User ramsey - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:36:33Z http://mathoverflow.net/feeds/user/12107 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104535/are-there-p-adic-modular-forms-for-non-congruence-subgroups Are there $p$-adic modular forms for non-congruence subgroups? Ramsey 2012-08-12T03:44:19Z 2013-05-02T23:22:00Z <p>My answer to the question:</p> <p><a href="http://mathoverflow.net/questions/104362/moduli-interpretations-for-modular-curves" rel="nofollow">http://mathoverflow.net/questions/104362/moduli-interpretations-for-modular-curves</a></p> <p>led me to wonder about the question in the present title.</p> <p>It seems that modular forms for non-congruence subgroups may not "arise" from algebraic geometry in the same way modular forms for congruence subgroups do (as is evidenced by the lack of a good moduli problem and this whole unbounded denominators thing). Nonetheless, they are clearly complex-analytic objects. I wonder if there's a $p$-adic analytic analog to be had.</p> <p>Here are two more precise questions:</p> <ol> <li><p>(less ambitious) Are there interesting congruences to be had between such modular forms? Interesting $p$-adic limits? Interesting $p$-adic families?</p></li> <li><p>(more ambitious) Is there some (perhaps inherently analytic) description of non-congruence forms in terms of moduli of elliptic curves that can be mimicked in the $p$-adic analytic setting?</p></li> </ol> <p>My rather vague feeling is that the Atkin and Swinnerton-Dyer congruences suggest that there's something to be said here, but I haven't been able to dig up much on these questions in particular. Does anyone know of such work?</p> http://mathoverflow.net/questions/122600/understanding-adjointness-of-sheaves-in-algebraic-geometry/122637#122637 Answer by Ramsey for Understanding Adjointness of Sheaves in Algebraic Geometry Ramsey 2013-02-22T15:16:10Z 2013-02-22T15:26:39Z <p>If you're just looking for some intuition, think about what each side is specifying fiber-by-fiber (think about the locally-free case, i.e. vector bundles, if this helps).</p> <p>The fiber of $f^\ast\mathcal{G}$ at $x\in X$ is the fiber of $\mathcal{G}$ at $f(x)$, and the fiber of $f_*\mathcal{F}$ at $y\in Y$ consists of the sections of the restriction of $\mathcal{F}$ to the pre-image $f^{-1}(y)$.</p> <p>Now think about what each side of the adjunction formula specifies point-by-point. The left side at $x\in X$ gives a linear map from $\mathcal{G}(f(x))$ to $\mathcal{F}(x)$. Now varying $x$ in the pre-image of $y\in Y$, this just amounts to a map $$\mathcal{G}(y)\to \mathcal{F}|_{f^{-1}(y)}(f^{-1}(y)),$$ which is what the right side gives at $y\in Y$.</p> http://mathoverflow.net/questions/109213/uniqueness-of-analytic-continuation-in-rigid-analytic-geometry/109228#109228 Answer by Ramsey for Uniqueness of analytic continuation in rigid analytic geometry Ramsey 2012-10-09T14:12:08Z 2012-10-09T14:12:08Z <p>Let me just complement xbnv's answer with a mild generalization. If $X$ is an <em>irreducible</em> rigid space (and let's suppose we're dealing with reduced spaces from the outset), then its normalization $\tilde{X}$ is connected and normal and is equipped with a finite surjective map $\tilde{X}\to X$. Using xbnv's answer, it follows that the statement holds for $X$ as well.</p> <p>In short, if you replace "connected" by "irreducible" then you're good.</p> http://mathoverflow.net/questions/108044/what-does-this-quotient-of-the-upper-half-plane-parametrize/108064#108064 Answer by Ramsey for What does this quotient of the upper half plane parametrize? Ramsey 2012-09-25T15:23:44Z 2012-09-25T15:23:44Z <p>Here's a very clunky answer:</p> <p>For a particular elliptic curve $E$, the collection of bases $\{P,Q\}$ of the $N$-torsion $E[N]$ is acted upon (diagonally) by the group $(\mathbb{Z}/N\mathbb{Z})^\times$. This group has a subgroup $G$ consisting of elements $x$ with a lifting to an element $y\in (\mathbb{Z}/N^2\mathbb{Z})^\times$ such that $y^2\equiv 1\pmod{N^2}$. (Of course, the group $G$ has a better description since this lifting is automatic at odd primes, etc. - but I'll stick with this one because its relation to the problem is more transparent).</p> <p>I think that the moduli problem classifies isomorphism classes of pairs $(E,O)$ where $O$ is a $G$-orbit of bases of the full $N$-torsion.</p> http://mathoverflow.net/questions/107566/fourier-expansion-of-eisenstein-series-at-various-cusps/107581#107581 Answer by Ramsey for Fourier expansion of Eisenstein series at various cusps Ramsey 2012-09-19T15:54:05Z 2012-09-19T23:33:48Z <p>Let $f$ be any modular form of weight $k$ whose coefficients at the cusp $\infty$ lie in $\mathcal{O}_K$. Then (using GAGA and the $q$-expansion principle - see for example Katz's article on $p$-adic forms) $f$ gives rise to a section of the sheaf $\omega^k$ on $X_0(N)$ (or $X_1(N)$ + fixed under the diamond operators if you want to keep you moduli spaces nice) that is defined over the ring $\mathcal{O}_K[1/N](\mu_N)$. </p> <p>Using the geometric description of $q$-expansions at all cusps in terms of the value of the geometric modular form on the Tate curve with level structure, one sees that in fact all $q$-expansions have coefficients in this ring (and are, as you say, generally expansions in $q^{1/N}$). This is simply because both $f$ and the Tate curve with these level structures are defined over this ring. This should answer your first question in the affirmative.