Betti numbers of complex "sphere"

Question

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.

Answer 1

This complex sphere deformation retracts onto the real sphere $S^{n-1}$ inside it, so they're homotopy equivalent, and in particular they have the same Betti numbers. To see this very explicitly, write $x_k = a_k + i b_k$, write $a = (a_1, a_2, \dots) \in \mathbb{R}^n$, and write $b = (b_1, b_2, \dots) \in \mathbb{R}^n$. The defining equation is a pair of equations

$$||a||^2 = 1 + ||b||^2$$ $$a \cdot b = 0$$

and we want to deformation retract this thing onto the subspace where $b = 0$. To do this we'll send $b$ to $(1 - t) b$, where $t \in [0, 1]$, and send $a$ to $f(t) a$ where $f(t)$ is a function chosen to have the property that the first equation still holds. This means that we want $||a||^2 = 1 + ||b||^2$ to imply

$$f(t)^2 ||a||^2 = f(t)^2 (1 + ||b||^2) = 1 + (1 - t)^2 ||b||^2$$

which gives

$$f(t) = \sqrt{ \frac{1 + (1 - t)^2 ||b||^2}{1 + ||b||^2} }.$$

As a sanity check, when $n = 0$ the space of solutions is two points, which is $S^0$, and when $n = 1$ the space of solutions is $\mathbb{C}^{\times}$ (rewrite the defining equation as $(x_1 + i x_2)(x_1 - i x_2) = 1$), which deformation retracts onto $S^1$.

Answer 2

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 $n=1$.