Propagator of four-dimensional Chern-Simons theory

Question

In https://arxiv.org/abs/1903.03601, on page 13, the propagator of 4d Chern-Simons theory is computed, in the gauge $D^iA_i=0$, where $D^i = (\partial_x,\partial_y,4\partial_z)$.

The gauge-fixed action is $$\begin{equation} S + S_{\text{gf}} = \frac{1}{2\pi}\int_{M}d^4w\, \Bigg(\varepsilon^{ijk}\textrm{Tr}\bigg(A_i\partial_jA_k + \frac{2}{3}A_iA_jA_k\bigg) -\xi^{-1}\textrm{Tr}\Big((D^iA_i)^2\Big)\Bigg)\,, \end{equation}\tag{3.1}$$ where $\xi$ is a gauge-fixing parameter. The propagator $\Delta_{ij}(w,w')$ on $M$ satisfies $$\begin{equation} \frac{1}{\pi}(\varepsilon^{ijk}\partial_j + \xi^{-1}D^iD^k)\Delta_{k\ell}(w,w') = c\,\delta^i_{\,\ell}\,\delta^{(4)}(w-w')\,, \end{equation}\tag{3.2}$$ where $c=t_a\otimes t_b\,(\kappa^{-1})^{ab} \, = t_a\otimes t^a \in \mathfrak{g}^{\otimes 2}$, where the Killing form is $\kappa_{ab}=\textrm{Tr}(t_at_b)$. In Landau gauge ($\xi=0$), the propagator is claimed to be $$\begin{equation} \label{PropagatoronM} \Delta_{ij}(w,w') = -\frac{c}{4\pi}\varepsilon_{ijk}D^k\bigg(\frac{1}{\|w-w'\|^2}\bigg) = -\frac{c}{4\pi}\varepsilon_{ijk}D^k\bigg(\frac{1}{(x-x')^2+(y-y')^2+(z-z')(\bar{z}-\bar{z}')}\bigg) \,. \end{equation}\tag{3.3}$$

Away from $w=w'$, this is obvious from the antisymmetry of the Levi-Civita tensor. However, it is not clear to me why this is the case for $w=w'$. In fact, by computing $D^k\Delta_{kl}$ explicitly, I obtain zero. What am I missing?

Answer 1

1. It is enough to consider the quadratic part of abelian Chern-Simons (CS) Lagrangian density in $R_{\xi}$-gauge $${\cal L}_2 ~=~ \frac{1}{2}\varepsilon^{ijk}A_i\partial_j A_k -\frac{1}{2\xi}(D^iA_i)^2. \tag{A}$$ The propagator $$ \Delta_{k\ell}(w)~=~\underbrace{\Delta^{(0)}_{k\ell}(w)}_{\text{Landau gauge}} +\xi \Delta^{(1)}_{k\ell}(w)\tag{B}$$ should satisfy $$\begin{align} \varepsilon^{ijk}&\partial_j\Delta^{(0)}_{k\ell}(w) + D^iD^k\Delta^{(1)}_{k\ell}(w)\cr ~=~&\left(\varepsilon^{ijk}\partial_j + \xi^{-1}D^iD^k\right)\Delta_{k\ell}(w)\cr ~=~&\left\{\begin{array}{rcl} 4\pi\delta^i_{\ell}\delta^3(w)&{\rm for}&d=3, \cr 4\pi^2\delta^i_{\ell}\delta^4(w)&{\rm for}&d=4. \end{array}\right. \end{align}\tag{C}$$

2. Remark: The CS level and the Lie algebra color factors only contribute by an overall normalization, which we have ignored for simplicity. Also we have taken the liberty to change other normalization conventions. It is straightforward to restore these.

3. Warm-up: The $d=3$ Euclidean propagator: $$\begin{align} \Delta^{(0)}_{k\ell}(w) ~=~&-\varepsilon_{k\ell m} D^m\frac{1}{(|w|^2+\epsilon)^{1/2}}\cr ~=~&\varepsilon_{k\ell m} \frac{w^m}{(|w|^2+\epsilon)^{3/2}},\end{align}\tag{D}$$ $$\begin{align} 2\Delta^{(1)}_{k\ell}(w)~=~&-\partial_k\partial_{\ell}(|w|^2+\epsilon)^{1/2}\cr ~=~&-\partial_k\frac{w_{\ell}}{(|w|^2+\epsilon)^{1/2}}\cr ~=~&-\frac{\delta_{k\ell}}{(|w|^2+\epsilon)^{1/2}}+\frac{w_kw_{\ell}}{(|w|^2+\epsilon)^{3/2}}. \end{align}\tag{E}$$ Here the regularization $\epsilon>0$ is an infinitesimal parameter.

