As an example of the application of what we learned in this section, let us consider the $\mathcal{\mathcal{N}}=1$ pure supersymmetric Yang-Mills theory with gauge group $SU\left(N\right)$. The content of this section will not be used much in the rest of the lecture note.

This theory has just the vector multiplet, with the Lagrangian

The one-loop running of the coupling is given by

$$E\frac{\partial}{\partial E}\tau \left(E\right)=3N,$$ | (3.3.2) |

and therefore we deﬁne the dynamical scale $\Lambda $ by the relation

$${\Lambda}^{3N}={e}^{2\pi i{\tau}_{UV}}{\Lambda}_{UV}^{3N}.$$ | (3.3.3) |

We assign R-charge zero to the gauge ﬁeld, and R-charge 1 to the gaugino ${\lambda}_{\alpha}$. The phase rotation ${\lambda}_{\alpha}\to {e}^{i\phi}{\lambda}_{\alpha}$ is anomalous, and needs to be compensated by $\mathit{\theta}\to \mathit{\theta}+2N\phi $. The shift of $\mathit{\theta}$ by $2\pi $ is still a symmetry, therefore the discrete rotation

$${\lambda}_{\alpha}\to {e}^{\pi i\u2215N}{\lambda}_{\alpha},\phantom{\rule{2em}{0ex}}\mathit{\theta}\to \mathit{\theta}+2\pi $$ | (3.3.4) |

is a symmetry generating ${\mathbb{Z}}_{2N}$. Note that under this symmetry, $\Lambda $ deﬁned above has the transformation

$$\Lambda \to {e}^{2\pi i\u2215\left(3N\right)}\Lambda .$$ | (3.3.5) |

This theory is believed to conﬁne, with nonzero gaugino condensate $\u27e8{\lambda}_{\alpha}{\lambda}^{\alpha}\u27e9$. What would be the value of this condensate? This should be of mass dimension 3 and of R-charge 2. The only candidate is

$$\u27e8{\lambda}_{\alpha}{\lambda}^{\alpha}\u27e9=c{\Lambda}^{3}$$ | (3.3.6) |

for some constant $c$. The symmetry (3.3.5) acts in the same way on both sides by the multiplication by ${e}^{2\pi i\u2215N}$. Assuming that the numerical constant $c$ is non-zero, this ${\mathbb{Z}}_{2N}$ is further spontaneously broken to ${\mathbb{Z}}_{2}$, generating $N$ distinct solutions

$$\u27e8{\lambda}_{\alpha}{\lambda}^{\alpha}\u27e9=c{e}^{2\pi i\ell \u2215N}{\Lambda}^{3}$$ | (3.3.7) |

where $\ell =0,1,\dots ,N-1$. Unbroken ${\mathbb{Z}}_{2}$ acts on the fermions by ${\lambda}_{\alpha}\to -{\lambda}_{\alpha}$, which is a $36{0}^{\circ}$ rotation. This ${\mathbb{Z}}_{2}$ symmetry is hard to break.

It is now generally believed that this theory has these $N$ supersymmetric vacua and not more. For other gauge groups, the analysis proceeds in the same manner, by replacing $N$ by the dual Coxeter number $C\left(\text{adj}\right)$ of the gauge group under consideration. For example, we have $N-2$ vacua for the pure $\mathcal{\mathcal{N}}=1$ $SO\left(N\right)$ gauge theory.

It is instructive to recall another way to compute the number of vacua in the $\mathcal{\mathcal{N}}=1$ pure Yang-Mills theory with gauge group $G$, originally discussed in [39]. We put the system in a spatial box of size $L\times L\times L$ with the periodic boundary condition in each direction. We keep the time direction as $\mathbb{R}$. By performing the Kaluza-Klein reduction along the three spatial directions, the system becomes supersymmetric quantum mechanics with inﬁnite number of degrees of freedom.

The box still preserves the translation generators ${P}^{\mu}$ and the supertranslations ${Q}_{\alpha}$ unbroken. We just use a linear combination $\mathcal{\mathcal{Q}}$ of ${Q}_{\alpha}$ and ${Q}_{\alpha}^{\u2020}$, satisfying

$$H={P}^{0}=\left\{\mathcal{\mathcal{Q}},{\mathcal{\mathcal{Q}}}^{\u2020}\right\}.$$ | (3.3.8) |

We also have the fermion number operator ${\left(-1\right)}^{F}$ such that

$$\left\{{\left(-1\right)}^{F},\mathcal{\mathcal{Q}}\right\}=0.$$ | (3.3.9) |

Consider eigenstates of the Hamiltonian $H$, given by

$$H|E\u27e9=E|E\u27e9.$$ | (3.3.10) |

In general, the multiplet structure under the algebra of $\mathcal{\mathcal{Q}}$, ${\mathcal{\mathcal{Q}}}^{\u2020}$, $H$ and ${\left(-1\right)}^{F}$ is of the form

involving four states. When $\mathcal{\mathcal{Q}}|E\u27e9=0$ or ${\mathcal{\mathcal{Q}}}^{\u2020}|E\u27e9=0$, the multiplet only has two states. If $\mathcal{\mathcal{Q}}|E\u27e9={\mathcal{\mathcal{Q}}}^{\u2020}|E\u27e9=0$, the multiplet has only one state, and $E$ is automatically zero due to the equality

We see that a bosonic state is always paired with a fermionic state unless $E=0$.

