We learned in the last section that the curve of $SU\left(N\right)$ theory with $2N$ ﬂavors is given by:

where $f$ is a complex number; the diﬀerential is $\stackrel{\u0303}{\lambda}=\stackrel{\u0303}{x}dz\u2215z$. This theory is superconformal, and $f$ is a function of the UV coupling constant ${\tau}_{UV}$. We would like to understand the strong-coupling limits of this theory.

As we did in Sec. 9, it is convenient to rewrite the curve in terms of the Seiberg-Witten diﬀerential $\lambda $, to the form

$${\lambda}^{N}+{\varphi}_{2}\left(z\right){\lambda}^{N-2}+\cdots +{\varphi}_{N}\left(z\right)=0.$$ | (12.1.2) |

We start from (12.1.1). First we gather terms with the same power of $\stackrel{\u0303}{x}$:

where

$${\u2661}_{1}=\frac{\sum {\stackrel{\u0303}{\mu}}_{i}}{\stackrel{\u0303}{z}}+f\stackrel{\u0303}{z}\sum \underset{i}{\overset{\prime}{\stackrel{\u0303}{\mu}}}.$$ | (12.1.4) |

We divide the whole equation by $\left(1-1\u2215\stackrel{\u0303}{z}-f\stackrel{\u0303}{z}\right)$ and deﬁne $x=\stackrel{\u0303}{x}+{\u2661}_{1}\u2215\left(1-1\u2215\stackrel{\u0303}{z}-f\stackrel{\u0303}{z}\right)\u2215N$. We now have

$${x}^{N}+{\u2663}_{2}{x}^{N-2}+\cdots +{\u2663}_{N}=0$$ | (12.1.5) |

where ${\u2663}_{k}$ has poles of order $k$ at two zeros ${\stackrel{\u0303}{z}}_{1,2}$ of $1-1\u2215\stackrel{\u0303}{z}-f\stackrel{\u0303}{z}=0$, due to the shift from $\stackrel{\u0303}{x}$ to $x$. We set $z=\stackrel{\u0303}{z}\u2215{\stackrel{\u0303}{z}}_{1}$ so that one zero is now at 1, and another is at $q={\stackrel{\u0303}{z}}_{2}\u2215{\stackrel{\u0303}{z}}_{1}$.

Introducing $\lambda =xdz\u2215z$, we have an equation of the form (12.1.2); ${\varphi}_{k}\left(z\right)$ has poles of order at most $k$ at $z=0$, $q$, $1$ and $\infty $. Consider the case when all ${\stackrel{\u0303}{\mu}}_{i}$ and ${\stackrel{\u0303}{\mu}}_{i}^{\prime}$ are generic, and assume $q\ll 1$. Then it is straightforward to determine how $\lambda $ behaves close to each of the singularity. As we are solving a degree-$N$ equation, we have $N$ residues at each singularity. They are given by

Here

$$\begin{array}{lllll}\hfill {\mu}_{i}& ={\stackrel{\u0303}{\mu}}_{i}-\frac{1}{N}\sum _{i}{\stackrel{\u0303}{\mu}}_{i}+O\left(q\right),\phantom{\rule{2em}{0ex}}& \hfill \sum {\mu}_{i}=0;& \phantom{\rule{2em}{0ex}}& \hfill \text{(12.1.7)}\\ \hfill \mu & =\frac{1}{N}\sum _{i}{\stackrel{\u0303}{\mu}}_{i}+O\left(q\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill \text{(12.1.8)}\end{array}$$and similarly for the ${\mu}_{i}^{\prime}$, ${\mu}^{\prime}$. Note that ${\mu}_{i}$ and $\mu $ are the mass parameters which enter the BPS mass formula. We found that they are related to the parameters ${\stackrel{\u0303}{\mu}}_{i}$ via a ﬁnite renormalization.

When $N=2$, the structure of the residues at all four punctures were of the same type, as they are all given by $\left(m,-m\right)$. For $N>2$, we see that the structure of the residues at $z=0,\infty $ and the structure at $z=q,1$ are diﬀerent. The former is of the form $\left({m}_{1},\dots ,{m}_{N}\right)$ with $\sum {m}_{i}=0$, and the latter is of the form $m\left(1,1,\dots ,1-N\right)$.

It is also instructive to consider the completely massless case, when we have ${\stackrel{\u0303}{\mu}}_{i}={\stackrel{\u0303}{\mu}}_{i}^{\prime}=0$ for all $i$. The original curve is just

$$\frac{{x}^{N}}{z}+f{x}^{N}z={x}^{N}+{u}_{2}{x}^{N-2}+\cdots +{u}_{N}.$$ | (12.1.9) |

After the same manipulation as above, we ﬁnd

$${\varphi}_{k}\left(z\right)=\frac{{u}_{k}}{\left(z-q\right)\left(z-1\right)}\frac{d{z}^{k}}{{z}^{k-1}}.$$ | (12.1.10) |

Therefore,

We observe here again that the behavior of the poles are all the same when $N=2$, while the behavior at $z=0,\infty $ and the behavior at $z=1,q$ are distinct when $N>2$.

