Is it known when a Nakajima quiver variety happens to be a local complete intersection? 


[For simplicity consider an affine quiver variety, i.e. the categorical quotient of the zero set of the moment map on the space of representations of a doubled quiver]



are quiver varieties local complete intersections?
By Proposition 1.3 of Beauville's original [article](http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=0CEkQFjAD&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.54.9491%26rep%3Drep1%26type%3Dps&ei=f0AZUp2wJOT54AOBxYDoAw&usg=AFQjCNEFQNxgm0ekulgib2n6CGTuvEq_sA&sig2=0ywjNlBzWaFeqipNNZ2FmA&bvm=bv.51156542,d.dmg) on symplectic singularities, any variety with symplectic singularities (in particular, a Nakajima quiver variety) is rational and Gorenstein, which is pretty close to a lci.


However, I don't think most non-smooth quiver varieties are actually lci's, since by 1.4 of the same paper of Beauville, any such quiver variety must have singular locus which is pure of dimension 2, a pretty weird condition. 


**EDIT**: Perhaps I should expand on this a little bit. In order for a conic symplectic singularity like an affine quiver variety to be an lci, any deformation of it must be lci as well. The deformation theory of conic affine symplectic varieties is very nice; there is a universal deformation (amongst those keeping the symplectic structure!) whose base is a vector space, and there are certain walls, which are codimension 1 pieces of the locus where the deformation is not Q-factorial and terminal (in most cases, this is means smooth, but not always). In essence the ``badness'' at these points is controlled by a single slice singularity. If for any of the walls, this singularity is not an lci, you're out of luck.


For a quiver variety, the walls correspond to roots, and the characteristic singularity you expect is the affinization of the cotangent bundle of a Grassmannian (which is the same as the closure of operators with square 0 of a given rank in the space of nilpotent matrices). The only one of these that the codimension 2 condition doesn't rule out is all matrices with square 0, or the point. These correspond to the weight associated by Nakajima being highest, lowest, or weight 0 for this root. That looks like a pretty weird condition to me.


