Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be your favorite simple complex Lie group, and $P\subset G$ your favorite parabolic subgroup. We can identify $T^*G/P$ with the space of pairs $$\{(gP,x)\in G/P\times \mathfrak g | x\perp \operatorname{Ad}_g(\mathfrak p)\}$$ where $\perp$ denotes perpendicularity in the Killing form. Thus, we have the second projection $p_2:T^*G/P\to \mathfrak g$; when $P=B$, this is the famous Springer map.

Now, let $e$ be your favorite nilpotent in $\mathfrak g$, and let $e,h,f$ be a completion of this to a $\mathfrak{sl}_2$ triple (which exists by Jacobson-Morozov). Then $S=e+\ker(\operatorname{ad}_f)\subset \mathfrak g$ is an affine subspace of $\mathfrak g$ transverse to the orbit $G\cdot e$ called the Slodowy slice to $e$.

It's a theorem that $p_2^{-1}(S)$ is a smooth symplectic variety (it's actually a symplectic reduction of $T^*G/P$ by the action of a nilpotent subgroup $M\subset G$ at a regular value of the moment map), and it's one that I like very much.

Does this variety have an agreed-upon name?

share|improve this question
    
Does your question make sense with suitable modifications over an algebraically closed field of arbitrary characteristic? Slodowy slices often come up in characteristic $p$ in work by Premet and others. Also, a small nitpick if someone's favorite nilpotent is the zero element; this is the (usually trivial) case where Jacobson-Morosov doesn't apply. –  Jim Humphreys Jan 10 '11 at 14:19
    
This should all work over any field where $P$ and $e$ are defined, though maybe I'm missing something silly. As for the case of the 0 nilpotent, if you take f to be 0, then everything works, and e=h=f=0 satisfy the Chevalley relations of $\mathfrak{sl}_2$. –  Ben Webster Jan 10 '11 at 17:47
add comment

1 Answer

In my paper "Singular blocks of parabolic category O and finite W-algebras", these are called "S3-varieties." S3 is for "Slodowy-Springer-Spaltenstein."

share|improve this answer
2  
So, I guess you can accept this answer as soon as someone else starts calling them S3-varieties as well. It's amazing you wrote an entire paper just for the possibility of extra MO reputation. –  Mike Skirvin Jun 15 '11 at 0:00
    
And since intent can be difficult to judge on the internet, let me say that my comment above should not be taken seriously. –  Mike Skirvin Jun 15 '11 at 0:03
    
This could be confused with Serre's S3 condition in commutative algebra. –  Joel Kamnitzer Jun 15 '11 at 1:42
1  
That's true. Serre, Springer, Spaltenstein and Steinberg really should have thought of that. –  Ben Webster Jun 16 '11 at 7:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.