</p> <p>As for your second, I'm a bit confused because your conclusion is roughly what I would take to be the definition of "the corresponding modular form over $\overline{\mathbb{F}}_\ell$ is cuspidal." </p> <p>Do you only mean to assume that the $q$-expansion at $\infty$ (making no assumptions at other cusps) has no constant term modulo $\ell$?</p> <p><strong>EDIT</strong> based on clarifications in the comments</p> <p>Addressing the clarified second question, the $q$-expansions one obtains from the geometric modular form by evaluating at the Tate curve with level structure really are the $q$-expansions of the original form (all this assumes one has embedded $K$ into $\mathbb{C}$ already, but I gather that the OP has done that from the phrasing of his question). In particular, their reductions modulo $\lambda$ coincide, so if the original form has the property that all of the constant terms in the $q$-expansions are divisible by $\lambda$, then the reduced mod $\lambda$ geometric modular form is cuspidal.</p> http://mathoverflow.net/questions/106135/visualizing-the-l-adic-fractal-in-the-partition-function-pn/106138#106138 Answer by Ramsey for Visualizing the l-adic fractal in the partition function p(n) Ramsey 2012-09-01T21:41:10Z 2012-09-02T01:56:34Z <p>I wouldn't take the term "fractal" too seriously (or at least too visually). </p> <p>Basically, they prove that the generating function of $p(n)$ (which happens to be a modular form) has nice congruence properties modulo powers of $p$ when hit with the $U_{p^2}$ operator repeatedly. This latter operator has the effect $\sum a_nq^n\mapsto \sum a_{p^2n}q^n$ on the generating series, and thus can be thought of loosely as "$p$-adically zooming in" on the expansion. Hence the $p$-adically fractal turn of phrase.</p> http://mathoverflow.net/questions/105226/lifting-invertible-functions-on-a-divisor-to-ambient-affine-variety/105231#105231 Answer by Ramsey for Lifting invertible functions on a divisor to ambient affine variety Ramsey 2012-08-22T12:58:06Z 2012-08-22T13:49:17Z <p><strong>Edit:</strong> As Jason points out, the following answers the original question, but not the revised question.</p> <p>Let $E$ be an elliptic curve in $\mathbb{P}^2_{\mathbb{C}}$ and $P\in E$ a point of order $2$. The tangent line $L$ to $E$ at $P$ meets $E$ at $P$ and the identity $O$. Now $E\setminus L$ is a divisor in $\mathbb{A}^2_{\mathbb{C}} = \mathbb{P}^2_{\mathbb{C}}\setminus L$ that carries a non-constant invertible function.</p> http://mathoverflow.net/questions/104362/moduli-interpretations-for-modular-curves/104369#104369 Answer by Ramsey for moduli interpretations for modular curves Ramsey 2012-08-09T17:03:17Z 2012-08-09T19:48:55Z <p>Over $\mathbb{C}$, elliptic curves with, say, a point of order $N$ can be identified with the quotient of the upper half plane $\mathbb{H}$ by $\Gamma_1(N)$ just by associating to the class of $\tau\in \mathbb{H}$ the isomorphism class of the air $(\mathbb{C}/(\mathbb{Z} \oplus \mathbb{Z}\tau), 1/N)$. Under this identification, modular forms of level $N$ can be realized as sections of a certain line bundle on "the space" (using the term loosely) of such isomorphism classes that is natural in a sense because is has only to do with this moduli interpretation (roughly speaking, the fiber at each point is a tensor power of the space of differentials on the associated elliptic curve).</p> <p>One can take this observation a lot further to get a good notion of modular forms over base rings other than $\mathbb{C}$ by studying sections of these natural invertible sheaves on modular curves over these more general bases. In particular, one gets a good notion of $p$-adic analytic modular forms by looking at rigid-analytic moduli spaces of elliptic curves.</p> <p>ADDED IN EDIT:</p> <p>Regarding your second question, here is perhaps one reason related to my answer above to "believe" that such a moduli interpretation shouldn't exist. I'm not sure how convincing it is...</p> <p>If there were such an interpretation, then one should be able to mimic the stuff in my answer above to get a geometric notion of modular forms for non-congruence subgroups over a more general class of rings, including, say, $\mathbb{Z}$. Then basic facts from algebraic geometry would kick in and tell you that the Fourier coefficients of such forms should have bounded denominators, as they do for congruence subgroups. This, however, is false for modular forms on non-congruence subgroups. I'm not sure of the history behind these results, but I know that Winnie Li and her collaborators have proven theorems in this area.</p> http://mathoverflow.net/questions/95914/2nd-eigenvalues-for-cusp-forms-for-gamma-04/95933#95933 Answer by Ramsey for 2nd eigenvalues for cusp forms for $\Gamma_0(4)$ Ramsey 2012-05-03T23:25:19Z 2012-05-03T23:25:19Z <p>I'm going to assume that the operator that you refer to as $T_2$ is the one often referred to as $U_2$ in this context: the one that has the effect $$\sum a_nq^n\mapsto \sum a_{2n}q^n$$ on $q$-expansions.</p> <p>For forms of level $\Gamma_0(4)$ and trivial character, this operator has the effect of decreasing the level, so the resulting form is a modular form for $\Gamma_0(2)$ with trivial character. In particular, if you had a newform on $\Gamma_0(4)$ that was an eigenform for $U_2$ with nonzero eigenvalue, then solving the equation $U_2f=\lambda f$ for $f$ would show that your form wasn't new at level $4$ at all.</p> <p>To see this "decreasing the level" statement, my tendency is to go to the geometric description of $U_2$, though there is a much more simple classical description using matrices (this must be in Miyaki's fine book). At the geometric level the $U_2$ operator involves replacing a cyclic subgroup of order $2^n$ of an elliptic curve by the collection of cyclic subgroups of order $2^{n+1}$ that contain it, so one can apply this to a level $2^{n+1}$ form and arrive at one of level $2^n$ in the construction.