# The spectral radius of unbalanced signed graphs without negative $C_{3}$ or
$C_{4}$

Yiting Cai111E-mail<EMAIL_ADDRESS>Bo Zhou222E-mail:
<EMAIL_ADDRESS>
School of Mathematical Sciences, South China Normal University,
Guangzhou 510631, P.R. China

###### Abstract

A signed graph is a graph in which every edge carries a $+$ or a $-$ sign. In
this paper, we determine the signed graphs with maximum spectral radius among
all unbalanced signed graphs with fixed order that contain neither negative
three-cycles nor negative four-cycles.

Key words: unbalanced signed graph, negative cycles, spectral radius

AMS Classification: 05C50

##  1 Introduction

A signed graph $\Gamma$ is a pair $\Gamma=(G,\sigma)$ where $G$ is a simple
graph, called the underlying graph, with vertex set $V(G)$ and edge set
$E(G)$, and $\sigma:E(G)\rightarrow\\{-1,1\\}$ is the sign function, called
the signature. The vertex (resp. edge) set of $\Gamma$ is just the vertex
(resp. edge) set of its underlying graph, denoted by $V(\Gamma)$ (resp.
$E(\Gamma)$). An edge $e$ is positive (resp. negative) if $\sigma(e)=1$ (resp.
$\sigma(e)=-1$). If all edges of $\Gamma=(G,\sigma)$ are positive (resp.
negative), then we denote it by $(G,+)$ (resp. $(G,-)$). $-\Gamma$ denotes the
signed graph obtained from $\Gamma$ by reversing the signs of all edges. The
adjacency matrix $A(\Gamma)$ of $\Gamma$ is obtained from the standard
adjacency matrix of the underlying graph $G$ by reversing the sign of all $1$s
which correspond to negative edges. That is,
$A(\Gamma)=(a_{uv}^{\sigma})_{u,v\in V(\Gamma)}$ with
$a_{uv}^{\sigma}=\sigma(uv)$ if $u$ and $v$ are adjacent, and $0$ otherwise.
The eigenvalues of $A(\Gamma)$ are the eigenvalues of $\Gamma$, which are
denoted by $\lambda_{1}(\Gamma)\geq\dots\geq\lambda_{n}(\Gamma)$, where $n$ is
the order of $\Gamma$. The spectral radius $\rho(\Gamma)$ of $\Gamma$ is the
maximum absolute values of the eigenvalues of $\Gamma$. That is,
$\rho(\Gamma)=\max\\{\lambda_{1}(\Gamma),-\lambda_{n}(\Gamma)\\}$. If
$\Gamma=(G,+)$, then $\rho(\Gamma)=\lambda_{1}(\Gamma)$ is the spectral radius
$\rho(G)$ of $G$.

For a signed graph $\Gamma$ with $\emptyset\neq U\subset V(\Gamma)$. Let
$\Gamma^{U}$ be the signed graph obtained from $\Gamma$ by reversing the sign
of each edge between a vertex in $U$ and a vertex in $V(\Gamma)\setminus U$.
The signed graphs $\Gamma$ and $\Gamma^{U}$ are said to be switching
equivalent. The switching equivalence is an equivalence relation that
preserves the eigenvalues. A cycle in $\Gamma$ is said to be positive (or
balanced) if it contains an even number of negative edges, otherwise the cycle
is negative (or unbalanced). $\Gamma=(G,\sigma)$ is said to be balanced if it
is switching equivalent to $(G,+)$. Otherwise, it is said to be unbalanced.
Equivalently, $\Gamma$ is balanced if every cycle is positive, see [16]. From
[7], for a signed graph $\Gamma=(G,\sigma)$, we have
$\lambda_{1}(\Gamma)\leq\lambda_{1}(G)$ with equality when it is connected if
and only if $\Gamma$ is balanced. As in [1, 2], we consider only unbalanced
signed graphs when we study the extremal problems about the spectral radius or
the largest eigenvalue.

Two signed graphs are called switching isomorphic if one of them is a
switching equivalent to an isomorphic copy of the other.

Given a set $\mathcal{F}$ of signed graphs, if a signed graph $\Gamma$
contains no signed subgraph isomorphic to any one in $\mathcal{F}$, then
$\Gamma$ is called $\mathcal{F}$-free.

Denote by $\mathcal{K}_{n}^{-}$ the set of unbalanced signed complete graphs
of order $n$, and for $n\geq 3$, $\mathcal{C}_{n}^{-}$ the set of unbalanced
signed cycles of order $n$.

For a graph $G$ with an edge $e$, $G-e$ denotes the graph obtained from $G$ by
removing the edge $e$. Denote by $K_{n}$ the complete graph of order $n$. For
$n\geq 3$, $C_{n}$ denotes the cycle of order $n$.

For $r\geq 3$, the $\mathcal{K}_{r}^{-}$-free unbalanced signed graphs of
fixed order with maximum spectral radius have been determined for $r=3$ by
Wang, Hou and Li [9], for $r=4$ by Chen and Yuan [5], for $r=5$ by Wang [12],
and for $r\geq 4$ by Xiong and Hou [14]. The $\mathcal{C}_{4}^{-}$-free (resp.
$\mathcal{C}_{2k+1}^{-}$-free with $3\leq k\leq\frac{n-11}{10}$) unbalanced
signed graphs of order $n$ with maximum largest eigenvalue (resp. spectral
radius) have been determined by Wang and Lin [13] (resp. Wang, Hou and Huang
[10]).

Let $\mathcal{C}_{3,4}^{-}=\mathcal{C}_{3}^{-}\cup\mathcal{C}_{4}^{-}$. In
this paper, we determine the signed graphs with maximum spectral radius among
all $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graphs of order $n\geq 5$.
For $n\geq 5$, let $\Gamma_{n}$ be the signed graph obtained from a copy of
$K_{n-2}$ by removing the edge $uw$ and adding two new vertices $v_{1}$ and
$v_{2}$ and three edges $v_{1}v_{2},v_{1}u$ and $v_{2}w$, where $v_{1}v_{2}$
is the unique negative edge, see Fig. 1. As the shortest negative cycles in
$\Gamma_{n}$ are $C_{5}$s, $\Gamma_{n}$ is a $\mathcal{C}_{3,4}^{-}$-free
unbalanced signed graph. It is evident that a $\mathcal{C}_{3,4}^{-}$-free
unbalanced signed graph of order $5$ is switching equivalent to $\Gamma_{5}$,
a negative $C_{5}$ with
$\rho(\Gamma_{5})=-\lambda_{5}(\Gamma_{5})=2>\lambda_{1}(\Gamma_{5})=1.6180$.

$K_{n-2}-uw$$u$$w$$v_{1}$$v_{2}$$-$ Figure 1: The signed graph $\Gamma_{n}$.

######  Theorem 1.1.

Let $\Gamma$ be a $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph of
order $n\geq 6$. Then

$\rho(\Gamma)\leq\gamma_{n}$

with equality if and only if $\Gamma$ is switching isomorphic to $\Gamma_{n}$,
where $\gamma_{n}$ is the largest root of the equation
$\lambda^{3}-(n-6)\lambda^{2}-3(n-4)\lambda-n+3=0$.

