For example, consider the $\phi^3$ theory in $d=8$, with Lagrangian:


In 8 dimensions, the box diagram now diverges. We would need to add a $\lambda_{4}\phi^{4}$ coupling to our Lagrangian. Then, this box divergence is thought of as the one-loop correction to our 4-point vertex.

Now, what if the $\phi^4$ term was not allowed in our Lagrangian due to some symmetry constraints? Can we argue that we wouldn't see the box divergence in the first place? That the same symmetry forbids the divergence?

This is not terribly realistic, but if you look at 't Hooft and Veltman's 1-loop Gravity paper, we see that we can't write down a counterterm due to Gauss-Bonnet. Could we stop there and immediately say that there will be no 1-loop graviton divergence without even calculating?

Punchline: No allowed counterterm $\Rightarrow$ No relevant divergence?

Edit: Here's a more relevant example: http://www.conferences.itp.phys.ethz.ch/lib/exe/fetch.php?media=qg11:ld.ethz.qg.pdf. Check out slides 6 and 7. It argues that in SUGRA at two loops in 4 dimensions, the only possible counterterm, $R^3$, "cannot be supersymmetrized." He implies that it follows that no divergence is allowed at two loops; however, at three loops, since we can write down a $R^4$ counterterm, a divergence is allowed but not necessarily present.

Edit 2: Signs seem to be pointing me to the BPHZ Theorem. Weinberg (Vol. 1, Chapter 12) says, "...the cancellation of ultraviolet divergences does not really depend on renormalizability; as long as we include every one of the infinite number of interactions allowed by symmetries, the so-called non-renormalizable theories are actually just as renormalizable as renormalizable theories." The disconnect may be coming when I try to relate the BPHZ subtractions method to the counterterm method, which appears to be explained in http://prd.aps.org/abstract/PRD/v25/i2/p392_1.

  • $\begingroup$ Can you be a bit more clear? The argument is very odd. You are trying to prove that a theory is finite (renormalizable) and you are supposing that you cannot write down a counterterm, for some unknown/topological/... reason. Somehow this implies that a generically (very) divergent integral is finite. Unless there is experimental evidence that the theory exists, I have no clue how your argument is supposed to fill the gap. $\endgroup$
    – Vibert
    Commented May 10, 2013 at 19:03
  • 1
    $\begingroup$ @Vibert My aim is a little more modest. Let me rephrase my question: Does a divergence ever arise that cannot be absorbed by a counterterm? I have pure gravity in mind. It's nonrenormalizable. There is a divergence at two loops. There is no divergence at one loop. Also, the one loop counterterm can be eliminated by a field redefinition. It seems that there is a correlation: If a counterterm can't be written down at say one loop, then it looks like there shouldn't be a divergence at one loop. $\endgroup$
    – jdn
    Commented May 10, 2013 at 19:54

1 Answer 1


I am not sure that one will appreciate this answer, but I will try.

I think that the punchline may not always be the case. It seems that you want some symmetry which forbids some counterterm. This means that this counterterm would have some unappreciated behaviour under the symmetry. Then you want the same symmetry to forbid the divergence. Why? Naively, because the correlator that gives rise to this diagram would have a very similar behavior to that of the counterterm. This may work, but only in the case when there is a regularization that respects your symmetry.

For example, in Schwinger model you have only one divergent diagram in the fermion determinant. Gauge and chiral symmetries tell us that every diagram in this determinant vanishes. What happens, however, is that any gauge-invariant regularization breaks the chiral symmetry and produces what is called chiral anomaly. Now, there is a divergence, and I guess (don't remember) that it should be canceled by a symmetry-breaking counterterm.

So, my point is that you have to do some work (e.g. ensure that the symmetry is not broken by some nasty quantum stuff) before trying to make such judgements.


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