# About the Lie brackets of vector fields

+ 2 like - 0 dislike
213 views

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?

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 $\phi^{-1 *}$.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.