most recent 30 from http://mathoverflow.net 2013-05-26T02:42:21Z http://mathoverflow.net/feeds/user/10427 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116104/functors-on-rigid-tensor-categories Auguste Hoang Duc 2012-12-11T18:21:21Z 2012-12-12T00:23:22Z

<p>This is a question about the proof of proposition 1.13 in Deligne and Milne, Tannakian Categories. Let $C,C'$ be two rigid tensor categories and $F,G : C \rightarrow C'$ be two tensor functors. Let $u : F \rightarrow G$ be a morphism of functors. Define the morphism $v : G \rightarrow F$ by $$ v(X) : G(X) \simeq G(X^\vee)^\vee \xrightarrow{{}^t u(X^\vee)} F(X^\vee)^\vee \simeq F(X).$$</p> <blockquote> <p>Why is $v$ the inverse of $u$ ?</p> </blockquote>

http://mathoverflow.net/questions/103676/maximal-subfields-in-a-division-algebra-over-a-local-field/103679#103679 Auguste Hoang Duc 2012-08-01T11:57:15Z 2012-08-01T11:57:15Z

<p>Corollaries 3.4 and 3.7 on pages 130 and 131 of Milne's notes (jmilne.org/math/CourseNotes/CFT.pdf) will give you the answers.</p>

http://mathoverflow.net/questions/65411/how-can-we-extend-galois-representations Auguste Hoang Duc 2011-05-19T09:17:55Z 2011-05-19T10:38:55Z

<p>Let $F$ be a number field and $E/F$ a Galois extension. Suppose we have a representation $\rho_E : Gal(\overline{F}/E) \rightarrow GL_n(\overline{Q}_p)$. My question is : what are sufficiant conditions so that $\rho_E$ can be extended to a representation $ Gal(\overline{F}/F) \rightarrow GL_n(\overline{Q}_p)$ ?</p> <p>A necessary condition is that $\rho_E$ is invariant under $Gal(E/F)$. This paper (last line of page 1)</p> <p><a href="http://www.institut.math.jussieu.fr/projets/fa/bpFiles/GaloisPatching_Harris.pdf" rel="nofollow">http://www.institut.math.jussieu.fr/projets/fa/bpFiles/GaloisPatching_Harris.pdf</a></p> <p>claims that such an extension exists if moreover $\rho_E$ is irreducible and $E/F$ is cyclic of prime ordre, but I don't know why it is true. </p>

http://mathoverflow.net/questions/53124/snake-lemma-in-category-of-groups/53249#53249 Auguste Hoang Duc 2011-01-25T17:40:03Z 2011-01-25T17:40:03Z

<p>I think one can have the Snake Lemma in the category of (non abelian) groups by the following way.</p> <p>First, I remind what is an exact sequence of pointed sets (see, for example, "Local fields" of Serre in the chapter about non abelian cohomology) : a pointed set is just a set $A$ with a based point $x_A$. A morphism of pointed sets is defined to be a map sending the based point to the based point. A sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is said to be exact if $f(A) = g^{-1}(x_C)$.</p> <p>Now let's go back to the snake Lemma. A cokernel is a pointed set with the class of $0$ as based point (I denote $Coker(f:A \rightarrow B)$ to be the left cosets $B/Im(f)$ but it also works for right cosets, and I call it cokernel even if it is NOT the cokernel in the theory of category). One can check that the Snake Lemma holds in the sense I have given above (with the same proof as Alex said).</p>

http://mathoverflow.net/questions/44850/what-are-the-p-adic-representations-of-hat-mathbbz Auguste Hoang Duc 2010-11-04T17:47:12Z 2010-11-19T14:08:22Z

<p>A continuous representation $\hat{\mathbb{Z}} \rightarrow GL_n(\mathbb{Q}_p)$ is determined by the image of $1$. But the image of $1$ does not always defines such a representation (consider for example the representation which sends $1$ on $p$ from $\mathbb{Z}$ to $GL_1(\mathbb{Q}_p)$). So my question is : what are the conditions on the image of $1$ ?</p> <p>For example if $n=1$, then I know that $1$ must be sent on an element of $\mathbb{Z}_p^\times$, but I don't know if the converse is true.</p> <p>EDIT: Correction about the example.</p>

http://mathoverflow.net/questions/114139/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h/111118#111118 Auguste Hoang Duc 2012-11-02T13:08:31Z 2012-11-02T13:08:31Z

@Fran&#231;ois: I think you are using the fact that '$a_i : G \rightarrow H_i$ ($i=1,2$) are surjective so $(a_1,a_2) : G \rightarrow H_1 \times H_2$ is surjective', which is wrong.

http://mathoverflow.net/questions/99426/books-and-or-papers-without-references Auguste Hoang Duc 2012-06-13T11:09:45Z 2012-06-13T11:09:45Z

such that <a href="http://www.springerlink.com/content/gtj5613410w64530/" rel="nofollow">springerlink.com/content/gtj5613410w64530</a> ?

http://mathoverflow.net/questions/72467/does-the-zariski-closure-of-a-maximal-subgroup-remain-maximal Auguste Hoang Duc 2011-08-09T11:21:20Z 2011-08-09T11:21:20Z

