Now let us quickly discuss the $SO\left(2N\right)$ gauge theories.

The vector multiplet scalar $\Phi $ is an $2N\times 2N$ antisymmetric matrix. Let us denote the hypermultiplets by $\left({Q}_{i}^{a},{\stackrel{\u0303}{Q}}_{a}^{i}\right)$ where $a=1,\dots ,2N$ and $i=1,\dots ,{N}_{f}$. We consider the branch of the supersymmetric vacuum given by

$$\left[\Phi ,{\Phi}^{\u2020}\right]=0.$$ | (11.4.1) |

As $\Phi $ is antisymmetric, the outcome of the diagonalization is

$$\Phi =diag\left({a}_{1},\dots ,{a}_{N},-{a}_{1},\dots ,-{a}_{N}\right).$$ | (11.4.2) |

In general the gauge group is broken to $U{\left(1\right)}^{N}$. The gauge invariant combination is given by

$${x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}=det\left(x+\Phi \right)$$ | (11.4.3) |

where $x$ is a dummy variable. Note that the odd powers automatically vanish due to the antisymmetry. In fact ${\u0169}_{N}$ deﬁned by the condition

$${u}_{2N}={\u0169}_{N}{\phantom{\rule{0.0pt}{0ex}}}^{2},\phantom{\rule{2em}{0ex}}{\u0169}_{N}={a}_{1}{a}_{2}\dots {a}_{N}$$ | (11.4.4) |

is also invariant under $SO\left(2N\right)$ but not under $O\left(2N\right)$.

The W-bosons have masses

$$|\pm {a}_{i}\pm {a}_{j}|$$ | (11.4.5) |

for $i\ne j$. Similarly, the monopole has the mass

$$\left|\tau \left(\pm {a}_{i}\pm {a}_{j}\right)\right|.$$ | (11.4.6) |

By expanding the superpotential

$$\sum _{i}\left({Q}_{i}\Phi {\stackrel{\u0303}{Q}}^{i}+{\mu}_{i}{Q}_{i}{\stackrel{\u0303}{Q}}^{i}\right),$$ | (11.4.7) |

classically we ﬁnd that there is a massless hypermultiplet charged under one of $U\left(1\right)$ gauge ﬁelds when ${\mu}_{s}=\pm {a}_{i}$ for some $i$ and $s$.

The one-loop running is given by

$$\Lambda \frac{d}{d\Lambda}\tau =-\frac{1}{2\pi i}\left(2\left(2N-2\right)-2{N}_{f}\right).$$ | (11.4.8) |

Therefore the theory is asymptotically free for ${N}_{f}<2N-2$, and is asymptotically conformal when ${N}_{f}=2N-2$.

The Seiberg-Witten curve of the pure theory is given by

$${x}^{2}\left(\frac{{\Lambda}^{2N-2}}{z}+{\Lambda}^{2N-2}z\right)={x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}$$ | (11.4.9) |

with the diﬀerential $\lambda =xdz\u2215z$. This is a $2N$-sheeted cover of the ultraviolet curve $C$, which is just a sphere with the complex coordinate $z$. By solving the equation, one ﬁnds $2N$ local solutions $\pm {x}_{i}\left(z\right)$. Correspondingly, we deﬁne $\pm {\lambda}_{i}=\pm {x}_{i}\left(z\right)dz\u2215z$.

Let us study the weakly-coupled regime. We introduce ${\underline{a}}_{i}$ by

$${x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}=\prod _{i=1}^{N}\left({x}^{2}-{\underline{a}}_{i}{\phantom{\rule{0.0pt}{0ex}}}^{2}\right).$$ | (11.4.10) |

The regime we are interested in is when $\left|{\underline{a}}_{i}\right|\gg \left|\Lambda \right|$.

