Periodic Table of SYK and supersymmetric SYK

TL;DR

This paper develops a unified random matrix theory to classify SYK and SUSY SYK models, revealing a periodic table structure of energy levels and confirming classifications through exact diagonalizations.

Contribution

It introduces a systematic classification scheme for SYK and SUSY SYK models using random matrix theory and identifies periodicity patterns in their energy spectra.

Findings

Energy level statistics match symmetry classifications

Identification of Bott periodicity in model parameters

Introduction of a new eigenvalue ratio for edge behavior

Abstract

We develop a systematic and unified random matrix theory to classify Sachdev-Ye-Kitaev (SYK) and supersymmetric (SUSY) SYK models and also work out the structure of the energy levels in one periodic table. The SYK with even $q$- and SUSY SYK with odd $q$-body interaction, $N$ even or odd number of Majorana fermions are put on the same footing in the minimal Hilbert space, $N\pmod 8$ and $q\pmod 4$ double Bott periodicity are identified. Exact diagonalizations are performed to study both the bulk energy level statistics and hard edge behaviours. A new moment ratio of the smallest positive eigenvalue is introduced to determine hard edge index efficiently. Excellent agreements between the ED results and the symmetry classifications are demonstrated. Our complete and systematic methods can be transformed to map out more complicated periodic tables of SYK models with more degree of freedoms, tensor models and symmetry protected topological phases. Possible classification of charge neutral quantum black holes are hinted.

Tables

Table 1. Table 1: The periodic table of SYK for even q > 2 𝑞 2 q>2 and SUSY SYK for odd q > 1 𝑞 1 q>1 . Only AIII in the 10-fold way is missing un . All the results in the table were achieved in a unified minimum scheme. The results for q ( mod 4 ) = 0 annotated 𝑞 pmod 4 0 q\pmod{4}=0 agree with those in You2017 ; hybrid for even N 𝑁 N and also odd N 𝑁 N in the extended scheme ( see SM C ), for q ( mod 4 ) = 1 , 2 , 3 annotated 𝑞 pmod 4 1 2 3 q\pmod{4}=1,2,3 for even N 𝑁 N agree with those in Li2017 ; Kanazawa2017 . Those for q ( mod 4 ) = 1 , 2 , 3 annotated 𝑞 pmod 4 1 2 3 q\pmod{4}=1,2,3 with odd N 𝑁 N are new. In the listed degeneracy, 1 + 1 1 1 1+1 means the two degenerate energy levels in the two opposite parities.

$N\left(mod8\right)$	0	1	2	3	4	5	6	7
$P^2$ value	$+$	$+$	$+$	$-$	$-$	$-$	$-$	$+$
$R^2$ value	$+$	$+$	$-$	$-$	$-$	$-$	$+$	$+$
$q\left(mod4\right)=0$	AI	AI	A	AII	AII	AII	A	AI
$Q_q$ degeneracy	1	1	1+1	2	2	2	1+1	1
$q\left(mod4\right)=2$	D	D	A	C	C	C	A	D
$Q_q$ degeneracy	1	1	1	1	1	1	1	1
$q\left(mod4\right)=1$	BDI	AI	CI	C	CII	AII	DIII	D
$Q_q$ degeneracy	1	1	1	1	2	2	2	1
$H$ degeneracy	2	1	2	2	4	2	4	2
$q\left(mod4\right)=3$	BDI	D	DIII	AII	CII	C	CI	AI
$Q_q$ degeneracy	1	1	2	2	2	1	1	1
$H$ degeneracy	2	2	4	2	4	2	2	1

Table 2. Table S1: Ten fold way of RMT classes, adapted from AZ ; Kitaev2009 ; four ; kane ; tenfold . T + subscript 𝑇 T_{+} ( T − subscript 𝑇 T_{-} ) denotes an anti-unitary operator that commutes (anti-commutes) with the Hamiltonian, Λ = T + ​ T − Λ subscript 𝑇 subscript 𝑇 \Lambda=T_{+}T_{-} is an unitary operator that anti-commutes with the Hamiltonian. The symbol — means such an operator does not exist. In the presence of conserved quantities, T + subscript 𝑇 T_{+} ( T − subscript 𝑇 T_{-} ) need also commute with all the compatible conserved quantities.

RMT Class	A	AI	AII	AIII	BDI	CII	C	CI	D	DIII
$T_{+}^2$	—	$+1$	$-1$	—	$+1$	$-1$	—	$+1$	—	$-1$
$T_{-}^2$	—	—	—	—	$+1$	$-1$	$-1$	$-1$	$+1$	$+1$
${\Lambda }^2$	—	—	—	$1$	$1$	$1$	—	$1$	—	$1$

Table 3. Table S2: RMT indices ( α , β ) 𝛼 𝛽 (\alpha,\beta) and the corresponding moment ratio A 𝐴 A for the 7 classes with a mirror symmetry (but no topological zero mode). A W subscript 𝐴 W A_{\text{W}} is that calculated from the exact P 1 ​ ( λ ) subscript 𝑃 1 𝜆 P_{1}(\lambda) of the 3 chiral Classes or our surmise of the 4 BdG classes. A num subscript 𝐴 num A_{\text{num}} is that calculated from the exact diagonization of corresponding random matrices of size N = 10 3 𝑁 superscript 10 3 N=10^{3} with Gaussian distributed entries, averaged over 10 5 superscript 10 5 10^{5} samples.

RMT $\alpha$	0		1			2	3
RMT $\beta$	1	2	1	2	4	2	4
Class	BDI	D	CI	AIII	DIII	C	CII
$A_{\text{W}}$	1.6018	1.5795	1.2732	1.2732	1.2732	1.1753	1.1291
$A_{\text{num}}$	1.6021	1.5832	1.2745	1.2738	1.2719	1.1735	1.1296

Table 4. Table S3: The Λ Λ \Lambda operators in the minimum scheme. There are only 3 possibilities: the conserved parity ( − 1 ) F superscript 1 𝐹 (-1)^{F} , the chirality parity ( − 1 ) F superscript 1 𝐹 (-1)^{F} and the trivial identity 1 1 1 .

$N\left(mod8\right)$	0	1	2	3	4	5	6	7
$\Lambda$	${\left(-1\right)}^F$	$1$	${\left(-1\right)}^F$	$1$	${\left(-1\right)}^F$	$1$	${\left(-1\right)}^F$	$1$
$q\left(mod2\right)=0$	$\Lambda$ conserved	trivial	$\Lambda$ conserved	trivial	$\Lambda$ conserved	trivial	$\Lambda$ conserved	trivial
$q\left(mod2\right)=1$	$\Lambda$ chirality	trivial	$\Lambda$ chirality	trivial	$\Lambda$ chirality	trivial	$\Lambda$ chirality	trivial
Any $q$	$[P,\Lambda ]=0$	$[P,\Lambda ]=0$	$\{P,\Lambda \}=0$	$[P,\Lambda ]=0$	$[P,\Lambda ]=0$	$[P,\Lambda ]=0$	$\{P,\Lambda \}=0$	$[P,\Lambda ]=0$
Any $q$	$[R,\Lambda ]=0$	$[R,\Lambda ]=0$	$\{R,\Lambda \}=0$	$[R,\Lambda ]=0$	$[R,\Lambda ]=0$	$[R,\Lambda ]=0$	$\{R,\Lambda \}=0$	$[R,\Lambda ]=0$

Table 5. Table S4: The energy level degeneracy of SYKq in the extended scheme at odd N 𝑁 N . When N 𝑁 N is even, there is no extended scheme, so we only list odd N 𝑁 N case here. When q 𝑞 q is even, 1 + 1 1 1 1+1 means the two energy levels at the two opposite parity ( − 1 ) F superscript 1 𝐹 (-1)^{F} sector. When q 𝑞 q is odd, for the supercharge Q q subscript 𝑄 𝑞 Q_{q} , it is at the two opposite parity ( − 1 ) F ​ i ​ χ ∞ superscript 1 𝐹 𝑖 subscript 𝜒 (-1)^{F}i\chi_{\infty} sector. There is always a mirror symmetry, so the degeneracy of H q subscript 𝐻 𝑞 H_{q} is always twice of Q q subscript 𝑄 𝑞 Q_{q} . How to compare this table with Table 1 in the main text is discussed in Sec. C6.

$N\left(mod8\right)$	1	3	5	7
$q\left(mod4\right)=0,Q_q$	1+1	2+2	2+2	1+1
$q\left(mod4\right)=2,Q_q$	1+1	1+1	1+1	1+1
$q\left(mod4\right)=1,Q_q$	1	1+1	2	1+1
$q\left(mod4\right)=1,H_q$	2	4	4	4
$q\left(mod4\right)=3,Q_q$	1+1	2	1+1	1
$q\left(mod4\right)=3,H_q$	4	4	4	2

Table 6. Table S5: The energy level degeneracy of 𝒩 = 1 𝒩 1 \mathcal{N}=1 supersymmetric SYK H q subscript 𝐻 𝑞 H_{q} in ( − 1 ) F superscript 1 𝐹 (-1)^{F} basis. When N 𝑁 N is even, 1 + 1 1 1 1+1 means in the two opposite parities of ( − 1 ) F superscript 1 𝐹 (-1)^{F} . When N 𝑁 N is odd, 1 + 1 = 1 + , + + 1 − , − , 1 + 1 + 1 + 1 = 1 + , + + 1 + , − + 1 − , + + 1 − , − , 2 + 2 = 2 + , + + 2 − , − formulae-sequence 1 1 subscript 1 subscript 1 formulae-sequence 1 1 1 1 subscript 1 subscript 1 subscript 1 subscript 1 2 2 subscript 2 subscript 2 1+1=1_{+,+}+1_{-,-},1+1+1+1=1_{+,+}+1_{+,-}+1_{-,+}+1_{-,-},2+2=2_{+,+}+2_{-,-} . The first of ( ± , ± ) plus-or-minus plus-or-minus (\pm,\pm) is ( − 1 ) F superscript 1 𝐹 (-1)^{F} , the second is i ​ χ ∞ ​ Q q 𝑖 subscript 𝜒 subscript 𝑄 𝑞 i\chi_{\infty}Q_{q} . It is compared to the previous two tables 1 and S4 in Sec. D6.

$N\left(mod8\right)$	0	1	2	3	4	5	6	7
$q\left(mod4\right)=1,H_q$	1+1	$1+1$	1+1	$1+1+1+1$	2+2	$2+2$	2+2	$1+1+1+1$
$q\left(mod4\right)=3,H_q$	1+1	$1+1+1+1$	2+2	$2+2$	2+2	$1+1+1+1$	1+1	$1+1$

Table 7. Table S6: The RMT classification and energy level degeneracy of complex fermion SYK models at even q 𝑞 q . The results for q ( mod 4 ) = 0 annotated 𝑞 pmod 4 0 q\pmod{4}=0 was achieved in You2017 ; hybrid . But the results for q ( mod 4 ) = 2 annotated 𝑞 pmod 4 2 q\pmod{4}=2 are new.

$N_c\left(mod4\right)$	0	1	2	3
$P^2$ value	$+$	$+$	$-$	$-$
$R^2$ value	$+$	$-$	$-$	$+$
$q\left(mod4\right)=0$ half-filling	AI	—	AII	—
degeneracy in $F=N_c/2$	1	—	2	—
non-half-filling	A	A	A	A
degeneracy in $F\ne N_c/2$	1	1	1	1
$q\left(mod4\right)=2$ half-filling	D	—	C	—
degeneracy in $F=N_c/2$	1	—	1	—
non-half-filling	A	A	A	A
degeneracy in $F\ne N_c/2$	1	1	1	1

