most recent 30 from http://mathoverflow.net 2013-05-22T07:57:03Z http://mathoverflow.net/feeds/question/50665 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50665/a-curious-construction-of-a-chain-complex-and-its-homology some guy on the street 2010-12-29T16:01:40Z 2010-12-29T21:56:23Z

<p>... curious to me, that is.</p> <p>Suppose two module filtrations $$ \cdots &lt; A_3 &lt; A_2 &lt; A_1 &lt; \cdots $$ and $$ \cdots &lt; B_3 &lt; B_2 &lt; B_1 &lt; \cdots $$ are comparable in the sense that for all $j$, $ B_{j+1} &lt; A_{j} &lt; 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 $$ \frac{A_i\cap B_i}{A_{i+1} + B_{i+1}} .$$</p> <p>My question is in two parts:</p> <ol> <li><p>this canonical isomorphism $H(A^+/B)\simeq H(B^+/A)$, has it got a name?</p></li> <li><p>is it <em>useful</em>?</p></li> </ol>

http://mathoverflow.net/questions/50665/a-curious-construction-of-a-chain-complex-and-its-homology/50681#50681 Noah Stein 2010-12-29T21:56:23Z 2010-12-29T21:56:23Z

<p>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</p> <p>\[ \cdots\to \frac{B_2}{B_4} \to \frac{B_1}{B_3}\to\frac{B_0}{B_2}\to\cdots \]</p> <p>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</p> <p>\[ 0\to \frac{A_2}{B_3}\to\frac{B_1}{B_3}\to\frac{B_1}{A_2}\to 0. \]</p> <p>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.</p>