[Paper Review] Hilbert Series for adjoint SQCD
Cheetah run!!
Introduction
Invention of Vector
Newton invent vector for represent velocity, momentum, and force.
$$\vec{F}=m\vec{a}$$
Vector developed to embed any kinds of data: Image, Word, etc.
However, they are different from physical vector.
Physical Quantities are Representation
Vector: fundamental representation of $SU(N)$ or $SO(N)$ operate
Spinor: spinor representation of $SO(N)$ operate
State: representation of $U(1)$ operate
Physical Laws are Invariant
Newtonian mechanics: Coordinate dependence
Lagrangian mechanics: Coordinate independence
Hamiltonian mechanics: Coordinate independence
Maxwell equation: Gauge Invariant
Dirac equation: Gauge Invariant
Yang-Mills equation: Gauge Invariant
Goal
General tool for finding invariant (tensor) expression under arbitrary group.
For example, we know these invariant expression are in Lagrangian or Hamiltonian.
$$(\partial \psi)^2,\ \frac{1}{2}m^2\vec{A}^2,\ \frac{1}{4}\vec{F}^2,\ \cdots$$
Representation
$SU(2)$ Lie algebra: Angular Momentum
We use $SU(2)$, because it's familiar and simple.
Lie algebra of $SU(2)$: $$[J_i,J_j]=i\epsilon_{ijk} J_k$$
Irreducible representation of $SU(2)$ is called spin-$j$ representation.
Spin-$\frac{1}{2}$ Representation
Act two-level object $\vec{\psi}$ which basis are $\left|\frac{1}{2}, -\frac{1}{2}\right\rangle,\ \left|\frac{1}{2}, +\frac{1}{2}\right\rangle$.
Then Lie algebra of $SU(2)$ have matrix representation with above basis.
$$J^\frac{1}{2}_1= \sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}, J^\frac{1}{2}_2=\sigma_2= \begin{pmatrix}0&-i\\i&0\end{pmatrix}, J^\frac{1}{2}_3=\sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
Usually represent spin up/down system
Spin-$1$ Representation
Act 3 dimensional object $\vec{E}$ which basis are $\left|1, -1\right\rangle, \left|1, 0 \right\rangle,\ \left|1, +1\right\rangle$
Then Lie algebra of $SU(2)$ have matrix representation with above basis.
$$J_1^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix},\ J_2^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\ J_3^1=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$
Same representation for $3$-dim real vector (ex: Electric field)
Adjoint Representation
Generator number(For $SU(N)$, $N^2-1$) dimentional representation.
It can represent in $N\times N$ matrix like QCD interaction term.
Let $\phi$ be adjoint representation, and thier components are $(\theta_1,\theta_2,\theta_3)$
Then $$\phi_{ab}\equiv \theta_i\sigma_i= \theta_1(\sigma_1)_{ab}+\theta_2(\sigma_2)_{ab}+\theta_3(\sigma_3)_{ab}$$
Tensor Representation
Act Multiple of objects.
For example, let's do $\vec{\psi} \otimes \vec{\psi}$: Sum of angular momentum
$$\left|+\right\rangle \left|+\right\rangle \rightarrow \begin{pmatrix}1\\0\\0\\0\end{pmatrix},\quad \left|+\right\rangle \left|-\right\rangle \rightarrow \begin{pmatrix}0\\1\\0\\0\end{pmatrix},\quad \left|-\right\rangle \left|+\right\rangle \rightarrow \begin{pmatrix}0\\0\\1\\0\end{pmatrix},\quad \left|-\right\rangle \left|-\right\rangle \rightarrow \begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$
Decompose into spin-$1$ (triplet) representation and spin-$0$ (singlet) representation.
Basis of vector space become
$$\left| 1,1\right\rangle = \left| +,+\right\rangle,\ \left| 1,0 \right\rangle = \frac{1}{\sqrt{2}}(\left| +,-\right\rangle + \left| -,+\right\rangle),\ \left| 1,-1\right\rangle = \left| -,-\right\rangle$$
$$\left| 0,0 \right\rangle =\frac{1}{\sqrt{2}}(\left| +,-\right\rangle - \left| -,+\right\rangle)$$
This express as $\frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0$
Lagrangian
Scalar
Lagrangian is scalar (invariance)
$$\mathcal{L}=\frac{1}{2}m^2A^2 + \cdots$$
However, each part are combination of vector.
$$ A^2 = A_\mu A^\mu $$
That components are singlet of tensor product of vectors. (dot product)
$$A^2 =\mbox{Tr}(A\otimes A) =\mbox{Tr}(A_\mu A^\nu) $$
Singlet
Tensor product of spin-$\frac{1}{2}$ representations and spin-$1$ representations:
$$\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0\quad \rightarrow |\vec{\psi}|^2 =\mbox{Tr}(\psi_a\psi^b) $$
$$1\otimes 1 = 2\oplus 1 \oplus 0 \quad \rightarrow |\vec{E}|^2= \mbox{Tr}(E_{ab}E^{bc}) $$
$$ 1 \otimes \frac{1}{2} \otimes \frac{1}{2} = 1 \otimes (1\oplus 0) = (1 \otimes 1)\oplus (1\otimes 0) = 2\oplus 1 \oplus 0 \oplus 1 \quad \rightarrow E_{ab} \psi^a \psi^b,\ E_{ab}=\vec{\sigma}_{ab}\cdot\vec{E},\ \epsilon^{ab}=\begin{pmatrix}0&1\\-1&0\end{pmatrix} $$
0 is scalar (singlet), so it can be part of Lagrangian.
For now, this looks easy but it is not in general.
Hilbert Series
Partition Function
Now we think every combinations of $\psi$ make a partition function.
