# Is the Springer resolution a blow-up?

Let's consider the Springer resolution of the nilpotent cone $\mathcal{N}$ of a complex semisimple Lie algebra $\mathfrak{g}$, which is $$\widetilde{\mathcal{N}}=T^*\mathcal{B}\rightarrow \mathcal{N}.$$ It is a resolution of singularity of $\mathcal{N}$. ( For the construction of Springer resolution you can see section 6 of "Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups" .)

My question is: is it a blow-up along some subscheme of $\mathcal{N}$?

Or for the simplest case, where $\mathfrak{g}=sl(2,\mathbb{C})$. Now $$\mathcal{N}= \{x^2=yz\} \in \mathbb{C}^3$$ In this case is $\widetilde{\mathcal{N}}$ the same as the blow-up at $0$?

-
In the simplest case of $\mathfrak{sl}_2$, the Springer resolution is easily seen to be the blow-up at the origin. In general, however, the Springer resolution is not a blow-up. – Mike Skirvin Jul 12 '12 at 6:12
A projective birational morphism $X\to Y$ (e.g., the Springer resolution) is the blow-up of $Y$ along some (non-unique) subscheme; see Hartshorne. But it can be hard to identify such a subscheme. The Springer resolution is an example where nature provides a resolution where it is hard to find an appropriate subscheme. Others are: the Hilbert scheme of $0$-dimensional subschemes of a smooth surface, resolving the corresponding symmetric product; determinantal varieties; theta divisors on Jacobians; vector bundles over homogeneous spaces $G/P$ collapsing to some cone in a representation of $G$. – inkspot Jul 12 '12 at 12:46
A subtlety which I hadn't realized before. Let $X$ be the subvariety $wy=wz=xy=xz=0$ of $\mathbb{A}^4$. This is the union of two coordinate $2$-planes, meeting transversely. Let $Y$ be the normalization of $X$, which separates the two planes. The map $Y \to X$ <b>is</b> a blowup. For example, it is the blowup at $\langle w,y \rangle$. However, note that the ideal $\langle w,y \rangle$ is not supported on the singular point of $X$; it is two crossing lines. The preimage of $\langle w,y \rangle$ is not the same thing as the exceptional locus. It is that preimage which is always Cartier. – David Speyer Jul 12 '12 at 14:00
The result from Hartshorne which inkspot is referencing is Theorem II.7.17. So $\widetilde{\mathcal{N}}$ is a blow up of $\mathcal{N}$, on very general grounds, but it is not yet clear to me whether we can take it to be the blow up of a subscheme supported on the singular locus. – David Speyer Jul 12 '12 at 14:04
David: iirc another, maybe more familiar, example of the same phenomenon is the small resolution of the 3-fold ODP (as features in the Atiyah flop). Here there are non-Cartier Weil divisors through the singular point, and blowing up such a Weil divisor produces a small resolution. (Sorry for the off-topic comment.) – user5117 Jul 12 '12 at 15:07