Covariant derivative in gauge theory

[ismaili]

Is the following formula correct?
Suppose we work in a 4D Euclidean space for a certain gauge theory,

\int d^4x~ \text{tr}\Big(D_i(\phi X_i )\Big) = \oint d^3S_i~ \text{tr}(\phi X_i)

and,

\int d^4x~\partial_j \text{tr}(\phi F_{mn}\epsilon_{mnij}) = \oint d^2S_j~ \text{tr}(\phi F_{mn}\epsilon_{ijmn})

where \phi is a scalar field.

If they are correct, how to prove them?
Thanks

 

[tom.stoer]

You try to use the Stokes theorem to transform a 4-volume integral to a 3-surface integral. Of course this works for an ordinary derivative, but not for a covariant derivative.

 

[ismaili]

You try to use the Stokes theorem to transform a 4-volume integral to a 3-surface integral. Of course this works for an ordinary derivative, but not for a covariant derivative.

I was worrying about this too.
For GR, we have corresponding Stoke's theorem for covariant derivative, i.e.
\int \sqrt{-g} d^4x ~D_\mu v^\mu = \int \sqrt{-g} d^3S_\mu ~ v^\mu
because, we have \partial_\mu(\sqrt{-g}v^\mu) = \sqrt{-g}D_{\mu} v^\mu

However, we have no corresponding formula in a gauge theory.
But I was confused by the following, can I say that
<br /> \int d^4x~\text{tr}\Big(D_i (\phi X_i)\Big)<br /> = \int d^4x~\text{tr} \Big( \partial_i (\phi X_i) \Big)<br /> - \int d^4x~\text{tr} \Big( [A_i , \phi X_i] \Big)<br />
where we can apply Stokes theorem to the first term, because it involves an ordinary derivative.
For the second term, however, due to the cyclic property of the trace, it seems to be zero too.
Hence, it seems we can also apply Stokes theorem on covariant derivative in a gauge theory?

As for the 2nd formula of my first post, it's more strange that it reduces the dimension of integral by two??

 

[tom.stoer]

Your idea regarding GR does no longer work for rank-2 tensors which is one reason that one is not able to define energy as a volume-integral based on the energy-momentum tensor.

Of course the trace does not kill the second integral. I guess X is a generator of the algebra. If you rewrite the trace using structure constants instead you'll see that immediately.

 

[tom.stoer]

Then there are some strange things in your formula:

1) what is the F in you first post? I haven't seen this in the literature
2) for a scalar field you usually have the square of D; where did you get you formula from?

 

[ismaili]

Your idea regarding GR does no longer work for rank-2 tensors which is one reason that one is not able to define energy as a volume-integral based on the energy-momentum tensor.

Of course the trace does not kill the second integral. I guess X is a generator of the algebra. If you rewrite the trace using structure constants instead you'll see that immediately.

I don't get your first point. Even for rank-2 tensors, we have
\partial_\mu(\sqrt{-g}F^{\mu\nu}) = \sqrt{-g}D_\mu F^{\mu\nu}
, so that we can always use Stoke's theorem for covariant derivative in GR.

I don't get the second point either, for that 2nd integral in my last post, we have
<br /> \int d^4x~\text{tr} \Big( [A_i , \phi X_i] \Big)<br /> = \int d^4x~\text{tr} \Big( A_i \phi X_i - \phi X_i A_i \Big) \\<br /> = \int d^4x~\text{tr} \Big( A_i \phi X_i - A_i\phi X_i \Big)<br /> = 0<br />
where we just used the cyclic property of trace, and X_i is some vector field.

Is the calculation of the integral wrong?
Thanks.

 

[ismaili]

Then there are some strange things in your formula:

1) what is the F in you first post? I haven't seen this in the literature
2) for a scalar field you usually have the square of D; where did you get you formula from?

Sorry, I forgot to say the F in the first post is the field strength of the the gauge field.
Scalar field usually involves square of D? why scalar field has something to do with the number of covariant derivative D?

I am reading about the instanton in gauge theory.
Those two formulas are central charges of the supersymmetric algebra.

 

[tom.stoer]

If you look at the Lagrangian of a scalar field theory the kinetic term is something like

(D_\mu \phi_i)^\dagger (D^\mu \phi_i)

So both the field and the covariant derivative are squared. Here I omitted the trace and introduced i as index in the fundamental rep.

If you look at the gauge current j the field is squared again:

j^a_\mu = \phi_i^\dagger (T^a)_{ik}\partial_\mu \psi_k + \text{h.c.}

Have a look at http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

Can you give me a reference for your formulas?

 

[tom.stoer]

I see you are studying 5D SUSY Yang-Mills; of course this is slightly different; I'll think about it.