Behavior in the strongly-coupled region

4.2 Behavior in the strongly-coupled region

Let us study what is going on in the strongly coupled region which is the interior of the $u$ -plane. There needs to be at least one singularity in this interior region to realize the monodromy $M_{\infty}$ of holomorphic functions $a$ and $a_{D}$ . So, most naively, we would expect the structure as in Fig. 4.3.

Figure 4.3: Naive guess which does not work

Where will the singularity be? Here the discrete unbroken $U (1)$ R-symmetry of the system is useful. Recall our $𝒩 = 2$ theory has an $S U {(2)}_{R}$ symmetry. Classically, we can also consider a $U {(1)}_{R}$ symmetry with the standard R-charge assignment given as follows:

\begin{matrix} R = 0 & A \\ 1 & λ & λ \\ 2 & Φ \end{matrix} .

(4.2.1)

Diﬀerent components in the same supersymmetry multiplet have diﬀerent charges, and therefore this is an R-symmetry.

Quantum mechanically, the rotation

λ \to e^{i φ} λ,

(4.2.2)

is anomalous, but can be compensated by

𝜃_{U V} \to 𝜃_{U V} + 8 φ,

(4.2.3)

as we learned in Sec. 3.2.1.

Therefore $φ = π ∕ 4$ is a genuine symmetry, which does

𝜃 \to 𝜃 + 2 π, Φ \to e^{π i ∕ 2} Φ .

(4.2.4)

This generates a $ℤ_{4}$ discrete R-symmetry of the system. In the low-energy variables, it acts as

𝜃_{I R} \to 𝜃_{I R} + 4 π, u \to - u .

(4.2.5)

Then, if there is a singularity at $u = u_{0}$ , there should be another at $u = - u_{0} .$ Therefore, if there is only one singularity, it is at $u = 0$ .

If this were really the case, we would ﬁnd that $τ (a)$ is given by

τ (a) = - \frac{8}{2 π i} log \frac{a}{Λ} + f (a) .

(4.2.6)

where $f (a)$ is a meromorphic function whose only singularity is at $a = 0$ . This does not sound right, however. The coupling is given by $I m τ (a)$ , which is the imaginary part of a holomorphic function. Then, it has no lower bound, and therefore it becomes negative for some value of $a$ . This means that the coupling $g^{2}$ is negative there, and the system becomes unstable. For example, supposing $f (a) = 0$ , the imaginary part is negative when $| a |$ is small enough. We conclude that our assumption of having just one singularity at $u = 0$ was too naive.

The next simplest possibility is then to suppose that there are two singularities at $u = \pm u_{0}$ , see Fig. 4.4.

Figure 4.4: Next guess which turns out to be correct

The only scale in the system is the dynamical scale $Λ$ , therefore $u_{0}$ should be given by $c Λ$ where $c$ is a number. Denoting the monodromies around two singularities $M_{\pm}$ , we should have

M_{\infty} = M_{+} M_{-},

(4.2.7)

since the path going around the inﬁnity of the $u$ -plane is topologically the same as the path which ﬁrst goes around $u = - u_{0}$ and then around $u = u_{0}$ . As two singularities are exchanged by a symmetry, the monodromies around them should be essentially the same, except for the relabeling of the charges. Or equivalently, they should be conjugate

M_{-} = X M_{+} X^{- 1}

(4.2.8)

by an $S L (2, ℤ)$ matrix $X$ . Note that this matrix $X$ can be thought of a half-monodromy associated to the symmetry operation (4.2.5), see Fig. 4.5

Figure 4.5: The relation between

M_{+}

and

M_{-}

A solution to these equations is given by

M_{+} = S T S^{- 1} = (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}), M_{-} = T^{2} S T S^{- 1} T^{- 2} = (\begin{matrix} - 1 & 4 \\ - 1 & 3 \end{matrix})

(4.2.9)

where $S$ and $T$ were given in (1.2.19), (1.2.20). Note that we have

X = T^{2},

(4.2.10)

which is roughly compatible with the fact that the discrete R-symmetry (4.2.5) shifts $𝜃_{I R}$ by $4 π$ .

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