Question

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.

Answer 1

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 $\geq232$.