8.5 $N_f=3$

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 $\sqrt{u}∕\Lambda$. We now put the $A$-cycle around $\left|z\right|=c$, where $1\ll c\ll \sqrt{u}∕\Lambda$. Then we see that the integral is given as before by

The $B$-cycle integral can be approximated by

From this we find

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 ${\tilde{\mu }}_i=0$ 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

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 $SO\left(2N_f\right)=SO\left(6\right)$ flavor symmetry.

To determine $c$, we set

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

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

The branch points collide when $u={\Lambda }^2$ or $u=\left(5∕4\right){\Lambda }^2$:

• When $u=\left(5∕4\right){\Lambda }^2$, two $u$-dependent branch points meet at $z=1∕2$. 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

Denoting the monodromies around $u={\Lambda }^2$, $u=\left(5∕4\right){\Lambda }^2$ 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

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 $u=\left(5∕4\right){\Lambda }^2$ 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 $SO\left(2N_f\right)$ 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.