User laurent berger - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:32:18Z http://mathoverflow.net/feeds/user/5743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126529/a-criterion-for-freeness-over-a-local-ring A criterion for freeness over a local ring Laurent Berger 2013-04-04T15:28:08Z 2013-04-04T16:03:53Z <p>Let $A=K[[X_1,\dots,X_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that</p> <ol> <li>for all $k$, the $A[1/X_k]$-module $M[1/X_k]$ is free of rank $d$;</li> <li>for every $i \neq j$, we have $M = M[1/X_i] \cap M[1/X_j]$.</li> </ol> <p>Does this imply that $M$ is free? </p> <p>It certainly does if $n=1$ (easy), and also if $n=2$ (reduce $M$ modulo $X$), but things seem trickier if $n \geq 3$. </p> <p>This question comes up when trying to prove that some $(\varphi,\Gamma)$-modules over rings in several variables are free.</p> http://mathoverflow.net/questions/125631/galois-descent-for-semilinear-endomorphisms/125703#125703 Answer by Laurent Berger for Galois descent for semilinear endomorphisms Laurent Berger 2013-03-27T08:53:37Z 2013-03-27T08:53:37Z <p>If $E$ is of dimension $1$, then $\phi$ is given in some basis by a scalar $x \in L$ and you are asking whether there is an element $y \in L^\times$ such that $x \cdot \sigma(y)/y$ belongs to $K$. I don't see why there should be a simple general criteria to decide whether such a $y$ exists. If $\sigma$ fixes $K$, then the norm of $x$ should be the norm of an element of $K$, and conversely this would be enough if in addition $Gal(L/K)$ was generated by $\sigma$.</p> <p>More generally, you can at least rewrite your problem in terms of cohomology: the group $\mathbb{Z}$ acts on $L$ by $n \cdot x = \sigma^n(x)$, and you're asking about the image of $H^1(\mathbb{Z},GL_d(K)) \to H^1(\mathbb{Z},GL_d(L))$. But without more specific information on $K$, $L$ and $\sigma$, I don't see what one could say in general.</p> http://mathoverflow.net/questions/116928/describing-the-ratio-of-uniformizers-in-b-dr/117074#117074 Answer by Laurent Berger for Describing the ratio of uniformizers in B_dR Laurent Berger 2012-12-23T08:56:36Z 2012-12-23T08:56:36Z <p>What you're asking for is a description of $$a_1 = \theta(\frac{[\tilde{p}]-p}{t}).$$ It is an element of $C_p$ and you can't really "write it down explicitly". What you can do is let $G_{Qp}$ act on it and see what happens. If $g \in G_{Qp}$ then $g(\tilde{p})=\tilde{p} \cdot \epsilon^{c(g)}$ where $c(\cdot)$ is the Kummer cocycle and $g(t)=\chi(g) t$ where $\chi(\cdot)$ is the cyclotomic character. This should imply that $$g(a_1) = \theta(\frac{[\tilde{p}][\epsilon^{c(g)}]-p}{\chi(g) t}) = \frac{a_1}{\chi(g)} + p \frac{c(g)}{\chi(g)}.$$ Now that you see this formula, you should recognize it. Let $V$ be the semistable extension of $Q_p$ by $Q_p(1)$. It has Hodge-Tate weights $0$ and $1$ and $a_1$ is basically the period corresponding to weight $0$, ie the period that tells you that $(V \otimes_{Qp} C_p)^{G_{Qp}} \neq 0$.</p> http://mathoverflow.net/questions/114018/fastest-way-to-factor-integers-260/114037#114037 Answer by Laurent Berger for Fastest way to factor integers < 2^60 Laurent Berger 2012-11-21T08:34:37Z 2012-11-21T08:34:37Z <p>A long long time ago I programmed my HP48 (4 MHz! Wow!) in assembly language to factor numbers. It would do trial division by all integers congruent to $\pm 1 \bmod{6}$, up to some bound B and then use Pollard's rho method. I guess in your case you need to figure out what B should be depending on your implementation. At the time, I was using "Prime Numbers and Computer Methods for Factorization" by Hans Riesel as a reference. Since $2^{60}$ is relatively small, I'm not sure that more powerful methods would really help. Pollard's rho method is believed to be in something like $O(p^{1/2})$ where $p$ is the smallest prime factor of your integer, which is quite small in your case.