Fermionic projected entangled-pair states and topological phases

Multiplying two fMPOs $O^L_a$ and $O^L_b$ gives rise to a new fMPO with a tensor that can be written as

$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle \sum_{\alpha,\alpha^{\prime},i,k,\beta,\beta^{\prime}} (B_{ab}^{ik})_{(\alpha,\alpha^{\prime}),(\beta,\beta^{\prime})} \vert \alpha) \vert \alpha^{\prime}) \vert i\rangle \langle k \vert (\beta^{\prime}\vert (\beta\vert \nonumber \end{align*}$

where the ordering was chosen such that the fMPO coefficients reduce to a matrix product of the matrices $B_{ab}^{ik}$, which are given by

$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle (B_{ab}^{ik})_{(\alpha,\alpha^{\prime}),(\beta,\beta^{\prime})} = (-1)^{\vert \alpha^{\prime}\vert (\vert \alpha\vert +\vert \beta\vert)}\sum_j (B_a^{ij})_{\alpha,\beta} (B_b^{\,jk})_{\alpha^{\prime},\beta^{\prime}}. \nonumber \end{align*}$

Similar to the bosonic case, the fact that $O^L_aO^L_b = \sum_{c = 1}^N N_{ab}^c O_c^L$ for every L implies the existence of a gauge transformation $X_{ab}$ that simultaneously brings the matrices $B_{ab}^{ik}$ into a canonical form (block upper triangular), where the diagonal blocks correspond to $B_c^{ik}$ appearing $N_{ab}^c$ times [34].

From the columns of the the gauge transform $X_{ab}$ and the rows of its inverse $X_{ab}^{-1}$, we can build fermionic splitting and fusion tensors $\mathsf{X^{c}_{ab, \mu}}$ and $\mathsf{X^{c+}_{ab, \mu}}$ ($\mu = 1, \ldots, N_{ab}^c$), such that

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{X^{c+}_{ab,\mu}}\otimesg\mathsf{B[a]}\otimesg\mathsf{B[b]}\otimesg \mathsf{X_{ab,\mu}^c}) = \mathsf{B[c]}.\label{eq:reduction} \nonumber \end{align} \tag{ 12 }$

We introduce the following graphical notation for the tensors $\mathsf{B[a]}$, $\mathsf{X_{ab, \mu}^c}$ and $\mathsf{X_{ab, \mu}^{c+}}$

where the red (horizontal) indices represent the internal fMPO indices and the black (vertical) indices represent the external fMPO indices. We can then denote the contraction in equation (12) graphically as

Note that although the fMPO tensors $\mathsf{B[a]}$ have even parity, the fusion tensors have a well defined parity that can be either even or odd. This parity depends on the degeneracy label μ and adds a $\mathbb{Z}_2$ grading denoted as $\vert \mu\vert$ to the degeneracy space.

The fusion tensors satisfy following properties:

where $P_{ab}$ is the projector onto the support of the internal indices of the fMPO tensor $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{B[a]}\otimesg\mathsf{B[b]})$. For our purposes we are interested in fMPOs that satisfy a slightly stronger condition than equation (14). Namely, we assume that the following zipper condition holds:

Up to this point, the properties of fMPO super algebras are very similar to those of bosonic MPO algebras. We will now discuss the implications of the presence of $\newcommand{\e}{{\rm e}} \epsilon_a = 1$ irreducible fMPOs. Because the graded center of the matrices $B[a]^{ij}$ for $\newcommand{\e}{{\rm e}} \epsilon_a =1$ contains the odd matrix Y, it is clear that we can contract $\mathsf{Y}$ onto any index of a fusion tensor corresponding to an irreducible fMPO with $\newcommand{\e}{{\rm e}} \epsilon =1$ to get another fusion tensor that also satisfies the defining equations (14) and (16). Because $\mathsf{Y}$ is odd this changes the parity of the fusion tensor $\mathsf{X_{ab, \mu}^c}$. Let us start with the situation $\newcommand{\e}{{\rm e}} \epsilon_a=\epsilon_b=1$ and consider the matrix

where without loss of generality we take $\mathsf{X_{ab, \mu}^c}$ and $\mathsf{X_{ab, \mu}^{c+}}$ to have even parity. Equation (17) represents an odd matrix that commutes with the matrices $B[c]^{ij}$ because of equation (16). But $\newcommand{\e}{{\rm e}} \epsilon_c =0$ so the center of the matrix algebra $B[c]^{ij}$ consists only of multiples of the identity. For this reason, the matrix in equation (17) is zero when $\newcommand{\e}{{\rm e}} \epsilon_a = \epsilon_b = 1$. Similar reasoning shows that also the odd matrix

is zero when $\newcommand{\e}{{\rm e}} \epsilon_a=\epsilon_b = 1$. On the other hand, the matrix

is an even matrix commuting with all matrices $B[c]^{ij}$, which implies that it is a multiple of the identity. Since $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} (\mathsf{Y}\otimesg \mathsf{Y}){\hspace{0pt}}^2 = -\mathds{1}\otimesg\mathds{1}$ we thus find that the matrix in equation (19) equals $\pm i\mathds{1}$. Combining all the properties just derived we can conclude that $N_{ab}^c$ is a multiple of two when $\newcommand{\e}{{\rm e}} \epsilon_a = \epsilon_b = 1$. The index μ labeling the fusion tensors $\mathsf{X}_{ab, \mu}^c$ has a natural tensor product structure $\mu = (\hat{\mu}, \vert \mu\vert)$, where $\hat{\mu} \in \{1, \dots, N_{ab}^c/2\}$ and $\vert \mu\vert$ also denotes the parity of the fusion tensor $\mathsf{X}_{ab, (\hat{\mu}, \vert \mu\vert)}^c$. We will adopt following graphical notation for the fusion tensors and the property derived from matrix (19):

where $\newcommand{\e}{{\rm e}} \eta^c_{ab, \hat{\mu}} \in \{0, 1\}$ are discrete quantities that are part of the algebraic structure defining the fMPO super algebra.

Let us revisit the matrix in equation (17) when $\newcommand{\e}{{\rm e}} \epsilon_a = 1$ and $\newcommand{\e}{{\rm e}} \epsilon_b = 0$. Now $\newcommand{\e}{{\rm e}} \epsilon_c = 1$ so the fact that this odd matrix commutes with all $B[c]^{ij}$ implies that it is a multiple of $\mathsf{Y}$. Since $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} (\mathsf{Y}\otimesg \mathds{1}){\hspace{0pt}}^2 = -\mathds{1}\otimesg\mathds{1}$ this implies that

Similar reasoning for the matrix in equation (18) when $\newcommand{\e}{{\rm e}} \epsilon_a = 0$ and $\newcommand{\e}{{\rm e}} \epsilon_b = 1$ shows that

So when $\newcommand{\e}{{\rm e}} \epsilon_c = 1$ there is no further restriction on $N_{ab}^c$ and the parity of the fusion tensor for each μ is completely arbitrary. We will keep the graphical notation introduced in equation (13) for the even parity fusion tensor $\mathsf{X_{ab, \mu}^c}$ and use the left hand sides of equations (21) and (22) as a graphical notation for the odd fusion tensors $\mathsf{X_{ab, \mu}^c}$. In appendix A we give a more detailed derivation of the fusion tensors and their properties.

Associativity of the product of three fMPOs $O^L_aO^L_bO^L_c$ clearly implies that $\sum_d N_{ab}^dN_{dc}^e = \sum_f N_{bc}^{\,f} N_{af}^e$. Associativity also allows one to derive an important property of the fusion tensors. The fMPO tensor of $O^L_aO^L_bO^L_c$, $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{B[a]}\otimesg\mathsf{B[b]}\otimesg\mathsf{B[c]})$, can be written as a sum in two different ways by either applying equation (16) first to $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{B[a]}\otimesg\mathsf{B[b]})$ or first to $\mathcal{C}(\mathsf{B[b]}\otimes\mathsf{B[c]})$. Let us first consider the case where $\newcommand{\e}{{\rm e}} \epsilon \equiv 0$. Equality of the two sums in this case implies that the fusion tensors satisfy³

where $\left[F^{abc}_e \right]^{d, \mu\nu}_{f, \lambda\kappa}$ is an invertible even matrix. We will often refer to this identity as an F-move and to the matrices $\left[F^{abc}_e \right]^{d, \mu\nu}_{f, \lambda\kappa}$ as the F-symbols.

As is familiar from bosonic fusion categories, the F-symbols have to satisfy a consistency equation called the (super) pentagon equation. This consistency condition arises from equating the two different paths one can follow to get from $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}((\mathsf{X^{\,f}_{ab, \mu}}\otimesg\mathds{1}\otimesg\mathds{1})\otimesg(\mathsf{X^g_{fc, \nu}}\otimesg\mathds{1})\otimesg\mathsf{X^e_{gd, \rho}})$ to $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}((\mathds{1}\otimesg\mathds{1}\otimesg\mathsf{X^{\,j}_{cd, \delta}})\otimesg(\mathds{1}\otimesg\mathsf{X^i_{bj, \gamma}})\otimesg\mathsf{X^e_{ai, \omega}})$ using F-moves. These different paths are shown in figure 1. Written down explicitly, the super pentagon equation is

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:pentagon} \sum_{h,\sigma,\lambda,\kappa}[F^{abc}_g]^{\,f,\mu\nu}_{h,\sigma\lambda}[F^{ahd}_e]^{g,\sigma\rho}_{i,\omega\kappa}[F^{bcd}_i]^{h,\lambda\kappa}_{j,\gamma\delta} = \sum_\sigma [F^{\,fcd}_e]^{g,\nu\rho}_{j,\sigma\delta}[F^{abj}_e]^{\,f,\mu\sigma}_{i,\omega\gamma}(-1)^{\vert \mu\vert \vert \delta\vert }\,, \nonumber \end{align} \tag{ 24 }$

where $\vert \mu\vert$ ($\vert \delta\vert$) denotes the parity of fusion tensor $\mathsf{X^{\,f}_{ab, \mu}}$ ($\mathsf{X^{\,j}_{cd, \delta}}$). We see that for $\newcommand{\e}{{\rm e}} \epsilon \equiv 0$, the only difference between the fermionic pentagon equation and the standard, bosonic pentagon equation is the minus sign depending on $\vert \mu\vert$ and $\vert \delta\vert$. This sign arises from the reordering of two fusion tensors so that a subsequent F-move can be applied. This step is also shown in figure 1. For $\newcommand{\e}{{\rm e}} \epsilon \equiv 0$ the super pentagon equation was previously derived in the construction of fermionic string-net models [23, 24].

