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I'm confused on a few things from Algebraic Number Theory by Serge Lang, Chap. VIII.

Lang

I assume that $\log$ means the principal value logarithm, which is defined on $\mathbb{C} - 0$, but only continuous and analytic on $\mathbb{C} - (-\infty,0]$. In that case, for a convergent product $\prod_n z_n$ of nonzero complex numbers with nonzero limit $z$, we only have $\log z = 2 \pi i k + \sum\limits_n \log z_n$ for some integer $k$. See for example Theorem 18 of these notes.

However, Lang immediately writes

$$\log \zeta(s) = \sum\limits_{p,m} \frac{1}{mp^{ms}}$$

where $\sum\limits_{p,m} \frac{1}{mp^{ms}} =\sum\limits_p \log (1 - \frac{1}{p^s})^{-1}$, skipping any discussion of the possibility of the $2\pi i k$. Why is this? Is he just skipping the proof that $k$ has to be zero?

I also don't understand what is meant by the comparison $$\log \zeta(s) \sim \log \frac{1}{s-1}$$

at the end. Neither of these functions are analytic in a punctured neighborhood of $1$, right? How does it make sense to talk about their difference being analytic with a singularity at $1$?

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Lang defines $\log$ as the series,

$$ \log(1 - z) = -\sum_{n \geq 1} \frac{z^n}{n},$$

which is valid for $\lvert z \rvert < 1$. This of course has all the regular properties of logarithms, and this is how he immediately jumps to $$ \log \zeta(s) = \sum_p \log \left( 1 - \frac{1}{p^{s}} \right) = \sum_{p, m} \frac{1}{mp^{ms}}.$$ The first equality is from the $\log(AB) = \log A + \log B$ property of logs, and the second comes from applying the definition-of-$\log$-as-series definition. There are some subtleties here coming from infinite expansions and infinite sums, but since initially $\mathrm{Re}(s) > 1$, everything in sight converges absolutely and this is sufficient to swap all orders of summation (and is sufficient for $\lvert p^{-s} \rvert < 1$ so that we can expand log in series).

Now as $$\sum_p \sum_{m \geq 2} \frac{1}{mp^{ms}}$$ converges absolutely for $\mathrm{Re}(s) > 1/2$, we can say that $$ \sum_p \sum_{m \geq 2} \frac{1}{mp^{ms}} = \sum_p \sum_{m \geq 1} \frac{1}{mp^{ms}} - \sum_p \frac{1}{p^s} = \log \zeta(s) - \sum_p \frac{1}{p^s} \tag{1}$$ converges for $\mathrm{Re}(s) > 1/2$. What exactly does this mean? The left hand side converges absolutely for $\mathrm{Re}(s) > 1/2$, and agrees with the right hand side for $\mathrm{Re}(s) > 1$. Thus the left hand side is the analytic continuation of the right hand side to $\mathrm{Re}(s) > 1/2$.

We have now shown that $$ \log \zeta(s) \sim \sum_p \frac{1}{p^s},$$ since their difference is analytic on a neighborhood of $s = 1$ (and whose analytic continuation is given in the LHS of $(1)$).

Lang doesn't address why $\zeta(s) \sim 1/(s-1)$ in this passage; presumably he showed earlier how to analytically continue $\zeta(s)$ past $\mathrm{Re}(s) > 1$ to $\mathrm{Re}(s) > 0$ and showed that the only pole is a simple pole at $s = 1$ with residue $1$. This implies that $\zeta(s) \sim 1/(s-1)$.

Taking this for granted, we have that $$ \zeta(s) = \frac{1}{s-1} + f(s),$$ where $f(s)$ is holomorphic on a neighborhood of $1$. It remains only to show that $$ \log \zeta(s) \sim \log \frac{1}{s-1}.$$ As written, the $\log$ term on the right doesn't naively agree with Lang's use of $\log$ earlier since near $1$ the fraction $\frac{1}{s-1}$ is not within the region of convergence for the series definition of $\log$. To make sense of this, we should first consider $$ \log \zeta(s) - \log \frac{1}{s-1} = \log \frac{\zeta(s)}{\frac{1}{s-1}} = \log \frac{\frac{1}{s-1} + f(s)}{\frac{1}{s-1}} = \log \Big( 1 + (s-1)f(s)\Big). \tag{2}$$ As $f(s)$ is holomorphic at $1$, the term $(s-1)f(s)$ can be made arbitrarily small by restricting to a small neighborhood of $1$, and in this neighborhood the series definition of the logarithm makes sense. Thus the RHS of $(2)$ is well-defined in a neighborhood of $1$.