</p> http://mathoverflow.net/questions/112545/absolute-galois-group-of-the-field-of-puiseux-series-over-overline-mathbbf/112560#112560 Answer by Laurent Berger for Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$ Laurent Berger 2012-11-16T09:32:01Z 2012-11-16T09:32:01Z <p>Let $E$ be the field $\overline{\mathbb{F}}_p((X))$. The field of Puiseux series whose exponents have denominators prime to $p$ is a subfield of $E^{sep}$, so the group you're asking about would then be the wild inertia subgroup of $Gal(E^{sep}/E)$. The group $Gal(E^{sep}/E)$ is quite complicated, and it comes up in arithmetic geometry, for example when studying the $\pi_1$ of curves. It also occurs as a closed subgroup of $Gal(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ by the theory of the field of norms of Fontaine and Wintenberger. Its representations on $\mathbb{Z}_p$-modules are described by $\varphi$-modules'' (like $(\varphi,\Gamma)$-modules without the $\Gamma$). If you want to include Puiseux series whose exponents have denominators divisible by $p$, then you're looking at the perfection of $E$. The group does not change, as $E^{sep}$ is dense in $E^{alg}$ by a theorem of Ax.</p> http://mathoverflow.net/questions/110544/is-there-any-theorem-like-implicit-function-theorem-in-mathbbq/110554#110554 Answer by Laurent Berger for Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? Laurent Berger 2012-10-24T16:02:45Z 2012-10-24T17:41:50Z <p>Here's what I think happens over $\mathbb{Q}$. Write your polynomial $P(X,Y)$ as a product of irreducible polynomials $P_i(X,Y)$. Hilbert's irreducibility theorem ( <a href="http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem</a> ) tells you that there are infinitely many $a$'s such that $P_i(X,a)$ is irreducible for every $i$. If one of them has a solution, it is therefore of degree $1$ in $X$. Some $P_i$ is therefore of degree $1$ in $X$, which answers your question.</p> <p>EDIT: it does not answer the question but rather shows that there is some polynomial $Q$ such that $P(Q(Y),Y)=0$ which is more reasonable, since then $P(Q(a)),a)=0$. This should have been the question.</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/110451#110451 Answer by Laurent Berger for Slick ways to make annoying verifications Laurent Berger 2012-10-23T17:33:22Z 2012-10-23T17:33:22Z <p>A surjective map between two modules (over a commutative ring) of the same rank is a bijection.</p> http://mathoverflow.net/questions/110057/lost-soul-loneliness-in-pursing-math-advice-needed/110221#110221 Answer by Laurent Berger for Lost soul: loneliness in pursing math. Advice needed. Laurent Berger 2012-10-21T07:15:49Z 2012-10-21T07:15:49Z <p>Hi Flora,</p> <p>it seems that many people have answered your question assuming that you're asking about becoming a research mathematician. But you can do a PhD in math and then go on to a different career, where you use your skills but don't necessarily go on to do original research. Many companies are hiring math PhDs and it seems to me that there are opportunities in very diverse fields. I'm not in the best position to discuss them ($p$-adic analysis doesn't seem to be very much in demand in the real world these days), but you should be able to find such information easily, maybe starting with the AMS?</p> <p>And don't let yourself be discouraged by the behavior of your colleagues. It's true that some mathematicians spend all their waking hours thinking and talking about maths, and a few of them are very successful but these are really a minority. It's my opinion that most of those who display this behavior would really be better off if they had a more balanced lifestyle. There may be cultural factors here: in the protestant anglo-saxon world, it's important to be seen working all the time, and bad form for people to enjoy themselves. </p> <p>Just do your own thing at your own pace. It's normal that you spend hours on a proof. Maths is not about being the first, but about having a deep understanding of what is going on. It certainly helps if you have flashes of insight, but there's nothing wrong about really understanding the situation first and finding the answer afterwards.</p> <p>Concerning the funding, it's true that being a PhD student does not pay much and that can be a real problem. I don't know about the country where you are but in France, some companies have programs where you work for them part-time and do a PhD in the rest of your time, the PhD being in some applied domain that is then useful for the company.