User grilo - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:39:19Z http://mathoverflow.net/feeds/user/25696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120972/realisation-functor Realisation functor Grilo 2013-02-06T13:49:40Z 2013-03-20T15:22:00Z <p>Let k be a field. Is there a realization functor</p> <p>$DM_{gm}(k,\mathbb{Z}/n)^{op} \to D^b_c(k, \mathbb{Z}/n)$</p> <p>from category of motives to category of complexes of étale sheaves of $\mathbb{Z}/n$ modules with bounded constructible cohomology sheaves?</p> http://mathoverflow.net/questions/120760/k-theory-and-tame-symbol K-Theory and Tame Symbol Grilo 2013-02-04T11:53:14Z 2013-02-04T12:20:41Z <p>This might be a little bit spesific but here it goes. While reading a paper (Brauer-Manin pairing...) by Yamazaki, I encountered this definition.</p> <p>Let $V$ be a variety. $y$ be a one dimensional point on $V$, i.e. $dim\overline{ { y } } = 1$. Then let $C(y)$ be its closure in $V$ and $\tilde{C}(y)$ be normalization and $\bar C(y)$ be smooth completion. Let $C_\infty : = \bar C(y) - \tilde{C}(y)$. Then he definied the following group:</p> <p>$UK^M_{r+1}:= ker [K_r^M(k(y)) \to \bigoplus_{x \in C_\infty}(K_{r-1}^M(k(x)) \oplus K_{r}^M(k(x))]$.</p> <p>First component is the tame symbol at $x$. He defined the second component as $a \to \partial_x(a \cup \pi_x)$ where $\partial_x$ is the tame symbol at $x$ and $\pi_x$ is a uniformizer at $x$. </p> <p>He says this group does not depend on the choice of the uniformizer, I couldn't see why it doesn't.</p> <p>Is there an easy way to tell this group does not depend on the choice of the uniformizer?</p> http://mathoverflow.net/questions/116868/fundamental-group-and-etale-cohomology Fundamental Group and Etale Cohomology Grilo 2012-12-20T14:33:03Z 2012-12-20T15:43:50Z <p>I encountered the following statement without a reference many times. For a smooth variety $X$ over a perfect field $k$.</p> <p>$Hom(H^1_{et}(X, \mathbb{Z}/n), \mathbb{Z}/n) \cong \pi^{ab}_1(X)/n$</p> <p>Is there any reference for this? Why this is true?</p> http://mathoverflow.net/questions/116770/reciprocity-map-and-cycle-class-map Reciprocity Map and Cycle Class Map Grilo 2012-12-19T08:56:34Z 2012-12-19T08:56:34Z <p>This might be a very naive question but here it goes. Let X be a smooth variety of dimension d over a p-adic field. We have the n part of the rerciprocity map:</p> <p>$rec/n: SK_1(X)/n \to \pi^{ab}_1(X)/n$</p> <p>Right hand side can be identified with $H^{2d+1}_{et}(X, \mathbb{Z}/n(d+1))$ by Poincare duality. Left hand side can be identified with $CH^{d+1}(X,1,\mathbb{Z}/n)$ (cf. Lemma 2.8, S. Landsburg, Relative Chow Groups)</p> <p>So I can see this as a map </p> <p>$rec/n: CH^{d+1}(X,1,\mathbb{Z}/n) \to H^{2d+1}_{et}(X, \mathbb{Z}/n(d+1))$ </p> <p>What I wonder is that if this map agrees with the higher cycle class map i.e. whether their kernels (cokernels) are isomorphic.</p> http://mathoverflow.net/questions/104681/comparing-spectral-sequences Comparing Spectral Sequences Grilo 2012-08-14T09:51:30Z 2012-08-17T11:40:34Z <p>There is a comparison theorem for spectral sequnces in Weibel's book (5.2.12) stating;</p> <p>Assume $E_{p,q}$ and $\bar E_{p,q}$ converge to $H_*$ $\bar H_*$ respectively. Furthermore we have given a map $h: H_{*} \to \bar H_{*}$ compatible with a morphism $f$ of spectral sequences. </p> <p>If $f^r: E^r_{p,q} \to \bar E^r_{p,q}$ is an isomorphism for all $p,q$ and some $r$ then $h$ is an isomorphism.</p> <p>What I want to ask is what happens if we have a milder situation than isomorphism. For example if they just differ on the border? </p> <p>To be precise let $E^2_{p,q}$ and $\bar E^2_{p,q}$ are two first quadrant spectral sequences converging to $H_*$ $\bar H_*$ respectively. Also there is a map $h$ compatible with a morphism $f$ of spectral sequences as above. Assume $E^2_{p,q} \cong \bar E^2_{p,q}$ if $q\neq0$ and $E^2_{p,0}$ vanishes. Can we calculate kernel and cokernel of $h$? </p> <p>Thanks for your help.