Construct vector space of every combination
$$V = \mbox{Span}\{1, \psi, \psi^2, \cdots\}$$
Partition function is $$Z(e^{-\beta})=\sum_n e^{-\beta E_n} = \mbox{Tr}_V q^H = \sum_n \left\langle n \right| q^H \left| n \right\rangle, \quad q\equiv e^{-\beta}$$
Set $E_n=n$, partition function is $$Z(q)=1+q + q^2 + q^3 + \cdots = \sum_n q^n = \frac{1}{1-q}$$
To add $\vec{E}$, use Grand Canonical Ensemble forms $$Z(e^{-\beta},e^{\mu})=\sum_{n,m} e^{-\beta E_n + \mu N_m} = \frac{1}{1-q}\frac{1}{1-p},\ q\equiv e^{-\beta},p\equiv e^\mu$$
Integrate over the group
Example of parity group: $G=\{I, P\} $, and group element denote $\sigma \in G$
Parameterize group element: $$\chi(I)=1,\ \chi(P)=-1$$
Find group invariant element:
$$Z[\chi(\sigma)t]=1+ \chi(\sigma)t + \chi(\sigma)^2t^2 + \chi(\sigma)^3t^3+\chi(\sigma)^4t^4 +\cdots\quad$$
$$\rightarrow Z[\chi(\sigma)^2t^2]= 1+ \chi(\sigma)^2t^2 + \chi(\sigma)^4t^4 +\cdots$$
Sum over the group. $$\frac{1}{2}\sum_{\sigma\in G} Z[\chi(\sigma)t] = Z[t^2] $$
This called Molien Formula.
For Lie group, it called 'Hilbert Series' (or Weyl-Molien Formula)
$$HS(t)=\int_{\sigma \in G} Z[\chi(\sigma) t]d\mu $$
Parameterize $SU(2)$
To parameterize group, we use character(trace of representation).
Spin-$\frac{1}{2}$ rep: $\chi_{fund} = z+\frac{1}{z}$
Spin-$1$ rep: $\chi_{adj} = z^2 + 1+\frac{1}{z^2}$
Haar Measure: $$\int_{SU(2)} d\mu_{SU(2)} = \frac{1}{2\pi i} \oint_{\left| z\right|=1} \frac{dz}{z} (1-z^2)$$
Result
Setting
Set spin-$\frac{1}{2}$ representation $t$ and spin-$1$ representation $s$.
$$HS(t,s)=\int_{SU(2)} d\mu_{SU(2)} Z[2\chi_\frac{1}{2}(z) t + \chi_1(z) s]$$
$$=\frac{1}{2\pi i} \oint_{\left| z \right|=1} \frac{dz}{z} (1-z^2)Z[2 (z+\frac{1}{z})t + (z^2 +1+ \frac{1}{z^2})s]$$
$$=\frac{1}{2\pi i} \oint_{\left| z \right|=1} \frac{dz}{z} (1-z^2) \frac{1-z^2}{(1-tz)^2(1-\frac{t}{z})^2(1-s)(1-sz^2)(1-\frac{s}{z^2})}$$
Computation
Use Residue theorem, since $0<\left| t\right|,\left| s\right| <1$, valid poles are $z=t,\sqrt{s},-\sqrt{s}$.
$$HS(s,t)=\frac{1+st^2}{(1-s^2)(1-t^2)(1-st^2)^2}$$
Analyze
Each term of Hilbert series means possible singlet.
$$HS(s,t)=1+s^2+s^4+t^2+3st^2+st^2+s^2t^2+4s^3t^2+s^4t^2+t^4+\cdots$$
To get basis, use inverse of Plethystic Exponential.
$$HS(s,t)=Z[s^2+t^2+3st^2-s^2t^4]$$
Which we see before.
1. Casimir invariant:
$$s^2 \rightarrow \mbox{Tr}(\phi^2):\ 1\otimes 1 = 2\oplus 1 \oplus 0$$
dim: $1: [0,\cdots,0]$
Let Adjoint representation be symmetric 2-tensor $\phi_{ab}=\begin{pmatrix} \theta_3&\theta_1-i\theta_2 \\ \theta_1+i\theta_2& -\theta_3\end{pmatrix}$(also spin-1 representation, 3-dim rotation matrices act)
$$\mbox{Tr}(\phi^2)= \theta_3^2+\theta_1^2+\theta_2^2+\theta_1^2+\theta_2^2+\theta_3^2=2(\theta_1^2+\theta_2^2+\theta_3^2)$$
How about Young Tableaux?
$\phi$ is symmteric, $\phi^\dagger=\phi$(Hermitian). Diagram is two horizental boxes.
$$\epsilon^{ab}\epsilon^{cd}\phi_{ac}\phi_{bd}=(\delta_a^c\delta_b^d-\delta_a^d\delta_c^b)\phi_{ac}\phi_{bd}=\phi_{11}\phi_{11}+\phi_{22}\phi_{11}+\phi_{11}\phi_{22}+\phi_{22}\phi_{22}-\phi_{11}\phi_{11}-\phi_{12}\phi_{21}-\phi_{21}\phi_{12}-\phi_{22}\phi_{22}\\ =\phi_{11}\phi_{22}+\phi_{22}\phi_{11}-\phi_{12}\phi_{21}-\phi_{21}\phi_{12}=-2(\theta_1^2+\theta_2^2+\theta_3^2)$$
2. Mesons:
$$t^2\rightarrow \epsilon^{ab}Q_aQ_b:\ \frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$$
dim: $2n C 2$
$$\epsilon^{ab}Q_aQ_b:\ Q^i_1 Q^j_2 - Q^i_2 Q^j_1$$ This is antisymmetric on $i,j$, so $SU(2N_f)$ representation is $[0,1,\cdots,0]$.
This is singlet because it means spin-$0$ state.
