Let $C \subset \mathbb{C}^n$ be a singular complex curve. Is there a way to compute its (singular) homology? (or at least its betti numbers / Euler characteristic).

If it were non-singular, then $C$ can be viewed as a topological surface, and thus it's homology can be computed using its genus, but singular curves are not topological manifolds.

Can it be computed via its resolution? I know that the genus is the same for $C$ and for the resolution of singularities of $C$, but is the homology preserved as well?


1 Answer 1


If $C'$ is the normalization of the curve $C$ there are finite sets of points $Z' \subset C'$ and $Z \subset C$ such that $$ C' \setminus Z' \cong C \setminus Z. $$ Now, you can write the excision exact sequences to express the cohomology of $C$ in terms of cohomology of $C'$ (which is a topological manifold) and the finite sets $Z'$ and $Z$.

  • $\begingroup$ I guess $Z$ would be the singular locus (finite as $C$ is a curve), but what are $Z'$? Above each point in $Z$ there is an exceptional divisor, which is not finite. $\endgroup$ Jul 6, 2023 at 11:31
  • $\begingroup$ @SergetheToaster If you construct the normalization explicitly as a sequence of blowups, it will be infinite as a subscheme of the blown up projective space. But the intersection of the exceptional divisor with the normalization $C'$ will of course be finite. $\endgroup$ Jul 6, 2023 at 12:13
  • $\begingroup$ I see, so the finite set can be computed by understanding the type of singularity. Many thanks! $\endgroup$ Jul 6, 2023 at 12:26

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