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Physical states in Gupta Bleuler quantization

I'm reading Timo Weigand notes for Gupta-Bleuler quantization of free EM field. .

On page 109, Author has made the following statements.

The Gupta-Bleuler condition for physical state is

$|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys} \leftrightarrow p^\mu\zeta_\mu = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.56)$

since $p^2 = 0 $ for massless photon, such $\zeta^\mu$ can be decomposed as

$\zeta^\mu = \zeta^\mu_T + \zeta^\mu_S \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.57)$

with $\zeta_S = c\cdot p, \zeta^2_S = 0$ and $\vec{\zeta}_T\cdot\vec{p} = 0, \zeta_T^2<0$. So $|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys}$ can be written as

$|\vec{p},\zeta\rangle = |\vec{p},\zeta_T\rangle + |\vec{p},\zeta_S\rangle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.58)$

where $|\vec{p},\zeta_T\rangle$ describes 2 transverse DOF of positive norm. and $|\vec{p},\zeta_S\rangle$ describes 1 combined timelike and longitudinal DOF of zero norm.

I have following doubts regarding this :

  1. How is such decomposition of $\zeta^\mu$ made ?
  2. why do $|\vec{p},\zeta_T\rangle$ have positive norm and $|\vec{p},\zeta_S\rangle$ have negative norm ?
  3. Why do the decomposed parts $\zeta^{\mu}_T$ and $\zeta^{\mu}_S$ follow the properties listed above the equation 4.58. ?

Please Help. Thanks.