# Differentiating a 'vector'

1. Feb 15, 2010

### 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.

2. Feb 16, 2010

### xepma

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]

This is simply a sum of differentiable stuff. So differentiating gives

[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]

Now you can identify the first term with $({\partial}_\mu U(g)) \phi$ and the second with $U(g){\partial}_\mu \phi$

3. Feb 16, 2010

### Frame Dragger

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

4. Feb 16, 2010

### vertices

thanks xempa - convincing explanation:)