</p> http://mathoverflow.net/questions/91656/uniqueness-of-modular-functions-with-a-certain-q-expansion/91662#91662 Answer by Ramsey for Uniqueness of modular functions with a certain $q$-expansion Ramsey 2012-03-19T20:17:01Z 2012-03-19T20:17:01Z <p>I <i>would</i> say that the basic fact underlying this is that the only regular functions on a compact Riemann surface are constant functions, but I'm guessing that this isn't the missing detail in your case.</p> <p>A modular function for $\mathrm{SL}_2(\mathbb{Z})$ is (basically by definition in your context) a holomorphic function on the quotient of the upper half-plane by its usual action of $\mathrm{SL}_2(\mathbb{Z})$ by fractional linear transformations. At the end of the day, this amounts to a holomorphic function on the Riemann sphere minus the point $\infty$ (if you haven't done so already, reading about the modular function known as the $j$-invariant will help to explain this). The $q$-expansion of such a function is its local expansion about $\infty$. In particular, if two such functions have $q$-expansions of the type you specify, then their difference is regular everywhere and vanishes at $\infty$, and hence must be the constant zero.</p> http://mathoverflow.net/questions/91077/how-to-prove-that-localisation-preserves-homs/91083#91083 Answer by Ramsey for how to prove that localisation preserves Hom's Ramsey 2012-03-13T14:23:47Z 2012-03-13T14:23:47Z <p>Yes. You can read a proof of this fact at:</p> <p><a href="http://math.stackexchange.com/questions/75812/does-localisation-commute-with-hom-for-finitely-generated-modules" rel="nofollow">http://math.stackexchange.com/questions/75812/does-localisation-commute-with-hom-for-finitely-generated-modules</a></p> http://mathoverflow.net/questions/90772/order-of-vanishing-at-the-cusps-for-the-modular-theta-function/90817#90817 Answer by Ramsey for Order of vanishing at the cusps for the modular theta function Ramsey 2012-03-10T14:10:59Z 2012-03-10T15:41:57Z <p>The usual way to investigate the order of vanishing of a modular form at a cusp other than $\infty$ is to find an element of $\mathrm{SL}_2(\mathbb{Z})$ that maps $\infty$ to your cusp and "recenter" your form at $\infty$ using this element. If your element is $\gamma$, then look at $j(\gamma,z)^{-1}f(\gamma z)$ (or something like this...) where $j$ is the cocyle by which the form $f$ transforms. With any luck you can work out a formula for this from which the behavior at $\infty$ (and hence the behavior of $f$ at your chosen cusp) is evident.</p> <p>For $\theta$ the easiest way to do this is to look at a table of "theta functions with characteristics" (like the one in Mumford's <i>Tata Lectures on Theta</i>, Volume I if I recall correctly). The result is that the $q$-expansions of $\theta$ at $\infty$, $0$, and $1/2$ are, respectively, $\sum q^{n^2}$, $\sum (q^{1/4})^{n^2}$, and $q^{1/4}\sum q^{n^2+n}$, where the sums are over $\mathbb{Z}$ and the latter two are really only defined up to some constant (perhaps just a fourth root of unity).</p> <p>Now each cusp on a modular curve comes with a "width" $h$ - its ramification index over $X(1)$, and usually the $q$-expansion of a modular form at a cusp is an expansion in $q^{1/h}$. In this case the cusp $1/2$ has width $1$, and clearly something else is going on here. It appears that $\theta$ vanishes to order $1/4$ at this cusp! </p> <p>While this turns out to be an oddly useful perspective (at least it has to me), it's nonsense on its face. To remedy this you need to be careful about setting up $\theta$ as a section of some complex-analytic line bundle on $X_0(4)$. As usual, this thing is just cooked up using the cocyle by which $\theta$ transforms and when you work out the local picture around the cusp $1/2$ you'll see that the section $\theta$ vanishes to order $1$ there as a section of this bundle.</p> <p>To read more about the goofy phenomenon at $1/2$ search around for "irregular cusp." The underlying reason for the problem is that, from a moduli point of view, the object classified by this cusp has non-trivial automorphisms (so complex-analytically $X_0(4)$ is better thought of as an orbifold). On the other hand, one can see this disparity on the power of $q$ needed to expand a modular form very explicitly in the integral weight case by messing with local calculations for the sheaf $\omega^{\otimes k}$ around the cusp. This was done very nicely in some early draft of Brian Conrad's book on the Ramanujan-Petersson conjectures. I don't know what the status of that book is these days though...</p> http://mathoverflow.net/questions/79428/zariski-closures-of-one-parameter-additive-maps-in-positive-characteristic/79433#79433 Answer by Ramsey for Zariski closures of one parameter additive maps in positive characteristic Ramsey 2011-10-28T22:12:09Z 2011-10-28T22:12:09Z <p>I'd like to understand the motivation for this question to see if I'm barking up the wrong tree here, but I think it's false (the later one-dimensional verbiage, that is) as stated.</p> <p>Let $k$ be an algebraically closed field of of characteristic $p$ and let $K$ be the function field in the single variable $t$ over $k$. So $K$ is an imperfect field of characteristic $p$.</p> <p>Consider the map $K\to K^2$ defined by $f(t)\mapsto (f(t),f(t+1))$. It's component functions are additive.