TITLE: Betti numbers of complex "sphere"
QUESTION [7 upvotes]: Let $X$ be the set of solutions to $x_1^2+\ldots+x_n^2=1$ in $\mathbb{C}^n$. This has real dimension $2(n-1)$, but since $X$ is an affine algebraic variety, the only possible non-zero topological Betti numbers of $X$ are $b_0,\ldots,b_{n-1}$. 

What is the top Betti number $b_{n-1}$?

The real sphere $S^{n-1}$ embeds in $X$, determining a class in $H_{n-1}(X,\mathbb{Q})$. I wonder if this class actually spans the homology group.
I have seem some results on Betti number for projective hypersurfaces, but not for affine ones.

REPLY [3 votes]: Here is the answer I wanted to write, but since Qiaochu posted his first I'll make mine Community Wiki and use his notation.
Recall that the tangent bundle to the sphere $S^{n-1}\subset \mathbb R^n$ is the submanifold $TS^{n-1}\subset S^{n-1}\times \mathbb R^n$ consisting of pairs $(u,v) \in S^{n-1}\times \mathbb R^n$ with $u\bullet v=0$ (usual scalar product in $\mathbb R^n$).
The amazing result is that we have a  diffeomorphism $$X\stackrel {\sim}{\to} TS^{n-1}: x=a+ib\mapsto (u=\frac {a}{||a||},v=b)$$ whose inverse diffeomorphism is $TS^{n-1}\stackrel {\sim}{\to} X:(u,v)\mapsto x+iy=(\sqrt {1+||v||^2}){u}+iv$.
Like all vector bundles $TS^{n-1}$ is homotopic to its base space $S^{n-1}$, and thus $X$ is homotopic to $S^{n-1}$ too, so that finally $b_{n-1}(X)=1$ for $n\geq2$ and $b_0(X)=2$ for $n=1$.