Zoom In Zoom Out Reset image size

Let us now also take fMPOs with $\newcommand{\e}{{\rm e}} \epsilon =1$ into account. As in equation (23), we want to relate $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{X}_{ab, \mu}^{d} \otimesg \mathsf{X}_{dc, \nu}^e)$ and $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{X}_{bc, \kappa}^{\,f} \otimesg \mathsf{X}_{af, \lambda}^e)$, which both reduce $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\mathsf{B[a]}\otimesg\mathsf{B[b]}\otimesg\mathsf{B[c]})$ to a direct sum of $\mathsf{B[e]}$. Since $\mathsf{B[e]}$ has a non-trivial center $\{\mathds{1}_e, \mathsf{Y}_e\}$ when $\newcommand{\e}{{\rm e}} \epsilon_e=1$ we find

$\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{X}_{ab,\mu}^{d} \otimesg \mathsf{X}_{dc,\nu}^e)= \left\{\begin{array}{@{}ll@{}} \sum_{f,\lambda,\kappa} \mathcal{C}(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \mathsf{X}_{af,\lambda}^e\otimesg ([F^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathds{1}_e)),&\epsilon_e=0 \nonumber \\ \sum_{f,\lambda,\kappa} \mathcal{C}(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \mathsf{X}_{af,\lambda}^e \otimesg ([F^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathds{1}_e + [G^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa}\mathsf{Y}_e)),&\epsilon_e=1 \end{array}\right. \nonumber \end{align*}$

From parity consideration, it follows for $\newcommand{\e}{{\rm e}} \epsilon_e=0$ that $[F^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=0$ if $\vert \mu\vert +\vert \nu\vert +\vert \kappa\vert +\vert \lambda\vert~$${\rm mod}~2 \neq 0$. For $\newcommand{\e}{{\rm e}} \epsilon_e=1$, we have $[F^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=0$ if $\vert \mu\vert +\vert \nu\vert +\vert \kappa\vert +\vert \lambda\vert~{\rm mod}~2 = \vert \mu\vert +\vert \kappa\vert~$$~{\rm mod}~2 \neq 0$ and $[G^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=0$ if $\vert \mu\vert +\vert \nu\vert +\vert \kappa\vert +\vert \lambda\vert~{\rm mod}~2 = \vert \mu\vert +\vert \kappa\vert~{\rm mod}~2 \neq 1$. If the fusion tensors are isometric, such that $\mathsf{X}^{c+}_{ab, \mu} = \mathsf{X}^{c\dagger}_{ab, \mu}$, we find that

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \left\{\begin{array}{@{}ll@{}} \sum_{f,\lambda\kappa} [\bar{F}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [F_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d,d^{\prime}}\delta_{\mu,\mu^{\prime}}\delta_{\nu,\nu^{\prime}},&\epsilon_e = 0, \nonumber \\ \sum_{f,\lambda\kappa}[\bar{F}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [F_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} + [\bar{G}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [G_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d,d^{\prime}}\delta_{\mu,\mu^{\prime}}\delta_{\nu,\nu^{\prime}},&\epsilon_e = 1, \nonumber \\ \sum_{f,\lambda\kappa} [\bar{F}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [G_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} - [\bar{G}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [F_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = 0,&\epsilon_e = 1. \end{array}\right. \nonumber \end{align} \tag{ 25 }$

This means that, for $\newcommand{\e}{{\rm e}} \epsilon_e=0$, $F^{abc}_{e}$ is itself a unitary matrix (note that it's square as $\sum_{d} N_{ab}^dN_{dc}^e= \sum_{f} N_{bc}^{\,f} N_{af}^e$), while for $\newcommand{\e}{{\rm e}} \epsilon_e=1$, the matrix $F^{abc}_e \otimes \mathds{1} + G^{abc}_e\otimes y$ is unitary and symplectic.

Having the F-move interact with the virtual fMPO indices is inconvenient in order to derive the super pentagon equation and to construct an explict fPEPS tensor satisfying the pulling through equation in the following section. Indeed, the latter requires that we have scalar coefficient $[F_{e}^{abc}]^{d, \mu\nu}_{f, \lambda\kappa}$ rather than a matrix. We can therefore switch to a different convention for the fusion tensors, where we redefine $\frac{1}{\sqrt{2}}\mathsf{X}_{ab, \mu}^{c} \to \tilde{\mathsf{X}}_{ab, (\hat{\mu}, 0)}^{c}$ and $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \frac{1}{\sqrt{2}}\mathcal{C}(\mathsf{X}_{ab, \mu}^{c} \otimesg \mathsf{Y}_c) \to \tilde{\mathsf{X}}_{ab, (\hat{\mu}, 1)}^{c}$ when $\newcommand{\e}{{\rm e}} \epsilon_c=1$, while $\tilde{\mathsf{X}}_{ab, (\hat{\mu}, \vert \mu\vert)}^{c} = \mathsf{X}_{ab, (\hat{\mu}, \vert \mu\vert)}^{c}$ when $\newcommand{\e}{{\rm e}} \epsilon_a=\epsilon_b=1$ and $\tilde{\mathsf{X}}_{ab, \mu}^{c} = \mathsf{X}_{ab, \mu}^{c}$ when $\newcommand{\e}{{\rm e}} \epsilon_a=\epsilon_b=0$. In all cases, $\vert \mu\vert$ denotes the parity of the fusion tensor $\tilde{\mathsf{X}}_{ab, \mu}^{c}$. The factors $\frac{1}{\sqrt{2}}$ are introduced such that

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \sum_{c,\mu} \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{c} \otimesg \tilde{\mathsf{X}}_{ab,\mu}^{c+}) = \sum_c \sum_{\hat{\mu}=1}^{N_{ab}^{c}}\sum_{\vert \mu\vert =0,1}\mathcal{C}(\tilde{\mathsf{X}}_{ab,(\hat{\mu},\vert \mu\vert)}^{c} \otimesg \tilde{\mathsf{X}}_{ab,(\hat{\mu},\vert \mu\vert)}^{c+}) \nonumber \end{align} \tag{ 26 }$

still defines a properly normalized projector onto the support subspace of the tensor $\mathsf{B}_{ab}$, while

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{c+} \otimesg \tilde{\mathsf{X}}_{ab,\nu}^{d}) = \delta_{c,d} \left\{\begin{array}{@{}ll@{}} \delta_{\mu,\nu} \mathds{1}_c, &\epsilon_c=0 \nonumber \\ \frac{1}{2} (\delta_{\mu,\nu} \mathds{1}_{c} - Y_{\mu,\nu} \mathsf{Y}_c),&\epsilon_c=1 \end{array}\right. \nonumber \end{align} \tag{ 27 }$

with $Y_{\mu, \nu} = \delta_{\hat{\mu}, \hat{\nu}}y_{\vert \mu\vert, \vert \nu\vert } = \delta_{\hat{\mu}, \hat{\nu}}( \vert \nu\vert - \vert \mu\vert)$. The latter expression for the case $\newcommand{\e}{{\rm e}} \epsilon_c$ is reminiscent of the pseudo-inverse of a Majorana fMPS.

The fusion tensors $\tilde{\mathsf{X}}_{ab, \mu}^c$ have the degeneracy structure $\mu=(\hat{\mu}, \vert \mu\vert)$ as soon as either $\newcommand{\e}{{\rm e}} \epsilon_a$, $\newcommand{\e}{{\rm e}} \epsilon_b$ or $\newcommand{\e}{{\rm e}} \epsilon_c$ is nonzero. Contraction with $\mathsf{Y}_c$ switches between $(\hat{\mu}, 0)$ and $(\hat{\mu}, 1)$ if $\newcommand{\e}{{\rm e}} \epsilon_c=1$, i.e.

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{c}\otimesg \mathsf{Y}_c) = \sum_{\nu} Y_{\nu,\mu} \tilde{\mathsf{X}}_{ab,\nu}^{c} \quad {\rm if}~\epsilon_c=1. \nonumber \end{align} \tag{ 28 }$

For the case with $\newcommand{\e}{{\rm e}} \epsilon_a = \epsilon_b=1$ we have:

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{Y}_a \otimesg \tilde{\mathsf{X}}_{ab,\mu}^{c}) = \sum_{\nu} (M_{ab}^c)_{\nu,\mu} \tilde{\mathsf{X}}_{ab,\nu}^{c} \nonumber \end{align} \tag{ 29 }$

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{Y}_b \otimesg \tilde{\mathsf{X}}_{ab,\mu}^{c}) = \sum_{\nu} (L_{ab}^c)_{\nu,\mu} \tilde{\mathsf{X}}_{ab,\nu}^{c}\,, \nonumber \end{align} \tag{ 30 }$

where $M_{ab}^{c}$ and $L_{ab}^{c}$ are odd (i.e. they are nonzero only for $\vert \mu\vert \neq \vert \nu\vert$). From the results of the previous section it follows that $\left(M_{ab}^c\right)_{\mu\nu} = \delta_{\hat{\mu}, \hat{\nu}}y_{\vert \mu\vert, \vert \nu\vert }$ and $\newcommand{\e}{{\rm e}} L_{ab}^c = (-1){\hspace{0pt}}^{\eta_{ab}^c+1}i\delta_{\hat{\mu}, \hat{\nu}} x_{\vert \mu\vert, \vert \nu\vert }$, with $x_{\vert \mu\vert \vert \nu\vert } = 1 - \delta_{\vert \mu\vert \vert \nu\vert }$.

When $\newcommand{\e}{{\rm e}} \epsilon_c=0$ we have $\mu=1, \ldots, N_{ab}^c$ whereas if $\newcommand{\e}{{\rm e}} \epsilon_c=1$, we have $\hat{\mu}=1, \ldots, N_{ab}^c$ and thus $\mu=1, \ldots, 2N_{ab}^c$. But here, $N_{ab}^c$ only represents the number of times O_c originates from multiplying O_a and O_b if these fMPOs are built from the fermionic tensors B_a, B_b and B_c without normalization factor. Since we take to act as a $\mathbb{Z}_2$ grading, we can define all Majorana fMPOs to have an additional global factor $1/\sqrt{2}$, so that in the case $\newcommand{\e}{{\rm e}} \epsilon_a=\epsilon_b=1$ we would also have $\mu=1, \ldots, 2N_{ab}^c$, i.e. $\hat{\mu}=1, \ldots, N_{ab}^c$, if we fix $N_{ab}^{c}$ in the relation $O_a O_b = \sum_{c} N_{ab}^{c} O_c$.