</p> <p>Certainly, this map isn't surjective. On the other hand, take a polynomial $P(X,Y)\in K[X,Y]$ vanishing on its image. Clearing denominators from $K$, we may regard $P$ as a polynomial $P(t,X,Y)\in k[t,X,Y]$ that has the property that $P(t,f(t),f(t+1))$ is the zero rational function in $t$ for all $f(t)\in K$. But, given any triple $(a,b,c)\in k^3$ it is easy to find (even a linear polynomial) $f(t)$ such that $f(a)=b$ and $f(a+1)=c$. It follows that $P(t,X,Y)$ is the zero polynomial (since $k$ is algebraically closed).</p> <p>Thus there is non-zero $P$ vanishing on the image, which means that the Zariski closure of the image is all of $K^2$.</p> <p>This seems to have nothing to do with the characteristic. The argument (if non-bogus) works fine for any algebraically closed $k$. I think the issue is that "additive" is an odd condition here from the point of view of algebraic geometry.</p> http://mathoverflow.net/questions/77180/motivation-behind-defining-the-ramification-divisor/77228#77228 Answer by Ramsey for Motivation behind defining the Ramification Divisor Ramsey 2011-10-05T12:22:45Z 2011-10-05T12:22:45Z <p>A few answers:</p> <ul> <li>As the comments mention, the "-1" is certainly needed to get a divisor in the first place, since $\nu_p$ is usually equal to 1 and exceeds 1 at the ramification points. Thus, the support of the ramification divisor as you define it is precisely the ramification locus.</li> <li>Ramification is a local phenomenon, so compactness is totally irrelevant.</li> <li>A meromorphic function on a Riemann surface $X$ can be interpreted as a map $X\to \mathbb{P}^1$ (the poles go to $\infty\in \mathbb{P}^1$ with ramification index equal to the degree of the pole).</li> <li>Here's how I think of/recall the Riemann-Hurwitz formula for $f:X\to X'$: Imagine that you have triangulated $X'$ such that all ramification points (or the images thereof if you think of them on $X$) occur at vertices. Now consider the "pullback" of this triangulation to $X$ (look a the preimages of the faces, edges, and vertices). If you compute the Euler characteristic of $X$ using this pullback triangulation you will see that it differs from the degree of $f$ times the Euler characteristic of $X'$ (computed using the original triangulation) exactly by the degree of your ramification divisor, and the Riemann-Hurwitz formula drops out!</li> </ul> http://mathoverflow.net/questions/74410/what-is-the-nature-of-the-locus-in-the-eigencurve-associated-to-some-conditions-o What is the nature of the locus in the eigencurve associated to some conditions on the associated automorphic representation (at $p$)? Ramsey 2011-09-03T00:47:16Z 2011-09-03T22:37:14Z <p>I've chatted informally with some folks about this question before and gotten some very nice insights, but I thought I'd toss it out to a wider audience because it is a continuing curiosity of mine.</p> <p>Roughly, here's what I have in mind: Let $\mathcal{E}$ denote the eigencurve of some tame level $N$. At classical points, one has attached to the associated form an automorphic representation $\pi = \otimes \pi_\ell$. For $\ell\neq p$, there are results (though I've forgotten the author(s) and couldn't dig up the reference in the course of posting this) about the loci in the eigencurve corresponding to various conditions on the $\pi_\ell$. By conditions here I'm rather open-minded - things such as special, supercuspidal, bounds on conductor, etc., are all of interest to me.</p> <p>My question is: What is known about the analogous questions for $\ell=p$?</p> http://mathoverflow.net/questions/71988/left-and-right-representations/72066#72066 Answer by Ramsey for Left and right representations Ramsey 2011-08-04T05:52:17Z 2011-08-04T15:02:38Z <p>As has been pointed out in comments, any particular representation of a group can be realized as a representation with the "opposite handedness," but of another group, namely the opposite group. As has also been mentioned, every group is canonically isomorphic to its opposite group via the inversion map. One might be inclined to combine these two comments to conclude that the distinction between left and right representations is moot. </p> <p>In my estimation: sometimes it is, and sometimes it isn't.</p> <p>For example:</p> <p>(1) When one defines the induction of a representation from a subgroup, one starts with a certain space of functions that transforms is a specified fashion on one side (depending on the original representation). The canonical representation of the original group (of the original handedness) on this space is on the other side, or equivalently, is a representation of the opposite group. However, we usually just trade it off for the same handedness representation of the same group through the inversion identification.</p> <p>(2) On the other hand, sometimes one naturally is led to deal with both kinds of actions at once - like when studying functions (better: forms of some sort) on double coset spaces of a pair of subgroups in a group, and it can be useful to keep the right and left actions as such. Natural notation such as $H\setminus G/ K$ and the bi-chiral transformation formulas play more nicely together this way, for example.</p> <p>I think that, in both cases the choice comes down to a convenience of one sort or another.</p> <p>For that matter, I suppose I feel like this convenience extends to the covariant/contravariant commentary here. One can trade off one for the other by trading off a category for its opposite. This can certainly make a number of statements/constructions more awkward, but again this seems a matter of convenience.</p> <p>All this aside, there is a lot to be said for convenience...