The curve for $SU\left(2\right)$ theory with ${N}_{f}=3$ ﬂavors was guessed in (6.4.11):

We see that $\lambda $ diverges at $z=0,1,\infty $ independent of $u$, and there are four branch points which move as $u$ changes, see Fig. 8.10. The reason we put tildes above the mass parameters will become clear soon.

Let us check the behavior when $\left|u\right|\gg |{\mu}_{i}{|}^{2},|\Lambda {|}^{2}$. Two branch points are at $z\sim O\left(1\right)$ and another branch point is at $\sqrt{u}\u2215\Lambda $. We now put the $A$-cycle around $\left|z\right|=c$, where $1\ll c\ll \sqrt{u}\u2215\Lambda $. Then we see that the integral is given as before by

$$a=\frac{1}{2\pi i}{\oint}_{A}x\frac{dz}{z}\simeq \sqrt{u}.$$ | (8.5.2) |

The $B$-cycle integral can be approximated by

$${a}_{D}\sim \frac{2}{2\pi i}{\int}_{\sqrt{u}\u2215\Lambda}^{1}a\frac{dz}{z}\simeq -\frac{2}{2\pi i}alog\frac{a}{\Lambda}.$$ | (8.5.3) |

From this we ﬁnd

$$\tau \left(a\right)=\frac{\partial {a}_{D}}{\partial a}=-\frac{2}{2\pi i}alog\frac{a}{\Lambda},$$ | (8.5.4) |

thus reproducing the ﬁeld-theoretical one-loop computation (8.1.6).

When ${\mu}_{1}={\mu}_{2}={\mu}_{3}=\mu $ and $\left|\mu \right|\gg \left|\Lambda \right|$, the coupling at the scale $\mu $ is still small, and the classical analysis using the superpotential (8.4.1) is almost valid. We expect that around $a\simeq \mu $, i.e. when $u\simeq {\mu}^{2}$, the gauge group $SU\left(2\right)$ is broken to $U\left(1\right)$ with three charge-1 hypermultiplets. This point on the $u$-plane counts as three singularities, since when ${\mu}_{1,2,3}$ are slightly diﬀerent, they should be at three slightly diﬀerent points $u\simeq {\mu}_{i}^{2}$. When $\left|u\right|\ll |\mu {|}^{2}$, the theory can be eﬀectively described by pure $SU\left(2\right)$ gauge theory, which have the monopole point and the dyon point. In total we expect ﬁve singularities on the $u$-plane, see Fig. 8.11.

We would like to study the massless case, $\mu =0$.
Here, we cannot just set ${\stackrel{\u0303}{\mu}}_{i}=0$
in the curve (8.5.1).^{11} We
already saw that, when ${N}_{f}=2$,
the vev $u$ can mix with
the one-instanton factor ${\Lambda}^{2}$.
Here, with ${N}_{f}=3$, the
one-instanton factor is $\Lambda $
and it can mix with any neutral chiral dimension-1 operator. The curve makes only
$U\left(3\right)$
ﬂavor symmetry manifest. The mass parameter corresponding to the
$U\left(1\right)$ ﬂavor
symmetry is neutral, chiral, and of dimension 1. Therefore there can be a mixing of the form

$${\stackrel{\u0303}{\mu}}_{i}={\mu}_{i}+c\Lambda $$ | (8.5.5) |

where $c$ is a constant. Here, we ﬁx the untilded mass parameter ${\mu}_{i}$ to transform linearly under the Weyl symmetry ${\mu}_{i}\to \pm {\mu}_{i}$ of the $SO\left(2{N}_{f}\right)=SO\left(6\right)$ ﬂavor symmetry.

To determine $c$, we set

$$\left({\stackrel{\u0303}{\mu}}_{1},{\stackrel{\u0303}{\mu}}_{2},{\stackrel{\u0303}{\mu}}_{3}\right)=\left(-\mu +c\Lambda ,\mu +c\Lambda ,\mu +c\Lambda \right)$$ | (8.5.6) |

and study the singularities in the $u$-plane. This is just the $SO\left(6\right)$ ﬂavor Weyl transform of the $SU\left(3\right)$ ﬂavor symmetric choice of masses, therefore three out of ﬁve singularities on the $u$-plane should still collide as in (8.11). By an explicit computation, one ﬁnds that this happens only when $c=1$.