##  2 Preliminaries

Most of the concepts like connectedness and degrees of vertices defined for
graphs are naturally extended to signed graphs. Let $\Gamma=(G,\sigma)$ be a
signed graph. For a vertex $u$ in $\Gamma$, we denote by $N_{\Gamma}(u)$ and
$d_{\Gamma}(u)$ the neighborhood and the degree $u$ in $\Gamma$, respectively.
For $\emptyset\neq U\subset V(\Gamma)$, let $\Gamma-U$ be the signed graph
obtained from $\Gamma$ be removing the vertices in $U$ and every edge incident
to a vertex of $U$, and $\Gamma[U]$ denotes the signed graph
$\Gamma-(V(\Gamma)\setminus U)$. The underlying graph of $\Gamma-U$ is $G-U$.
A chord of a cycle $C$ is an edge between two non-consecutive vertices of the
cycle.

Let $M$ be an $n\times n$ matrix whose rows and columns are indexed by
elements in $X=\\{1,\dots,n\\}$. Let $\pi=\\{X_{1},\dots,X_{s}\\}$ be a
partition of $X$. For $1\leq i,j\leq s$, let $M_{ij}$ be the submatrix of $M$
whose rows and columns are indexed by elements of $X_{i}$ and $X_{j}$. The
partition $\pi$ is equitable if the row sum of each $M_{ij}$ is a constant for
$1\leq i,j\leq s$. The $s\times s$ matrix whose $(i,j)$-entry is the average
row sum of $M_{ij}$ with $1\leq i,j\leq s$ is called the quotient matrix of
$M$.

######  Lemma 2.1.

[3, Lemma 2.3.1] Let $M$ be a real symmetric matrix. Then the spectrum of the
quotient matrix of $M$ with respect to an equitable partition is contained in
the spectrum of $M$.

For integers $n$ and $\tau$ with $n\geq 6$ and $0\leq\tau\leq n-5$, we define
a signed graph $\Gamma_{n,\tau}$ on vertices $v_{1},\dots,v_{n}$ as displayed
in Fig. 2, where $V(\Gamma_{n,\tau})=\\{v_{1},v_{2},v_{3},v_{5}\\}\cup X\cup
Y$ with $|X|=\tau$, $|Y|=n-4-\tau$ and $v_{4}\in Y$, $v_{1}v_{2}$ is the
unique negative edge, $v_{1}\dots v_{5}v_{1}$ is an induced cycle,
$d_{\Gamma_{n,\tau}}(v_{1})=2+\tau$, $d_{\Gamma_{n,\tau}}(v_{2})=2$,
$\Gamma_{n,\tau}[X\cup\\{v_{1},v_{5}\\}]=(K_{\tau+2},+)$,
$\Gamma_{n,\tau}[Y\cup\\{v_{3},v_{5}\\}]=(K_{n-\tau-2}-v_{3}v_{5},+)$, and
$\Gamma_{n,\tau}[X\cup Y]=(K_{n-4},+)$. With a proper labeling of the
vertices, $\Gamma_{n}$ is (isomorphic to) $\Gamma_{n,0}$.

Denote by $I_{n}$ the identity matrix of order $n$.

$X$$Y$$\underbrace{~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}}$$K_{n-4}$$-$$v_{1}$$v_{5}$$v_{3}$$v_{2}$
Figure 2: The signed graph $\Gamma_{n,\tau}$.

######  Lemma 2.2.

Let $\Gamma_{n,\tau}$ be the signed graph in Fig. 2 with $n\geq 6$ and
$0\leq\tau\leq n-5$.

(i)
$\lambda_{1}(\Gamma_{n,\tau})\leq\lambda_{1}(\Gamma_{n,0})=\lambda_{1}(\Gamma_{n,n-5})$
with equality if and only if $\tau=0,n-5$.

(ii) $\lambda_{1}(\Gamma_{n,0})>n-4$ and $\lambda_{1}(\Gamma_{n,0})$ is equal
to the largest root of $\lambda^{3}-(n-6)\lambda^{2}-3(n-4)\lambda-n+3=0$.

###### Proof.

First, we show the following claim.

######  Claim 2.1.

For $0\leq\tau\leq n-5$, the eigenvalues of $\Gamma_{n,\tau}$ are $-1$ with
multiplicity $n-5$ and the five roots of $f_{\tau}(\lambda)=0$, where

$\displaystyle f_{\tau}(\lambda)$
$\displaystyle=\lambda^{5}-(n-5)\lambda^{4}-(2n-5)\lambda^{3}+(2n\tau+3n-2\tau^{2}-10\tau-15)\lambda^{2}$
$\displaystyle\quad+(n\tau+4n-\tau^{2}-5\tau-15)\lambda+n-3.$

###### Proof.

Observe that the $\tau+1$ rows of $I_{n}+A(\Gamma_{n,\tau})$ corresponding to
vertices in $\\{v_{5}\\}\cup X$ are equal and the $n-\tau-4$ rows of
$I_{n}+A(\Gamma_{n,\tau})$ corresponding to vertices in $Y$ are equal. So the
rank of $I_{n}+A(\Gamma_{n,\tau})$ is at most $5$, implying that $-1$ is an
eigenvalue of multiplicity at least $n-5$. Let $V_{i}=\\{v_{i}\\}$ for
$i=1,2,3$, $V_{4}=\\{v_{5}\\}\cup X$ and $V_{5}=Y$. Then the quotient matrix
of $A(\Gamma_{n,\tau})$ with respect to the equitable partition
$V(\Gamma_{n,\tau})=V_{1}\cup\dots\cup V_{5}$ is

$\begin{pmatrix}0&-1&0&\tau+1&0\\\ -1&0&1&0&0\\\ 0&1&0&0&n-\tau-4\\\
1&0&0&\tau&n-\tau-4\\\ 0&0&1&\tau+1&n-\tau-5\\\ \end{pmatrix},$

whose characteristic polynomial is $f_{\tau}(\lambda)$. As
$f_{\tau}(-1)=(\tau+1)(n-\tau-4)\neq 0$, $-1$ is not a root of
$f_{\tau}(\lambda)=0$. By Lemma 2.1, the eigenvalues of $\Gamma_{n,\tau}$ are
$-1$ with multiplicity $n-5$ and the five roots of $f_{\tau}(\lambda)=0$. ∎

By Claim 2.1, $\lambda_{1}(\Gamma_{n,\tau})$ is equal to the largest root of
$f_{\tau}(\lambda)=0$.