We have $2N$ mass terms in the system. First of all we split them into $N$ mass terms encoded in the region $z\sim 0$, and another $N$ mass terms in the region $z\sim \infty $. Correspondingly, we started from the ﬂavor symmetry $U\left(2N\right)$ and decomposed it into $U\left(N\right)\times U\left(N\right)$. We further decompose each of $U\left(N\right)$ into $SU\left(N\right)$ and $U\left(1\right)$. Combined, we use the decomposition of the ﬂavor symmetry of the form

The residues of $\lambda $ at the puncture $A$ at $z=0$ and those at the puncture $D$ at $z=\infty $ encode the mass terms for $SU{\left(N\right)}_{A,D}$ respectively, whereas those at the puncture $B$ at $z=q$ and those at the puncture $C$ at $z=1$ encode the mass terms for $U{\left(1\right)}_{B,C}$; compare (12.1.6).

We then say that the singularity at $z=0$ carry the $SU\left(N\right)$ symmetry, the one at $z=q$ carry the $U\left(1\right)$ symmetry, and similarly for those at $z=1$, $=\infty $. We can visualize the situation as in Fig. 12.1. We call the punctures at $z=0,\infty $ the full punctures, and those at $z=q,1$ the simple punctures. In the 6d viewpoint, these are four-dimensional defect objects extending along the Minkowski ${\mathbb{R}}^{3,1}$, and they carry respective ﬂavor symmetries on them.

When $N=2$, the original symmetry is not just $U\left(2\right)$ but $SO\left(4\right)$. Accordingly, the split $U\left(2\right)\simeq SU\left(2\right)\times U\left(1\right)$ is enhanced to the following structure

and therefore the distinction of the types of punctures is gone.

Clearly $f\sim q\sim {e}^{2\pi i{\tau}_{UV}}$ in the weak coupling region, see Fig. 12.2. When the coupling is extremely weak, we can think that the four-punctured sphere on the left is composed of two three-punctured spheres. In the tube region connecting the two, the behavior of $\lambda $ is essentially given just by

$${\varphi}_{k}\left(z\right)\sim {u}_{k}\frac{d{z}^{k}}{{z}^{k}}.$$ | (12.1.14) |

Writing

$$\prod \left(x-{a}_{i}\right)={x}^{N}+{u}_{2}{x}^{N-2}+\cdots +{u}_{N},$$ | (12.1.15) |

we ﬁnd that the residues of $\lambda $ in the tube region is given by ${a}_{1},\dots ,{a}_{N}$. Therefore, we ﬁnd full punctures after we split oﬀ two spheres.

The resulting three-punctured sphere has one simple puncture and two full punctures. Therefore it should carry $U\left(1\right)\times SU\left(N\right)\times SU\left(N\right)$ symmetry. The four-punctured sphere represents the $SU\left(N\right)$ theory with $2N$ ﬂavors. The tube region carries the $SU\left(N\right)$ vector multiplet. Then each three-punctured sphere just represents $N$ ﬂavors, i.e. hypermultiplets $\left({Q}_{i}^{a},{\stackrel{\u0303}{Q}}_{a}^{i}\right)$ where $a,i=1,\dots ,N$. Then two $SU\left(N\right)$ symmetries can be identiﬁed with those acting on the index $a$ and $i$ respectively, and the $U\left(1\right)$ symmetry is such that $Q$ has charge $+1$ while $\stackrel{\u0303}{Q}$ has charge $-1$.

The ultraviolet curve of the $SU\left(N\right)$ theory with $2N$ ﬂavors, shown in Fig. 12.1, is composed of two copies of this three-punctured sphere. The $2N$ hypermultiplets are split into $N$ hypermultiplets $\left({Q}_{i}^{a},{\stackrel{\u0303}{Q}}_{a}^{i}\right)$ charged under $SU{\left(N\right)}_{A}$ and $U{\left(1\right)}_{B}$, and another $N$ hypermultiplets $\left({Q}^{\prime}{\phantom{\rule{0.0pt}{0ex}}}_{i}^{a},{\stackrel{\u0303}{Q}}^{\prime}{\phantom{\rule{0.0pt}{0ex}}}_{a}^{i}\right)$ charged under $SU{\left(N\right)}_{D}$ and $U{\left(1\right)}_{C}$.

Let us consider what happens when $q\to \infty $. As shown in Fig. 12.3, it just ends up exchanging the puncture $B$ and $C$, at the same time redeﬁning the coupling $q$ via ${q}^{\prime}=1\u2215q$. This means that this strongly-coupled limit turns out to be another weakly-coupled $SU\left(N\right)$ gauge theory with $2N$ ﬂavors. This time, the $2N$ hypermultiplets are split into $N$ hypermultiplets $\left({q}_{i}^{a},{\stackrel{\u0303}{q}}_{a}^{i}\right)$ and another $N$ hypermultiplets $\left({q}^{\prime}{\phantom{\rule{0.0pt}{0ex}}}_{i}^{a},{\stackrel{\u0303}{q}}^{\prime}{\phantom{\rule{0.0pt}{0ex}}}_{a}^{i}\right)$, but notice that the ﬁrst $N$ are charged under $SU{\left(N\right)}_{A}$ and $U{\left(1\right)}_{C}$ while the second $N$ are charged under $SU{\left(N\right)}_{D}$ and $U{\left(1\right)}_{B}$. As we learned for the case of the $SU\left(2\right)$ theory with four ﬂavors in Sec. 9.4, the new quarks are magnetic from the point of view of the original theory.

We would like to understand the limit $q\to 1$ too. We need to split the four-punctured sphere as shown in Fig. 12.4. But the conﬁguration of punctures are not what we already know: we have two full punctures on one side, and two simple punctures on the other side. We need to study more about the 6d construction before answering what happens in the limit.