Finally we can set $\mu =0$. The curve is now

$$\frac{{\left(x-\Lambda \right)}^{2}}{z}+2\Lambda \left(x-\Lambda \right)z={x}^{2}-u.$$ | (8.5.7) |

There is an $u$-independent branch point of $x\left(z\right)$ at $z=1$. Two other branch points move with $z$, and are at the solutions of

$${\Lambda}^{2}{z}^{2}-{\Lambda}^{2}z+u-{\Lambda}^{2}=0.$$ | (8.5.8) |

The branch points collide when $u={\Lambda}^{2}$ or $u=\left(5\u22154\right){\Lambda}^{2}$:

- When $u=\left(5\u22154\right){\Lambda}^{2}$, two $u$-dependent branch points meet at $z=1\u22152$. The local physics there is just $U\left(1\right)$ gauge theory with one charged hypermultiplet.
- When $u={\Lambda}^{2}$, one branch point moves to $z=0$ and the other branch point collides with the $u$-independent branch point at $z=1$. From the general analysis we know that there are ﬁve singularities on the $u$-plane, therefore this point should count as four colliding singularities, see Fig. 8.12.

The monodromy at inﬁnity is

$${M}_{\infty}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right).$$ | (8.5.9) |

Denoting the monodromies around $u={\Lambda}^{2}$, $u=\left(5\u22154\right){\Lambda}^{2}$ by ${M}_{+}$ and ${M}_{-}$, we have

By going to the S-dual frame at $u={\Lambda}^{2}$, we ﬁnd that the running of the dual coupling is

$$\tau \left(E\right)=+\frac{4}{2\pi i}logE$$ | (8.5.11) |

where the scale is set by $E\sim \left(u-{\Lambda}^{2}\right)$. Comparing with (8.3.13), the low energy physics can be guessed to be a $U\left(1\right)$ gauge theory, coupled either (i) to just one charge-2 hypermultiplets or (ii) to four charge-1 hypermultiplets.

Recall that the classical theory has a Higgs branch. The choice (i) does not have a Higgs branch at $u={\Lambda}^{2}$. It does not have one at $u=\left(5\u22154\right){\Lambda}^{2}$ either. The Higgs branch should be preserved by the quantum correction, and thus this choice is ruled out.

The choice (ii) does have a Higgs branch at $u={\Lambda}^{2}$. We have four charge-1 hypermultiplets coupled to the $U\left(1\right)$ gauge multiplet. Then the complex dimension of the Higgs branch is $2\cdot 4-2\cdot 1=6$. This is acted on by the $SU\left(4\right)$ ﬂavor symmetry rotating four hypermultiplets.

Classically, we have three hypermultiplets in the doublet of $SU\left(2\right)$. Then the complex dimension of the Higgs branch is $4\cdot 3-2\cdot 3=6$. This agrees with the computation above. Recall that three hypermultiplets in the doublet of $SU\left(2\right)$ count as six half-hypermultiplets of $SU\left(2\right)$ doublet, with $SO\left(6\right)$ ﬂavor symmetry. As $SO\left(6\right)\simeq SU\left(4\right)$, we see that the symmetry of the Higgs branch also agrees. We should recall that the monopole in this theory transforms as the spinor representation of the $SO\left(2{N}_{f}\right)$ ﬂavor symmetry. In our case the spinor of $SO\left(6\right)$ is the fundamental four-dimensional representation of $SU\left(4\right)$. This is also consistent with our choice that at $u={\Lambda}^{2}$ there are four charged hypermultiplets electrically coupled to the dual $U\left(1\right)$. By a more detailed analysis we can check that the Higgs branches agree as hyperkähler manifolds.