Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

... curious to me, that is.

Suppose two module filtrations $$ \cdots < A_3 < A_2 < A_1 < \cdots $$ and $$ \cdots < B_3 < B_2 < B_1 < \cdots $$ are comparable in the sense that for all $j$, $ B_{j+1} < A_{j} < B_{j-1} $; then there are natural complexes $$ \cdots \to \frac{A_3}{B_4} \to \frac{A_2}{B_3} \to \frac{A_1}{B_2} \to \cdots $$ and $$ \cdots \to \frac{B_3}{A_4} \to \frac{B_2}{A_3} \to \frac{B_1}{A_2} \to \cdots $$ both of which have homology groups $$ \frac{A_i\cap B_i}{A_{i+1} + B_{i+1}} .$$

My question is in two parts:

  1. this canonical isomorphism $H(A^+/B)\simeq H(B^+/A)$, has it got a name?

  2. is it useful?

share|improve this question
add comment

1 Answer 1

up vote 7 down vote accepted

I don't know that this qualifies as an answer, but I wanted to point out that your isomorphism arises as the boundary map in a short exact sequence of chain complexes. Note that

\[ \cdots\to \frac{B_2}{B_4} \to \frac{B_1}{B_3}\to\frac{B_0}{B_2}\to\cdots \]

is a long exact sequence. When viewed as a chain complex, it fits together with your other two chain complexes (with one of them shifted) into a short exact sequence of chain complexes with columns looking like

\[ 0\to \frac{A_2}{B_3}\to\frac{B_1}{B_3}\to\frac{B_1}{A_2}\to 0. \]

The induced long exact sequence in homology has one of every three terms zero because the added chain complex is exact. Thus the boundary maps are isomorphisms (I haven't checked that they are the isomorphisms you mention, but it would be somewhat odd if they weren't...). As for question 2, I'm not sure, but at least it's a special case of something useful.

share|improve this answer
oh, nifty! As it happens, I had actually started by looking at that long exact sequence, and only stumbled on the other two in trying to get a non-exact complex, and asking what the homology was... I will ponder this more! –  some guy on the street Dec 30 '10 at 0:46
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.