# what are the singularities of a normal crossings divisor?

This is probably a very stupid question. I'm sorry.

Let $D$ be a simple normal crossings divisor on some smooth projective variety $D$. By this I mean that the irreducible components $D_i$ are smooth and all possible intersections $\bigcap D_{i_1} \cap \cdots \cap D_{i_k}$ are transversal.

I don't understand what people mean by the singular locus of $D$. In view of the definition, I would say $D$ is smooth. Could anybody help me clarify this point?

• this is usually called a strict or simple normal crossings divisor – Niels Apr 9 at 20:11

## 2 Answers

The points of intersection of distinct reducible components are not smooth. Consider the equation $xy = 0$ in $A^2$, for example. The intersection of the coordinate axes is certainly transversal, but the variety is not smooth at the origin.

• Ok, so what happens is that the intersections are smooth when considered as subvarieties (in your example the point is smooth) but $D$ is not smooth itself. That's it? – div90 Jun 20 '13 at 14:32
• The points on $D$ are smooth as points on the ambient variety; in this example the ambient variety is some affine space, and the divisor $D$ is the union of the x-axis and y-axis. The origin is the only singular point on the divisor $D$ in this example. – Ariyan Javanpeykar Jun 20 '13 at 15:31
• For a divisor with irreducible components having multiplicity one, we can identify it with a reducible subvariety. As a point on this subvariety, a point of intersection is singular, since the dimension of the cotangent space $\frac{m_x}{m_x^2}$ is not equal to the dimension of the subvariety. This is the same reason that an irreducible nodal curve is singular. – Jesse Silliman Jun 21 '13 at 3:18

In this case, you consider $$D$$ a divisor (with normal crossing) on a nonsingular variety $$X$$ of dimension $$n$$. So, with this notation, the singular set of $$D$$ is an analitic subvariety of $$X$$ of codimension $$2$$. More precisely, if you write $$D = \cup_{i} D_{i}$$ with $$D_{i}$$ its irreducible components, the singular set of $$D$$ is given by

$$sing(D) = \bigcup_{i \neq j} D_{i} \cap D_{j}.$$