The advantage of working with an overcomplete basis of fusion tensors is that we can now write the F-move as an even transformation acting purely on the degeneracy spaces and not on the virtual indices of fMPOs, exactly as in the bosonic case, i.e. we can write

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{d} \otimesg \tilde{\mathsf{X}}_{dc,\nu}^e)= \sum_{f,\lambda,\kappa} [\tilde{F}^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathcal{C}(\tilde{\mathsf{X}}_{bc,\kappa}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e). \nonumber \end{align} \tag{ 31 }$

Let us explain this in more detail by providing an explicit recipe for going from the F-symbols to the $\tilde{F}$-symbols.

• 1.  
  Step 1: We first write

  $\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{X}_{ab,\mu}^{d} \otimesg \tilde{\mathsf{X}}_{dc,\nu}^e)= \sum_{f,\lambda,\kappa} [(\,f_1)^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathcal{C}(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e)\,, \nonumber \end{align} \tag{ 32 }$

  where $[(\,f_1){\hspace{0pt}}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=[F^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}$ if $\newcommand{\e}{{\rm e}} \epsilon_e=0$, and

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {[(\,f_1)^{abc}_e]}^{d,\mu(\hat{\nu},0)}_{f,(\hat{\lambda},0)\kappa} = {[F^{abc}_e]}^{d,\mu\hat{\nu}}_{f,\hat{\lambda}\kappa} \qquad {[(\,f_1)^{abc}_e]}^{d,\mu(\hat{\nu},1)}_{f,(\hat{\lambda},0)\kappa} = {[G^{abc}_e]}^{d,\mu\hat{\nu}}_{f,\hat{\lambda}\kappa} \nonumber \end{align} \tag{ 33 }$

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {[(\,f_1)^{abc}_e]}^{d,\mu(\hat{\nu},0)}_{f,(\hat{\lambda},1)\kappa} = -{[G^{abc}_e]}^{d,\mu\hat{\nu}}_{f,\hat{\lambda}\kappa} \qquad {[(\,f_1)^{abc}_e]}^{d,\mu(\hat{\nu},1)}_{f,(\hat{\lambda},1)\kappa} = {[F^{abc}_e]}^{d,\mu\hat{\nu}}_{f,\hat{\lambda}\kappa} \nonumber \end{align} \tag{ 34 }$

  if $\newcommand{\e}{{\rm e}} \epsilon_e=1$. From the properties of F and G, we can check that $(\,f_1){\hspace{0pt}}^{abc}_e$ is still a unitary matrix, and is even, i.e. its elements $[(\,f_1){\hspace{0pt}}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}$ vanish if $\vert \mu\vert +\vert \nu\vert +\vert \kappa\vert +\vert \lambda\vert~{\rm mod}~2 \neq 0$. Furthermore, in the isometric case, $(\,f_1){\hspace{0pt}}^{abc}_e$ is unitary, i.e.

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{f,\lambda\kappa} [(\bar{f}_1)_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [(\,f_1)_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d^{\prime},d}\delta_{\mu^{\prime},\mu}\delta_{\nu^{\prime},\nu}\,, \nonumber \end{align} \tag{ 35 }$

  from which also follows

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{d,\mu,\nu} [(\,f_1)_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} [(\bar{f}_1)_e^{abc}]^{d,\mu\nu}_{f^{\prime},\lambda^{\prime}\kappa^{\prime}} = \delta_{f,f^{\prime}}\delta_{\kappa,\kappa^{\prime}}\delta_{\lambda,\lambda^{\prime}}\, . \nonumber \end{align} \tag{ 36 }$
• 2.  
  Step 2:

  $\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{d} \otimesg \tilde{\mathsf{X}}_{dc,\nu}^e)= \sum_{f,\lambda,\kappa} [(\,f_2)^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathcal{C}(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e) \nonumber \end{align} \tag{ 37 }$

  with $[(\,f_2){\hspace{0pt}}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=[(\,f_1){\hspace{0pt}}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}$ if $\newcommand{\e}{{\rm e}} \epsilon_d=0$. If $\newcommand{\e}{{\rm e}} \epsilon_d=1$, we obtain

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle [(\,f_2)^{abc}_e]^{d,(\hat{\mu},0)\nu}_{f,\lambda\kappa} = \frac{1}{\sqrt{2}}[(\,f_1)^{abc}_e]^{d,\hat{\mu}\nu}_{f,\lambda\kappa}, \nonumber \end{align} \tag{ 38 }$

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {[(\,f_2)^{abc}_e]}^{d,(\hat{\mu},1)\nu}_{f,\lambda\kappa} = \frac{1}{\sqrt{2}}\sum_{\nu^{\prime}} (M_{dc}^{e})_{\nu^{\prime},\nu} {[(\,f_1)^{abc}_e]}^{d,\hat{\mu}\nu^{\prime}}_{f,\lambda\kappa}. \nonumber \end{align} \tag{ 39 }$

  Note that $(\,f_2){\hspace{0pt}}^{abc}_e$ is still even, because $M_{dc}^{e}$ is odd. Furthermore, in the isometric case, we obtain

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{f,\lambda\kappa} [(\bar{f}_2)_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [(\,f_2)_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d^{\prime},d}\left\{\begin{array}{@{}ll@{}} \delta_{\mu^{\prime},\mu}\delta_{\nu^{\prime},\nu},&\epsilon_d = 0 \nonumber \\ \left[\delta_{\mu^{\prime},\mu} \delta_{\nu^{\prime},\nu} + Y_{\mu^{\prime},\mu} (M_{dc}^{e})_{\nu^{\prime},\nu} \right]/2,&\epsilon_d = 1 \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 40 }$

  and

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{d,\mu,\nu} [(\,f_2)_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa}[(\bar{f}_2)_e^{abc}]^{d,\mu\nu}_{f^{\prime},\lambda^{\prime}\kappa^{\prime}} = \delta_{f,f^{\prime}}\delta_{\kappa,\kappa^{\prime}}\delta_{\lambda,\lambda^{\prime}}. \nonumber \end{align} \tag{ 41 }$

  Note that if there is a d with $\newcommand{\e}{{\rm e}} \epsilon_d=1$ present, the matrix $(\,f_2){\hspace{0pt}}^{abc}_e$ has more columns than rows and can therefore no longer be unitary. However, the above expression shows that it is still isometric and defines a projector upon premultiplication with its hermitian conjugate.
• 3.  
  Step 3:

  $\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\tilde{\mathsf{X}}_{ab,\mu}^{d} \otimesg \tilde{\mathsf{X}}_{dc,\nu}^e)= \sum_{f,\lambda,\kappa} [\tilde{F}^{abc}_e]^{d,\mu\nu}_{f,\lambda\kappa} \mathcal{C}(\tilde{\mathsf{X}}_{bc,\kappa}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e) \nonumber \end{align} \tag{ 42 }$

  with $[\tilde{F}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}=[(\,f_2){\hspace{0pt}}^{abc}_e]^{d, \mu\nu}_{f, \lambda\kappa}$ if $\newcommand{\e}{{\rm e}} \epsilon_f=0$. If $\newcommand{\e}{{\rm e}} \epsilon_f=1$, we obtain the required relation by the following substitution:

  $\begin{align*} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e)&= \mathcal{C}\left(\mathsf{X}_{bc,\kappa}^{\,f} \otimesg \frac{1}{2}(\mathds{1}_f - \mathsf{Y}_f\otimesg \mathsf{Y}_f)\otimesg \tilde{\mathsf{X}}_{af,\lambda}^e\right)\nonumber \\ &= \frac{1}{\sqrt{2}} \mathcal{C}\left(\tilde{\mathsf{X}}_{bc,(\hat{\kappa},0)}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e\right)-\frac{1}{\sqrt{2}} \mathcal{C}\left(\tilde{\mathsf{X}}_{bc,(\kappa,1)}^{\,f} \otimesg (\mathsf{Y}_f\otimesg \tilde{\mathsf{X}}_{af,\lambda}^e)\right)\nonumber \\ &= \frac{1}{\sqrt{2}} \mathcal{C}\left(\tilde{\mathsf{X}}_{bc,(\hat{\kappa},0)}^{\,f} \otimesg \tilde{\mathsf{X}}_{af,\lambda}^e\right)-\frac{1}{\sqrt{2}} \sum_{\lambda^{\prime}} (L_{af}^e)_{\lambda^{\prime},\lambda} \mathcal{C}\left(\tilde{\mathsf{X}}_{bc,(\kappa,1)}^{\,f}\otimesg \tilde{\mathsf{X}}_{af,\lambda^{\prime}}^e)\right). \nonumber \end{align*}$

  So we get for the final $\tilde{F}$-symbols

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle [\tilde{F}^{abc}_e]^{d,\mu\nu}_{f,\lambda(\hat{\kappa},0)} = \frac{1}{\sqrt{2}}[(\,f_2)^{abc}_e]^{d,\mu\nu}_{f,\lambda\hat{\kappa}}, \nonumber \end{align} \tag{ 43 }$

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {[\tilde{F}^{abc}_e]}^{d,\mu\nu}_{f,\lambda(\hat{\kappa},1)} = -\frac{1}{\sqrt{2}}\sum_{\lambda^{\prime}} (L_{af}^{e})_{\lambda,\lambda^{\prime}} [(\,f_2)^{abc}_e]^{d,\mu\nu}_{f,\lambda^{\prime}\hat{\kappa}}. \nonumber \end{align} \tag{ 44 }$

  The resulting $\tilde{F}^{abc}_e$ is even (because $L_{af}^{e}$ is odd), not necessarily square and in the isometric case satisfies

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:isometric1} \sum_{f,\lambda\kappa} [\bar{\tilde{F}}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [\tilde{F}_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d^{\prime},d}\left\{\begin{array}{@{}ll@{}} \delta_{\mu^{\prime},\mu}\delta_{\nu^{\prime},\nu^{\prime}},&\epsilon_d = 0, \nonumber \\ \left[\delta_{\mu^{\prime},\mu} \delta_{\nu^{\prime},\nu} + Y_{\mu^{\prime},\mu} (M_{dc}^{e})_{\nu^{\prime},\nu} \right]/2,&\epsilon_d = 1, \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 45 }$

  and

  $\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:isometric2} \sum_{d,\mu,\nu} [\tilde{F}_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa}[\bar{\tilde{F}}_e^{abc}]^{d,\mu\nu}_{f^{\prime},\lambda^{\prime}\kappa^{\prime}} = \delta_{f,f^{\prime}} \left\{\begin{array}{@{}ll@{}} \delta_{\kappa,\kappa^{\prime}}\delta_{\lambda,\lambda^{\prime}},&\epsilon_f = 0, \nonumber \\ \left[\delta_{\kappa,\kappa^{\prime}} \delta_{\lambda^{\prime},\lambda^{\prime}} + Y_{\kappa,\kappa^{\prime}} (L_{af}^{e})_{\lambda,\lambda^{\prime}} \right]/2,&\epsilon_f = 1. \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 46 }$

Fusing the product of four MPOs using these fusion tensors in two different ways gives rise to the super pentagon equation for $\tilde{F}$.