More detail, $L^+\epsilon^{ab}Q_aQ_b = 0$, $L^-\epsilon^{ab}Q_aQ_b = 0$: only $1$-dim
More more detail, 
Left one is singlet, and antisymmetric 2-tensor.
3. Adjoint mesons:
$$st^2 \rightarrow \epsilon^{ab}\epsilon^{cd}Q_a\phi_{bc} Q_b:\ 1 \otimes \frac{1}{2} \otimes \frac{1}{2}=2\oplus 1 \oplus 0 \oplus 1$$
$$\epsilon^{ab}\epsilon^{cd}Q_a\phi_{bc} Q_b:\ Q^i_1\phi_{21}Q^j_2-Q^i_1\phi_{22}Q^j_1- Q^i_2\phi_{11}Q^j_2+Q^i_2\phi_{12}Q^j_1$$
$$=Q_1^{i*}(\theta_1+i\theta_2)Q_2^j-Q_1^{i*}(-\theta_3)Q_1^j-Q_2^{i*}(\theta_3)Q_2^j+Q_2^{i*}(\theta_1-i\theta_2)Q_1^j$$
symmetric on $i$, $j$, so$[2,0,\cdots,0]$
This is singlet because singlet on Young Tableaux form
More detail, group element action $Q\rightarrow U^{J^1_{1/2}\theta_1+J^2_{1/2}\theta_2+J^3_{1/2}\theta_3}Q$ and $\phi\rightarrow U^{J^1_1\theta_1+J^2_1\theta_2+J^3_1\theta_3}\phi$ gives $I$ or different saying, $J^\pm=J^\pm_{1/2}\otimes I_1 \otimes I_{1/2} + I_{1/2} \otimes J^\pm_1 \otimes I_{1/2} + I_{1/2} \otimes I_1 \otimes J^\pm_{1/2}$ act give $0$.
Appendix
Lie group and Lie algebra
Then group element can represented by $$e^{i\vec{\alpha} \cdot \vec{\sigma}/2}$$
which are the most general $2\times 2$ unitary matrices with determinant 1.
Lie algebra is a tangent vector of Lie group.
$SO(3)$ fundamental rep. and $SU(2)$ adjoint rep.
$$J_1^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&1\\1&0&1\\0&1&0\end{pmatrix},\ J_2^1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\ J_3^1=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$
$$\tilde{J}_1^1=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix},\ \tilde{J}_2^1=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix},\ \tilde{J}_3^1=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$$
We can define matrix $$U=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & -1 \\ i & 0 & i \\ 0 & -\sqrt{2} & 0\end{pmatrix}$$
and use similarity transformation $$iJ_1 = U\tilde{J}_1U^{-1} \quad iJ_2 = U \tilde{J}_2U^{-1}\quad iJ_z=U\tilde{J}_3U^{-1}$$
Spin-$\frac{1}{2}$ $\otimes$ spin-$\frac{1}{2}$
This $4$-dim vector space with operator
$$\sigma_{3,tot} = \sigma_3 \otimes I + I \otimes \sigma_3 = \frac{1}{\sqrt{2}} \begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&-1\end{pmatrix}$$
$$\sigma^2_{tot} = (\sigma_{1,tot})^2+(\sigma_{2,tot})^2+(\sigma_{3,tot})^2=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{pmatrix}$$
This matrix can be diagonals as $$\sigma_{3,tot}=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}\oplus\begin{pmatrix}0\end{pmatrix}$$
$$\sigma^2_{tot}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\oplus\begin{pmatrix}0\end{pmatrix}$$
Each are spin-$1$ representation and spin-$0$(singlet) representation.
Clebsh-Gordan rule
Formula for tensor product of $SU(2)$ representation
Tensor product of spin $j$ and spin $j'$ become $$ j \otimes j' = (j+j') \oplus \cdots \oplus \left| j-j' \right|$$
This is simple only $SU(2)$.
$SU(2)$ Haar Measure
Integral over the group express (Weyl integral formula)
$$\int_Gf(g)d\mu = |W|^{-1}\int_T \left| \Delta(t)\right|^2 \int_{G/T} f(yty^{-1})d[y]dt$$
Where $d[y]$ is the normalized volume measure on the quotient manifold $G/T$ and $dt$ is the normalized Haar measure on $T$.
Here $\Delta$ is given by the Weyl denominator formula and $|W|$ is the order of the Weyl group.
When $f$ is a class function, $$\int_G f(g)dg = |W|^{-1}\int_T f(t) \left| \Delta (t) \right|^2 dt$$
$$\int_{SU(2)} f(g)dg = \frac{1}{2} \int_0^{2\pi} f(\mbox{diag}(e^{i\theta},e^{-i\theta})) 4\sin^2 (\theta) \frac{d\theta}{2\pi}$$
Paramterize $z=e^{i\theta}$,
$$\int_{SU(2)} f(g)dg = \frac{1}{2\pi i} \oint_{\left| z\right|=1} \frac{dz}{z} (1-z^2)$$
$SU(2)$ Character
Character is trace over representation.
Trace characterize conjugacy class, so trace is enough to consider group element.
Because one group element can become another group element in same conjugacy class with similarity transformation.
Integral over conjugacy class cover all group by Weyl integral formula.
