## Does exist a $\varepsilon$-tubular neighborhood of a smooth complex quasi-affine algebraic variety

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...

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 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 2011 at 19:53 Do you want the projection to be holomorphic? – Qfwfq Feb 11 2011 at 21:55 no no i want the projection only smooth – Italo Feb 12 2011 at 0:10

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.

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 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 2011 at 0:38