7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Good question! Do you have a source on the statement of the vanishing of the Betti numbers for $i \geq n$? I'd very much like to know. $\endgroup$ – Fredrik Meyer Apr 14 '16 at 15:28
  • 3
    $\begingroup$ @FrederikMeyer The argument I had in mind was that the Betti numbers are the dimensions of the algebraic de Rham cohomology groups. Since $X$ is affine, the sheaves of differentials are acyclic, and the $i$-th cohomology group is {global closed algebraic $i$-forms} / {exact forms}. Because $X$ has algebraic dimension $n-1$, there are no algebraic $i$-forms for $i\geq n$. $\endgroup$ – User Apr 14 '16 at 15:32
  • 2
    $\begingroup$ @Fredrik: in fact you can show that a (complex) $n$-dimensional smooth affine variety is homotopy equivalent to a (real) $n$-dimensional CW complex using Morse theory. This is the Andreotti-Frankel theorem: en.wikipedia.org/wiki/Andreotti%E2%80%93Frankel_theorem $\endgroup$ – Qiaochu Yuan Apr 14 '16 at 16:48
7
$\begingroup$

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 $S^1$.

$\endgroup$
3
$\begingroup$

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 $n=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.