How can we show the Lorentz symmetry is not anomalous in $\phi^{4}$ theory? how can I show in a lagrangian with scalar fields and $\phi^{4}$ interaction, the energy-momentum tensor isn't anomalous?
 A: In general, when you have a global symmetry group, say $G$ you can detect an anomaly as follows; Promote the symmetry to background gauge. I.e. make the symmetry transformations local, but add a gauge field, say $A$, to absorb them. However, that field is a background gauge field, namely, it does not have any curvature and you do not sum over it in the path integral. Now your partition function is a function of $A$, $Z[A]$. Now do the suggested gauge transformation: $A\mapsto A^\lambda$. If the partition function does not return to itself but picks up a phase, $$Z\!\left[A^\lambda\right] = \mathrm{e}^{\mathrm{i}\,\Phi[A,\lambda]}\;Z[A],$$ then $G$ is anomalous. If it does not pick up a phase, it is not anomalous (modulo some subtle cases, where the anomaly is hidden). This is equivalent to the question of whether the path integral measure is invariant or not. For example, in the ABJ anomaly, you can either go the standard way looking at how the fermionic measure transforms under the two $\mathrm{U}(1)$s or turn on background gauge fields for both of them and do gauge transformations. In both cases, you pick up the same phase.
Now in the example of $\phi^4$ and Lorentz symmetry, the way to go is make Lorentz symmetry into a background gauge symmetry -- essentially couple the theory to non-dynamical gravity, so you have:
$$ Z[g] = \int \mathrm{D}\phi \ \exp\!\left(-\int\mathrm{d}^dx \sqrt{g} \left[\tfrac{1}{2}g_{\mu\nu}\nabla^\mu\phi\nabla^\nu\phi + m^2 \phi^2 + \lambda \phi^4 \right]\right).$$
Obviously doing a gauge transformation, i.e. a local Lorentz transformation does not change the partition function, as every term is manifestly gauge invariant. Hence
$$ Z\!\left[g^\lambda\right] = Z[g], $$
and thus you do not have an anomaly.
By contrast, in the example of (bosonic) string theory (to connect to Andrew's comment on the question), when you take the gauge-fixed Polyakov path integral, couple to a background metric, $h$, you can always go to conformal gauge $h = \mathrm{e}^{2\omega} \hat{h}$, and your background gauge transformations become Weyl transformations. Doing such a gauge transformation, gives $Z\!\left[\omega+\delta\omega\right] = \mathrm{e}^{\Phi[\omega,\delta\omega]} Z[\omega]$, where $\Phi$ is proportional to $c$, the central charge. Therefore there there is an anomaly, unless $c=0$, which in turn implies that $d=d_\text{crit}=26$.
A: The easiest way (particularly for $\phi^4$ theory where you don't have to worry about any subtleties like gauge invariance) is just to use a Lorentz-invariant method to compute and renormalize correlation functions. For example, use the path integral formalism, and use dimensional regularization. Then every step of the calculation is manifestly Lorentz invariant, and in the end you get a finite answer.