As a final point on fMPO super algebras, we want to consider the irreducible fMPOs for which $a^*=a$, i.e. the irreducible fMPOs satisfying $(O^L_a){\hspace{0pt}}^\dagger = O^L_a$. It was shown in [13] that in the bosonic case one can associate an invariant $\varkappa_a \in \{-1, 1\}$ to such MPOs, which coincides with the Frobenius–Schur indicator from fusion categories. In the fermionic case, this invariant has a natural generalization. A crucial observation to obtain the correct generalization is that Hermitian conjugation involves a reordering of the basis vectors for operators that act on the graded tensor product of super vector spaces. Hermitian conjugation is most naturally defined in the following basis, where contraction coincides with matrix multiplication of the components:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \left(\sum_{i_1,i_2,j_1,j_2}M_{i_1,i_2,j_1,j_2}\vert i_1\rangle \vert i_2\rangle \langle\,j_2\vert \langle\,j_1\vert \right)^\dagger = \sum_{i_1,i_2,j_1,j_2}\bar{M}_{i_1,i_2,j_1,j_2}\vert\,j_1\rangle \vert\,j_2\rangle \langle i_2\vert \langle i_1\vert \, . \nonumber \end{align} \tag{ 47 }$

However, the natural basis in which fMPOs are expressed is of the form $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \vert i_1\rangle\langle\,j_1\vert \otimesg \vert i_2\rangle\langle\,j_2\vert$, on which Hermitian conjugation then acts as

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle (\vert i_1\rangle\langle\,j_1\vert \otimesg \vert i_2\rangle\langle\,j_2\vert)^\dagger = (-1)^{(\vert i_1\vert + \vert\,j_1\vert)(\vert i_2\vert +\vert\,j_2\vert)} \vert\,j_1\rangle\langle i_1\vert \otimesg\vert\,j_2\rangle\langle i_2\vert . \nonumber \end{align} \tag{ 48 }$

So Hermitian conjugation does not only result in complex conjugation for the components but also produces additional signs. For this reason it might not be clear at first sight that $(O_a^L){\hspace{0pt}}^\dagger$ is actually also an fMPO. However, the minus sign produced by Hermitian conjugation is the same as the minus sign one gets from reordering of fermion modes under reflection symmetry, and we know this sign can be absorbed in the fMPO tensors by redefining them as $B^{i, j} \rightarrow P^{\vert i\vert +\vert\,j\vert }B^{i, j}$ (or equivalently as $B^{i, j}P^{\vert i\vert +\vert\,j\vert }$) [15], where P is the matrix containing the components of $\mathsf{P}$ as defined earlier. One can check this by explicitly evaluating the redefined fMPO components:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle &\qquad {\rm tr}(P^{\vert i_1\vert +\vert\,j_1\vert }B^{i_1,j_1}\dots P^{\vert i_{N-1}\vert +\vert\,j_{N-1}\vert }B^{i_{N-1},j_{N-1}}P^{\vert i_{N}\vert +\vert\,j_{N}\vert }B^{i_N,j_N}) \nonumber \\ &= (-1)^{(\vert i_{N}\vert +\vert\,j_{N}\vert)(\vert i_1\vert +\dots \vert\,j_{N-1}\vert)} \nonumber \\ &\qquad {\rm tr}(P^{\vert i_1\vert +\vert\,j_1\vert +\vert i_{N}\vert +\vert\,j_{N}\vert }B^{i_1,j_1}\dots P^{\vert i_{N_1}\vert +\vert\,j_{N-1}\vert }B^{i_{N-1},j_{N-1}}B^{i_N,j_N}) \nonumber \\ &= (-1)^{(\vert i_{N}\vert +\vert\,j_{N}\vert)(\vert i_1\vert +\dots \vert\,j_{N-1}\vert) + (\vert i_{N-1}\vert +\vert\,j_{N-1}\vert)(\vert i_1\vert +\dots \vert\,j_{N-2}\vert)} \nonumber \\ & \qquad{\rm tr}(P^{\vert i_1\vert +\vert\,j_1\vert +\vert i_{N}\vert +\vert\,j_{N}\vert + \vert i_{N_1}\vert +\vert\,j_{N-1}\vert }B^{i_1,j_1}\dots B^{i_{N-1},j_{N-1}}B^{i_N,j_N}) \nonumber \\ &= \dots \nonumber \\ &= (-1)^{(\vert i_{N}\vert +\vert\,j_{N}\vert)(\vert i_1\vert +\dots \vert\,j_{N-1}\vert) + (\vert i_{N-1}\vert +\vert\,j_{N-1}\vert)(\vert i_1\vert +\dots \vert\,j_{N-2}\vert) +\dots + (\vert i_2\vert +\vert\,j_2\vert)(\vert i_1\vert +\vert\,j_1\vert)} \nonumber \\ & \qquad {\rm tr}(P^{\sum_{\alpha=1}^N (\vert i_\alpha\vert +\vert\,j_\alpha\vert)}B^{i_1,j_1}\dots B^{i_{N-1},j_{N-1}}B^{i_N,j_N})\, . \nonumber \end{align} \tag{ 49 }$

Since we work with anti-periodic boundary conditions all irreducible fMPOs are even so $\sum_{\alpha=1}^N (\vert i_\alpha\vert +\vert\,j_\alpha\vert)= 0$ mod 2, which indeed shows that $P^{\vert i\vert +\vert\,j\vert }B^{i, j}$ produces the original fMPO with the desired minus sign.

The property $(O_a^L){\hspace{0pt}}^\dagger = O_a^\dagger$ now implies that the matrices of tensor components $B^{i, j}$ satisfy [34]

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle P^{\vert i\vert +\vert\,j\vert }\bar{B}_a^{\,j,i} = Z_a^{-1} B_a^{i,j}Z_a\,, \nonumber \end{align} \tag{ 50 }$

where Z_a is an invertible matrix with parity $\mu_a$. Iterating this relation twice we find

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:Za} (-1)^{\mu_a(\vert i\vert +\vert\,j\vert)}B^{i,j}_a = \left(\bar{Z}^{-1}_aZ^{-1}_a \right)B^{i,j}_a\left(Z_a\bar{Z}_a \right)\, . \nonumber \end{align} \tag{ 51 }$

If $\newcommand{\e}{{\rm e}} \epsilon_a = 0$ the center of the algebra spanned by $B^{i, j}_a$ consists only of multiples of the identity. Therefore, if $\mu_a = 0$, we can conclude from (51) that $Z_a\bar{Z}_a = \alpha \mathds{1}$ and thus $\bar{Z}_aZ_a = \bar{\alpha}\mathds{1}$, where without loss of generality we can take α to be a phase by rescaling Z_a. Combining these two equations gives $\alpha^2 = 1$ and thus $Z_a\bar{Z}_a = (-1){\hspace{0pt}}^{\rho_a} \mathds{1}$, where $\rho_a\in \{0, 1\}$. If $\mu_a = 1$, we similarly find that $Z_a\bar{Z}_a = (-1){\hspace{0pt}}^{\rho_a}i P$. For $\newcommand{\e}{{\rm e}} \epsilon_a = 1$ the center of the algebra spanned by $B_a^{i, j}$ contains the odd matrix Y, so that both Z_a and YZ_a are valid gauge transformations satisfying (51). This implies that the parity of Z_a is ambiguous and we can take it to be even. In this case we find similarly to the situation with $\newcommand{\e}{{\rm e}} \epsilon_a =0$ that $Z_a\bar{Z}_a = (-1){\hspace{0pt}}^{\hat{\rho}_a}\mathds{1}$. By defining $\newcommand{\e}{{\rm e}} Z^1_a \equiv YZ_a$ one can obtain another invariant by $Z_a^1\bar{Z}_a^1 = (-1){\hspace{0pt}}^{\tilde{\rho}_a}iP$. One can check that these two invariants are independent. The invariant obtained from the odd gauge transformation Z_aY, however, is not independent. So in total we have found eight different possibilities. For $\newcommand{\e}{{\rm e}} \epsilon_a = 0$ we have four possibilities labeled by $\mu_a$ and $\rho_a$. When $\newcommand{\e}{{\rm e}} \epsilon_a =1$ we also find four possibilies, labeled by $\hat{\rho}_a$ and $\tilde{\rho}_a$. Using similar techniques as for fermionic matrix product states with time reversal symmetry or reflection symmetry one can show that these eight possibilies form a $\mathbb{Z}_8$ group where the group structure corresponds to taking the graded tensor product of fMPOs [15]. So if we take the invariant $\newcommand{\e}{{\rm e}} \epsilon_a$ as part of the definition of the Frobenius–Schur indicator we see that it is isomorphic to $\mathbb{Z}_8$ in the fermionic case, while it is only isomorphic to $\mathbb{Z}_2$ in the bosonic case.

