User tom lovering - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:58:39Z http://mathoverflow.net/feeds/user/8080 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127726/integer-multiples-of-a-irrational-dense-in-r-z/127728#127728 Answer by Tom Lovering for Integer multiples of a irrational dense in R/Z ? Tom Lovering 2013-04-16T18:08:06Z 2013-04-16T18:08:06Z <p>Yes. For elementary reasons.</p> <p>Suppose it weren't dense. Then there would be some little interval not hit, of some positive length say $1/N$.</p> <p>But this cannot happen. Divide the circle into N little equal intervals and consider $0,r,2r,...,Nr$. By pigeonhole some two of them lie in the same interval, say $ar,br$ (and these elements are distinct since $r$ is not rational). Then modulo $\mathbb{Z}$, $-1/N &lt; (a-b)r &lt; 1/N$, and so some multiple of this will lie in the interval we supposed contained no element.</p> http://mathoverflow.net/questions/122855/is-strong-multiplicity-one-obviously-stronger-than-multiplicity-one Is strong multiplicity one (obviously) stronger than multiplicity one? Tom Lovering 2013-02-25T03:49:59Z 2013-02-25T14:26:00Z <p>In the theory of automorphic representations one says that G satisfies a "multiplicity one" property if every cuspidal representation occurs with multiplicity one in $L^2(G(F)\backslash G(A))$.</p> <p>One also says that G satisfies "strong multiplicity one" if a cuspidal representation is uniquely determined up to isomorphism by its behaviour at cofinitely many places of F. </p> <p>My question is: does the latter always imply the former (as the name would suggest), and is there a decent reason?</p> <p>This question was prompted while thinking about Jacquet-Langlands, whose proof (via the trace formula) seems to give multiplicity one as a by-product and whose statement also directly implies strong multiplicity one (for inner forms of GL2 using the result for GL2 itself). However, I didn't see any way to deduce multiplicity one directly from the usual statement.</p> <p>I also can't see how strong multiplicity one for classical modular forms should directly imply the q-expansion principle (which I guess is roughly the same as multiplicity one). Apologies if I'm missing something extremely obvious on this dozy Sunday evening.</p> <p>Thanks, Tom.</p> http://mathoverflow.net/questions/108953/motivic-generalisation-of-neron-ogg-shaferevich-criterion Motivic generalisation of Neron-Ogg-Shaferevich criterion Tom Lovering 2012-10-05T19:51:27Z 2012-12-03T19:06:56Z <p>Given a variety $X$ over $\mathbb{Q}$ with good reduction at $p$, proper smooth base change tells us that its $l$-adic cohomology groups are unramified at $p$ (and I'd guess some $p$-adic Hodge theory tells us its p-adic cohomology is crystalline).</p> <p>My question is to what extent it's possible to find a converse to this statement. More precisely, I have yet to see a counterexample to the following "conjecture" (though I still suspect it's wrong).</p> <p><strong>"Conjecture"</strong>: Let $K$ be a number field, $p$ and $l$ primes, and $V$ a geometric (say, coming from the variety $Y$) $l$-adic representation of $G_K$ that is unramified/crystalline at $\mathfrak{p}|p$. Then there exists a smooth proper variety $X$ such that $X$ has good reduction at $\mathfrak{p}$ and $V$ can be cut out of the cohomology of $X$.</p> <p>From googling around, the things I know so far are (at least for $l \not= p$):</p> <ul> <li>If $Y$ is an abelian variety, the classical Neron-Ogg-Shafarevich condition means that $Y$ itself is a witness to the conjecture.</li> <li>We can take torsors for abelian varieties with no $K$-rational points, and these can have the same representations, but fail to have good reduction (in this paper <a href="http://arxiv.org/abs/math/0605326" rel="nofollow">http://arxiv.org/abs/math/0605326</a> of Dalawat).</li> <li>There exist curves which have bad reduction, but whose Jacobians have good reduction.</li> </ul> <p>If anyone knows any more about this story I'd be interested to hear. Ultimately I guess it would be nice to have a definition for when a motive is unramified/has good reduction, and cohomologically this surely has to mean unramified/crystalline, but it would be nice if this could always be realised "geometrically".</p> <p>Thanks, Tom.