BFSS$_2$

The Euclidean action of  super-Yang-Mills  BFSS$_2$  matrix quantum mechanics in two dimensions is given by

     \begin{eqnarray}S_{\rm BFSS_2}=N\int_0^{\beta} dt Tr\bigg(\frac{1}{2}(D_tX)^2-\frac{\Lambda}{2}X^2+\frac{1}{2}\psi D_t\psi+\frac{\alpha}{2}\psi[X,\psi] \bigg)~,\bar{\psi}=\psi~.    \end{eqnarray}

We recall that the two models with $\alpha=0$ and $\alpha\neq 0$ are completely equivalent. 

The lattice action (gauge-fixed + pseudo-fermions and $\Lambda(t)=\lambda={\rm constant}$) reads explicitly 

\begin{eqnarray}       S_{\rm eff}  &=&N\big[\frac{1}{a}-\frac{a}{2}\lambda\big]\sum_{n=0}^{\Lambda-1} TrX(n)X(n)-\frac{N}{a}\sum_{n=0}^{\Lambda-2} TrX(n)X(n+1)-\frac{N}{a}\sum_{i,j}e^{-i(\theta_i-\theta_j)}X_{ij}(\Lambda-1)X_{ji}(0)\nonumber\\       &-&\frac{1}{2}\sum_{i\ne j}\ln\sin^2\frac{\theta_i-\theta_j}{2}+a_0\xi^{\dagger}\xi+\sum_{\sigma=1}^Ma_{\sigma}\xi^{\dagger}\zeta_{\sigma}.          \end{eqnarray}

The coefficients $a_{\sigma}$ and $b_{\sigma}$ are obtained by running the Remez algorithm (corresponding to the rational approximation $r(x)$ of $x^{-1/4}$).

The new pseudo-fermion $\zeta_{\sigma}$ is obtained from the pseudo-fermion $\xi$ by solving (using the multi-mass conjugate gradient method) the linear system:

\begin{eqnarray}(\Delta+b_{\sigma})_{{\cal A}{\cal B}}\zeta_{\sigma}^{\cal B}=\xi^{\cal A}.\end{eqnarray}

The pseudo-fermion $\xi$ itself is obtained from the Gaussian pseudo-fermion $\eta$ by solving (using again the multi-mass conjugate gradient method) another linear system (corresponding to the rational approximation $r^{\prime}(x)$ of $x^{1/8}$) given by

\begin{eqnarray}  \xi&=&a_0^{\prime}\eta+\sum_{\sigma=1}^{M^{\prime}}a_{\sigma}^{\prime}h_{\sigma}~,~(\Delta+b_{\sigma}^{\prime})h_{\sigma}=\eta.\end{eqnarray}

The Laplacian $\Delta$ is given in terms of the Dirac operator ${\cal M}$ by

\begin{eqnarray}\Delta={\cal M}^{\dagger}{\cal M}.\end{eqnarray}    

\begin{eqnarray}  {\cal M}_{{\cal A}{\cal B}}  &=&\delta^{AB}(\hat{\delta}_{n_A+1,n_B}-\hat{\delta}_{n_A-1,n_B})\pm d^{AB}\delta_{n_B,0}\delta_{n_A,\Lambda-1}\mp d^{BA} \delta_{n_B,\Lambda-1}\delta_{n_A,0}\nonumber\\ &-&2a \alpha [t^A,t^B]_{ij}X_{ji}(n_A)\delta_{n_A,n_B}.   \end{eqnarray}