0
$\begingroup$

Let $X$ be a simplicial complex and let $A,B\subset X$ be subcomplexes such that $C=A\cap B$ is a non-empty simplicial complex. Finally, let $C_{\cdot}(X)$, $C_{\cdot}(A)$, $C_{\cdot}(B)$, $C_{\cdot}(C)$ be the corresponding chain complexes which coefficients in a field $\mathbb{F}$.

Question: Under what conditions does there exist a chain map $m:C_{\cdot}(A)\to C_{\cdot}(B)$ such that $m$ restricts to the identity on $C_{\cdot}(C)$? What if $A,B,C,$ and $X$ are flag complexes?

Some Basic Observations:

It seems like simple sufficient condition is that there exist a retraction $r:A\to C$. However, this does not appear to be necessary. For instance, if we take

  • $X$ to be the simplicial complex given by the faces $(1,2,4),(2,3,4),(3,1,4),(1,2,5),(2,3,5),(3,1,5)$
  • $A$ to be the subcomplex given by the faces $(1,2,4),(2,3,4),(3,1,4)$
  • $B$ to be the subcomplex given by the faces $(1,2,5),(2,3,5),(3,1,5)$

Then $C$ is the simplicial comlex given by the edges $(1,2)$, $(2,3)$, and $(1,3)$. While $C$ is not a retraction of $A$, an $m$ satisfying the above conditions comes from exchanging the vertices $4$ and $5$.

In general one can write down linear systems arising from the relevant commuting diagrams and solve them. The solutions haven't been very illuminating. However, if one considers small random complexes, then more often than not I have been able to solve the linear system. In a very non-rigorous way this seems to suggest that maybe there are some weak sufficient conditions for such maps to exist.

Improve this question
$\endgroup$
7
  • $\begingroup$ In what terms do you want an answer? If it's just the existence, without any good reason or naturality, this has nothing to do with $X$ and you are just speaking about comparing two short exact sequences in the category of complexes. $\endgroup$
    – Alex Degtyarev
    Feb 26, 2015 at 22:05
  • $\begingroup$ I agree, X doesn't really play any role. I included it simply so people wouldn't complain about $A\cap B$. Categorical proofs, notions, etc. are fine by me and I would be content with (non-constructive) existence statements. $\endgroup$
    – Paul
    Feb 26, 2015 at 22:07
  • 1
    $\begingroup$ The kernel of the map $H_*(A \cap B) \to H_*(A)$ must be contained in the kernel of the map $H_*(A \cap B) \to H_*(B)$. I believe that this condition is also sufficient. $\endgroup$
    – Tyler Lawson
    Feb 26, 2015 at 23:01
  • $\begingroup$ There must be more here that you have in mind but haven't written down: the intersection could just be a single vertex without imposing any seriois constraints on the two subcomplexes! $\endgroup$
    – Vidit Nanda
    Feb 27, 2015 at 14:39
  • $\begingroup$ @TylerLawson If $A$ and $B$ are arbitrary aside from being connected, and if $C$ is a point, then your condition is immediately satisfied since both kernels are trivial, no? I don't think one can deduce the existence of a chain map in this case, but perhaps I have misunderstood your comment. $\endgroup$
    – Vidit Nanda
    Feb 27, 2015 at 15:02

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.