Equations

$$Q_q=i^{\lfloor q/2\rfloor }\sum _{i_1<\cdots <i_q}{C}_{i_1\cdots i_q}{\chi }_{i_1}{\chi }_{i_2}\cdots {\chi }_{i_q}$$

$$P{Q}_q{P}^{-1}$$

$$R{Q}_q{R}^{-1}$$

$$P^2=(-1{)}^{\lfloor \lceil N/2\rceil /2\rfloor },R^2=(-1{)}^{\lfloor 3\lfloor N/2\rfloor /2\rfloor }.$$

$$P_{\text{AIII}}(\lambda )$$

$$P_{\text{BDI}}(\lambda )$$

$$P_{\text{CII}}(\lambda )$$

$$P({\lambda }_1,{\lambda }_2)\propto |{\lambda }_1^2-{\lambda }_2^2{|}^{\beta }|{\lambda }_1{\lambda }_2{|}^{\alpha }{e}^{-({\lambda }_1^2+{\lambda }_2^2)}$$

$$P_1(\lambda )={\int }_{\lambda }^{\infty }d{\lambda }_{2}P(\lambda ,{\lambda }_2)$$

$$P_{\text{D}}(\lambda )=\frac{1}{\pi }{e}^{-2{\lambda }^2}[6\lambda -4{\lambda }^3+\sqrt{\pi }{e}^{{\lambda }^2}\mathrm{erfc}(\lambda )(3-4{\lambda }^2+4{\lambda }^4)]$$

$$P_{\text{C}}(\lambda )=\frac{2}{3\pi }{e}^{-2{\lambda }^2}{\lambda }^2[30\lambda -4{\lambda }^3+\sqrt{\pi }{e}^{{\lambda }^2}\mathrm{erfc}(\lambda )(15-12{\lambda }^2+4{\lambda }^4)]$$

$$P_{\text{DIII}}(\lambda )=4\lambda e^{-2{\lambda }^2}$$

$$P_{\text{CI}}(\lambda )=4\lambda e^{-2{\lambda }^2}$$

$$\displaystyle P_{1}(\lambda_{1})=\frac{1}{(N-1)!}\Big{(}\prod_{n=2}\int_{\lambda_{1}}^{\infty}d\lambda_{n}\Big{)}P(\{\lambda\})\propto\Big{(}\prod_{n=2}\int_{\lambda_{1}}^{\infty}d\lambda_{n}\Big{)}\Big{(}\prod_{i<j}|\lambda_{i}^{2}-\lambda_{j}^{2}|^{\beta}\Big{)}\Big{(}\prod_{n=1}\lambda_{n}e^{-\lambda_{n}^{2}}\Big{)}$$

$$P_1({\lambda }_1)$$

$$\displaystyle=\lambda_{1}e^{-N\lambda_{1}^{2}}\Big{(}\prod_{n=2}\int_{0}^{\infty}|\lambda_{n}^{2}|^{\beta}e^{-\lambda_{n}^{2}}d\lambda_{n}^{2}\Big{)}\Big{(}\prod_{1<i<j}|\lambda_{i}^{2}-\lambda_{j}^{2}|^{\beta}\Big{)}=C_{N,\beta}\lambda_{1}e^{-N\lambda_{1}^{2}}$$

$$\displaystyle A=\frac{\langle\lambda_{\text{min}}^{2}\rangle}{\langle\lambda_{\text{min}}\rangle^{2}}=\frac{\int_{0}^{\infty}d\lambda\lambda^{2}P(\lambda)}{\Big{[}\int_{0}^{\infty}d\lambda\lambda P(\lambda)\Big{]}^{2}}\geq 1$$

$$c_i=({\chi }_{2i}-i{\chi }_{2i-1})/2,c_i^{†}=({\chi }_{2i}+i{\chi }_{2i-1})/2,i=1,\cdots N_c=(N+1)/2$$

$$P=K\prod _{i=1}^{N_c}(c_i^{†}+c_i),R=K\prod _{i=1}^{N_c}(c_i^{†}-c_i)$$

$$P=Z{\chi }_{\infty },Z=P{\chi }_{\infty }=K\prod _{i=1}^{N_c-1}(c_i^{†}+c_i).$$

$$P{\chi }_i{P}^{-1}=(-1{)}^{N_c-1}{\chi }_i,$$

$$P{\chi }_{\infty }{P}^{-1}=(-1{)}^{N_c-1}{\chi }_{\infty },$$

$$P{Q}_q{P}^{-1}=(-1{)}^{\lfloor q/2\rfloor }(-1{)}^{q(N_c-1)}{Q}_q,R{Q}_q{R}^{-1}=(-1{)}^{\lfloor q/2\rfloor }(-1{)}^{q{N}_c}{Q}_q,Z{Q}_q{Z}^{-1}=(-1{)}^{\lfloor q/2\rfloor }(-1{)}^{q{N}_c}{Q}_q,$$

$$P^2=(-1{)}^{\lfloor N_c/2\rfloor },R^2=(-1{)}^{\lfloor 3N_c/2\rfloor },Z^2=(-1{)}^{\lfloor (N_c-1)/2\rfloor }.$$

$$P(-1{)}^F{P}^{-1}=(-1{)}^{N_c}(-1{)}^F,R(-1{)}^F{R}^{-1}=(-1{)}^{N_c}(-1{)}^F,Z(-1{)}^F{Z}^{-1}=(-1{)}^{N_c-1}(-1{)}^F$$

$$PR=i^{N_c}{\chi }_2{\chi }_4\cdots {\chi }_{2N_c}{\chi }_1{\chi }_3\cdots {\chi }_{2N_c-1},(-1{)}^F=(i{\chi }_1{\chi }_2)(i{\chi }_3{\chi }_4)\cdots (i{\chi }_{2N_c-1}{\chi }_{2N_c})$$

$$P(-1{)}^{F}i{\chi }_{\infty }{P}^{-1}=+(-1{)}^F{\chi }_{\infty },R(-1{)}^{F}i{\chi }_{\infty }{R}^{-1}=-(-1{)}^F{\chi }_{\infty },Z(-1{)}^{F}i{\chi }_{\infty }{Z}^{-1}=-(-1{)}^F{\chi }_{\infty }$$

$$ZiZR{Z}^{-1}=-iZR,RiZR{R}^{-1}=-iZR,PiZR{P}^{-1}=iZR$$

$$PiZP{Q}_q{P}^{-1}=(-1{)}^{\lfloor q/2\rfloor +1}(-1{)}^{(q+1)(N_c-1)}iZP{Q}_q,$$

$$ZiZP{Q}_q{Z}^{-1}=(-1{)}^{\lfloor q/2\rfloor }(-1{)}^{(q+1)N_c}iZP{Q}_q,$$

$$PiZP{Q}_q{P}^{-1}=-iZP{Q}_q,$$

$$PiZP{Q}_q{P}^{-1}=+iZP{Q}_q,$$

$$H_c=\sum J_{i_1{i}_2\cdots i_{q/2};i_{q/2+1}\cdots i_q}{c}_{i_1}^{†}{c}_{i_2}^{†}\cdots c_{i_{q/2}}^{†}{c}_{i_{q/2+1}}\cdots c_{i_q}-\mu F$$

$$P{c}_i{P}^{-1}=(-1{)}^{N_c-1}{c}_i^{†},P{c}_i^{†}{P}^{-1}=(-1{)}^{N_c-1}{c}_i,R{c}_i{R}^{-1}=(-1{)}^{N_c}{c}_i^{†},R{c}_i^{†}{R}^{-1}=(-1{)}^{N_c}{c}_i$$

Full text

1. Introduction.— Extensive research efforts are currently being devoted to investigate quantum chaos and quantum information scramblings in Sachdev-Ye-Kitaev (SYK) model Sachdev-Ye ; Georges2001 ; KitaevTalks ; JT-2 ; Sachdev2015 ; Ye and its supersymmetric generalization Fu2017 . A salient feature of the SYK models is that it is quantum chaotic in both early time (Ehrenfest time) Polchinski2016 ; Maldacena2016 ; Gross2017 ; Qi ; Bagrets2016 ; liu1 ; liu2 and late time (Heisenberg time) Fu2016 ; You2017 ; Cotler2017 ; Li2017 ; Kanazawa2017 ; Garcia2016 ; Garcia2018 ; hybrid .

In the early time, an exponential grow in the out of time correlation functions can be used to characterize the quantum information scramblings. It was found that the SYK models show the maximal quantum chaos with the largest possible Lyapunov exponent $\lambda_{L}=2\pi/\beta$ saturating the quantum chaos bound chaosbound . This feature ties that of the quantum black holes which are the fastest quantum information scramblers in Nature. This remarkable feat suggests that the SYK models may be a boundary theory of some sort of bulk dilaton gravity theory such as the well known 2d Jackiw-Teitelboim gravity JT-1 ; JT-2 .

In the late time, despite its sparse nature in randomness, the SYK models can still be described by random matrix theory (RMT) Fu2016 ; You2017 ; Cotler2017 ; Li2017 ; Kanazawa2017 ; Garcia2016 ; Garcia2018 ; hybrid ; caution . The RMT has also been employed to study the quantum chaotic behaviours of event horizon fluctuations of black holes Cotler2017 . The RMT classifications of the SYK models have been applied to $q\pmod{4}=0$ for all $N$ You2017 ; Cotler2017 ; Garcia2016 ; hybrid and $q\pmod{4}=1,2,3$ for even $N$ case Li2017 ; Kanazawa2017 only. It was generally believed that the odd $N$ case need to be treated quite differently than the even $N$ case. In fact, $N$ odd case was first discussed in the context of the 1d SPT phase Fidkowski2011 , and then applied to the SYK model with $q=4$ in You2017 ; hybrid . The authors developed a special procedure to treat $N$ odd case by adding an extra Majorana fermion at infinity into the system. While such a procedure is not necessary in the even $N$ case.

In this work, by employing the Clifford algebra representation of Majorana fermions in a minimal Hilbert space, we develop a systematic and unified random matrix theory to classify SYK models with generic $q$-body interaction and general $N$-sites. After identifying complete set of conserved (or chiral) quantities and anti-unitary (or unitary) symmetry operators, we also work out the fine structure of the energy levels explicitly. The SYK with even $q$ and SUSY SYK with odd $q$, $N$ even or odd are treated on the same footing. There are 9 classes, except class AIII(chGUE) with $N\pmod{8}$ and $q\pmod{4}=0$ double Bott periodicity identified in one periodic table 1. Our systematic approach not only reproduces the previously known results in a much more efficient way, but also lead to new results on $q\pmod{4}=1,2,3$ with odd $N$, thus complete the whole periodic table for the SYK models. Therefore it provides additional deep and global insights into the family of SYK models showing maximal quantum chaos.

For the 7 RMT classes with a spectral mirror symmetry, we derive a Wigner-like surmise for the probability function for the lowest positive energy level, which can be used to describe the quantum chaotic behaviours near the hard edge. A new moment ratio $A$ ( See SM-A ) is introduced to determine the hard edge exponent effectively. When combining with the $r$-parameter Atas2013 ; You2017 ; hybrid used to determine the bulk energy level statistics (ELS), the parameter pair $(r,A)$ can be used to determine the 10-fold way uniquely and efficiently. Exact diagonalizations (EDs) are performed to study both the bulk ELS and hard edge behaviours. Excellent agreements are found between the ED results and the symmetry classifications. We also establish intricate and constructive connections between the unified minimum scheme and various other extended schemes in both classification and degeneracy counting in odd $N$ case. Our unified scheme and the new moment ratio $A$ can be easily transformed to study other SYK-like models with more symmetries, tensor models and classifications of SPT phases.

2. A unified and systematic scheme to classify Majorana SYK models.— Consider a generic all-to-all $q$-body Majorana interacting Hamiltonian

[TABLE]

where $C_{i_{1}\cdots i_{q}}$ is real and satisfy the Gaussian distributions with mean $\langle C_{i_{1}\cdots i_{q}}\rangle=0$ and variance $\langle C_{i_{1}\cdots i_{q}}^{2}\rangle=J_{\text{SYK}}^{2}(q-1)!/N^{q-1}$ for even $q$ or $\langle C_{i_{1}\cdots i_{q}}^{2}\rangle=J_{\mathcal{N}=1}(q-1)!/N^{q-1}$ for odd $q$. The overall factor $i^{\lfloor q/2\rfloor}$ ensures the Hermitian of $Q_{q}$, and $\lfloor q/2\rfloor$ (integer floor) denotes the biggest integer smaller than $q/2$. When $q$ is even, $Q_{q}$ is a bosonic operator which is the SYK Model with $q$-fermion interactions KitaevTalks ; JT-2 ; Maldacena2016 ; when $q$ is odd, $Q_{q}$ is a fermionic operator which is the supercharge of $\mathcal{N}=1$ supersymmetric SYK model Fu2017 ; Li2017 ; Kanazawa2017 . Below we choose $J_{\text{SYK}}=1$ and $J_{\mathcal{N}=1}=1$ when performing numerical calculations.

The Hamiltonian Eq.(1) contains $N$ Majorana fermions satisfying the Clifford algebra $\{\chi_{i},\chi_{j}\}=2\delta_{ij}$ and $\chi_{i}^{\dagger}=\chi_{i}$, thus admitting a $2^{\lfloor N/2\rfloor}$-dimensional matrix representations Clifford . In this representation, one can choose $\chi_{i}$ with odd $i$ to be real and symmetric, and $\chi_{i}$ with even $i$ to be pure imaginary and skew-symmetric footnote1 . By collecting real and imaginary represented Majoranas, one can define two anti-unitary particle-hole symmetry operators: $P=K\prod_{i=1}^{\lceil N/2\rceil}\chi_{2i-1}$ and $R=K\prod_{i=1}^{\lfloor N/2\rfloor}i\chi_{2i}$, where $\lceil N/2\rceil$ (integer ceiling) denotes the smallest integer greater than $N/2$, therefore $\lfloor N/2\rfloor+\lceil N/2\rceil=N$ holds for any integer $N$.

It can be shown that $P\chi_{i}P^{-1}=-(-1)^{\lceil N/2\rceil}\chi_{i}$ and $R\chi_{i}R^{-1}=(-1)^{\lfloor N/2\rfloor}\chi_{i}$, thus

[TABLE]

One can also find their squared values

[TABLE]

Multiplying two equations in Eq.(2) leads to the following unified classification: When $q(1+N)$ is odd, one of $P$, $R$ commutes with $Q_{q}$ and the other anti-commutes with $Q_{q}$, thus $\Lambda=PR$ is a chirality operator satisfying $\{\Lambda,Q_{q}\}=0$; when $q(1+N)$ is even, $P$ and $R$ either both commute with $Q_{q}$ or both anti-commute with $Q_{q}$, thus $\Lambda=PR$ is a conserved quantity (fermion number parity) satisfying $[\Lambda,Q_{q}]=0$ if $N$ is even, or an identity operator if $N$ is odd.