</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71726#71726 Answer by Ramsey for What are your favorite instructional counterexamples? Ramsey 2011-07-31T15:12:44Z 2011-07-31T15:12:44Z <p>A standard result in introductory calculus classes is that, if a function has positive derivative on an open interval, then it's increasing there.</p> <p>Based on this, students tend to think that, if $f'(a)>0$, then $f$ must be increasing "near $a$."</p> <p>However, the example $f(x) = 2x^2\sin(1/x)+x$ (set $f(0)=0$) shows that this is quite false!</p> http://mathoverflow.net/questions/71435/which-conjectures-only-need-the-grand-riemann-hypothesis-to-become-genuine-theore/71454#71454 Answer by Ramsey for Which conjectures only need the Grand Riemann Hypothesis to become genuine theorems? Ramsey 2011-07-28T01:54:14Z 2011-07-28T13:27:59Z <p>I like the phrase "only need the grand Riemann hypothesis"...</p> <p>One of my favorite results known contingent on this result (rather, the weaker generalized Riemann hypothesis), is that the ring of integers in a number field (<strong>EDIT</strong>: with infinite unit group) is Euclidean with respect to <em>some</em> Euclidean algorithm if an only if is is a PID. Interestingly, the "amount" of GRH needed here far exceeds that of the field in question. One must assume GRH for an infinite number of extension fields as well.</p> http://mathoverflow.net/questions/70140/applications-of-full-integral-weight-modular-forms-in-elementary-number-theory/70180#70180 Answer by Ramsey for Applications of full integral weight modular forms in elementary number theory Ramsey 2011-07-12T21:29:03Z 2011-07-12T21:29:03Z <p>Perhaps this is "out of bounds" given the phrasing of the question, but those Eisenstein series you mention don't just have divisor sums as coefficients - the constant term is a special value of the Riemann zeta function.</p> <p>This implies all sorts of neat stuff. The various relations between divisor sums that you mention come with relations between zeta values. These give very nice congruences, in particular. </p> <p>You can take this to reasoning pretty far to deduce things like $p$-adic interpolation of zeta values a la Kubota-Leopoldt from the much simpler interpolation properties of the divisor sum functions. This was done by Serre in the 1973 paper ("Formes modulaires et fontiones zeta $p$-adiques") that gave birth to the theory of $p$-adic modular forms.</p> http://mathoverflow.net/questions/69278/hartshornes-associated-scheme-for-a-variety/69285#69285 Answer by Ramsey for Hartshorne's associated scheme for a variety Ramsey 2011-07-01T18:22:41Z 2011-07-01T18:22:41Z <p>To show that $(t(V),\alpha_*\mathcal{O}(V))$ is a scheme, you must show that $t(V)$ has an open cover on which this ringed space is isomorphic to an affine scheme.</p> <p>Take an affine open cover $\{U_i\}$ of $V$. Since you believe the affine case, it suffices to show that $\{t(U_i)\}$ is an open cover of $t(V)$, and</p> <p><code>$(t(V),\alpha_*\mathcal{O}(V))|_{t(U_i)} \cong (t(U_i),\alpha_*\mathcal{O}(U_i))$</code></p> <p>for each $i$. Given your last paragraph, it sounds like the first of these points is your difficulty. Let $Y$ be a nonempty irreducible closed subset $Y\subseteq U_i$. For each $j$, $Y\cap U_j$ is (when nonempty) a nonempty irreducible closed subset of $U_i\cap U_j$ (since an open subset of an irreducible is irreducible). The intersection $U_i\cap U_j$ is an affine open subset of $U_j$, and it's not hard to see (look at the pre-image of the corresponding prime ideals!) that $Y\cap U_j$ extends in a natural way to an irreducible closed subset of $U_j$. These extensions glue for varying $j$ to give an irreducible closed subset of $V$, since a locally irreducible subset of a (connected) space is irreducible. This furnishes the map $t(U_i)\to t(V)$ (which, in particular, I think addresses the issue you raise in the last paragraph). </p> <p>It remains to see that this is an open subset and gives an open cover of $t(V)$, and to prove the above isomorphism. So now try from here...</p> http://mathoverflow.net/questions/69064/do-coarse-moduli-spaces-respect-galois-actions/69071#69071 Answer by Ramsey for Do coarse moduli spaces respect Galois actions? Ramsey 2011-06-29T00:01:54Z 2011-06-29T00:01:54Z <p>If $X$ is the coarse moduli scheme associated to a functor $F$ on schemes, then in particular, there is a natural transformation $F\to h_X$, where $h_X$ is the functor of points of $X$. </p> <p>Unless I am missing something, if you apply the naturality of this transformation to the map $\mathrm{Spec}(\mathbb{C}) \to \mathrm{Spec}(\mathbb{C})$ associated to your automorphism $\sigma$, this implies that the two actions coincide in your case, since the associated map $F(\mathrm{Spec}(\mathbb{C})) \to F(\mathrm{Spec}(\mathbb{C}))$ is given by the fiber product of the curve, and the associated map $X(\mathrm{Spec}(\mathbb{C})) \to X(\mathrm{Spec}(\mathbb{C}))$ is just the action on the points. I wish I could make the obvious 2-by-2 commutative diagram...</p> http://mathoverflow.net/questions/69062/binomial-coefficients-and-derivatives-of-modular-forms/69069#69069 Answer by Ramsey for Binomial coefficients and derivatives of modular forms Ramsey 2011-06-28T23:33:29Z 2011-06-28T23:33:29Z <p>This isn't really an answer, but a long comment with a bit of LaTeX that thought would render poorly in the comment box.</p> <p>The following fact may be lurking in the background here: While the derivative of a modular form of weight $k$ is not generally modular, the map $D$ on modular forms of weight $k$ defined by </p> <p>$$D(f) = q\frac{df}{dq} - \frac{k}{12}f\cdot E_2$$ actually <em>does</em> preserve modularity.