What set do you apply Zorn's lemma to ? Because a union of strict subgroups may not be strict.

http://mathoverflow.net/questions/69620/k-th-powers-in-the-field-of-p-adics/69626#69626 Auguste Hoang Duc 2011-07-06T19:45:05Z 2011-07-06T19:45:05Z

I am sorry, there are typos in my previous comment. I wanted to say that $(O_K^\times)^n$ is open. And a $\times$ is missing.

http://mathoverflow.net/questions/69620/k-th-powers-in-the-field-of-p-adics/69626#69626 Auguste Hoang Duc 2011-07-06T19:14:35Z 2011-07-06T19:14:35Z

@L Spice : Because $Z_p = F_p^\times \times Z_p \times Z$ (as topological groups). It is aslo true that for $K$ a p-adic field $O_K^\times$ is open. Milne proves it with Newton's Lemma in his notes on Class Field Theory (look at page 22, (1.7) after corollary 1.5) <a href="http://www.jmilne.org/math/CourseNotes/CFT.pdf" rel="nofollow">jmilne.org/math/CourseNotes/CFT.pdf</a>

http://mathoverflow.net/questions/69620/k-th-powers-in-the-field-of-p-adics/69626#69626 Auguste Hoang Duc 2011-07-06T12:26:29Z 2011-07-06T12:26:29Z

You can notice that $(Z_p^\times)^k$ is open. It avoids the use of series.

http://mathoverflow.net/questions/65411/how-can-we-extend-galois-representations Auguste Hoang Duc 2011-05-19T10:23:29Z 2011-05-19T10:23:29Z

It means that : for all $\sigma \in G(\overline{{F})/F$, the representation $\rho_E^\sigma := \rho_E(\sigma \cdot \sigma^{-1})$ is isomorphic to $\rho_E$. Note that the isomorphic class of $\rho_E^\sigma$ does not depend on $\sigma \in G(E/F)$.

http://mathoverflow.net/questions/60457/elementaryshortuseful/60573#60573 Auguste Hoang Duc 2011-04-12T08:30:45Z 2011-04-12T08:30:45Z

@L Spice : Can you tell me about your method ? (sorry for the split, but the comment was too long).

http://mathoverflow.net/questions/60457/elementaryshortuseful/60573#60573 Auguste Hoang Duc 2011-04-12T08:25:48Z 2011-04-12T08:25:48Z

@L Spice : I call it Dunford because it is the French term. My method is the following : let $P(x)$ be the caracteristic polynomial of a matrix $A$ and let $Q(X):=P(X)/(gcd(P'(X),P(X)))$ (assume $caract(k)=0$ otherwise the formula for $Q(X)$ is more complicated). Consider the sequence defined by $A_0:=A$ and $A_{n+1}:=A_n-Q(A_n)/Q'(A_n)$. Then for all $n \geq log_2(dimension)$, the matrix $A_n$ is the semisimple part of $A$ (the key point is to notice that the semi-simple part is a zero of $Q(X)$ in the vector space $k[A]$). I don't know about your method, so I can't tell if it is the same.

http://mathoverflow.net/questions/60979/what-is-the-definition-of-an-invariant-of-elliptic-curve Auguste Hoang Duc 2011-04-07T18:14:28Z 2011-04-07T18:14:28Z

The equation of $C'$ IS an elliptic curve, but it is not in the Weierstrass form. So the formula of $j$ is more complicated in this case. In Silvermann's book, transformation doesn't involve $y'$ in $x'$.

http://mathoverflow.net/questions/60457/elementaryshortuseful/60573#60573 Auguste Hoang Duc 2011-04-06T16:36:03Z 2011-04-06T16:36:03Z

I was amazed when I learned that it also works on spaces other than $R^n$, e.g. : - In $M_n(k)$, to find the Dunford decomposition. - In $Z/nZ$, to solve congruence. - In $Z_p$, to prove Hensel lemma, but this situation is quite similar from $R^n$.

http://mathoverflow.net/questions/60076/on-maximal-ideals Auguste Hoang Duc 2011-03-30T15:13:07Z 2011-03-30T15:13:07Z

What is Zpn[X] ?

http://mathoverflow.net/questions/58847/polynomial-with-galois-group-d-2n/58849#58849 Auguste Hoang Duc 2011-03-18T21:22:21Z 2011-03-18T21:22:21Z

I have found this one, which may be interesting : <a href="http://library.msri.org/books/Book45/files/book45.pdf" rel="nofollow">library.msri.org/books/Book45/files/book45.pdf</a>

http://mathoverflow.net/questions/54550/the-third-axiom-in-the-definition-of-infinite-dimensional-vector-bundles-why/54552#54552 Auguste Hoang Duc 2011-02-06T21:03:05Z 2011-02-06T21:03:05Z

Shouldn't $h$ goes into $R$ instead of $L^2[0,1]$ ?

http://mathoverflow.net/questions/52848/on-rational-functions-with-rational-power-series/52869#52869 Auguste Hoang Duc 2011-01-27T22:06:58Z 2011-01-27T22:06:58Z

There is something which bothers me. You use the fact that the subfield of $\C$ fixed by $Aut(\C)$ is $\Q$. Why is it true ? For exemple it is false if you replace $\C$ by $\R$.