Now we prove (i). Let $U=\\{v_{2}\\}$. Then $\Gamma_{n,n-5}^{U}$ is isomorphic
to $\Gamma_{n,0}$, so $\Gamma_{n,n-5}$ is switching equivalent to
$\Gamma_{n,0}$. So $\lambda_{1}(\Gamma_{n,0})=\lambda_{1}(\Gamma_{n,n-5})$. If
$n=6$, then (i) is trivial as $\tau=0,1$. Suppose that $n\geq 7$ and
$1\leq\tau\leq n-6$. It suffices to show that
$\lambda_{1}(\Gamma_{n,\tau})<\lambda_{1}(\Gamma_{n,0})$. Note that

$f_{\tau}(\lambda)-f_{\tau-1}(\lambda)=(n-2\tau-4)\lambda(2\lambda+1).$

If $n-2\tau-4>0$, then $f_{\tau}(\lambda)>f_{\tau-1}(\lambda)$ for
$\lambda>0$, so
$\lambda_{1}(\Gamma_{n,\tau})<\lambda_{1}(\Gamma_{n,\tau-1})<\dots<\lambda_{1}(\Gamma_{n,0})$.
If $n-2\tau-4<0$, then $f_{\tau}(\lambda)<f_{\tau-1}(\lambda)$ and so
$f_{\tau+1}(\lambda)<f_{\tau}(\lambda)$ for $\lambda>0$, implying that
$\lambda_{1}(\Gamma_{n,\tau})<\lambda_{1}(\Gamma_{n,\tau+1})<\dots<\lambda_{1}(\Gamma_{n,n-5})$.
Suppose that $n-2\tau-4=0$. Then $f_{\tau}(\lambda)=f_{\tau-1}(\lambda)$. In
this case $f_{\tau-1}(\lambda)-f_{\tau-2}(\lambda)=2\lambda(2\lambda+1)>0$ for
$\lambda>0$, so $f_{\tau-1}(\lambda)>f_{\tau-2}(\lambda)$, implying that
$\lambda_{1}(\Gamma_{n,\tau})=\lambda_{1}(\Gamma_{n,\tau-1})<\lambda_{1}(\Gamma_{n,\tau-2})<\dots<\lambda_{1}(\Gamma_{n,0})$.
Now (i) follows.

To prove (ii), note that

$f_{0}(\lambda)=(\lambda^{2}-\lambda-1)g(\lambda),$

where $g(\lambda)=\lambda^{3}-(n-6)\lambda^{2}-3(n-4)\lambda-n+3$. It can be
checked that $g(n-4)=-n^{2}+7n-13<0$ and $g(n-3)=2n-6>0$, so $g(\lambda)=0$
has a root larger than $n-4$, which is also larger than the larger root of
$\lambda^{2}-\lambda-1=0$. So $\lambda_{1}(\Gamma_{n,0})>n-4$, and
$\lambda_{1}(\Gamma_{n,0})$ is equal to the largest root of $g(\lambda)=0$. ∎

######  Lemma 2.3.

[15] Two signed graphs with the same underlying graph are switching equivalent
if and only if they have the same class of positive cycles.

Stanić [8] noted that for an eigenvalue $\lambda$ of a signed graph $\Gamma$,
there is a switching equivalent signed graph for which the
$\lambda$-eigenspace contains an eigenvector whose non-zero entries are of the
same sign.

######  Lemma 2.4.

[8, 6] A signed graph $\Gamma$ is switching equivalent a signed graph
$\Gamma^{\prime}$ such that $\lambda_{1}(\Gamma^{\prime})$ has a non-negative
eigenvector.

The balanced clique number of $\Gamma$, denoted by $\omega_{b}(\Gamma)$, is
the maximum order of a balanced complete subgraph.

######  Lemma 2.5.

[11] Let $\Gamma$ be a signed graph of order $n$. Then

$\lambda_{1}(\Gamma)\leq n\left(1-\frac{1}{\omega_{b}(\Gamma)}\right).$

Let $\Gamma$ be a signed graph on vertices $v_{1},\dots,v_{n}$. A real vector
$\mathbf{x}=(x_{1},\dots,x_{n})^{\top}$ is viewed as a function on
$\\{v_{1},\dots,v_{n}\\}$ that maps vertex $v_{i}$ to $x_{i}$ , i.e.,
$\mathbf{x}(v_{i})=x_{i}$, for $i=1,\dots,n$.

######  Lemma 2.6.

[4] Let $\mathbf{x}=(x_{1},\dots,x_{n})^{\top}$ be an eigenvector associated
with the largest eigenvalue of a signed graph $\Gamma$ and let $v_{r},v_{s}$
be fixed vertices of $\Gamma$. If $x_{r}x_{s}\geq 0$, at least one of
$x_{r},x_{s}$ is nonzero, and $v_{r}$ and $v_{s}$ is are not adjacent (resp.
$v_{r}v_{s}$ is a negative edge), then for a signed graph $\Gamma^{\prime}$
obtained by adding a positive edge $v_{r}v_{s}$ (resp. removing $v_{r}v_{s}$
or reversing its sign) we have
$\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma)$.

##  3 Proof of Theorem 1.1

###### Proof of Theorem 1.1.

For a $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph of order $6$, the
length of a shortest negative cycle is $5$ or $6$. We display all
$\mathcal{C}_{3,4}^{-}$-free unbalanced signed graphs of order $6$ under
switching equivalence in Fig. 3, where the largest and the smallest
eigenvalues are listed below the corresponding signed graphs, and the last
signed graph is just $\Gamma_{6,0}$. It is easy to see that $\Gamma_{6,0}$ is
the unique one with maximum spectral radius, which, by Lemma 2.2 (ii), is the
largest root of the equation $\lambda^{3}-6\lambda-3=0$.

$-$$1.6180$$-2$$-$$1.7321$$-1.7321$$-$$1.8608$$-2.1149$$-$$2.3028$$-2$$-$$2.1642$$-2.3914$$-$$2.6691$$-2.1451$
Figure 3: The $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graphs of order
$6$ together with the largest and the smallest eigenvalues.

Suppose in the following that $n\geq 7$. Let $\Gamma=(G,\sigma)$ be a
$\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph with order $n$ that
maximizes the spectral radius. Since $\Gamma_{n,0}$ is
$\mathcal{C}_{3,4}^{-}$-free and unbalanced, we have
$\rho(\Gamma)\geq\lambda_{1}(\Gamma_{n,0})>n-4$ by Lemma 2.2 (ii). By a direct
calculation, we have $\rho(\Gamma_{7,0})=3.7136>3.7$.

######  Claim 3.1.

$\rho(\Gamma)=\lambda_{1}(\Gamma)$.

###### Proof.

Suppose that $\rho(\Gamma)\neq\lambda_{1}(\Gamma)$. Since
$\rho(\Gamma)=\max\\{\lambda_{1}(\Gamma),-\lambda_{n}(\Gamma)\\}$, we have
$\lambda_{1}(\Gamma)<-\lambda_{n}(\Gamma)=\rho(\Gamma)$. As $\Gamma$ is
$\mathcal{C}_{3}^{-}$-free, we have $\omega_{b}(-\Gamma)\leq 2$. By Lemma 2.5,
we have

$\rho(\Gamma)=-\lambda_{n}(\Gamma)=\lambda_{1}(-\Gamma)\leq
n\left(1-\frac{1}{\omega_{b}(-\Gamma)}\right)\leq\frac{n}{2}\leq\begin{cases}n-4&\mbox{if
}n\geq 8,\\\ 3.7&\mbox{if }n=7,\end{cases}$

a contradiction. ∎

By Lemma 2.4, $\Gamma$ is switching equivalent to a signed graph
$\Gamma^{*}=(G,\sigma^{*})$ such that $A(\Gamma^{*})$ has a non-negative unit
eigenvector $\mathbf{x}=(x_{1},\dots,x_{n})^{\top}$ associated with
$\lambda_{1}(\Gamma^{*})$. By Lemma 2.3, $\Gamma^{*}$ is a
$\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph of order $n$. By Claim
3.1, we have
$\rho(\Gamma^{*})=\lambda_{1}(\Gamma^{*})=\lambda_{1}(\Gamma)=\rho(\Gamma)>n-4$.

