# Explicit dual basis for the cotangent bundle for complex projective space

The module of sections of the cotangent bundle for complex projective space is finitely generated projective as the cotangent bundle is locally trivial. Thus the module of sections has a dual basis. But what is the dual basis explicitly? Say, on the coordinate chart where $$(u_1,\dotsc,u_n)$$ is sent to the equivalence class of $$(u_1,\dotsc,u_n,1)$$ what is the formula for the dual basis? This problem is of course solvable, but the answer must be known and in a reasonably simple form. I would be very grateful for someone sending me in the right direction.

• The dual basis of the module of sections (i.e. the global 1-forms) needs to be defined in terms of globally defined 1-forms. It is this which makes the problem messy. To convert local to global would need an appropriate partition of unity. Jun 29 at 14:16
• The dual basis exists for all finitely generated projective modules, and therefore for all modules of sections of locally trivial bundles by the Serre-Swan theorem. It is the reason why projection matrices are associated to both locally trivial bundles and finitely generated projective modules, and therefore the basis of algebraic K-theory. It consists of elements $e_i$ of the module (sections) and sections $e^i$ of the dual bundle. The usual projection matrix is then $p_{ij}=e^i(e_j)$. Jul 1 at 8:58
• Perhaps I misunderstand the question, since I am a differential and not algebraic geometer. When you say "module of sections", do you mean holomorphic sections of the cotangent bundle over $\mathbb{C}^n$. But with respect to what ring? I don't see how to find a basis of that module with respect to the ring of holomorphic functions on $\mathbb{C}^n$. If on the other hand, if the ring is the space of smooth functions, then $du^1, \dots, du^n$ form a basis. Aug 2 at 13:58