</p> <p>I think this is sometimes called the Halperin-Fricke operator or something like this. It's also a derivation, for what it's worth. It certainly directly explains the first of your equations above, and I wonder if some cleverness iterating it would yield your more general observations.</p> http://mathoverflow.net/questions/65850/are-there-analytic-p-adic-modular-forms/65852#65852 Answer by Ramsey for Are there 'analytic' $p$-adic modular forms. Ramsey 2011-05-24T14:25:48Z 2011-05-24T14:25:48Z <p>There is such a theory, but the analytic object that the forms live on is an analytified modular curve, not simply $\mathbb{C}_p$ (though there is a "$p$-adic upper-half plane" that can be used to uniformize some similar moduli spaces, but as far as I know not the usual modular curves).</p> <p>Basically, if $f$ is a classical modular form of some weight $k$, $f$ can be realized as a section of a sheaf $\omega^{\otimes k}$ on a complex-analytic modular curve such as $X_1(N)$ obtained via quotient from the complex upper-half plane. These curves and sheaves have algebraic models defined over $\mathbb{Q}$, and (under mild hypotheses) the form $f$ actually arises from a section of the associated sheaf $\omega^{\otimes k}$ on the modular curve defined over $\mathbb{Q}$ (or some finite extension).</p> <p>Now you can go in another direction and consider the rigid-analytic space over $\mathbb{Q}_p$ associated to the smooth algebraic curve $X_1(N)$ and your form $f$ gives rise to a $p$-adic analytic object on this curve. Now one can play games like considering subspaces obtained by removing disks around supersingular points to obtain general $p$-adic modular forms (such as limits of classical forms of varying weight) and overconvergent modular forms.</p> <p>For what it's worth, these curves are $p$-adic analytic moduli spaces, and this point of view on $p$-adic modular forms essentially differs from Katz's by thinking about rigid spaces as opposed to formal schemes (the Raynaud point of view on $p$-adic analytic geometry).</p> <p>Look at the papers of Coleman (such as his $p$-adic Banach spaces paper or the eigencurve paper with Mazur) if you would like to read more on this point of view.</p> http://mathoverflow.net/questions/64833/semisimplicity-of-p-adic-galois-representations/64842#64842 Answer by Ramsey for semisimplicity of p-adic Galois representations Ramsey 2011-05-12T21:40:21Z 2011-05-12T21:40:21Z <p>Regarding the natural related question: looking for a $p$-adic valued Haar measure on $p$-adic stuff is generally doomed to failure. The problem is that small sets tend to have large $p$-power index and hence large $p$-adic measure. To see this explicitly in the simplest case, look at the description of the ring of distributions on $\mathbb{Z}_p$ given by Amice in terms of power series. Using this description, it's easy to verify that there is no nontrivial translation-invariant $p$-adic valued distribution on $\mathbb{Z}_p$. </p> http://mathoverflow.net/questions/63258/seemingly-emergent-structures-in-mathematics Seemingly emergent structures in mathematics Ramsey 2011-04-28T05:52:54Z 2011-04-28T09:59:18Z <p>I rather suspect that this must have come up here on MO already, but my handful of searches didn't turn up the thread, so...</p> <p>I'm curious about examples of mathematical structure that seems to arise "from nothing." The example that motivates this is one that I was teaching today, namely, the central limit theorem. </p> <p>I was trying to convey to my (business math) students how astounding it is that the sampling distributions of the mean of a uniformly distributed random variable approach a normal distribution as the sample size increases. </p> <p>Out of complete randomness, very specific and rather subtle structure arises (if in the limit). </p> <p>I'd be amused to see other examples of this perceived phenomenon in different areas of mathematics. Not just structure where it wasn't expected (which is quite cool, but ubiquitous), but structure that seems to "arise from a vacuum."</p> http://mathoverflow.net/questions/62429/a-small-doubt-about-jargon/62434#62434 Answer by Ramsey for a small doubt about jargon Ramsey 2011-04-20T16:33:58Z 2011-04-20T16:33:58Z <p>A Dirichlet character $\chi$ is a character of the group $(\mathbb{Z}/n\mathbb{Z})^\times$ for some positive integer $n$. This group is canonically isomorphic to the Galois group $Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$. The kernel of $\chi$ is a subgroup of this Galois group whose fixed field $K$ is by definition the subfield of $\mathbb{Q}(\mu_n)$ "cut out" by $\chi$.</p> <p>In particular, $\chi$ identifies $Gal(K/\mathbb{Q})$ with the image of $\chi$ in $\mathbb{C}^\times$ (if $\chi$ is taken to be complex-valued) which is cyclic, so $K$ is a cyclic extension of $\mathbb{Q}$.</p> http://mathoverflow.net/questions/62193/what-do-poles-of-differentials-on-a-curve-mean/62269#62269 Answer by Ramsey for What do poles of differentials on a curve mean? Ramsey 2011-04-19T13:47:52Z 2011-04-19T19:30:26Z <p>I think of them pretty simply as differential forms with zeros in the denominator of the "coefficient function" upon choosing a local uniformizing parameter (which is really just the definition).</p> <p><strong>Edit: The following isn't quite right, as pointed out in the comments. See below for attempt at fixing it.</strong></p> <p><strong><em>_</em>____</strong></p> <p>However, if you want something more akin to your "map to $\mathbb{P}^1$" description, you might try something like this: to the invertible sheaf $\Omega$ you can associate a projective bundle $\mathbb{P}(\Omega)$ equipped with a map $\pi:\mathbb{P}(\Omega)\to C$ whose fibers are the projectivizations of the fibers of $\Omega$. Then a meromorphic differential form should correspond to a section $s$ to $\pi$ and the poles of the form are the points where $s(x)=\infty$.