Is $\phi^4$ theory in 4d conformally invariant at the classial level? I used to believe the three following statements to be true (at the classical level only):


*

*From scale invariance full conformal invariance follows.

*Scale invariance is present if there are no dimensional parameters in the Lagrangian. 

*The energy-momentum tensor for scale or conformally-invariant theory is traceless.
However, when looking at the particular example of the $\phi^4$ theory in 4d I begin to doubt. The Lagrangian is, of course,
$$\mathcal{L}=\frac12(\partial \phi)^2-g \phi^4,\quad S=\int d^4x \mathcal{L}$$
In 4d field $\phi$ is of mass dimension 1 and $g$ is dimensionless. The theory is scale invariant (if under $x'=\lambda x$ field transforms as $\phi'(x')=\lambda^{-1}\phi(x)$), in accordance with statement (2).
However, it seems to me that the theory is not invariant under inversions (I won't bother you with my failed attempts here) and its energy-momentum tensor 
$$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-\delta_{\mu\nu}\left(\frac12(\partial\phi)^2-g\phi^4\right)$$ 
is not traceless $$T^\mu_\mu=(1-d/2)(\partial\phi)^2+dg\phi^4.$$


*

*My question is which of the assertions (1,2,3) are in fact true and how all this works at the example of the $\phi^4$ theory? 


I stress once again that here I'm interested in the classical aspects only.
 A: (1) is not true. Typical counterexamples are Maxwell's theory in dimension $d \neq 4$ or the theory of elasticity in 2 dimensions. See also the other answer.
(3) is not completely true either. The correct statement is that a theory is invariant under scale transformations if
$$
T^\mu_\mu = \partial_\mu K^\mu
$$
for some operator $K^\mu$ (the virial current mentioned in the other answer), and it is invariant under special conformal transformations if
$$
T^\mu_\mu = \partial_\mu \partial_\nu L^{\mu\nu}
$$
for some $L^{\mu\nu}$. This is nicely explained in a paper by Polchinski. Equivalently, if $T^{\mu\nu}$ satisfies the above condition, it can be "improved" by adding a term that does not affect its conservation property but cancel the trace.
Explicitly, in your example the canonical energy-momentum tensor is
$$
T_c^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial^\nu \phi - \eta^{\mu\nu} \mathcal{L}
= \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi + g \eta^{\mu\nu} \phi^4
$$
and the improved energy-momentum tensor
\begin{align}
T^{\mu\nu} &= T_c^{\mu\nu} - \frac{d-2}{4(d-1)} (\partial^\mu \partial^\nu - \eta^{\mu\nu} \square) \phi^2
\\
&= \frac{1}{2(d-1)} \left[ d \partial^\mu \phi \partial^\nu \phi - \eta^{\mu\nu} \partial_\rho \phi \partial^\rho \phi - (d-2) \phi \partial^\mu \partial^\nu \phi + (d-2) \eta^{\mu\nu} \phi \square \phi \right]
+ g \eta^{\mu\nu} \phi^4
\end{align}
Both tensors satisfy the conservation property $\partial_\nu T^{\mu\nu} = \partial_\nu T^{\mu\nu}_c = 0$ when imposing the equation of motion
$$
\square \phi + 4 g \phi^3 = 0
$$
But now you can check that the trace of the improved energy-momentum tensor
$$
T^\mu_\mu = \frac{d-2}{2} \phi \square \phi + d g \phi^4
$$
also vanishes by the equation of motion in $d = 4$.
(note: this is a slightly edited duplicate of my answer to another question)
A: (2) and (3) are true. (1) is not known in general (for unitary theories). 
It is true in 2d (I remember seeing a paper about a similar result for 3d but I can't find it) that scale invariance and unitarity imply conformal invariance. In general dimensions, this is only true when the Virial 
$$V^\mu=\frac{\delta L}{\delta(\partial^\rho \phi)}(\eta^{\mu\rho}\Delta+i S^{\mu\rho})\phi$$
can be written as a divergence
$$V^\mu=\partial_\alpha \sigma^{\alpha\mu}$$
If that is true then you can come up with a modification of the stress-tensor so that
$$T^\mu_\mu=\partial_\mu j^\mu_D$$ where $j_D$ is the dilation current. Then scale invariance implies conformal invariance. See section 4.2.2 of DiFrancesco's book to see the full derivation.
Basically if you can modify the energy-momentum tensor to make it traceless, in a way that you don't spoil it's conservation nor the Ward identity, then the theory will be conformal by (3).
In your example you can compute the Virial and see whether it can be written as the divergence of something else, if you succeed then it is indeed conformal.
Further reading concerning new developments in the subject of scale invariance $\Rightarrow$ conformal invariance:
http://arxiv.org/abs/1309.2921
http://arxiv.org/abs/1402.6322
http://arxiv.org/abs/1505.01152
