1. Introduction
i | (i) Obtain a general ansatz explicitly satisfying the form invariance condition. |
ii | (ii) Insert this ansatz into the Yang–Mills equation. |
2. Form invariance condition
3. General procedure
1. Construct the general ansatz that is Lie algebra valued and invariant under the pair of Lorentz transformations. | |
2. Impose the form invariance condition ( | |
3. Insert the ansatz satisfying the form invariance condition ( | |
4. (optional) Introduce a topological term in the action, which does not affect the theory at quantum level but is proportional to the winding number and makes the nontrivial topology manifest. It also provides boundary conditions for the solutions to the Yang–Mills equation. | |
5. Solve the simplified Yang–Mills equation under certain boundary conditions, both of which are obtained before. Even if in some cases it is difficult to find analytical solutions, this well-defined differential equation system will make it easier to systematically analyze the nontrivial solutions to the Yang–Mills equation. |
4. Flat space with gauge group SU(2)
5. Topological term and boundary condition
• | Vacuum solution: $ \begin{eqnarray}{A}_{\mu ,a}=0.\end{eqnarray}$ |
• | Wu-Yang monopole: $ \begin{eqnarray}{A}_{\mu ,a}=-\displaystyle \frac{{\epsilon }_{\mu {ai}}{x}_{i}}{{r}^{2}}.\end{eqnarray}$ |
• | Pure gauge solution: $ \begin{eqnarray}{A}_{\mu ,a}=-2\displaystyle \frac{{\epsilon }_{\mu {ai}}{x}_{i}}{{r}^{2}}.\end{eqnarray}$ |
• | Vacuum solution: $ \begin{eqnarray}{A}_{\mu ,a}=0.\end{eqnarray}$ |
• | Meron solution: $ \begin{eqnarray}{A}_{\mu ,a}=\displaystyle \frac{1}{{r}^{2}}\,{\eta }_{a\mu \nu }{x}_{\nu }.\end{eqnarray}$ |
• | Instanton solution: $ \begin{eqnarray}{A}_{\mu ,a}=\displaystyle \frac{2}{{r}^{2}+c}\,{\eta }_{a\mu \nu }{x}_{\nu },\end{eqnarray}$ with a positive real constant c. |
• | Anti-instanton solution: $ \begin{eqnarray}{A}_{\mu ,a}=\displaystyle \frac{2c}{{r}^{2}({r}^{2}+c)}\,{\eta }_{a\mu \nu }{x}_{\nu },\end{eqnarray}$ with a positive real constant c. |
• | Pure gauge solution: $ \begin{eqnarray}{A}_{\mu ,a}=\displaystyle \frac{2}{{r}^{2}}\,{\eta }_{a\mu \nu }{x}_{\nu }.\end{eqnarray}$ |
6. 3D curved space with gauge group SU(2)
1. $G={cH},G\ne 0$ and $H\ne 0$; | |
2. $G\ne 0,H=0$; | |
3. $G=0,H\ne 0$; | |
4. $G=0,H=0$. |
• | Vacuum solution: $ \begin{eqnarray}G=0,H=1\quad \Rightarrow \quad {A}_{\mu ,a}=0.\end{eqnarray}$ |
• | Pure gauge solution: $ \begin{eqnarray}G=0,H=-1\quad \Rightarrow \quad {A}_{\mu ,a}=-2\displaystyle \frac{{\epsilon }_{\mu {ai}}{x}_{i}}{{r}^{2}}.\end{eqnarray}$ |
• | Nontrivial solution: $ \begin{eqnarray}G=0,H=0\quad \Rightarrow \quad {A}_{\mu ,a}=-\displaystyle \frac{{\epsilon }_{\mu {ai}}{x}_{i}}{{r}^{2}}.\end{eqnarray}$ |
7. 4D curved space with gauge group SU(2)
• | Vacuum solution: $ \begin{eqnarray}p=0\quad \Rightarrow \quad {A}_{\mu ,a}=0.\end{eqnarray}$ |
• | Pure gauge solution: $ \begin{eqnarray}p=1\quad \Rightarrow \quad {A}_{\mu ,a}=\displaystyle \frac{2}{{r}^{2}}{\eta }_{a\mu \nu }{x}^{\nu }.\end{eqnarray}$ |
• | Nontrival solution: $ \begin{eqnarray}p=\displaystyle \frac{1}{2}\quad \Rightarrow \quad {A}_{\mu ,a}=\displaystyle \frac{1}{{r}^{2}}{\eta }_{a\mu \nu }{x}^{\nu }.\end{eqnarray}$ |
