup$ 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) $\endgroup$ – reuns Jan 19 '17 at 22:33

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 $m$.

  • $\begingroup$ 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. $\endgroup$ – Richard Jan 20 '17 at 14:35
  • $\begingroup$ I've added these points. $2s-1 = m - 2r \iff r = \frac{m+1}{2} - s$. $\endgroup$ – Daniel Fischer Jan 20 '17 at 14:45
  • $\begingroup$ What I meant is that if $s\ge 1$ then $r\le (m-1)/2$ which is equivalent to say that $r<m/2$, since $r$ is an integer and $m$ is odd. Thank you again. $\endgroup$ – Richard Jan 20 '17 at 14:49
  • $\begingroup$ If $1 \leqslant s \leqslant \frac{m-1}{2}$, then $1 = \frac{m+1}{2} - \frac{m-1}{2} \leqslant \frac{m+1}{2} - s \leqslant \frac{m+1}{2} - 1 = \frac{m-1}{2}$. $\endgroup$ – Daniel Fischer Jan 20 '17 at 14:54
  • $\begingroup$ To Daniel Fischer: please, see also this very related doubt: math.stackexchange.com/questions/2106173/… Thank you very much. $\endgroup$ – Richard Jan 20 '17 at 15:35

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