# Conformal vs. scale invariance of ${\cal N} = 4$ Supersymmetric Yang-Mills theory

I will quote the following from the Wikipedia article on Supersymmetry Nonrenormalization theorems.

"In ${\cal N} = 4$ super Yang–Mills the $\beta$-function is zero for all couplings, meaning that the theory is conformal."

When we say that the $\beta$-function vanishes for a QFT, we conclude that scale invariance remains preserved at the quantum level.

However scale invariance doesn't necessarily imply conformal invariance, e.g. this paper has two such examples. Hereby I am getting confused, this paper by Sohnius and West originally shows why the $\beta$- function for ${\cal N} = 4$ super Yang–Mills vanishes. But then why is the claim for conformal invariance justified?

The fact that all $\beta$-functions vanish in a theory is equivalent to the statement that the energy-momentum tensor is traceless by the operator identity $$T^\mu_\mu(x) = \sum_\mathcal{O} \beta_\mathcal{O} \mathcal{O}(x) = 0.$$ This is in turn sufficient to conclude that the theory is conformal: the charge $$K^\mu \equiv \int d\Sigma_\nu \left( 2 x^\mu x_\rho T^{\nu\rho} - x^2 T^{\mu\nu} \right)$$ associated with special conformal transformations is conserved.
In $\mathcal{N} = 4$ super Yang-Mills, all $\beta$-functions vanish, and therefore the charge $K^\mu$ is conserved and the theory is conformal.
In a theory that is scale invariant but not conformal, not all $\beta$-function vanish: there is a particular operator (or a linear combination of operators) that is the divergence of a current $V^\mu$ (called the "virial" current) and whose $\beta$-function is non-zero: $$T^\mu_\mu(x) = \partial_\mu V^\mu(x) \neq 0.$$ You can check that the charge $D$ associated with scale transformations, $$D \equiv \int d\Sigma_\nu x_\mu T^{\mu\nu}$$ is conserved in this case, which means that the theory is scale invariant. However, the charge $K^\mu$ is not conserved: the theory is not conformal.
The other possibility for a theory to be scale invariant but not conformal is to have no energy-momentum tensor. In this case, the construction of the charge $K^\mu$ fails from the start. But we know that $\mathcal{N} = 4$ super Yang-Mills does have an energy-momentum tensor, since we can construct it explicitly from the Lagrangian.