We draw the A-cycle on the ultraviolet curve at $\left|z\right|=1$, see Fig. 11.6. On the A-cycle, the equation (11.4.9) can be solved approximately to give

$${x}_{i}\left(z\right)={\underline{a}}_{i}+O\left(\Lambda \right).$$ | (11.4.11) |

We lift the A-cycle on $C$ to the sheets of $\Sigma $. We have $N$ pairs of cycles $\pm {A}_{i}$. Then

$${a}_{i}=\frac{1}{2\pi i}{\oint}_{{A}_{i}}\lambda =\frac{1}{2\pi i}\oint {\lambda}_{i}={\underline{a}}_{i}+O\left(\Lambda \right).$$ | (11.4.12) |

We can now suspend ring-like membranes between sheets. They clearly have masses

$$|\pm {a}_{i}\pm {a}_{j}|.$$ | (11.4.13) |

We ﬁnd that we need to impose the constraint that M2-brane cannot be suspended between the $i$-th sheet and the $\left(-i\right)$-th sheet, to forbid the W-boson with mass $|\pm 2{a}_{i}|$. As for the monopoles, the branch points are at around

$${z}^{+}\sim {\left(\frac{E}{\Lambda}\right)}^{2N-2},\phantom{\rule{1em}{0ex}}{z}^{-}\sim {\left(\frac{\Lambda}{E}\right)}^{2N-2}.$$ | (11.4.14) |

Then the monopole mass can be approximately computed as in the case of $SU\left(N\right)$ gauge theory: we ﬁnd

$$\sim \left|\left({a}_{i}-{a}_{j}\right)\frac{1}{2\pi i}{\int}_{{z}^{-}}^{{z}^{+}}\frac{dz}{z}\right|=|\left({a}_{i}-{a}_{j}\right)\frac{2\left(2N-2\right)}{2\pi i}log\frac{E}{\Lambda}|.$$ | (11.4.15) |

From this we see that the running coupling is

$$\tau \left(E\right)=\frac{2\left(2N-2\right)}{2\pi i}log\frac{E}{\Lambda},$$ | (11.4.16) |

correctly reproducing the one-loop analysis.

Let us study the low-energy coupling matrix ${\tau}^{ij}$. The branch points are at $z=0,\infty $ together with $N$ pairs on generic places of the $z$-sphere. At $z=\infty $, there are $N-2$ solutions behaving as $x\sim {z}^{1\u2215\left(2N-2\right)}$ and two solutions behaving as $x\sim {z}^{-1\u22152}$. Therefore it counts as a branch point of degree $2N-2$ and another of degree 2. The structure of the branching at $z=0$ is the same. Next, consider one of $N$ pairs of branch points of these latter type. When the sheets $i$ and $j$ meet there, the sheets $-i$ and $-j$ meet at the same time. Slightly moving them apart, we ﬁnd that there are $4N$ branch points of degree 2 in total. Using the Riemann-Hurwitz theorem, we see

$$\chi \left(\Sigma \right)=2N\chi \left(C\right)-2\left(2N-3\right)-2-4N.$$ | (11.4.17) |

Therefore the genus of the Seiberg-Witten curve is $g=2N-1$. Therefore, the independent 1-cycles on $\Sigma $ can be labeled as ${\xc3}_{1}$, …, ${\xc3}_{2N-1}$ and ${\stackrel{\u0303}{B}}^{1}$,…, ${\stackrel{\u0303}{B}}^{2N-1}$ with the intersection

$${\xc3}_{i}\cdot {\xc3}_{j}=0={\stackrel{\u0303}{B}}^{i}\cdot {\stackrel{\u0303}{B}}^{j},\phantom{\rule{2em}{0ex}}{\xc3}_{i}\cdot {\stackrel{\u0303}{B}}^{j}={\delta}_{i}^{j}.$$ | (11.4.18) |

Note that the curve $\Sigma $ has the symmetry ${\mathbb{Z}}_{2}$ acting by $x\to -x$. Under this symmetry, the diﬀerential is odd: $\lambda \to -\lambda $. Correspondingly, only the 1-cycles $L$ odd under this ${\mathbb{Z}}_{2}$ action can have ${\oint}_{L}\lambda \ne 0$. The cycles ${A}_{i}$ for $i=1,\dots ,N$ obtained by lifting the A-cycle on the ultraviolet curve $C$ to $\Sigma $ are indeed odd. The period matrix ${\tau}^{ij}$ computed as in (11.2.23) is an $\left(2N-1\right)\times \left(2N-1\right)$ matrix, which is symmetric and whose imaginary part is positive deﬁnite. By restricting to the subspace odd under ${\mathbb{Z}}_{2}$ action, we end up having $N\times N$ matrix, which is again symmetric and whose imaginary part is positive deﬁnite.

