The pure $SU\left(2\right)$ theory contains only an $\mathcal{\mathcal{N}}=2$ vector multiplet for the $SU\left(2\right)$ gauge group, with its Lagrangian given by (2.1.11). For reference we reproduce it here:

A supersymmetric vacuum is classically characterized by the solution to the D-term constraint

$$\left[{\Phi}^{\u2020},\Phi \right]=0.$$ | (4.1.2) |

This means that $\Phi $ can be diagonalized by a gauge rotation. Let

$$\Phi =diag\left(a,-a\right).$$ | (4.1.3) |

Roughly speaking, the gauge coupling $\tau $ runs from a very high energy scale down to the energy scale $a$ according to the one-loop renormalization of the $SU\left(2\right)$ theory. Then the vev $a$ breaks the gauge group $SU\left(2\right)$ to $U\left(1\right)$. There are massive excitations charged under the unbroken $U\left(1\right)$, but they will soon decouple, and the coupling remains almost constant below the energy scale $a$. This evolution is shown in Fig. 4.1.

Let us describe it slightly more quantitatively. Our normalization of the $U\left(1\right)$ Lagrangian and the gauge coupling was given in (1.2.7) and (1.2.15), which we reproduce here:

In the broken vacuum, the low-energy $U\left(1\right)$ and the high-energy $SU\left(2\right)$ are related as in (1.3.3), which we also reproduce here

$${F}_{\mu \nu}^{SU\left(2\right)}=diag\left({F}_{\mu \nu}^{U\left(1\right)},-{F}_{\mu \nu}^{U\left(1\right)}\right).$$ | (4.1.5) |

Plugging this in to the high-energy Lagrangian (4.1.1) and comparing the deﬁnitions of $\tau $s, we ﬁnd

$${\tau}^{U\left(1\right)}=2{\tau}^{SU\left(2\right)}.$$ | (4.1.6) |

This relation gets modiﬁed by the quantum corrections.

Let us denote by $\tau \left(a\right)$ the low-energy coupling of the $U\left(1\right)$ gauge ﬁeld when the vev is given by (4.1.3), and by ${\tau}_{UV}$ the high-energy coupling of the $SU\left(2\right)$ gauge ﬁeld at the high-energy renormalization point ${\Lambda}_{UV}$. The one-loop running (3.1.6) then gives

$$\begin{array}{lll}\hfill \tau \left(a\right)& =2{\tau}_{UV}-\frac{8}{2\pi i}log\frac{a}{{\Lambda}_{UV}}+\cdots \phantom{\rule{2em}{0ex}}& \hfill \text{(4.1.7)}\\ \hfill & =-\frac{8}{2\pi i}log\frac{a}{\Lambda}+\cdots \phantom{\rule{2em}{0ex}}& \hfill \text{(4.1.8)}\end{array}$$where we deﬁned

$${\Lambda}^{4}={\Lambda}_{UV}^{4}{e}^{2\pi i{\tau}_{UV}}.$$ | (4.1.9) |

The dual variable ${a}_{D}$ can be obtained by integrating (4.1.8) once, and we ﬁnd

$${a}_{D}=-\frac{8a}{2\pi i}log\frac{a}{\Lambda}+\cdots \phantom{\rule{0.3em}{0ex}}.$$ | (4.1.10) |

As long as we keep $\left|a\right|\gg \left|\Lambda \right|$, the coupling $\tau \left(a\right)$ remains weak, and the computation above gives a reliable approximation.

A gauge-invariant way to label the supersymmetric vacua is to use

$$u=\frac{1}{2}\u27e8tr\phantom{\rule{0.3em}{0ex}}{\varphi}^{2}\u27e9={a}^{2}+\cdots $$ | (4.1.11) |

where $\cdots \phantom{\rule{0.3em}{0ex}}$ are quantum corrections. Let us consider adiabatically rotating the phase of $u$ by $2\pi $:

$$u={e}^{i\mathit{\theta}}\left|u\right|,\phantom{\rule{2em}{0ex}}\mathit{\theta}=0\sim 2\pi $$ | (4.1.12) |

We have $a\mapsto -a$. From the explicit form of ${a}_{D}$ we ﬁnd ${a}_{D}\to -{a}_{D}+4a$. We denote it as

$$\left(a,{a}_{D}\right)\to \left(a,{a}_{D}\right)\left(\begin{array}{cc}\hfill -1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right).$$ | (4.1.13) |

The mass formula of BPS particles is

$$M=|na+m{a}_{D}|=\left|\left(a,{a}_{D}\right)\left(\begin{array}{c}\hfill n\hfill \\ \hfill m\hfill \end{array}\right)\right|.$$ | (4.1.14) |

Therefore, the transformation (4.1.16) can also be ascribed to the transformation of the charges:

We call this matrix

$${M}_{\infty}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$$ | (4.1.16) |

the monodromy at inﬁnity. The situation is schematically shown in Fig 4.2. The space of the supersymmetric vacua, parametrized by $u$, is often called the $u$-plane.

In our argument, the matrix (4.1.13) could have had non-integral entries, as we read the matrix elements oﬀ from an approximate formula of $a$ and ${a}_{D}$. However, the transformation (4.1.15) should necessarily map integral vectors to integral vectors, which guarantees that the matrix (4.1.15) is integral. Not only that, this transformation is just a relabeling of the charges and should not change the Dirac pairing

$$n{m}^{\prime}-m{n}^{\prime}=det\left(\begin{array}{cc}\hfill n\hfill & \hfill {n}^{\prime}\hfill \\ \hfill m\hfill & \hfill {m}^{\prime}\hfill \end{array}\right)$$ | (4.1.17) |

which measure the angular momentum carried in the space when we have two particles with charges $\left(n,m\right)$ and $\left({n}^{\prime},{m}^{\prime}\right)$, respectively. A transformation given by

$$\left(\begin{array}{c}\hfill n\hfill \\ \hfill m\hfill \end{array}\right)\to M\left(\begin{array}{c}\hfill n\hfill \\ \hfill m\hfill \end{array}\right)$$ | (4.1.18) |

aﬀects the Dirac pairing as

Therefore, $M$ should necessarily has unit determinant. A $2\times 2$ integral matrix with unit determinant is called an element of $SL\left(2,\mathbb{Z}\right)$. It is reassuring that the matrix (4.1.16) satisﬁes this condition.