The question.
Let (X,J) be a complex manifold and u a holomorphic vector field, i.e. LuJ=0. The holomorphicity of u implies that the Lie derivative Lu on forms preserves the (p,q) decomposition and also that it commutes with barpartial. From this it follows that u acts infinitesimally on the Dolbeault cohomology groups Hp,q(X) of X. My question is, does anyone know of an example in which this action is non-trivial?
Some context.
To give some context, first note that the analgous action for de Rham cohomology is always trivial: If M is any smooth manifold and v any vector field, then the formula Lv=dcirciv+ivcircd shows that the infinitesimal action of v on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field v generates diffeomorphisms which are by construction isotopic to the identity map.)
Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a d-closed form. (For example, this is true if X is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let alpha be a barpartial-closed (p,q)-form which is also d-closed. Then we know that Lualpha=d(iualpha) is also of type (p,q). So,
Lualpha=barpartialleft((iualpha)p,q−1right)+partialleft((iualpha)p−1,qright)
and the other contributions barpartial((iualpha)p−1,q) and partial((iualpha)p,q−1) vanish. Now the fact that barpartial((iualpha)p−1,q)=0 and our hypothesis imply that there is a (p-1, q-1)-form beta such that
(iualpha)p−1,q+barpartialbeta
is closed. Hence
partialleft((iualpha)p−1,qright)=barpartialpartialbeta
and so
Lualpha=barpartialleft((iualpha)p,q−1+partialbetaright)
which proves the action of u on Hp,q(X) is trivial.
No comments:
Post a Comment