For simplicity we will restrict our construction to the honeycomb lattice. To specify fermionic tensors one does not only have to specify the coefficients, but also in what ordering of the basis vectors these coefficients are defined. For the fermionic PEPS tensors on the A-sublattice we will choose the following internal ordering:

where ν is the physical index and $\lambda, \gamma, \beta$ are the virtual ones. Note that the arrows in the graphical notation denote which indices correspond to bra's, and which to kets. In the basis just specified, the tensor components are

This graphical notation requires some explanation. Each index is specified by four labels: three labels are denoted by Latin letters and one label is denoted by a Greek letter, which is also exactly the data that specified a fusion tensor $\mathsf{X^c_{ab, \mu}}$ in the previous section. Each external line in the graphical notation carries a label denoted by a Latin letter. The tensor components are zero when lines that are connected in the body of the tensor carry a different label. This is taken into account by the delta tensors in equation (53). However, in the remainder of this paper these delta conditions will be implicit in our definition of fixed-point tensor components and should be clear from the graphical notation. The physical index is labeled by the three labels carried by the lines that end in the body of the tensor (in the figure these are labels $b, c$ and f) and a corresponding Greek label (κ in the figure). The possibly non-zero tensor components are given by the $\tilde{F}$-symbols of the previous section, where each of the four tensor indices maps to a fusion tensor that defines the $\tilde{F}$-symbol. The parity of the index also equals the parity of the corresponding fusion tensor.

The tensors on the B-sublattice are defined with following internal ordering:

And in this basis, the tensor coefficients are analogously represented as

where the bar denotes complex conjugation. All PEPS tensor components are given in terms of $\tilde{F}$ symbols. When $\newcommand{\e}{{\rm e}} \epsilon\equiv 0$ the $\tilde{F}$-symbols are equivalent to the standard F-symbols and the fermionic PEPS is very closely related to the bosonic string-net PEPS. However, when taking Majorana fMPOs into account, the $\tilde{F}$ symbols are a particular choice of associators, and their explicit construction is given in section 3.2.

Let us also comment on the choice of arrows in our definition of the PEPS tensors. Reversing the arrows interchanges bra's with kets and for fermionic PEPS this has a non-trivial effect for the simple reason that $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}((\alpha\vert \otimesg \vert \beta)) = \delta_{\alpha, \beta}$ while $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathcal{C}(\vert \beta)\otimesg (\alpha\vert)=(-1){\hspace{0pt}}^{\vert \alpha\vert }\delta_{\alpha, \beta}$. From this we see that reversing the arrow on a link is equivalent to inserting a parity matrix $P = \mathds{1}_0\oplus -\mathds{1}_1$ on the corresponding virtual index in the contracted network, where $\mathds{1}_0$ ($\mathds{1}_1$) is the identity on the parity even (odd) subspace. So if we would flip all the arrows surrounding a vertex, the three resulting parity matrices on the neighbouring virtual indices can be intertwined to a parity matrix on the physical index since the PEPS tensors are even. This shows that to every fermionic PEPS we can actually associate an entire family of PEPS, that are related to the original one by on-site parity actions, by flipping the arrows surrounding vertices. For this reason, the choice of arrows is very reminiscent of a lattice spin structure.

We will define two types of tensors to construct fMPOs on the virtual level of the fermionic PEPS. The first, right-handed type, defined with the internal ordering

has components which are again determined by the $\tilde{F}$-symbols in the following way:

In appendix B we show that the fMPOs constructed from tensors (56) and (57) form an explicit representation of the fMPO algebra whose $\tilde{F}$-symbols we took to define the tensor components. To place the fMPO on the virtual level of the fermionic PEPS we will introduce an additional convention. The closed fMPO should be interpreted as a polygon, i.e. as a closed collection of straight lines and angles between them. On every angle we place a diagonal matrix that inserts some weights, depending on the labels carried by the outer lines. The rule to add the weights is the following: to each label a we associate a positive number d_a (the choice of d_a is not arbitrary as we will see further on) and the weights are then given by $d_a^{\frac{1}{2}(1-\frac{\alpha}{\pi})}$, where α is the inner (outer) angle in radians for the inner (outer) line. For example, when the fMPO contains an angle of $\frac{2\pi}{3}$ the weights are:

For notational simplicity this convention will always be implicit in our graphical notation from now on.

The reason to define the right-handed fMPO tensors as in (56) and (57) is that the pentagon equation now implies that the following pulling through identity holds:

Note that equation (59) is only equivalent to the pentagon equation when we use the $\tilde{F}$-symbols in defining the tensor components. The underlying reason is as follows. Every index of the fixed point tensors coresponds to a fusion tensor, and the four fusion tensors from every index in a tensor together correspond to an F move whose $\tilde{F}$-symbol determines the tensor component. Since the indices are defined in a super vector space, an even and an odd vector are necessarily orthogonal. However, as explained in section 3, when $\newcommand{\e}{{\rm e}} \epsilon_c=1$, the even and odd version of the fusion tensor $\mathsf{X}_{ab, \mu}^c$ correspond to the same fusion channel. Because of this, equation (59) would only be equal to the pentagon equation up to factors of two when the tensors are defined in terms of the F symbols.

Let us now define the second, left-handed, type of fMPO tensor with the internal ordering

and components

We will now restrict to $\tilde{F}$-symbols that are unitary or isometric matrices, i.e. $\tilde{F}$-symbols that satisfy equations (45) and (46), which we restate here for convenience:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{f,\lambda\kappa} [\bar{\tilde{F}}_e^{abc}]^{d^{\prime},\mu^{\prime}\nu^{\prime}}_{f,\lambda\kappa} [\tilde{F}_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa} = \delta_{d^{\prime},d}\left\{\begin{array}{@{}ll@{}} \delta_{\mu^{\prime},\mu}\delta_{\nu^{\prime},\nu^{\prime}},&\epsilon_d = 0, \nonumber \\ \left[\delta_{\mu^{\prime},\mu} \delta_{\nu^{\prime},\nu} + Y_{\mu^{\prime},\mu} (M_{dc}^{e})_{\nu^{\prime},\nu} \right]/2,&\epsilon_d = 1, \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 62 }$

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{d,\mu,\nu} [\tilde{F}_e^{abc}]^{d,\mu\nu}_{f,\lambda\kappa}[\bar{\tilde{F}}_e^{abc}]^{d,\mu\nu}_{f^{\prime},\lambda^{\prime}\kappa^{\prime}} = \delta_{f,f^{\prime}} \left\{\begin{array}{@{}ll@{}} \delta_{\kappa,\kappa^{\prime}}\delta_{\lambda,\lambda^{\prime}},&\epsilon_f = 0, \nonumber \\ \left[\delta_{\kappa,\kappa^{\prime}} \delta_{\lambda^{\prime},\lambda^{\prime}} + Y_{\kappa,\kappa^{\prime}} (L_{af}^{e})_{\lambda,\lambda^{\prime}} \right]/2,&\epsilon_f = 1. \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 63 }$

In this case, one sees that with our definition of the left-handed fMPO tensors the following properties are satisfied

where we used approximate equality to denote that these are not strict tensor identities, but are only satisfied on the relevant subspaces. In other words, these identities should only hold when the fMPO is embedded within the fermionic PEPS. One can check that this is indeed the case for the fMPOs and fermionic PEPS just defined. As a final step, we require that the $\tilde{F}$-symbols satisfy

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:pivotal2} \left[\tilde{F}^{abc}_e\right]^{d,\mu\nu}_{f,\lambda\sigma}\frac{\sqrt{d_ed_b}}{\sqrt{d_fd_d}} = \left[\tilde{F}^{adc}_f \right]^{b,\mu\sigma}_{e,\lambda\nu}\frac{\theta^{ac,\mu}_b\theta^{ac,\lambda}_e}{\theta^{ac,\mu}_d\theta^{ac,\lambda}_f}\frac{\theta^{ac}_\sigma}{\theta^{ac}_\nu}t^{ac}_\mu s^{ac}_\lambda\,, \nonumber \end{align} \tag{ 65 }$

where $\theta\in$ U(1) and $t, s \in \{1, -1\}$. It is this condition that fixes the positive numbers d_a. Equation (65) is a generalization of the pivotal property for bosonic fusion categories, which together with the isometric property implies that the fMPOs also satisfy following properties:

where the black dot is a graphical notation for the parity matrix $\mathsf{P} = \sum_\alpha (-1){\hspace{0pt}}^{\vert \alpha\vert }\vert \alpha)(\alpha\vert$. The reason for requiring unitarity and a generalization of the pivotal property is that from the pulling through identity (59) we can now derive the complete set of pulling through identities for the A-sublattice:

In a similar way one can derive the pulling through identities for the B-sublattice:

where the identity in the top left corner follows from the (complex conjugate of the) super pentagon equation, and all other identities can be derived from this one using properties (64) and (66). Note that the pulling through identities (67) and (68) imply that closed fMPOs on the virtual level of the PEPS contain parity matrices on their internal indices. They encode the rules of how these parity matrices move or change in their total number by a multiple of two when the fMPO moves through the PEPS tensors. One can check that these rules completely determine the position of the parity matrices on every closed fMPO and imply that their number is always odd for every fMPO along a contractible cycle. This implies that our formalism survives an important consistency check. In [15] is was explained that an $\newcommand{\e}{{\rm e}} \epsilon=1$ fMPO evaluates to zero when it is closed with an even number of parity matrices $\mathsf{P}$ inserted on its internal indices; in particular we cannot close it without inserting any parity matrix. But we just argued that the pulling through identities imply that every fMPO along a contractible cycle contains an odd number of parity matrices, thus preventing the fermionic PEPS with Majorana symmetry fMPOs from contracting to the zero vector. fMPOs along non-contractible cycles require a more detailed analysis. We will come back to this point in section 6.1.

The tensor networks we have constructed here involve a particular choice of spin structure. Apart from the spin structures related by flipping arrows around a vertex, there are still many more choices one can make. However, not all of them will be consistent with the pulling through identities (67) and (68), in the sense that these local identities will not imply that the fMPOs can be moved freely through the entire tensor network. All spin structures we have found to be compatible are of the Kasteleyn type [25, 26], which means that when going around a plaquette in a particular direction the number of arrows on the edges bounding that plaquette pointing in the opposite direction is odd.

In this section we have constructed fermionic PEPS tensors on the honeycomb lattice and fMPO tensors, both right- and left-handed, such that the pulling through identities hold. The pulling through identities are a fingerprint of non-trivial topological order in PEPS, which can—for example—be seen by defining the fermionic PEPS on a torus. In this situation, one can place fMPOs on the virtual level along non-contractible cycles. This will lead to PEPS that are locally indistuinguishable from each other, since the fMPOs can move freely on the virtual level. This results in a topological ground state degeneracy.

