Completeness relation of Polarisation tensors I was studying certain topics related to asymptotic symmetries and soft theorems and was mainly looking at "Lectures on Infrared Structure of gauge theories and gravity" By Strominger (https://arxiv.org/abs/1703.05448). I was confused in the section in massless QED (Section 2), which I am trying to explain as follows:
Consider a photon $p^{\mu}$. The momenta can be expressed in terms of its Energy $\omega$, and the direction on the celestial sphere labeled by the stereographic coordinates ($z,\bar{z}$). The four momenta $p^{\mu}$ in these coordinates can be expressed as (this is explained in eqns 2.8.13 and 2.8.14)
\begin{align}
p^{\mu}=\frac{1}{1+z\bar{z}}\Big(1+z\bar{z},z+\bar{z},-i(z-\bar{z}),1-z\bar{z}\Big)
\end{align}
It is mentioned that the two polarisation (positive and negative) can be expressed as
\begin{align}
\epsilon^{\mu}_{+}=\frac{1}{\sqrt{2}}\Big(\bar{z},1,-i,-\bar{z}\Big)\\
\epsilon^{\mu}_{-}=\frac{1}{\sqrt{2}}\Big(z,1,i,-z\Big)
\end{align}
I was able to show that each of the polarisation vectors are orthogonal to $p^{\mu}$, i.e,
\begin{align}
\epsilon_{\pm}\cdot p=0
\end{align}
But I was not sure what is the completeness relation that should be satisfied by these polarisation vectors. As far as I know, if one considers the spatial part of the polarisation vectors they satisfy the relation
\begin{align}
\sum_{\alpha=\pm}\epsilon^{\alpha}_{i}\epsilon^{*\alpha}_{j}=\delta_{ij}-\frac{p_{i}p_{j}}{\vec{p}^{2}}
\end{align}
But when I substitute $i,j$ to any of the spatial indices in the l.h.s and r.h.s of the above equation, they do not seem to match. for e.g if $a,b=1$ then l.h.s gives the answer $1$, but the r.h.s gives $\frac{(z^{2}-1)(\bar{z}^{2}-1)}{(1+z\bar{z})^2}$, which do not seem to match.
I am not understanding where I am going wrong. I suspect that the problem is in the definition of the completeness relation and r.h.s of the completeness relation depends upon a particular gauge in which the polarisation vector is expressed. If that is so, there how do one figure out the correct completeness relation in this coordinates?
 A: Suppose we normalize the polarization tensors so that
$$
\epsilon_\alpha(q) \cdot {\bar \epsilon}_\beta(q) = g_{\alpha\beta}(q)
$$
The completeness relations for the polarization vector always takes the form
$$
\Pi^{\mu\nu}(q) =  g^{\alpha\beta}(q) \epsilon^\mu_\alpha(q) {\bar \epsilon}^\nu_\beta (q) = g^{\mu\nu} + t^\mu(q) q^\nu + t^\nu(q) q^\mu
$$
where $t^\mu(q)$ depends on the gauge choice. The only thing we know it must satisfy is
$$
q \cdot t(q) = - 1 .
$$
For example, consider axial gauge $n \cdot \epsilon_\alpha(q) = 0$. Then $t(q)$ takes the form
$$
t^\mu(q) = \frac{n^2}{2(n\cdot q)^2} q^\mu - \frac{n^\mu}{n \cdot q}
$$
It follows that
$$
\Pi^{\mu\nu}(q) = g^{\mu\nu} + \frac{n^2}{(n\cdot q)^2} q^\mu q^\nu - \frac{1}{n\cdot q}(n^\mu q^\nu + n^\nu q^\mu)
$$
It is easy to check that $q_\mu \Pi^{\mu\nu}(q) = n_\mu \Pi^{\mu\nu}(q) = 0$.

*

*The case you are mentioning in your post is the Coulomb gauge where $n^\mu = (1,0,0,0)$. It follows that $\Pi^{0\mu}(q) = 0$ and
$$
\Pi^{ij}(q) = \delta^{ij} - \frac{q^i q^j}{(q^0)^2}  = \delta^{ij} - \frac{q^i q^j}{|\vec{q}\,|^2} 
$$


*In Strominger's notes, he is using a null gauge $n^\mu = (1,0,0,-1)$ so you can check that the following is true
$$
\Pi^{\mu\nu}(q) = g^{\mu\nu} - \frac{1}{n\cdot q}(n^\mu q^\nu + n^\nu q^\mu)
$$