In the above case when $\Lambda$ is a conserved parity, namely, when $(q,N)=$ (even, even), the Hamiltonian can be block-diagonal decomposed into even and odd parity sectors. Using the fact $PR=(-1)^{\lceil N/2\rceil\lfloor N/2\rfloor}RP$, one can find out the relation between the anti-unitary operators and the conserved parity: $[P,\Lambda]=[R,\Lambda]=0$ if $N\pmod{8}=0,4$, which means $P$ and $R$ preserve the parity; $\{P,\Lambda\}=\{R,\Lambda\}=0$ if $N\pmod{8}=2,6$, which means $P$ and $R$ swap the parity. If $\Lambda$ is a chirality operator, namely, when $(q,N)=$ (odd, even), which is also a parity operator, then there is no need to worry about these commutation relations. Of course, if $\Lambda=1$, namely, when $(q,N)=$ (odd, odd) or $(q,N)=$ (even, odd), then it can be ignored anyway.

The classification can be systematically done by evaluating Eq.(2), Eq.(3) and their relations with $\Lambda$ presented above. The periodicity of the classification can be directly inferred from that of Eq.(2) and Eq.(3) which are invariant under $q\to q+4$ or $N\to N+8$. So the Bott periodicity in $q$ and in $N$ is $4$ and $8$ respectively. So in the following, one need only apply the scheme to $q\pmod{4}$ and $N\pmod{8}$ cases.

3. SYK models with $q\pmod{4}=0$.— Since $q(1+N)$ is even, $[P,Q_{q}]=[R,Q_{q}]=0$, $\Lambda=(-1)^{F}$ is a conserved parity when $N$ is even or $\Lambda=1$ when $N$ is odd. Depending on the commutation relation between $P$ and $\Lambda$, the value of $P^{2}$, one reach the following classification: When $N\pmod{8}=2,6$, both $P$ and $R$ swap the parity, thus $Q_{q}$ is in Class A(GUE), the energy level degeneracy is $1+1$ which means there are two degenerate energy levels with opposite parities. When $N\pmod{8}\neq 2,6$, either $P$ preserves the parity ( $N\pmod{8}=0,4$ ) or no conserved quantity ( $N$ is odd ), thus $P^{2}=+1$ means $Q_{q}$ belongs to Class AI(GOE) with no degeneracy, $P^{2}=-1$ means $Q_{q}$ belongs to Class AII(GSE) with double degeneracy. The classification and the energy level degeneracy are summarized in the 4th and 5th row of Table 1 respectively

All these results for both even and odd $N$ have been achieved before in You2017 ; hybrid in the enlarged Hilbert space by adding an extra fermions at $\infty$ when $N$ is odd. However, here we achieved the same results in the minimal Hilbert space in both even and odd $N$ in a unified scheme. Our new method maybe the most powerful and convenient way to achieve new results as shown in the following cases.

4. SYK models with $q\pmod{4}=2$.— Since $q(1+N)$ is even, two operators always anti-commute with Hamiltonian $\{P,Q_{q}\}=\{R,Q_{q}\}=0$, and $\Lambda=PR$ is the conserved parity if $N$ is even or identity operator if $N$ is odd. Depending on commutation relation between $P$ and $\Lambda$, the value of $P^{2}$, one find the following classification: When $N\pmod{8}=2,6$, both operators swap the parity, thus $Q_{q}$ is in Class A(GUE); when $N\pmod{8}\neq 2,6$, either $P$ preserves parity ( $N\pmod{8}=0,4$ ) or no conserved quantity ( $N$ is odd ), thus $P^{2}=+1$ means $Q_{q}$ belongs to Class D(BdG) and $P^{2}=-1$ means $Q_{q}$ belongs to Class C(BdG). There is no degeneracy in all these cases. The classification and the energy level degeneracy is listed in the 6th and 7th row of Table 1 respectively.

The classifications and degeneracy with even $N$ are consistent with those in Kanazawa2017 . However, our results with odd $N$ are new. We present an ED study of ELS for $q=6$ SYK model with $N=17,19,21,23$ in Fig.1. The random matrix indices $\beta$ and $\alpha$ are extracted from the probability distribution function $P(r)$ and $P(\lambda_{1})$, or $\langle\tilde{r}\rangle$ and $A$ respectively. Both methods lead to $(\beta,\alpha)=(2,0),(2,2),(2,2),(2,0)$ for $N=17,19,21,23$ respectively. We also checked the degeneracy is always 1. These results match the classification and degeneracy listed in Table I.

5. SYK supercharge with $q\pmod{4}=1$.— Now $q(1+N)$ is odd for even $N$ and even for odd $N$, the former leads to the chiral operator $\Lambda=(-1)^{F}$, the latter leads to $\Lambda=1$, so neither will lead to a conserved quantity. In the following, we discuss even $N$ and odd $N$ respectively.

When $N\pmod{8}=0,4$, $\{P,Q_{q}\}=[R,Q_{q}]=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we find: $N\pmod{8}=0$ is class BDI(chGOE), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy; $N\pmod{8}=4$ is class CII(chGSE), thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $H=Q_{q}^{2}$ has 4-fold degeneracy. When $N\pmod{8}=2,6$, $[P,Q_{q}]=\{R,Q_{q}\}=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=2$ is class CI(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy; $N\pmod{8}=6$ is class DIII(BdG), thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $H=Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=1,5$, $[P,Q_{q}]=[R,Q_{q}]=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=1$ is Class AI(GOE), thus $Q_{q}$ has no degeneracy, no mirror symmetry either, so $H=Q_{q}^{2}$ has no degeneracy. $N\pmod{8}=5$ is Class AII(GSE), thus $Q_{q}$ has double degeneracy and no mirror symmetry, so $H=Q_{q}^{2}$ have 2-fold degeneracy. When $N\pmod{8}=3,7$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=3$ is Class C(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy. $N\pmod{8}=7$ is Class D(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy.

The classification for $q\pmod{4}=1$ and the energy level degeneracy are summarized in the 8th, 9th (for $Q$) and 10th row (for $H=Q^{2}$) of Table 1 respectively. The results with even $N$ are consistent with those in Kanazawa2017 . However, the results with odd $N$ are new. We present an ED study of energy level statistics for $q=5$ SYK supercharge with $N=17,19,21,23$ in Fig.2. Since the index $\alpha$ is only defined for the cases with mirror symmetry, when a mirror symmetry is absent, we plot the spectral density averaged from $1$ sample and $10^{4}$ samples. It shows the mirror symmetry is absent for every single realization of $C_{i_{1},\cdots,i_{q}}$, but still emerges after making disorder averages. The random matrix indices $\beta$ and $\alpha$, if exist, are extracted from probability distribution $P(r)$ and $P(\lambda_{1})$ or $\langle\tilde{r}\rangle$ and $A$. Both methods lead to $(\beta,\alpha)=(1,-),(2,2),(4,-),(2,0)$ for $N=17,19,21,23$ respectively. Both the indices $(\alpha,\beta)$ and the energy level degeneracy match our classification and degeneracy listed in Table I.

6. SYK supercharge with $q\pmod{4}=3$.— Now $q(1+N)$ is odd for even $N$ and even for odd $N$. The former leads to the chiral operator $\Lambda=(-1)^{F}$, the latter leads to $\Lambda=1$, so neither will lead to conserved quantity. In the following, we discuss even $N$ and odd $N$ respectively.

When $N\pmod{8}=0,4$, $[P,Q_{q}]=\{R,Q_{q}\}=0$, From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=0$ is class BDI(chGOE), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy; $N\pmod{8}=4$ is class CII(chGSE), thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy. When $N\pmod{8}=2,6$, $\{P,Q_{q}\}=[R,Q_{q}]=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=2$ is class DIII(BdG), thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $H=Q_{q}^{2}$ has 4-fold degeneracy; $N\pmod{8}=6$ is class CI(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=1,5$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=1$ is Class D(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy; $N\pmod{8}=5$ is Class C(BdG), thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ have 2-fold degeneracy. When $N\pmod{8}=3,7$, $[P,Q_{q}]=[R,Q_{q}]=0$. From the values of $P^{2}$ and $R^{2}$ in Table 1, we obtain: $N\pmod{8}=3$ is Class AII(GSE), thus $Q_{q}$ has double degeneracy, but no mirror symmetry, so $H=Q_{q}^{2}$ has 2-fold degeneracy; $N\pmod{8}=7$ is Class AI(GOE), thus $Q_{q}$ has no degeneracy and no mirror symmetry either, so $H=Q_{q}^{2}$ have no degeneracy.

The classification for $q\pmod{4}=3$ and the energy level degeneracy are summarized in the 11th, 12th and 13th row of Table 1 respectively. Classifications and degeneracy with even $N$ are consistent with Kanazawa2017 . However, or results with odd $N$ are new. We present an ED study of $q=3$ SYK supercharge for $N=17,19,21,23$ in Fig.3. These data show $(\beta,\alpha)=(2,0),(4,-),(2,2),(1,-),$ for $N=17,19,21,23$ respectively. Both the indices $(\alpha,\beta)$ and the energy level degeneracy match our classification and degeneracy listed in Table I.

7. Discussion and Summary.— In the $1/N$ expansion of the SYK model Polchinski2016 ; Maldacena2016 ; Gross2017 ; Qi and SUSY SYK model Fu2017 with any $q$-body interaction, a $1/q$ expansion follows which can be used to derive 2- or 4-point functions at any $\beta J$ scale. At a double scaling limit $N\rightarrow\infty$, $q\rightarrow\infty$, but keeps $q^{2}/N$ at a fixed ratio, one may map the effective action to a Liouville conformal field theory Bagrets2016 ; liu1 ; liu2 ; Cotler2017 . Here, we put SYK with even $q$ and SUSY SYK with odd $q$, $N$ even or odd on the same footing and find the $N\pmod{8}$ and $q\pmod{4}=0$ double Bott periodicity. Unfortunately, the double Bott periodicity can not be seen in the $1/N$, $1/q$ expansion or the double scaling limit. This maybe due to the huge time scale separations between the $1/N$ expansion method and the RMT classifications outlined in the introduction. Indeed, the Lyapunov exponent can be extracted by evaluating out of time ordered correlation (OTOC) function in the large $N$ limit where there is a wide separation between the dissipation time $t_{d}\sim 1/\beta$ and the scrambling time $t_{s}\sim t_{d}\log N$. At any finite size $N=23$ in Fig.1-3, this window may be still to narrow to extract the Lyapunov exponent Fu2016 . Non-perturbative effects maybe involved to see the double Bott periodicity in the $1/N$ expansion. The reason why one can put SYK and SUSY SYK in the same periodic table maybe also due to the fact that the ${\cal N}=1$ SUSY is broken at any finite $N$, but only non-perturbatively. Indeed, the ground state energy is lifted from zero to $e^{-N}$ which is in accessible to the perturbative $1/N$ expansion. So one expect this SUSY breaking at any finite $N$ is a non-perturbative instanton tunneling effects. Similarly, in the perturbative $1/N$ expansion of the Dicke model KAM in the super-radiant phase, one must consider the non-perturbative instanton tunneling events to find the $e^{-N}$ splitting between the two states with opposite parities.

It is interesting to make an analogy to the periodic tables of TI and TSC first put forward in Kitaev2009 ; four . Despite using the same Cartan labels and same sets of anti-unitary operators, it is on topological equivalent classes in Bott periodicity in the space dimension $d\pmod{8}$. It is also for a gapped bulk states with gapless edge modes. While ours is for gapless quantum spin liquids whose quantum chaos can be characterized by bulk energy level statistics and edge exponents. Later, adding more symmetries such as translational symmetries, point group symmetrise or non-isomorphic symmetries lead to more rich periodic tables for topological crystalline symmetries tenfold . However, for many body interacting electron systems, the anti-unitary operators are not enough, the SPT or SET phases maybe classified by using more advanced mathematical tools such as co-homology, cobordism and tensor categories tenfold ; wen which were already used in rational conformal field theory (RCFT) and also topological quantum computing.

The most compact, systematic and complete classification method developed in this work can be applied to all kinds of generalizations of Majorana SYK models such as the Brownian SYK Brown , the colored-SYK model Gross2017 ; color , two indices SYK model Ye or other Majorana SYK-like model with global $O(M),U(M)$ symmetry Qi ; Yoon . This is just putting more symmetries lead to more conserved quantities, also more operators, therefore generating more complicated periodic tables ( for the color degree of freedoms, see hybrid , for $U(1)$ complex SYK or ${\cal N}=1$ SUSY SYK, see SM section E ). Its advantages over the extended schemes ( see SM ) may be more evident and dramatic as the periodic tables get more rich and complicated. It may also be applied to study quantum chaos in the colored (Gurau-Witten) Tensor model tensor1 ; tensor2 or uncolored Klebanov-Tarnopolsky tensor model tensorrev . A new moment ratio $A$ proposed in this work is a powerful and efficient tool to identify the edge exponent in the 3 chiral Classes and 4 BdG Classes. It may also be used to characterize the edge transition between different random matrix ensembles, such as the edge transitions in Hybrid ${\cal N}=1$ SUSY SYK model un .

The $q=4$ SYK models may be experimentally realized in various cold atom and solid state systems cold ; solid . The periodic table tells that an isolated $q=4$ SYK island should show sensitive even/odd effects in the number of particles inside the island which is an interesting mesoscopic quantum phenomena. Due to its quantum chaotic nature, the SYK islands always satisfy the Eigenstate Thermalization Hypothesis (ETH), therefore is an equilibrium system with a well defined temperature ( so is a black hole ). The number of particles $N$ in an isolated island can be easily adjusted, so the $N\pmod{8}$ Bott periodicity may be observed experimentally. However, it is experimentally more challenging to adjust $q$-body interaction to observe $q\pmod{4}$ Bott periodicity. The Coulomb interaction in a solid system automatically leads to $q=4$ body interaction. The $q=6$ body interaction and the SUSY SYK with an odd $q$-body such as $q=3$ interaction maybe realized in cold atom systems.

