Consider a connected (possibily reducible) proper algebraic variety $X$ over $\mathbb{C}$ of pure dimension $n$. Is it true that $H^{2n}(X,\mathbb{Q})\cong \mathbb{Q}^r$ with $r=$#irreducible components of $X$, where the left-hand side is the singular cohomology with respect to the analytic topology?

  • $\begingroup$ That is not proper. $\endgroup$ Aug 14 '17 at 14:46
  • $\begingroup$ @MarianoSuárez-Álvarez Of course. $\endgroup$ Aug 14 '17 at 15:03

Each complete (compact in the analytic topology) irreducible complex $n$-dimensional algebraic variety admits a triangulation making it a closed oriented connected pseudomanifold of dimension $2n$. (Anosov in the article does not give a reference, just lists this as one of the examples. However: The most difficult part of the proof is the existence of a triangulation, see e.g. Hironaka's paper "Triangulations of Algebraic Sets", 1974, although the original proof is due to Lojasiewicz. Checking that irreducibility implies oriented pseudomanifold is a good exercise in definitions.) If $X$ is an $m$-dimensional closed oriented connected psedomanifold, then $H^m(X; {\mathbb Z})\cong {\mathbb Z}$. This is an exercise in Spanier's "Algebraic Topology" (p. 206); if you want to see a proof, take a look at Seifert and Threlfall "Topology", section 24. The rest (reducible case) follows from the Mayer-Vietoris sequence since intersections of irreducible components have real codimension $\ge 2$.


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