I am studying loop quantum gravity using the book by Pullin and Gambini. I am having some trouble understanding and getting past the chapter on Yang Mills theory, mainly because I am confused about some of the notation and concepts. I am hoping someone can shed some light on this for me.

First of all, the Yang Mills covariant derivative is defined and written as:

$$ D_{\mu} \equiv \partial_{\mu} - ig/2\sigma^iA^i_{\mu} $$

I understand that the superscripted indices, $i$, on $\sigma^i$ and $A^i_{\mu}$ are internal indices of the theory that run from 1 to 3. But is the $i$ in front of the coupling parameter, $g$, meant to indicate this internal index as well, or is this the imaginary unit? I am guessing the second, but the notation is a little ambiguous to me so I would like to have some clarification.

Secondly, I don't understand the relation between the commutator of the covariant derivative with itself and the field tensor. In the book the following is written:

$$ \left[D_{\mu},D_{\nu}\right] = -ig/2F^i_{\mu\nu}\sigma^i $$

Where I understand that $F^i_{\mu\nu}$ is the field tensor of the theory. I have a few questions about this. Firstly: if I write out the commutator explicitly, using the definition of the covariant derivative from above, I get:

$$\begin{align} \left[D_{\mu},D_{\nu}\right] = D_{\mu}D_{\nu} - D_{\nu}D_{\mu} &= (\partial_{\mu} - ig/2\sigma^iA^i_{\mu})(\partial_{\nu} - ig/2\sigma^iA^i_{\nu}) - (\partial_{\nu} - ig/2\sigma^iA^i_{\nu})(\partial_{\mu} - ig/2\sigma^iA^i_{\mu}) \\ &=\partial_{\mu}\partial_{\nu} - \partial_{\mu}ig/2\sigma^iA^i_{\nu} - ig/2\sigma^iA^i_{\mu}\partial_{\nu}+i^2g^2/4\sigma^iA^i_{\mu}\sigma^iA^i_{\nu} \\ &- \partial_{\nu}\partial_{\mu}+\partial_{\nu}ig/2\sigma^iA^i_{\mu}+ig/2\sigma^iA^i_{\nu}\partial_{\mu}-i^2g^2/4\sigma^iA^i_{\nu}\sigma^iA^i_{\mu} \end{align}$$

Now, I understand that the first and fifth terms on the right cancel, because the partial derivatives, $\partial_{\mu}$ and $\partial_{\nu}$, commute. But I don't understand how to manipulate the remaining terms to get to a something of the form $-ig/2F^i_{\mu\nu}\sigma^i$. Can someone please show me the full derivation for this?

Then the book goes on to say that if one indeed works out the commutator explicitly, one gets that the field tensor is given by

$$ F^i_{\mu\nu} = \partial_{\mu}A^i_{\nu}-\partial_{\nu}A^i_{\mu}+g\epsilon^{ijk}A^j_{\mu}A^k_{\nu}, $$

where, I assume, $\epsilon^{ijk}$ are the Levi-Civita symbols, right? Maybe the full derivation of the commutator will already explain this too, but how does one get from the commutator to this last equation.

Finally, and this is a purely conceptual question. Why is it that the commutator of the covariant derivatives yields the field tensor? Is this a just a definition in gauge theory?

Any help would be greatly appreciated!

  • $\begingroup$ FWIW, an $i$ that is not an index is the imaginary unit. $\endgroup$
    – Qmechanic
    Commented Jan 20, 2019 at 18:35
  • $\begingroup$ So I could write the second term of the covariant derivative as $-ig/2\sigma^kA^k_{\mu}$, where $i$ is the imaginary unit and $k$ is an internal index then, right? $\endgroup$ Commented Jan 20, 2019 at 18:49
  • $\begingroup$ Yes, that would avoid the notational clash. $\endgroup$
    – Qmechanic
    Commented Jan 20, 2019 at 18:55
  • $\begingroup$ When you computed the commutator of the covariant derivatives, you should have gotten 10 terms, not 8. Since these derivatives are acting on something, there are two terms that come from $\partial_\mu$ and $A_\nu$: $\partial_\mu A_\nu$ and also $A_\nu \partial_\mu$. Six of the ten terms cancel in pairs. The two A-A terms reduce to one using the commutator of the sigma matrices. $\endgroup$
    – G. Smith
    Commented Jan 20, 2019 at 20:38
  • 1
    $\begingroup$ I asked about the sigmas because you left the partial derivatives to the left of them. You should move the partials as far to the right as possible, so that they come immediately before what they are differentiating. And make sure that you understand that $\partial_\mu A_\nu$ in the final result really means $(\partial_\mu A_\nu)\psi$, not $\partial_\mu(A_\nu \psi)$. $\endgroup$
    – G. Smith
    Commented Jan 21, 2019 at 1:28

2 Answers 2


Let's put in a $\psi$ to give the covariant derivatives something to differentiate. This will make clear why you got 8 terms instead of 10.

