2
$\begingroup$

In Henneaux's Lectures on the Antifield BRST Formalism for Gauge Theories, it is claimed in Exercise 1 that diffeomorphisms $\delta_\xi A_\mu=\xi^\rho\partial_\rho A_\mu+\partial_\mu\xi^\rho A_\rho$ differ from ordinary gauge transformations $\delta_\Lambda A_\mu=\partial_\mu\Lambda$ by a trivial gauge transformation $\delta_\mu A=\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\frac{\delta S}{\delta A_\nu(y)}$, for some $\mu_{\mu\nu}(x,y)=-\mu_{\nu\mu}(y,x)$. In here $S=\int\text{d}x\epsilon^{\mu\nu\rho}F_{\mu\nu}A_\rho$ is the action for Abelian Chern-Simons theory.

I tried to proof this. The equations of motion are $\frac{\delta S}{\delta A_\mu(x)}=\epsilon^{\mu\alpha\beta}\partial_\alpha A_\beta(x)$. Therefore, what we want to proof is that there is a $\Lambda$ and $\mu_{\mu\nu}(x,y)$ such that $$\delta_\xi A_\mu-\delta_\Lambda A_\mu=\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(y).$$ Choosing $\mu_{\mu\nu}(x,y)=\epsilon_{\mu\nu\alpha}\xi^\alpha(x)\delta(x-y)$ yields $$\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(y)=\epsilon_{\mu\nu\gamma}\xi^\gamma(x)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(x)=\xi^\alpha\partial_\alpha A_\mu+\partial_\mu \xi^\beta A_\beta-\partial_\mu (\xi^\beta A_\beta)=\delta_\xi A_\mu-\delta_{\xi^\nu A_\nu}A_\mu.$$ Thus, this choice almost works. However, the parameter $\Lambda$ shouldn't depend on $A$. Is there another choice of $\mu$ and $\Lambda$ so that $\Lambda$ doesn't depend on $A$?

Investigating the problem further , I noticed the following. If indeed there is such a $\mu$ and $\Lambda$, then, whenever the equations of motion are satisfied we have that $\delta_\xi A_\mu-\delta_\Lambda A_\mu=0$. The equations of motion imply that $A$ is pure gauge, i.e. $A_\mu=\partial_\mu \Omega$ for some $\Omega$. Then $$\delta_\xi A_\mu-\delta_\Lambda A_\mu=\xi^\rho\partial_\rho\partial_\mu\Omega+\partial_\mu\xi^\rho\partial_\rho\Omega-\partial_\mu\Lambda=\partial_\mu(\xi^\rho\partial_\rho\Omega-\Lambda).$$ There is no way that this vanishes identically unless we choose $\Lambda=\xi^\rho\partial_\rho\Omega=\xi^\rho A_\rho$ (up to a constant). We conclude that the $\Lambda$ must depend on $A$ as we found above. Then, isn't the statement in the exercise wrong? Or are we allowed to use a different gauge parameter for every field?

$\endgroup$

1 Answer 1

0
$\begingroup$

I don't have Henneaux's lectures on hand, but I assume the context is classical (not quantum) gauge theory.

The paradox comes from thinking that the gauge-transform parameter $\Lambda$ is not allowed to depend on the gauge field $A$ that is being transformed, but that is incorrect. Within any contractible patch, we can take the quantities $A_a$ (and also $\xi^a$) to be ordinary smooth functions, and then $\Lambda=\xi^a A_a$ is also a smooth function. The transform $A_a\to A_a+\partial_a\Lambda$ is a legal gauge transform for any smooth function $\Lambda$, including one of the form $\Lambda=\xi^a A_a$.

On a general manifold, the gauge field $A_a$ is defined only patchwise, and that's fine. A gauge transform $A_a\to A_a+\partial_a\Lambda$ is also defined patchwise, and setting $\Lambda=\xi^a A_a$ (defined patchwise) still gives a legal gauge transform because $A_a$ and $\partial_a\Lambda$ are both affected the same way by the transition functions that relate the different patches to each other.

$\endgroup$
4
  • $\begingroup$ Thank you very much for your response! I understand that $\Lambda=\xi^\mu A_\mu=A(\xi)$ is a well defined gauge-transform parameter. My problem is however that for the vector fields to be equivalent one needs that $\delta_\xi-\delta_\Lambda=\int\text{d}^Dy\,\frac{\delta S}{\delta A_\mu(x)}\mu_{\mu\nu}(x,y)\frac{\delta}{\delta A_\nu(y)}$ should hold as a vector field equation. $\endgroup$ May 21, 2020 at 21:39
  • $\begingroup$ @IvánMauricioBurbano You're welcome, and sorry if I misunderstood your question. On the right-hand side of the equation in your comment, the integrand depends on $A$ through $\delta S/\delta A_\mu\sim \epsilon^{\mu\nu\rho}\partial_\nu A_\rho$. Similarly, if $\Lambda=\xi^\mu A_\mu$, then the $\delta_\Lambda$ on the left-hand side can be written $\int dx\ \partial_\nu (\xi^\mu A_\mu)\delta/\delta A_\nu$, if I'm not mistaken. If the $A$-dependence is allowed on the right-hand side, then shouldn't it also be allowed on the left-hand side? Or am I still misunderstanding your question? $\endgroup$ May 22, 2020 at 1:09
  • $\begingroup$ I didn't make myself clear. It is fine for vector fields to depend on the field $A$. However the parameters associated to them shouldn't. This is more clearly seen in the case of a non abelian YM or CS theory. In there, the vector field corresponding to a gauge transformation is $\delta_\Lambda=\int\text{d}^Dy(f^a_{bc}A^b_\mu \Lambda^c+\partial_\mu\Lambda^a)\frac{\delta}{\delta A^a_\mu}$. One, wouldn't consider however this vector field a gauge transformation if $\Lambda$ depended on $A$. Well, as I write this now I've found myself thinking, why not? $\endgroup$ May 22, 2020 at 2:27
  • $\begingroup$ I will think about this and get back to you. $\endgroup$ May 22, 2020 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.