May be $TM$ a tangent bundle of the manifold $M$ and $\wedge^n TM$ the set of all $n$-forms. The map $d: \bigwedge^n TM \rightarrow \bigwedge^{n+1}TM$ is called the exterior derivative and it holds $d^2=0$ such there can be built the DeRham complex

$$\cdots \overset{d_{n-1}}\longrightarrow \bigwedge^n TM \overset{d_n}\longrightarrow \bigwedge^{n+1}TM \overset{d_{n+1}}\longrightarrow \cdots$$

with the $n$-th cohomology class $H_n =\ker(d_n)/\operatorname{im}(d_{n-1})$. May be $A$ a 1-form field on $M$. Then the first cohomology class is given by all $A \in \wedge^1 TM$ with $dA=0$ but with the restriction $A \neq d \phi$ for a 0-form field $\phi$. How I can compute the most general expression for the first cohomology class (or higher cohomology classes)?


The function $A(\vec{x}) = \int_{\partial M} d^nx'|\vec{x}-\vec{x'}|^{1-n} K(\vec{x'})n(\vec{x'})$ with the 1-form unit normal vector field $n(\vec{x'})$ and an arbitrary function $K(\vec{x'})$ (or similar functions?) satisfies above conditions.

  • 3
    $\begingroup$ De Rham cohomology is built out of the cotangent bundle $T^*M \to M$, not $TM.$ Specifically, the space $\Omega^p(M)$ of $p$-forms on $M$ is the space of (smooth) sections of $\Lambda^n T^*M$, and $\Omega^*(M)$ becomes a cochain complex with operator $d:\Omega^p(M) \to \Omega^{p+1}(M)$; the cohomology of this complex is the de Rham cohomology of $M$. $\endgroup$ – anomaly May 21 '15 at 18:57
  • 2
    $\begingroup$ Also, cohomology is usually written with upper indices (that is, $H^n$ rather than $H_n$). $\endgroup$ – anomaly May 21 '15 at 18:58

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.