User olivier - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:05:55Z http://mathoverflow.net/feeds/user/2284 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127769/applications-of-govorov-lazard-theorem/127806#127806 Answer by Olivier for Applications of Govorov-Lazard Theorem? Olivier 2013-04-17T08:37:11Z 2013-04-18T14:33:29Z <p>Interesting is always in the eye of the beholder but this theorem plays a crucial role in the proof of the proper base change theorem in SGA4 Exposé XVII (by P.Deligne). One wishes to prove an isomorphism between complexes and the proof is pure EGA/SGA reduction style: first reduce to complexes concentrated in a single degree, then to modules over $\mathbb{Z}/n\mathbb Z$, then to flat modules and then (by the theorem of Govorov-Lazard, which I knew solely as a theorem of Lazard) to free modules by the commutativity of $H^{i}(X_{s},-)$ with direct limits. For free modules, the result is obvious. The moral of the story to me as always been that whenever a functor commutes with direct limits, then it is enough to consider free modules if one wishes to prove something about flat modules. </p> http://mathoverflow.net/questions/124595/automorphism-group-of-regular-graph/124597#124597 Answer by Olivier for Automorphism group of regular graph Olivier 2013-03-15T08:37:58Z 2013-03-15T08:37:58Z <p>Under these weak hypotheses, the answer is: it could be anything. The trivial group is possible, as well as $\mathfrak S_{n}$ and essentially any group in between by Frucht's theorem (which realizes any group as the automorphism group of a regular graph). Of course, there are trivial obstructions on $k$ and $n$ for some groups to appear (for instance $\mathfrak S_{n}$ of course appears only if $k=0$ or $k=n-1$) but I don't see we can say much more beyond trivialities in the generality you are considering. </p> http://mathoverflow.net/questions/123081/should-one-attack-hard-problems/123089#123089 Answer by Olivier for Should one attack hard problems? Olivier 2013-02-27T10:28:06Z 2013-02-27T10:28:06Z <p>I think Markus Redeker's answer captures the essential point. If the problem is hard and famous (at least in the relevant sub-field), so <em>a fortiori</em> for a problem like P≠NP, I would add the further restriction then you should consider attacking it only if that new idea you have allows you to solve an easy or average (but still new) related problem or at the very least allows you to reprove in a completely different way a known result. If this works, then 1) you now know that this new idea is not completely crazy or just a variant of an old one 2) you have a worthy PhD. 3) you can think about making math history. In fact, if you skim through math history, recent or otherwise, you will see that because of point 1) (testing that you idea is indeed new and worth pursuing) many historical breakthroughs were preceded by an easy (or at least much easier) variant relying on similar techniques. </p> http://mathoverflow.net/questions/74707/mathematical-habits-of-thought-and-action-which-would-be-of-use-to-non-mathematic/74847#74847 Answer by Olivier for Mathematical habits of thought and action which would be of use to non-mathematicians Olivier 2011-09-08T08:52:17Z 2013-02-27T08:21:26Z <p>Keep in mind that it is easy to make mistakes.</p> <p>The most striking thing I learned from doing mathematics is that even in an environment entirely devoid of ambiguities and <em>characterized</em> by precise axiomatic constraints to the point that it became synonymous with it, even when I am doing my absolute best to be completely careful and precise, even when I double check each of my words, then show it to two careful colleagues, then let it simmer for a while, then go through it again with a critical eye, then show it to an authority in the field, then re-read it again; even after this excruciating process of constant self-examination, even after the strength of my arguments has confounded (perhaps in the two meanings of the word) my utmost critical self as well as the objections of several knowledgeable observers, I know that dozens of mistakes, inaccuracies and outright errors still remain.</p> <p>Doing math is certainly not the only way to come to this bitter conclusion-simply interacting with people is usually enough, as Philipp Roth once famously remarked-yet I can't help to shudder when I sometimes contemplate how many things I <em>must</em> be getting completely and obviously wrong whenever I am outside my tiny bubble of professional rigour, where a prompt and witty remark is more than often enough to obtain general assent.</p> http://mathoverflow.net/questions/122795/unramified-base-change-in-characteristic-p-0/122867#122867 Answer by Olivier for unramified base change in characteristic p > 0? Olivier 2013-02-25T09:39:08Z 2013-02-25T14:59:54Z <p>Maybe I should reproduce my comment as an answer, in order for MO not to treat the question has unanswered.</p> <p>The full local Langlands correspondence is a theorem of G.Laumon, M.Rappoport and U.Stuhler in $\mathcal D$-elliptic sheaves and the Langlands correspondence (Invent. Math. 113).</p> <p>Here is the Mathscinet review:</p> <p><a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=110755&amp;vfpref=html&amp;r=29&amp;mx-pid=1228127" rel="nofollow">http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=110755&amp;vfpref=html&amp;r=29&amp;mx-pid=1228127</a></p> <p>The result for $n=2$ is due to Deligne. </p> http://mathoverflow.net/questions/121688/definition-of-cm-modular-form/121693#121693 Answer by Olivier for Definition of CM modular form Olivier 2013-02-13T09:29:56Z 2013-02-14T08:49:41Z <p>Let $f$ be a newform of level $N$ and weight $k\geq 2$. We say $f$ has CM by the quadratic field $K$ is there exists a quadratic extension $K/\mathbb Q$ such that if $\eta_{K/\mathbb Q}$ is the quadratic character whose kernel is $G_{K}$ then the automorphic representation $\pi(f)$ of $\operatorname{GL}(2,\mathbb A_{\mathbb Q})$ is isomorphic to $\pi(f)\otimes\eta_{K/\mathbb Q}$. If this is true, then $K$ has to be an imaginary quadratic extension. More generally, if $F$ is a totally real field and $\pi$ is an automorphic representation (EDIT: as wccanard points out, here again the condition that the weight should be greater than $2$ has to be included) of $\operatorname{GL}(2,\mathbb A_{F})$ isomorphic to $\pi\otimes\eta_{K/F}$ for $K/F$ quadratic then $K$ is a CM extension (a totally imaginary quadratic extension of $F$).