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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 about the general case?

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?

flag	asked Jul 20 2011 at 21:26 Narumi 21●1

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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 2011 at 22:03

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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 2011 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 2011 at 4:16

Karl, actually I didn't see your comment when I wrote mine. – Donu Arapura Jul 21 2011 at 11:08

Sándor Kovács · Answer 1 · 2012-01-07 17:16:14Z UTC

No, this is not true. For instance, any smooth projective irreducible curve can be the exceptional locus. Take your favorite curve and take a cone over that. The exceptional locus will be isomorphic to the original curve. If the curve was smooth and projectively normal (e.g., by a complete linear system), then the singularity will be normal. And it (the exceptional locus) could be non-irreducible. There are many possibilities.

You can read more about this for instance in Laufer's book: Normal two dimensional singularities. There is also a lot about two and higher dimensional singularities in Kollár-Mori's book.

roy smith · Answer 2 · 2011-07-21 03:27:57Z UTC

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If C is any non hyperelliptic curve, and J(C) its Jacobian, consider the difference map of CxC-->Pic^0(C) ≈ J(C). This map collapses the diagonal curve isomorphic to C in CxC to a point in J(C), and is otherwise an embedding. Hence this map is a resolution of the (projective) image surface with exceptional locus ≈ C. I presume it is minimal.

answered Jul 21 2011 at 3:27

roy smith
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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 2011 at 5:50

The exceptional locus of a minimal resolution of singularities

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