A question about partial Euler products - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:01:17Z http://mathoverflow.net/feeds/question/95205 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95205/a-question-about-partial-euler-products A question about partial Euler products Ramin 2012-04-25T22:08:39Z 2012-05-30T18:53:33Z <p>Let $K/{\mathbb Q}$ be an extension of degree $d$. Let $S$ be the set of primes $p$ which split completely in $K$. What can one say about the analytic properties of $$\zeta_{K, S}(s) : = \prod_{p \in S} \frac{1}{1-p^{-s}}.$$ More generally, one can define a similar partial Euler product for any splitting type, and ask the same question about the analytic properties of the resulting function. Any advice would be greatly appreciated. </p> http://mathoverflow.net/questions/95205/a-question-about-partial-euler-products/95218#95218 Answer by unknown (google) for A question about partial Euler products unknown (google) 2012-04-26T01:52:29Z 2012-04-26T01:52:29Z <p>I hope this could be relevant to your nice question. There is a paper by K. Williams where a Mertens' type theorem for arithmetic progressions was proved,</p> <p>K. Williams, Mertens' Theorem for Arithmetic Progressions, J. Number Theory 6 (1974) 353--359.</p> <p>In this paper, the relation</p> <p><code>$$\sum_{\chi \bmod q}\chi(p)\bar{\chi}(a) = \begin{cases} \varphi(q), &amp; p \equiv a \bmod q, \\ &amp; \\ 0, &amp; \mbox{otherwise.} \end{cases}$$</code></p> <p>for Dirichlet characters is used to derive an expression for</p> <p><code>$$\prod_{\substack{p \leq x\\p \equiv a \bmod q}}\left(1-\frac{1}{p}\right).$$</code></p> <p>Letting $G$ be the Galois group of $K/\mathbb{Q}$, $C_G(g)$ be the centralizer of $g$ in $G$, $\mathcal{C}(g)$ the set of conjugates of $g$ in $G$, $\sigma_p$ the Frobenius element of $p$, the <a href="http://en.wikipedia.org/wiki/Schur_orthogonality_relations" rel="nofollow">Schur orthogonality relations</a></p> <p><code>$$\sum_{\chi}\chi(\sigma_p)\overline{\chi(g)} = \begin{cases} \left| C_G(\sigma_p)\right| = \frac{|G|}{|\mathcal{C}(\sigma_p)|}, &amp; \sigma_p, g \mbox{ are conjugate}, \\ &amp; \\ 0, &amp; \mbox{otherwise} \end{cases}$$</code></p> <p>where the sum is over all irreducible characters $\chi$ of $G$, can be used in a similar way to derive a formula for the partial Euler product in your nice question by writing, for some fixed $g\in G$,</p> <p><code>$$\prod_{\substack{p\\\sigma_p \in \mathcal{C}(g)}}(1-p^{-s})^{-C_G(g)} = \prod_{\chi}\left(\prod_{p}\left(1-p^{-s}\right)^{-\chi(p)}\right)^{\overline{\chi(g)}}$$</code></p> <p>where the $\chi$ are irreducible characters of $G$. In the mentioned paper, where the $\chi$ are Dirichlet characters, Williams wrote</p> <p><code>$$\prod_{p \leq x}\left(1-\frac{1}{p}\right)^{\chi(p)} = \left(\frac{1}{L(1,\chi)}+O(1/\log x)\right)\left(K(1,\chi)+O(1/x)\right)$$</code></p> <p>where $K(1,\chi)$ had some properties that could be stated in the paper. Hence in the case of the product in your nice question, maybe something can be worked out along such lines too. I guess it's more complicated as there are now Artin L-functions.</p> http://mathoverflow.net/questions/95205/a-question-about-partial-euler-products/95243#95243 Answer by Filippo Alberto Edoardo for A question about partial Euler products Filippo Alberto Edoardo 2012-04-26T10:38:55Z 2012-04-26T23:27:48Z <p>I do not know if you are willing to restrict to abelian $K/\mathbb{Q}$: if not, please throw away my whole answer. But let me assume that your base field is any number field $F$, as it does not change anything: and, accordingly, that $K/F$ is abelian.</p> <p>Suppose first that $K/F$ is a ray class field for some conductor $\mathfrak{f}$ (for $F=\mathbb{Q}$ it means $K$ is either $\mathbb{Q}(\zeta_f)$ or $\mathbb{Q}(\zeta_f)^+$ for some $f$). Then you are simply considering the partial zeta function $\zeta({\bf{1}},s)$ where $\bf{1}$ is the trivial class in the ray class group modulo $\mathfrak{f}$. Those are well-studied and can be interpreted as suitable Mellin transform, pretty much as the usual Dedekind zeta function and admits the usual analytic continuation (see J. Neukirch <em>Algebraic Number Theory</em>, Chapter VII, Section 5, theorem 5.5 for the Mellin transform statement, Theorem 5.9 for analytic continuation <em>caveat: I only have the German version and I ignore if numbering has been changed in the translation</em>). This settles also the case in which you consider some different splitting behaviour which would be controlled by class field theory pretty much the same way – <em>i. e.</em> your Euler product would be that of $\zeta({\bf{c}},s)$ for the class $\bf{c}$ in the ray class field containing precisely the integral ideals whose splitting time is your favorite one.</p> <p>If now your <em>abelian</em> extension is only contained in a ray class field, you need to follow more closely Tate's thesis (see the last chapter of Cassels and Fröhlich's <em>Algebraic Number Theory</em>) and chose the right measure to put on the unit ball inside $\mathbb{Q}_p$ for the primes you want to throw away – I imagine that you should chose the measure giving volume $1$ to some suitable subgroup depending on your local extension, but I am not checking. I guess that this would not affect anything at all, because Tate's strategy is somehow ''formal'' in the sense that it just depends on some very nice property of adelic Fourier transform. As for a reference, I suggest D. Ramakrishan and R. Valenza <em>Fourier Analysis on Number Fields</em>.</p> <p>Let me also observe that what you are asking for is ''natural'' in the sense that when looking at zeta functions as being suitable measures on Galois group, your Euler product at negative integers $k$ gives the special values of the measure agains the function $\chi_S|\cdot|^k$ where $\chi_S$ is the characteristic function of the set of primes $S$ seen as elements of the Galois group via class field theory – this is discussed very well in the introduction to Deligne and Ribet's paper about special values of $L$-functions at negative integers.</p> <p>If your extension is not abelian, I think the situation is much more involved because it is not enough to play the same game simply on adeles group: and, as said before, you'd rather forget everything I have written...</p> http://mathoverflow.net/questions/95205/a-question-about-partial-euler-products/98407#98407 Answer by Ramin for A question about partial Euler products Ramin 2012-05-30T18:53:33Z 2012-05-30T18:53:33Z <p>This question turned out to be not too difficult. Please see <a href="http://www.math.uic.edu/~rtakloo/euler-product.pdf" rel="nofollow">http://www.math.uic.edu/~rtakloo/euler-product.pdf</a> for a (casual) writeup of an answer. Thanks for your comments and hints. </p>