The one-loop contribution to a time ordered product of conserved currents In two dimensions one can define for a Lagrangian describing free Dirac fermions with $N$ associated flavours by
$$\mathcal{L}=i\bar{\psi}_i\gamma^\mu \partial_\mu \psi^i $$
and associate vector currents 
$$V^{\mu a}=\bar{\psi}\gamma^\mu M^a\psi $$
where $M^a$ is a $SU(N)$ matrix generator.
How can one conclude that the expression for the one-loop diagram contributing to the time ordered product
$$\Pi^{\mu\nu,ab}=T(V^{\mu a}(x),V^{\nu b}(0) )$$
is 
$$\Pi^{\mu\nu,ab}(\text{loop})=\frac{1}{4\pi^2}\delta^{ab}\frac{1}{x^4}(x^2 g^{\mu\nu}-2x^\mu x^\nu) \quad ?$$
The corresponding diagram is 

(This is material coming from IETP Lectures on Particle Physics and Field Theory volume II by M.Shifman chapter VII Section 3.)
And how do you know that the subheading diagram looks like 

 A: We are interested in computing
$$
\Pi^{\mu\nu,ab}(x) = \langle V^{\mu,a}(x)  V^{\nu,b}(0) \rangle = \langle : {\bar \psi}(x) \gamma^\mu M^a \psi(x) : : {\bar \psi}(0) \gamma^\nu M^b \psi(0) : \rangle
$$
Several structures of this quantity can be directly computed. Firstly, let us study the index structure. Firstly, due to translational invariance we have $\Pi^{\mu\nu,ab}(x) = \Pi^{\nu\mu,ba}(-x)$.
We will first show that $\Pi^{\mu\nu,ab}(x)  \propto Tr(M^a M^b) = \delta^{ab}$. To see this, let us write out the index structure explicitly
$$
\Pi^{\mu\nu,ab} =   \gamma_{ij}^\mu \gamma_{i'j'}^\nu M_{rs}^a  M_{r's'}^b \langle : {\bar \psi}_{i,r}(x) \psi_{j,s}(x) : : {\bar \psi}_{i',r'}(0) \psi_{j',s'}(0) : \rangle
$$
The quantity in the correlation function is proportional to $\delta_{rs'}\delta_{r's}$, since ${\bar \psi}$ contracts with $\psi$. Thus, the full quantity is proportional to $M^a_{rs} M^b_{r's'} \delta_{rs'} \delta_{r's} = Tr(M^a M^b) = \delta^{ab}$. Thus, we have $\Pi^{\mu\nu,ab}(x) = \Pi^{\nu\mu,ab}(-x)$.
Next, this quantity has Lorentz indices $\mu\nu$. The only Lorentz quantities in the game here are $g^{\mu\nu}$ and $x^\mu$, $x^\nu$. The only tensor that can be constructed is
$$
A(x^2) g^{\mu\nu} + B(x^2) x^\mu x^\nu
$$
Next, we can study the quantity under scaling of $x$. Recall that in the $d=2$ free theory, $\psi$ has mass dimension $\frac{1}{2}$. Thus, if $x^\mu \to \lambda x^\mu \implies \psi \to \lambda^{1/2} \psi$. Thus, we must have
$$
\Pi^{\mu\nu,ab}(\lambda x) = \lambda^{-2} \Pi^{\mu\nu,ab}(\lambda x) 
$$
Thus,
$$
A(x^2) \propto \frac{1}{x^2},~~~B(x^2) \propto \frac{1}{x^4}
$$
Putting all this together, we find the structure
$$
\Pi^{\mu\nu,ab}(x) = A \delta^{ab} \frac{1}{x^4} \left( x^2 g^{\mu\nu} + B x^\mu x^\nu \right)
$$
Finally, we can fix $B$, by requiring that since $V^{\mu,a}$ is a conserved current, we must have
$$
\partial_\mu \Pi^{\mu\nu,ab}(x) =0,~~~\text{if}~~~ x \neq 0
$$
A quick computation gives
$$
\partial_\mu \Pi^{\mu\nu,ab}(x) = - A \delta^{ab} \left( B+2 \right) \frac{x^\nu}{(x^2)^2}    \implies B = - 2 
$$
Putting all of this together
$$
\boxed{ \Pi^{\mu\nu,ab}(x) = A \delta^{ab} \frac{1}{x^4} \left( x^2 g^{\mu\nu} - 2 x^\mu x^\nu \right) } 
$$
This is all we can say about this quantity without explicitly computing diagrams. The constant $A$ can be determined only by actually computing the relevant Feynman diagrams. 
