I'm trying to understand how the near boundary expansion of a field in AdS$_{d+1}$ is related to the conformal dimension of the corresponding operator in the dual CFT$_d$. I use coordinates in which the boundary is at $r\to\infty$.
What I understand: scalar field case
A scalar field near the AdS boundary behaves as \begin{equation} \phi(r\to\infty)\sim \frac{A}{r^{\Delta_+}}+\frac{B}{r^{\Delta_-}} \end{equation} where $\Delta_{\pm}=\frac{d}{2}\pm\sqrt{\frac{d^2}{4}+m^2L^2}$ are the solutions of $$ m^2L^2=\Delta(\Delta-d) $$and $\Delta_+$ is found to be the conformal dimension of the dual operator. Do you agree so far?
What I do NOT understand: vector field case
For a vector field we have [ref: Zaanen's book pag 16] $$ \Delta^{vec}_\pm=\frac{d}{2}\pm\sqrt{\frac{(d-2)^2}{4}+m^2L^2}, $$ solutions of $m^2L^2=(\Delta^{vec}-1)(\Delta^{vec}-(d-1))$. Now, since there are again two solutions to the algebraic equation above, I would expect the vector field near the boundary to behave like $$ A_\mu(r\to\infty)\sim \frac{a_\mu}{r^{\Delta^{vec}_+}}+\frac{b_\mu}{r^{\Delta^{vec}_-}} $$ Why does this not happen? How are the two solutions $\Delta^{vec}_\pm$ related to the conformal dimension of the boundary vector operator?
Example
As an example, I propose you a massless vector field. In this case the two solutions of the algebraic equation are $$ \Delta^{vec}_+=d-1\;\;\;\;, \;\;\; \Delta^{vec}_-=1 $$ and so I would expect the vector field to behave as something like $$A_\mu\sim\frac{a_\mu}{r^{d-1}}+\frac{b_\mu}{r^{1}}$$ However, for an AdS-RN black hole the $A_t$ field is $$ A_t=\mu\left(1-\frac{r_h^{d-2}}{r^{d-2}}\right)\:. $$
Can someone explain this? Can this be related to gauge symmetry? Of course, there is no reason why my expectation should be satisfied, but then I don't understand the point of calling out the two solutions $\Delta^{vec}_\pm$.
$\small{\text{Notice: in the reference I cited, they use $d+1$ instead of $d$, because they work with AdS}_{d+2}\text{ and CFT}_{d+1}.}$
