I have been watching the lecture https://youtu.be/lrikIt9MXpQ from the school LACES 2020 by INFN.

The $\mathcal{N}=1$ SYM is investigated, with the action: $$ \mathcal{L} = \frac{1}{32\pi} \mathrm{Im} \left[\tau \int d^2 \theta \ W^{\alpha} W_\alpha\right] $$ And there is claim, that the holomorphy constraints the dependence of coupling constant on the holomorphic scale $\bar{\Lambda}$ to be: $$ \tau(\mu) = \underbrace{\frac{b_1}{2 \pi} \log \frac{\bar{\Lambda}}{\mu}}_{\text{1-loop}} + \underbrace{\sum_n c_n \left(\frac{\bar{\Lambda}}{\mu} \right)^{n b_1}}_{\text{instantons}} $$ Where holomorphic scale $\bar{\Lambda}$ is defined as: $$ \bar{\Lambda} = \Lambda e^{i \theta / b_1} \qquad \Lambda = \mu_0 e^{-8 \pi^2 / b_1 g^2 (\mu)} $$ It claimed (approx 6:10 on the video) that there can be no corrections beyond 1-loop, because instanton dependence on $g^2 (\mu)$ has to be via $\tau$, and the higher loop corrections would come a powers of $g^2 (\mu)$ and they have to be combined with the corresponding power of $\theta$, but these terms cannot be generated, because $\theta$ enters the Lagrangian only as a term, proportional to the derivative of something ($F \wedge \tilde{F} = d(A \wedge d A + \frac{2}{3} A \wedge A \wedge A)$), and does not contribute to the perturbation theory.

I do not understand, how the instanton dependence is connected with the reasoning on the perturbation theory, instanton contributions are in some sense, by definition, non-perturbative phenomena. How does this constrain the perturbative expansion? I guess, that the crucial part in the reasoning should be some holomorphy properties, but I do not see how, they emerge in the present case.

I would be grateful for comments and clarification


1 Answer 1


The point is that the $U(1)_{R}$ current and the conformal current, at the classical level, lives in the same super-multiplet, so an anomaly of one current is related to the other. Indeed at classical level we have

$$ \partial_{m}j^{m} = \sigma^{m}_{\alpha\dot\alpha}\bar S_{m}^{\dot\alpha} = \sigma^{m}_{\alpha\dot\alpha}S_{m}^{\alpha}=T^{m}\,_{m}=0 $$

where $j^{m}$ is the $U(1)_{R}$ current, $(S_{m}^{\alpha},\bar S_{m}^{\dot\alpha})$ is the supersymmetry current and $T_{mn}$ is the energy-momentum tensor. Conformal symmetry means $T^{m}\,_{m}=0$.

Since the anomaly of the $U(1)_{R}$ current is one-loop exact we know how this symmetry is broken: it is broken by instanton effects. This is so because the anomaly is of the form

$$ dj \propto F\wedge F \implies (Q_{2}-Q_{1})=\int_{\partial \Sigma} j = \int_{\Sigma}F\wedge F $$

where $j$ is the $3$-from current, $F$ is the $2$-form field strength, $(Q_{2}-Q_{1})$ is the charge violation and $\Sigma$ is the history of the space-time from time $1$ to time $2$. Since, by definition, instantons are the only thing that couples with $\int _{\Sigma}F\wedge F$, the only thing that couples to the $U(1)_{R}$ anomaly are the instantons.

Once we know how the $U(1)_{R}$ is broken, by supersymmetry you can learn about how the conformal symmetry is broken and this is why instantons are the only non-perturbative objects that shows up. The relation between these two anomalies is not so straightforward so I will not discuss it here.

You can see more about it in Supersymmetry and the Adler-Bardeen theorem - M. T. Grisaru, P. C. West and also in Supersymmetry and Nonperturbative beta Functions - Nathan Seiberg.

(particularly in the first paragraph bellow equation $13$ of the last reference).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.