The classical black holes can be classified just by 3 classical quantities the mass $M$, charge $Q$ and the angular momentum $L$ ( No hair theorem ). So far, the charge neutral$Q=0$ quantum black holes are only discussed in the AI (GUE) class Cotler2017 ; Brown ; genus . It remains unknown if the periodic Table 1 can be applied to the JT gravity side JT-1 ; JT-2 and its genus expansion genus . It remains to be seen if it also implies the classifications of quantum natures of charge neutral black holes.

We thank S Shenker for helpful discussions. We acknowledge AFOSR FA9550-16-1-0412 for supports. This research at KITP was supported in part by the National Science Foundation under Grant No. NSF PHY-1748958.

I A. Ten-fold way, bulk energy level statistics and edge exponents

In the pioneering works Wigner ; Dyson , Wigner and Dyson classified 3 symmetry classes by an anti-unitary operator $T_{+}$, i.e. time-reversal symmetry operator, which commutes with Hamiltonian. In the absence of $T_{+}$, the symmetry class is A(GUE); while $T_{+}^{2}=+1$, it is AI(GOE) and $T_{+}^{2}=-1$, it is AII(GSE). These 3 symmetry classes also known as 3 Wigner-Dyson classes. Later, 7 additional symmetry classes were found by considering another anti-unitary operator $T_{-}$ and an unitary operator $\Lambda$ which anti-commutes with Hamiltonian. All the 10 symmetry classes AZ can be completely determined by squared values of $T_{+}^{2}$, $T_{-}^{2}$ and $\Lambda^{2}$. Dropping the $T_{-}$ and $\Lambda$ operators recovers the original 3 Wigner-Dyson classes. The symmetry classification scheme also applies to topological insulators and superconductors Kitaev2009 ; four ; tenfold in the context of topological equivalent classes. To be self-contained, we list the 10 fold way of RMT in the Table S1 which was heavily consulted in the main text and in the SM.

In the presence of conserved quantities color , one need to identify all the anti-unitary operators which preserve all the compatible conserved quantities. The classification need be applied to the each block of the Hamiltonian.

**A1. Bulk Energy level statistics and bulk index $\beta$. **

For the 3 Wigner-Dyson classes, no mirror symmetry exists in the energy levels, the joint probability distribution for all the eigen-values can be described by $P(\{\lambda_{i}\})\propto\prod_{i<j}|\lambda_{i}-\lambda_{j}|^{\beta}\prod_{n}e^{-\lambda_{n}^{2}}$, where $\beta$ is the Wigner-Dyson index characterizing the strength of level repulsion. For class A(GUE), AI(GOE), AII(GSE), $\beta=2,1,4$ respectively.

Among the 7 additional classes: AIII(chGUE), BDI(chGOE), CII(chGSE), C(BdG), D(BdG), CI(BdG), DIII(BdG) which owns a mirror symmetry, the first three are chiral ensembles relevant to the systems with Dirac fermions such as QCD; the last four are BdG class, arising via Bogolioubov-de Gennes (BdG) mean field approximation of superconductors. Due to the mirror symmetry, every energy level appears in pair $\pm\lambda$ and the joint probability density distributions for all the eigen-values can be described by $P(\{\lambda_{i}\})\propto\prod_{i<j}|\lambda_{i}^{2}-\lambda_{j}^{2}|^{\beta}\prod_{n}|\lambda_{n}|^{\alpha}e^{-\lambda_{n}^{2}}$, where $\alpha$ is the edge exponent characterizing the repulsion from “hard edge” at $\lambda=0$. (On the contrary, the soft edge is referred to the spectral edge at the maxima $|\lambda|$.) The $\alpha$ index of the 7 classes with the mirror symmetry is listed in Table S2 . Note that the $\alpha$ index is not defined for the 3 original Wigner-Dyson classes due to the lacking of mirror symmetric, thus no special point in the spectrum after unfolding procedure.

In performing exact Diagonizations (ED), unfolding procedure is needed when comparing numerical data with RMT predictions, so it may not be convenient to directly use $P(\{\lambda_{i}\})$. To avoid this kind of procedures, a universal ratio $r_{n}=(\lambda_{n+2}-\lambda_{n+1})/(\lambda_{n+1}-\lambda_{n})$ of the consecutive energy level spacings can be constructed to describe the bulk energy level statistics. Due to the cancelation on the dependence on the density of states in the ratio, the unfolding procedure can be avoided. By performing the exact calculation for $n=3$ case, the authors in Ref.Atas2013 derived a Wigner-like surmises for the distribution of $r_{n}$: $P_{W}(r)\propto(r+r^{2})^{\beta}/(1+r+r^{2})^{1+3\beta/2}$. The level repulsion is reflected in the asymptotic behaviour $P(r)\sim r^{\beta}$ at small $r$. The most efficient way to extract $\beta$ index is to study the expectation $\langle r\rangle$ or its cousin $\langle\tilde{r}\rangle$ with $\tilde{r}_{n}=\min(r_{n},1/r_{n})$. It was documented that $\langle\tilde{r}\rangle=0.5359,0.6027,0.6762$ for $\beta=1,2,4$ respectively. Thus, $\langle\tilde{r}\rangle$-parameter was also used to determine transitions between random matrix ensembles with different $\beta$ indexes hybrid . However, in the presence of energy level degeneracy in the same conserved quantity sector, the ratio of next nearest neighbor energy level spacing $r^{\prime}_{n}=(\lambda_{n+2}-\lambda_{n})/(\lambda_{n}-\lambda_{n-2})$, its expectation value $\langle r^{\prime}\rangle$ or its cousin $\langle\tilde{r}^{\prime}\rangle$ with $\tilde{r}^{\prime}_{n}=\min(r^{\prime}_{n},1/r^{\prime}_{n})$ must be used to characterize the transitions. The values of $\langle r^{\prime}\rangle$ and $\langle\tilde{r}^{\prime}\rangle$ for $\beta=1,2,4$ are documented in hybrid .

**A2. Wigner surmises and a new moment ratio for the 7 classes with a mirror symmetry to determine the edge exponent $\alpha$. **

For the seven classes with a mirror symmetry, in addition to the bulk energy level statistics, one may also need to focus on the universality near the hard edge $\lambda=0$. One way is to integrate out all the energy levels in $P(\{\lambda_{i}\})$ except the smallest positive one which result in the probability distribution function of the smallest positive eigenvalues $P_{1}(\lambda)$. The probability distribution function of $k$-th positive eigenvalues $P_{k}(\lambda)$ can be similarly obtained. Exact results of $P_{k}(\lambda)$ were known for the three chiral classes describing quarks in QCD rmt-1 ; rmt-2 . Here, we only list the exact results of distribution function of the smallest positive eigenvalue (denotes as $\lambda$) for the three chiral classes, AIII(chGUE), BDI(chGOE) and CII(chGSE):

[TABLE]

where $I_{n}(\lambda)$ is the Modified Bessel functions of the first kind.

However, the exact results of distribution function of the smallest positive eigenvalue of 4 BdG classes, C, D, CI, and DIII, are not known yet. Here, instead of deriving the exact results, we propose Wigner surmises for the distribution function of the 4 BdG class. This method was proved to be very accurate for Hermitian and non-Hermitian chiral random matrices ensemble to describe QCD WS . Here, we will show it can be used to determine the edge exponent $\alpha$ very effectively also for the 4 BdG classes.

Since $P_{1}(\lambda)$ describes the energy level closest to the hard edge, it shows universal RMT behaviours. Other $P_{k}(\lambda),k>1$ can be treated in a similar way, but it is believed that $P_{k>1}$ is less universal than $P_{1}(\lambda)$. This is also reflected in the fact that $P_{k}$ with large $k$ may show large deviations when compared with numerical data. So in the following, we will mainly focus on $P_{1}(\lambda)$. Since $\lambda$ has dimension of energy, thus a rescaling $P_{1}(\lambda)\to cP_{1}(c\lambda)$ may be needed when comparing with numerical data. As shown below, typically, one choose $c\int d\lambda\lambda P_{1}(\lambda)=\langle\lambda_{\text{min}}\rangle$ to fix the rescaling factor which need to be the same for all $P_{k}$.

Starting from the joint probability density for the energy levels with $n=2$

[TABLE]

where $\propto$ means the normalization constant is ignored. The distribution function of smallest positive eigenvalue can be obtained from

[TABLE]

which can be done analytically.

For BdG class D, we need choose $\beta=2$, $\alpha=0$ and obtain

[TABLE]

where $\operatorname{erfc}(\lambda)$ is the complementary error function. In order to use $P_{\text{D}}(\lambda)$ to compare with numerical data, we need to rescale $P_{\text{D}}(\lambda)\to\tilde{P}(\tilde{\lambda})=cP_{\text{D}}(c\lambda)$ such that $\int_{0}^{\infty}d\tilde{\lambda}\tilde{\lambda}\tilde{P}(\tilde{\lambda})=\langle\lambda_{\min}\rangle$ with $\tilde{\lambda}=c\lambda$. This requirement sets the scale factor to be $c=\frac{7-4\sqrt{2}}{2\sqrt{\pi}\langle\lambda_{\min}\rangle}$ where $\langle\lambda_{\min}\rangle$ is the mean value of the smallest positive eigenvalue from the numerical data.

For BdG class C, we need choose $\beta=2$, $\alpha=2$ and obtain

[TABLE]

In order to use $P_{\text{C}}(\lambda)$ to compare with numerical data, we need again to rescale $P_{\text{C}}(\lambda)\to\tilde{P}(\lambda)=cP_{\text{C}}(c\lambda)$ where the scale factor is found to be $c=\frac{5(2-\sqrt{2})}{2\sqrt{\pi}\langle\lambda_{\min}\rangle}$.

For BdG class DIII, we need choose $\beta=4$, $\alpha=1$ and obtain

[TABLE]

In order to use $P_{\text{DIII}}(\lambda)$ to compare with numerical data, we need rescale $P_{\text{DIII}}(\lambda)\to\tilde{P}(\lambda)=cP_{\text{DIII}}(c\lambda)$ where the scale factor is found to be $c=\frac{\sqrt{\pi}}{2\sqrt{2}\langle\lambda_{\min}\rangle}$.

For BdG class CI, we need choose $\beta=1$, $\alpha=1$ and obtain

[TABLE]

which is the same as class DIII.

In fact, the independence of $P_{1}(\lambda)$ at $\alpha=1$ on $\beta$ index upto to a rescaling factor can be understood by a direct evaluation of the integral:

[TABLE]

A change of variable and shift the integral domain lead to:

[TABLE]

So it is obvious no $\beta$ dependence in the final result after a normalization and a suitable rescale of $\lambda_{1}$ ( For example, in the $N\to\infty$, one just need to scale $\lambda\to c\lambda/\sqrt{N}$ ). If one choose a suitable scale, all $\alpha=1$ ensemble will have the same $P_{1}$ as Class AIII. A rescale of $\lambda_{1}$ will not affect the moment ratio $A$ defined in Eq.S12, so $\alpha=1$ will lead to the same $A_{W}$ in Table S2.

Using the fact $\operatorname{erfc}(\lambda)=1-2\lambda/\sqrt{\pi}+O(\lambda^{2})$, one can check all of these results have the asymptotic behaviour $P(\lambda)\sim\lambda^{\alpha}$ when $\lambda$ is small. In principle, the edge exponent $\alpha$ can be obtained by studying the small $\lambda$ behaviour of $P(\lambda)$, But in practice, although this method can be used to easily distinguish between $\alpha=0$ and $\alpha>0$ cases, it may not be efficiently used to tell the differences among $\alpha=1,2,3$ cases. Inspired by the $r$-parameter introduced to get rid of the dependence on the local density of state and determine the bulk energy level statistics efficiently, we find it is more convenient to introduce a new moment ratio $A$ which is a dimensionless quantity to get rid of the dependence on the rescaling factor:

[TABLE]

which is obviously indeed independent of the rescaling factor.

Since the Majorana SYK model do not contain any topological protected zero modes, there is a one-to-one correspondence between RMT indices $(\beta,\alpha)$ and RMT classes. We also perform EDs on random matrices of size $10^{3}$ from these 7 symmetry classes. The moment ratio $A$ calculated from the analytical functions for the 7 classes S1,S2,S3,S6,S7,S8,S9 and the numerical data are listed in Table S2. A comparison between our Winger-like surmise and numerical data is shown in Fig. S1. The combination of two dimensionless $\tilde{r}$ and $A$-values can also be used to determine the transition between random matrix ensembles with different $\beta,\alpha$ indices such as that in the hybrid ${\cal N}=1$ SUSY SYK model un .

II B. Constructing the energy levels by anti-unitary operators in the minimum scheme

In the main text, we counted the energy level degeneracy just from the block structure of the corresponding 10-fold way RMT classes. Here, we construct the energy levels systematically and explicitly in terms of the anti-unitary operators, so in addition to counting the energy level degeneracy, it also provides the detailed structures of all the energy levels. We also discuss the relations among various different basis.