This guarantees that the Witten index

$$tr\phantom{\rule{0.3em}{0ex}}{e}^{-\beta H}{\left(-1\right)}^{F}=tr\phantom{\rule{0.3em}{0ex}}{\left|\right.}_{E=0}{\left(-1\right)}^{F}$$ | (3.3.13) |

is a robust quantity independent of the change in the size $L$ of the box: when a perturbation makes a number of zero-energy states to non-zero energy $E\ne 0$, the states involved are necessarily composed of pairs of a fermionic state and a bosonic state. Thus it cannot change $tr\phantom{\rule{0.3em}{0ex}}{\left(-1\right)}^{F}$.

Therefore, we can compute the Witten index in the limit where the box size $L$ is far smaller than the scale ${\Lambda}^{-1}$ set by the dynamics. Then the system is weakly coupled, and we can use perturbative analysis. To have almost zero energy, we need to have ${F}_{\mu \nu}=0$ in the spatial directions, since magnetic ﬁelds contribute to the energy. Then the only low-energy degrees of freedom in the system are the holonomies

$${U}_{x},{U}_{y},{U}_{z}\in SU\left(N\right),$$ | (3.3.14) |

which commute with each other. Assuming that they can be simultaneously diagonalized, we have

$$\begin{array}{lll}\hfill {U}_{x}& =diag\left({e}^{i{\mathit{\theta}}_{1}^{x}},\dots ,{e}^{i{\mathit{\theta}}_{N}^{x}}\right),\phantom{\rule{2em}{0ex}}& \hfill \text{(3.3.15)}\\ \hfill {U}_{y}& =diag\left({e}^{i{\mathit{\theta}}_{1}^{y}},\dots ,{e}^{i{\mathit{\theta}}_{N}^{y}}\right),\phantom{\rule{2em}{0ex}}& \hfill \text{(3.3.16)}\\ \hfill {U}_{z}& =diag\left({e}^{i{\mathit{\theta}}_{1}^{z}},\dots ,{e}^{i{\mathit{\theta}}_{N}^{z}}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(3.3.17)}\end{array}$$together with gaugino zero modes

$${\lambda}_{1}^{\alpha =1},\dots ,{\lambda}_{N}^{\alpha =1},\phantom{\rule{1em}{0ex}}{\lambda}_{1}^{\alpha =2},\dots ,{\lambda}_{N}^{\alpha =2}$$ | (3.3.18) |

with the condition that

The wavefunction of this truncated quantum system is given by a linear combination of states of the form

which is invariant under the permutation acting on the index $i=1,\dots N$. To have zero energy, the wavefunction cannot have dependence on ${\mathit{\theta}}_{i}^{x,y,z}$ anyway, since the derivatives with respect to them are the components of the electric ﬁeld, and they contribute to the energy. Thus the only possible zero energy states are just invariant polynomials of $\lambda $s. We ﬁnd $N$ states with the wavefunctions given by

$$1,\phantom{\rule{1em}{0ex}}S,\phantom{\rule{1em}{0ex}}{S}^{2},\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{S}^{N-1}$$ | (3.3.21) |

where $S={\sum}_{i}{\lambda}_{i}^{\alpha =1}{\lambda}_{i}^{\alpha =2}$. They all have the same Grassmann parity, and contribute to the Witten index with the same sign. Thus we found $N$ states in the limit of small box, too.

The construction so far, when applied to other groups, only gives $1+rankG$ states. For example, let us consider for $G=SO\left(N\right)$ for $N>4$. Then the method explained so far only gives $\lfloor N\u22152\rfloor +1$ states

$$1,\phantom{\rule{1em}{0ex}}S,\phantom{\rule{1em}{0ex}}{S}^{2},\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{S}^{\lfloor N\u22152\rfloor},$$ | (3.3.22) |

and does not agree with $C\left(\text{adj}\right)=N-2$
when $N\ge 7$.
This conundrum was already pointed out in [39] and resolved later in the Appendix I of [40] by the same
author.^{6}
What was wrong was the assumption that three commuting matrices
${U}_{x,y,z}$
can be simultaneously diagonalized as in (3.3.17). It is known that there is another
component where they cannot be simultaneously diagonalized into the Cartan torus. For
$SO\left(7\right)$, an
example is given by the triple

These three matrices might look diagonal, but not in the same Cartan subgroup. This component adds one supersymmetric state. Then, in total, we have $\left(\lfloor 7\u22152\rfloor +1\right)+1=5=7-2$, reproducing $C\left(\text{adj}\right)$.

For larger $N$, one can consider ${U}_{x,y,z}$ given by the form

where ${U}_{x,y,z}^{\prime}$ are in the Cartan subgroup of $SO\left(N-7\right)$. Applying the analysis leading to (3.3.21) in both components, i.e. in the component where ${U}_{x,y,z}$ are in the Cartan subgroup of $SO\left(N\right)$, and in the component where ${U}_{x,y,z}$ has the form (3.3.26), we ﬁnd in total

$$\left(\lfloor N\u22152\rfloor +1\right)+\left(\lfloor \left(N-7\right)\u22152\rfloor +1\right)=N-2$$ | (3.3.27) |

zero-energy states, thus reproducing $C\left(\text{adj}\right)$ states. This analysis has been extended to arbitrary gauge groups [41, 42].