By the way, you want to make sure that a contracted index only appears twice. For example, your terms with four $i$ indices don't make sense.

$$\begin{align} \left[ D_\mu, D_\nu \right] \psi &= (D_\mu D_\nu - D_\nu D_\mu) \psi \\ &= ( \partial_\mu - \frac{ig}{2} \sigma^i A^i_\mu ) ( \partial_\nu - \frac{ig}{2} \sigma^j A^j_\nu ) \psi - ( \partial_\nu - \frac{ig}{2} \sigma^j A^j_\nu ) ( \partial_\mu - \frac{ig}{2} \sigma^i A^i_\mu ) \psi \\ &= \partial_\mu \partial_\nu \psi - \frac{ig}{2} \sigma^j [ ( \partial_\mu A^j_\nu ) \psi + A^j_\nu \partial_\mu \psi] - \frac{ig}{2} \sigma^i A^i_\mu \partial_\nu \psi - \frac{g^2}{4} \sigma^i \sigma^j A^i_\mu A^j_\nu \psi \\ &- \partial_\nu \partial_\mu \psi + \frac{ig}{2} \sigma^i [ ( \partial_\nu A^i_\mu )\psi + A^i_\mu \partial_\nu \psi] + \frac{ig}{2} \sigma^j A^j_\nu \partial_\mu \psi + \frac{g^2}{4} \sigma^j \sigma^i A^j_\nu A^i_{\mu}\psi \end{align}$$

There are 10 terms. Terms 1 and 6 cancel. So do terms 3 and 9 (because the name of a contracted index doesn't matter) and terms 4 and 8. This leaves

$$\begin{align} &\left[ D_\mu, D_\nu \right] \psi = -\frac{ig}{2}\sigma^i ( \partial_\mu A^i_\nu - \partial_\nu A^i_\mu ) \psi - \frac{g^2}{4} [\sigma^i, \sigma^j] A^i_\mu A^j_\nu \psi \end{align}$$

The commutator of the Pauli matrices is

$$[\sigma^i, \sigma^j] = 2i \epsilon^{ijk} \sigma^k$$


$$\begin{align} \left[ D_\mu, D_\nu \right] \psi &= -\frac{ig}{2}\sigma^i ( \partial_\mu A^i_\nu - \partial_\nu A^i_\mu ) \psi -\frac{i g^2}{2} \epsilon^{ijk} \sigma^k A^i_\mu A^j_\nu \psi \\ &= -\frac{ig}{2}\sigma^i ( \partial_\mu A^i_\nu - \partial_\nu A^i_\mu + g \epsilon^{ijk} A^j_\mu A^k_\nu) \psi \\ &=-\frac{ig}{2}\sigma^i F^i_{\mu\nu} \psi \end{align}$$

In the next to the last line, the contracted indices in the third term were renamed so that the sigma matrix had index $i$ rather than $k$, and I made use of the fact that you can cycle the indices of the Levi-Civita tensor.


"But is the i in front of the coupling parameter, g, meant to indicate this internal index as well, or is this the imaginary unit?"

It's the imaginary unit. Notice that $D_\mu$ has no $i$ index, so it must be that $i$ on the right is summed over. When using this Einstein summation convention, indices should never be repeated more than twice. Also, there is a general rule of thumb that 'no index is special', so an expression that is somehow 'weighted' by the index is treating different indices preferentially and picking out a particular ordering of the indices.

"Can someone please show me the full derivation for this?"

It's often easier to first work out some more 'elementary' commutators and then use properties of the commutator. E.g. $$ \left[\sigma,\partial\right]=0 $$ $$ \left[\sigma^i,\sigma ^j\right]= 2i \epsilon ^{ijk} \sigma ^k $$ $$ \left[A^i,A^j\right]=0$$ $$ \left[\partial,A\right]= (\partial A)$$

etc. Now for instance:

$$\left[-ig/2 \sigma^i A^i _\mu,-ig/2 \sigma^j A^j _\nu \right] = -i g^2 /2 \epsilon ^{ijk} \sigma^k A^i _\mu A^j _\nu$$

"Why is it that the commutator of the covariant derivatives yields the field tensor? Is this a just a definition in gauge theory?"

It is basically the definition, but here's why: In gauge theory the meaning of the gauge field is to act as a 'connection form'. It encodes how the covariant derivative 'parallel transports' (that is, moves without changing) objects that transform under the gauge group.

Then one may ask, does parallel transporting from A to B depend on the path taken? This can be phrased infinitesimally by comparing a path: $x \to x + dx^\mu \to x + dx^\mu+dx^\nu$ with the path $x \to x + dx^\nu \to x + dx^\nu +dx^\mu$.

Equivalently, one may take the closed loop $x \to x + dx^\mu \to x + dx^\mu +dx^\nu \to x+ dx^\nu \to x$. The parallel transport along this closed loop will be $\left[D_\mu ,D_\nu\right]$. This is a gauge invariant "local" question and so has a gauge invariant "local" answer. The only gauge invariant local operator that we know of is the field strength, so this identification is natural.

  • $\begingroup$ Thank you very much for this! This is already helping me lot. However, I would really like to see how to do the full derivation without commutator identities as to fully be able to follow what is going on. Furthermore, why are the elementary commutators you give straightforward? For example: $\left[\sigma^i,\partial_{\mu}\right] = \sigma^i\partial_{\mu} - \partial_{\mu}\sigma^i = 0$. Why is this the case? I see why the last of these terms is zero, because $\sigma^i$ is constant, but why is the first? $\endgroup$ Commented Jan 20, 2019 at 23:15
  • 1
    $\begingroup$ You’re making the same mistake again that I pointed out in a comment to your question. Whenever you are dealing with a operator like a partial derivative, you have to let it operate on the invisible thing to the right of the commutator. There are three terms here. Just put a position-dependent vector $\psi$ to the right of the commutator and it should become obvious. $\endgroup$
    – G. Smith
    Commented Jan 21, 2019 at 1:13
  • $\begingroup$ @G.Smith Ah, now I understand what you meant! Thank you for clarifying that for me! $\endgroup$ Commented Jan 21, 2019 at 15:00

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