B1. The $\Lambda$ operator in the minimum scheme

Here we first summarize the nature of the operator $\Lambda=PR$. Because $PR=(-1)^{\lceil N/2\rceil\lfloor N/2\rfloor}RP$, so it is enough to just focus on $\Lambda=PR$.

For SYK, $q$ is even, then $q(N+1)$ is always even: when $N$ is even, $\Lambda=(-1)^{F}$ is the conserved parity, when $N$ is odd, $\Lambda=1$.

For $\mathcal{N}=1$ SYK, $q$ is odd, when $N$ is even, $q(N+1)$ is odd, $\Lambda=(-1)^{F}$ is the parity operator which anti-commutes with $Q$, so is a chirality operator. When $N$ is odd, $q(N+1)$ is even, $\Lambda=1$.

In short, when $N$ is even, $\Lambda$ is the conserved parity operator when $q$ is even, but a chirality operator when $q$ is odd. When $N$ is odd, $\Lambda=1$ is trivial in any case. Its complete properties is listed in the table S3.

B2. SYK models with $q\pmod{4}=0$.

When $N$ is even, $\Lambda=PR$ is the conserved parity $(-1)^{F}$. For $N\pmod{8}=0,4$, no relation between the two opposite parity blocks. For $N\pmod{8}=2,6$, $P$ swaps between the two opposite parity blocks, thus there is a one to one correspondence between the energy levels in the two opposite parities. When $N$ is odd, $\Lambda=1$, the Hamiltonian has no block structure. In the following, we will determine the energy level degeneracy when $N$ is even or odd respectively.

1. $N$ is even.

When $N$ is even, $\Lambda=PR=(-1)^{F}$ is the conserved parity. If choosing $\psi_{+}$ to be an eigenstate with the eigenenergy $E$ and even parity, because $PR\psi_{+}=\psi_{+}$, $R\psi_{+}\propto P^{2}R\psi_{+}=P\psi_{+}$, then one only need clarify the relations between $\psi_{+}$ and $P\psi_{+}$.

When $N\pmod{8}=0$, $P$ and $R$ commute with $Q_{q}$ and preserve parity, $P^{2}=R^{2}=+1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}\propto\psi_{+}$, thus $Q_{q}$ has no degeneracy.

When $N\pmod{8}=2$, $P$ and $R$ commute with $Q_{q}$ and swap parity, $P^{2}=-R^{2}=+1$. If $\psi_{+}$ is an eigenstate with the eigenenergy $E$ and even parity, then $P\psi_{+}$ is an eigenstate with eigenenergy $E$, but odd parity, thus $Q_{q}$ has 1+1-fold degeneracy which means the two degenerate states have opposite parities.

When $N\pmod{8}=4$, $P$ and $R$ commute with $Q_{q}$ and preserve parity, and $P^{2}=R^{2}=-1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstate with eigenenergy $E$ and even parity, thus $Q_{q}$ has 2-fold degeneracy.

When $N\pmod{8}=6$, $P$ and $R$ commute with $Q_{q}$ and swap parity, and $P^{2}=-R^{2}=-1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}$ is also an eigenstate with eigenenergy $E$, but odd parity, thus $Q_{q}$ has 1+1-fold degeneracy.

2. $N$ is odd.

When $N$ is odd, $\Lambda=PR=1$, if choosing $\psi$ to be an eigenstate with eigenenergy $E$, because $PR\psi=\psi$ and $R\psi\propto P^{2}R\psi=P\psi$. Then one still only clarify the relations between $\psi$ and $P\psi$.

When $N\pmod{8}=1,7$, $P$ and $R$ commute with $Q_{q}$, and $P^{2}=R^{2}=+1$. If $\psi$ is an eigenstate with eigenenergy $E$, then $P\psi\propto\psi$, thus $Q_{q}$ has no degeneracy.

When $N\pmod{8}=3,5$, $P$ and $R$ commute with $Q_{q}$, and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate with eigenenergy $E$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with the same eigenenergy $E$, thus $Q_{q}$ has 2-fold degeneracy.

The energy level degeneracy confirms that listed in the 5th row of Table 1.

B3. SYK models with $q\pmod{4}=2$.

To determine the level degeneracy, we can use the same method as $q\pmod{4}=0$ case discussed above. For any $N$, $P$ and $R$ always anti-commute with $Q_{q}$. If $\psi$ is an eigenstate with eigenenergy $E$, then $P\psi$ is an eigenstate with eigenenergy $-E$, thus $Q_{q}$ has no degeneracy, but a spectral mirror symmetry in any case. In the following, we just spell out the detailed energy level structure.

1. $N$ is even.

When $N\pmod{8}=0$, $P$ and $R$ anti-commute with $Q_{q}$, preserve parity, and $P^{2}=R^{2}=+1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}$ is an eigenstate with eigenenergy $-E$ and even parity, thus $Q_{q}$ has no degeneracy.

When $N\pmod{8}=2$, $P$ and $R$ anti-commute with $Q_{q}$, swap parity, and $P^{2}=-R^{2}=+1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}$ is an eigenstate with eigenenergy $-E$ and odd parity, thus $Q_{q}$ has no degeneracy.

When $N\pmod{8}=4$, $P$ and $R$ anti-commute with $Q_{q}$, preserve parity, and $P^{2}=R^{2}=-1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}$ is a eigenstate with eigenenergy $-E$ and even parity, thus $Q_{q}$ has no degeneracy.

When $N\pmod{8}=6$, $P$ and $R$ anti-commute with $Q_{q}$, swap parity, and $P^{2}=-R^{2}=-1$. If $\psi_{+}$ is an eigenstate with eigenenergy $E$ and even parity, then $P\psi_{+}$ is an eigenstate with eigenenergy $-E$ and odd parity, thus $Q_{q}$ has no degeneracy.

2. $N$ is odd.

There is no conserved quantity, $\Lambda=PR=1$, $P$, $R$ anti-commute with $Q_{q}$. If $\psi$ is an eigenstate with eigenenergy $E$, then $P\psi=R\psi$ is an eigenstate with eigenenergy $-E$, thus $Q_{q}$ has no degeneracy.

The energy level degeneracy confirms that listed in the 7th row of Table 1.

B4. SYK supercharge with $q\pmod{4}=1$.

In the following, we will discuss the $N$ even or odd respectively.

1. $N$ is even.

To determine the energy level degeneracy for $N$ even case, $\Lambda=PR=(-1)^{F}$ is the chirality operator which anti-commutes with $Q_{q}$, no conserved quantity exists. If choosing $\psi$ to be an eigenstate of $Q_{q}$ with eigenenergy quasi $\sqrt{E}$, then one need clarify the relations among $\psi$, $P\psi$, $R\psi$, and $PR$.

When $N\pmod{8}=0$, $\{P,Q_{q}\}=[R,Q_{q}]=0$ and $P^{2}=R^{2}=+1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, ( thus $H=Q_{q}^{2}$ haa an eigenenergy $E$,) then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, $R\psi\propto\psi$ and $PR\psi\propto P\psi$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=2$, $[P,Q_{q}]=\{R,Q_{q}\}=0$ and $P^{2}=-R^{2}=+1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $R\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, $P\psi\propto\psi$ and $PR\psi\propto R\psi$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=4$, $\{P,Q_{q}\}=[R,Q_{q}]=0$ and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $\psi$ and $R\psi$ are two orthogonal eigenstates (namely, $(\psi,R\psi)=0$ ) with eigenenergy $\sqrt{E}$, $P\psi$ and $PR\psi$ are two orthogonal eigenstates (namely, $(P\psi,PR\psi)=0$ ) with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=6$, $[P,Q_{q}]=\{R,Q_{q}\}=0$ and $P^{2}=-R^{2}=-1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, $R\psi$ and $PR\psi$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

2. $N$ is odd.

To determine level degeneracy for $N$ odd case, since $\Lambda=PR=1$, so $\psi=PR\psi$ and $R\psi\propto P^{2}R\psi=P\psi$, one only need clarify relations between $\psi$ and $P\psi$.

When $N\pmod{8}=1$, $[P,Q_{q}]=[R,Q_{q}]=0$ and $P^{2}=R^{2}=+1$. In this case, if $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $P\psi\propto\psi$, thus $Q_{q}$ has no degeneracy and no mirror symmetry either, so $Q_{q}^{2}$ has no degeneracy.

When $N\pmod{8}=3$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=5$, $[P,Q_{q}]=[R,Q_{q}]=0$ and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, thus $Q_{q}$ has double degeneracy and no mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=7$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

So the energy level degeneracy of $Q_{q}$ and $H_{q}=Q^{2}_{q}$ confirm those listed in the 9th and 10th rows of Table 1.

B5. SYK supercharge with $q\pmod{4}=3$.

To determine the level degeneracy, we can use the similar procedure as that in the above $q\pmod{4}=1$ case. So we also discuss $N$ even and odd respectively.

1. $N$ is even.

When $N\pmod{8}=0$, $[P,Q_{q}]=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=+1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, (thus $H=Q_{q}^{2}$ haa an eigenenergy $E$,) then $R\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, $P\psi\propto\psi$ and $PR\psi\propto R\psi$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=2$, $\{P,Q_{q}\}=[R,Q_{q}]=0$ and $P^{2}=-R^{2}=+1$. If $\psi$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$, then $\psi$ and $R\psi$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, $P\psi$ and $PR\psi$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has double degeneracy and a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=4$, $[P,Q_{q}]=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=-1$. if $\psi$ is an eigenstate with eigenenergy $\sqrt{E}$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, $R\psi$ and $PR\psi$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has double degeneracy and a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=6$, $\{P,Q_{q}\}=[R,Q_{q}]=0$ and $P^{2}=-R^{2}=-1$. In this case, if $\psi$ is an eigenstate with eigenenergy $\sqrt{E}$, then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, $R\psi\propto\psi$ and $PR\psi\propto P\psi$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

2. $N$ is odd.

When $N\pmod{8}=1$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=+1$. If $\psi$ is an eigenstate with eigenenergy $\sqrt{E}$, then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=3$, $[P,Q_{q}]=[R,Q_{q}]=0$ and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate with eigenenergy $\sqrt{E}$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, thus $Q_{q}$ has double degeneracy, but no mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=5$, $\{P,Q_{q}\}=\{R,Q_{q}\}=0$ and $P^{2}=R^{2}=-1$. If $\psi$ is an eigenstate with eigenenergy $\sqrt{E}$, then $P\psi$ is an eigenstate with eigenenergy $-\sqrt{E}$, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=7$, $[P,Q_{q}]=[R,Q_{q}]=0$ and $P^{2}=R^{2}=+1$. If $\psi$ is a eigenstate with eigenenergy $\sqrt{E}$, then $P\psi\propto\psi$, thus $Q_{q}$ has no degeneracy and no mirror symmetry either, so $Q_{q}^{2}$ has no degeneracy.

So the energy level degeneracy of $Q_{q}$ and $H_{q}=Q^{2}_{q}$ confirm those listed in the 12nd and 13rd rows of Table 1.

III C. Classifying SYK models with odd $N$ in enlarged Hilbert space by adding a Majorana fermion: odd $N$ case

When $N$ is odd, following You2017 ; hybrid , one may add a Majorana fermion $\chi_{\infty}=\chi_{N+1}$ to the system. Then the Hilbert space has dimension $2^{N_{c}}$ which is twice that of the minimum scheme used in the main text. In fact, as shown below, there are two more conserved quantities which all commute with the Hamiltonian, but not commute with each other. So there is only one more compatible conserved quantity.

In the minimal scheme used in the main text, one must choose $\chi_{2i-1}$ to be real, $\chi_{2i}$ to be imaginary footnote1 . In the enlarged Hilbert space, one can choose either way. To be contrasted and also complementary to that used in the main text, we choose $\chi_{2i}$ to be real, while $\chi_{2i-1}$ to be imaginary. The two different choices lead to quite different classification schemes, but to the same results after interpreting the results carefully. In this section, especially Sec.C2-1, we will establish the main differences and also intricate connections between the two choices. Then the $N_{c}=(N+1)/2$ complex fermion creation and annihilation operators $c_{i}^{\dagger}$ and $c_{i}$ are defined as:

[TABLE]

Then the fermion number operator $F=\sum_{i=1}^{N_{c}}c_{i}^{\dagger}c_{i}$. One can define two particle-hole symmetry operators to be Fu2016 ; You2017 ; Cotler2017 ; hybrid

[TABLE]

There is also another useful anti-unitary operator $Z$ by factoring out the extra Majorana fermion from the particle-hole symmetry operator $P$, which can be written as You2017 ; hybrid

[TABLE]

It can be shown that hybrid ; color

[TABLE]

which leads to:

[TABLE]

One may also find their squared values

[TABLE]

In fact, one may also construct other anti-unitary operators such as adding $\chi_{\infty}$ to $R$, which leads to another anti-unitary operator Z^{\prime}=R\chi_{\infty}=K\Big{[}\prod_{i=1}^{N_{c}}(c_{i}^{\dagger}-c_{i})\Big{]}\chi_{\infty}\>. However, $Z^{\prime}$ can be expressed by a product of $R$, $P$, and $Z$, so $R$, $P$, and $Z$ will be the complete set of anti-unitary operators to perform the classifications. The relation between the anti-unitary operators and RMT classes is listed in Table S1.

**C1. The systematic approach in the enlarged Hilbert space. **

Because $N$ is fixed to be odd in this section, so in the following, we will discuss $q$ even and odd respectively.

