Transversals to singular subvarieties

Question

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.)

Answer 1

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.