######  Claim 3.2.

$\mathbf{x}$ contains at most two zero entries.

###### Proof.

Suppose that this is not true. Assume that $x_{i}=x_{j}=x_{k}=0$ with $1\leq
i<j<k\leq n$. Let $\mathbf{y}$ be the vector obtained from $\mathbf{x}$ by
removing the entries $x_{i},x_{j}$ and $x_{k}$. It is well known that among
graphs with fixed order, the spectral radius is uniquely maximized by the
complete graph. By Rayleigh’s principle,

$\displaystyle\lambda_{1}(\Gamma^{*})$
$\displaystyle=\mathbf{x}^{\top}A(\Gamma^{*})\mathbf{x}=\mathbf{y}^{\top}A(\Gamma^{*}-\\{v_{i},v_{j},v_{k}\\})\mathbf{y}$
$\displaystyle\leq\lambda_{1}(\Gamma^{*}-\\{v_{i},v_{j},v_{k}\\})\leq\lambda_{1}(G-\\{v_{i},v_{j},v_{k}\\})\leq\lambda_{1}(K_{n-3})=n-4,$

a contradiction. ∎

######  Claim 3.3.

$\Gamma^{*}$ is connected.

###### Proof.

Suppose that $\Gamma^{*}$ is not connected. Note that $\Gamma^{*}$ contains at
least one negative cycle. Assume that $v_{1}v_{2}$ is an edge in one component
with $x_{1}\geq x_{2}$ and $v_{3}$ is a vertex in another component. By Claim
3.2, we have $x_{3}>0$ or $x_{1}>0$. Let $\Gamma^{\prime}$ be the signed graph
formed by adding a positive edge $v_{1}v_{3}$ to $\Gamma^{*}$. Evidently,
$\Gamma^{\prime}$ is a $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph of
order $n$. By Lemma 2.6, we have
$\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction. ∎

Since $\Gamma^{*}$ is unbalanced, $\Gamma^{*}$ contains at least one negative
cycle. Let $C:=v_{1}\dots v_{r}v_{1}$ be a shortest negative cycle in
$\Gamma^{*}$. Then $r\geq 5$.

######  Claim 3.4.

$C$ is an induced cycle.

###### Proof.

Suppose that $C$ is not an induced cycle. Then there is a chord $v_{s}v_{t}$
of $C$ with $1\leq s<t\leq r$. As $C$ is negative, $v_{s}$ and $v_{t}$
separate $C$ into two paths connecting $v_{s}$ and $v_{t}$, one path $P$ with
an even number of negative edges, and the other path $Q$ with an odd number of
negative edges. If $v_{s}v_{t}$ is an negative edge, then $P$ and $v_{s}v_{t}$
form a negative cycle that is shorter than $C$, otherwise, $Q$ and
$v_{s}v_{t}$ form a negative cycle that is shorter than $C$, which is a
contradiction in either case. ∎

######  Claim 3.5.

If $v_{s}v_{t}$ is a negative edge of $\Gamma^{*}$ and it is not an edge of
$C$, then $x_{s}=x_{t}=0$.

###### Proof.

Construct a signed graph $\Gamma^{\prime}$ by removing the edge $v_{s}v_{t}$
from $\Gamma^{*}$. As $C$ is a negative cycle in $\Gamma^{\prime}$,
$\Gamma^{\prime}$ is a $\mathcal{C}_{3,4}^{-}$-free unbalanced signed graph.
If $x_{s}\neq 0$ or $x_{t}\neq 0$, then we have by Lemma 2.6 that
$\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction. ∎

By Claims 3.2 and 3.5, there is at most one negative edge outside $C$.

######  Claim 3.6.

Any negative edge on $C$ does not lie on a positive $C_{3}$.

###### Proof.

Suppose that a positive $C_{3}$ contains a negative edge, say $e=v_{1}v_{2}$,
on $C$. Let $v_{s}$ be the vertex of the positive $C_{3}$ different from
$v_{1}$ and $v_{2}$. By Claim 3.4, $v_{s}\notin V(C)$. Note that $v_{1}v_{s}$
and $v_{2}v_{s}$ have opposite signs, say $v_{1}v_{s}$ is negative. By Claims
3.5 and 3.2, $x_{1}=x_{s}=0$ and $x_{2}>0$. Let $\Gamma^{\prime}$ be the
signed graph formed by removing the edge $v_{1}v_{2}$ from $\Gamma^{*}$. Then
$v_{2}\dots v_{r}v_{1}v_{s}v_{2}$ is a negative cycle in $\Gamma^{\prime}$, so
$\Gamma^{\prime}$ is obviously $\mathcal{C}_{3,4}$-free and unbalanced. By
Lemma 2.6, we have $\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a
contradiction. ∎

######  Claim 3.7.

If there is a negative edge outside $C$, then it does not lie on a positive
$C_{4}$.

###### Proof.

Suppose that there is a negative edge $v_{s}v_{t}$ outside $C$ but it lies on
some positive $C_{4}$. As $v_{s}v_{t}$ is the unique negative edge outside
$C$, there is a negative edge, say $v_{1}v_{2}$, on $C$ so that it lies on the
positive $C_{4}$ containing $v_{s}v_{t}$.

Suppose first that $\\{s,t\\}\cap\\{1,2\\}\neq\emptyset$, say $s=1$. Then
$v_{t}\notin V(C)$. By Claims 3.5 and 3.2, $x_{1}=x_{t}=0$ and $x_{2}>0$. Let
$\Gamma^{\prime}$ be the signed graph formed by removing the edge $v_{1}v_{2}$
from $\Gamma^{*}$. Let $u$ be the vertex of the positive $C_{4}$ different
from $v_{1},v_{2}$ and $v_{t}$. If $u\in V(C)$, then $u=v_{3}$, so
$v_{2}v_{3}$, $v_{t}v_{3}$ have the same sign. Then $\Gamma^{\prime}$ contains
a negative cycle $v_{3}\dots v_{r}v_{1}v_{t}v_{3}$ if $u\in V(C)$ and a
negative cycle $v_{2}\dots v_{r}v_{1}v_{t}uv_{2}$ if $u\notin V(C)$. So
$\Gamma^{\prime}$ is obviously $\mathcal{C}_{3,4}$-free and unbalanced. By
Lemma 2.6, we have $\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a
contradiction.

