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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.

Mark Grant
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Grilo
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  • Is it a first quadrant SS? – Fernando Muro Aug 14 '12 at 10:06
  • Sorry, I fixed the question. – Grilo Aug 14 '12 at 10:54
  • Consider the mapping cone of a chain map. It may be viewed as the total complex of a double complex with two rows. Say $\bar{E}^r_{p,q}$ is the spectral sequence for this double complex, living in the region $q\leq1$. Take for $E^r_{p,q}$ the spectral sequence corresponding with the $q=1$ row. Then the question is about the long exact sequence of homology for a mapping cone. A lot can happen. – Wilberd van der Kallen Aug 14 '12 at 15:40
  • Wilberd -- I am not quite sure this works: the spectral sequence you mention (I usually think of it as having two non-zero columns rather than rows) is mapped to by the 1-column spectral sequence that computes the homology of the source chain complex shifted by 1, and is maps to by the 1-column spectral sequence that computes the homology of the target chain complex. In neither case is the 0-th column of $E_{p,q}$ (0-th row in your version) zero. – algori Aug 14 '12 at 16:21
  • Grilo -- where do the differentials of your spectral sequences go? Since you use $E_{pq}$, I presume you consider something like homology (and not cohomology) spectral sequences of fibrations with differential $d^r$ going from $E_{pq}$ to $E_{p-r,q+r-1}$; but is this case it is a bit strange that it is the source $E_{pq}$ and not the target $\bar E_{pq}$ spectral sequence whose 0-th row is 0. – algori Aug 14 '12 at 16:47
  • 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. – Grilo Aug 14 '12 at 18:05
  • Grilo -- We really would like to know how your differentials go. For me they go from (p,q) to (p-r,q+r-1). And please use a double complex with two rows, not one with two columns, to represent a mapping cone. There are of course two spectral sequences associated with a double complex. One of them works. I am sure. – Wilberd van der Kallen Aug 15 '12 at 07:30
  • Wilberd -- Yes the differentials go as you said from (p,q) to (p-r, q+r-1). – Grilo Aug 15 '12 at 08:30
  • Grilo -- Oops! Now I see why you object to my "example". Back to the drawing board. – Wilberd van der Kallen Aug 15 '12 at 11:16

1 Answers1

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The cokernel is entirely due to $\bar{E}^\infty_{*,0}$ but the kernel is more mysteriuous.

First observe that if $g:C_\cdot\to D_\cdot$ is a chain map and $i$ is such that $g_i$ is surjective and $g_{i-1}$ is injective, then $H_i(g)$ is surjective and $H_{i-1}(g)$ is injective. (Exercise.)

Using this, one sees by induction on $r$ that $f^r_{p,q}$ is surjective for $q\geq1$ and injective (hence bijective) for $q\geq r-1\geq1$.

Now take $r=\infty$. One sees that $h$ hits all layers except the top one in the filtration of $\bar{H}_*$.

Let me include Grilo's formulas as I now believe they should read: We have exact sequences

$$0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^{r+1}_{n+1,0}\to E^{r+2}_{n-r,r}\to\bar{E}^{r+2}_{n-r,r}\to0$$

and then $$0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^{r+1}_{n+1,0}\to E^{\infty}_{n-r,r}\to\bar{E}^{\infty}_{n-r,r}\to0$$

Putting $r=n$ it becomes

$$0\to{\bar{E}}^{\infty}_{n+1,0} \to\bar{E}^{n+1}_{n+1,0}\to E^{\infty}_{0,n+1}\to\bar{E}^{\infty}_{0,n+1}\to0$$

or

$$H_{n+1}\to \bar{H}_{n+1} \to\bar{E}^{n+1}_{n+1,0}\to E^{\infty}_{0,n+1}\to\bar{E}^{\infty}_{0,n+1}\to0$$

So far so good.

Wilberd van der Kallen
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  • 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}

     – Grilo Aug 17 '12 at 06:55
  • 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}

     – Grilo Aug 17 '12 at 06:57
  • 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}

     – Grilo Aug 17 '12 at 07:02
  • I can not follow. You mean $\bar{E}^2_{n+1,0}\to E^3_{n-1,1}\to \bar{E}^3_{n-1,1}$? The map $E^2_{n-1,1}\to \bar{E}^2_{n-1,1}$ is an isomorphism after all. – Wilberd van der Kallen Aug 17 '12 at 08:30
  • And how do you get from $\bar{E}^{r+1}{n-r+2,0}$ to $\bar{E}^{n+1}{n+1,0}$? – Wilberd van der Kallen Aug 17 '12 at 08:54
  • "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}

     – Grilo Aug 17 '12 at 09:19
  • 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}

     – Grilo Aug 17 '12 at 09:23
  • 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.

     – Grilo Aug 17 '12 at 10:51
  • Right, and your $\bar{H}_{n-1}\to \bar{E}^2_{n+1,0}$ should be $\bar{H}_{n+1}\to \bar{E}^2_{n+1,0}$ – Wilberd van der Kallen Aug 17 '12 at 11:47