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.

1

# Is an irreducible holomorphic symplectic manifold a simple Lie algebra?

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.