There is still a question of what exactly the LHS of $(2)$ means, as Lang's definition of the logarithm is a bit restrictive in this passage. But if we define the LHS of $(2)$ as the log of the ratio (appearing after the first equality in $(2)$), then everything is well-defined and we are done. It is not clear from this passage whether this is what Lang intended or not, but fortunately all descriptions of analytic objects must agree.

As an alternative course of action (not at all implied by Lang), we might note that the series usual definition of $\log$ is a definition of the principal branch, and we know that as long as $\mathrm{Re}(z) > 0$ then we can safely and naively define $\log z$ as the principal branch of the logarithm. Then for $\mathrm{Re}(s) > 1$ (and at least restricting to $\mathrm{Im}(s)$ small), it's clear that $\zeta(s)$ will be a complex number with positive real part. And thus $\log \zeta(s)$ is well-defined in this region. Similarly $\frac{1}{s-1}$ will have positive real part, and thus $\log(\frac{1}{s-1})$ is well-defined.

These each use the principal branch, and thus their difference agrees with the RHS of $(2)$ in the neighborhood of $s=1$ with $\mathrm{Re}(s) > 1$. Then $(2)$ gives an analytic continuation of the LHS, showing that $$ \log \zeta(s) \sim \log \frac{1}{s-1},$$ and therefore that $$ \log \zeta(s) \sim \log \sum_p \frac{1}{p^s} \sim \log \frac{1}{s-1},$$ as desired.

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  • $\begingroup$ Thank you. Can you explain more about $\zeta(s)$ having positive real part for $\operatorname{Im}(s)$ small? Does this follow from looking at the terms $$\frac{1}{k^s} = \frac{1}{e^{s\ln k}} = e^{-\sigma \ln k} e^{-it\ln k }$$ in the Dirichlet series? $\endgroup$ – D_S Apr 10 '18 at 16:25
  • $\begingroup$ In the beginning, how do we know we can evaluate $\log \zeta(s)$ using the power series definition? $\endgroup$ – D_S Apr 10 '18 at 16:30
  • $\begingroup$ We know that $\zeta(s)$ is continuous, and that (for instance) $\zeta(1.1)$ is a finite real number. Thus for all $s$ near $1.1$, $\zeta(1.1)$ has positive real part. Generically it's apparent that $\zeta(s)$ for $\mathrm{Re}(s) > 1$ and small imaginary part always has positive real part. I have never really thought about how small the real part can get near near the line $\mathrm{Re}(s) = 1$, so I don't know there. But nonetheless the series $\zeta(s)$ and product agree for $\mathrm{Re}(s) > 1$ with small imaginary part, and thus are the same function. (This answers your second comment too) $\endgroup$ – davidlowryduda Apr 10 '18 at 17:19
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You're right that he's being sloppy about the branch of the logarithm. It doesn't really matter, though, because $2\pi i k$ would just be some constant and would not affect the final conclusion (which only cares about the value modulo adding functions that are analytic at $1$). In fact, though, it is easy to see that $k$ must be $0$. When $s$ is real, it is obvious that $k=0$, since all the terms are real. Now just note by continuity that $k$ must be locally constant, and therefore $k=0$ for all $s$ (here we are always assuming that $\operatorname{Re}(s)>1$ so everything converges).

As for the comparision $$\log \zeta(s) \sim \log \frac{1}{s-1},$$ nothing in the definition requires these functions to be analytic in a deleted neighborhood of $1$: a "singularity" at $1$ does not have to be an isolated singularity! All this statement means is that the difference $\log\zeta(s)-\log\frac{1}{s-1}$ (which is defined for $\operatorname{Re}(s)>1$, say) can be analytically continued over a neighborhood of $1$.

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