Suppose next that $\\{s,t\\}\cap\\{1,2\\}=\emptyset$. By Claims 3.5 and 3.2,
$x_{1},x_{2}>0$. Assume that the positive $C_{4}$ is
$v_{1}v_{2}v_{t}v_{s}v_{1}$ (if it is $v_{1}v_{t}v_{s}v_{2}v_{1}$, then the
proof is similar). Then $v_{1}v_{s}$ and $v_{2}v_{t}$ have the same sign. If
$\\{v_{s},v_{t}\\}\cap V(C)\neq\emptyset$, then $\\{v_{s},v_{t}\\}\cap
V(C)=\\{v_{s}\\}$ with $s=r$, or $\\{v_{s},v_{t}\\}\cap V(C)=\\{v_{t}\\}$ with
$t=3$. Let $\Gamma^{\prime\prime}$ be the signed graph formed by removing the
edge $v_{1}v_{2}$ from $\Gamma^{*}$. If $\\{v_{s},v_{t}\\}\cap
V(C)=\emptyset$, then $\Gamma^{\prime\prime}$ contains a negative cycle
$v_{2}\dots v_{r}v_{1}v_{s}v_{t}v_{2}$; Otherwise, $\Gamma^{\prime\prime}$
contains a negative cycle obtained by replacing $v_{r}v_{1},v_{1}v_{2}$ with
$v_{r}v_{t},v_{t}v_{2}$ in $C$ if $\\{v_{s},v_{t}\\}\cap V(C)=\\{v_{s}\\}$
(resp. by replacing $v_{1}v_{2},v_{2}v_{3}$ with $v_{1}v_{s},v_{s}v_{3}$ in
$C$ if $\\{v_{s},v_{t}\\}\cap V(C)=\\{v_{t}\\}$). So $\Gamma^{\prime\prime}$
is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. Then we have By Lemma 2.6 that
$\lambda_{1}(\Gamma^{\prime\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction.
∎

######  Claim 3.8.

$C$ contains exactly one negative edge.

###### Proof.

Suppose that $C$ contains more than one negative edges. Choose two negative
edges $v_{1}v_{2}$ and $v_{i}v_{i+1}$ on $C$, where $2\leq i\leq r$. If $i=r$,
then $v_{i+1}=v_{1}$. Let $\widetilde{\Gamma}$ be a signed graph from
$\Gamma^{*}$ by reversing the sign of $v_{1}v_{2}$ and $v_{i}v_{i+1}$ from
$\Gamma^{*}$.

We claim that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free.

Suppose that there is a negative $C_{3}$ in $\widetilde{\Gamma}$. Then it
contains one of the positive edges $v_{1}v_{2}$ and $v_{i}v_{i+1}$. Claim 3.4,
it contains exactly one of the positive edges $v_{1}v_{2}$ and $v_{i}v_{i+1}$,
say $v_{1}v_{2}$. Then reversing the sign of $v_{1}v_{2}$ in this negative
$C_{3}$ in $\widetilde{\Gamma}$, we obtain a positive $C_{3}$ containing the
negative $v_{1}v_{2}$ in $\Gamma^{*}$, contradicting Claim 3.6.

Suppose there is a negative $C_{4}$ in $\widetilde{\Gamma}$. Then it contains
one of $v_{1}v_{2}$ and $v_{i}v_{i+1}$. If it contains both $v_{1}v_{2}$ and
$v_{i}v_{i+1}$, then $i=2$ or $i=r$ by Claim 3.4, so by reversing the sign of
$v_{1}v_{2}$ and $v_{i}v_{i+1}$ in this negative $C_{4}$ in
$\widetilde{\Gamma}$, we obtain a negative $C_{4}$ in $\Gamma^{*}$, which is
impossible. So it contains exactly one of $v_{1}v_{2}$ and $v_{i}v_{i+1}$, say
$v_{1}v_{2}$. Assume that it is $v_{1}v_{s}v_{t}v_{2}v_{1}$. By reversing the
sign of $v_{1}v_{2}$ in this negative $C_{4}$ of $\widetilde{\Gamma}$, we
obtain a positive $C_{4}$ of $\Gamma^{*}$ containing the negative edge
$v_{1}v_{2}$. By Claim 3.7, it does not contain any negative edge outside $C$,
so it contains the negative edge $v_{2}v_{3}$ or $v_{1}v_{r}$, i.e., $t=3$ or
$s=r$, say $t=3$. By Claim 3.2, one of $x_{1}$, $x_{2}$ and $x_{3}$ is
nonzero. Construct a signed graph $\Gamma^{\prime}$ by removing the edge
$v_{1}v_{2}$ from $\Gamma^{*}$ if $x_{1}\neq 0$ or $x_{2}\neq 0$, and by
removing an edge $v_{2}v_{3}$ if $x_{3}\neq 0$. Note that replacing two edges
$v_{1}v_{2}$ and $v_{2}v_{3}$ on $C$ by two edges $v_{1}v_{s}$ and
$v_{s}v_{3}$, we get a negative cycle in $\Gamma^{\prime}$, so
$\Gamma^{\prime}$ is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. By Lemma
2.6, we have $\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a
contradiction.

It follows that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free, as
desired.

As the cycle obtained from $C$ by reversing the sign of $v_{1}v_{2}$ and
$v_{i}v_{i+1}$ is a negative cycle in $\widetilde{\Gamma}$,
$\widetilde{\Gamma}$ is unbalanced. By Rayleigh’s principle, we have

$\lambda_{1}(\widetilde{\Gamma})-\lambda_{1}(\Gamma^{*})\geq\mathbf{x}^{T}(A(\widetilde{\Gamma})-A(\Gamma^{*}))\mathbf{x}=4(x_{1}x_{2}+x_{i}x_{i+1})\geq
0,$

so $\lambda_{1}(\widetilde{\Gamma})=\lambda_{1}(\Gamma^{*})$, implying that
$\mathbf{x}$ is an eigenvector of $A(\widetilde{\Gamma})$ associated with
$\lambda_{1}(\widetilde{\Gamma})$. If $i=2,r$, say $i=2$, then

$\lambda_{1}(\widetilde{\Gamma})x_{1}=\lambda_{1}(\Gamma^{*})x_{1}+2x_{2}$

and

$\lambda_{1}(\widetilde{\Gamma})x_{2}=\lambda_{1}(\Gamma^{*})x_{2}+2(x_{1}+x_{3}),$

so $x_{2}=0$ and $x_{1}+x_{3}=0$, implying that $x_{1}=x_{2}=x_{3}=0$,
contradicting Claim 3.2. If $3\leq i\leq r-1$, then
$x_{1}x_{2}=x_{i}x_{i+1}=0$, so assuming $x_{1}=x_{i}=0$, we have

$\lambda_{1}(\widetilde{\Gamma})x_{i}=\lambda_{1}(\Gamma^{*})x_{i}+2x_{i+1},$

so $x_{i+1}=0$, contradicting Claim 3.2. ∎

By Claim 3.8, $C$ contains exactly one negative edge, say $v_{1}v_{2}$.