1. $q$ is even

The total fermion number parity $(-1)^{F}$ is a conserved quantity. The commutation relations between the anti-unitary operators and the conserved parity can be easily worked out

[TABLE]

From the expressions,

[TABLE]

one can see $(-1)^{F}=(-1)^{N_{c}(N_{c}+1)/2}PR=(-1)^{\lceil N_{c}/2\rceil}PR$ which can be simply written as $(-1)^{F}\propto PR$.

When doing the classification, one need use the anti-unitary operators which preserves the parity $(-1)^{F}$, then obtain the classification from their commutation relations with $Q_{q}$, also their squared values by comparing with Table. S1. When counting the energy level degeneracy, if $\psi_{+}$ is an eigenstate of $Q_{q}$ with the eigenenergy $E$ and even parity, then the degeneracy can be obtained by finding the relations among $\psi_{+}$, $P\psi_{+}$, $Z\psi_{+}$, and $ZP\psi_{+}$. Because $(-1)^{F}=(-1)^{\lceil N_{c}/2\rceil}PR$, it is easy to show that $PR\psi_{+}\propto\psi_{+}$, $R\psi_{+}\propto P\psi_{+}$, $ZR\psi_{+}\propto ZP\psi_{+}$, and $ZPR\psi_{+}\propto Z\psi_{+}$. Due to $PR=\pm RP$ and $ZPR=\pm PZR$, different orderings of $P,R,Z$ in these states will not lead to any new states.

2. $q$ is odd

The total fermion number parity $(-1)^{F}$ and $\chi_{\infty}$ anti-commute with $Q_{q}$, thus $(-1)^{F}i\chi_{\infty}$ commutes with $Q_{q}$. Notice, here $i$ is a necessary factor to make $(-1)^{F}i\chi_{\infty}$ Hermitian. Since $[(-1)^{F}i\chi_{\infty}]^{2}=(-1)^{2F}=1$, we will still call it “parity”. One can check its relation with $P$, $R$ and $Z$

[TABLE]

For the classification, one need use the anti-unitary operators which commute with $(-1)^{F}i\chi_{\infty}$, then obtain the classification by checking their commutation relations with $Q_{q}$, their squared value by comparing with Table. S1. For the energy level degeneracy, if $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $E$ and even “parity”, then the degeneracy can be obtained by clarifying the relations among $\psi_{+}$, $P\psi_{+}$, $Z\psi_{+}$, and $ZP\psi_{+}$. Note that due to the fact $(-1)^{F}i\chi_{\infty}=(-1)^{\lfloor N_{c}/2\rfloor}iZR$, one can see that $ZR\psi_{+}\propto\psi_{+}$, $R\psi_{+}\propto Z\psi_{+}$, $PR\psi_{+}\propto ZP\psi_{+}$, and $ZPR\psi_{+}\propto P\psi_{+}$.

In the following, we are ready to discuss the classification and level degeneracy.

**C2. $q\pmod{4}=0$ and $N$ odd **

Then $P,R,Z$ all commute with $Q_{q}$, $[P,Q_{q}]=[R,Q_{q}]=[Z,Q_{q}]=0$, thus play the role as $T_{+}$.

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $Z$ preserves parity, $P^{2}=-R^{2}=Z^{2}=+1$, one must use $Z$ to do the classification, thus $Q_{q}$ belongs to Class AI(GOE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with an eigenenergy $E$ and even parity, then $\psi_{+}$ and $Z\psi_{+}$ are the same eigenstate with eigenenergy $E$ and even parity, while $P\psi_{+}$ and $ZP\psi_{+}$ are the same eigenstate with eigenenergy $E$ and odd parity, thus $Q_{q}$ has 1+1-fold degeneracy.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $Z$ swaps parity, $P^{2}=R^{2}=-Z^{2}=-1$, one can use either $P$ or $R$ to do the classification, thus $Q_{q}$ belongs to Class AII(GSE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, $Z\psi_{+}$ and $ZP\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $Q_{q}$ has 2+2-fold degeneracy.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $Z$ preserves parity, $P^{2}=-R^{2}=Z^{2}=-1$, one must use $Z$ to do the classification, thus $Q_{q}$ belongs to Class AII(GSE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, $P\psi_{+}$ and $ZP\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $Q_{q}$ has 2+2-fold degeneracy.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $Z$ swaps parity, $P^{2}=R^{2}=-Z^{2}=+1$, one can use either $P$ or $R$ to do the classification, thus $Q_{q}$ belongs to Class AI(GOE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with an eigenenergy $E$ and even parity, then $\psi_{+}$ and $P\psi_{+}$ are the same eigenstate with eigenenergy $E$ and even parity, $Z\psi_{+}$ and $ZP\psi_{+}$ are the same eigenstate with eigenenergy $E$ and odd parity, thus $Q_{q}$ has 1+1-fold degeneracy.

These results are listed in the 2nd row in Table S4. Here, we perform the classification with the conserved parity $(-1)^{F}$ which is also Hermitian. In fact, there is another conserved quantity $\chi_{\infty}$ which is also Hermitian. So one can also do the classification by using $\chi_{\infty}$ and indeed reach the same conclusions. Note that choosing different conserved quantity, one may need to use different anti-unitary symmetry operators to do classification. However, because $\{\chi_{\infty},(-1)^{F}\}=0$, so they do not commute. One can only use one of the two or their combination to do the classification as shown in the following.

1. The impossibility to do the classifications without touching $\chi_{\infty}$ in the basis S13.

One may would like to just use $R$ and $Z$ to classify which do not involve $\chi_{\infty}$. However, in the basis defined by Eq.S13, the classification can not be done without touching $\chi_{\infty}$ as shown in the following.

As shown above, there are two mutually anti-commuting conserved quantities $\chi_{\infty},(-1)^{F}$, one can form another conserved quantity $iZR=(-1)^{\lfloor N_{c}/2\rfloor}(-1)^{F}i\chi_{\infty}$ which is just the product of the two. But $\chi_{\infty}$ gets canceled out in the product. So it is tempting to see if one can do the classification using the anti-unitary operators $R,Z$ and this conserved quantity, all of which do not involve $\chi_{\infty}$.

When $N$ is odd, we have $[Z,R]=0$ which lead to the commutation relation between $Z$, $R$, $P$ and the conserved quantity $iZR$

[TABLE]

Thus $Z$ and $R$ can not be used to classify. One must use $P$, which preserves $iZR$, commute with Hamiltonian and $P^{2}=(-1)^{\lfloor N_{c}/2\rfloor}=(-1)^{\lfloor(N+1)/4\rfloor}$. So we reach the same classification. This is expected as stated in footnote1 , so $\chi_{\infty}$ is unavoidable in the basis IR.

However, if one redefine Eq.S13 as $c_{i}=(\chi_{2i-1}-i\chi_{2i})/\sqrt{2}$ by choosing $\chi_{2i-1}$ real and $\chi_{2i}$ imaginary, one may just use $R$ and $Z$ to classify which do not touch $\chi_{\infty}$. Of course, this is expected, because that is the basis footnote1 we used in the main text where one can perform the classification in the minimal Hilbert space without adding the Majorana fermions at $\infty$. It was designed not to touch ( in fact, no need to add ) $\chi_{\infty}$ at first place.

**C3. $q\pmod{4}=2$ and odd $N$. **

Then $P,R,Z$ all anti-commute with $Q_{q}$, $\{P,Q_{q}\}=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, thus play role as $T_{-}$.

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $Z$ preserves parity, $P^{2}=-R^{2}=Z^{2}=+1$. So one must use $Z$ to do the classification, thus $Q_{q}$ belongs to Class D(BdG). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with the eigenenergy $E$ and even parity, then $\psi_{+}$ and $ZP\psi_{+}$ are two orthogonal eigenstates with the same eigenenergy $E$, but opposite parities, $P\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstate with the same eigenenergy $-E$, but opposite parities, thus $Q_{q}$ has 1+1-fold degeneracy and also a mirror symmetry.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $Z$ swaps parity, $P^{2}=R^{2}=-Z^{2}=-1$, So one can use either $P$ or $R$ to do the classification, thus $Q_{q}$ belongs to Class C(BdG). The relations among $\psi_{+}$, $P\psi_{+}$, $Z\psi_{+}$, and $ZP\psi_{+}$ are the same as those in the $N\pmod{8}=1$ case, thus $Q_{q}$ has 1+1-fold degeneracy and also a mirror symmetry.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $Z$ preserves parity, and $P^{2}=R^{2}=Z^{2}=-1$. So one must use $Z$ to do the classification, thus $Q_{q}$ belongs to Class C(BdG). The relations among $\psi_{+}$, $P\psi_{+}$, $Z\psi_{+}$, and $ZP\psi_{+}$ are the same as those in the $N\pmod{8}=1$ case, thus $Q_{q}$ has 1+1-fold degeneracy and a mirror symmetry.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $Z$ swap parity, $P^{2}=R^{2}=-Z^{2}=+1$. So one can use either $P$ or $R$ to do the classification, thus $Q_{q}$ belongs to Class D(BdG). The relations among $\psi_{+}$, $P\psi_{+}$, $Z\psi_{+}$, and $ZP\psi_{+}$ are the same as those in the $N\pmod{8}=1$ case, thus $Q_{q}$ has 1+1-fold degeneracy and a mirror symmetry.

These results are listed in the 3rd row in Table S4.

**C4. $q\pmod{4}=1$ and odd $N$ **

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ preserves parity, $R$ and $Z$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=-R^{2}=Z^{2}=+1$, thus $Q_{q}$ belongs to Class AI(GOE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $P\psi_{+}$ are the same eigenstate with an eigenenergy $\sqrt{E}$ and even “parity”, $ZP\psi_{+}$ and $Z\psi_{+}$ are the same eigenstate with eigenenergy $-\sqrt{E}$ and odd “parity”, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ preserves parity, $R$ and $Z$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=R^{2}=-Z^{2}=-1$, thus $Q_{q}$ belongs to Class C(BdG). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, but opposite “parity”, $ZP\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, but opposite “parity”, thus $Q_{q}$ has 1+1 degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ preserves parity, $R$ and $Z$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=-R^{2}=Z^{2}=-1$, thus $Q_{q}$ belongs to Class AII(GSE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$ and even “parity”, $ZP\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$ and odd “parity”, thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ preserves parity, $R$ and $Z$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=R^{2}=-Z^{2}=+1$, thus $Q_{q}$ belongs to Class D(BdG). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with and eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $\sqrt{E}$, but opposite “parity”, $ZP\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, but opposite “parity”, thus $Q_{q}$ has 1+1 degeneracy and a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

These results for $Q_{q}$ and $H_{q}=Q^{2}_{q}$ are listed in the 4th and 5th rows in Table S4 respectively.

**C5. $q\pmod{4}=3$ and odd $N$. **

The situation is similar to $q\pmod{4}=1$ case.

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ preserves parity, $R$ and $Z$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=-R^{2}=Z^{2}=+1$, thus $Q_{q}$ belongs to Class D(BdG). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with an eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $+\sqrt{E}$, but opposite “parity”, $P\psi_{+}$ and $ZP\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, but opposite “parity”, thus $Q_{q}$ has 1+1 degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ preserves parity, $R$ and $Z$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=R^{2}=-Z^{2}=-1$, thus $Q_{q}$ belongs to Class AII(GSE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $+\sqrt{E}$ and even “parity”, $Z\psi_{+}$ and $ZP\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$ and odd “parity”, thus $Q_{q}$ has double degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ preserves parity, $R$ and $Z$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=-R^{2}=Z^{2}=-1$, thus $Q_{q}$ belongs to Class C(BdG). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $Z\psi_{+}$ are two orthogonal eigenstates with eigenenergy $+\sqrt{E}$, but opposite “parity”, $ZP\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $-\sqrt{E}$, but opposite “parity”, thus $Q_{q}$ has 1+1 degeneracy and also a mirror symmetry, so $Q_{q}^{2}$ has 4-fold degeneracy.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ preserves parity, $R$ and $Z$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=R^{2}=-Z^{2}=+1$, thus $Q_{q}$ belongs to Class AI(GOE). If $\psi_{+}$ is an eigenstate of $Q_{q}$ with eigenenergy $\sqrt{E}$ and even “parity”, then $\psi_{+}$ and $P\psi_{+}$ are the same eigenstate with eigenenergy $+\sqrt{E}$ and even “parity”, $Z\psi_{+}$ and $ZP\psi_{+}$ are the same eigenstates with eigenenergy $-\sqrt{E}$ and odd“parity”, thus $Q_{q}$ has no degeneracy, but a mirror symmetry, so $Q_{q}^{2}$ has 2-fold degeneracy.

These results for $Q_{q}$ and $H_{q}=Q^{2}_{q}$ are listed in the 6th and 7th rows in Table S4 respectively.

**C6. Comparison of Table S4 of the extended scheme at odd $N$ to Table.1 of the minimum scheme in the main text **

When the total number of Majorana fermion $N$ is even, it is straightforward and convenient to pair two Majorana fermions to form a Dirac fermion Fidkowski2011 ; You2017 ; Cotler2017 ; Fu2016 . This construction has the same Hilbert space dimension $2^{N/2}$ and also same sets of symmetry operators as that with the Majorana fermions in the minimum scheme used in the main text, thus the two schemes are equivalent. This is why we only list odd $N$ in Fig.S4.

