Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi!

By a complex quasi-affine variety i mean the complement of an affine algebraic variety with respect to another algebraic variety, more precisely a quasi-affine algebraic variety is

$$V= V_{1}\setminus V_{2}$$ with $$V_{1}:=V\left(P_{1},\ldots, P_{k}\right\)$$ $$V_{2}:=V\left(Q_{1},\ldots, Q_{l}\right\)$$ and $P_{i},Q_{j}\in \mathbb{C}\left\[x_{1},\ldots, x_{n}\right\]$. Clearly, if $V_{2}=\emptyset$, then $V=V_{1}$ is an affine variety. Suppose $V$ is smooth and connected, so it is moreover a smooth real not compact and connected differentiable manifold. Is it possible to find a smooth embedding
$$i: V\hookrightarrow \mathbb{R}^{N}$$ for $N$ large enough s.t. $i\left(V\right)$ has an $\varepsilon$-tubular neighborhood $\mathcal{T}_{\varepsilon}V\subset \mathbb{R}^{N}$? By an $\varepsilon$-tubular neighborhood i mean $$\mathcal{T}_{\varepsilon}V=\bigcup_{x\in i\left(V \right)}B_{\mathbb{R}^{N}}\left(x,\varepsilon\right)\qquad \varepsilon>0$$ together a smooth minimal point projection $\pi$ $$\pi: \mathcal{T}_{\varepsilon}V\rightarrow i\left( V \right)$$
that associates to any point $y\in \mathcal{T}_{\varepsilon}V$ its closest point $x\in i\left(V \right)$.

The case i have in mind $V$ is the variety of $k\times k$ complex matrices of rank exactly $r$, i don't know if this helps...

Thank you in advance!

share|improve this question
    
Someone should correct me if I'm wrong - but I think the Atiyah class gives an infinitesimal obstruction to the existence of tubular neighborhoods. I don't know what to say about your specific example. –  Paul Siegel Feb 11 '11 at 19:53
    
Do you want the projection to be holomorphic? –  Qfwfq Feb 11 '11 at 21:55
    
no no i want the projection only smooth –  Italo Feb 12 '11 at 0:10

1 Answer 1

If I understand the question correctly, no because $V \subset \mathbb{C}^2$ could be the union of $xy = 1$ with $x=0$. You can see from looking at the real solutions that it does not have a tubular neighborhood of fixed width. And it can be modified to make a connected example.

The case that interests you, $V$ is actually only quasiaffine in $M_r^k(\mathbb{C})$, not affine, so it's a strange question.

Now I suspect that if you are allowed to change the embedding, then every affine (or quasiaffine) variety is isomorphic to an affine variety with such a tubular neighborhood. If that is your real question, then I suspect yes, but I would have to think about it some more. One thing that is definitely true, for any smooth manifold in any Euclidean space which is a closed subset, is that it has a tubular neighborhood which is allowed to get narrower and narrower as you go to infinity.

share|improve this answer
    
Thanks, i discarded that example because it isn't connected and i wasn't able to find one connected. I said affine but i meant quasi affine too. Anyway i modified my question, i hope it is more clear now. The tubular neighborhood that gets narrower and narrower going to infinity is exactly the situation i want to avoid and it is the motivation of this question. I suspected that smooth algebraic varieties are "rigid" enough to hope to find such a neighborhood but i'm not so sure any more. –  Italo Feb 12 '11 at 0:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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