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13

Fundamental? Power counting. The SM is renormalizable, that is, without dimensionful couplings. The lagrangian must have dimension 4, and any vertex with 5 legs or more would dictate a coupling of dimension -1 or less. So the gauge couplings g,g', the Yukawa y, etc, are all dimensionless. (You do have Feynman diagrams of higher dimension, like a 4-fermion ...

6

There is no intuition, really. In this particular theory, the first order contribution to the wave strength renormalization just happens to vanish. Why does it vanish? Well, this constant comes from the kinetic term, so it is associated to the momentum of the diagram. And, to first order, the only diagram is the slug It is clear that the diagram is momentum ...

3

The terms $K$ and $W$ are not kinetic and potential terms, rather $K$ is a K"ahler potential and $W$ is a superpotential. Neither term enters the Lagrangian directly, but they are used to construct it. For details see, for example, Cyril Closset's lecture notes on supersymmetry.

2

't Hooft's argument, if I recall correctly, only shows that the higher order parts of the beta function can be made to vanish at one chosen value of the renormalization scale $\mu$. Seiberg's argument shows that these values vanish at all values of $\mu$, but only for the $\mathcal{N}=2$ QFTs.

2

Your statement (*) is valid only for $d > 2$ (you have to take care when dividing an inequality by a non-positive number). Based on the original formula for $[\lambda_n]$, renormalizability holds for all $n$ if $d \le 2$. You are correct that the "simplest" (free) scalar field theory is renormalizable for all $d$.

1

After watching a colloquium by Alexander Zamolodchikov on this topic, I think I can answer this question in broad strokes now. At a basic level, $T\overline{T}$ deformation refers to a flow on the space of 2D QFTs in a direction given by the determinant of the stress-energy tensor $T_{\mu \nu}$. Explicitly, the infinitesimal change in the Lagrangian density ...

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