For $SU(2)$, only $J_3$ have trace elements.
$$\chi(J(\theta))=\chi(J_3(\theta))=e^{2ij\theta}+e^{i(2j-1)\theta}+\cdots+e^{-i(2j+1)}\theta$$
$$\chi(\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix})=\frac{e^{i(2j+1)\theta}-e^{-i(2j+1)\theta}}{e^{i\theta}-e^{-i\theta}}=\frac{\sin ((2j+1)\theta}{\sin \theta}$$
Paramterize $z=e^{i\theta}$,
$$\chi(z)=z^{2j}+z^{2j-1}+\cdots + z^{-2j}$$
$SU(3)$ Representation
Two kinds of spin exists: $[n,m]$
Ladder operator called Cartan matrix: $[[2,-1],[-1,2]]$
Fundamental rep: $[1,0]$, $(1,0) \rightarrow (-1,1) \rightarrow (0,-1)$
Antifundamental rep: $[0,1]$, $(0,1) \rightarrow (1,-1) \rightarrow (-1,0)$
Character of $[1,0]$: $z_1 + \frac{z_2}{z_1} + \frac{1}{z_2}$, $[0,1]$: $z_2 + \frac{z_1}{z_2} + \frac{1}{z_1}$
Adjoint rep: $[1,1]$ : $(1,1)\rightarrow (-1,2),(2,-1) \rightarrow (0,0) \rightarrow (1,-2),(-2,1)\rightarrow (-1,-1)$
Character of $[1,1]$: $2+\frac{z_1}{z_2^2}+\frac{1}{z_1z_2}+\frac{z_1^2}{z_2}+\frac{z_2}{z_1^2}+z_1z_2+\frac{z_2^2}{z_1}$
Haar Measure of $SU(3)$: $$\int_{SU(3)} d\mu_{SU(3)} = \frac{1}{(2\pi i)^2} \oint_{\left| z_1 \right|=1} \frac{dz_1}{z_1} \oint_{\left| z_2 \right|=1} \frac{dz_2}{z_2} (1-z_1z_2)(1-\frac{z_1^2}{z_2})(1-\frac{z_2^2}{z_1})$$
Clebsch-Gordan rule for $SU(3)$: Young Tableau
1. A row must not be longer than the one before it.
2. The quantum labels (numbers in the rectangle) should not decrease while going left to right in a row.
3. The quantum labels must strictly increase while going down in a column.
For example, $[1,0]\otimes [1,1]=[2,1]\oplus [0,2]\oplus [1,0]$
Canonical Ensemble
Now we think every combinations of $\psi$ make a partition function.
Construct \textcolor{blue}{vector space} of every combination $$V = \mbox{Span}\{1, \psi, \psi^2, \cdots\}$$
Partition function is $$Z(e^{-\beta})=\sum_n e^{-\beta E_n} = \mbox{Tr}_V q^H = \sum_n \left\langle n \right| q^H \left| n \right\rangle, \quad q\equiv e^{-\beta}$$
When we can define energy of each state as degree of $\psi$: $$H=\begin{pmatrix}0&0&0&\cdots\\ 0&1&0&\cdots\\0&0&2&\cdots \\ \vdots&\vdots &\vdots& \ddots\end{pmatrix}, \mbox{or diag} (H)=[0,1,2,\cdots]$$
Then, partition function is simply $$Z(q)=1+q + q^2 + q^3 + \cdots = \sum_n q^n = \frac{1}{1-q}$$
Grand Canonical Ensemble
To add $\vec{E}$, use Grand Canonical Ensemble forms
$$Z(e^{-\beta},e^{-\mu})=\sum_{n,m} e^{-\beta E_n + \mu N_m} = \mbox{Tr}_{V''}q^Hp^L$$
$$= \sum_{n,m}\left\langle n,m\right| q^Hp^L\left| n,m\right\rangle = \frac{1}{1-q}\frac{1}{1-p}$$
$$q\equiv e^{-\beta},p\equiv e^\mu$$
where $V''=V\otimes V'$, $V'=\mbox{Span}\{1,\vec{E},\vec{E}^2,\cdots\}$
Also $H$ and $L$ defined as \textcolor{blue}{degree of $\psi$ and $\vec{E}$} of each state.
$$\mbox{diag}(H)=[0,1,2,\cdots],\ \mbox{diag}(L)=[0,1,2,\cdots]$$
We can generalize this.
Plethystic Exponential
Tool for construct muti-variable partition function
$$PE[x] =Z(x) = 1 + x + x^2 + x^3 +\cdots = \frac{1}{1-x}$$
Follows partition function's condition $$PE[x+y]=1+x+y+x^2+y^2+xy+\cdots$$
$$=(1+x+x^2+\cdots)(1+y+y^2+\cdots)=PE[x]PE[y]$$
For general, $$PE[g(t_1, \cdots, t_n)] = \exp (\sum g(t_1^r, \cdots, t_n^r)/r)$$
Haar Measure
Weyl's Integration Formula shows Jacobian.
$$\int_G f d\mu_G = \frac{1}{\left|\mathfrak{W}\right|} \int_T f(y) \prod_{\alpha \in R} (1 - e^{2\pi i \alpha (y)})d\mu_T$$
where $\alpha$ is root of root system $R$, $\mathfrak{W}$ is Weyl group.
$T$ become torus, it called 'maximal torus' of compact Lie group.
Weyl Character Formula
Character is trace of representation.
$$\chi_R(\sigma)=\mbox{Tr}(e^{R(\sigma)})=\frac{\sum_{w\in W}\epsilon(w)e^{w(\lambda+\rho)(\sigma)}}{\Pi_{\alpha\in \Delta^+} (e^{\alpha(\sigma)/2}-e^{-\alpha(\sigma)/2})}$$
$$\chi_R(\sigma)=\frac{\sum_{w\in W}\epsilon(w)e^{w(\lambda+\rho)(\sigma)}}{\sum_{w\in W}\epsilon(w)e^{w(\rho)(\sigma)}}$$
When $W$ is the Weyl group, $\Delta^+$ is the set of the positive roots of the root system $\Delta$, $\rho$ is the half-sum of the positive roots(Weyl vector), $\lambda$ is the highest weight of the irreducible representation $V$, $\epsilon(w)$ is the determinant of the action of $w$ on the Cartan subalgebra. This is equal to $(-1)^{l(w)}$, where $l(w)$ is the length of the Weyl group element, defined to be the minimal number of reflections with respect to simple roots such that $w$ equals the product of those reflections.