</p> http://mathoverflow.net/questions/98418/what-are-the-truly-global-methods-in-number-theory What are the truly 'global methods' in number theory? Tom Lovering 2012-05-30T21:16:29Z 2012-05-31T08:48:23Z <p>I have spent some time being confused by the nature of global methods in number theory. It seems that there are in some sense (for my purposes) three levels at which algebraic number theorists operate: local (at one prime), everywhere local (at all primes simultaneously, including the infinite ones) and global (actually playing with the number field). When we talk about things like local class field theory we mean the first one, and when we talk about global class field theory I guess we mean the last one (but the extent to which the ideles seem to get involved suggests to my naive mind a strong whiff of the second one also).</p> <p>When we talk about local to global principles we most definitely mean the passage from the second to the third. I guess my question can therefore by crystallised in terms of elementary number theory as follows: what global techniques do we have for proving the non-existence of solutions to diophantine equations? In other words, given a failure of the local-global principle, what are the techniques one can use to demonstrate it independently of local information?</p> <p>If I am given a diophantine equation and asked to show it has no solutions, I can think of very few methods that are not in some sense local', certainly if we count working at the infinite primes also as local (which surely we should?). </p> <p>One possibility I have been considering is that something to do with heights/descent is perhaps a global method. However, the height of a point is still measurable by concatenating local data, and descent is normally via some trick involving congruences, but perhaps the well-orderedness' of the process is a truly global trick.</p> <p>Also, returning to a pessimistic analysis, in an important classical result that I have seen called a `measure of the failure of the local-global principle': the finiteness of the class number, it seems to me that the statement doesn't involve the infinite primes in any way, so to prove it we milk the infinite primes for all the have got and get the Minkowski bound by directly studying local behaviour above infinity. Is my opinion in this regard incorrect? If so, which of the arguments are truly global?</p> <p>So to conclude, are there such things as "global methods", and if there are, what are they? Apologies for posing what is probably a naive, overly-simplistic and absurdly general question, but I am hoping several people may have thought about this and have interesting things to say.</p> <p>Thanks, Tom.</p> http://mathoverflow.net/questions/90837/comparison-between-singular-and-etale-cohomology-in-batyrevs-paper-on-birational Comparison between singular and etale cohomology in Batyrev's paper on Birational Calabi-Yau varieties Tom Lovering 2012-03-10T17:59:38Z 2012-03-10T18:05:12Z <p>My question refers to the paper <a href="http://arxiv.org/pdf/alg-geom/9710020.pdf" rel="nofollow">http://arxiv.org/pdf/alg-geom/9710020.pdf</a> where Batyrev proves that birational Calabi-Yau algebraic varieties have equal Betti numbers by counting points over finite fields using p-adic integration and so computes the Betti numbers using the Weil conjectures.</p> <p>It seems that he is doing the following. Given a variety $X$ over $\mathbb{C}$, we can actually write it (and all the associated data we care about) as a variety $\mathcal{X}$ over $\mathcal{R}$, some finite-type $\mathbb{Z}$-algebra: i.e. such that $\mathcal{X} \otimes_\mathcal{R} \mathbb{C} = X$. We then fix an approriate maximal ideal $J(\pi)$ of $\mathcal{R}$ which lies above $p \in \mathbb{Z}$. I think we then turn our attention to the variety $\mathcal{X}\otimes_\mathcal{R} (\mathcal{R}/J(\pi))$, and using a ring of integers $R$ of a local number field with this as special fibre, we can count the number of points this variety has over every finite field extension of $\mathbb{F}_q = \mathcal{R}/J(\pi)$ using p-adic integration.</p> <p>So the Weil conjectures give us the Betti numbers of this variety, and by proper smooth base change these Betti numbers are the same as those of $\mathcal{X} \otimes_\mathcal{R} \mathbb{C}$, but where the map $\mathcal{R} \rightarrow \mathbb{C}$ is not the natural inclusion but rather $\mathcal{R} \rightarrow R \hookrightarrow \mathbb{C}$. Since he is trying to compute the cohomology of the former, this doesn't make sense to me.</p> <p>Can anyone see if I'm making a mistake somewhere? (or how my issue can be resolved?)</p> <p>Thanks, Tom.</p> http://mathoverflow.net/questions/82260/class-field-theory-using-only-ideles-of-norm-1 Class field theory using only ideles of norm 1 Tom Lovering 2011-11-30T10:06:34Z 2011-11-30T15:46:50Z <p>I am a total non-expert, so the answer to this question may be obvious, but here goes.</p> <p>In Chevalley's formulation of CFT we get Artin maps $J_k \rightarrow Gal(L/k)$, where $J_k$ is the group of all ideles of $k$. However, we know there is a nice subgroup $J_k^1$ of the ideles obtained by taking only those satisfying the product formula $\prod_{v} |x_v| = 1$. Note that this still contains all the principal ideles, still surjects onto $I_k$ and has additional attractive properties like the compactness of $J_k^1/k^*$. Is there a way to set up CFT using quotients of this nicer group, and if so, what are the advantages of working with the superficially more unwieldy $J_k$?</p> <p>Thanks.</p> http://mathoverflow.net/questions/50004/l-functions-and-higher-dimensional-eichler-shimura-relation/50015#50015 Comment by Tom Lovering Tom Lovering 2013-04-06T19:16:16Z 2013-04-06T19:16:16Z A good reference for the method Kevin is talking about seems to be the paper of Scholze <a href="http://arxiv.org/pdf/1003.1935.pdf" rel="nofollow">arxiv.org/pdf/1003.1935.pdf</a> In this paper he does rather more (also checking the equivalent of an Eichler Shimura relation at bad places), but it isn't too difficult to fish out the pieces you want, in particular he checks the boundary components line up. http://mathoverflow.net/questions/122855/is-strong-multiplicity-one-obviously-stronger-than-multiplicity-one Comment by Tom Lovering Tom Lovering 2013-02-25T14:11:10Z 2013-02-25T14:11:10Z @Nosr When I say 'q-expansion principle' here I just mean the fact that a q-expansion uniquely determines its modular form, sorry. @Dror Ah, that would define my question nicely out of existence (and possibly give anecdotal evidence that with my definition it's not true...). http://mathoverflow.net/questions/91496/dimension-of-the-basis-of-a-vector-space Comment by Tom Lovering Tom Lovering 2012-03-18T13:56:13Z 2012-03-18T13:56:13Z Isn't the infinite case much easier than the finite case (and in fact almost trivial)? Given two sets $A,B$ of basis elements just write each element of $A$ in terms of finitely many elements in $B$, and since this spans we must use every element of $B$ somewhere, so $B$ is a union of finite sets indexed by $A$, hence if they're both infinite, $|B| \leq |A|$. Similarly $|A| \leq |B|$. Maybe when set theory wasn't well-developed this proof was less obvious, but to me it seems likely that it's a sufficiently easy result it might not have been formally published but just remarked somewhere. http://mathoverflow.net/questions/90837/comparison-between-singular-and-etale-cohomology-in-batyrevs-paper-on-birational Comment by Tom Lovering Tom Lovering 2012-03-12T14:22:03Z 2012-03-12T14:22:03Z Thanks for the quick reply. I think it's helped focus in on my confusion. If $R$ were constructed as you say, I'm suspicious that its function field might have some transcendence degree over $\mathbb{Q}_p$. On the other hand, the paper seems to say that $R$ is &quot;the maximal compact subring in a local $p$-adic field,&quot; which I took to mean an <i>algebraic</i> extension of $\mathbb{Q}_p$, and I think actually being a local field is necessary to make the rest of the argument work. So I guess my question now is: is $Frac(R)$ (as you define it) obviously algebraic $/\mathbb{Q}_p$? http://mathoverflow.net/questions/90837/comparison-between-singular-and-etale-cohomology-in-batyrevs-paper-on-birational Comment by Tom Lovering Tom Lovering 2012-03-12T12:43:24Z 2012-03-12T12:43:24Z I don't think there's any kind of natural embedding of $\mathcal{R}$ into $R$ :the paper seems to go via $\mathcal{R} \rightarrow \mathcal{R}\otimes_\mathbb{Z} \mathbb{Z}_p \rightarrow R$ which feels unlikely to be injective. For example, if $\mathcal{R}$ is something like $\mathbb{Z}[x_1,...,x_n]$. So do you mean I should artificially construct some kind of injection $\mathcal{R} \hookrightarrow R$,using the fact $\mathbb{Z}_p$ has lots of transcendental elements,and then play around to make it compatible with the inclusion of $\mathcal{R}$ in $\mathbb{C}$?