# The exceptional locus of a minimal resolution of singularities

Let X be a surface. (A surface is an excellent integral normal separated 2-dimensional scheme.)

Let $\psi:Y\longrightarrow X$ be a minimal resolution of singularities and let $E$ be an irreducible component of the exceptional locus of $\psi$.

Is $E$ of genus zero? That is, do we have that $E$ is isomorphic to $\mathbf{P}^1_k$ for some field $k$?

Is the exceptional locus of $\psi$ a chain of rational curves? (This means that $(E_i,E_i) <0$, $(E_i, E_{i+1}) = (E_i, E_{i-1}) = 1$ and $(E_i,E_j) = 0$ if $j \neq i-1, i,i+1$. Here $E_i$ denotes an exceptional component.)

I know this is true if $X$ has "tame cyclic quotient singularities".

I also know that, by Lipman's theorem and what is stated on Wikipedia about it, this is true if $X$ has pseudo-rational singularities.

What else can we say in general about the "shape" of the exceptional locus. If it's not a chain, does it have a loop? Can things get arbitrarily complicated? Where can I find the theory behind this?

• Just a quick comment. If the exceptional locus forms a loop of $\mathbb{P}^1$'s, then it is called a cusp singularities. For rational singularities, the chain of $\mathbb{P}^1$'s is a tree and of course a large number of singularities that appear in nature are rational'', so I suppose the take home message is that trees appear in nature. – Karl Schwede Jul 20 '11 at 22:03
• If $X$ has worse than rational singularities, then the components can have positive genera (as in Sandor's examples). Also the dual graph is usually not a chain even for rational singularities (e.g. $E_6,\ldots$), and for more general singularities you can get loops as well. – Donu Arapura Jul 20 '11 at 22:07
• That's a good point. I shouldn't have used the word chain, bur rather I should have said graph above. – Karl Schwede Jul 21 '11 at 4:16
• Karl, actually I didn't see your comment when I wrote mine. – Donu Arapura Jul 21 '11 at 11:08

• Nice example! I suppose one might want to say that $C$ is non-rational, but I am only adding that so I can say this: This resolution is obviously minimal, since $C\times C$ does not contain any rational curves, so it can't contain a $(-1)$-curve either. – Sándor Kovács Jul 21 '11 at 5:50