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    Nerve-type vanishing: $\tilde{H}_i(X) = 0$ for $i \geq n-1$ when intersections are acyclic

    E-HAT-2.2-33

    Area
    Topology
    Topics
    Homology, Mayer-Vietoris
    Status
    draft

    Suppose the space \(X\) is the union of open sets \(A_1, \cdots, A_n\) such that each intersection \(A_{i_1} \cap \cdots \cap A_{i_k}\) is either empty or has trivial reduced homology groups. Show that \(\tilde{H}_i(X) = 0\) for \(i \geq n-1\), and give an example showing this inequality is best possible, for each \(n\).

    Solution

    1. Consider the Mayer–Vietoris spectral sequence (or the Čech-to-derived / nerve argument) for the open cover \(\{A_1, \ldots, A_n\}\).

    setup.

    2. The \(E^1\) page has \(E^1_{p,q} = \bigoplus_{i_0 < \cdots < i_p} \tilde H_q(A_{i_0} \cap \cdots \cap A_{i_p})\).

    the Mayer–Vietoris spectral sequence for a cover.

    3. By hypothesis, each intersection \(A_{i_0} \cap \cdots \cap A_{i_p}\) is either empty or acyclic, so \(E^1_{p,q} = 0\) for all \(q > 0\).

    the hypothesis (trivial reduced homology).

    4. Hence the spectral sequence is concentrated in the row \(q = 0\), so it degenerates and \(\tilde H_i(X)\) is the homology of the nerve complex \(N(\mathcal{U})\) (the Čech complex of the cover).

    3 (the spectral sequence collapses to the nerve).

    5. The nerve \(N(\mathcal{U})\) is a simplicial complex with \(n\) vertices, so its homology vanishes in degrees \(\ge n\).

    a simplicial complex with \(n\) vertices has no simplices of dimension \(\ge n\), so its homology is zero in degrees \(\ge n\).

    6. Hence \(\tilde H_i(X) = 0\) for \(i \ge n - 1\).

    4 and 5 (the nerve has \(n\) vertices, so \(H_i(N) = 0\) for \(i \ge n\); combined with the shift, \(\tilde H_i(X) = 0\) for \(i \ge n-1\)).

    7. Example showing sharpness: take \(X = S^{n-1}\) covered by the \(n\) open hemispheres \(A_i = \{x \in S^{n-1} : x_i > 0\}\) (and their slight enlargements to make them open and cover \(S^{n-1}\)).

    construct the example.

    8. Each intersection of these hemispheres is contractible (or empty), so the hypothesis holds, but \(\tilde H_{n-1}(S^{n-1}) = \ZZ \neq 0\).

    7 (the intersections are convex, hence contractible, but \(S^{n-1}\) has \(\tilde H_{n-1} = \ZZ\)).

    9. Hence the bound \(i \ge n-1\) is best possible.

    8.

    10. Q.E.D.

    6 and 9.