1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Jun 29 at 14:16
  • $\begingroup$ 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 $p_{ij}=e^i(e_j)$. $\endgroup$ Jul 1 at 8:58
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Aug 2 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.