# Background Gauge Condition In Moduli Space

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I'm really confused on the background gauge condition for the moduli space of BPS-monopoles:

$$D_i \delta A_i + e [\phi , \delta \phi]=0$$

I can see that this conditions is the same as saying that $\delta A_i$, $\delta \phi$ are orthogonal to small gauge transformations by using the inner product. But I don't understand why this condition ensures that deformations of the field $\{A_i,\phi\}$ (in the temporal gauge) are again in the moduli space.

For instance, equation (2.21) in http://arxiv.org/abs/hep-th/9603086 (note that they have used a different notation than I've used in the above equation).

I was wondering if anybody could explain this to me or give me a source where they explain this?

Best regards,

Hunter

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user Hunter
$\{A, \phi\}$ satisfies the Bogomolny equations. We want $\{A+\delta A, \phi+\delta \phi\}$ to satisfy them too. $\delta$ is infinitesimal, so $\{\delta A, \delta \phi \}$ has to satisfy the linearized Bogomolny equations. The requirement for orthogonality to the small gauge transformations is there because of eq (2.13) - i.e. the moduli space arises as a quotient of the space of fields by these gauge transformations.

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user twistor59
Ok, thanks for your reply. I understand that we start out with the time-independent fields $\{A_i(\vec{x}), \phi(\vec{x})\}$ and consider a time-dependent deformation $\{A_i(\vec{x})+ \delta A_i(x), \phi(\vec{x}) + \delta \phi(x)\}$. This has to satisfy the linearised Bogomolny equation: $\varepsilon_{ijk} \partial_j \delta A_k + \varepsilon_{ijk} [ A_j, \delta A_k] = \partial_i \delta \phi + [ A_i, \delta \phi] + [\delta A_i, \phi]$. But I really don't see how to get equation (2.13) from that. I've tried to derive it from Gauss's constraint, but with no luck.

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user Hunter
For the infinitesimally perturbed solution to satisfy (2.13), the tangent vector shouldn't have a component in the "gauge transformation direction" on the space of fields, hence the orthogonality statement. Integrating $\int (\delta_{\alpha}A_a) (D_a \Lambda) dx=0$ (i.e. the orthogonality condition, where $D_a \Lambda$ is the gauge transformation (2.19)) by parts, you get $D_a(\delta_{\alpha}A_a)=0$ i.e. (2.21).

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user twistor59
I don't really understand your first statement. Basically, you are saying that first we require the orthogonality statement, from which we can derive the background gauge condition (whereas I thought that we first somehow derived the background conditions from which we can show the orthogonality statement). But why exactly do we require the orthogonality statement?

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user Hunter
The way I understand equation (2.13) is that the true configuration space, $\mathcal{C}$, is the space of all finite energy configurations, $\mathcal{A}$, with the small gauge transformations, $\mathcal{G}$, divided out (i.e. the coset space). Therefore, we would like to find out what the group of small gauge transformations is (which are the transformations that tend to the identity at spatial infinity). And somehow we can divide out all the small transformations by imposing orthogonality, but I don't understand why.

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user Hunter
According to (2.13), the moduli space $\mathcal{C}$ is the space of solutions modulo gauge transformations, so if we can find tangent vectors to $\mathcal{A}$ which are orthogonal to the action of $\mathcal{G}$ on $\mathcal{A}$, then we have succeeded in identifying tangent vectors to $\mathcal{C}$. In other words if we're at a point $X$ of $\mathcal{A}$ and make an infinitesimal gauge transformation in $\mathcal{G}$, we move "up" the gauge orbit through $X$, but we're not interested in this type of tangent vectors, so we take the ones orthogonal to them.

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user twistor59
Thanks for your answer, I've never thought about this space in a geometric way. If I understand you correctly, any vector in $\mathcal{A}$ can be decomposed into components that are tangent to $\mathcal{G}$ (which we are not interested in) and components to $\mathcal{C}$. To ensure we only obtain the tangent vectors to $\mathcal{C}$, we take the vectors orthogonal to $\mathcal{G}$, because tangent vectors to $\mathcal{G}$ cannot be orthogonal to each other. Is this correct?

