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Let $M$ be a closed, triangulated manifold of dimension $m$ and $K(M)$ be its triangulation. Let $f_i$ denote the number of $i$-simplices of $K(M)$. As proved by Klee the face numbers satisfy the following Dehn-Sommerville relations $$ f_k = \sum_{i=k}^m (-1)^{i+m} \binom{i+1}{k+1} f_i,$$ for $k = 0, 1, \dots, m$.

The above formula holds true for more general spaces called the semi-Eulerian complexes (i.e., these are simplicial complexes such that the Euler characteristic of the link of every non-empty face is equal to that of an appropriate-dimensional sphere).

Recall that the $h$-vector of a simplicial complex is defined as $h_i = \sum_{j=0}^i (-1)^{i-j} \binom{m-j}{m-i} f_{j-1}$ for $0\leq i\leq m+1$. The Dehn-Sommerville relations have a particularly beautiful expression in terms of these $h$-vectors: $$h_{m+1-i} - h_i = (-1)^i\binom{m}{i}(\chi(M) - \chi(S^m)) $$ for $0\leq i\leq m+1$.

My question: Do these relations hold if one were to have a $\Delta$-complex (i.e., semi-simplicial complex) instead of simplicial triangulation of the manifold (one might have to assume the structure to be regular, i.e., attaching maps for homeomorphisms) ?

Is there an analogue of semi-Eulerian complex for $\Delta$-complexes ?

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  • $\begingroup$ I am already confused by the first line, since there is generally no unique simplicial complexes whose geometric realiazation is $M$. Are you saying that the DS equations are satisfied by any simplicial complex $K$ with $|K|$ homeomorphic to $M$? $\endgroup$ Commented Dec 15, 2014 at 17:45
  • $\begingroup$ Sorry for the confusion, I think I have fixed it now. Here is the link to Klee's paper cms.math.ca/cjm/a145460. $\endgroup$ Commented Dec 16, 2014 at 1:55

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I think the answer is yes and the following is a sketch of proof:

  • The second barycentric subdivision of a $\Delta$-complex is a triangulation.

  • The f-vector of a $\Delta$-complex and of its barycentric subdivision are related by a matrix that preserves the Dehn-Sommerville relations.

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  • $\begingroup$ Right. The $(i,j)$-th entry of the matrix that takes a given $f$-vector to the $f$-vector of the subdivision is $$\sum_{k=0}^i (k-i-1)^{k+j+1} \binom{i+1}{k}.$$ $\endgroup$ Commented May 17, 2015 at 5:44

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