(1) Closed Friedmann universe: this was firstly studied in [11]. The metric reads $ \begin{eqnarray}{\rm{d}}{s}^{2}={R}^{2}(\chi )\left({\rm{d}}{\chi }^{2}-{a}^{2}{\rm{d}}{{\rm{\Omega }}}_{3}^{2}\right),\end{eqnarray}$ which is conformal to the Einstein static universe. The Euclidean version of this spacetime has the metric $ \begin{eqnarray}{\rm{d}}{s}^{2}={\rm{d}}{r}^{2}+{a}^{2}{\rm{d}}{{\rm{\Omega }}}_{3}^{2},\end{eqnarray}$ which is conformal to $ \begin{eqnarray}{\rm{d}}{s}^{2}=\displaystyle \frac{{r}^{2}}{{a}^{2}}{\rm{d}}{r}^{2}+{r}^{2}{\rm{d}}{{\rm{\Omega }}}_{3}^{2},\end{eqnarray}$ and it is a special case of ( $ \begin{eqnarray}p(r)=\displaystyle \frac{1}{1+{\rm{\exp }}\left[\tfrac{2}{a}({r}_{0}\mp r)\right]}.\end{eqnarray}$ | |
(2) Euclidean AdS4: this case was discussed in [12]. We choose the metric for the Euclidean AdS space to be $ \begin{eqnarray}{\rm{d}}{s}^{2}={a}^{2}\left({\rm{d}}{\xi }^{2}+{{\rm{\sinh }}}^{2}\xi \,{\rm{d}}{{\rm{\Omega }}}_{3}^{2}\right),\end{eqnarray}$ which is a special case of ( $ \begin{eqnarray}{\rm{d}}{s}^{2}=\displaystyle \frac{{\rm{d}}{r}^{2}}{1+\tfrac{{r}^{2}}{{a}^{2}}}+{r}^{2}\,{\rm{d}}{{\rm{\Omega }}}_{3}^{2},\end{eqnarray}$ i.e. $h{(r)=(1+{r}^{2}/{a}^{2})}^{-1}$ in this case. Using the Ansatz ( $ \begin{eqnarray}p(r)=\displaystyle \frac{{r}^{2}}{{r}^{2}+{\lambda }^{2}{\left(a+\sqrt{{r}^{2}+{a}^{2}}\right)}^{2}},\end{eqnarray}$ where λ is a free constant. This solution can be interpreted as the instanton solution on AdS4, and it can be generalized to certain modified AdS spaces [13]. | |
(3) Lorentzian dS4: this case was considered recently in [14–16]. The metric for the Lorentzian dS4 space is given by $ \begin{eqnarray}{\rm{d}}{s}^{2}={a}^{2}\left(-{\rm{d}}{\eta }^{2}+{{\rm{\cosh }}}^{2}\eta \,{\rm{d}}{{\rm{\Omega }}}_{3}^{2}\right),\end{eqnarray}$ which is different from the class ( $ \begin{eqnarray}{\rm{d}}{s}^{2}={a}^{2}\left({\rm{d}}{\eta }_{E}^{2}+{{\rm{\cos }}}^{2}{\eta }_{E}\,{\rm{d}}{{\rm{\Omega }}}_{3}^{2}\right).\end{eqnarray}$ Defining $r\equiv a\,{\rm{\cos }}\,{\eta }_{E}$, we can rewrite the metric ( $ \begin{eqnarray}{\rm{d}}{s}^{2}=\displaystyle \frac{{\rm{d}}{r}^{2}}{1-\tfrac{{r}^{2}}{{a}^{2}}}+{r}^{2}\,{\rm{d}}{{\rm{\Omega }}}_{3}^{2},\end{eqnarray}$ i.e. $h{(r)=(1-{r}^{2}/{a}^{2})}^{-1}$ in this case. Using the ansatz ( $ \begin{eqnarray}p(r)=\displaystyle \frac{{r}^{2}}{{r}^{2}+{\lambda }^{2}{\left(a+\sqrt{{a}^{2}-{r}^{2}}\right)}^{2}},\end{eqnarray}$ where $r\in [0,a]$ in order for the solution to be real-valued. There is an alternative way of finding solutions to the Yang–Mills equation on the Lorentzian dS4 ( $ \begin{eqnarray}\begin{array}{rcl}{A}_{\mu }{\rm{d}}{x}^{\mu } & = & p(t)\left({U}^{-1}{\partial }_{\mu }U\right){\rm{d}}{x}^{\mu }=p(t)\left(2{e}_{\mu }^{a}{T}^{a}\right){\rm{d}}{x}^{\mu }\\ & = & 2\,p(t){T}^{a}\,{e}^{a},\end{array}\end{eqnarray}$ where the generators Ta are given by ( $ \begin{eqnarray}p(t)=\displaystyle \frac{1}{2}\left[1+\sqrt{2}\,{\rm{{\rm{sech}} }}\left(\sqrt{2}(t-{t}_{0}\right)\right],\end{eqnarray}$ where t0 is a constant. |