# When is the intersection of an isolated normal singularity with a generic linear subspace through that singularity normal?

Suppose I have an affine subvariety $A \subset {\mathbb C}^N$ of dimension $n \geq 3$ which has an isolated singularity at $0$ (lets say for the sake of simplicity that it is non-singular everywhere else). Suppose that this variety $A$ is normal. In order to study singularities it often seems like a good idea to study hyperplane sections or intersections with lower dimensional linear subspaces passing through that singularity.

Let $L$ be a generic linear subspace of ${\mathbb C}^N$ of dimension $N - n + 2$ passing through the origin. If $L$ is generic enough then $A \cap L$ has dimension $2$ and has an isolated singularity at the origin. Is it true that $A \cap L$ is normal for $L$ generic enough?

-

I don't think so. There are examples of isolated normal threefold singularities that are not Cohen-Macaulay. A hyperplane section is not Cohen-Macaulay, hence it can not be normal, because a normal surface is Cohen-Macaulay.

-
E.g., affine cone over a general embedding of an Abelian surface. –  Jason Starr May 20 '13 at 20:34
Excellent thanks! –  Mark McLean May 20 '13 at 21:09

I'm going to assume your singularity is dimension $\geq 3$. Angelo beat me to the answer but he is right, this is not true. But it is true sometimes (including the Cohen-Macaulay case as he implied).

A singularity is normal if it is $R1$ and $S2$. In your case, an isolated singularity is normal if the depth at the singular point is at least 2.

Now, a general hyperplane section will be $R1$ by Bertini. So we just need to check that the general hyperplane is $S2$. Well, for this we just need the depth to be at least 2 again, and hence we just need the original singularity to have depth $\geq 3$.

Conclusion: If your singularity is $S3$ (in your case just $\text{depth} \geq 3$), then what you want holds after cutting down by ONE hyperplane

EDIT: As Angelo pointed out, the actual question didn't cut down by just one hyperplane. In that case you can't just have depth $\geq 3$, you need $X$ to be Cohen-Macaulay.

Of course, not all singularities satisfy this, for example a cone over an Abelian surface.

You might also look at this preprint which seems to have some related results: Tadashi Ochiai, Kazuma Shimomoto

-
Actually, the way the question was formulated so that the intersection has dimension 2. In this case, to get a positive answer you need the singularity to be Cohen-Macaulay. –  Angelo May 21 '13 at 13:18
Angelo, of course you are right, I misread the question. –  Karl Schwede May 21 '13 at 13:46