$SU(3)$ Singlet
Hilbert Series of $SU(3)$ with $N_f$ flavor ($N_f$ kinds of (anti)fund. rep. objects):
$$HS(s,t,\tilde{t})=\int_{SU(3)}d\mu_{SU(3)} PE[N_f[1,0]t + N_f [0,1]\tilde{t} + [1,1]s]$$
With Residue theorem, basis become (When $N_f=1$)
$$PE^{-1}[HS(s,t,\tilde{t})]=s^2 + s^3 + t\tilde{t} + st\tilde{t}+s^2 t\tilde{t}+s^3t^3+s^3\tilde{t}^3-s^6t^3\tilde{t}^3$$
Casimir invariants: $s^k \rightarrow \mbox{Tr}(\phi^k),\ k=2,3$: $1$ dimensional
$$\phi=\lambda_i \theta_i=\begin{pmatrix}\theta_3+\frac{1}{\sqrt{3}}\theta_8 & \theta_1-i\theta_2& \theta_4-i\theta_5 \\ \theta_1+i\theta_2 & -\theta_3+\frac{1}{\sqrt{3}}\theta_8& \theta_6-i\theta_7\\ \theta_4+i\theta_5& \theta_6+i\theta_7& -\frac{2}{\sqrt{3}}\theta_8\end{pmatrix}$$
$$\mbox{Tr}(\phi^2)=\theta_3^2+\frac{1}{3}\theta_8^2+\theta_1^2+\theta_2^2+\theta_4^2+\theta_5^2+\theta_1^2+\theta_2^2+\theta_3^2+\frac{1}{3}\theta_8^2+\theta_6^2+\theta_7^2+\theta_4^2+\theta_5^2+\theta_6^2+\theta_7^2+\frac{4}{3}\theta_8^2=2\theta_i^2$$
$$\mbox{Tr}(\phi^3)=3\theta_3 \theta_4^2+3\theta_3\theta_5^2+6\theta_1\theta_4\theta_6+6\theta_2\theta_5\theta_6-3\theta_3\theta_6^2-6\theta_2\theta_4\theta_7+4\theta_1\theta_5\theta_7-3\theta_3\theta_7^2+2\sqrt{3}\theta_1^2\theta_8+2\sqrt{3}\theta_2^2\theta_8+2\sqrt{3}\theta_3^2\theta_8-\sqrt{3}\theta_4^2\theta_8-\sqrt{3}\theta_5^2\theta_8-\sqrt{3}\theta_6^2\theta_8-\sqrt{3}\theta_7^2\theta_8-\frac{2}{\sqrt{3}}\theta_8^3=d_{ijk}\theta_i\theta_j\theta_k$$
($d_{ijk}$ is completely symmetric coefficient constants, $d_{ijk}=\frac{1}{4}\mbox{Tr}(\{\lambda_i,\lambda_j\}\lambda_k)=d_{jki}=d_{jik}$)
Measons: $t\tilde{t}\rightarrow \epsilon^{ab}Q^i_a\tilde{Q}^j_b$: $N_f^2$ dimensional
$$\epsilon^{ab}Q^i_a\tilde{Q}^j_b= Q^i_1\tilde{Q}^j_2-Q^i_2\tilde{Q}^j_1$$
Adjoint Measons: $s^lt\tilde{t} \rightarrow Q_a^i(\phi^l)_b^a\tilde{Q}_j^b,\ l=1,2$: $N_f^2$ dimensional
$$Q_a^i(\phi^1)_b^a\tilde{Q}_j^b = Q^i_1\phi_1^1\tilde{Q}^1_j+Q^i_1\phi_2^1\tilde{Q}^2_j+Q^i_1\phi_3^1\tilde{Q}^3_j +Q^i_2\phi_1^2\tilde{Q}^1_j+Q^i_2\phi_2^2\tilde{Q}^2_j+Q^i_2\phi_3^2\tilde{Q}^3_j +Q^i_3\phi_1^3\tilde{Q}^1_j+Q^i_3\phi_2^3\tilde{Q}^2_j+Q^i_3\phi_3^3\tilde{Q}^3_j$$
$$=\tilde{Q}_j^2(Q_1^i(\theta_1-i\theta_2)+Q_3^i(\theta_6+i\theta_7)+Q_2^i(-\theta_3+\frac{1}{\sqrt{3}}\theta_8))\\ +\tilde{Q}_j^3(Q_1^i(\theta_4-i\theta_5)+Q_2^i(\theta_6-i\theta_7)-Q_3^i\frac{2}{\sqrt{3}}\theta_8)\\ +\tilde{Q}^1_j(Q_2^i(\theta_1+i\theta_2)+Q_3^i(\theta_4+i\theta_5)+Q_1^i(\theta_3+\frac{1}{\sqrt{3}}\theta_8))$$
$$Q_a^i(\phi^2)_b^a\tilde{Q}_j^b = Q^i_1\phi_{11}\phi_{11}\tilde{Q}^1_j+Q^i_1\phi_{11}\phi_{12}\tilde{Q}^2_j+Q^i_1\phi_{11}\phi_{13}\tilde{Q}^3_j+ Q^i_1\phi_{12}\phi_{21}\tilde{Q}^1_j+Q^i_1\phi_{12}\phi_{22}\tilde{Q}^2_j+Q^i_1\phi_{12}\phi_{23}\tilde{Q}^3_j+ Q^i_1\phi_{13}\phi_{31}\tilde{Q}^1_j+Q^i_1\phi_{13}\phi_{32}\tilde{Q}^2_j+Q^i_1\phi_{13}\phi_{33}\tilde{Q}^3_j+ Q^i_2\phi_{21}\phi_{11}\tilde{Q}^1_j+Q^i_2\phi_{21}\phi_{12}\tilde{Q}^2_j+Q^i_2\phi_{21}\phi_{13}\tilde{Q}^3_j+ Q^i_2\phi_{22}\phi_{21}\tilde{Q}^1_j+Q^i_2\phi_{22}\phi_{22}\tilde{Q}^2_j+Q^i_2\phi_{22}\phi_{23}\tilde{Q}^3_j+ Q^i_2\phi_{23}\phi_{31}\tilde{Q}^1_j+Q^i_2\phi_{23}\phi_{32}\tilde{Q}^2_j+Q^i_2\phi_{23}\phi_{33}\tilde{Q}^3_j+ Q^i_3\phi_{31}\phi_{11}\tilde{Q}^1_j+Q^i_3\phi_{31}\phi_{12}\tilde{Q}^2_j+Q^i_3\phi_{31}\phi_{13}\tilde{Q}^3_j+ Q^i_3\phi_{32}\phi_{21}\tilde{Q}^1_j+Q^i_3\phi_{32}\phi_{22}\tilde{Q}^2_j+Q^i_3\phi_{32}\phi_{23}\tilde{Q}^3_j+ Q^i_3\phi_{33}\phi_{31}\tilde{Q}^1_j+Q^i_3\phi_{33}\phi_{32}\tilde{Q}^2_j+Q^i_3\phi_{33}\phi_{33}\tilde{Q}^3_j$$
