4
$\begingroup$

Let $K$ be a simplicial complex: it consists from the set (called the set of vertices) and a family of subsets of set of vertices satisfying the property of being closed under taking subsets (those set are called simplices: condition translates that the subset of a simplex is again simplex). Homology of simplicial complex is defined as follows: we consider $C_q(K)$, the free $R$-module ($R$ is a ring) generated by all oriented $q$ simplices with relation $\sigma+\sigma'=0$ where $\sigma'$ is $\sigma$ with reversed orientation. The boundary is defined with as $\partial=\sum_{i=1}^q (-1)^i\partial_i$ where $\partial_i[v_0,...,v_q]=[v_0,...,v_{i-1},v_{i+1},...,v_q]$.

Now given any abstract simplicial set $X=(X_q)_{q=0}^{\infty}$ one can consider $S_q(X)$=free module generated by $X_q$. The boundary is defined as $\partial=\sum_{i=0}^q(-1)^i\partial_q$ where $\partial_i$ are part of simplicial set data.

If we start with simplicial complex one can associate simplicial set $SimpSet(K)=(Ss_q(K))_{q=0}^{\infty}$ where $Ss_q(K)=\{(v_0,...,v_q): [v_0,...,v_q] \in K \}$ where $(v_0,...,v_q)$ is an ordered tuple and we allow repetitions. Face and degeneracies are defined as follows: $$\partial_i (v_0,...,v_q)=(v_0,...,v_{i-1},v_{i+1},...,v_q)$$ $$\sigma_i (v_0,...,v_q)=(v_0,...,v_i,v_i,v_{i+1},...,v_q).$$ This defines simplicial set.

Therefore we arrive at two a priori different homology theories but at the end of the day they should coincide. My guess for a natural (chain) map yielding isomorphism in cohomology would be $[v_0,...,v_q] \mapsto (v_0,...,v_q)$.

How to prove that this map induces isomorphism in homology? If it is not the case, how this isomorphism should be defined?

$\endgroup$
1
  • 3
    $\begingroup$ The chain complex of the $C_q(K)$ should agree up to isomorphism with the normalized Moore complex of the associated simplicial R-module under the Dold-Kan correspondence, I believe, but I'm not certain enough to put it as an answer. If this is the case, it's actually a point-set isomorphism rather than just a quasi-isomorphism. $\endgroup$ Nov 30, 2017 at 0:21

1 Answer 1

8
+50
$\begingroup$

There is a problem with your definition of the boundary map $\partial : C_q(K)\to C_{q-1}(K)$. The formula $\partial _i[v_0,\cdots,v_q]=[v_0,\cdots,v_{i-1},v_{i+1},\cdots,v_q]$ depends on an ordering of the vertices of the simplex $[v_0,\cdots,v_q]$, but such an ordering is not part of the definition of a simplicial complex. One can try to avoid this problem by including orientation data in the definition of $C_q(K)$ as you have done, but then one has to specify how an orientation of a $q$-simplex induces orientations of each of its $(q-1)$-dimensional faces, in order for the alternating sum formula for the boundary map to make sense. This can be done, but it takes a little work. (For example, one would have to verify that $\partial^2=0$, which takes more work if one only has orientations and not orderings of vertices.)

Your proposed chain map $[v_0,\cdots,v_q]\mapsto (v_0,\cdots,v_q)$ is also not well defined for the same reason. If one chooses a partial ordering of all the vertices of $K$ that restricts to a linear ordering on the vertices of each simplex, then one gets a well-defined chain map, but it is not natural since it depends on the choice of the ordering. To get a natural map one needs to replace simplicial complexes with ordered simplicial complexes or more generally semi-simplicial sets, which are simplicial sets without degeneracies.

After making these adjustments to the definitions one does obtain a well-defined chain map. This can be shown to induce an isomorphism on homology by first treating the case of finite complexes by induction using the five lemma applied to Mayer-Vietoris sequences. Then one passes to infinite complexes using a direct limit argument.

$\endgroup$
0

Your Answer

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

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