How Ward Identity indicate vacuum polarization correction? In Peskin & Schroeder Chapter 7.5 Renormalization of The Electric Charge, they mention that vacuum polarization correction is
$$
iM= (-ie)^2(-1)\int_{}{}\frac{d^4k}
{(2\pi)^4}Tr\bigg[\gamma^{\mu}\frac{(\require{cancel}\cancel{p}+\cancel{k}+m)}{(\cancel{k}+\cancel{p})^2-m^2}\gamma^{\nu}\frac{(\cancel{k}+m)]}{k^2-m^2}\bigg]=i\Pi^{\mu\nu}_2(p)
$$ 
and continue


The only tensors that can appear in $\Pi^{\mu\nu}_2(p)$ are $g^{\mu\nu}$ and $p^{\mu}p^{\nu}$. The Ward Identity, however, tells us that $p^{\mu}\Pi^{\mu\nu}_2(p)=0$. This implies that $\Pi^{\mu\nu}_2(p)$ is proportional to the projector $(g^{\mu\nu}-\frac{p^{\mu}{p^{\nu}}}{p^2})$.


I can not see the connection of Ward identity and "the projector", indeed. What do they mean even by projector? 
 A: (i) "The only tensors that can appear in $\Pi_{\mu\nu}$ are $g_{\mu \nu}$ and $p_\mu p_\nu$": this is because these are the only rank-2 tensors available in the problem. $p_\mu$ is a vector on which the problem depends and $g_{\mu \nu}$ is the invariant tensor. Using only this two tensors (one rank-1 and one rank-2), you cannot construct any rank-2 tensors other than the specified ones.
(ii) "$p_{\mu}\Pi^{\mu\nu}_2(p)=0$ implies that $\Pi^{\mu\nu}_2(p)$ is proportional to the projector $(g^{\mu\nu}-\frac{p^{\mu}{p^{\nu}}}{p^2})$": because of (i), we have that $\Pi^{\mu\nu}_2(p)=(Ag^{\mu \nu}+Bp^\mu p^\nu)\Pi(p)$, and it is easy to convince yourself that $p_\mu\Pi^{\mu\nu}_2(p)=p_\mu(Ag^{\mu \nu}+Bp^\mu p^\nu)\Pi(p)=0$ implies that $A=1$ and $B=-1/p^2$ (if you don't absorb any arbitrary global factor into the $\Pi(p)$ as defined above).
(iii) "What do they mean even by projector?" You can write any vector as $p_\mu=\frac{k_\mu k_\nu}{k^2}p^\nu+\left(g_{\mu\nu}-\frac{k_{\mu}{k_{\nu}}}{k^2}\right)p^\nu$, where the first term is the part of $p$ longitudinal to $k$ (you can see this by verifying that if $p_\nu=k_\nu$, we have $\frac{k^{\mu}{k^{\nu}}}{k^2}k_\nu=k^\mu$) and the second term is the part of $p$ transverse to $k$. In this sense, $\left(g^{\mu\nu}-\frac{k^{\mu}{k^{\nu}}}{k^2}\right)$ is the projector to the direction transverse to $k$.