######  Claim 3.9.

There is no negative edge of $\Gamma^{*}$ outside $C$.

###### Proof.

Suppose that this is not true. Then there is a negative edge, say
$v_{s}v_{t}$, of $\Gamma^{*}$ outside $C$. By Claim 3.5, $x_{s}=x_{t}=0$. By
Claims 3.5 and 3.2, $v_{s}v_{t}$ is the unique negative edge of $\Gamma^{*}$
that is not on $C$. By Claim 3.8, $v_{1}v_{2}$ and $v_{s}v_{t}$ are the only
negative edges of $\Gamma^{*}$.

Suppose that $\\{s,t\\}\cap\\{1,2\\}\neq\emptyset$, say $s=1$. Then $v_{t}$
lies outside $C$. By Claim 3.6, $v_{2}v_{t}\notin E(\Gamma^{*})$. Let
$\Gamma^{\prime}$ be a signed graph formed by adding a positive edge
$v_{2}v_{t}$ to $\Gamma^{*}$. Evidently, there is no negative $C_{3}$
containing $v_{2}v_{t}$ in $\Gamma^{\prime}$. If there is a negative $C_{4}$
containing $v_{2}v_{t}$ in $\Gamma^{\prime}$, then it contains one of
$v_{1}v_{2}$ and $v_{1}v_{t}$, say $v_{1}v_{2}$, then it is
$v_{1}v_{2}v_{t}uv_{1}$ for some common neighbor $u$ of $v_{1}$ and $v_{t}$ in
$\Gamma^{*}$, so $v_{1}v_{t}uv_{1}$ is a negative $C_{3}$ in $\Gamma^{*}$, a
contradiction. So $\Gamma^{\prime}$ is $\mathcal{C}_{3,4}^{-}$-free, and as
$C$ is a negative cycle of $\Gamma^{\prime}$, it is unbalanced. By Claim 3.2,
$x_{2}>0$. Then by Lemma 2.6, we have
$\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction. Thus
$\\{s,t\\}\cap\\{1,2\\}=\emptyset$. By Claim 3.2, $x_{1},x_{2}>0$.

Case 1. $\\{v_{s},v_{t}\\}\cap V(C)\neq\emptyset$.

Assume that $v_{s}\in V(C)$ with $\lceil\frac{r+1}{2}\rceil+1\leq s\leq r$.
Then $v_{t}\notin V(C)$. By Claim 3.4, $C$ is an induced cycle.

Case 1.1. $s\leq r-2$.

In this case, $r\geq 7$. By Claim 3.4, $v_{2}v_{s}\not\in E(\Gamma^{*})$. Let
$\widetilde{\Gamma}$ be a signed graph formed by adding a positive
$v_{2}v_{s}$. It is obvious that $\widetilde{\Gamma}$ is unbalanced as $C$ is
also the negative cycle in $\widetilde{\Gamma}$.

We claim that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free. If there
is a negative $C_{3}$ in $\widetilde{\Gamma}$, then it contains $v_{2}v_{s}$
and exactly one of $v_{1}v_{2}$ and $v_{s}v_{t}$, so $v_{1}v_{s}\in
E(\Gamma^{*})$ or $v_{2}v_{t}\in E(\Gamma^{*})$. By Claim 3.4,
$v_{1}v_{s}\not\in E(\Gamma^{*})$. If $v_{2}v_{t}\in E(\Gamma^{*})$, then we
would have a negative cycle $v_{2}\dots v_{s}v_{t}v_{2}$ that is shorter than
$C$ in $\Gamma^{*}$, a contradiction. So $\widetilde{\Gamma}$ is
$\mathcal{C}_{3}^{-}$-free. Suppose that there is a negative $C_{4}$ in
$\widetilde{\Gamma}$. Then it contains $v_{2}v_{s}$ and exactly one of
$v_{1}v_{2}$ and $v_{s}v_{t}$. Suppose that this negative $C_{4}$ of
$\widetilde{\Gamma}$ contains $v_{1}v_{2}$. Then it is $v_{1}v_{2}v_{s}uv_{1}$
for some common neighbor $u$ of $v_{1}$ and $v_{s}$. By Claim 3.4 and $s\leq
r-2$, $u\not\in V(C)$, so $v_{1}\dots v_{s}uv_{1}$ is a negative cycle in
$\Gamma^{*}$ that is shorter than $C$, which is a contradiction. Suppose that
this negative $C_{4}$ of $\widetilde{\Gamma}$ contains $v_{s}v_{t}$. Then it
is $v_{2}v_{s}v_{t}wv_{2}$ for some common neighbor $w$ of $v_{2}$ and
$v_{t}$. If $w\in V(C)$, then $w=v_{3}$. So $\Gamma^{*}$ contains a negative
cycle $v_{3}\dots v_{s}v_{t}v_{3}$ if $w\in V(C)$, and $v_{2}\dots
v_{s}v_{t}wv_{2}$ if $w\not\in V(C)$, which are shorter than $C$, a
contradiction. It follows that $\widetilde{\Gamma}$ is
$\mathcal{C}_{3,4}^{-}$-free.

As $x_{2}>0$, we have by Lemma 2.6 that
$\lambda_{1}(\widetilde{\Gamma})>\lambda_{1}(\Gamma^{*})$, a contradiction.

Case 1.2. $s=r-1$.

Note that $v_{1}v_{t}\notin E(\Gamma^{*})$, as otherwise
$v_{1}v_{r}v_{r-1}v_{t}v_{1}$ is a negative $C_{4}$ in $\Gamma^{*}$, a
contradiction. Let $\widetilde{\Gamma}$ be a signed graph formed by removing a
negative edge $v_{r-1}v_{t}$ and adding a positive $v_{1}v_{t}$ from
$\Gamma^{*}$.

We claim that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free. If there
is a negative $C_{3}$ in $\widetilde{\Gamma}$, then it contains both
$v_{1}v_{t}$ and $v_{1}v_{2}$, so $v_{2}v_{t}\in E(\Gamma^{*})$, implying that
there is a negative cycle $v_{2}\dots v_{r-1}v_{t}v_{2}$ that is shorter than
$C$ in $\Gamma^{*}$, a contradiction. So $\widetilde{\Gamma}$ is
$\mathcal{C}_{3}^{-}$-free. Suppose that there is a negative $C_{4}$ in
$\widetilde{\Gamma}$. Then it contains both $v_{1}v_{t}$ and $v_{1}v_{2}$, so
it is $v_{t}v_{1}v_{2}uv_{t}$ for some common neighbor $u$ of $v_{2}$ and
$v_{t}$ in $\Gamma^{*}$. If $u\in V(C)$, then $u=v_{3}$, so $v_{3}\dots
v_{r-1}v_{t}v_{3}$ is negative cycle shorter than $C$ in $\Gamma^{*}$, a
contradiction. So $u\notin V(C)$. Let $\Gamma^{\prime\prime}$ be a signed
graph formed by removing $v_{1}v_{2}$ from $\Gamma^{*}$. As $v_{2}\dots
v_{r-1}v_{t}uv_{2}$ is a negative cycle in $\Gamma^{\prime\prime}$,
$\Gamma^{\prime\prime}$ is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. By
Lemma 2.6, we have
$\lambda_{1}(\Gamma^{\prime\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction.
It follows that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free.
Evidently, $\widetilde{\Gamma}$ is unbalanced. By Rayleigh’s principle, we
have

$\lambda_{1}(\widetilde{\Gamma})-\lambda_{1}(\Gamma^{*})\geq\mathbf{x}^{T}(A(\widetilde{\Gamma})-A(\Gamma^{*}))\mathbf{x}=2x_{t}(x_{1}+x_{r-1})=0,$

so $\lambda_{1}(\widetilde{\Gamma})=\lambda_{1}(\Gamma^{*})$. Then
$\lambda_{1}(\widetilde{\Gamma})x_{t}=\lambda_{1}(\Gamma^{*})x_{t}+x_{1}+x_{r-1}$,
which implies that $x_{1}=0$, a contradiction.