Baryons: $t^3 \rightarrow \epsilon^{a_1a_2a_3}Q_{a_1}^{i_1}Q_{a_2}^{i_2}Q_{a_3}^{i_3}$: ${N_f \choose 3}$ dimensional
$$\epsilon^{a_1a_2a_3}Q_{a_1}^{i_1}Q_{a_2}^{i_2}Q_{a_3}^{i_3} = Q^{i_1}_1Q^{i_2}_2Q^{i_3}_3 + Q^{i_1}_2Q^{i_2}_3Q^{i_3}_1 + Q^{i_1}_3Q^{i_2}_1Q^{i_3}_2 - Q^{i_1}_3Q^{i_2}_2Q^{i_3}_1 - Q^{i_1}_1Q^{i_2}_3Q^{i_3}_2 - Q^{i_1}_2Q^{i_2}_1Q^{i_3}_3$$
Antibaryons: $\tilde{t}^3 \rightarrow \epsilon^{a_1a_2a_3}\tilde{Q}_{a_1}^{i_1}\tilde{Q}_{a_2}^{i_2}\tilde{Q}_{a_3}^{i_3}$: ${N_f \choose 3}$ dimensional
Adjoint (anti)baryons are nontrivial singlet.
: $[1,1,0,\cdots,0;0,\cdots,0]$ $\frac{1}{3}(N_f-1)N_f(N_f+1)$ dimensional
$$\epsilon^{a_1a_2b_1}Q_{a_1}^{i_1}Q_{a_2}^{i_2}\phi_{b_1}^{z_1}Q^{j_1}_{z_1} = Q_1^{i_1}Q_2^{i_2}\phi_3^{z_1}Q^{j_1}_{z_1} + Q_2^{i_1}Q_3^{i_2}\phi_1^{z_1}Q^{j_1}_{z_1} + Q_3^{i_1}Q_1^{i_2}\phi_2^{z_1}Q^{j_1}_{z_1} - Q_3^{i_1}Q_2^{i_2}\phi_1^{z_1}Q^{j_1}_{z_1} - Q_1^{i_1}Q_3^{i_2}\phi_2^{z_1}Q^{j_1}_{z_1} - Q_2^{i_1}Q_1^{i_2}\phi_3^{z_1}Q^{j_1}_{z_1}$$
$s^2t^3 \rightarrow \epsilon^{a_1b_1b_2}Q_{a_1}^{i_1}(P_1)_{b_1}^{j_1}(P_1)_{b_2}^{j_2},\ \epsilon^{a_1a_2b_1}Q_{a_1}^{i_1}Q_{a_2}^{i_2}(P_2)_{b_1}^{j_1}$
: $[1,1,0,\cdots,0;0,\cdots,0]$ $\frac{1}{3}(N_f-1)N_f(N_f+1)$ dimensional
$$\epsilon^{a_1b_1b_2}Q_{a_1}^{i_1}\phi_{b_1}^{z_1}Q^{j_1}_{z_1}\phi_{b_2}^{z_2}Q^{j_2}_{z_2} = Q_1^{i_1}\phi_2^{z_1}Q^{j_1}_{z_1}\phi_3^{z_2}Q^{j_2}_{z_2} + Q_2^{i_1}\phi_3^{z_1}Q^{j_1}_{z_1}\phi_1^{z_2}Q^{j_2}_{z_2} + Q_3^{i_1}\phi_1^{z_1}Q^{j_1}_{z_1}\phi_2^{z_2}Q^{j_2}_{z_2} - Q_3^{i_1}\phi_2^{z_1}Q^{j_1}_{z_1}\phi_1^{z_2}Q^{j_2}_{z_2} - Q_1^{i_1}\phi_3^{z_1}Q^{j_1}_{z_1}\phi_2^{z_2}Q^{j_2}_{z_2} - Q_2^{i_1}\phi_1^{z_1}Q^{j_1}_{z_1}\phi_3^{z_2}Q^{j_2}_{z_2}$$
$$\epsilon^{a_1a_2b_1}Q_{a_1}^{i_1}Q_{a_2}^{i_2}\phi_{b_1}^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} = Q_1^{i_1}Q_2^{i_2}\phi_3^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} + Q_2^{i_1}Q_3^{i_2}\phi_1^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} + Q_3^{i_1}Q_1^{i_2}\phi_2^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} - Q_3^{i_1}Q_2^{i_2}\phi_1^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} - Q_1^{i_1}Q_3^{i_2}\phi_2^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2} - Q_2^{i_1}Q_1^{i_2}\phi_3^{z_1}\phi_{z_1}^{z_2}Q^{j_1}_{z_2}$$
$s^3t^3 \rightarrow \epsilon^{abc}(P_1)_a^i(P_1)_b^j(P_1)_c^k,\ \epsilon^{a_1b_1b_2}Q_{a_1}^{i_1}(P_1)_{b_1}^{j_1}(P_2)_{b_2}^{j_2},\ \epsilon^{a_1a_2b_1}Q_{a_1}^{i_1}Q_{a_2}^{i_2}(P_3)_{b_1}^{j_1}$
: $[3,0,\cdots,0;0,\cdots,0]$ $\frac{1}{3!}N_f(N_f+1)(N_f+2)$ dimensional