However, when the total number of Majorana fermion $N$ is odd, As shown in this section, it is typical and intuitive to define a enlarged Hilbert space by adding an extra Majorana fermion that is decoupled from the ones appearing in the Hamiltonian Fidkowski2011 ; You2017 ; hybrid , thus the dimension of an enlarged Hilbert space is twice that used in the main text. The relation between the two can be understood from the $2^{(N+1)/2}$-dimensional representation of Majoranas, then the $Q_{q}$ is automatically in a block diagonal form for even $q$ or block off-diagonal form for odd $q$, each block is identical to our $2^{(N-1)/2}$-dimensional minimal representation of $Q_{q}$, thus doubled the level degeneracy of $Q_{q}$ (for even $q$) and $Q_{q}^{2}$ (for odd $q$). In the following, we will show that this is indeed the case when comparing Table 1 in the minimum scheme to Table S4 in the extended scheme.

(1) even $q$

When $q\pmod{4}=0$, the level degeneracy $d=1,2,2,1$ in the minimum scheme at $N=1,3,5,7$ respectively in Table S4 which is identical to $d=1,2,2,1$ in the same parity sector in extended scheme in Table 1.

When $q\pmod{4}=2$, the level degeneracy $d=1,1,1,1$ in the minimum scheme at $N=1,3,5,7$ respectively in Table S4 which is identical to $d=1,1,1,1$ in the same parity sector in extended scheme in Table 1.

As a result, the degeneracy of $Q_{q}$ in the enlarged Hilbert space is twice that in the minimum Hilbert space for even $q$.

(2) odd $q$

For odd $q$, there is always an extra partner of eigenenergy with negative energy. If $Q_{q}$ with odd $q$ has no mirror symmetry in the minimum scheme, then adding an extra Majorana fermions makes $Q_{q}$ to have a spectral mirror symmetry in enlarged Hilbert space. Then the degeneracy of $Q_{q}$ is the same in both schemes, but the degeneracy of $Q_{q}^{2}$ will be doubled in enlarged Hilbert space. If $Q_{q}$ with odd $q$ has a mirror symmetry already in the minimum scheme, then the degeneracy of $Q_{q}$ in the minimum scheme is the same as that projected to the same parity $(-1)^{F}i\chi_{\infty}$ sector. The degeneracy of $Q_{q}^{2}$ will also be doubled in enlarged Hilbert space.

When $q\pmod{4}=1$, the level degeneracy $d=1,1,2,1$ of $Q_{q}$ with ( no,yes,no,yes) mirror symmetry at $N=1,3,5,7$ in the minimum scheme respectively is identical to $d=1,2,2,1$ of $Q_{q}$ with ( no,yes,no,yes) projection in the extended scheme. Then the degeneracy of $H_{q}=Q^{2}_{q}$ is $d_{H}=1,2,2,2$ in the minimum scheme which is twice enlarged to $d_{H}=2,4,4,4$ in the extended scheme.

When $q\pmod{4}=3$, the level degeneracy $d=1,2,1,1$ of $Q_{q}$ with ( yes,no,yes, no ) mirror symmetry at $N=1,3,5,7$ in the minimum scheme respectively is identical to $d=1,2,1,1$ of $Q_{q}$ with ( yes,no,yes,no) projection in the extended scheme. Then the degeneracy of $H_{q}=Q^{2}_{q}$ is $d_{H}=2,2,2,1$ in the minimum scheme which is twice enlarged to $d_{H}=4,4,4,2$ in the extended scheme.

As a comparison, we also perform exact diagonalization in the enlarged Hilbert which lead to the same results presented in Fig.1-3 in the main text.

IV D. Counting the degeneracy of the $\mathcal{N}=1$ supersymmetric SYK model in $(-1)^{F}$ basis: Odd $q$ case

For $\mathcal{N}=1$ supersymmetric SYK model, for even $N$ or odd $N$ in the enlarged Hilbert space, $(-1)^{F}$ is always a conserved quantity $[H,(-1)^{F}]=0$. In the previous section, we classified the eigenstates of $H_{q}=Q^{2}_{q}$ in terms of those of $Q_{q}$. Due to $\{(-1)^{F},Q_{q}\}=0$, so any eignestate of $Q_{q}$ is NOT an eigen-state of the parity $(-1)^{F}$, but a linear combination of the two with opposite parities quasi .

In this section, we will classify the eigenstate of $H_{q}=Q^{2}_{q}$ in terms of parity $(-1)^{F}$. Taking $\psi_{+}$ as the ground state with a positive parity, Then $H$ always has at least a doubly degenerate ground state with two opposite parities $(\psi_{+},Q\psi_{+})$. So in the following, if the degeneracy at a given parity sector is $d$. Then the total degeneracy $d_{t}=2d$ is in-dependent of which basis one is using, so it should be the same as that achieved in the last section using $Q_{q}$ basis.

Note that the anti-unitary operators $P$, $R$, $Z$ ( if exists ) and the fermion number parity $(-1)^{F}$ only depend on $N$. So the commutation relations between these anti-unitary operators and the parity listed in Eq.(S18) are independent of choice of Hamiltonian. The squared values of these anti-unitary operators Eq.S17 also hold independent of choice of Hamiltonian.

**D1. The systematic approach. **

Because $q$ is fixed to be odd, so in the following, we discuss even $N$ and odd $N$ respectively.

1. $N$ is even

When $N$ is even, if $\psi_{+}$ is a even parity state, then we need discuss the relations among all the states generated from all the anti-unitary symmetry operators acting on $\psi_{+}$: $\psi_{+},P\psi_{+},R\psi_{+},PR\psi_{+}$ and also their corresponding opposite parity partners $Q_{q}\psi_{+},PQ_{q}\psi_{+},RQ_{q}\psi_{+},PRQ_{q}\psi_{+}$. From $\Lambda=PR\propto(-1)^{F}$, one can see the relations $\psi_{+}\propto PR\psi_{+}$, $P\psi_{+}\propto R\psi_{+}$ and also the relations among their opposite parity partners $Q_{q}\psi_{+}\propto PRQ_{q}\psi_{+}$, $PQ_{q}\psi_{+}\propto RQ_{q}\psi_{+}$. So one only need to clarify the relations among the four states $\psi_{+},P\psi_{+}$, and their corresponding opposite parity partners $Q_{q}\psi_{+},PQ_{q}\psi_{+}$.

2. $N$ is odd

When $N$ is odd, if $\psi_{+}$ is a even parity state, then we need discuss relation among all the states generated from all symmetry operators acting on $\psi_{+}$, $\psi_{+}$, $P\psi_{+}$, $R\psi_{+}$, $Z\psi_{+}$, $PR\psi_{+}$, $ZR\psi_{+}$, $ZP\psi_{+}$, $ZPR\psi_{+}$ and their corresponding opposite parity partners $Q_{q}\psi_{+}$, $PQ_{q}\psi_{+}$, $RQ_{q}\psi_{+}$, $ZQ_{q}\psi_{+}$, $PRQ_{q}\psi_{+}$, $ZRQ_{q}\psi_{+}$, $ZPQ_{q}\psi_{+}$, $ZPRQ_{q}\psi_{+}$. From $PR\propto(-1)^{F}$, one can see the relations $\psi_{+}\propto PR\psi_{+}$, $P\psi_{+}\propto R\psi_{+}$, $Z\psi_{+}\propto ZPR\psi_{+}$, $ZP\psi_{+}\propto ZR\psi_{+}$, and also the relations among their opposite parity partners $Q_{q}\psi_{+}\propto PRQ_{q}\psi_{+}$, $PQ_{q}\psi_{+}\propto RQ_{q}\psi_{+}$, $ZQ_{q}\psi_{+}\propto ZPRQ_{q}\psi_{+}$, and $ZPQ_{q}\psi_{+}\propto ZRQ_{q}\psi_{+}$. So one only need to clarify relations among the 8 states $\psi_{+},P\psi_{+},Z\psi_{+},ZP\psi_{+}$ and their corresponding opposite parity partners $Q_{q}\psi_{+}$, $PQ_{q}\psi_{+}$, $ZQ_{q}\psi_{+}$, and $ZPQ_{q}\psi_{+}$.

Now, it is important to observe that there is an extra conserved quantity $ZPQ_{q}\propto\chi_{\infty}Q_{q}$ which also commutes with the parity $[ZPQ_{q},(-1)^{F}]=0$, so one can construct the extra “conserved quantity” $(iZPQ_{q})^{\dagger}=iZPQ_{q}$, and $(iZPQ_{q})^{2}=(i\chi_{\infty}Q_{q})^{2}=\chi_{\infty}^{2}Q_{q}^{2}=Q_{q}^{2}=H$. Now the situation is similar to previous discussion in $[H,Q_{q}]$ basis where one find an extra conserved quantity $[(-1)^{F}i\chi_{\infty},Q_{q}]=0$ in the $Q_{q}$ basis. Here one just shift $\chi_{\infty}$ to $Q_{q}$ to make the extra conserved quantity $[i\chi_{\infty}Q_{q},(-1)^{F}]=0$ in the $(-1)^{F}$ basis. In this basis, if choosing $\psi_{+,s}$ to satisfy $(-1)^{F}\psi_{+,s}=\psi_{+,s}$, $iZPQ_{q}\psi_{+,s}=s\sqrt{E}\psi_{+,s}$ where $s=\pm$, then $H\psi_{+,s}=E\psi_{+,s}$. The extra conserved quantities also leads to $ZPQ_{q}\psi_{+,s}\sim\psi_{+,s},ZQ_{q}\psi_{+,s}\sim P\psi_{+,s},PQ_{q}\psi_{+,s}\sim Z\psi_{+,s},ZP\psi_{+,s}\sim Q_{q}\psi_{+,s}$, so the 8 states reduce to 4 states $\psi_{+,+}$, $P\psi_{+,+}$, $Z\psi_{+,+}$, and $Q_{q}\psi_{+,+}$. In the following, we will name the sign of eigenvalue of $iZPQ_{q}$ as $s$.

The relations among $P$, $R$, $Z$, $Q$ and $iZPQ_{q}$, can be easily worked out

[TABLE]

Thus for odd $q$:

[TABLE]

where we dropped the $R$ commutation relation related to Eq.IV, because, as demonstrated above, we only need to tell the relations among $\psi_{+,+}$, $P\psi_{+,+}$, $Z\psi_{+,+},Q_{q}\psi_{+,+}$. So $P$, $Z$, and $Q_{q}$ are enough.

Here, it is easy to see that the degeneracy of $H$ in extended scheme is twice that of the degeneracy of $H$ in minima scheme, due to two mutually commuting parities.

In the following, we are ready to discuss the classification and level degeneracy. All the results in the following sections D2-D5 are listed in Table S5.

**D2. Case $q\pmod{4}=1$ and even $N$ **

When $N\pmod{8}=0$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $\{P,Q_{q}\}=[R,Q_{q}]=0$, and $P^{2}=R^{2}=+1$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $P\psi_{+}\propto\psi_{+}$ is the same eigenstate with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}\propto PQ_{q}\psi_{+}$ is another eigenstate with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 1+1-fold degeneracy.

When $N\pmod{8}=2$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=0$, and $P^{2}=-R^{2}=+1$ and $(PQ_{q})^{2}=Q_{q}^{2}$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $PQ_{q}\psi_{+}\propto\psi_{+}$ is the same eigenstate with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}\propto P\psi_{+}$ is another eigenstate with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 1+1-fold degeneracy.

When $N\pmod{8}=4$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $\{P,Q_{q}\}=[R,Q_{q}]=0$, and $P^{2}=R^{2}=-1$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}$ and $PQ_{q}\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 2+2-fold degeneracy.

When $N\pmod{8}=6$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $[P,Q_{q}]=\{R,Q_{q}\}=0$, and $P^{2}=-R^{2}=-1$ and $(PQ_{q})^{2}=-Q_{q}^{2}$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $PQ_{q}\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, ( This is because $(\psi_{+},PQ_{q}\psi_{+})=(PQ_{q}\psi_{+},(PQ_{q})^{2}\psi_{+})^{*}/E=-(Q_{q}^{2}\psi_{+},PQ_{q}\psi_{+})/E=-(\psi_{+},PQ_{q}\psi_{+})=0)$, $Q_{q}\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 2+2-fold degeneracy.

**D3. Case $q\pmod{4}=3$ and even $N$ **

When $N\pmod{8}=0$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $[P,Q_{q}]=\{R,Q_{q}\}=0$, and $P^{2}=R^{2}=+1$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $P\psi_{+}\propto\psi_{+}$ is the same eigenstate with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}\propto PQ_{q}\psi_{+}$ is an eigenstate with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 1+1-fold degeneracy.

When $N\pmod{8}=2$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=0$, and $P^{2}=-R^{2}=+1$ and $(PQ_{q})^{2}=-Q_{q}^{2}$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $PQ_{q}\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 2+2-fold degeneracy.

When $N\pmod{8}=4$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $[P,Q_{q}]=\{R,Q_{q}\}=0$, and $P^{2}=R^{2}=-1$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $\psi_{+}$ and $P\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}$ and $PQ_{q}\psi_{+}$ are two orthogonal eigenstates with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 2+2-fold degeneracy.