We define the fusion tensors $\mathsf{X_{g_2, g_1}}$ associated to the fMPO group representation constructed from tensors (71) and (72) with components

in the basis

Note that the parity of this fusion tensor is $Z(g_1, h) + Z(g_2, g_1h) + Z(g_2g_1, h) = Z(g_2, g_1)$ mod 2. At this point we would like to note that the parity of the internal fMPO indices has no physical value, we could as well interchange even with odd for the internal fMPO indices for any of the O_g. If we denote with $x(g) \in \{0, 1\}$ whether or not we have interchanged even and odd for the fMPO O_g, then the parity of the fusion tensors changes as $Z(g_2, g_1) \rightarrow Z(g_2, g_1)+x(g_2) +x(g_1)+x(g_2g_1)$ mod 2. So we see that the only invariant information associated to the fMPO is the second cohomology class $H^2(G, \mathbb{Z})$ represented by $Z(g_2, g_1)$. One can also check that PEPS constructed from different $Z(g_2, g_1)$ in the same cohomology class are equivalent in the following way: after taking the tensor product with product states such that the local physical super vector spaces are the same, there exists a strictly on-site unitary that maps one PEPS to the other and intertwines both left regular symmetry actions. Similarly to the bosonic case, multiplying $\mathsf{X_{g_2, g_1}}$ with the phase $\gamma(g_2, g_1)$ changes $\alpha(g_3, g_2, g_1)$ by a coboundary $\gamma(g_3, g_2)\gamma(g_3g_2, g_1)\bar{\gamma}(g_2, g_1)\bar{\gamma}(g_3, g_2g_1)$. This implies that only $\alpha(g_3, g_2, g_1)$ modulo coboundaries contains invariant information.

The super cocycle relation (70) implies that the fusion tensors defined above indeed satisfy the zipper condition:

Again applying the super cocycle relation shows that the F-move for these fusion tensors produces the super cocycle that we used to construct the PEPS:

This F-move is written down as an equation in the following way

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \mathcal{C}\left(\mathsf{X_{g_3,g_2}}\otimesg \mathsf{X_{g_3g_2,g_1}} \right) = \alpha(g_3,g_2,g_1) \mathcal{C}\left(\mathsf{X_{g_2,g_1}} \otimesg \mathsf{X_{g_3,g_2g_1}} \right)\,, \nonumber \end{align} \tag{ 80 }$

where $\mathcal{C}$ represents the proper fermionic contraction as depicted in (79).

In the previous section we showed that the Frobenius–Schur indicator associated to an $\newcommand{\e}{{\rm e}} \epsilon\equiv 0$ fMPO group representation is completely fixed by the supercocycle. The invariant algebraic data associated to the fMPO representation is therefore given by $Z(g_2, g_1)$ and $\alpha(g_3, g_2, g_1)$. Since the fMPO describes all possible anomalous symmetry actions on the boundary of the two-dimensional system, this data should directly classify the SPT phase of the short-range entangled bulk. Let us now ask the question of what happens to this data when we stack different SPT phases, i.e. when we take the graded tensor product of PEPS with the same global symmetry. It is clear that the stacked PEPS has a virtual symmetry given by the graded tensor product of the original fMPO representations. The fusion tensor of the graded tensor product of two fMPOs is also just the graded tensor product of the individual fusion tensors, which we denote by $\mathsf{X_{g_2, g_1}^1}$ and $\mathsf{X_{g_2, g_1}^2}$. Using the rules of fermionic contraction with super vector spaces we can now easily obtain the supercocycle for the stacked PEPS by evaluating the F-move for $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathsf{X_{g_2, g_1}^1}\otimesg \mathsf{X_{g_2, g_1}^2}$:

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle & \mathcal{C}\left[\left( \mathsf{X_{g_3,g_2}^1}\otimesg \mathsf{X_{g_3,g_2}^2}\right)\otimesg \left(\mathsf{X_{g_3g_2,g_1}^1}\otimesg \mathsf{X_{g_3g_2,g_1}^2}\right)\right] \nonumber \\ &= (-1)^{Z_1(g_3g_2,g_1)Z_2(g_3,g_2)}\mathcal{C}\left[\mathsf{X_{g_3,g_2}^1}\otimesg \mathsf{X_{g_3g_2,g_1}^1}\otimesg \mathsf{X_{g_3,g_2}^2}\otimesg \mathsf{X_{g_3g_2,g_1}^2}\right] \nonumber \\ &= (-1)^{Z_1(g_3g_2,g_1)Z_2(g_3,g_2)}\alpha_1(g_3,g_2,g_1)\alpha_2(g_3,g_2,g_1) \nonumber \\ & \mathcal{C}\left[\mathsf{X_{g_2,g_1}^1}\otimesg \mathsf{X_{g_3,g_2g_1}^1}\otimesg\mathsf{X_{g_2,g_1}^2}\otimesg \mathsf{X_{g_3,g_2g_1}^2}\right] \nonumber \\ &= (-1)^{Z_1(g_3g_2,g_1)Z_2(g_3,g_2)+Z_1(g_3,g_2g_1)Z_2(g_2,g_1)}\alpha_1(g_3,g_2,g_1)\alpha_2(g_3,g_2,g_1) \nonumber \\ & \mathcal{C}\left[\left(\mathsf{X_{g_2,g_1}^1}\otimesg\mathsf{X_{g_2,g_1}^2}\right)\otimesg\left( \mathsf{X_{g_3,g_2g_1}^1}\otimesg\mathsf{X_{g_3,g_2g_1}^2}\right)\right] \, . \nonumber \end{align} \tag{ 81 }$

The parity of $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \mathsf{X_{g_2, g_1}^1}\otimesg \mathsf{X_{g_2, g_1}^2}$ is of course just given by $Z_1(g_2, g_1) + Z_2(g_2, g_1)$. We therefore find that the stacked SPT PEPS is described by the following algebraic data:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{\alpha}(g_3,g_2,g_1) &= (-1)^{Z_1(g_3g_2,g_1)Z_2(g_3,g_2)+Z_1(g_3,g_2g_1)Z_2(g_2,g_1)}\alpha_1(g_3,g_2,g_1)\alpha_2(g_3,g_2,g_1) \nonumber \\ \tilde{Z}(g_2,g_1) &= Z_1(g_2,g_1) + Z_2(g_2,g_1)\;\;\;{\rm mod } ~2\, . \nonumber \end{align} \tag{ 82 }$

This shows how the algebraic data changes under stacking and allows one to calculate the group structure of Gu–Wen SPT phases.

One of the characterizing physical properties of SPT phases is that symmetry defects can carry fractional quantum numbers. In this section we will discuss how the projective nature of defects in Gu–Wen phases is derived from the defining algebraic data $Z(g_2, g_2)$ and $\alpha(g_3, g_2, g_1)$.

5.2.1. π-flux defects.

In section 4 we explained how the fixed-point PEPS obtained via the bootstrap method incorporate a lattice spin structure. Different spin structures can be obtained by choosing a closed path on the dual lattice and putting a parity matrix $\mathsf{P}$ on every virtual index that crosses this cut. It is important to note that internal fMPO indices crossing the path should also gain a parity matrix. In figure 2 we show a part of such a path and the associated parity matrices in the PEPS.

Zoom In Zoom Out Reset image size

If we now choose an open path on the dual lattice and again insert parity matrices on links that cross the path we have created π-flux defects on the plaquettes where the path ends. Symmetry fMPOs on the virtual level of the PEPS that encircle one of these π-flux defects contain an even number of parity matrices on their internal indices. One can verify that group fMPOs O_g as constructed above satisfy

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{O}_g\tilde{O}_h = (-1)^{Z(g,h)}\tilde{O}_{gh}\,, \nonumber \end{align} \tag{ 83 }$

where the tilde denotes the fact that the fMPOs contain an even number of parity matrices. This gives an explicit physical interpretation to $Z(g, h)$: it is the projective representation under which π-flux defects transform.

A convenient way to think about symmetry defects is the following: if we put the PEPS on a cylinder and we twist the boundary conditions along the periodic direction by the group element g, then there are symmetry defects at both ends of the cylinder. This boils down to simply placing a fMPO O_g on the virtual level going from one end of the cylinder to the other. Figure 3 contains a graphical representation of this situation. There is one subtlety if we apply this reasoning to π-flux defects. Let us consider the PEPS on the cylinder with periodic boundary conditions along the periodic direction. In this case a fMPO wrapping the non-contractible cycle will contain an even number of parity matrices on its internal indices. But as explained above, in this case the fMPOs form a projective representation. We can also define the PEPS with anti-periodic boundary conditions by choosing a path on the dual lattice extending from one end of the cylinder to the other and again inserting the appropriate parity matrices. Now the fMPOs wrapping the cylinder contain an odd number of parity matrices and form a non-projective representation. This shows that the PEPS with periodic boundary conditions should be intepreted as having a π-flux through the cylinder. The cylinder with anti-periodic boundary conditions contains no flux and can in principle be 'capped off' to a sphere. This is the tensor network analogue of the fact that the Neveu-Schwarz spin structure on the circle can be extended to the unique spin structure on a disc, while the Ramond spin structure does not have this property.

Zoom In Zoom Out Reset image size

5.2.2. General defects.

To study general symmetry defects we have to use a second type of fusion tensor, which in the basis

has following components:

This second type of fusion tensor can be obtained by reducing a right-handed and a left-handed fMPO tensor to a right-handed one. We now again consider the cylinder with boundary conditions twisted by g as in figure 3. We also impose anti-periodic boundary conditions such that there is no π-flux through the cylinder. One can check that the physical symmetry action of elements in $\mathcal{Z}_g$, the center of g, gets intertwined to an action on the left virtual indices and the right virtual indices. The action on the left virtual indices is given by the following fMPO:

A tedious, but straightforward calculation shows that this fMPO is a projective representation of $\mathcal{Z}_g$ with 2-cocycle

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:2cocycle} \omega_g(h,k) = (-1)^{Z(h,k)(Z(hk,g)+Z(g,hk))}\frac{\alpha(h,g,k)}{\alpha(h,k,g)\alpha(g,h,k)}. \nonumber \end{align} \tag{ 87 }$

For $h, k$ and l commuting with g the supercocycle relation implies that this phase indeed satisfies the 2-cocycle relation:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \omega_g(h,k)\omega_g(hk,l) = \omega_g(h,kl)\omega_g(k,l)\, . \nonumber \end{align} \tag{ 88 }$

The virtual symmetry action on the right boundary indices is of course also projective, but with 2-cocycle $\bar{\omega}_g(h, k)$. Equation (87) thus describes the fractionalization of symmetry defects in Gu–Wen SPT phases. It is a generalization of the slant product for bosonic SPT phases.

