Bott and Tu Spectral Sequence of a Double Complex

Question

This page has confused me a great deal, I have a few questions:

1. Is $A^k$ the same thing as $C^k = \oplus_{p + q = n} K^{p,q}$ as defined in example 14.2 or is it different because $A = \oplus K_p$ has a different grading?

2. What is $B^k$ in this scenario? Isn't each part of the direct sum of $B$ just a column? How do you get the diagonal part of that?

3. Why is $[b]$ a cocyle in $B^k \cap K_p / K_{p+1}$? I thought that $B^k \cap K_p / K_{p+1}$ is just part of the short exact sequence, how is it a cohomology group?

Answer 1

After discussing with colleagues I am led to believe that the following answers are correct:

1. Yes, $A^k = C^k$.
2. $B = \oplus K_p / K_{p+1}$, it is just a direct sum of the columns of the spreadsheet, it retains the bidegrees of $K = \oplus K^{p,q}$, i.e. we can reference any element $\phi \in K^{p,q}$ as an element of $B$ and still say $\phi \in B$ has bidegree $p,q$. Then when we take the total complex of $K$, we also consider $B$ with the total complex single degree, but in this case, since we are only considering the columns, the single degree degenerates to just being the second degree $q$. So $B^k$ tells you to look at $C^k$ and then divide it into its columns and consider them all together.
3. The notation of the bracket $[b]$ here does not mean "take the cohomology class", it means take the equivalence class of going from $K_p \to K_p / K_{p+1}$, so it can certainly be a cocycle.