$$\epsilon^{abc}\phi_a^{z_1}Q^i_{z_1}\phi_b^{z_2}Q^j_{z_2}\phi_c^{z_3}Q^k_{z_3} = \phi_1^{z_1}Q^i_{z_1}\phi_2^{z_2}Q^j_{z_2}\phi_3^{z_3}Q^k_{z_3} + \phi_2^{z_1}Q^i_{z_1}\phi_3^{z_2}Q^j_{z_2}\phi_1^{z_3}Q^k_{z_3} + \phi_3^{z_1}Q^i_{z_1}\phi_1^{z_2}Q^j_{z_2}\phi_2^{z_3}Q^k_{z_3} - \phi_3^{z_1}Q^i_{z_1}\phi_2^{z_2}Q^j_{z_2}\phi_1^{z_3}Q^k_{z_3} - \phi_1^{z_1}Q^i_{z_1}\phi_3^{z_2}Q^j_{z_2}\phi_2^{z_3}Q^k_{z_3} - \phi_2^{z_1}Q^i_{z_1}\phi_1^{z_2}Q^j_{z_2}\phi_3^{z_3}Q^k_{z_3}$$
$$\epsilon^{a_1b_1b_2}Q_{a_1}^{i_1}\phi_{b_1}^{z_1}Q^{j_1}_{z_1}\phi_{b_2}^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} = Q_1^{i_1}\phi_2^{z_1}Q^{j_1}_{z_1}\phi_3^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} + Q_2^{i_1}\phi_3^{z_1}Q^{j_1}_{z_1}\phi_1^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} + Q_3^{i_1}\phi_1^{z_1}Q^{j_1}_{z_1}\phi_2^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} - Q_3^{i_1}\phi_2^{z_1}Q^{j_1}_{z_1}\phi_1^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} - Q_1^{i_1}\phi_3^{z_1}Q^{j_1}_{z_1}\phi_2^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3} - Q_2^{i_1}\phi_1^{z_1}Q^{j_1}_{z_1}\phi_3^{z_2}\phi_{z_2}^{z_3}Q^{j_2}_{z_3}$$
$$\epsilon^{a_1a_2b_1}Q_{a_1}^{i_1}Q_{a_2}^{i_2}\phi_{b_1}^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} = Q_1^{i_1}Q_2^{i_2}\phi_3^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} + Q_2^{i_1}Q_3^{i_2}\phi_1^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} + Q_3^{i_1}Q_1^{i_2}\phi_3^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} - Q_3^{i_1}Q_2^{i_2}\phi_1^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} - Q_1^{i_1}Q_3^{i_2}\phi_2^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3} - Q_2^{i_1}Q_1^{i_2}\phi_3^{z_1}\phi_{z_1}^{z_2}\phi_{z_2}^{z_3}Q^{j_1}_{z_3}$$
When $(P_m)^i_a = \phi_a^{b_1}\phi_{b_1}^{b_2}\cdots Q_{b_m}^i$
Adjoint Barions: A Combinatorial Problems of Partitions
$$\mathcal{B}_{\alpha_1,\cdots,\alpha_{N_c}}$$
with $0\le \alpha_1 \le \cdots \le \alpha_{N_c}\le N_c$, $N_A\equiv \alpha_1 + \cdots + \alpha_{N_c}$ then $0\le N_A \le \frac{1}{2} N_c (N_c-1)$.
There is cancellation between them.
Partition function: $\mathcal{Z}_{N_c}(t) = \sum^\infty_{N_A=0} a_{N_A,N_c} t^{N_A}$
where $a_{N_A,N_c}$ is number of adjoint baryons.
Let $n_i$ be the number of slots which contain $i$ adjoint fiields. Then total number of adjoint fields is $n_1 + 2n_2 + \cdots + N_c n_{N_c}$,
$$\mathcal{Z}_{N_c}(t) = \sum_{\{n_i\}} t^{n_1 + 2n_2 +\cdots+N_c n_{N_c}} = 1/(1-t)(1-t^2)\cdots(1-t^{N_c})=\prod^{N_c}_{j=1} \frac{1}{1-t^j}$$
Then result is $\sum_{N_A=0}^\infty a_{N_A,N_c}t^{N_A} = \prod_{j=1}^{N_c} \frac{1}{1-t^j}$, then $a_{N_A,N_c} = \frac{1}{2\pi i} \oint_{\left| t \right| = 1} \frac{dt}{t^{N_A+1}}\prod_{j=1}^{N_c} \frac{1}{1-t^j}$
Asymptotic Formula
Since the upper bound of the number of adjoint fields $N_A$ is of order $N_c^2$, consider the limit of large $N_c$ and fixed ratio $x\equiv N_A/N_c^2$.
In $a_{N_A,N_c} = \frac{1}{2\pi i} \oint_{\left| t \right| = 1} \frac{dt}{t^{N_A+1}}\prod_{j=1}^{N_c} \frac{1}{1-t^j}$, the main contribution to the integral comes from $t \lesssim 1$, so let $t=1-\frac{a}{N_c}$.