</p> http://mathoverflow.net/questions/108842/the-significance-of-modularity-for-all-galois-representations/108922#108922 Answer by Laurent Berger for The significance of modularity for all Galois representations Laurent Berger 2012-10-05T14:27:48Z 2012-10-05T14:27:48Z <p>You would lose all the p-adic families of Galois representations, that are necessary to prove properties of the modular ones. Hida's family was necessary for the proof ofMTT, for example.</p> http://mathoverflow.net/questions/108402/decomposition-of-matrices-in-semisimple-and-nilpotent-parts/108408#108408 Answer by Laurent Berger for Decomposition of Matrices in Semisimple and Nilpotent Parts Laurent Berger 2012-09-29T14:27:23Z 2012-09-29T15:17:44Z <ol> <li><p>You don't need to conjugate if you want $R$ to be diagonalizable (as opposed to diagonal).</p></li> <li><p>I assume you want $R$ and $M$ to have coefficients in $K$, otherwise just work in the algebraic closure. </p></li> <li><p>The statement is then true if $K$ is perfect and possibly false otherwise, as you can see by taking $A=[[0, 1], [t, 0]]$ in $K=F_2(t)$.</p></li> </ol> <p>EDIT : see <a href="http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition" rel="nofollow">http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition</a></p> http://mathoverflow.net/questions/108391/are-d-dr-and-d-st-potentially-comparable/108401#108401 Answer by Laurent Berger for Are D_dR and D_st "potentially comparable"? Laurent Berger 2012-09-29T11:44:50Z 2012-09-29T12:17:36Z <p>Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.</p> <p>Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset" (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.</p> <p>EDIT : I should have said that $V$ is of dimension 2 here.</p> http://mathoverflow.net/questions/106468/local-galois-representation-with-higher-coefficient/106484#106484 Answer by Laurent Berger for local galois representation with higher coefficient Laurent Berger 2012-09-06T07:06:19Z 2012-09-06T11:47:57Z <p>If $F$ is not finite but rather equal to $C_p$ then this is really Sen's theory (see for instance Fontaine's course notes in Astérisque 295). If $F$ is merely a finite extension of $K$, then I'm not sure that you need to introduce a lot of machinery: restrict your representation to $G_F$ so that it's linear, do what you have to do, and then take in account the extra structure that you had. Alternatively, a semilinear representation is the same as an element of $H^1(G,GL_d(F))$, so you could use Galois-cohomological techniques, especially the inflation-restriction sequence with $G_F$ and $G_K$.</p> <p>EDIT : this answered the question for $F$-semilinear representations of $G_K$ with $F$ an extension of $K$. Since then the OP has modified his question so my answer is not relevant anymore :(</p> http://mathoverflow.net/questions/103911/phi-gamma-module-of-ordinary-elliptic-curve/103959#103959 Answer by Laurent Berger for (phi, Gamma) module of ordinary elliptic curve Laurent Berger 2012-08-04T16:25:23Z 2012-08-04T16:25:23Z <p>If your elliptic curve has good ordinary reduction, then the attached Galois representation is reducible : it is an extension of $\eta_2$ by $\eta_1 \chi$ where $\eta_{1,2}$ are unramified characters and $\chi$ is the cyclotomic character. The $(\phi,\Gamma)$-modules of $\eta_2$ and $\eta_1 \chi$ are easy to compute, so it remains to say something about the extension, ie the upper right star in the matrices of $\phi$ and $\gamma \in \Gamma$. This is less easy; one can't simply write general formulas, but by using the results of Cherbonnier-Colmez (JAMS) and Colmez (eg his paper on trianguline representations), you can say a number of interesting things.</p> http://mathoverflow.net/questions/101893/incidences-of-rigorous-proofs-used-in-legal-proceedings/101935#101935 Answer by Laurent Berger for Incidences of rigorous proofs used in legal proceedings Laurent Berger 2012-07-11T09:52:19Z 2012-07-11T09:52:19Z <p>In France some roadside apparatus will take a picture of you at point A and another apparatus will take a picture at point B on the same highway. They will then apply the mean value theorem to determine if you deserve a ticket for speeding.</p> http://mathoverflow.net/questions/100323/whats-the-name-for-the-analogue-of-divided-power-algebras-for-xi-i/100586#100586 Answer by Laurent Berger for What's the name for the analogue of divided power algebras for x^i/i? Laurent Berger 2012-06-25T11:56:15Z 2012-06-25T11:56:15Z <p>One possible way to generalize divided powers is to model sequences $x^n/a_n$ where $a_{n+m}/(a_na_m)$ is integral (an integer, or at least integral in some sense). You can model these in the obvious way, like divided powers. Choosing $a_n = n!$ gives you divided powers, and choosing $a_n=p^n$ gives you a sequence which, from the point of view of $p$-adic analysis, is much more "regular" in its growth. You can come up with other examples.</p> <p>As you may know, one uses divided powers to construct Fontaine's ring of periods $B_{cris}$ and using $a_n=p^n$ instead of $a_n=n!$ gives a ring $B_{max}$ that is similar in nature to $B_{cris}$ but much better behaved (see III.2 of Colmez' 1998 paper "Théorie d'Iwasawa des représentations de de Rham d'un corps local").</p> <p>More generally, tweaking divided powers to get various convergence conditions is something that happens often in $p$-adic Hodge theory. See for instance 5.2.3 of Fontaine's "Le corps des périodes $p$-adiques" for one of many examples of custom made convergence conditions.</p> <p>In the $p$-adic setting, I would say that your log series belongs to the "space of holomorphic functions on the $p$-adic open unit disk, having order of growth $\leq 1$". Using precise analytic conditions rather than "divided powers" seems the right way to study the analytic series that interest you.</p> http://mathoverflow.net/questions/98103/1-dimensional-semi-stable-galois-representations-with-coefficients/98115#98115 Answer by Laurent Berger for 1-dimensional semi-stable Galois representations with coefficients Laurent Berger 2012-05-27T14:14:23Z 2012-05-27T14:30:11Z <p>If $E \neq Q_p$ then there may be more $1$-dimensional crystalline representations that the ones you mention. By Lubin-Tate theory, every character of $G_K$ can be written as an unramified character times a character of $O_K^\times$ (after making proper choices and identifications). The algebraic characters of $O_K^\times$ are then crystalline and if $E$ contains $K$, then these provide examples of crystalline characters.</p> <p>EDIT : oops sorry, I did not read the question carefully enough, I did not see that you were asking for an explicit description of <em>all</em> such representations. As David Loeffler pointed out, the answer is in Brian Conrad's paper (now in appendix B, the paper having grown since last time I mentioned it). </p> <p>Here is what it basically says, using the same identifications as above: assume that $E$ contains $K^{gal}$ (it's usually harmless to assume that the coefficient field is large enough). If $s$ runs through the set of embeddings $s : K \to E$ and the $a_s$ are integers, then $x \mapsto \prod_s s(x)^{a_s}$ gives rise to a crystalline character of $G_K$ and they're all of this type times an unramified character.</p> http://mathoverflow.net/questions/96257/where-to-find-asterisque-online/96305#96305 Answer by Laurent Berger for Where to find Asterisque online? Laurent Berger 2012-05-08T07:33:04Z 2012-05-08T07:53:09Z <p>The issue of providing online access to Asterisque is a difficult one. The SMF offers electronic versions of its other journals, and the subscription rates for these has apparently significantly dropped. Asterisque is now the most profitable publication of the SMF and there is a lot of reluctance to join the 21st century and jeopardize this profitability. I agree with Obelisque that the best solution is to buy the old issues, but the ones that are most in demand are also out of print. All of these topics are currently being discussed (electronic versions, reprinting the back issues, ...). It is also not impossible that the Bourbaki seminars will soon be available online, maybe with a five year buffer, as the contract beween Bourbaki and the SMF gives total freedom to Bourbaki.</p> <p>EDIT : of course, one reason why the standard model for journals does not quite apply to Asterisque is because it's part journal, part book series...</p> http://mathoverflow.net/questions/91104/are-the-smooth-vectors-of-a-frechet-space-dense/91152#91152 Answer by Laurent Berger for are the smooth vectors of a Frechet space dense? Laurent Berger 2012-03-14T08:10:56Z 2012-03-14T08:10:56Z <p>In full generality, the answer is "no". If you have a unitary irreducible representation of a $p$-adic Lie group on a $p$-adic Banach space, then the locally constant vectors are usually not dense. There may even be no nonzero locally algebraic vector. The locally constant vectors are dense however if you have an $\ell$-adic Lie group acting on a $p$-adic space, with $\ell \neq p$ (this is a theorem of Vigneras). </p> http://mathoverflow.net/questions/83953/can-a-p-adic-representation-and-its-twist-by-a-non-crystalline-character-both-hav/83959#83959 Answer by Laurent Berger for Can a p-adic representation and its twist by a non-crystalline character both have nontrivial $D_{cris}$? Laurent Berger 2011-12-20T16:36:05Z 2011-12-20T16:36:05Z <p>Let me use Colmez' article "Representations triangulines" as a reference. Let $V$ be a repn which satisfies your condition.</p> <p>By proposition 4.3, $V$ is trianguline. By proposition 4.10, the HT weight of $\chi$ has to be an integer. You can then assume that $\chi$ has finite order, and this implies that $V$ is potentially crystalline on an abelian extension of $Q_p$ (aka crystabelline).</p> <p>Conversely, it seems likely that one can give examples of irreducible crystabelline representations $V$ such that $D_{cris}(V)$ and $D_{cris}(V \otimes \chi)$ are both nonzero, with $\chi$ of finite order (see the examples in 2.4 of Berger-Breuil's "Sur quelques representations potentiellement cristallines de $GL_2(Q_p)$").</p> http://mathoverflow.net/questions/82720/banach-mazur-applied-to-a-hilbert-space Banach-Mazur applied to a Hilbert space Laurent Berger 2011-12-05T18:40:39Z 2011-12-06T01:13:46Z <p>The Banach-Mazur theorem says that every separable Banach space is isometric to a subspace of $C^0([0;1],R)$, the space of continuous real valued functions on the interval $[0;1]$, with the sup norm.</p> <p>If we apply this to $\ell^2(R)$, then we see that $C^0([0;1],R)$ has a subspace which is a Hilbert space for the sup norm. </p> <p>My question is can one write down explicitly such a subspace of $C^0([0;1],R)$?</p> <p>I'm just curious, that's all.</p> http://mathoverflow.net/questions/80637/p-adic-representations/81861#81861 Answer by Laurent Berger for P-adic representations Laurent Berger 2011-11-25T10:11:27Z 2011-11-25T10:11:27Z <p>The whole thing is done with more details in Tate's original article "p-divisible groups", section 3.2. Tate proves that one can approximate a cocyle in $C_p(i)$ by cocyles with values in $Q_p^{alg}(i)$ and this is how he reduces the computation to the "discrete case". </p> <p>I would suggest that it's better to prove the result by $p$-adic approximation. This way, you can basically work with cocycles with values in $O_{C_p}(i)/p^n$, also a discrete space.</p> http://mathoverflow.net/questions/81610/is-an-identity-that-is-true-for-matrix-lie-groups-true-for-all-lie-groups/81612#81612 Answer by Laurent Berger for Is an identity that is true for matrix Lie groups true for all Lie groups? Laurent Berger 2011-11-22T14:44:54Z 2011-11-22T14:44:54Z <p>Every finite dimensional Lie algbera over a field of characteristic $0$ is a subalgebra of some matrix Lie algebra (Ado's theorem), so every Lie group is locally isomorphic to a group of matrices.</p> http://mathoverflow.net/questions/81342/elementary-results-with-p-adic-numbers/81403#81403 Answer by Laurent Berger for Elementary results with p-adic numbers Laurent Berger 2011-11-20T08:29:35Z 2011-11-20T08:29:35Z <blockquote> <p>In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis?</p> </blockquote> <p>These results have very different levels of sophistication. I would propose the Weil conjectures as an application : <a href="http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures" rel="nofollow">http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures</a></p> <p>The first statement (rationality of the Zeta function) was originally proved by Dwork using purely $p$-adic methods. It's a beautiful application of $p$-adic analysis and $p$-adic functional analysis. In addition, it should not be too hard to state the theorem, since it's about counting solutions of polynomials in finite fields.</p> <p>Finally, you can also say that Kedlaya now has given a purely $p$-adic proof of the complete conjecture (previously proved by Deligne using other methods).</p> http://mathoverflow.net/questions/78328/enumerating-non-abelian-extensions-of-mathbbq-p/78329#78329 Answer by Laurent Berger for Enumerating non-abelian extensions of $\mathbb{Q}_p$? Laurent Berger 2011-10-17T11:59:17Z 2011-10-17T11:59:17Z <p>This is standard stuff. Here is (in French) the solution as an exercise, copy-pasted from the final exam of a course I gave on local fields.</p> <p>Soit $K$ une extension totalement ramifiée de degré $n$ de $Q_p$ et $\pi_K$ une uniformisante de $K$. On suppose pour l'instant que $p \nmid n$.</p> <ol> <li><p>Montrer que si $w \in Q_p$ et $w^n=1$, alors $w^m=1$ où $m = n \wedge (p-1)$ (si $p \neq 2$) et $m=2$ si $p=2$.</p></li> <li><p>Montrer que l'application $x \mapsto x^n$ de $1+M_K$ dans lui-même est surjective.</p></li> <li><p>Montrer que dans $O_K$, on peut écrire $\pi_K^n = p w (1+z)$ où $w^{p-1} = 1$ et $z \in M_K$.</p></li> <li><p>En déduire que $Q_p$ admet exactement $n$ extensions totalement ramifiées de degré $n$.</p></li> </ol> http://mathoverflow.net/questions/70262/is-there-a-trianguline-period-ring-or-is-one-expected/70299#70299 Answer by Laurent Berger for Is there a "trianguline period ring", or is one expected? Laurent Berger 2011-07-14T06:45:05Z 2011-07-14T06:45:05Z <p>The category of trianguline representation is stable under all the usual representation-theoretic operations (subs, quotients, $\oplus$, $\otimes$), so by some general tannakian formalism, there does exist a ring $B_{tri}$. The rough idea is to look at $Q_p^{alg} \otimes B_{st} \langle \langle \log(t) \rangle \rangle$ where "$\langle \langle \log(t) \rangle \rangle$" means "power series with some non zero radius of convergence" and $t$ is the usual $t$ in this business. It's interesting to note that I first heard about this ring from Fontaine (around 2003-04 maybe - I was still at Harvard) when trianguline representations had not yet been defined. Fontaine told me at the time that repns admissible for this ring should be interesting! A few comments are in order: </p> <ul> <li>$B_{st}$ does not have the structure of a Banach space, so you need to figure out what "radius of convergence" means</li> <li>$\exp(\log(t))=t$, so there are relations in the definition of your ring</li> <li>you need to decide if you want a ring for "trianguline" or "split-trianguline" repns</li> </ul> <p>I thought about this again a few weeks ago and, if I remember correctly, came to the conclusion that if you take $B = Q_p(\mu_p) \otimes \hat{Q}_p^{nr} \otimes B_e \otimes Q_p \langle \langle \log(t) \rangle \rangle \otimes Q_p[\log(\tilde{p})]$ (whew!), where $B_e=B_{cris}^{\phi=1}$, then $B$-adm reps of $G_{Q_p}$ are trianguline, and conversely split trianguline reps of $G_{Q_p}$ with integer slopes are $B$-adm. This hopefully gives an idea of the kind of ring which one should be looking for.</p> http://mathoverflow.net/questions/52101/longest-coinciding-pair-of-integer-sequences-known/69409#69409 Answer by Laurent Berger for Longest coinciding pair of integer sequences known Laurent Berger 2011-07-03T18:04:13Z 2011-07-03T18:04:13Z <p>There is also, of course, what comes out of the answer which was given to this question:</p> <p><a href="http://mathoverflow.net/questions/11517/computer-algebra-errors" rel="nofollow">http://mathoverflow.net/questions/11517/computer-algebra-errors</a></p> http://mathoverflow.net/questions/65840/tamagawa-numbers-of-crystalline-galois-representations/66037#66037 Answer by Laurent Berger for Tamagawa numbers of crystalline Galois representations Laurent Berger 2011-05-26T08:57:48Z 2011-05-26T08:57:48Z <p>I think that for $K_n=\mathbf{Q}_p$ what you're looking for is in my (unpublished) paper</p> <p><a href="http://perso.ens-lyon.fr/laurent.berger/autrestextes/tamag0919.pdf" rel="nofollow">http://perso.ens-lyon.fr/laurent.berger/autrestextes/tamag0919.pdf</a></p> <p>see proposition II.2 for instance.</p> <p>This paper is unpublished because it was rewritten and massively expanded with/by Denis Benois. There is some stuff in the paper with Benois about going up the cyclotomic tower which may help. I'm sorry I don't have time to look in Benois-Berger to see if there's an answer to your full question.</p> http://mathoverflow.net/questions/62972/resources-for-mathematics-advising/63023#63023 Answer by Laurent Berger for Resources for mathematics advising. Laurent Berger 2011-04-26T10:56:54Z 2011-04-26T10:56:54Z <p>Since the OP is asking for resources, I'll give two references that I've found to be useful.</p> <ul> <li>The Survival of a Mathematician, by Steven Krantz</li> <li>I Want to Be a Mathematician: An Automathography, by Paul Halmos</li> </ul> <p>Both books have something to say on the subject.</p> <p>And it can't hurt to add the AMS ethical guidelines:</p> <p><a href="http://www.ams.org/about-us/governance/policy-statements/sec-ethics" rel="nofollow">http://www.ams.org/about-us/governance/policy-statements/sec-ethics</a> </p> http://mathoverflow.net/questions/61998/crystalline-characters/62004#62004 Answer by Laurent Berger for Crystalline Characters Laurent Berger 2011-04-17T07:18:26Z 2011-04-17T07:18:26Z <p>Hi Kevin,</p> <ol> <li>Go to <a href="http://math.stanford.edu/~conrad/" rel="nofollow">http://math.stanford.edu/~conrad/</a></li> <li>Download "Grunwald--Wang for global character groups"</li> <li>Read appendix A, especially prop A.3</li> </ol> <p>The answer (note that $K$ and $L$ are switched in Brian's paper) is that once you've identified your character as a character of $K^\times$ via local class field theory, it should be "algebraic" on $O_K^\times$.</p> http://mathoverflow.net/questions/57952/how-irregular-can-a-p-adic-galois-representation-be/57955#57955 Answer by Laurent Berger for how irregular can a $p$-adic Galois representation be? Laurent Berger 2011-03-09T14:29:29Z 2011-03-09T14:29:29Z <p>Here are two things that can occur:</p> <p>1) If V is the representation attached to an overconvergent modular form f, then V will be unramified at almost every prime but will not be de Rham at p (unless f is classical).</p> <p>2) Ramakrishna has written an article "Infinitely ramified Galois representations". Here's part of the introduction: "In this paper we show how to construct [...] representations [...] that are ramified at an infinite number of primes." Under GRH, these repns are crystalline at p.</p> <p>So both conditions in FM can fail independently.</p> http://mathoverflow.net/questions/126529/a-criterion-for-freeness-over-a-local-ring/126533#126533 Comment by Laurent Berger Laurent Berger 2013-04-04T16:36:08Z 2013-04-04T16:36:08Z Great, thank you! http://mathoverflow.net/questions/111187/commuting-invariants-and-duals-of-c-p-vector-spaces Comment by Laurent Berger Laurent Berger 2012-12-17T08:41:20Z 2012-12-17T08:41:20Z Concerning your first question: a $G_K$-equivariant map from V to $C_p$ exists, for example, whenever $V$ is of Hodge-Tate type with one weight equal to $0$. This happens for lots of repns for which $V^{G_K} = 0$. http://mathoverflow.net/questions/115375/algebraic-maximal-extension-and-algebraic-closure/115385#115385 Comment by Laurent Berger Laurent Berger 2012-12-04T15:49:59Z 2012-12-04T15:49:59Z $\mathbf{C}_p$ has plenty of immediate extensions, that is extensions that have the same value group and the same residue field. The compositum of all these is its spherical completion. Now lots of intermediate fields between $\mathbf{C}_p$ and its spherical completion would not be algebraic maximal. http://mathoverflow.net/questions/114245/i-know-that-you-know Comment by Laurent Berger Laurent Berger 2012-11-23T15:35:36Z 2012-11-23T15:35:36Z There is a (not serious) mention of this in &quot;A canticle for Leibowitz&quot;, but sadly I don't remember where exactly. Can somebody find the exact quotation? http://mathoverflow.net/questions/112545/absolute-galois-group-of-the-field-of-puiseux-series-over-overline-mathbbf/113884#113884 Comment by Laurent Berger Laurent Berger 2012-11-20T12:27:22Z 2012-11-20T12:27:22Z @beginner If I can be allowed to advertize my own stuff: I wrote a survey on p-adic representations a few years ago, and one chapter concerns $(\varphi,\Gamma)$-modules. See [05] of <a href="http://perso.ens-lyon.fr/laurent.berger/publications.php" rel="nofollow">perso.ens-lyon.fr/laurent.berger/publications.php</a> http://mathoverflow.net/questions/112545/absolute-galois-group-of-the-field-of-puiseux-series-over-overline-mathbbf/113884#113884 Comment by Laurent Berger Laurent Berger 2012-11-20T08:59:50Z 2012-11-20T08:59:50Z @Filippo You can do it in a few words: Colmez recalls the theory of $(\varphi,\Gamma)$-modules in a footnote, footnote 86 of his Bourbaki seminar on Kato's Euler system ([21] of <a href="http://www.math.jussieu.fr/~colmez/publications.html" rel="nofollow">math.jussieu.fr/~colmez/publications.html</a>). http://mathoverflow.net/questions/112627/why-is-gauss-credited-with-his-connection/112639#112639 Comment by Laurent Berger Laurent Berger 2012-11-17T14:48:14Z 2012-11-17T14:48:14Z Ah, so the story about Gauss inventing the least squares method to compute orbital parameters of asteroids is bogus? http://mathoverflow.net/questions/112545/absolute-galois-group-of-the-field-of-puiseux-series-over-overline-mathbbf Comment by Laurent Berger Laurent Berger 2012-11-16T09:25:10Z 2012-11-16T09:25:10Z @Spice: if you replace $\overline{\mathbb{F}}_p$ iwth $\mathbb{C}$, then the field you get is algebraically closed. http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/110451#110451 Comment by Laurent Berger Laurent Berger 2012-10-24T07:44:51Z 2012-10-24T07:44:51Z @Charles : yes, sorry, I forgot to say &quot;free&quot;, so the modules should be free of the same rank. Thanks. http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/32984#32984 Comment by Laurent Berger Laurent Berger 2012-10-23T17:29:20Z 2012-10-23T17:29:20Z And the closed graph criteria also works in a compact space! http://mathoverflow.net/questions/108391/are-d-dr-and-d-st-potentially-comparable/108401#108401 Comment by Laurent Berger Laurent Berger 2012-09-29T11:56:59Z 2012-09-29T11:56:59Z By the way, these &quot;thin subsets&quot; are called &quot;parties fines&quot; in our article. I suspect that my coauthor was playing a joke on me (and the readers) since I later found out that in French, &quot;partie fine&quot; also means &quot;orgy&quot;. http://mathoverflow.net/questions/106468/local-galois-representation-with-higher-coefficient Comment by Laurent Berger Laurent Berger 2012-09-06T11:50:00Z 2012-09-06T11:50:00Z If you're looking at linear representations of $G$ with coefficients, then everything works &quot;the same&quot;. See for example 3.1 of Breuil-M&#233;zard's 2002 Duke paper. http://mathoverflow.net/questions/106468/local-galois-representation-with-higher-coefficient Comment by Laurent Berger Laurent Berger 2012-09-06T07:15:18Z 2012-09-06T07:15:18Z You also changed the setting completely. Is $F$ an extension of $K$ or a subfield of $K$?? http://mathoverflow.net/questions/70262/is-there-a-trianguline-period-ring-or-is-one-expected/70299#70299 Comment by Laurent Berger Laurent Berger 2012-08-16T16:01:55Z 2012-08-16T16:01:55Z This ring has not been studied. At some point, my student Di Matteo was interested in it, so you could ask him. http://mathoverflow.net/questions/103945/valuations-on-tensor-products Comment by Laurent Berger Laurent Berger 2012-08-04T16:32:32Z 2012-08-04T16:32:32Z Isn't this just a matter of saying $v(b \otimes c) = v(b) + v(c)$ and letting the valuation of an element of $B \otimes C$ be the sup, over all possible ways to write the element as $\sum b_i \otimes c_i$, of $\inf_i v(b_i \otimes c_i)$?