Case 1.3. $s=r$.

By Claim 3.7, $v_{2}v_{t}\notin E(\Gamma^{*})$. Let $\widetilde{\Gamma}$ be a
signed graph formed by adding a positive $v_{2}v_{t}$ from $\Gamma^{*}$.

If there is a negative $C_{3}$ in $\widetilde{\Gamma}$, then it contains
$v_{2}v_{t}$ and exactly one of $v_{1}v_{2}$ and $v_{r}v_{t}$, so
$v_{1}v_{t}\in E(\Gamma^{*})$ or $v_{2}v_{r}\in E(\Gamma^{*})$. But
$v_{1}v_{t}\in E(\Gamma^{*})$ leads to a negative
$C_{3}:=v_{1}v_{r}v_{t}v_{1}$ in $\Gamma^{*}$, and $v_{2}v_{r}\in
E(\Gamma^{*})$ contradicts Claim 3.4. So $\widetilde{\Gamma}$ is
$\mathcal{C}_{3}^{-}$-free. Suppose that there is a negative $C_{4}$ in
$\widetilde{\Gamma}$. Then it contains $v_{2}v_{t}$ and exactly one of
$v_{1}v_{2}$ and $v_{r}v_{t}$. If it contains $v_{1}v_{2}$, then it is
$v_{1}v_{2}v_{t}uv_{1}$ for some common neighbor $u$ of $v_{1}$ and $v_{t}$ in
$\Gamma^{*}$ with $u\neq v_{r}$, as $u\not\in V(C)$ by Claim 3.4,
$v_{1}v_{r}v_{t}uv_{1}$ is a negative $C_{4}$ in $\Gamma^{*}$, a
contradiction. Similarly, we have a contradiction if it contains $v_{r}v_{t}$.
It follows that $\widetilde{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free.
Evidently, $\widetilde{\Gamma}$ is unbalanced. As $x_{1},x_{2}>0$, we have by
Lemma 2.6, $\lambda_{1}(\widetilde{\Gamma})>\lambda_{1}(\Gamma^{*})$, a
contradiction.

Case 2. $\\{v_{s},v_{t}\\}\cap V(C)=\emptyset$.

By Claim 3.7, $\\{v_{1}v_{s},v_{2}v_{t}\\}\not\subseteq E(\Gamma^{*})$, and
$\\{v_{1}v_{t},v_{2}v_{s}\\}\not\subseteq E(\Gamma^{*})$. Assume that
$v_{1}v_{s}\notin E(\Gamma^{*})$. There are two possibilities:
$v_{1}v_{t}\notin E(\Gamma^{*})$ or $v_{2}v_{s}\notin E(\Gamma^{*})$.

If $v_{1}v_{t}\notin E(\Gamma^{*})$, then
$\\{v_{2}v_{t},v_{2}v_{s}\\}\not\subseteq E(\Gamma^{*})$, as otherwise,
$v_{2}v_{s}v_{t}v_{2}$ would be a negative $C_{3}$. So one of the following
must hold:

  1. (i)

$v_{2}v_{s}\notin E(\Gamma^{*})$ and $v_{2}v_{t}\notin E(\Gamma^{*})$,

  2. (ii)

$v_{2}v_{t}\in E(\Gamma^{*})$ and $v_{2}v_{s}\notin E(\Gamma^{*})$,

  3. (iii)

$v_{2}v_{s}\in E(\Gamma^{*})$ and $v_{2}v_{t}\notin E(\Gamma^{*})$.

Suppose first that (i) holds. Suppose that $u\in N_{\Gamma^{*}}(v_{1})\cap
N_{\Gamma^{*}}(v_{t})\neq\emptyset$ and $w\in N_{\Gamma^{*}}(v_{2})\cap
N_{\Gamma^{*}}(v_{s})\neq\emptyset$. Note that $u\neq w$, as otherwise,
$v_{1}v_{2}uv_{1}$ is a negative $C_{3}$ in $\Gamma^{*}$, a contradiction.
Construct a signed graph $\Gamma^{\prime\prime}$ by removing $v_{1}v_{2}$ from
$\Gamma^{*}$. As $v_{2}\dots v_{r}v_{1}uv_{t}v_{s}wv_{2}$ is a negative cycle
in $\Gamma^{\prime\prime}$, it is easily to see that $\Gamma^{\prime\prime}$
is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. By Lemma 2.6, we have
$\lambda_{1}(\Gamma^{\prime\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction.
It follows that $N_{\Gamma^{*}}(v_{1})\cap N_{\Gamma^{*}}(v_{t})=\emptyset$ or
$N_{\Gamma^{*}}(v_{2})\cap N_{\Gamma^{*}}(v_{s})=\emptyset$. Assume that
$N_{\Gamma^{*}}(v_{1})\cap N_{\Gamma^{*}}(v_{t})=\emptyset$.

Let $\widehat{\Gamma}$ be a signed graph formed by removing the edge
$v_{s}v_{t}$ and adding a positive edge $v_{2}v_{t}$ from $\Gamma^{*}$. If
there is a negative $C_{3}$ in $\widehat{\Gamma}$, then it contains both
$v_{2}v_{t}$ and $v_{1}v_{2}$, so $v_{1}v_{t}\in E(\Gamma^{*})$, a
contradiction. If there is a negative $C_{4}$ in $\widehat{\Gamma}$, then it
contains both $v_{2}v_{t}$ and $v_{1}v_{2}$, so $v_{1}$ and $v_{t}$ have a
common neighbor in $\Gamma^{*}$, a contradiction. It follows that
$\widehat{\Gamma}$ is $\mathcal{C}_{3,4}^{-}$-free. Evidently,
$\widetilde{\Gamma}$ is unbalanced. Note that

$\lambda_{1}(\widehat{\Gamma})-\lambda_{1}(\Gamma^{*})\geq
2x_{t}(x_{2}+x_{s})=0.$

Then $\lambda_{1}(\widehat{\Gamma})=\lambda_{1}(\Gamma^{*})$, implying that
$\lambda_{1}(\widehat{\Gamma})x_{t}=\lambda_{1}(\Gamma^{*})x_{t}+x_{2}+x_{s}$,
so $x_{2}=0$, a contradiction.

