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Post Undeleted by Peter Michor
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Peter Michor
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Near a point where $X\ne 0$, chooseIf you can find a coordinate systemclosed 1-form $(u_i)$$\alpha\in \Omega^1(M\setminus Z(X))$ with $\partial_{u_1} = X$$i_X\alpha=1$, then $i_Xdu_1=1$ does it. It$\eta_n=\alpha\wedge i_X\eta_n = \alpha\wedge d\eta_{n-2}= -d(\alpha\wedge\eta_{n-2})$ is even exact. But thenMaybe, the source that you are reading has such $\eta_n = du_1\wedge i_X\eta_n = du_1\wedge d\eta_{n-1}$ is exact$\alpha$.

Near a point where $X\ne 0$, choose a coordinate system $(u_i)$ with $\partial_{u_1} = X$, then $i_Xdu_1=1$ does it. It is even exact. But then $\eta_n = du_1\wedge i_X\eta_n = du_1\wedge d\eta_{n-1}$ is exact.

If you can find a closed 1-form $\alpha\in \Omega^1(M\setminus Z(X))$ with $i_X\alpha=1$, then $\eta_n=\alpha\wedge i_X\eta_n = \alpha\wedge d\eta_{n-2}= -d(\alpha\wedge\eta_{n-2})$ is exact. Maybe, the source that you are reading has such $\alpha$.

Post Deleted by Peter Michor
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Peter Michor
  • 25.3k
  • 2
  • 64
  • 112

Near a point where $X\ne 0$, choose a coordinate system $(u_i)$ with $\partial_{u_1} = X$, then $i_Xdu_1=1$ does it. It is even exact. But then $\eta_n = du_1\wedge i_X\eta_n = du_1\wedge d\eta_{n-1}$ is exact.