Use Euler-Maclaurin formula ($\sum_{n=a}^b f(n) \sim \int_a^b f(s) ds + \frac{1}{2}(f(a)+f(b))$) to approximate $\mathcal{Z}_{N_c}(t)$.
$$\log \mathcal{Z}_{N_c}(t) = \log (\prod_{j=1}^{N_c} \frac{1}{1-t^j}) = -\sum_{j=1}^{N_c} \log (1-t^j) $$
$$\sim -\int_1^{N_c} \log (1-t^s) ds - \frac{1}{2} \log[(1-t)(1-t^{N_c})] $$
$$= -\frac{\mbox{Li}_2(t) - \mbox{Li}_2(t^{N_c})}{\log t} - \frac{1}{2} \log[(1-t)(1-t^{N_c})]$$
$$\sim N_c (\frac{\pi^2}{6a}-\frac{\mbox{Li}_2(e^{-a})}{a}) - \frac{1}{2} \log N_c + [\frac{1}{2} \log (\frac{a}{(e^a-1)(1-e^{-a})^a}) - \frac{pi^2}{12} + \frac{a}{2} -1]$$
Finally, $\mathcal{Z}_{N_c}(t) \sim \frac{C(a)}{\sqrt{N_c}} \exp [N_c (\frac{\pi^2}{6a} - \frac{\mbox{Li}_2(e^{-a})}{a})]$ where $C(a) = \sqrt{\frac{a}{(e^a-1)(1-e^{-a})^a}} \exp (-\frac{\pi^2}{12} + \frac{a}{2} - 1)$.
We use Polylogarithm function(dilogarithm) to integral $\log (1-a^x)$, with integral by substitution $a^x\equiv \beta$, $dx = \frac{1}{\log a} \frac{1}{\beta} d\beta$, then $$\int_0^t \log (1-a^x)dx=\frac{1}{\log a} \int_1^{a^t} \frac{\log (1-\beta)}{\beta} d\beta = \frac{Li(\beta}{\log a}$$
Saddle Point Method
Replace$\mathcal{Z}$.
$$\frac{1}{t^{N_A+1}} \sim \exp (-N_A \log t) \sim \exp [N_A (1-t)] = \exp (\frac{N_A}{N_c} a) = \exp (axN_c)$$.
Then $a_{N_A,N_c} \sim \frac{1}{2\pi i} \oint_\mathcal{C} f(a, N_c) e^{N_c \phi(a)} da$,
where contour $\mathcal{C}$ is small circle enclosing the origin $a=0$ in the anticlockwise direction and we define
$$f(a,N_c) = \frac{C(a)}{N_c^{3/2}},\ \phi(a) = ax + \frac{1}{a}(\frac{\pi^2}{6} - \mbox{Li}_2 (e^{-a}))$$
Deform the contour $\mathcal{C}$ to a new contour passing through the location $a_0$ of the saddle point in the direction of steepest descent. Calculate $a_0$ from the relation $\phi'(a_0)=0$:
$$x = \frac{1}{a_0^2}(\frac{\pi^2}{6}-\mbox{Li}_2(e^{-a_0}))+\frac{1}{a_0}\log(1-e^{-a_0})\equiv \xi(a_0)$$
Since $\xi(a_0)$ is transcendental, it is difficult to obtain an anayltical expression of $a_0$ in terms of $x$. However, given a numerical value of $x$, it is possible to determine the numerical value of $a_0$.
Second derivative $\phi''(a_0)>0$, so the steepest descent direction is parallel to the imaginary axis.
$$a_{N_A,N_c} \sim \frac{1}{2\pi i} \int_{a_0-i\epsilon}^{a_0+i\epsilon} da f(a_0,N_c)\exp(N_c\phi(a_0) + \frac{1}{2} N_c \phi''(a_0)(a-a_0)^2))$$
$$=\frac{1}{2\pi} e^{N_c\phi(a_0)}f(a_0,N_c)\int_{-\epsilon}^\epsilon d\alpha e^{-\frac{1}{2}N_c \phi''(a_0)\alpha^2} \quad (a=a_0+i\alpha)$$
$$\sim \frac{1}{\pi} \frac{e^{N_c\phi(a_0)}f(a_0,N_c)}{\sqrt{2N_c \phi''(a_0)}} \int_{-\infty}^\infty ds e^{-s^2} \quad (s=\alpha\sqrt{\frac{1}{2}N_c\phi''(a_0)})$$
$$\sim \frac{e^{N_c\phi(a_0)}f(a_0,N_c)}{\sqrt{2\pi N_c \phi''(a_0)}}$$
Therefore, substitute $x$ and $f(a_0,N_c)$:
$$a_{N_A,N_c} = \frac{F(a_0)}{N_c^2} \exp (N_c[\frac{2}{a_0}(\frac{\pi^2}{6} - \mbox{Li}_2(e^{-a_0}))+\log (1-e^{-a_0})])$$
where
$$F(a_0) = \frac{C(a_0)}{\sqrt{2\pi \phi''(a_0)}}$$
$$\phi''(a_0)=\frac{2}{a_0^3}(\frac{\pi^2}{6}-\mbox{Li}_2(e^{-a_0})) + \frac{2}{a_0^2}\log(1-e^{-a_0}) - \frac{1}{a_0(e^{a_0}-1)}$$
$$\phi''(a_0)=\frac{2}{a_0^3}(\frac{\pi^2}{6}-\mbox{Li}_2(e^{-a_0})) + \frac{2}{a_0^2}\log(1-e^{-a_0}) - \frac{1}{a_0(e^{a_0}-1)}$$
Applicable Topics
SMEFT:
Use of SMEFT:
Add gravity:
Use for String:
Use for Supergravity:
Use for SCFT: