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 \(\phi^{-1 *}\).