The curve of the $SO\left(2N\right)$ theory with one hypermultiplet in the $2N$-dimensional representation is

$${x}^{2}\left(\frac{{\Lambda}^{2N-4}\left({x}^{2}-{\mu}^{2}\right)}{z}+{\Lambda}^{2N-2}z\right)={x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}.$$ | (11.4.19) |

Let us just see that there is a singularity in the Coulomb branch when ${a}_{i}=\pm \mu $ for some $i$. As always, we assume $\left|{\underline{a}}_{i}\right|,\left|\mu \right|\gg \left|\Lambda \right|$ , make the redeﬁnition $\stackrel{\u0303}{z}=z\u2215{\Lambda}^{2N-4}$ and take the limit of the curve:

$${x}^{2}\frac{\left({x}^{2}-{\mu}^{2}\right)}{z}={x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}.$$ | (11.4.20) |

This equation is factorized when $\pm {\underline{a}}_{i}=\mu $ or ${\underline{a}}_{i}=0$ for some $i$. The latter choice does not ﬁt the assumption that $\left|{\underline{a}}_{i}\right|\gg \left|\Lambda \right|$. Then we ﬁnd the singularities when $\pm {\underline{a}}_{i}\sim \mu $ in the weakly-coupled region.

In general, the curve of the $SO\left(2N\right)$ with ${N}_{f}={N}_{R}+{N}_{L}$ hypermultiplets in the vector representation is given by

$$\begin{array}{cc}& {x}^{2}\left(\frac{{\Lambda}^{2\left(N-{N}_{R}-1\right)}\prod _{i=1}^{{N}_{R}}\left({x}^{2}-{\mu}_{i}^{2}\right)}{z}+{\Lambda}^{2\left(N-{N}_{L}-1\right)}z\prod _{i=1}^{{N}_{L}}\left({x}^{2}-{\mu}_{i}^{\prime}{\phantom{\rule{0.0pt}{0ex}}}^{2}\right)\right)\\ & ={x}^{2N}+{u}_{2}{x}^{2N-2}+{u}_{4}{x}^{2N-4}+\cdots +{u}_{2N}.& \text{(11.4.21)}\end{array}$$Strictly speaking, this is only for ${N}_{L}+{N}_{R}<2N-2$. When ${N}_{L}={N}_{R}=N-1$, we need to put two complex numbers $f$ and ${f}^{\prime}$ instead of the powers of $\Lambda $, much as in (11) for the case of the $SU\left(N\right)$ theory with $2N$ ﬂavors.

Let us check the one-loop running when ${\mu}_{i}={\mu}_{i}^{\prime}=0$. Assume $\left|{\underline{a}}_{i}\right|\gg \left|\Lambda \right|$. As always we ﬁnd ${a}_{i}={\underline{a}}_{i}+O\left(\Lambda \right)$. The branch points on the ultraviolet curve are at around

$${z}^{+}\sim {\left(\frac{E}{\Lambda}\right)}^{2N-2-2{N}_{L}},\phantom{\rule{1em}{0ex}}{z}^{-}\sim {\left(\frac{\Lambda}{E}\right)}^{2N-2-2{N}_{R}}.$$ | (11.4.22) |

Then the monopole mass can be approximately computed as in the case of $SU\left(N\right)$ gauge theory: we ﬁnd

From this we see that the running coupling is

$$\tau \left(E\right)=\frac{2\left(2N-2\right)-2\left({N}_{L}+{N}_{R}\right)}{2\pi i}log\frac{E}{\Lambda},$$ | (11.4.24) |

correctly reproducing the one-loop analysis. Again, the condition that the theory is asymptotically free or conformal is related to the fact that the left hand side of the equation of the curve has lower degree than or equal degree to the right hand side.