Following the conventions of "Quantum Field Theory and the Standard Model" by Schwartz, we have that for Yang-Mills Theory, an infinitesimal gauge transformation acts like

$$\delta_{\alpha} A = d\alpha - i\left[A, \alpha\right].$$

I am trying to compute the commutator of two gauge transformations, which I expect to give

$$\left[\delta_{\alpha},\delta_{\beta}\right]A = i\delta_{\left[\alpha, \beta\right]}A.$$

However, this is not what I find. Carrying out the computation, I find that

$\left[\delta_{\alpha},\delta_{\beta}\right]A = \delta_{\alpha}\delta_{\beta}A - \delta_{\beta}\delta{_\alpha}A = \delta_{\alpha}\left(d\beta - i\left[A, \beta\right]\right) - \delta_{\beta}\left(d\alpha - i\left[A, \alpha\right]\right) = d\alpha - i\left[d\beta - i\left[A, \beta\right], \alpha \right] - d\beta + i\left[d\alpha - i\left[A, \alpha\right],\beta\right] = d\alpha -d\beta - i\left[ d\beta, \alpha\right] - \left[\left[A,\beta\right],\alpha\right] + i\left[d\alpha, \beta\right] + \left[ \left[A,\alpha\right], \beta \right]$

Some of the commutators can be simplified by realizing that

$\left[d\alpha, \beta\right] - \left[d\beta, \alpha\right] = d\alpha\beta - \beta d\alpha - d\beta \alpha + \alpha d\beta = d\left(\alpha\beta\right) - d\left(\beta\alpha\right) = d\left[\alpha, \beta\right]$.

We can also use the Jacobi identity to see that

$\left[\left[A,\alpha\right], \beta\right] - \left[\left[A, \beta\right], \alpha\right] = \left[\left[A,\alpha\right], \beta\right] + \left[\left[\beta, A\right], \alpha\right] = -\left[\left[\alpha, \beta\right], A\right]$.

Putting everything together, we have that

$\left[\delta_{\alpha},\delta_{\beta}\right]A = d\alpha - d\beta + id\left[\alpha, \beta\right] + \left[A, \left[\alpha, \beta\right]\right] = i\delta_{\left[\alpha, \beta\right]}A + d\alpha - d\beta$.

My question is why has the extra $d\alpha - d\beta$ appeared? Am I carrying out some step in the computation incorrectly, or am I missing something conceptually? As I check, I also computed the commutator by starting with the identity

$e^{i\delta_{\alpha}}e^{i\delta_{\beta}}e^{-i\delta_{\alpha}}e^{-i\delta_{\beta}}A = (1 - \left[\delta_{\alpha}, \delta_{\beta}\right])A + \mathcal{O}(\alpha^2)$.

Here I applied the finite gauge transformations on the left hand side, expanded to second order in $\alpha$ and $\beta$, and matched terms with the right hand side. After doing so I found $\left[\delta_{\alpha},\delta_{\beta}\right]A = i\delta_{\left[\alpha, \beta\right]}A$, as expected, so I'm fairly certain the extra $d\alpha - d\beta$ terms should not be present, but I don't understand where my mistake is when I start from the infinitesimal case.

  • $\begingroup$ Hi bittermania. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Oct 26, 2018 at 17:44

1 Answer 1


Try this: $$ \left[\delta_{\alpha},\delta_{\beta}\right]A = (A + \delta_{\beta}A + \delta_{\alpha}\left(A + \delta_{\beta}A\right)) - (A + \delta_{\alpha}A + \delta_{\beta}\left(A + \delta_{\alpha}A\right)). $$


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