When $N\pmod{8}=6$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $\{P,Q_{q}\}=[R,Q_{q}]=0$, and $P^{2}=-R^{2}=-1$ and $(PQ_{q})^{2}=Q_{q}^{2}$. If $\psi_{+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$ and even parity, then $PQ_{q}\psi_{+}\propto\psi_{+}$ is the same eigenstate with eigenenergy $E$ and even parity, $Q_{q}\psi_{+}\propto P\psi_{+}$ is an eigenstate with eigenenergy $E$ and odd parity, thus $H=Q_{q}^{2}$ has 1+1-fold degeneracy.

**D4. Case $q\pmod{4}=1$ and odd $N$ **

Note that $P$ always maps $s\to-s$, $Z$ always maps $s\to+s$, $Q_{q}$ always maps $s\to-s$. In this section, we will classify the energy levels in the parity basis of $((-1)^{F},i\chi_{\infty}Q_{q})$ which splits the Hilbert space into 4 sectors.

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $Z$ preserve parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=-R^{2}=Z^{2}=+1$ and $(PQ_{q})^{2}=+Q_{q}^{2}$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $Z\psi_{+,+}\propto\psi_{+,+}$ is an eigenstate with eigenenergy $E$, even parity and $s=+$, $Q_{q}\psi_{+,+}\propto P\psi_{+,+}=\psi_{-,-}$ is an eigenstate with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{-,-}$-fold degeneracy.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $Z$ swaps parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=R^{2}=-Z^{2}=-1$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $P\psi_{+,+}=\psi_{+,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=-$, $Z\psi_{+,+}=\psi_{-,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=+$, $Q\psi_{+,+}=\psi_{-,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{+,-}+1_{-,+}+1_{-,-}$-fold degeneracy.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $Z$ preserves parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=-R^{2}=Z^{2}=-1$ and $(PQ_{q})^{2}=-Q_{q}^{2}$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $\psi_{+,+}$ and $Z\psi_{+,+}=\tilde{\psi}_{+,+}$ are two orthogonal eigenstates of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, $P\psi_{+,+}=\psi_{-,-}$ and $Q\psi_{+,+}=\tilde{\psi}_{-,-}$ are two orthogonal eigenstates of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $2_{+,+}+2_{-,-}$-fold degeneracy.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $Z$ swaps parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=R^{2}=-Z^{2}=+1$. In this case, if $\psi_{++}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $P\psi_{+,+}=\psi_{+,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=-$, $Z\psi_{+,+}=\psi_{-,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=+$, $Q\psi_{+,+}=\psi_{-,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{+,-}+1_{-,+}+1_{-,-}$-fold degeneracy.

**D5. $q\pmod{4}=3$ and odd $N$ **

Note that $P$ always maps $s\to+s$, $Z$ always maps $s\to-s$, $Q_{q}$ always maps $s\to-s$. In this section, we will classify the energy levels in the parity basis of $((-1)^{F},i\chi_{\infty}Q_{q})$ which splits the Hilbert space into 4 sectors.

When $N\pmod{8}=1$, $N_{c}\pmod{4}=1$, $P$ and $R$ swap parity, $Z$ preserve parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=-R^{2}=Z^{2}=+1$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $P\psi_{+,+}=\psi_{-,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=+$, $Z\psi_{+,+}=\psi_{+,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=-$, $Q\psi_{+,+}=\psi_{-,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{+,-}+1_{-,+}+1_{-,-}$-fold degeneracy.

When $N\pmod{8}=3$, $N_{c}\pmod{4}=2$, $P$ and $R$ preserve parity, $Z$ swaps parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=R^{2}=-Z^{2}=-1$ and $(ZQ_{q})^{2}=-Q_{q}^{2}$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $\psi_{+,+}$ and $P\psi_{+,+}=\tilde{\psi}_{+,+}$ are two orthogonal eigenstates of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, $Z\psi_{+,+}=\psi_{-,-}$ and $Q\psi_{+,+}=\tilde{\psi}_{-,-}$ are two orthogonal eigenstates of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $2_{+,+}+2_{-,-}$-fold degeneracy.

When $N\pmod{8}=5$, $N_{c}\pmod{4}=3$, $P$ and $R$ swap parity, $Z$ preserves parity, $\{P,Q_{q}\}=[R,Q_{q}]=[Z,Q_{q}]=0$, and $P^{2}=-R^{2}=Z^{2}=-1$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $P\psi_{+,+}=\psi_{-,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, odd parity and $s=+$, $Z\psi_{+,+}=\psi_{+,-}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=-$, $Q\psi_{+,+}=\psi_{-,-}$ is an eigenstate of $H=Q_{q}^{2}$ with odd parity, $s=-$, and eigenenergy $E$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{+,-}+1_{-,+}+1_{-,-}$-fold degeneracy.

When $N\pmod{8}=7$, $N_{c}\pmod{4}=0$, $P$ and $R$ preserve parity, $Z$ swaps parity, $[P,Q_{q}]=\{R,Q_{q}\}=\{Z,Q_{q}\}=0$, and $P^{2}=R^{2}=-Z^{2}=+1$ and $(ZQ_{q})^{2}=+Q_{q}^{2}$. If $\psi_{+,+}$ is an eigenstate of $H=Q_{q}^{2}$ with eigenenergy $E$, even parity and $s=+$, then $P\psi_{+,+}\propto\psi_{+,+}$ is an eigenstate with eigenenergy $E$, even parity and $s=+$, $Q_{q}\psi_{+,+}\propto Z\psi_{+,+}=\psi_{-,-}$ is an eigenstate with eigenenergy $E$, odd parity and $s=-$, thus $H=Q_{q}^{2}$ has $1_{+,+}+1_{-,-}$-fold degeneracy.

**D6. Comparison of Table 1 in the main text and Tables S4, S5. **

When $q\pmod{4}=1$, at even $N=0,2,4,6$, $d_{H}=2,2,4,4$ in $Q_{q}$ basis in Table 1 which matches $d_{H}=1+1,1+1,2+2,2+2$ in the $(-1)^{F}$ basis in Table S5. Because $\{Q_{q},(-1)^{F}\}=0$, so the eigen-states in the $Q_{q}$ basis has no definite parity. However, the degeneracy should be independent of the basis. At odd $N=1,3,5,7$, $d_{H}=2,4,4,4$ in the extended $(Q_{q},(-1)^{F}i\chi_{\infty})$ basis in Table S4 matches $d_{H}=1+1,1+1+1+1,2+2,1+1+1+1$ in the $((-1)^{F},i\chi_{\infty}Q_{q})$ basis Table S5. Again, because $\{Q_{q},(-1)^{F}\}=0$, so the eigen-states in the former are different than those in the latter footnote1 . However, the total degeneracy should be independent of the basis.

When $q\pmod{4}=3$, at even $N=0,2,4,6$, $d_{H}=2,4,4,2$ in $Q_{q}$ basis in Table 1 matches $d_{H}=1+1,2+2,2+2,1+1$ in the $(-1)^{F}$ basis in Table S5. Because $\{Q_{q},(-1)^{F}\}=0$, so the eigen-states in the $Q_{q}$ basis has no definite parity footnote1 . At odd $N=1,3,5,7$, $d_{H}=4,4,4,2$ in the extended $(Q_{q},(-1)^{F}i\chi_{\infty})$ basis in Table S4 matches $d_{H}=1+1+1+1,2+2,1+1+1+1,1+1$ in the extended $((-1)^{F},i\chi_{\infty}Q_{q})$ basis Table S5. Again, because $\{Q_{q},(-1)^{F}\}=0$, so the eigen-states in the former are different than those in the latter. However, the total degeneracy should be independent of the basis.

V E. RMT Classification of complex fermion SYK models

The Hamiltonian for the SYK model with complex fermions is

[TABLE]

where $q$ is an even number, $J_{i_{1}i_{2}\cdots i_{q/2},i_{q/2+1}\cdots i_{q}}$ is complex Gaussian random variable antisymmetry in $i_{1}i_{2}\cdots i_{q/2}$ and $i_{q/2+q}\cdots i_{q}$ and $J_{i_{1}i_{2}\cdots i_{q/2};i_{q/2+1}\cdots i_{q}}=J_{i_{q/2+1}\cdots i_{q};i_{1}i_{2}\cdots i_{q/2}}^{*}$, $\mu$ is the chemical potential for the fermion number operator $F=\sum_{i=1}^{N_{c}}c_{i}^{\dagger}c_{i}$. $H_{c}$ has one conserved quantity $F$, thus acquire a block diagonal structure labeled by $F=0,1,\cdots,N_{c}$.

The anti-unitary operators $P$, $R$ defined in Eq.(S14) satisfy:

[TABLE]

thus $P$ and $R$ always map $F$ to $N_{c}-F$.

One can also work out their squared values $P^{2}=(-1)^{\lfloor N_{c}/2\rfloor}$ and $R^{2}=(-1)^{\lfloor 3N_{c}/2\rfloor}$ and

[TABLE]

where $q$ is even.

So $PH_{c}P^{-1}=(-1)^{q/2}H_{c}$ only holds in the half-filling sector $F=N_{c}/2$ (canonical ensemble) or $\mu=0$ (grand-canonical ensemble). Similarly, $RH_{c}R^{-1}=(-1)^{q/2}H_{c}$ also only holds in the half-filling sector.

The classification of complex fermion SYK models is determined by $q$ and $N_{c}$. When $N_{c}$ is odd, no half-filling sector exists, thus all sectors belong to Class A(GUE). When $N_{c}$ is even, all the non-half-filling sectors still belong to Class A(GUE). But in the half-filling sector, $[P,H]=0$ if $q\pmod{4}=0$ and the classification is given by $P^{2}$ or $R^{2}$ (both $=T_{+}^{2}$); $\{P,H\}=0$ if $q\pmod{4}=2$ and the classification is given by $P^{2}$ or $R^{2}$ ( both $=T_{-}^{2}$). The classification results are listed in Table. S6.

Now we determine the energy level degeneracy. If $\psi_{F}$ is a eigenstate with eigenenergy $E$ in sector $F$, then the degeneracy can be determined by clarifying the relations between $\psi_{F}$ and $P\psi_{F}$, Note that due to $(-1)^{F}=(-1)^{\lceil N_{c}/2\rceil}PR$, $PR\psi_{F}\propto\psi_{F}$ and $R\psi_{F}\propto P\psi_{F}$.

(1) When $q\pmod{4}=0$

In this case, $[P,H_{c}]=0$. If $\psi_{F}$ is an eigenstate with eigenenergy $E$ in non-half-filling sector $F$ (namely, $F\neq N_{c}/2$), then $P\psi_{F}$ is an eigenstate with eigenenergy $E$ in another non-half-filling sector $N_{c}-F$. Thus $H_{c}$ has no degeneracy in non-half-filling sector.

If $N_{c}\pmod{4}=0$ and $\psi$ is an eigenstate with eigenenergy $E$ in half-filling sector $F=N_{c}/2$, then $P\psi\propto\psi$ is the same eigenstate as $\psi$ in half-filling sector $F=N_{c}/2$. Thus $H_{c}$ has no degeneracy in half-filling sector. If $N_{c}\pmod{4}=2$ and $\psi$ is an eigenstate with eigenenergy $E$ in half-filling sector $F=N_{c}/2$, then $\psi$ and $P\psi$ are two orthogonal eigenstates with the same eigenenergy $E$ in half-filling sector $F=N_{c}/2$. Thus $H_{c}$ has double degeneracy in half-filling sector.

(2) When $q\pmod{4}=2$

In this case, $\{P,H_{c}\}=0$. If $\psi_{F}$ is an eigenstate with eigenenergy $E$ in non-half-filling sector $F$ (namely, $F\neq N_{c}/2$), then $P\psi_{F}$ is an eigenstate with eigenenergy $-E$ in another non-half-filling sector $N_{c}-F$. Thus $H_{c}$ has no degeneracy in non-half-filling sector.

If $N_{c}\pmod{4}=0$ or $N_{c}\pmod{4}=2$, $\psi_{F}$ is an eigenstate with eigenenergy $E$ in half-filling sector $F=N_{c}/2$, then $P\psi_{F}$ is an eigenstate with eigenenergy $-E$ also in half-filling sector $F=N_{c}/2$. Thus $H_{c}$ has no degeneracy in half-filling sector, no matter $N_{c}\pmod{4}=0$ or $2$, but enjoys a spectral mirror symmetry.

The pattern of energy levels of $H_{c}$ is listed in Table S6. When comparing with Table 1, as expected, it is easy to see that at half-filling case in Table S6, the two tables match in both classification and degeneracy when setting $N_{c}=2N$ in both $q\pmod{4}=0$ and $q\pmod{4}=2$. It is tempting to put the complex SYK with even $q$ and the ${\cal N}=2$ SUSY SYK with odd $q$ on the same periodic table S6. It was known that the complex fermion SYK may be due to a charged ( Reissner-Nordstrom ) black hole which still has no angular momentum Sachdev2015 . So a periodic table of complex SYK and ${\cal N}=2$ SUSY SYK may also lead to a classification of charged quantum black holes. However, in sharp contrast to the ${\cal N}=1$ SUSY SYK, the ${\cal N}=2$ SUSY is not broken at any finite $N$, the Witten index $W$ is non-vanishing, there is an extensive ground state degeneracy equal to $W$ at the exact zero energy. It is also not known if this extensive zero mode characterized by the Witten index can be identified as the topological zero modes in the RMT. However, as elucidated in this work, there is no zero modes in the periodic table 1 of SYK and ${\cal N}=1$ SUSY SYK.