Let us now consider the Gu–Wen tensor network on a torus. In this case we can twist the boundary conditions in both the x and y direction with group elements h and g, provided that $[g, h] = ghg^{-1}h^{-1} = e$, by putting fMPOs along the non-contractible cycles. These fMPOs labeled by h and g meet in one point, where they have to be connected using fusion tensors. There are many different possibilities to connect the fMPOs in this way, but using the pivotal properties of Gu–Wen fusion tensors discussed in appendix C one can show that all these different choices only differ by a phase factor for the twisted wavefunction. In this section we find it convenient to work in the following basis for the twisted Gu–Wen states:

where opposite sides should be identified. This figure shows the position of the fMPOs O_h and O_g on the virtual level of the Gu–Wen tensor network on the torus and how they are connected using fusion tensors. S_x and S_y denote whether the boundary conditions are periodic or anti-periodic along the two non-contractible cycles of the torus, where $S=0\, (1)$ means anti-periodic (periodic). Note that since all PEPS and fMPO tensors are even, the parity of the twisted state (89) is determined by the parity of the fusion tensors, which gives $Z(g, h)+Z(h, g)$ mod 2.

We can now define the $\mathcal{S}$ transformation on these states as

Using the F move (79) and the pivotal properties (C.7) introduced in appendix C the twisted state after the $\mathcal{S}$ transformation can be expressed back in the standard basis (89). If we denote the basis state (89) as $\vert (h, g);(S_x, S_y)\rangle$ then the $\mathcal{S}$ transformation takes following matrix form

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S = & \mathop{\sum_{g,h}}\limits_{{[g,h]=e}}\sum_{S_x,S_y} (-1)^{Z(g,g^{-1})S_x + (Z(g,h)+Z(h,g))S_y}\,\frac{\alpha(g^{-1},h,g)}{\alpha(h,g^{-1},g)\alpha(g^{-1},g,h)} \nonumber \\ & \vert (g^{-1},h);(S_y,S_x)\rangle\langle (h,g);(S_x,S_y)\vert \, . \nonumber \end{align} \tag{ 91 }$

From the supercocycle relation it follows that the S matrix satisfies $S^4 = (-1){\hspace{0pt}}^{Z(g, h)+Z(h, g)}\mathds{1}$, which is to be expected since S⁴ represents a $2\pi$ rotation and $Z(g, h)+Z(h, g)$ is the fermion parity of the twisted state $\vert (h, g);(S_x, S_y)\rangle$.

We can now define the $\mathcal{T}$ transformation, corresponding to a Dehn twist on the twisted states:

The state after the $\mathcal{T}$ transformation can again be brought back into the standard basis (89) using F-moves and the pivotal properties of the fusion tensors. This gives following expression for the T matrix:

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T = \mathop{\sum_{g,h}}\limits_{{[g,h]=e}}\sum_{S_x,S_y} (-1)^{Z(g,h)(S_y+Z(g,h)+Z(h,g))} \alpha(g,h,g)\vert (gh,g);(S_x+S_y,S_y)\rangle\langle (h,g);(S_x,S_y)\vert \, . \nonumber \end{align} \tag{ 93 }$

The S and T matrices obviously depend on the representative cocycles $Z(g, h)$ and $\alpha(g, h, k)$. However, under a coboundary transformation

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle Z(g,h) & \rightarrow Z(g,h) + x(g)+x(h)+x(gh)~ {\rm with }~ x(g) \in \{0,1\}\nonumber \\ \alpha(g,h,k) & \rightarrow \alpha(g,h,k)\frac{\gamma(g,h)\gamma(gh,k)}{\gamma(g,hk)\gamma(h,k)} \nonumber \end{align} \tag{ 94 }$

the S matrix transforms as $USU^\dagger$, with U a diagonal unitary matrix. The T matrix does not have this property under coboundary transformations of $Z(g, h)$. This seems to imply that T is not an object containing universal information about the Gu–Wen phase. However, T² does have the desired property $T^2\rightarrow UT^2U^\dagger$ under general coboundary transformations, implying that its eigenvalues are relevant invariants. This ambiguity has a physical meaning if we interpret the eigenvalues of T as ${\rm e}^{{\rm i}2\pi h}$, where h are the topological spins of the defects. Because the transparant particle in Gu–Wen phases is a fermion, the topological spins are only defined modulo $1/2$. This sign ambiguity in the eigenvalues of T can be avoided by looking at T². As explained above, the coboundary transformation on $Z(g, h)$ can be interpreted as attaching a fermion to the virtual fMPO indices. Since g defects are connected via fMPOs O_g such a coboundary transformation indeed has the net effect of attaching fermions to the defects, changing the topological spin by $1/2$. The ambiguity in T also manifests itself in the relation $(ST){\hspace{0pt}}^3 = (-1){\hspace{0pt}}^{Z(g, h)}S^2$, which follows from the supercocycle relation. This shows that S and T only form a representation of $SL(2, \mathbb{Z})$ up to a minus sign which changes under coboundary transformations.

Finally, we want to point out that the S and T matrices for the gauged, topologically ordered PEPS can be obtained from those of the SPT phase [6, 36]. It was shown in [6] that the S and T matrices for the gauged theory are obtained by applying $\mathcal{S}$ and $\mathcal{T}$ on the states

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:gaugedstate} \frac{1}{\vert G\vert }\sum_{x\in G}U(x)^{\otimes L_xL_y}\vert (h,g);(S_x,S_y)\rangle\,, \nonumber \end{align} \tag{ 95 }$

where $U(x)$ is the on-site physical symmetry action and $L_xL_y$ is the size of the torus. If $k \in \mathcal{Z}_{h, g}$, the centralizer of both h and g, then the twisted states satisfy $\newcommand{\e}{{\rm e}} U(k){\hspace{0pt}}^{\otimes L_xL_y}\vert (h, g);(S_x, S_y)\rangle = \epsilon_{h, g}^{S_x, S_y}(k)\vert (h, g);(S_x, S_y)\rangle$. Using the results of section 5.2.2 it follows that the one-dimensional representation $\newcommand{\e}{{\rm e}} \epsilon_{h, g}^{S_x, S_y}(k)$ of $\mathcal{Z}_{h, g}$ is given by

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \epsilon_{h,g}^{S_x,S_y}(k) = (-1)^{(Z(h,k)+Z(k,h))S_y + (Z(g,k)+Z(k,g))S_x + (Z(g,h)+Z(h,g))(Z(g,k)+Z(k,g))}\frac{\omega_g(h,k)}{\omega_g(k,h)}\, . \nonumber \end{align} \tag{ 96 }$

Since the states (95) are obtained by a projection on the symmetric subspace, only those states for which $\newcommand{\e}{{\rm e}} \epsilon_{h, g}^{S_x, S_y}(k) = 1$ for all $k \in \mathcal{Z}_{h, g}$ are non-zero. Both $\mathcal{S}$ and $\mathcal{T}$ commute with the global symmetry action $U(x){\hspace{0pt}}^{\otimes L_xL_y}$. For $\mathcal{S}$, this is immediate, but for $\mathcal{T}$ this follows from the results in [13]. From the commutativity of $\mathcal{S}, \mathcal{T}$ and $U(x){\hspace{0pt}}^{\otimes L_xL_y}$ one can easily infer the S and T matrices of the gauged theory from those of the Gu–Wen SPT. However, note that the S and T matrices obtained in this way are not expressed in the basis that has a definite anyon flux through one of the holes of the torus. To compare the S and T matrices before and after gauging, note that the action of $\mathcal{S}$ and $\mathcal{T}$ on the states $\sum_{x\in G} \Gamma^\mu_{ij}(x)U(x){\hspace{0pt}}^{\otimes L_xL_y}\vert (h, g);(S_x, S_y)\rangle$, with $\Gamma^\mu(g)$ an irrep of G, is independent of $\mu, i$ and j. This shows that the S and T matrices of the SPT phase consist of multiple copies of the same block (up to diagonal unitary similarity transformations), and the gauging proces selects only one of these identical blocks.

As explained in section 4, it is essential that every fMPO closed along a contractible loop has an odd number of parity matrices $\mathsf{P}$ inserted on its internal indices. In section 5 we explained that the number of parity matrices modulo two on fMPOs along non-contracible cycles is determined by the boundary conditions, or equivalently, by the spin structure. Here we will show that this leads to a non-trivial interplay between spin structure, ground state degeneracy and ground state parity for the topologically ordered fermionic PEPS contructed from the fMPO superalgebra $\{O^L_1, O^L_\sigma\}$ with $\newcommand{\e}{{\rm e}} \epsilon_\sigma = 1$ via equations (53) and (55).

We start by showing that the fermionic PEPS on the torus with periodic boundary conditions in both directions (PP) evaluates to zero if no fMPO $O_\sigma$ is inserted on the virtual level. To see this, first construct a tensor $\mathsf{\tilde{C}_P}$ by contracting all PEPS tensors that lie in the same column, where P denotes that we use periodic boundary conditions in the direction along the column. The ordering convention for the indices of $\mathsf{\tilde{C}_P}$ is as follows: first the virtual indices corresponding to the left hand side of the column, then the physical indices and lastly the virtual indicices on the right hand side. The virtual indices are ordered such that contracting neighboring columns corresponds to matrix multplication of the components of $\mathsf{\tilde{C}_P}$. The procedure just described is of course just the fermionic version of standard reinterpretation of a PEPS on the cylinder as a matrix product state with tensors $\mathsf{\tilde{C}_P}$. We will denote the fMPO $O^{L_y}_\sigma$ going along the periodic direction, with the external indices reordered in the same way as the virtual indices of $\mathsf{\tilde{C}}$, as $\mathsf{X_{\sigma, P}}$. It is crucial to note that $\mathsf{X_{\sigma, P}}$ has odd parity while $\mathsf{X_{\sigma, A}}$ is even. This is because $\newcommand{\e}{{\rm e}} \epsilon_\sigma =1$ and it was shown in [15] that such fMPOs have to be closed with $\mathsf{Y}$ on the internal indices under periodic boundary conditions in order to be non-zero. With anti-periodic boundary conditions $O^{L_y}_\sigma$ has to be closed without $\mathsf{Y}$ and is therefore even. Figure 6 gives a graphical representation of the tensors just defined.

