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

The tangent bundle of a hyper-Kahler manifold gives a quadratic Lie algebra in the derived category. Can this be regarded as a simple Lie algebra according to Vogel's definition?

A point of view that came from studying Rozansky-Witten invariants is that the tangent bundle of a holomorphic symplectic manifold or hyper-Kahler manifold is a Lie algebra with a non-degenerate invariant symmetric bilinear form. Here the tangent bundle is taken as an object in the derived category and then shifted. The Atiyah class is interpreted as a Lie bracket and the Bianchi identity as the Jacobi identity. The symplectic form is interpreted as a symmetric form since we shifted. Some references are (and please add or request any reference I have omitted)

MR2024627 (2004m:57026) Roberts, Justin . Rozansky-Witten theory. Topology and geometry of manifolds (Athens, GA, 2001), 1--17, Proc. Sympos. Pure Math., 71, Amer. Math. Soc., Providence, RI, 2003.

MR2110899 (2005h:53070) Nieper-Wißkirchen, Marc . Chern numbers and Rozansky-Witten invariants of compact hyper-Kähler manifolds. World Scientific Publishing Co., Inc., River Edge, NJ, 2004. xxii+150 pp. ISBN: 981-238-851-6

MR2472137 (2010d:14020) Markarian, Nikita . The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem. J. Lond. Math. Soc. (2) 79 (2009), no. 1, 129--143.

Now Vogel has constructed a universal simple Lie algebra. The question is whether the tangent bundle of an irreducible holomorphic symplectic manifold meets Vogel's criteria for a simple Lie algebra. This question is for algebraic geometers so I will expand on this. The first condition is that End(L)=End(I) where I is the trivial representation so End(I) is the commutative ring of scalars. In this example Ext(O). This obviously fails for the product of two manifolds so I have naively excluded this by imposing the irreducible condition. The second condition is that $\mathrm{Hom}(\bigwedge^2L,L)$ is a free End(I)-module with basis the Lie bracket.

One reason I find this confusing is that End(I) has nilpotent elements whereas I am used to a field.

If the answer to both questions is Yes then we get a character of Vogel's universal ring. I would expect this to be of interest to both subjects.

Edit The paper http://arxiv.org/abs/1205.3705 has now been posted on the arxiv and this proves that $K3$-surfaces do give a character of Vogel's ring.

share|improve this question
    
Vogel's paper: math.jussieu.fr/~vogel/A299.ps.gz –  Allen Knutson Mar 5 '10 at 19:45
    
I have a question. In his paper Vogel does not require that $Hom(\wedge^2 L,L)$ is one dimensional (I am refering to the definition of simple on the bottom of page 11 in his paper "The universal Lie algebra"). Why do you need it here? –  DamienC Aug 24 '11 at 13:44
    
Vogel has two conditions. He defines simple by $End(L)$ is one dimensional. He then has a second condition in 3.2 Theorem that a certain commutative square is Cartesian. My intention is that the two second conditions are equivalent. The second condition is needed to define a character of the universal ring. –  Bruce Westbury Aug 24 '11 at 16:31
    
If I'm not mistaken this point of view on RW invariants is due to Kontsevich (arXiv:dg-ga/9704009) and Kapranov (arXiv:alg-geom/9704009). –  David Ben-Zvi Dec 2 '11 at 17:14
add comment

1 Answer

I believe this is a very interesting question, that I have been asking myself for quite a long time.

Nevertheless, I have been told by Prof. Beauville that even in the irreducible case one does not have that $$ Ext_X(\mathcal O_X,\mathcal O_X)=Ext_X(T_X,T_X) $$

Namely, consider $X$ being the Hilbert scheme of two points on a $K3$ surface.

Then $Ext_X(\mathcal O_X,\mathcal O_X)=\mathbb{C}\oplus\mathbb{C}[-2]\oplus\mathbb{C}[-4]$.

But $Ext_X(T_X,T_X)=Ext_X(\mathcal O_X,(T^*_X)^{\otimes 2})$ contains $Ext_X(\mathcal O_X,\Omega^2_X)$, which is huge ($h^{2,2}=232$).

Anyway, I must say that this does not kill the question (this just tells we have to reformulate it). I hope to be able to write more about it soon.

EDIT: it seems that the answer to the question is NO. The point is that 232 is also the dimension of $H^1(X,S^3(T_X))$ ($X$ is again a $K3$), therefore $Ext_X^1(S^2(T_X),T_X)=RHom_X(\wedge^2(T_X[-1]),T_X[-1]))$ has dimension $\geq232$.

share|improve this answer
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.