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I want someone to clarify me the next step in page 11. They write for $\delta=\prod p^l$ ,$k=\prod p^{\lambda}$ the sum: $\sum_{\delta |k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p|\delta}(l+1-lp^{-s})$ is $$k^{1-s}\prod_{p|k}(\lambda+1-\lambda p^{-s})-\sum_{p|k} (\frac{k}{p})^{1-s}(\lambda-(\lambda-1)p^{-s})\prod_{p'|k, p'\neq p} (\lambda+1-\lambda (p')^{-s})+\sum_{pp'|k}(\frac{k}{pp'})^{1-s} (\lambda-(\lambda-1)p^{-s})(\lambda-(\lambda-1)(p')^{-s})\prod_{p'|k, p''\neq p,p'} (\lambda+1-\lambda(p'')^{-s})-...= k^{1-s}\prod_{p|k}((\lambda+1-\lambda p^{-s})-\frac{1}{p^{1-s}}(\lambda-(\lambda-1)p^{-s}))$$

If I understand correctly, the LHS of the above is from exclusion-inclusive principle, but I don't understand the RHS, I don't see how did they get the second term in the brackets of the product, i.e $\frac{1}{p^{1-s}}(\lambda-(\lambda-1)p^{-s})$.

Any hints?

Thanks.

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The notation used in the book is somewhat confusing as, although there is nothing to show it, $\lambda$ is a function of $p$, and $l$ is a function of $p$ and $\delta$. To indicate this, I will write $l_\delta(p)$ and $\lambda(p)$ instead of $l$ and $\lambda$.

In the sum we start with, $$\Xi:=\sum_{\delta \mid k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s}),$$ only values of $\delta$ for which $\mu(k/\delta)$ is nonzero contribute to the sum. Now, recall that $\mu(i)$ is nonzero only if $i$ is squarefree. In this case, if $i$ has $j$ prime factors, $\mu(i)$ equals $(-1)^j$. If $i$ is not squarefree, $\mu(i)$ is $0$. Let the long sum starting $k^{1-s} \prod_{p\mid k}\ldots$ be called $\Sigma$, and call its first piece $\Sigma_0$, its second piece $\Sigma_1$, and so on, so that $$\Sigma=\Sigma_0+\Sigma_1+\Sigma_2+\cdots.\qquad (*)$$ The expansion $(*)$ comes from setting $i:=k/\delta$ in $\Xi$ and looking successively at the cases where $j=0$, $1$, $2$, and so on. For example, if $j$ is $0$, we must have $i=1$, so $\delta=k$ and $$ \Sigma_0=\delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s})= k^{1-s}\prod_{p\mid k}(\lambda(p)+1-\lambda(p)p^{-s}). $$ This gives the first term of $\Sigma$. The second term of $\Sigma$ comes from the case when $j=1$. In this case, $i$ is prime, so $\delta=k/p$, for some $p$ dividing $k$. Therefore, the contribution to $\Xi$ is $$ \Sigma_1=\sum_{p\mid k, \delta=k/p} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s})$$ $$ =-\sum_{p\mid k} (\frac{k}{p})^{1-s} (\lambda(p)-(\lambda(p)-1)p^{-s}) \prod_{p'\mid k,\ p'\ne p}(\lambda(p')+1-\lambda(p')p'^{-s}). $$ The third term of $\Sigma$ comes from the case where $j=2$. In this case, $i=p p'$ for distinct primes $p$ and $p'$ dividing $k$, so $\delta=k/(p p')$. The contribution to $\Xi$ is then $$ \Sigma_2=\sum_{p, p'\mid k,\ p\ne p'} (\frac{k}{p p' })^{1-s} (\lambda(p)-(\lambda(p)-1)p^{-s}) (\lambda(p')-(\lambda(p')-1)p'^{-s})$$ $$ \prod_{p''\mid k,\ p''\ne p,\ p''\ne p'}(\lambda(p'')+1-\lambda(p'')p''^{-s}). $$ Continuing in this way, we can write down all of the expansion $(*)$. This proves that $\Xi=\Sigma$.

To see that the product that we finally end up with, $$\Pi= k^{1-s}\prod_{p\mid k}((\lambda(p)+1-\lambda(p) p^{-s})-\frac{1}{p^{1-s}}(\lambda(p)-(\lambda(p)-1)p^{-s})),$$ is equal to $\Sigma$, we can write $$\Pi=k^{1-s} \prod_{p\mid k} (A_p + B_p),\qquad (**)$$ where $$A_p=\lambda(p)+1-\lambda(p) p^{-s},\ \ \ B_p=-\frac{1}{p^{1-s}}(\lambda(p)-(\lambda(p)-1)p^{-s}).$$ Then, if we expand $(**)$ and sum together all terms which contain exactly $j$ factors of the form $B_p$, we recover $\Sigma_j$. For example, if we take $j=0$, we get $k^{1-s} \prod_{p\mid k} A_p$, which equals $\Sigma_0$. If we take $j=1$, we get $$k^{1-s} \sum_{p\mid k} B_p \prod_{p'\mid k, \ p'\ne p} A_{p'},$$ which equals $\Sigma_1$, and so on. This shows that $\Sigma=\Pi$.

Another way to derive $\Pi$ is to go back to the sum we started with, $$ \Xi:=\sum_{\delta \mid k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s}). $$ If we rewrite this in terms of $\delta':=k/\delta$, and write $\delta'=\prod_{p\mid k} p^{l_{\delta'}(p)}$, we get $$ \Xi=\sum_{\delta'\mid k} (\frac{k}{\delta'})^{1-s} \mu(\delta') \prod_{p\mid k} (\lambda(p)-l_{\delta'}(p)+1-(\lambda(p)-l_{\delta'}(p))p^{-s}). $$ We can rewrite this as $$ \Xi=k^{1-s} \prod_{p\mid k} (\lambda(p)+1-\lambda(p) p^{-s}) \sum_{\delta'\mid k} G(\delta'),\qquad (***)$$ where $$G(\delta')= \frac{1}{\delta'^{1-s}} \mu(\delta') \prod_{p\mid \delta'}\frac{\lambda(p)-l_{\delta'}(p)+1-(\lambda(p)-l_{\delta'}(p))p^{-s}}{\lambda(p)+1-\lambda(p) p^{-s}}. $$ But now $G(\delta')$ is multiplicative, $G(1)=1$, and $G(\delta')$ vanishes unless $\delta'$ is squarefree. Therefore $$ \sum_{\delta'\mid k} G(\delta')=\prod_{p\mid k} (1 + G(p)) =\prod_{p\mid k} (1 - \frac{1}{p^{1-s}} \frac{\lambda(p)-(\lambda(p)-1)p^{-s}}{\lambda(p)+1-\lambda(p) p^{-s}}). $$ Plugging this into $(***)$ gives $\Xi=\Pi$.

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