In Chow's You Could Have Invented Spectral Sequences (3rd page, left column) appears the following isomorphism of vector spaces: $$\frac{Z_d}{B_d}\cong \frac{Z_d+C_{d,1}}{B_d+C_{d,1}}\oplus \frac{Z_d\cap C_{d,1}}{B_d\cap C_{d,1}}$$ The context is a 2-degree filtered chain complex $C_\bullet$.

Why does this relation hold? I tried the second isomorphism theorem but didn't manage to get anything out of it.

After that, the author says the naive hope is that $$E_{d,2}^{1}\overset{?}{\cong}\frac{Z_{d}+C_{d,1}}{B_{d}+C_{d,1}}=\frac{Z_{d,2}+C_{d,1}}{B_{d,2}+C_{d,1}}$$ But that this does not, in general, hold. Now, I know relative homology groups are isomorphic to the quotient of relative cycles by relative boundaries. Isn't this precisely the quotient $\frac{Z_{d}+C_{d,1}}{B_{d}+C_{d,1}}$? Hence, isn't there an isomorphism $$H_{d}(C_{\bullet}/C_{1\bullet})\cong\frac{Z_{d}+C_{d,1}}{B_{d}+C_{d,1}}$$ where $C_\bullet$ is the original complex and $C_{1\bullet}$ is the complex in the first filtration degree? Now by definition $E^1_{d,2}=H_d(C_{\bullet}/C_{1\bullet})$, so it seems the isomorphism below does hold. $$E_{d,2}^{1}\cong\frac{Z_{d}+C_{d,1}}{B_{d}+C_{d,1}}=\frac{Z_{d,2}+C_{d,1}}{B_{d,2}+C_{d,1}}$$

What is my error? Why doesn't the isomorphism above always hold?

Finally, the author says that generally $E^1_{d,1} \ncong \frac{Z_{d,1}}{B_{d,1}}$. But isn't $E^1_{d,1} = H(C_{1\bullet}/0)\cong H(C_{1\bullet})$ by definition? ($C_{1\bullet}$ is the complex at the first filtration degree.) But $H(C_{1\bullet})=\frac{Z_{d,1}}{B_{d,1}}$ by definition, so again, I get the isomorphism does hold.

What is my mistake here?

Added: I don't understand what the boundary maps $\partial ^0$ are. Help is appreciated.

Update: I think I have detected my first mistake. Although the definition of relative boundaries is indeed $B_d$ relative to $C_{d-1}$ equals $B_d+C_{d-1}$, this is not how relative cycles are defined. Instead, $Z_d$ relative to $C_{d-1}$ equals $ \left\{ \gamma \in C_d:\partial _d \gamma \in C_{d-1} \right\}\neq Z_d+C_{d-1}$.

Note - for context for these relative boundaries and cycles, see Rotman's Introduction to Algebraic Topology, p.99. This resolves most of my issues because now I really don't except those isomorphisms. Of course I would like someone to confirm I'm not spewing nonsense, since Rotman only defined these for pairs of spaces - I just figured they work for filtrations aswell (by the third isomorphism theorem).

I think the only things bugging me now are (see Chow's article):

Why does the first relation hold, and what exactly is $\partial ^0$?

Update 2: I think I have realized what $\partial ^0$ is. For each quotient $E_{d,p}^0$ we have an induced boundary map $\partial_{d,p}^0:E_{d,p}^0\rightarrow E_{d-1,p}^0$ defined by $$\partial_{d,p}^0:[\gamma]\mapsto[\partial_{d,p} \gamma]$$ where $\partial _{d,p}$ is the $d^{\text{th}}$ boundary of the complex $C_{p\bullet}$. $\partial ^0=\partial ^0_d:\bigoplus _{p=1}^nE^0_{d,p}\rightarrow \bigoplus _{p=1}^nE^0_{d-1,p}$ is simply the direct sum of all these induced boundary maps.

Is this correct?


The first relation: take any vector space with subspaces $C, Z\supseteq B$. The natural surjection $Z/B \to (Z+C)/(B+C)$ has kernel $(Z\cap C +B)/B$ which is isomorphic to $Z\cap C /B\cap C$ so the isomorphism you want follows from the first isomorphism theorem.

$\partial^0$ is the map formed as the direct sum of the maps $C_{dp}/C_{d,p-1}\to C_{d-1,p}/C_{d-1,p-1} $ given by $c + C_{d,p-1}\mapsto \partial(c)+ C_{d-1,p-1}$

  • $\begingroup$ Thank you for your answer regarding the first relation! I have just figured out the same answer for $\partial ^0$ (see edit) :) $\endgroup$ – user153312 Apr 23 '15 at 14:26
  • $\begingroup$ Then the answer is "yes, this is correct" ;) $\endgroup$ – Matthew Towers Apr 23 '15 at 14:29

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