Does the Chain Rule Apply to Gauge Transformations in Lie Groups?

[vertices]

Again, I'm not sure whether this is the best place to post this question but its to do with gauge transformations, etc.

The question itself is rather stupid...

If we have a matrix U(g) (a Lie Group) and a vector φ in C (which is a scalar in spacetime) - does it make sense to use the chain rule thus:

$${\partial}_\mu (U(g) \phi) = U(g){\partial}_\mu \phi + ({\partial}_\mu U(g)) \phi$$

We are separately differentiating a matrix and vector - this seems very odd to me.

 

[xepma]

Again, I'm not sure whether this is the best place to post this question but its to do with gauge transformations, etc.

The question itself is rather stupid...

If we have a matrix U(g) (a Lie Group) and a vector φ in C (which is a scalar in spacetime) - does it make sense to use the chain rule thus:

$${\partial}_\mu (U(g) \phi) = U(g){\partial}_\mu \phi + ({\partial}_\mu U(g)) \phi$$

We are separately differentiating a matrix and vector - this seems very odd to me.

Look at it from a component point of view. The i'th component of the vector $\phi' = U(g)\phi$ is

[tex]\phi'_{i} = \sum_j U(g)_{ij}\phi_j[/itex]<br /> <br /> This is simply a sum of differentiable stuff. So differentiating gives<br /> <br /> [tex]{\partial}_\mu \phi'_{i} = {\partial}_\mu\left(\sum_j U(g)_{ij}\phi_j\right) = \sum_j \left({\partial}_\mu U(g)_{ij}\right)\phi_j + \sum_j U(g)_{ij}\left({\partial}_\mu\phi_j\right)[/itex]<br /> <br /> Now you can identify the first term with $({\partial}_\mu U(g)) \phi$ and the second with $U(g){\partial}_\mu \phi$[/tex][/tex]

 

[Frame Dragger]

That makes sense, and doesn't seem stupid to me.

 

[vertices]

Look at it from a component point of view. The i'th component of the vector $\phi' = U(g)\phi$ is

[tex]\phi'_{i} = \sum_j U(g)_{ij}\phi_j[/itex]<br /> <br /> This is simply a sum of differentiable stuff. So differentiating gives<br /> <br /> [tex]{\partial}_\mu \phi'_{i} = {\partial}_\mu\left(\sum_j U(g)_{ij}\phi_j\right) = \sum_j \left({\partial}_\mu U(g)_{ij}\right)\phi_j + \sum_j U(g)_{ij}\left({\partial}_\mu\phi_j\right)[/itex]<br /> <br /> Now you can identify the first term with $({\partial}_\mu U(g)) \phi$ and the second with $U(g){\partial}_\mu \phi$[/tex][/tex]

[tex]$$<br /> thanks xempa - convincing explanation:)$$[/tex]