Skip to main content
4 of 6
added 11 characters in body
Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

The answer is yes, already for an affine variety.

The following example is taken from Dimca's book Singularities and topology of hypersurfaces, see page 102 and page 105. We work over $\mathbb{C}$.

Let $V \subset \mathbb{P}^n$ be a hypersurface and $U:=\mathbb{P}^n \setminus V$ its complement. Since $V$ is very ample, $U$ is an affine variety. Then we have the following

Proposition. Assume that $V$ has $k$ irreducible components $V_1, \ldots, V_k$ with $\deg V_i =d_i$. Then $$H_1(U, \mathbb{Z})=\mathbb{Z}^{k-1} \oplus \mathbb{Z}/d \mathbb{Z},$$ where $d$ denotes the greatest common divisor of the integers $d_1, \ldots, d_k$.

If moreover $V$ has only normal crossing singularities in codimension $1$, then $\pi_1(U)$ is abelian, hence it is isomorphic to the group $H_1(U, \mathbb{Z})$ given above.

In particular, if $V$ is irreducible and normal of degree $d$, one has $$\pi_1(U)=\mathbb{Z}/d \mathbb{Z}.$$

Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283