# I Commutator of covariant derivatives

1. Nov 19, 2016

### naima

Hi there
I came across this paper.
the author defines a covariant derivative in (1.3)
$D_\mu = \partial_\mu - ig A_\mu$
He defines in (1.6)
$F_{jk} = i/g [D_j,D_k]$
Why is it equal to $\partial_j A_k - \partial_k A_j - ig [A_j, A_k]$?
I suppose that it comes from a property of Lie derivatives.
Thanks

2. Nov 19, 2016

### ShayanJ

Just put the definition of $D_\mu$ in $[D_\mu,D_\nu]$ and calculate!
You can apply the above operator on a scalar field, i.e. calculate $[D_\mu,D_\nu]\phi$ to avoid confusion.

3. Nov 19, 2016

### naima

Why does $A_j \partial_k$ disappear?

4. Nov 19, 2016

### Orodruin

Staff Emeritus
What you have is a commutator of operators, in order to compute the commutator - which is another operator - you need to consider the action of the commutator.

I suggest you start with the following example with two operators acting on functions on the real line:

Operator $A$ is such that $Af(x) = A(x)f(x)$, where $A(x)$ is a function.
Operator $B$ is the derivative operator such that $Bf(x) = f'(x)$.

The commutator $[A,B]$ is the operator such that $[A,B]f(x) = (AB - BA)f(x)$ for all $f$. What is this operator?

5. Nov 21, 2016

### naima

Thank you

This shows that when you have product of operators D and B acting on a vector V, the notation DBV may be considered as a compositiion law
D(BV) or else (DB) V if you can give sense to (DB)
IF D and B are patrices the internal law of multiplication gives
(DB)V = D(BV)
here (DB) = D(B .)
If D is a derivative we have D(BV) = (DB)V + B (DV)
This gives a sense to (DB)
here (DB) = DB - BD

So the equality in the paper comes from an implicit choice for the product of operators.
Unless it is based on a more intrinsic way to mix covariant derivatives.

6. Nov 21, 2016

### Orodruin

Staff Emeritus
It does not matter whether D and B are matrices or linear operators on a general function space (including derivatives). The commutator [D,B] is defined as the linear operator such that [D,B]f = D(Bf) - B(Df) for all f in the vector space that the operators act on.

If D is a derivative and B is multiplication by a function b, you have D(Bf) = D(bf) = f (Db) + b(Df) = f(Db) + B(Df) and therefore D(Bf) - B(Df) = [D,B]f = f (Db), i.e., the commutator [D,B] is multiplication by the function Db. Note that I have used capital and lower case Bs to distinguish DB (operator product) from Db (a function).