I am trying to implement Theorem 1.1 in the paper "Poset Fiber Theorems" by Bjorner, Wachs, and Welker. https://www.researchgate.net/publication/228782786_Poset_fiber_theorems

I am pretty excited to learn about the theorem, which is a generalization of the well-known Quillen fiber lemma and should be very useful for computation. But when I implement the theorem to the following example, the result seems inconsistent with my knowledge:

Poset map

The posets are as shown in their Hasse diagrams and how the map is defined is: fixing $3,6,9,10$, mapping $11$ to a new minimum point $0$, and mapping other points downward by 1 level to the unique choice in $\{3,6,9,10\}$. From the construction, it is order-preserving and hence a poset map.

The conditions of Theorem 1.1 are satisfied as shown in the table on the right. Since $Q$ is the cone over the order complex $\Delta(\{3,6,9,10\})$, it is contractible. Thus, applying Theorem 1.1 in the paper, $\Delta(P)$, the order complex of $P$, is homotopic equivalent to a wedge of $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$, where $\star$ denotes the join operation.

I am only interested in the homology group over fields, so I will use Betti numbers for the rest of the discussion. I also use Kunneth's formula for the join of simplicial complexes (see, e.g., the very beginning of https://arxiv.org/abs/math/0412552) in the case of field coefficients:

For $q = 0$, the reduced Betti numbers for $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$ are all zeros.

For $q = 9$ or $10$, the reduced Betti numbers are $0,0,1$ (i.e. $H_k$ for $k = 0, 1, 2$).

For $q = 3$ or $6$, $f^{-1}(Q_{\leq q})$ is isomorphic to the face poset of a $2$-simplex. Hence, the reduced Betti numbers of $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$ are all zeros.

Thus, by Theorem 1.1 in the paper, the reduced Betti numbers of $\Delta(P)$ are 0, 0, 2 (i.e. $H_k$ for $k = 0, 1, 2$) and $0$ for other dimensions.

However, it can be noted that $P$ is isomorphic to the face poset of the CW complex obtained by taking two copies of a $2$-simplex and identifying only their corresponding vertices. (You may visualize it as a pair of panties.) Thus, the reduced Betti numbers are 0, 2, 0 (i.e. $H_k$ for $k = 0, 1, 2$) and $0$ for other dimensions. I also verify the result by computer.

I have been checking logical bugs the whole day and cannot find a mistake. I am relatively new to these poset topology things and afraid that I have made some naive/stupid mistakes. Any comments, questions, suspicion about my computations, or pointing out my mistakes are welcome. Thank you.

  • 3
    $\begingroup$ The preimage of $Q_{\le 3}$ should contain everything NOT mapped to $6$. Aren't you missing $7$ and $8$? (same for $6$ instead of $3$) $\endgroup$ – j.p. Jul 29 '20 at 9:01
  • $\begingroup$ I think you are right. LOL Thanks for pointing out. This saves me a lot of time. $\endgroup$ – Min Wu Jul 29 '20 at 13:45

I corrected my computation and it turns out the assumptions required by Theorem 1.1 in the paper are not satisfied. Many thanks to everyone, who spent time reading my long writing. Cheers.

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