finiteness of class number: a bound for semi-simple groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:52:51Z http://mathoverflow.net/feeds/question/82951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82951/finiteness-of-class-number-a-bound-for-semi-simple-groups finiteness of class number: a bound for semi-simple groups? genshin 2011-12-08T10:27:29Z 2011-12-10T18:05:47Z <p>Let $F$ be a number field, and $G$ a connected semi-simple linear algebraic $F$-group, which does not contain anisotropic (simple) $F$-factors. Write $\hat{F}$ for the ring of finite adeles $F\otimes\hat{\mathbb{Z}}$. Then the strong approximation theorem implies that the double coset $G(F)\backslash G(\hat{F}) / K_G$ is finite for any compact open subgroup $K_G\subset G(\hat{F})$. In fact it is even equal to one (i.e. trivial double quotient) if $G$ is simply connected as a semi-simple group.</p> <p>And in general, for $G$ semi-simple but not simply-connected, how should one bound the growth of the size of the double quotient? At least we know that there is an isogeny $G'\rightarrow G$ with $G'$ semi-simple and simply connected. Can we expect the double quotient to be bounded by some function in terms of the degree of $G'\rightarrow G$ and the set of finite places where $K_G$ is not a maximal compact open subgroup?</p> <p>At least it seems that one could not expect the double quotient to be uniformly bounded when $K_G$ shrinks to the neutral element.</p> <p>Thanks!</p> http://mathoverflow.net/questions/82951/finiteness-of-class-number-a-bound-for-semi-simple-groups/83133#83133 Answer by Peter McNamara for finiteness of class number: a bound for semi-simple groups? Peter McNamara 2011-12-10T18:05:47Z 2011-12-10T18:05:47Z <p>The size of the double quotient can be bounded by a function exponential in the number of places where K<sub>G</sub> is not maximal compact. The base of the exponential will of course depend on the kernel of the central isogeny, but all the dependence on the field F can be moved into the implied constant.</p> <p>The part about this that I don't know how to prove is to show that the double quotient is finite in the first place. I can only do this for PGL<sub>n</sub>, which I think is a worthwhile group to think about for this problem since you can see everything explicitly.</p> <p>Let Z be the kernel of the central isogeny G'-->G. The natural map $G(\hat F)\rightarrow \prod'_v H^1(F_v,Z)$ has kernel $G'(\hat F)$ in which the rational points are dense by strong approximation. This map is surjective, so we can rewrite the adelic double quotient as a double quotient of this restricted direct product of H<sup>1</sup>'s.</p> <p>Now consider what happens when we shrink K<sub>G</sub> at one place v from a maximal compact to something smaller. Then the cardinality of the double quotient can increase by at most a factor of |H<sup>1</sup>(F<sub>v</sub>,Z)|. To show the desired exponential growth, we simply need a bound on |H<sup>1</sup>(F<sub>v</sub>,Z)| that is independent of v.</p> <p>For Z=&mu;<sub>n</sub>, we know explicitly $H^1(k,\mu_n)=k^\times/(k^\times)^n$ for any field k and the bound is easy. If G is split, then Z is a product of &mu;<sub>n</sub>'s and there is no more work to be done. For general G, pick a finite Galois extension K/F over which G splits. Then the inflation-restriction exact sequence together with the result for split groups implies the necessary uniform bound on |H<sup>1</sup>(F<sub>v</sub>,Z)|.</p>