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Say $\mathbb{C}^d \subset Y^{N-k} \subset \mathbb{C}^N$ are closed imbeddings of complex analytic subvarieties of the indicated dimension, $Y$ is not smooth. At a point $y \in Y$, a generic, sufficiently small polydisc $\mathbb{D}^k \ni y$ will satisfy $\mathbb{D}^k \cap Y = y$. I would like to do this continuously along the $\mathbb{C}^d$.

For $p \in \mathbb{C}^d$, does there exist an analytic neighborhood $p \in U \subset \mathbb{C}^d$, and a subvariety $\widetilde{U} \cong U \times \mathbb{D}^k \subset \mathbb{C}^N$ such that $\widetilde{U} \cap Y = U$?

(I am not really sure what the right tag for this question is. Actually if someone would clue me in as to where to find some basic treatment of whatever subject this question belongs to, that would be great.)

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up vote 2 down vote accepted

I don't think this can be done.

It seems to me that your $\widetilde U$ would be a submanifold of $\mathbb C^N$, so it should be a local complete intersection and then $U$ would be a local complete intersection in $Y$. However, that does not have to be the case.

Let's say that $N=3$, $d=1$, $k=1$. Or even more specifically, let $Y$ be a quadric cone, $p\in Y$ the vertex and $L\simeq \mathbb C\subset Y$ a line through $p$. Then $L$ is not a Cartier divisor on $Y$, so it cannot be "cut out" by a single equation. Actually this example may not be the best as $2L$ is a Cartier divisor.

So, let's take $N=4$, $d=2$, $k=1$, $Y$ the cone over $\mathbb P^1\times \mathbb P^1$, and $L\subset Y$ the cone over one of the rulings of the $\mathbb P^1\times \mathbb P^1$, then $L\simeq C^2$ and no multiple of $L$ is a Cartier divisor.

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Hi Sandor, thanks for answering. In the question, I only meant set-theoretic intersection; but I guess it looks like your second answer answers that too. For a disc just passing through the vertex, I thought you should just take any general disc so the intersection with the $Y$ is finitely many points, then shrink to avoid those points. Does that not work? – Vivek Shende Jun 13 '11 at 8:06
I see. That seems to be OK as long as you only want it set-theoretically. My problem was that for instance in the first case I think you can get twice the point, but not the point itself. In general, it may be hard to get the point with multiplicity less than its multiplicity on $Y$, but (again) this does not seem to be an issue for you. – Sándor Kovács Jun 13 '11 at 8:37

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