About the Lie brackets of vector fields

Question

Let $M$ be a manifold and $\phi$ a smooth function. I define a twisted Lie bracket:

$$[X,Y]_{\phi}= X \phi Y-Y \phi X=(X\phi)Y-(Y\phi )X+\phi [X,Y]$$

It is a vector field. If $\phi$ is inversible, we have:

$$[X,Y]_{\phi}= \phi^{-1}[\phi X,\phi Y]$$

We have the Jacobi identities for the twisted Lie brackets.

Then, for a connection $\nabla$, we can define a twisted curvature:

$$R_{\phi}(\nabla)(X,Y)=\nabla_X \phi \nabla_Y - \nabla_Y \phi \nabla_X-\nabla_{[X,Y]_{\phi}}=\phi R(X,Y)$$

We can also define a differential over the exterior forms:

$$d_{\phi} (\alpha)(X,Y)=X\phi \alpha(Y)-Y\phi \alpha (X)-\alpha ([X,Y]_{\phi})=\phi d\alpha (X,Y)$$

Can we have a twisted De Rham cohomology?

Comments

 It appears that for invertible $\phi$, your twisted Lie Bracket is nothing but the pushforward of the bracket onto whatever $\phi$ maps onto.
So you can pullback your cohomology along $\phi^{-1 *}$.