Zoom In Zoom Out Reset image size

Now we can easily show that the fermionic PEPS in the PP sector without any fMPO is zero. Its coefficients on a torus consisting of L_x columns are given by tr($P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}$), where i_j represents the collection of all physical indices in the jth column and $P^{\otimes L_y}$, the tensor product of L_y parity matrices, is generated as a supertrace by the fermionic contraction (see [15] for more detail). In appendix D we show that $\newcommand{\e}{{\rm e}} \mathsf{X_{\sigma, P}}\mathsf{X_{\sigma, P}} = (-1){\hspace{0pt}}^\eta i \mathsf{X_{1, P}}$, which can now be used to show that

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {\rm tr}(P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}) &= (-1)^{\eta+1}i\, {\rm tr}(P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}X_{\sigma,P}X_{\sigma,P}) \nonumber \\ &=(-1)^{\eta}i\, {\rm tr}(P^{\otimes L_y}X_{\sigma,P}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}X_{\sigma,P}) \nonumber \\ &= (-1)^{\eta}i\, {\rm tr}(P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}X_{\sigma,P}X_{\sigma,P})\nonumber \\ &= - {\rm tr}(P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}})\,, \nonumber \end{align} \tag{ 100 }$

where the second equality follows from the fact that $\mathsf{X_{\sigma, P}}$ is odd and the third equality follows from the pulling through property.

The non-zero states in the PP sector can be schematically represented as

where the torus is depicted as a rectangle with opposite sides identified. The red line represents a fMPO $\mathsf{X_\sigma}$ on the virtual level of the PEPS wrapping a non-contractible cycle. The state on the left has coefficients ${\rm tr}(X_{\sigma, P}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}})$, where the fermionic contraction now does not generate a matrix $P^{\otimes L_y}$ because $X_{\sigma, P}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}}$ has odd parity. For this reason we cannot conclude that this state is zero. Similar reasoning shows that the other two states in the PP sector may also be non-zero. Note that the three ground states in the PP sector all have odd fermion parity because of the matrix $\mathsf{Y}$ on the internal fMPO indices.

In the AP sector, with anti-periodic boundary conditions in the x-direction, one can show that the following state is zero:

where the dashed line represents the anti-periodic boundary conditions, i.e. along that line on the dual lattice we have inserted parity matrices P on the virtual indices. Note that the periodic boundary conditions in the y-direction imply that the fMPO $\mathsf{O^{L_y}_\sigma}$ is odd. The coefficients for this state are ${\rm tr}(P^{\otimes L_y}X_{\sigma, P}\tilde{C}^{i_1}\tilde{C}^{i_2}\dots\tilde{C}^{i_{L_x}})$, where $P^{\otimes L_y}$ is now not generated by the fermionic contraction because $\mathsf{X_{\sigma, P}}$ is odd but is inserted by hand because of the anti-periodic boundary conditions. This trace expression for the coefficients can easily be seen to be zero. The three non-zero states in the AP sector are

where both σ-fMPOs are even because they cross the dashed line an odd number of times. Note that the coefficients of the state in the AP sector without any fMPO are ${\rm tr}(P^{\otimes L_y}\tilde{C}^{i_1}\tilde{C}^{i_2}P^{\otimes L_y}\dots\tilde{C}^{i_{L_x}})$, where one $P^{\otimes L_y}$ is generated by the supertrace of an even tensor and the second $P^{\otimes L_y}$ comes from the anti-periodic boundary conditions. Analogously, one can show that the three non-zero states in the PA sector are:

In the AA sector one can show that the following state is zero:

where the fMPO is odd because it crosses a dashed line an even number of times. The state above is zero because the two graphical expressions given for it differ by a minus sign, as can easily be seen by using the pulling through property and the fact that $\mathsf{Y}$ is odd. The non-zero states in the AA sector are then given by

So to conclude, we have found that the fermionic PEPS constructed from the fMPO algebra $\{O^L_1, O^L_\sigma\}$ with $\newcommand{\e}{{\rm e}} \epsilon_\sigma = 1$ has three non-zero ground states in each spin structure sector. In the PP sector these states have odd parity, while in the AP, PA and AA sectors they have even parity. This agrees with the results of [25], where an explict commuting projector Hamiltonian was constructed for the topological phases captured by the fermionic PEPS described in this section.

Above we used the fMPO group representations $\{O^L_1, O^L_\sigma\}$ with $\newcommand{\e}{{\rm e}} \epsilon_\sigma =1$ to construct fermionic PEPS with non-trivial topological order. Here we will discuss applications of these fMPOs for $\mathbb{Z}_2$ symmetry-protected phases. In analogy to section 5, where we treated the case $\newcommand{\e}{{\rm e}} \epsilon\equiv 0$, we construct the short-range entangled PEPS on the hexagonal lattice using the tensors

for the A-sublattice and a similar modification of (55) for the B-sublattice. The resulting PEPS has a global $\mathbb{Z}_2$ symmetry, where the physical on-site symmetry action gets intertwined to a fMPO $O_\sigma$ on the virtual indices, where $O_\sigma$ is the same fMPO as before constructed from the tensor shown in figure 4.

The PEPS obtained via the tensors (107) describes a wave function where Majorana chains are bound to domain walls of the plaquette variables. An explicit commuting projector Hamiltonian with this type of ground state was constructed in [26]. A physical property of this SPT phase is that $\mathbb{Z}_2$ symmetry defects bind Majorana modes. In the tensor network language this can easily be seen by defining the PEPS on a cylinder with twisted boundary conditions along the non-contracible cycle. This is done by simply placing the fMPO $O_\sigma$ along the cylinder on the virtual level, going from one end of the cylinder to the other. At the two boundaries of the cylinder this results in a symmetry defect. Because $O_\sigma$ is of the type $\newcommand{\e}{{\rm e}} \epsilon_\sigma = 1$, the resulting fMPS on the cylinder has a non-trivial center corresponding to $\mathsf{Y}$ acting on the internal fMPO index. One can use similar reasoning as in [15] to conclude that there will be Majorana modes at the ends of the cylinder.

Other immediate concequences of the results in [15] involve the entanglement spectrum and the physical systems that can realize this phase. First, the PEPS on a cylinder with periodic boundary conditions has at least a two-fold degeneracy in its entanglement spectrum for cuts wrapping the non-contractible cycle. This follows from the fact that the $\mathbb{Z}_2$ symmetry action $\tilde{O}_\sigma$ on the boundary is odd and therefore anti-commutes with fermion parity. If we twist the boundary conditions with $O_\sigma$, then there will again be at least a two-fold degeneracy, both in the periodic and the anti-periodic sector. This degeneracy follows from the fact that $\newcommand{\e}{{\rm e}} \epsilon_\sigma = 1$. By interpreting the PEPS on the cylinder with boundary conditions twisted by $O_\sigma$ as a MPS and applying the results of [15] it follows that the $\mathbb{Z}_2$ Majorana SPT phases cannot occur in systems with (unbroken) particle number conservation.

In appendix D we show that under periodic boundary conditions, the fMPOs $\tilde{O}_\sigma$ satisfy $\newcommand{\e}{{\rm e}} \tilde{O}_\sigma \tilde{O}_\sigma = (-1){\hspace{0pt}}^\eta i \tilde{O}_1$. This gives a physical interpretation to the invariant η: it determines the projective representation of the global $\mathbb{Z}_2$ symmetry on π-flux defects. Note that since $\tilde{O}_\sigma$ is odd, this projective representation is consistent with the fact that two π-flux defects fuse to the vacuum. The combined action on two π-flux defects is given by $\newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \tilde{O}_\sigma\otimesg \tilde{O}_\sigma$, which satisfies

$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\otimesg}{{\otimes_{\mathfrak{g}}}} \displaystyle \left(\tilde{O}_\sigma\otimesg \tilde{O}_\sigma\right)\left(\tilde{O}_\sigma\otimesg \tilde{O}_\sigma\right) = -\tilde{O}_\sigma\tilde{O}_\sigma\otimesg \tilde{O}_\sigma\tilde{O}_\sigma = -(-1)^\eta {\rm i}(-1)^\eta {\rm i} \tilde{O}_1\otimesg \tilde{O}_1 = \tilde{O}_1\otimesg \tilde{O}_1\, . \nonumber \end{align} \tag{ 108 }$

This shows that the symmetry action on two π-flux defects is indeed non-projective.

For the fMPO $O_\sigma$ we readily determine the Frobenius–Schur indicator as defined in the general theory of fMPO super algebras. We find that $Z_\sigma$ as defined as in section 3 takes the following form:

Note that $Z_\sigma$ is even. The first invariant associated to the Frobenius–Schur indicator can now easily be obtained from

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle Z_\sigma \bar{Z}_\sigma = (-1)^\rho \mathds{1}\, . \nonumber \end{align} \tag{ 110 }$

The second invariant determining the Frobenius–Schur indicator we get from

$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (YZ_\sigma)\left(\bar{Y}\bar{Z}_\sigma\right) = (-1)^{\rho+\eta+1}{\rm i}P\, . \nonumber \end{align} \tag{ 111 }$

From this we see that the value of the Frobenius–Schur indicator uniquely determines the F-symbols for the $\{O_1, O_\sigma\}$ group representation with $\newcommand{\e}{{\rm e}} \epsilon_\sigma =1$. The same holds for the case $\newcommand{\e}{{\rm e}} \epsilon_\sigma = 0$, where the Frobenius–Schur indicator completely fixes one of the four supercohomology classes for $\mathbb{Z}_2$. So in total we have eight different fermionic SPT phases with a global $\mathbb{Z}_2$ symmetry. In section 3 we also mentioned that the Frobenius–Schur indicator is isomorophic to $\mathbb{Z}_8$, implying that the $\mathbb{Z}_2$ SPT phases form a $\mathbb{Z}_8$ group under stacking, agreeing with previous studies [41–44]. Since the Frobenius–Schur indicator has the same mathematical origin as the invariants associated to time-reversal or reflection invariant fMPS, we have thus connected the classification of two-dimensional unitary $\mathbb{Z}_2$ SPT phases to the classification of one-dimensional SPT phases with time-reversal or reflection symmetry. This is the tensor network manifestation of the Smith isomorphism, which relates the cobordism groups conjectured to describe both types of SPT phases [45].