</p> <p>As Marc Palm writes, when $f$ has CM by $K$ there exists a character $\chi$ of $\mathbb A_{K}^{\times}/K^{\times}$ such that for all finite place $v$, the $L$-factor $L_{v}(f,s)$ of $f$ is equal to the product $\underset{w|v}{\prod}L_{w}(\chi,s)$ of $L$-factors of $\chi$ over places of $K$ above $v$. A highbrow version of this last statement is that $\pi(f)$ is isomorphic to the automorphic induction of $\chi$ from $K$ to $F$.</p> http://mathoverflow.net/questions/120913/periods-for-2-variable-p-adic-l-functions/120954#120954 Answer by Olivier for Periods for 2-variable p-adic L-functions Olivier 2013-02-06T08:43:56Z 2013-02-06T08:43:56Z <p>Assuming you are writing about Mok's Compositio 2009 article, the answer is easy: it's a question of quantifier ordering. There are two statements which you could call the interpolation property for a several variables $p$-adic $L$-function $L_{p}$ (which I assume to be living in $R[[X]]$ where $R$ is some ring and $X$ is the cyclotomic variable). If $x$ is a classical point of $R$, I denote by $L_{p}^{cyc}(V_{x},-)$ the usual cyclotomic $p$-adic $L$-function of the Galois representation $V_{x}$. </p> <p>(1) For all classical points of $R$, there exists a period $\Omega_{x}$ such that for all character $\chi$, $L_{p}(x,\chi)=L_{p}(V_{x},\chi)/\Omega_{x}$.</p> <p>(2) There exists $\Omega$ in $R$ such that for all classical points $x$ of $R$ and for all character $\chi$, $L_{p}(x,\chi)=L_{p}(V_{x},\chi)/\Omega_{x}$.</p> <p>Type (1) results guarantee interpolation only locally at a classical point and thus typically require only the rather weak condition of non-vanishing of $L$-values. On the other hand, they provide only local informations, so they are particularly suitable for problems which are local at a classical point: the main example being the problem of trivial zeroes. This is why type (1) is what Greenberg-Stevens and Mok construct.</p> <p>Type (2) results are much stronger and much more precise but they typically require $p$-adic interpolation of comparison theorem between Betti and De Rham cohomology or De Rham and étale cohomology. Under current technology, the proofs of these theorems relies on techniques formally similar to the complete intersection and freeness criterion of Taylor-Wiles and Fujiwara so tend to require comparable hypotheses. It seems likely to me that Kisin's amelioration of these techniques could have a bearing on these questions but I am not sure how this could be done.</p> <p>You could also read the recent works of T.Ochiai (appeared in Documenta) and M.Dimitrov (to appear in American Journal of Math) on the topic: they contain very lucid explanations of the relevant points (both easily available online). </p> http://mathoverflow.net/questions/110023/level-lowering-for-weight-1-forms/110030#110030 Answer by Olivier for Level lowering for weight 1 forms Olivier 2012-10-18T18:03:42Z 2012-10-18T18:03:42Z <p>Everything you wish for is true for modular forms over $\mathbb Q$ even at $p=2$, as it follows from refined forms of Serre's conjecture; here I am assuming of course that $\bar{\rho}$ is absolutely irreducible (I think that you meant to include this explicitly in your set-up, of course otherwise level-lowering can fail). In particular your argument is correct. I don't think there is a significantly more direct general argument incorporating $p=2$ as it is my understanding that it is only with Khare-Wintenberger's proof that the last cases of "Weak Serre implies Refined Serre" were proved. </p> <p>Over totally real field, this is the whole business of Serre's weight; a topic which has seen exponential development these last 10 years. I am vey far from the most superficial understanding of the current literature but I guess Buzzard-Diamond-Jarvis is a good place to start. Other key players are Toby Gee, Matt Emerton and Florian Herzig. I don't think much is actually known about level-lowering in weight 1, though I could very well be wrong.</p> http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/109180#109180 Answer by Olivier for Axiomatizing Gross-Zagier formulae Olivier 2012-10-08T20:42:53Z 2012-10-09T11:20:26Z <p>UPDATE: I have updated this answer slightly to take into account Victor's remark.</p> <p>I think that the precise questions being asked admit a straightforward answer. At the moment, no such formula is known and the proofs of Gross-Zagier, Waldspurger, Zhang et al. and Howard all absolutely and crucially require the hypothesis of self-duality. The reason for this is that the representation-theoretic part of the proof requires an understanding of test-vectors, as in the works of Tunell and Saito or as in the conjecture of Gross-Prasad. This is explained for instance in the article of Gross entitled Heegner points and representation theory as well as in Non-triviality of CM points in ring class field towers. Aflalo, Esther and Nekovář, Jan. Israel J. Math. 175 (2010), 225--284 (in which the formal setting is explored).</p> <p>As for whether similar formula could hold, I am not too optimistic. A Gross-Zagier formula should involve the $\psi$-eigenpart of the action of $\textrm{Gal}(H/\mathbb Q)$ on the projection of a CM point on the $\pi(f)$-component of the Jacobian of a Shimura curve. However, comparing the Galois action on CM points with the adelic action on the Jacobian, we see that this $\chi$-eigenpart can be non-trivial only when the restriction of $\psi$ to $\mathbb A_{\mathbb Q}$ is equal to $\chi$, or equivalently only in the self-dual case. This is proved for instance in Cornut, Christophe; Vatsal, Vinayak Nontriviality of Rankin-Selberg L-functions and CM points. Lemma 4.6.</p> <p>Note also that this is what we should expect from the conjectures on special values of $L$-functions applied to $L(f/K,\psi,s)$ when $f$ is not self-dual. In that case, the conjecture implies that $L'(f/K,\psi,s)$ should be related to cohomology classes in the <em>dual</em> of the motive of $f$. Only in the self-dual case does these collapse in a formula involving the height of a point. Finally, in the situation you describe, even though $L(f,\psi,s)$ might vanish at 1, it is is expected that it doesn't generically (say in a relevant $\mathbb Z_{p}$-extension), so the conjectural relation between $L(f/K,\psi,s)$ (or its Selmer group) and putative point could hold only "locally at the specialization corresponding to $f$" in a $p$-adic family of automorphic representation containing $\pi(f)\otimes\psi$. All the arguments that I know relating these objects would then simply vanish.</p> <p>Now an argument from ignorance is not a very good one, and I would very much like to be proven false, if only to learn something. Hidden behind all this is the question of the link between Kato's Euler system and rational points on modular varieties. The link is mysterious to me, but David Loeffler and Sarah Zerbes have some ideas.</p> http://mathoverflow.net/questions/108093/topics-for-an-undergraduate-expository-paper-in-number-theory/108094#108094 Answer by Olivier for Topics for an Undergraduate Expository Paper in Number Theory Olivier 2012-09-25T20:20:13Z 2012-09-25T20:20:13Z <p>Just on top of my head, I can think of the following themes, all having in common that they can be approached with a minimal knowledge and that there exists a continuous path from them to current research (those with a star could be linked to real analysis):</p> <p>Quadratic reciprocity*, higher reciprocity laws (starting with the biquadratic character of 2), sample cases of Fermat's last theorem, primality testing*, the last entry of Gauss diary, counting solutions of polynomial equations modulo various primes*, factorization of Fermat and Mersenne numbers, representing integers by quadratic forms, representing integers by sum of squares*, classifying integral quadratic forms, cyclotomy...</p> <p>Generally speaking, I think perusing the table of content of Gauss' Disquisitiones is a very good source of inspiration for projects like that.</p> http://mathoverflow.net/questions/43147/citing-papers-that-are-in-a-language-that-you-do-not-read/43151#43151 Answer by Olivier for Citing papers that are in a language that you do not read. Olivier 2010-10-22T09:08:19Z 2012-07-22T19:34:25Z <p>I think a common-sense approach is to cite the original paper (whatever the language) in order to give credit and attribution but only rely on arguments from papers you can understand in your proofs (so you don't violate the golden rule). </p> <p>Regarding reviewers, the worst that can happen (I think) is that you use a crucial argument from a paper you can understand but that the reviewer cannot understand. In that case, I think the problem is the same whether the reviewer cannot understand it because he is unfamiliar with the math or because he is unfamiliar with the natural language. In both cases, you, as the author, should try to present relatively clear references, and that includes translations when appropriate I guess, but ultimately this is a failure of the reviewer. If I were reviewing a paper and found myself in this situation, I would politely ask the author if there is a translation available. If not, I would tell the editors I am not competent, but wouldn't blame the author.</p> <p>It is a bad idea to upset reviewers, but banning reference in languages other than English (or any other language) even for attribution purpose is an outrageous suggestion that should not be complied with.</p> http://mathoverflow.net/questions/91546/arithmetic-geometry-examples/91697#91697 Answer by Olivier for Arithmetic geometry examples Olivier 2012-03-20T09:11:30Z 2012-06-05T14:49:29Z <p>The residual representation of $G_{\mathbb Q_{p}}$ attached to an eigencuspform is markedly different depending on whether $p$ divides the coefficient $a_{p}$, the non-ordinary case, or not, the ordinary case (the representation is reducible if and only if $p$ does not divide $a_{p}$; this translates into very different behaviors for $p$-adic families of cuspforms). But what does $p$ divides $a_{p}$ mean? It means more precisely that, after a choice of an embedding $i_{p}$ of $\bar{\mathbb Q}$ inside <code>$\bar{\mathbb Q}_{p}$</code>, the $p$-adic norm of $i_{p}(a_{p})$ is not 1.</p> <p>The eigencuspform $f=q+\alpha q^{2}-\alpha q^{3}+(\alpha^{2}-2)q^{4}+(-\alpha^{2}+1)q^{5}+\cdots\in S_{2}(\Gamma_{0}(389))$ where $\alpha$ is a root of $x^{3}-4x-2$ is $5$-ordinary for two of the embeddings of $\mathbb Q[X]/(X^3-4X-2)$ into $\bar{\mathbb Q}_{5}$ but not for the third one (because 1 is a root of $x^{3}-4x-2$ modulo 5).</p> http://mathoverflow.net/questions/62265/characterization-of-the-poisson-law Characterization of the Poisson law Olivier 2011-04-19T13:21:47Z 2012-04-12T11:26:21Z <p>This semester, I teach an introduction to probability course tailored for students with no science background and so with very <em>very</em> little prerequisites. We started with the basics of analytic combinatorics then moved on to random variables and the study of common laws (binomial, hypergeometric, geometric, Poisson). The audience being what it is, I try to avoid as much as possible calculus derivations of probability facts. </p> <p>For some aspects of the course, it worked out well (for instance the derivation of the expectation of the binomial law) but because I am barely more knowledgeable than my students when it comes to probability, I have been unable to answer this question:</p> <blockquote> <p>Is there a set of natural probability properties which characterize the discrete Poisson law?</p> </blockquote> <p>If yes, then I could use this as a definition of the Poisson law, which would suit my students better than saying "it's the law such that $P(X=k)=e^{-\lambda}\frac{\lambda^{k}}{k!}$". By natural above, I want to convey the meaning that I hope they can be formulated using natural language (like, say, memorylessness) rather than using analytic objects.</p> <p>More precisely, what I have in mind is the following:</p> <blockquote> <p>Is there such a set of properties which would make it at least a little intuitively plausible that the sum of two variables following Poisson law also follows Poisson law?</p> </blockquote> <p>Of course, the proof of the above fact is completely elementary, but it would still be above the level of everyone in the audience except perhaps the 3 top students.</p> <p>Note that I would be happy even if proving that this set of properties characterize Poisson law turned out to be much harder than anything I will do in this course (or even much harder than anything I know myself about probability), because what I am looking for is not logical rigour but rather psychological efficiency: in 10 years, my students will have completely forgotten what a derivative is, but I would like them to be able to recollect something if confronted with an epidemiological survey using random variables (at least my most successful students use this course to strengthen their math knowledge before studying medicine).</p> <p>I realize this question is very elementary, and would understand if it is deemed inappropriate, but the standard references I might consult on the subject will invariably (and with good reasons) develop much more calculus that my students will ever know before dealing with such questions (typically, they will characterize the Poisson law as the limit of the binomial law via Stirling's formula).</p> http://mathoverflow.net/questions/93544/geometric-interpretation-of-hida-isomorphism/93747#93747 Answer by Olivier for Geometric interpretation of Hida isomorphism Olivier 2012-04-11T09:48:20Z 2012-04-11T11:10:35Z <p>I am not sure what your criteria would be for a proof to be given a geometric interpretation, but the reason why weights "disappear" when we take the inverse limit on the level stems from the contraction property of Hecke operators (at $p$), or informally from the fact that Hecke operators at $p$ diminish the level.</p> <p>As you know, the proof of the isomorphism between the two different Hecke algebras requires the definition of a map between Hecke algebras acting on forms of weight 2 and forms of weight $k$. Because theses two algebras are sub-algebras of endomorphisms generated by the same abstract elements but acting on different objects, this amounts to constructing a map between the cohomology of (one of the level of) the modular tower with coefficients in the constant sheaf and (one of the level of) the modular tower with coefficients in a sheaf of weight $k$ (or the same thing with the modular tower replaced by the Igusa tower, as in Kevin's answer). This last map is really no big deal: if memory serves, on the sheaves it is just projection on the last component. The remarkable fact is that the map on cohomology then is surjective with a finite kernel (and is an isomorphism in the ordinary case); the proof of this assertion being exactly the contraction property. Note that the proof necessarily requires the choice of a level at some point; how else would you even state the result?</p> <p>Note for instance that for a tower of more general Shimura varieties, it is not at all obvious how the contraction property will play out: group-theoretic properties of <code>$\operatorname{GL}_{2}(\mathbb Q_{p})$</code> really do play an important role in the proof. See the reference below though, for an answer to these questions.</p> <p>So in the end, the isomorphism between the two Hecke algebras seems to me to come from the interplay between the cohomology of modular varieties and group-theoretic properties of the Hecke algebra. A very general formulation of this fact can be found in D.Mauger Algèbres de Hecke quasi-ordinaires universelles. Ann. Sci. École Norm. Sup. (4) 37 (2004) (section 2.4 to be precise)</p> http://mathoverflow.net/questions/88106/a-p-adic-analogue-for-a-formula-of-riemann/88111#88111 Answer by Olivier for A p-adic analogue for a formula of Riemann? Olivier 2012-02-10T15:42:43Z 2012-02-10T15:42:43Z <p>Yes, there is.</p> <p>And in surmising correctly that finding a $p$-adic analogue of this formula will provide an understanding of the functional equation of $p$-adic $L$-functions, you have just got a glimpse of what are now called the Coleman map, the Block-Kato exponential map, explicit reciprocity laws, and from then on the $p$-adic Langlands program. The article <em>Théorie d'Iwasawa des représentations de de Rham d'un corps local</em> Annals of Math 148 or the survey <em>Fonctions $L$ $p$-adiques</em> Séminaire Bourbaki 851 will tell you (much much) more.</p> http://mathoverflow.net/questions/87886/decomposition-of-k-10-in-copies-of-the-petersen-graph Decomposition of $K_{10}$ in copies of the Petersen graph Olivier 2012-02-08T12:51:08Z 2012-02-08T15:05:24Z <p>It is a well-known and cute exercise in algebraic graph theory to show that $K_{10}$ cannot be written as the edge-disjoint union of three copies of the Petersen graph $P$. Indeed, the graph $G$ whose edge-set is the complementary of the two copies of $P$ in $K_{10}$ is a $3$-regular <em>bipartite</em> graph. When I taught this, the classroom discussion went as follows:</p> <p>Q: Can we compute the spectrum of $G$. A: Well we could always compute a 10x10 determinant but (I am going to regret this if I don't know how to do it) I think we can do much better. Q: How? A: OK, let me be honest, I have no idea, but I am hoping that $G$ will turn out to be a well-known graph.</p> <p>And indeed, we wrote down a decomposition of $K_{10}$ for which $G$ turned out to be the connected bipartite 3-regular circulant graph on ten vertices $Z$ (the spectrum then being very easy to compute).</p> <p>Left silent in this discussion was whether this was the only possibility. I do believe it is. Indeed, unless I am mistaken (something which is alas entirely possible), $G$ is by construction a 3-regular bipartite connected graph on ten vertices. Beside, the two copies of $P$ share an eigenvector $v$ for the eigenvalue 1 so $v$ is an eigenvector for the eigenvalue $-3$ of $G$. Thus, $v$ has values in ${\pm 1}$ and gives the bipartition on $G$. From this, it follows that the set of vertices on which $v$ takes the value $1$ (resp. $-1$) is a 5-cycle. On each vertex, the third edge of each copy of $P$ thus connects the first $C_{5}$ to the second. Let $H$ be the graph whose edge set is given by the edges of the $P$ between the cycles. The graph $G$ is then the complementary graph of $H$ in the complete bipartite graph $K_{5,5}$. The graph $H$ is by construction 2-regular and bipartite, hence either $C_{10}$ or the union of $C_{4}$ and $C_{6}$. In both cases, there is a bijection between vertices of $H$ and triplets of vertices of $H$ which sends a vertex $w$ to the three vertices in the other bipartition class which are not adjacent to $w$. Hence, $G$ does not have two pairs of vertices with the same neighborhoods. But there are only two 3-regular bipartite connected graph on ten vertices, one has two pairs of vertices with the same neighborhoods and the other is $Z$.</p> <p>However, the above is deeply unsatisfying to me, if only because I don't trust my capacities to really enumerate all the possible ways to fit two copies of $P$ in $K_{10}$ at all, so that I am unconvinced that I did not make a mistake in the above. Moreover, the punchline of the argument is a classification of bipartite regular 3-connected graphs on ten vertices, something I can do only via a tedious enumeration (or by looking it up).</p> <blockquote> <p>Is there a conceptual way to show that the decomposition as two copies of $P$ and $Z$ is the only possible one (provided the above is correct)?</p> </blockquote> <p>More specifically, is it possible to compute the spectrum, or the automorphism group of $G$, or perhaps even a large subgroup of the automorphism group of $G$ without relying on long(ish) enumerations?</p> http://mathoverflow.net/questions/87656/r-noetherian-is-factorial/87664#87664 Answer by Olivier for R noetherian is factorial Olivier 2012-02-06T13:56:51Z 2012-02-06T13:56:51Z <p>This is well-known and references are presumably easy to find. Let me just give a few pointers.</p> <p>It is enough to show that prime ideals of height one are principal. Let $I$ be such a prime ideal. Thanks to our hypotheses on $R$, the ideal $I$ has a finite free resolution. Hence, if $I$ is a projective module, then it is stably free, and thus principal. In order to show that $I$ is projective, we look at the cokernel $X$ of $\operatorname{Hom}(I,M)\rightarrow\operatorname{Hom}(I,N)$ when $M\rightarrow N$ is a surjection, and to do so it is enough to localize at a maximal ideal $\mathfrak{m}$. The residual field $k$ of the local ring $R_{\mathfrak{m}}$ admits a finite free resolution by $R$-modules by our hypothesis so admits a finite free resolution by $R_{\mathfrak{m}}$-modules. So the ring $R_{\mathfrak{m}}$ is regular so the ideal $IR_{\mathfrak{m}}$ is principal so $IR_{\mathfrak{m}}$ is a projective $R_{\mathfrak{m}}$-module so $X_{\mathfrak{m}}$ vanishes. So $X$ is zero, so $I$ is projective and hence principal. And so $R$ is factorial.</p> http://mathoverflow.net/questions/87407/pd-finite-for-finite-module-over-local-cm-ring/87411#87411 Answer by Olivier for pd finite for finite module over local CM ring? Olivier 2012-02-03T08:52:55Z 2012-02-03T08:52:55Z <p>No. Put $k=R/m$. Then $k$ is of finite projective dimension if and only if $R$ is regular. This is the famous theorem of Auslander-Buchsbaum, Serre (see for instance Bruns-Herzog Cohen-Macaulay rings theorem 2.2.7 for a proof). So the residual field of a non-regular CM ring will give a counter-example. </p> http://mathoverflow.net/questions/85523/the-historical-development-of-automorphic-geometry/85567#85567 Answer by Olivier for The historical development of automorphic geometry Olivier 2012-01-13T09:39:35Z 2012-01-13T14:49:05Z <p>A common answer to question 1 is to mention the entries of Gauss's diary from 1814, including famously (but not restricted to) the last one, in which he studies some properties of biquadratic reciprocities and their links with the division of the lemniscate. Gauss understood these properties to derive from properties of Punktgitter (lattices of points) in the complex planes and his formulation is surprisingly close to the statement that (some) elliptic curves with CM are automorphic objects. </p> <p>Though I am not sure that this is the best way to think about the work of Gauss himself (he himself seems to have been unsatisfied with his work and to have considered it of relatively low value compared to the latter approach of Einsenstein) and though it had no impact on mathematics at the time for the simple reason that it was kept private for over 80 years, it certainly counts as work done on "automorphic geometry" done in the early XIX° century.</p> http://mathoverflow.net/questions/7678/free-subquotient-of-galois-representations-coming-from-hida-theory Free subquotient of Galois representations coming from Hida theory Olivier 2009-12-03T15:44:09Z 2011-12-19T04:58:49Z <p>Let $\mathbf{T}$ be the reduced nearly ordinary Hecke algebra of level $N$ of Hida theory for <code>$\operatorname{GL}_{2}$</code> over <code>$\mathbb{Q}$</code> (or more generally over a totally real field $F$). Then $\mathbf{T}$ is finitely generated over a regular ring $\Lambda$ of dimension 3. Let $\mathfrak{m}$ be a maximal non-Eisentein ideal of $\mathbf{T}$.</p> <p>By patching pseudo-representations attached to algebraic modular forms, Wiles (and Hida) have constructed a two-dimensional <code>$G_{\mathbb{Q}}$</code>-representation $(V,\rho)$ with coefficients in <code>$\mathbf{T}_{\mathfrak{m}}\otimes_{\Lambda}\operatorname{Frac}(\Lambda)$</code>. This representation admits a 1-dimensional sub-space <code>$V^{+}$</code> and a free 1-dimensional quotient <code>$V^{-}$</code> both stable under the action of <code>$G_{\mathbb{Q}_{p}}$</code>. Because $\mathfrak{m}$ is non-Eisenstein, there exists a choice of basis of $V$ such that $\rho$ has values in <code>$\operatorname{GL}_{2}(\mathbf{T}_{\mathfrak{m}})$</code>. The lattice $L\subset V$ corresponding to this choice of basis admits a free sub-module <code>$L^{+}=L\cap V^{+}$</code> of rank 1 stable under <code>$G_{\mathbb{Q}_{p}}$</code>. However, it is unclear to me whether $L$ admits a free rank 1 quotient stable under <code>$G_{\mathbb{Q}_{p}}$</code>. This is true if $\rho$ modulo $\mathfrak{m}$ is of the form <code>$$\rho\sim\begin{pmatrix}\chi_{1}&amp;*\\ 0&amp;\chi_{2}\end{pmatrix}$$</code> with <code>$\chi_{1}\neq\chi_{2}$</code> because then <code>$L/L^{+}$</code> is generated by a single element according to Nakayama lemma. However, without this hypothesis, I don`t see an obvious proof of this fact, nor have I good reasons to believe it should be true. Does anyone know for sure?</p> http://mathoverflow.net/questions/78954/modular-form-fourier-coefficients-and-associated-automorphic-representation/78959#78959 Answer by Olivier for modular form Fourier coefficients and associated automorphic representation Olivier 2011-10-24T06:47:43Z 2011-10-24T06:47:43Z <p>No. </p> <p>A supercuspidal representation, a Steinberg twisted by a ramified character and a principal series ramified at both characters at $p$ will all have zero $a_{p}$. A reference for this is Jacquet-Langlands LN 114 Proposition 3.5, 3.6. I am also not sure one can distinguish a priori a simply ramified principal series and a Steinberg twisted by an unramified character purely using $a_{p}$, you might need to look at the order of the central character at $p$ to do this (in the latter case, the central character is trivial at $p$ while it is not in the former). </p> http://mathoverflow.net/questions/78443/cm-abelian-varieties-and-potential-good-reduction/78447#78447 Answer by Olivier for CM abelian varieties and potential good reduction Olivier 2011-10-18T12:27:53Z 2011-10-18T12:27:53Z <p>No, absolutely not</p> <p>In fact, the hypotheses you discuss are rather weak. Take $F$ a totally real number field. If $A/F$ is the abelian variety attached to an eigenform $f$ of weight $2$ and level $N$, then the representation $\rho$ attached to the $p$-adic Tate module of $A$ is crystalline at $p$ for $p\nmid N$ . At $\ell≠p$, the representation $\rho$ is potentially unramified if and only if the automorphic representation $\pi(f)_{\ell}$ is principal series of supercuspidal (i.e not Steinberg). This certainly happens a lot. All such eigenforms do not have potential CM.</p> <p>I'm not sure I understand your question about the Fontaine-Mazur conjecture but much is known in the set-up that you seem to be interested in. For instance, if $F$ is totally real and $p\geq 7$ then an odd $G_F$-representation in <code>$\operatorname{GL}_{2}(\mathbb F_{p})$</code> which stays irreducible after restriction to $G_{F(\zeta_{p})}$ comes from the Tate module of an abelian variety $A/F$.</p> http://mathoverflow.net/questions/77950/cohomology-of-sl-2-mathbbf-p-acting-on-trace-zero-matrices-over-mathbbf/78007#78007 Answer by Olivier for Cohomology of $SL_2(\mathbb{F}_p)$ acting on trace zero matrices over $\mathbb{F}_p$ Olivier 2011-10-13T08:13:09Z 2011-10-13T08:13:09Z <p>It seems to me that Sah's lemma will do the trick.</p> <blockquote> <p>(Sah's lemma) Let $G$ be a group, $M$ a $G$-representation and $g\in Z(G)$. Then $x\mapsto (g-1)x$ is the zero map on $H^{1}(G,M)$.</p> </blockquote> <p>The proof is a computation of $(g-1)f(h)$ using the cocycle relation and the fact that $f(h)=f(ghg^{-1})$. Applying the lemma to <code>$\operatorname{SL}_{2}(\mathbb F_{p})$</code>, $M$ and a non-trivial homotethy, for instance $-Id$, gives the result.</p> http://mathoverflow.net/questions/74749/unmathematical-habits-of-thought-and-action-which-would-be-of-use-to-mathematicia/76648#76648 Answer by Olivier for Unmathematical habits of thought and action which would be of use to mathematicians Olivier 2011-09-28T15:25:59Z 2011-09-28T15:25:59Z <p>A long quote, from which one can extrapolate trivially a tentative answer.</p> <p>"[T]echnical treatises in science do not generally receive such a license for explicitly personal expression. I believe that this convention in technical writing has been both harmful and more than a bit deceptive. Science, done perforce by ordinary human beings expressing ordinary motives and foibles of the species, cannot be grasped as an enterprise without some acknowledgment of personal dimensions in preferences and decisions – for, although a final product may display logical coherence, other decisions, leading to other formulations of equally tight structure, could have been followed, and we do need to know why an author proceeded as he did if we wish to achieve our best understanding of his accomplishments, including the general worth of his conclusions.</p> <p>Logical coherence may remain formally separate from ontogenetic construction, or psychological origin, but a full understanding of form does require some insight into intention and working procedure. Perhaps some presentations of broad theories in the history of science – Newton's Principia comes immediately to mind – remain virtually free of personal statement (sometimes making them, as in this case, virtually unreadable thereby). But most comprehensive works, in all fields of science, from Galileo's Dialogo to Darwin's Origin, gain stylistic strength and logical power by their suffusion with honorable statements about authorial intents, purposes, prejudices and preferences."</p> <p>SJ Gould The structure of evolutionary theory p. 34.</p> http://mathoverflow.net/questions/75335/what-is-the-reason-for-modularity-results/75354#75354 Answer by Olivier for What is the "reason" for modularity results? Olivier 2011-09-13T21:08:56Z 2011-09-13T21:08:56Z <p>I don't think it is too much an overstatement to say that nobody has any idea why the most general conceivable form of the modularity conjectures-say a combination of Langlands program and the Fontaine-Mazur conjecture-should be true. As in the case of conjectures on special values of $L$-function, the most one could probably say is that their inner consistency is absolutely impressive so that in some sense, they feel too good to be not true.</p> <p>That said, not all is lost, I think, in your quest to get a philosophical understanding of this topic, especially if you set yourself a more modest goal at first. Because why things should be true is probably inherently subjective, I will only offer my personal experience with modularity results for $\operatorname{GL}_2$. I think that the first significative experience I had towards a modicum of understanding of the deep reasons why these should be true was to realize how utterly surprising they were. The more I understood about abstract universal deformation rings and the less I could see why they should be Hecke algebras. The Taylor-Wiles method, I still don't claim any deep or philosophical understanding of, but this is mostly because I never read closely enough the literature. Some papers from Kisin, for instance, do explain that there seems to be a trade-off between how singular a deformation ring can be and the local behaviour of the Galois representation at p. The next big step for me was to read carefully Taylor's paper on potential modularity. This paper makes it very clear that modularity results are very amenable to bootstrapping: prove one, and you may get a lot for free. So to recap: modularity results should be true because (in certain settings), one can reduce them to much simpler modularity results and then get rid of the singularities of the universal deformation ring (provided you have what you need to do so).</p> <p>Not very philosophical perhaps but since the name of Weil has appeared in the answers by Emerton and Joël, let me conclude by quoting his magnificent (if slightly depressing) words on finding philosophical understanding of mathematical theories.</p> <p>Rien n’est plus fécond, tous les mathématiciens le savent, que ces obscures analogies, ces troubles reflets d’une théorie à une autre, ces furtives caresses, ces brouilleries inexplicables ; rien aussi ne donne plus de plaisir au chercheur. Un jour vient où l’illusion se dissipe ; le pressentiment se change en certitude ; les théories jumelles révèlent leur source commune avant de disparaître ; comme l’enseigne la Gita, on atteint à la connaissance et à l’indifférence en même temps. La métaphysique est devenue mathématique, prête à former la matière d’un traité dont la beauté froide ne saurait plus nous émouvoir.</p> http://mathoverflow.net/questions/74974/characterization-of-prime-ideals-in-regular-local-rings/74981#74981 Answer by Olivier for Characterization of prime ideals in regular local rings Olivier 2011-09-09T07:47:21Z 2011-09-09T07:47:21Z <p>The answer is no. It is easy enough to construct counter-examples, but to convince yourself that such a statement is hopeless, here is a salient point. </p> <p>Any ideal $P$ such that $P$ is a "colon ideal" of a regular system of parameters is such that $R/P$ is itself a regular ring (indeed $R/P$ is of dimension $d-h$ and its maximal ideal is generated by $x_{h+1},\cdots,x_{d}$). On the other hand, Cohen's structure theorem tells you that any equicharacteristic complete noetherian local domain is a quotient of a regular ring by a prime ideal (and quite a bit more than that, of course). So any equicharacteristic complete noetherian local domain which is not regular would provide a counterexample. </p> http://mathoverflow.net/questions/65516/why-does-h1g-p-i-p-mathbbf-delta-epsilon-1-vanish/65525#65525 Answer by Olivier for Why does $H^1(G_p/I_p,\mathbb{F}(\delta\epsilon^{-1}))$ vanish? Olivier 2011-05-20T10:59:24Z 2011-05-20T11:09:59Z <p>A variant of David Loeffler's answer...</p> <p>If more generally $V$ is a $G_{\mathbb Q_{p}}$-representation with coefficients in a field, then the dimension of $H^{1}(G_{p}/I_{p},V)$ (EDIT : or rather $H^{1}(G_{p}/I_{p},V^{I_{p}})$) is equal to the dimension of $H^{0}(G_{p},V)$ because the former is the cokernel of the map $Fr_{p}-1$ acting on $V^{I_{p}}$ and the latter is the kernel of this map. </p> <p>In the situation you described, $\mathbb F(\delta\epsilon^{-1})$ has no non-trivial invariants under $G_{p}$ because $\delta≠\epsilon$ (and this is why you need your form to be $p$-distinguished) so its $H^{1}$ is trivial as well.</p> http://mathoverflow.net/questions/65451/do-tamagawa-numbers-of-galois-representations-stabilise-in-the-cyclotomic-tower/65469#65469 Answer by Olivier for Do Tamagawa numbers of Galois representations stabilise in the cyclotomic tower? Olivier 2011-05-19T19:40:48Z 2011-05-19T20:21:58Z <p>As stated, the answer to your question is certainly no. </p> <p>For instance, an elliptic curve $E/\mathbb Q$ with split multiplicative ordinary reduction at $p$ will have unbounded Tamagawa number at $p$ in the cyclotomic extension of $\mathbb Q$. To see this, you can check that the Tamagawa number is the order of $H^{2}({\mathbb Q_{\infty,p}},(T_{p}E)_{p}^{+})$. This is what B.Mazur calls the phenomenon of anomalous primes in his Inventiones 18 paper, which is the algebraic counterpart of the existence of exceptional zeroes for $p$-adic $L$-functions.</p> <p>However, perhaps you meant for the Tamagawa factors outside $p$ to be eventually constant? In that case, I think the answer is yes.</p> <p>First, $\operatorname {Gal}(K(\mu_{p^{\infty}})/K)$ has finite prime-to-$p$ part so there exists a large enough $n$ so that $\operatorname {Gal}(K(\mu_{p^{\infty}})/K(\mu_{p^n}))$ is unramified outside $p$ (EDIT: Of course this first step is unnecessary). Let $v\nmid p$ be a place of $K(\mu_{p^{n}})$ and let $w|v$ be a place of $K(\mu_{p^{n'}})$ with $n'≥n$. Then <code>$H^{1}(I_{v},T)\simeq H^{1}(I_{w},T)$</code> so the Tamagawa number is constant.</p> http://mathoverflow.net/questions/63969/what-is-an-explicit-example-of-a-variety-x-which-is-finite-over-spec-f-p-but-whic/63984#63984 Answer by Olivier for What is an explicit example of a variety X which is finite over Spec F_p but which does not lift to a scheme Y which is finite and flat over Spec Z_p? Olivier 2011-05-05T11:42:43Z 2011-05-05T11:42:43Z <p>Am I missing something or is this the classical question of Serre? A class of examples is given in Exemples de variétés projectives en caractéristique p non relevables en caractéristique zéro. Proc. Nat. Acad. Sci. U.S.A. 47 1961 108–109. </p> <p>They come from the quotient of some complete intersections by some finite groups, but if you read closely the proof (by the theory of the étale fundamental group, the impossibility of constructing a lift is reduced to the impossibility of constructing some group representations), you see that the ideas are in fact quite general. In order to give a non vacuous answer, let me also draw your attention to the letter of Serre in the appendix of the document illusie_trieste.pdf on Luc Illusie's website.</p> http://mathoverflow.net/questions/62607/how-is-etale-cohomology-of-integer-rings-related-to-galois-cohomology/62658#62658 Answer by Olivier for How is etale cohomology of integer rings related to Galois cohomology? Olivier 2011-04-22T19:54:34Z 2011-04-23T07:28:39Z <p>By usual (sometimes not so trivial) homological arguments, one can reduce to the case where $M$ is a finite discrete module over an artinian ring of residual characteristic $p$. In that case, I think you want $S$ to contain places above $p$ as well, even if your $M$ is unramified at $p$, so let me assume this. </p> <p>The module $M$ induces an étale sheaf $M_{et}$ on $\operatorname{Spec}\mathcal O_{L,S}$ for all finite extension $L/K$. The spectral sequence UPDATE (converging to $H^{i+j}(\operatorname{Spec}\mathcal O_{K,S},M_{et})$) $$E_{2}^{i,j}=\underset{\longrightarrow}{\operatorname{\lim}}\ H^{i}(\operatorname{Gal}(L/K),H^{j}(\operatorname{Spec}\mathcal O_{L,S},M_{et}))$$ then induces isomorphisms between $E_{2}^{i,0}$ and $H^{i}(\operatorname{Spec}\mathcal O_{L,S},M_{et})$ or in other words $H^{i}(G_{K,S},M)$ is isomorphic to $H^{i}(\operatorname{Spec}\mathcal O_{K,S},M_{et})$. So you can assume that you are working with Galois cohomology throughout $provided$ you use Galois cohomology with restricted ramification.</p> <p>Because the Tamagawa Number Conjectures are formulated only in the setting above, Bloch and Kato could have used Galois cohomology instead of étale cohomology everywhere without changing anything. To touch upon your last question, I think there are two reasons why they chose étale cohomology.</p> <p>First, at least at the time they wrote, Galois cohomology was not the most familiar object of the two. In fact, many classical well-known results were given correct complete proofs only very late (in the late 90s in some cases). On the other hand, SGA (and works of Bloch and Kato themselves) existed as references for étale cohomology.</p> <p>Second, using étale cohmology, one can formulate the TNC over more general bases than $\operatorname{Spec}\mathcal O_{K,S}$ (for instance any scheme of finite type of $\mathbb Z[1/p]$). This kind of generalization had been the key idea of previous works of Kato and Bloch-Kato on higher class field theory so it is not surprising that they decided to at least allow the same kind of generality in their subsequent works.</p> http://mathoverflow.net/questions/129092/the-topology-of-power-series-ring Comment by Olivier Olivier 2013-04-29T14:21:27Z 2013-04-29T14:21:27Z I'm sure an answer can be very quickly found at <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/124595/automorphism-group-of-regular-graph/124597#124597 Comment by Olivier Olivier 2013-03-15T12:39:47Z 2013-03-15T12:39:47Z @Chris Godsil Frucht's <i>original</i> construction does not produce regular graphs but the first regular construction is also due to Frucht as far as I know (<i>Graphs of degree three with a given abstract group</i> 1949) so might reasonably be called Frucht's theorem as well. Do you disagree? http://mathoverflow.net/questions/122795/unramified-base-change-in-characteristic-p-0 Comment by Olivier Olivier 2013-02-24T13:40:52Z 2013-02-24T13:40:52Z Doesn't that follow from the proof of the local Langlands correspondence for function fields (by Laumon, Rappoport and Stuhler)? http://mathoverflow.net/questions/122351/grothendieck-monodromy-theorem-for-l-adic-sheaves Comment by Olivier Olivier 2013-02-20T08:57:42Z 2013-02-20T08:57:42Z This question looks very interesting to me, but there are a few things I don't quite understand. Are you looking at the $G_{F}$ action the &#233;tale cohomology of $X$ with coefficients in $C$? Would you give an example of the typical situation you have in mind? http://mathoverflow.net/questions/121688/definition-of-cm-modular-form/121693#121693 Comment by Olivier Olivier 2013-02-14T08:48:14Z 2013-02-14T08:48:14Z Dear wccanard (!), I did write that $k$ should be greater than 2, but now I realize I did not repeat the condition when passing to $F$. Thanks for pointing this out. http://mathoverflow.net/questions/121688/definition-of-cm-modular-form/121693#121693 Comment by Olivier Olivier 2013-02-13T10:31:48Z 2013-02-13T10:31:48Z Let me add a reference. You can read Motives and automorphic forms: the potentially abelian case, available on L.Fargues webpage. This is a modern exploration of the topic (which contains much much more than the answer to your question). http://mathoverflow.net/questions/121171/which-finite-group-is-not-the-automorphism-group-of-some-rooted-finite-trees/121176#121176 Comment by Olivier Olivier 2013-02-08T12:10:06Z 2013-02-08T12:10:06Z I think this indeed a theorem of Jordan. http://mathoverflow.net/questions/120913/periods-for-2-variable-p-adic-l-functions/120990#120990 Comment by Olivier Olivier 2013-02-06T17:17:56Z 2013-02-06T17:17:56Z Of course, the freeness over the localized Hecke algebra is much easier to prove (the real freeness being false under suitable hypotheses) and this is why Greenberg/Stevens type results require much weaker hypotheses. http://mathoverflow.net/questions/120913/periods-for-2-variable-p-adic-l-functions/120990#120990 Comment by Olivier Olivier 2013-02-06T17:13:14Z 2013-02-06T17:13:14Z Dear Jo&#235;l, The point is that different hypotheses will yield different conclusions about the nature of $c$. Does $c$ vary analytically or not on $U$ for instance? You are right to say that $A(x)$ is a necessary condition, but as usual, the devil is in the details. You say that $A(x)$ means that the space of modular symbols is free on the Hecke algebra, but which Hecke algebra? If it is the Hecke algebra localized at $x$, then you get Greenberg/Stevens, Mok style results. If it is the Hecke algebra before localization, then you get Kitagawa style results (type (1) and (2) of my answer). http://mathoverflow.net/questions/120913/periods-for-2-variable-p-adic-l-functions/120954#120954 Comment by Olivier Olivier 2013-02-06T16:51:03Z 2013-02-06T16:51:03Z @Filipo: I don't think there is a subscript $\chi$ in (2). There is a subscript $x$ as there should be. @Jo&#235;l: Kitagawa-Mazur give type (2) results, maybe not in Kitagawa's and Mazur's work, but in subsequent works. http://mathoverflow.net/questions/119062/hamiltonian-cycles-in-power-graphs Comment by Olivier Olivier 2013-01-17T09:40:43Z 2013-01-17T09:40:43Z Dear Gerhard, Can you explain why you think this implication is reasonable? I don't see it myself (but I don't know much about the topic). Moreover, the data in the linked note show that there exists values of $n$ such that G2(n+2) has less hamiltonian cycles than G2(n). http://mathoverflow.net/questions/116771/on-topology-of-p-adic-numbers Comment by Olivier Olivier 2012-12-19T09:34:11Z 2012-12-19T09:34:11Z Both Wikipedia and Google know the answer to this (very elementary) question. If you don't find it there or by yourself, you should try <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/116526/generalized-euler-sequence-on-a-projective-scheme Comment by Olivier Olivier 2012-12-18T08:55:34Z 2012-12-18T08:55:34Z <i>I don't want to put so much spam in my thesis, and would like to cite this result, which probably has been proven in the 60s</i> Giving appropriate credit is nice and important of course. Then again, a thesis seems to me to be a good place to write up things that are supposedly well-known to the experts but for which a good reference is lacking. http://mathoverflow.net/questions/112122/a-weaker-concept-of-graph-homomorphism Comment by Olivier Olivier 2012-11-12T09:40:45Z 2012-11-12T09:40:45Z The usual definition also has attractive features from the point of view of algebraic graph theory: existence and uniqueness of cores,cores of vertex-transitive graphs are vertex-transitive of cardinal dividing the the cardinal of the graph etc... If you allow contraction of edges, this seems to disappear. http://mathoverflow.net/questions/111324/where-in-the-literature-does-the-anticyclotomic-mathbfz-p-extension-of-an-im Comment by Olivier Olivier 2012-11-03T08:37:07Z 2012-11-03T08:37:07Z No examples are given, but from his description of the general construction and say the content of Weber Algebra, you get the anticyclotomic extension for free.