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user Hunter

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The standard way to introduce BPS monopoles is via the Georgi Glashow model on $\mathbb{R}^4$, which is defined by the Lagrangian density $$\mathcal{L}=-\frac{1}{2}Tr(F^{\mu\nu}F_{\mu\nu})+Tr(D^{\mu}\Phi D_{\mu}\Phi)-\frac{1}{2}(Tr((\Phi)^2)-\alpha^2)^2$$(I won't bother defining all the terms, anyone interested enough to read this will be familiar with them). The E.O.M. are $$D_{\mu}F^{\mu\nu}=[\Phi, D^{\nu}\Phi]$$ $$D^{\mu}D_{\mu}\Phi=-\Phi Tr(\Phi^2-\alpha^2)$$Taking the $\nu=0$ component of the first EOM, we get $$D_iD_iA_0-D_i\dot{A_i}=[\Phi, [A_0, \Phi]]+[\Phi, \dot{\Phi}]$$This would be something like the equivalent of the Gauss constraint in Maxwell gauge theory. In the temporal gauge $A_0=0$ this becomes $$D_i\dot{A_i}+[\Phi, \dot{\Phi}]=0 \ \ \ (1)$$Now if we imagine perturbing one minimum energy monopole configuration into a nearby one, the components of the tangent vector to the space of solutions are just the time derivatives of the fields in the perturbation. An example of such a perturbation would be a bodily shift of the monopole from one point to an infinitesimally separated point.

Starting with a solution $(A_i, \Phi)$ of the Bogomolny equations, the author of the reference in the question considers a perturbed solution $(A_i+\delta_{\alpha}A_i, \Phi+\delta_{\alpha}\Phi)$ where $\alpha$ parametrizes the perturbation. Since $\delta$ is small, the perturbation $(\delta_{\alpha}A_i, \delta_{\alpha}\Phi)$ satisfies the linearized Bogomolny equation which would be something like $$\epsilon_{ijk}(D_j\delta_{\alpha}A_k-D_k\delta_{\alpha}A_i ) = D_i\delta_{\alpha}\Phi + [\delta_{\alpha}A_i, \Phi]$$ Now lots of solutions to this equation will arise from perturbations which arise from applying infinitesimal small gauge transformations to the starting point $(A_i, \Phi)$. However physically meaningful solutions must satisfy (1), i.e. $$D_i(\delta_{\alpha}A_i)+[\Phi, \delta_{\alpha}\Phi]=0 \ \ \ (2)$$(Here it is understood that the covariant derivative is evaluated using the connection at the point at which the tangent vector is attached, i.e $(A_i, \Phi)$). (2) is the background gauge condition.

Now, for the purposes of identifying tangent vectors to the moduli space (fields modulo small gauge transformations), we want a way to filter out any infinitesimal perturbation of a solution $(A_i, \Phi)$ which is just given by an infinitesimal gauge transformation. An unwanted perturbation of this type would be of the form $$(D_i\Lambda, [\Lambda, \Phi]) \ \ \ (3)$$ where $\Lambda$ is some arbitrary Lie algebra valued function.

Now we can define an inner product on the space of fields. For a pair of infinitesimal perturbations $(\delta_{\alpha}A_i, \delta_{\alpha}\Phi)$ and $(\delta_{\beta}A_i, \delta_{\beta}\Phi)$ of a field (i.e. tangent vectors at the point representing the field) we can define their inner product $$(\delta_{\alpha}A_i, \delta_{\alpha}\Phi)\cdot(\delta_{\beta}A_i, \delta_{\beta}\Phi)\equiv \int d^3x \{Tr(\delta_{\alpha}A_i\delta_{\beta}A_i)+Tr(\delta_{\alpha}\Phi\delta_{\beta}\Phi)\}$$Suppose then we look for perturbations orthogonal to the unwanted transformations of the form (3) - see the diagram below.$$0=(\delta_{\alpha}A_i, \delta_{\alpha}\Phi)\cdot(D_i\Lambda, [\Phi,\Lambda])$$ $$=\int d^3x \{Tr(\delta_{\alpha}A_iD_i\Lambda)+Tr(\delta_{\alpha}\Phi[\Phi,\Lambda])\}$$Integrating the first term by parts (assuming $\Lambda$ vanishes at spatial infinity), and using the cyclic property of trace on the second, our condition for orthogonality becomes $$\int d^3x \{Tr(D_i(\delta_{\alpha}A_i)\Lambda)+Tr([\Phi,\delta_{\alpha}\Phi]\Lambda)\} =0$$But this is ensured by our background gauge condition (2), so this is indeed the condition for filtering out the unwanted transformations.

This post imported from StackExchange Physics at 2014-03-22 17:11 (UCT), posted by SE-user twistor59
answered Jun 26, 2013 by (2,490 points)

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