4. Sketched proof of the $d=3$ case: $$\begin{align} \varepsilon^{ijk}&\partial_j\Delta^{(0)}_{k\ell}(w)\cr ~=~&\varepsilon^{ijk}\varepsilon_{k\ell m} \partial_j\frac{w^m}{(|w|^2+\epsilon)^{3/2}}\cr ~=~&\left(\delta^i_{\ell}\delta^j_m- (i\leftrightarrow j) \right) \left(\frac{\delta^m_j}{(|w|^2+\epsilon)^{3/2}} -\frac{3 w^m w_j}{(|w|^2+\epsilon)^{5/2}}\right)\cr ~=~&\frac{3\epsilon\delta^i_{\ell}}{(|w|^2+\epsilon)^{5/2}} -\frac{\delta^i_{\ell}}{(|w|^2+\epsilon)^{3/2}} +\frac{3 w^i w_{\ell}}{(|w|^2+\epsilon)^{5/2}}\cr ~=~&\underbrace{\frac{3\epsilon\delta^i_{\ell}}{(|w|^2+\epsilon)^{5/2}}}_{=~4\pi\delta^i_{\ell} \delta^3(w)} -D^i\frac{w_{\ell}}{(|w|^2+\epsilon)^{3/2}} , \end{align}\tag{F}$$ $$\begin{align} D^iD^k&\Delta^{(1)}_{k\ell}(w)\cr ~=~&D^i\left(\frac{5w_{\ell}}{2(|w|^2+\epsilon)^{3/2}}-\frac{3|w|^2w_{\ell}}{2(|w|^2+\epsilon)^{5/2}}\right)\cr ~=~&D^i\left(\frac{w_{\ell}}{(|w|^2+\epsilon)^{3/2}}+\underbrace{\frac{3\epsilon w_{\ell}}{2(|w|^2+\epsilon)^{5/2}}}_{=~0}\right).\end{align}\tag{G}$$ Here we have represented$^1$ the 3D Dirac delta distribution as a generalized function. $\Box$

5. OP's question: The $d=4$ Euclidean propagator: $$\begin{align} \Delta^{(0)}_{k\ell}(w) ~=~&-\varepsilon_{k\ell m} D^m\frac{1}{|w|^2+\epsilon}\cr ~=~&\varepsilon_{k\ell m} \frac{2W^m}{(|w|^2+\epsilon)^2},\end{align}\tag{H}$$ $$\begin{align} 4\Delta^{(1)}_{k\ell}(w) ~=~& -\partial_k\partial_{\ell}\ln(|w|^2+\epsilon) \cr ~=~& -\frac{2\partial_k w_{\ell}}{|w|^2+\epsilon} +\frac{4w_kw_{\ell}}{(|w|^2+\epsilon)^2}.\end{align}\tag{I}$$ Note the unusual definitions $$\begin{align} \varepsilon^{xy\bar{z}}~=~&1~=~\varepsilon_{xy\bar{z}},\cr D^m~:=~&(\partial_x,\partial_y,4\partial_z), \cr \partial_j~:=~&(\partial_x,\partial_y,\partial_{\bar{z}}),\cr \Box_w~:=~&\partial_jD^j~=~\partial_x^2+\partial_y^2+4\partial_z\partial_{\bar{z}}, \cr W^m~:=~&\frac{1}{2}D^m|w|^2 ~=~(x,y,2\bar{z}), \cr w_j~:=~&\frac{1}{2}\partial_j|w|^2 ~=~(x,y,\frac{z}{2}),\cr |w|^2~:=~&x^2+y^2+z\bar{z}. \end{align}\tag{J}$$ The unusual factor of 4 ensures that the linearized EOMs $$F_{xy}~\approx~0 , \qquad F_{x\bar{z}}~\approx~0 , \qquad F_{y\bar{z}}~\approx~0,\tag{K}$$ become that the gauge fields $A_x$, $A_y$ & $A_{\bar{z}}$ each satisfy the wave equation with the Laplacian $\Box_w$, cf. Ref. 2.

6. Sketched proof of the $d=4$ case: $$\begin{align} \varepsilon^{ijk}&\partial_j\Delta^{(0)}_{k\ell}(w)\cr ~=~&\varepsilon^{ijk}\varepsilon_{k\ell m} \partial_j\frac{2W^m}{(|w|^2+\epsilon)^2}\cr ~=~&\left(\delta^i_{\ell}\delta^j_m- (i\leftrightarrow j) \right) \left(\frac{2\partial_jW^m}{(|w|^2+\epsilon)^2} -\frac{8 W^m w_j}{(|w|^2+\epsilon)^{3}}\right)\cr ~=~&\frac{8\epsilon\delta^i_{\ell}}{(|w|^2+\epsilon)^3} -\frac{2\partial_{\ell}W^i}{(|w|^2+\epsilon)^2} +\frac{8 W^i w_{\ell}}{(|w|^2+\epsilon)^3}\cr ~=~&\underbrace{\frac{8\epsilon\delta^i_{\ell}}{(|w|^2+\epsilon)^3}}_{=~4\pi^2\delta^i_{\ell} \delta^4(w)} -D^i\frac{2w_{\ell}}{(|w|^2+\epsilon)^2}, \end{align}\tag{L}$$ $$\begin{align} D^iD^k&\Delta^{(1)}_{k\ell}(w)\cr ~=~&D^i\left(\frac{W^k\partial_k w_{\ell}+w_kD^kw_{\ell}}{(|w|^2+\epsilon)^2}\right.\cr &\left.+\frac{4w_{\ell}}{(|w|^2+\epsilon)^2} -\frac{4|w|^2w_{\ell}}{(|w|^2+\epsilon)^3} \right)\cr ~=~&D^i\left(\frac{2w_{\ell}}{(|w|^2+\epsilon)^2}+\underbrace{\frac{4\epsilon w_{\ell}}{(|w|^2+\epsilon)^3}}_{=~0}\right).\end{align}\tag{M}$$ $\Box$

References:

1. R. Bittleston & D. Skinner, arXiv:1903.03601; eqs. (3.1)-(3.3).

2. K. Costello, E. Witten & M. Yamazaki, arXiv:1709.09993; eq. (4.4).

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$^1$ See e.g. this Math.SE post for a 1D example. Such representations generalize to any dimension. E.g. the 3D representation is proven in my Phys.SE answer here.