# How to derive the noether current for q-form symmetries

+ 1 like - 0 dislike
201 views

Hi,

I want to explore generalised Noether currents obtained from q-form symmetries in an action.

The regular way we obtain Noether currents is fairly straightforward: We have a 0-form symmetry $\phi \to \phi + \delta \phi$, and the action $S(\phi, \partial_\mu \phi)\to S(\phi + \delta \phi, \partial_\mu \phi + \partial_\mu \delta \phi) \approx S(\phi, \partial_\mu \phi) + \partial_\mu j^\mu$ where $j^\mu$ is the Noether current.

I want to perform this calculation for q-form symmetries, or even just 1-form symmetries. It should result in a d-q-1 form (Or equivalently q+1 form) current. The key steps in my opinion are to understand how to write down the Euler Lagrange equation that results when you vary an action with respect to forms rather than fields.

So it would be something like: $\psi \to \psi + \omega$ where $\psi, \omega \in \Omega^1(M^d)$, and then we should be able to write down $S\to S + \delta S$ where $\delta S$ is a total derivative of a q form?

Any further insight would be very appreciated!

Are you anticipating that the solution is different from the usual derivation? If I understood the question correctly I don't see where or why there would be deviation (e.g. for the Euler-Lagrange equation Yang-Mills theories have 1-forms as the fundamental field), but I would have to carry it out or see a calculation to know for sure.

@Qghost I'm anticipating the difficulty being that I tried following the derivation for a one form ($A_\mu \to A_\mu + e_\mu$), but it yielded a 1 form current following the same steps. I don't know how I would arrive at a two form current, since EL equations only work for the linear approximation, not the higher order terms in the Taylor expansion.

Could you try out an example calculation to see if you face any difficulties in doing this calculation?

+ 1 like - 0 dislike

EDIT: aside from sign errors, I had realized that the notation I used implied something completely different from what I had derived (I don't think the resulting expressions had a definition!). It should be fixed now, as well as slightly less step-by-step, so the reader can understand the process of the derivation.

Let $S=\displaystyle\int_{\Omega} \mathcal{L}$ be the action for a Lagrangian D-form $\mathcal{L}=\mathcal{L}(\omega,\star\omega,d\omega,d\star\omega,\star d\omega, \star d\star\omega)$ where D is the dimenstion of space(time), $\omega$ is a q-form and $\star$ is the Hodge star.

Defining derivatives with respect to a form as $\partial_{\alpha_i} (\alpha_0 \wedge \ldots \wedge \alpha_n) = \alpha_0 \wedge \ldots \wedge \alpha_i \! \! \! \! \! \Big/ \wedge \ldots \wedge \alpha_n$, and denoting the position of the form within the wedge products as $l(\alpha_i)=i$, the variation of the action becomes

$$\frac{d}{d\epsilon}S[\omega+\epsilon\chi]\Big|_{\epsilon=0}$$
$$=\int_\Omega (-1)^{l(\omega)}\chi \wedge \partial_\omega\mathcal{L} + (-1)^{l(\star\omega)}\star\chi \wedge \partial_{\star\omega}\mathcal{L} + (-1)^{l(d\omega)}d\chi \wedge \partial_{d\omega}\mathcal{L}\\ + (-1)^{l(d\star\omega)}d\star\chi \wedge \partial_{d\star\omega}\mathcal{L} + (-1)^{l(\star d\omega)}\star d\chi \wedge \partial_{\star d\omega}\mathcal{L} + (-1)^{l(\star d\star\omega)}\star d\star \chi \wedge \partial_{\star d\star\omega}\mathcal{L}$$

Exterior derivatives can then be "donated" to the Lagrangian from the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^q \alpha \wedge d\beta$, and likewise for the Hodge star operator given the identities $\alpha\wedge\star\beta =\langle\alpha,\beta\rangle\nu=\star\alpha\wedge\beta$ (if the order of alpha and beta sum to D) and $\star \star \alpha = (-1)^{q(D-q)}s\alpha$.

$$\int_\Omega \left(\alpha\partial_\omega\mathcal{L} + \beta\star\partial_{\star\omega}\mathcal{L} + \gamma d\partial_{d\omega}\mathcal{L} + \delta\star d\partial_{d\star\omega}\mathcal{L} + \zeta d\star\partial_{\star d\omega}\mathcal{L} + \xi \star d\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\chi \\ + d\left[\left(\gamma^* \partial_{d\omega}\mathcal{L} + \zeta^*\star\partial_{\star d\omega}\mathcal{L} \right)\wedge\chi +\left(\delta^*\partial_{d\star\omega}\mathcal{L} + \xi^*\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\star\chi\right]$$

To avoid clutter I have written sign prefactors with greek letters. They can be found using the exterior derivative and hodge star's identities.

Since $\chi$ is a symmetry of the action, it follows that $\delta\mathcal{L}[\omega + \epsilon \chi] = \delta\mathcal{L}[\omega] + d\Lambda$ for a (D-1)-form $\Lambda$. On shell, then, requiring $\delta S=0$, the conservation law manifests:
$$dJ=0$$
$$J=\left(\gamma^* \partial_{d\omega}\mathcal{L} + \zeta^*\star\partial_{\star d\omega}\mathcal{L} \right)\wedge\chi +\left(\delta^*\partial_{d\star\omega}\mathcal{L} + \xi^*\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\star\chi-\Lambda$$

answered Oct 10, 2020 by (55 points)
edited Oct 11, 2020

Everything in parentheses should be wedge multiplied by chi or its hodge dual. I think there is a formatting error because I have that in the LaTeX code, but it is not showing..

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.