</p> <p>I can't say that I've ever seen exactly this written down, but it seems quite reasonable...</p> <p><strong><em>_</em>____</strong></p> <p>Perhaps one should consider the projectivization of the bundle $\Omega\oplus\mathcal{O}$ instead. Clearly if I take a regular differential form $\omega$, it gives rise to a section of $\mathbb{P}(\Omega\oplus \mathcal{O})$ by considering the image of $\omega\oplus 1$ in the projectivization. </p> <p>Arguing locally, it seems that if I take a meromorphic differential form of the form $t^{-n}udt$ where $u$ is a local unit, I can associate to it the image of $udt\oplus t^n$ in the projectivization. Now glue over the curve to associate a section of this projective bundle to your chosen meromorphic differential form. If everything glues without incident, it seems that the resulting section should have the desired property that the poles are the points mapping to $\infty$ in the fiber, just by construction. </p> <p>Hopefully this makes more sense than my first attempt :)</p> http://mathoverflow.net/questions/61632/what-is-the-high-concept-explanation-on-why-real-numbers-are-useful-in-number-the/61651#61651 Answer by Ramsey for What is the high-concept explanation on why real numbers are useful in number theory? Ramsey 2011-04-14T05:15:45Z 2011-04-14T05:15:45Z <p>To continue on a theme of a couple answers thus far:</p> <p>The real numbers, along with the $p$-adics for each rational prime $p$ are the completions of the rational numbers. Regarding arithmetic objects (such as quadratic forms or algebraic varieties in general, among others) defined over $\mathbb{Q}$ as being defined over these completions endows them with additional structure that one can exploit.</p> <p>This "exploitation" comes in various forms, including the Hasse principle mentioned in other posts (points over $\mathbb{Q}$ beget points over completions and the latter are some kind of evidence for the former).</p> <p>Then there is also the Birch and Swinnerton-Dyer (BSD) conjecture for elliptic curves and its various generalizations. These purport to relate global ("over $\mathbb{Q}$") things like values of a global $L$-function to a host of data that is germane to the various local completions of $\mathbb{Q}$. The period that turns up in the BSD conjecture, for example, is really a facet of the elliptic curve regarded over the reals where one can integrate forms in the usual sense. The Tamagawa numbers, on the other hand, come from the various $p$-adic completions of $\mathbb{Q}$.</p> <p>I guess the theme in all of this is that global fields lack structure that their various completions possess, and embedding them in these completions gives extra insight and "handles" to grab onto in order to study the global field and the objects defined over it.</p> http://mathoverflow.net/questions/59498/bounded-denominators-for-modular-forms/59501#59501 Answer by Ramsey for Bounded denominators for modular forms Ramsey 2011-03-24T22:48:23Z 2011-03-24T22:48:23Z <p>One direction is known. A modular form for a congruence subgroup has bounded denominators. </p> <p>This is because such a form can be interpreted as a section of a natural line bundle on a moduli space of elliptic curves with "level structure." Equivalently, this form can be regarded as a rule that assigns to each relative elliptic curve over a ring $A$ with level structure an element in an a certain $A$-module. The $q$-expansion of such a form arises by considering the Tate elliptic curve and its level structures. Roughly speaking (i.e. ignoring the level structure and the bad fiber), these are defined over the ring $\mathbb{Z}[[q]]$. The result is that a modular form for a congruence subgroup that has rational coefficients moreover has a $q$-expansion in $\mathbb{Z}[[q]]\otimes\mathbb{Q}$, and hence has bounded denominators ($\mathbb{Z}[[q]]\otimes \mathbb{Q}$ being much smaller than $\mathbb{Q}[[q]]$). </p> http://mathoverflow.net/questions/59290/why-is-p2-special-if-we-want-to-classify-cplx-representation-of-gl2zp/59301#59301 Answer by Ramsey for Why is p=2 special, if we want to classify cplx. representation of GL2(Zp)? Ramsey 2011-03-23T13:49:28Z 2011-03-23T13:49:28Z <p>Shalika constructs a bunch of representations of $p$-adic $SL_2$ using the Weil representation. In order to get genuine (non-projective) representations he needs to work with an even-dimensional vector space in this construction, so one is naturally led to look at quadratic extensions of the $p$-adic field. If the residue characteristic $p$ is not $2$, then these are quite simple to understand (there are three of the them, they're easy to write down in a pretty uniform way independent of the field, etc.). If $p=2$ there can be gobs of quadratic extensions (if I recall correctly the number grows exponentially with the degree over $\mathbb{Q}_2$), so the situation is more complicated.</p> <p>I don't know if this is really a direct obstruction to the construction of these representations, but I do know that Shalika and Sally did many calculations with the explicit realizations of these representations associated to the three easy-to-write-down quadratic extensions when $p\neq 2$. In the handful of these that I've looked at or carried out, the analysis is greatly simplified by the assumption that $p\neq 2$.