User grilo - MathOverflow

Realisation functor

2013-02-06T13:49:40Z

Let k be a field. Is there a realization functor

$DM_{gm}(k,\mathbb{Z}/n)^{op} \to D^b_c(k, \mathbb{Z}/n)$

from category of motives to category of complexes of étale sheaves of $\mathbb{Z}/n$ modules with bounded constructible cohomology sheaves?

K-Theory and Tame Symbol

2013-02-04T11:53:14Z

This might be a little bit spesific but here it goes. While reading a paper (Brauer-Manin pairing...) by Yamazaki, I encountered this definition.

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:

$UK^M_{r+1}:= ker [K_r^M(k(y)) \to \bigoplus_{x \in C_\infty}(K_{r-1}^M(k(x)) \oplus K_{r}^M(k(x))]$.

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 $x$.

He says this group does not depend on the choice of the uniformizer, I couldn't see why it doesn't.

Is there an easy way to tell this group does not depend on the choice of the uniformizer?

Fundamental Group and Etale Cohomology

2012-12-20T14:33:03Z

I encountered the following statement without a reference many times. For a smooth variety $X$ over a perfect field $k$.

$Hom(H^1_{et}(X, \mathbb{Z}/n), \mathbb{Z}/n) \cong \pi^{ab}_1(X)/n$

Is there any reference for this? Why this is true?

Reciprocity Map and Cycle Class Map

2012-12-19T08:56:34Z

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:

$rec/n: SK_1(X)/n \to \pi^{ab}_1(X)/n$

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)

So I can see this as a map

$rec/n: CH^{d+1}(X,1,\mathbb{Z}/n) \to H^{2d+1}_{et}(X, \mathbb{Z}/n(d+1))$

What I wonder is that if this map agrees with the higher cycle class map i.e. whether their kernels (cokernels) are isomorphic.

Comparing Spectral Sequences

2012-08-14T09:51:30Z

There is a comparison theorem for spectral sequnces in Weibel's book (5.2.12) stating;

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.

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.

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?

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 $h$?

Thanks for your help.

Comment by Grilo

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 $a$.

Comment by Grilo

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.

Comment by Grilo

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}

Comment by Grilo

2012-08-17T09:19:35Z

"Please think this as the second comment" 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}

Comment by Grilo

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}

Comment by Grilo

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}

Comment by Grilo

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}

Comment by Grilo

2012-08-15T08:30:24Z

Wilberd -- Yes the differentials go as you said from (p,q) to (p-r, q+r-1).

Comment by Grilo

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.

Comment by Grilo

2012-08-14T10:54:14Z

Sorry, I fixed the question.