</p> http://mathoverflow.net/questions/116770/reciprocity-map-and-cycle-class-map Comment by Grilo Grilo 2012-12-20T01:17:00Z 2012-12-20T01:17:00Z $SK_1(X)$ is defined as $coker(\bigoplus_{y \in X_{(1)}}K_2(k(y)) \to \bigoplus_{x \in X_{(0)}}k(x)^*)$ where this map arises from localization theory of $K$ groups. Here $X_{(a)}$ denotes the points of dimension $a$. http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T10:51:54Z 2012-08-17T10:51:54Z I think there is a mistake in your edit. The first sequence should be \begin{eqnarray} 0 \to \bar E^{r+2}_{n+1,0} \to \bar E^{r+1}_{n+1,0} \to E^{r+2}_{n-r,r} \to \bar E^{r+2}_{n-r,r} \to 0 \nonumber \end{eqnarray} Since we have a map from $\bar E^{r+1}_{n+1,0}$ to $E^{r+2}_{n-r,r}$. And the rest follows accordingly. http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T09:23:26Z 2012-08-17T09:23:26Z Also as you said it should be \begin{eqnarray} \bar E^2_{n+1,0} \to E^3_{n-1,1} \to \bar E^3_{n-1,1} \to 0 \nonumber \end{eqnarray} I now got confused about if the correspoding kernels for graded pieces gives the exact sequence \begin{eqnarray} 0 \to \bar E^\infty_{n+1,0} \to \bar E^{2}_{n+1,0} \to H_n \to \bar H_n \nonumber \end{eqnarray} http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T09:19:35Z 2012-08-17T09:19:35Z &quot;Please think this as the second comment&quot; and the kernel of the first map is $\bar E^3_{n+1,0}$. This implies with simple calculation \begin{eqnarray} 0 \to \bar E^3_{n+1,0} \to \bar E^2_{n+1,0} \to E^\infty_{n-1,1} \to {\bar E}^\infty_{n-1,1} \to 0 \nonumber \end{eqnarray} Also in general the same method gives \begin{eqnarray} 0 \to \bar E^{r+2}_{n+1,0} \to \bar E^{r+1}_{n+1,0} \to E^\infty_{n-r,r} \to {\bar E}^\infty_{n-r,r} \to 0 \nonumber \end{eqnarray} http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T07:02:04Z 2012-08-17T07:02:04Z And the interesting result comes combining with what you write about cokernel: we have a long exact sequence \begin{eqnarray} .... \to H_{n-1} \to \bar H_{n-1} \to \bar E^{2}_{n+1,0} \to H_n \to \bar H_n \to \bar E^{2}_{n,0} \to ... \nonumber \end{eqnarray} http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T06:57:36Z 2012-08-17T06:57:36Z Finally if we look at the last one we get \begin{eqnarray} 0 \to \bar E^\infty_{n+1,0} \to \bar E^{n+1}_{n+1,0} \to E^\infty_{0,n} \to {\bar E}^\infty_{0,n} \to 0 \nonumber \end{eqnarray} since $\bar E^\infty_{n+1,0} = \bar E^{n+2}_{n+1,0}$. Combining these one gets an exact sequence \begin{eqnarray} 0 \to \bar E^\infty_{n+1,0} \to \bar E^{2}_{n+1,0} \to H_n \to \bar H_n \nonumber \end{eqnarray} http://mathoverflow.net/questions/104681/comparing-spectral-sequences/104786#104786 Comment by Grilo Grilo 2012-08-17T06:55:34Z 2012-08-17T06:55:34Z Wilberd - Thanks a lot for the input. I think I also calculated the kernel. Since I am new to SS it is probable that I might missed something. Anyways here it goes: First note that $E^3_{n-1,1} = Ker (d^2_{n-1,1})$ and ${\bar E}^3_{n-1,1} = Ker ({\bar d}^2_{n-1,1})/ Im ({\bar d}^2_{n+1,0})$ and since $d^2_{n,1}$'s are the same we have the following exact sequence \begin{eqnarray} \bar E^2_{n+1,0} \to E^2_{n-1,1} \to {\bar E}^2_{n-1,1} \to 0 \nonumber \end{eqnarray} http://mathoverflow.net/questions/104681/comparing-spectral-sequences Comment by Grilo Grilo 2012-08-15T08:30:24Z 2012-08-15T08:30:24Z Wilberd -- Yes the differentials go as you said from (p,q) to (p-r, q+r-1). http://mathoverflow.net/questions/104681/comparing-spectral-sequences Comment by Grilo Grilo 2012-08-14T18:05:35Z 2012-08-14T18:05:35Z Algori - My main concern is Niveau spectral sequences related with a homology theory. There is a map between the homology theories which is compatible with the morphism between spectral sequences. And I have exactly the above situation namely they only differ at 0th row and the first one vanishes there. http://mathoverflow.net/questions/104681/comparing-spectral-sequences Comment by Grilo Grilo 2012-08-14T10:54:14Z 2012-08-14T10:54:14Z Sorry, I fixed the question.