Higher homotopy groups of the complement of an affine variety? Let $f$ be a polynomial in $n$ complex variables and let $Z(f)$ be the zero set of $f$ in $\mathbb{C}^n$.  I am wondering if $\pi_i(\mathbb{C}^n - Z(f))$ is ever nonzero for any $i \geq 2$?  Does the answer change over $\mathbb{R}$?
 A: Let $f(x,y)=xy-1$ and let $X=\mathbb{C}^2-Z(f)$.  Note that $Z(f)\cong\mathbb{C}\setminus\{0\}$, so the one-point compactification of $Z(f)$ is a $2$-sphere with two points identified.  In particular, $H_1$ of this one-point compactification is isomorphic to $\mathbb{Z}$, and so by Alexander duality, $H^2(X)\cong\mathbb{Z}$.
Let's now compute $\pi_1(X)$.  We cover $X$ by the open sets $U=X-Z(x)$ and $V=\{(x,y)\in\mathbb{C}^2:|xy|<1\}$ and use van Kampen's theorem.  We have $U\cong (\mathbb{C}\setminus\{0\})^2$, by mapping $(x,y)$ to $(x,y-1/x)$.  Also, $V$ is contractible, since it deformation-retracts to $\mathbb{C}\times\{0\}$ by moving $y$ along the straight line path to $0$.  Similarly, $U\cap V$ deformation-retracts to $(\mathbb{C}\setminus\{0\})\times\{0\}$.  A generator of $\pi_1(U\cap V)\cong\mathbb{Z}$ is a counterclockwise loop around $0$ in the $x$-coordinate while $y$ is fixed at $0$.  Under the homeomorphism $U\cong (\mathbb{C}\setminus\{0\})^2$ sending $(x,y)$ to $(x,y-1/x)$, this loop maps to a loop which winds around $0$ counterclockwise in the $x$ coordinate while simultaneously winding clockwise around $0$ in the $y$-coordinate, representing the element $(1,-1)\in\mathbb{Z}^2\cong\pi_1(U)$.  So, by van Kampen's theorem, $\pi_1(X)\cong\mathbb{Z}^2/\langle (1,-1)\rangle\cong \mathbb{Z}$.
Now, if the higher homotopy groups of $X$ were trivial, then $X$ would be a $K(\pi_1(X),1)=K(\mathbb{Z},1)$, i.e. it would be homotopy equivalent to $S^1$.  But this is impossible since $H^2(X)$ is nontrivial.  Thus $X$ must have nontrivial higher homotopy groups.

Over $\mathbb{R}$ it is much easier to get an example: just observe that $V(x_1^2+x_2^2+\dots+x_n^2)$ is a point so its complement is homotopy equivalent to $S^{n-1}$.
