4
$\begingroup$

I am trying to read and understand this paper by Monteiro, Stark-Muchao, and Wikeley about self-dual yang-mills and self-dual gravity.

In the introduction to this paper, they review a way to formulate both theories using a single scalar degree of freedom. They give an explicit derivation for self dual yang mills but not for self dual gravity. I am trying to find an elementary derivation for self dual gravity along the same lines. The paper they cite does not provide a proof in any way I understand.

For self-dual yang-mills, $A_\mu$ is a $\mathfrak{g}$-valued 1-form and the field strength tensor is $$ F_{\mu \nu} = \partial_\mu A_\nu + \partial_\nu A_\mu + [A_\mu, A_\nu] $$ and the self duality equation is $$ F_{\mu \nu} = \frac{i}{2} \epsilon_{\mu \nu \alpha \beta} F^{\alpha \beta}. $$

If one defines a set of coordinates $$ u = \frac{t - z}{\sqrt{2}}, \quad v = \frac{t + z}{\sqrt{2}}, \quad w = \frac{x + i y}{\sqrt{2}}, \quad \bar{w} = \frac{x - i y}{\sqrt{2}} $$ the metric and $\Box$ become $$ ds^2 = 2 (- du dv + dw d \bar{w}), \hspace{2 cm} \Box = 2( -\partial_u \partial_v + \partial_w \partial_{\bar{w}} ). $$ The self-duality equation then reduces to

$$F_{uw} = 0, \qquad F_{u v } = F_{w \bar{w}}, \qquad F_{v \bar{w} } = 0. $$

We can pick a gauge where we can set $A_u = 0$, then the first two of these equations give us $$ F_{u w} = 0 \implies \partial_u A_{w} = 0 \implies A_{w} = 0 $$ $$ F_{uv} = F_{w \bar{w}} \implies \partial_u A_v = \partial_{w} A_{\bar{w}}. $$ The second equation is solved by \begin{equation} A_v = \partial_{w} \Phi, \hspace{1 cm} A_{\bar{w}} = \partial_u \Phi. \end{equation} The only remaining equation is then \begin{equation} 0 = F_{v \bar{w}} = \Box \Phi - [\partial_u \Phi, \partial_w \Phi ]. \end{equation}

This equation gives the equation of motion for self-dual yang-mills in terms of a $\mathfrak{g}$-valued scalar $\Phi$.

The equation for self-dual gravity is $$ R_{\mu \nu \rho \sigma} = \frac{i}{2} \epsilon_{\mu \nu \alpha \beta} R^{\alpha \beta}_{\quad \rho \sigma}. $$

According to the paper, using "steps analogous to" those above for self-dual yang mills show that one can always write a metric in the form $$ ds^2 = -2 du dv + 2 dw d \bar{w} + \partial_w^2 \phi dv^2 + \partial_u^2 \phi d \bar{w}^2 + 2 \partial_u \partial_v \phi $$

where $\phi$ is a single scalar degree of freedom with equation of motion

$$ 0 = \Box \phi - [ \partial_u \phi, \partial_w \phi ]_{ P.B.} $$

where we have defined the Poisson bracket

$$ [ f, g ]_{P.B.} = \frac{\partial f}{\partial u} \frac{\partial g}{\partial w} - \frac{\partial g}{\partial u} \frac{\partial f}{\partial w} . $$

My question is for someone to essentially point me to a place where I can find a derivation of these equations (that one can always write the metric in the above form, and that the equation of motion for $\phi$ is given above) from first principles, where by "first principles" I mean the self duality equation $R_{\mu \nu \rho \sigma} = \frac{i}{2} \epsilon_{\mu \nu \alpha \beta} R^{\alpha \beta}_{\quad \rho \sigma}$.

For some reason I am having a lot of difficulty finding a source which does this and assumes no background knowledge in areas of mathematics I do not understand. I would also accept an answer which gives a "roadmap" of (digestible) sources necessary to understand the derivation of these equations where no step is left out along the way.

$\endgroup$
1
  • $\begingroup$ I can't even see the action (2.8) anywhere in reference 40... $\endgroup$ Commented Nov 30, 2023 at 3:12

1 Answer 1

0
$\begingroup$

One way to find this is the following. Perturb the metric with a graviton: $$ g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} $$

In the coordinates defined in your question, the line element is

\begin{aligned} ds^2 &= -2 \, du \, dv + 2 \, dw \, d\bar{w} \\ &\quad + h_{uu} \, (du)^2 + 2 \, h_{uv} \, du \, dv + 2 \, h_{uw} \, du \, dw + 2 \, h_{u\bar{w}} \, du \, d\bar{w} \\ &\quad + h_{vv} \, (dv)^2 + 2 \, h_{vw} \, dv \, dw + 2 \, h_{v\bar{w}} \, dv \, d\bar{w} \\ &\quad + h_{ww} \, (dw)^2 + 2 \, h_{w\bar{w}} \, dw \, d\bar{w} + h_{\bar{w}\bar{w}} \, (d\bar{w})^2. \end{aligned}

Plugging in the metric to the self-duality condition and picking a metric leads to an equation solved by the following components

$$h_{v v}=\partial_w^2 \phi, \quad h_{\bar{w} \bar{w}}=\partial_u^2 \phi, \quad h_{v \bar{w}}=h_{\bar{w} v}=\partial_w \partial_u \phi .$$ with all others vanishing. This gives the line element:

$$ds^2 = -2 du dv + 2 dw d \bar{w} + \partial_w^2 \phi dv^2 + \partial_u^2 \phi d \bar{w}^2 + 2 \partial_u \partial_v \phi dvd\bar{w}$$ along with the equations you presented.

$\endgroup$

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