</p> http://mathoverflow.net/questions/122600/understanding-adjointness-of-sheaves-in-algebraic-geometry/122637#122637 Comment by Ramsey Ramsey 2013-02-22T15:19:32Z 2013-02-22T15:19:32Z Oy. I was thinking about finite maps. I'm going to edit. http://mathoverflow.net/questions/20283/how-do-you-pronounce-hartshorne/20284#20284 Comment by Ramsey Ramsey 2013-02-08T20:37:45Z 2013-02-08T20:37:45Z I once heard somebody quip that the man's name is pronounced &quot;Hart's Horn&quot; but the book is pronounced &quot;Hart Shorn.&quot; http://mathoverflow.net/questions/120215/what-are-the-interesting-quartic-number-fields-for-calculation-of-torsion-groups Comment by Ramsey Ramsey 2013-01-29T14:45:37Z 2013-01-29T14:45:37Z I'm guessing this will get the &quot;not a real question&quot; hammer, but you might try to do yourself and your question a favor by first asking something more pointed. Saying what constitutes &quot;interesting&quot; to you here would be a start. http://mathoverflow.net/questions/119825/constructible-topology-on-schemes Comment by Ramsey Ramsey 2013-01-25T15:01:02Z 2013-01-25T15:01:02Z I don't know if this question will survive, but I'll admit that, as a non-expert who's never really had occasion to work with constructible sets, this is something I've been idly curious about on a handful of occasions. I'd like to hear the short version of why they're useful and maybe see a quick example or two. http://mathoverflow.net/questions/119202/variant-of-leopoldts-conjecture Comment by Ramsey Ramsey 2013-01-18T15:43:04Z 2013-01-18T15:43:04Z I think what David meant was that CM fields are characterized by the property that, under any embedding into the complex numbers, the image is preserved by complex conjugation and the resulting involution on the field is independent of the embedding. http://mathoverflow.net/questions/119202/variant-of-leopoldts-conjecture Comment by Ramsey Ramsey 2013-01-17T20:53:53Z 2013-01-17T20:53:53Z Curious: have you checked in any examples? http://mathoverflow.net/questions/118399/moonshine-and-the-riemann-zeta-hypothesis-or-function Comment by Ramsey Ramsey 2013-01-08T22:06:36Z 2013-01-08T22:06:36Z I don't think that the Riemann Hypothesis was inspired by physics. The statement, along with a handful of other (now proven) conjectures, occurs in Riemann's original paper on the distribution of primes. http://mathoverflow.net/questions/118201/simpler-way-to-define-modular-forms/118212#118212 Comment by Ramsey Ramsey 2013-01-06T17:53:28Z 2013-01-06T17:53:28Z Also - I think that constant functions are perfectly good modular forms of weight zero. Of course, the only cuspidal one is the constant zero. http://mathoverflow.net/questions/118201/simpler-way-to-define-modular-forms/118212#118212 Comment by Ramsey Ramsey 2013-01-06T17:49:37Z 2013-01-06T17:49:37Z I would agree with this if the OP wanted to define cusp forms, but it appears that (s)he wants all holomorphic modular forms - including the Eisenstein series. http://mathoverflow.net/questions/118201/simpler-way-to-define-modular-forms/118207#118207 Comment by Ramsey Ramsey 2013-01-06T16:31:19Z 2013-01-06T16:31:19Z I don't think that I understand this. What is meant by &quot;the true local coordinate at infinity&quot;? As sections of powers of the usual sheaf $\omega$, Eisenstein series don't have poles at the cusps. Are you perhaps thinking of modular forms as differentials on the modular curve itself? The Kodaira-Spencer isomorphism identifies such regular differentials with weight $2$ cusp forms, and more generally weight two forms with differentials with at worst simple poles at the cusps. http://mathoverflow.net/questions/117904/elementary-examples-of-the-weil-conjectures Comment by Ramsey Ramsey 2013-01-02T21:40:04Z 2013-01-02T21:40:04Z I'm not sure if this is too elementary, but the case of elliptic curves can be worked out pretty concretely. Silverman does this in his first book on elliptic curves, for example. http://mathoverflow.net/questions/113898/david-hilbert-on-complex-multiplication Comment by Ramsey Ramsey 2012-11-20T03:46:31Z 2012-11-20T03:46:31Z You could always ask Milne. My impression is that he's quite receptive such queries. http://mathoverflow.net/questions/113811/do-l-functions-exist-for-half-integral-weight-modular-forms/113814#113814 Comment by Ramsey Ramsey 2012-11-19T13:51:25Z 2012-11-19T13:51:25Z Just to emphasize something touched on in the last sentence here: I've always thought of the natural &quot;$L$-function type object&quot; of a half-integral weight form as the entire family of quadratic twists of its Shimura image form. I don't know enough about the metaplectic/automorphic picture to understand if there's something intrinsic about this point of view, but I've been idly curious about this. For that matter: is there a Langlands correspondence for metaplectic groups? Is the Galois side just the family of quadratic twists of the Galois representation of the Shimura lifting in this case? http://mathoverflow.net/questions/109227/how-to-prove-this-equation Comment by Ramsey Ramsey 2012-10-09T14:34:19Z 2012-10-09T14:34:19Z I'm guessing that this question won't last long in its current form, but let me point out that you at very least need to add some hypotheses. If the scope of $K$ is &quot;compact topological abelian group&quot; then your statement is false. http://mathoverflow.net/questions/109213/uniqueness-of-analytic-continuation-in-rigid-analytic-geometry/109221#109221 Comment by Ramsey Ramsey 2012-10-09T13:20:45Z 2012-10-09T13:20:45Z Brian Conrad's papers a re a great resource for these things. <i>Irreducible Components of Rigid Spaces</i> is fantastic. His paper (preprint at present, I suppose) on Moishezon spaces has a bunch of stuff about meromorphic functions in the rigid context.