$\phi^4$ theory is not perturbatively renormalizable in 5 dimensions.
I have come across literature where renormalizability is discussed w.r.t $N$, for fields obeying $O(N)$ symmetry. But it is not clear to me if renormalizability w.r.t $N$ implies renormalizability overall?
If I think of it naively, one can expand in terms of both the coupling constant $\lambda$, and the parameter, $N$.
So, even if one proves renormalizability w.r.t $N$, the final observables may still depend on the UV cutoff for some powers of $\lambda$. Alternatively, one has to assume that the theory is renormalizable for all powers of $\lambda$ and $N^1$. And then show the theory is renormalizable for all values of $N$. Is the initial assumption valid?
Does my argument(s) even make sense?
Best Answer
Indeed, when we say a theory is renormalizable, it is tacitly assumed that there is only one perturbative expansion of the theory that we know how to do. When this is not the case, renormalizability is really a property of the parameter we are perturbing in. To say that the $\frac{1}{N}$ expansion is renormalizable means that a finite number of counter-terms added to the action will be enough to absorb all divergences. The $\lambda$ expansion being non-renormalizable, on the other hand, means that as we go to arbitrarily high powers of $\lambda$, the number of counter-terms one must add will grow without bound.
Aside
The trick that makes it easy to see why the two approaches can be so different is called the Hubbard-Stratonovich transformation. This introduces an auxiliary field in order to get a more explicit dependence on $N$. Namely a $\frac{1}{\sqrt{N}} \sigma \phi_i \phi_i$ vertex. Once you use this to build Feynman diagrams, it is no longer necessarily true that loop diagrams are higher order than tree diagrams. A $\phi$ loop in the $\sigma$ propagator, for instance, will involve $\delta^i_j \delta^j_i = \delta^i_i = N$ which cancels the extra two vertices and gives a diagram which is just as important as the one without the $\phi$ loop.