Suppose next that (ii) holds. Let $\widetilde{\Gamma}$ be a signed graph
formed by adding a positive edge $v_{1}v_{s}$ from $\Gamma^{*}$. If there is a
negative $C_{3}$ in $\widetilde{\Gamma}$, then it contains $v_{1}v_{s}$ and
exactly one of $v_{1}v_{2}$ and $v_{s}v_{t}$, so $v_{2}v_{s}\in E(\Gamma^{*})$
or $v_{1}v_{t}\in E(\Gamma^{*})$, a contradiction. If there is a negative
$C_{4}$ in $\widetilde{\Gamma}$, then it contains $v_{1}v_{s}$ and exactly one
of $v_{1}v_{2}$ and $v_{s}v_{t}$, say $v_{1}v_{2}$. Then $v_{2}$ and $v_{s}$
have a common neighbor $u$ in $\Gamma^{*}$ with $n\neq v_{t}$, so
$v_{2}uv_{s}v_{t}v_{2}$ is a negative $C_{4}$ in $\Gamma^{*}$, a
contradiction. It follows that $\widetilde{\Gamma}$ is
$\mathcal{C}_{3,4}^{-}$-free. Evidently, $\widetilde{\Gamma}$ is unbalanced.
As $x_{1}>0$, we have by Lemma 2.6,
$\lambda_{1}(\widetilde{\Gamma})>\lambda_{1}(\Gamma^{*})$, a contradiction.

Suppose that (iii) holds. Let $\widetilde{\Gamma}$ be a signed graph formed by
adding a positive $v_{1}v_{t}$ from $\Gamma^{*}$. If there is a negative
$C_{3}$ in $\widetilde{\Gamma}$, then it contains $v_{1}v_{t}$ and exactly one
of $v_{1}v_{2}$ and $v_{s}v_{t}$, so $v_{2}v_{t}\in E(\Gamma^{*})$ or
$v_{1}v_{s}\in E(\Gamma^{*})$, a contradiction. Suppose that there is a
negative $C_{4}$ in $\widetilde{\Gamma}$. Then it contains $v_{1}v_{t}$ and
exactly one of $v_{1}v_{2}$ and $v_{s}v_{t}$, say $v_{1}v_{2}$. Then $v_{2}$
and $v_{t}$ have a common neighbor $v\neq v_{1},v_{s}$, so
$v_{2}vv_{t}v_{s}v_{2}$ is a negative $C_{4}$ in $\Gamma^{*}$, a
contradiction. It follows that $\widetilde{\Gamma}$ is
$\mathcal{C}_{3,4}^{-}$-free. Evidently, $\widetilde{\Gamma}$ is unbalanced.
As $x_{1}>0$, we have by Lemma 2.6,
$\lambda_{1}(\widetilde{\Gamma})>\lambda_{1}(\Gamma^{*})$, a contradiction.

Suppose that $v_{2}v_{s}\notin E(\Gamma^{*})$. Then it is impossible that both
$v_{t}v_{1}\in E(\Gamma^{*})$ and $v_{t}v_{2}\in E(\Gamma^{*})$. By similar
argument as above, we also have a contradiction. ∎

By Claim 3.9, $v_{1}v_{2}$ is the unique negative edge of $\Gamma^{*}$, which
lies on $C$.

######  Claim 3.10.

$x_{i}>0$ for $3\leq i\leq n$.

###### Proof.

By Claim 3.3, $d_{\Gamma^{*}}(v_{i})\geq 1$ for $i=1,\dots,n$.

First, we show that $x_{i}>0$ for any $3\leq i\leq n$ with
$d_{\Gamma^{*}}(v_{i})\geq 2$. Otherwise, $x_{i}=0$ for some $i$ with $3\leq
i\leq n$ and $d_{\Gamma^{*}}(v_{i})\geq 2$. By Claim 3.2, one entry of
$\mathbf{x}$ at some vertex of $N_{\Gamma^{*}}(v_{i})$ is nonzero. As
$v_{1}v_{2}$ is the unique negative edge of $\Gamma^{*}$, all edges of
$\\{v_{s}v_{i}:v_{s}\in N_{\Gamma^{*}}(v_{i})\\}$ are positive, so
$\sum\limits_{v_{s}\in N_{\Gamma^{*}}(v_{i})}\sigma(v_{s}v_{i})x_{s}>0$.
However, from $A(\Gamma^{*})\mathbf{x}=\lambda_{1}(\Gamma^{*})\mathbf{x}$, we
have $\sum\limits_{v_{s}\in
N_{\Gamma^{*}}(v_{i})}\sigma(v_{s}v_{i})x_{s}=\lambda_{1}(\Gamma^{*})x_{i}=0$,
a contradiction.

Next, we show that $x_{i}>0$ for any $r+1\leq i\leq n$ with
$d_{\Gamma^{*}}(v_{i})=1$. Otherwise, $x_{i}=0$ for some $i$ with $r+1\leq
i\leq n$ and $d_{\Gamma^{*}}(v_{i})=1$. Let

$j=\begin{cases}r-1&\mbox{if $N_{\Gamma^{*}}(v_{i})\neq\\{v_{r-1}\\}$},\\\
3&\mbox{otherwise}.\end{cases}$

By the above argument, $x_{j}>0$. Construct a signed graph $\Gamma^{\prime}$
by adding a positive $v_{i}v_{j}$ from $\Gamma^{*}$ . As $v_{1}v_{2}$ is the
unique negative edge of $\Gamma^{*}$, it may be easily checked that
$\Gamma^{\prime}$ is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. By Lemma
2.6, we have $\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a
contradiction. ∎

######  Claim 3.11.

$C$ is $5$-cycle, i.e., $r=5$.

###### Proof.

Note that $r\geq 5$. If $r\geq 6$, then we construct a signed graph
$\Gamma^{\prime}$ by adding a positive $v_{3}v_{r-1}$, which may be easily
checked is $\mathcal{C}_{3,4}^{-}$-free and unbalanced. By Lemma 2.6 and Claim
3.10, $\lambda_{1}(\Gamma^{\prime})>\lambda_{1}(\Gamma^{*})$, a contradiction.
∎

By Claims 3.4 and 3.11, $C=v_{1}v_{2}v_{3}v_{4}v_{5}v_{1}$ and it is an
induced cycle in $\Gamma^{*}$. Note that it is a shortest negative cycle
containing the unique negative edge $v_{1}v_{2}$ of $\Gamma^{*}$. Let
$X=N_{\Gamma^{*}}(v_{1})\setminus\\{v_{2},v_{5}\\}$,
$X^{\prime}=N_{\Gamma^{*}}(v_{2})\setminus\\{v_{1},v_{3}\\}$ and
$Y=V(\Gamma^{*})\setminus(X\cup X^{\prime}\cup\\{v_{1},v_{2},v_{3},v_{5}\\})$.
Since $\Gamma^{*}$ is $\mathcal{C}_{3,4}^{-}$-free, $X\cap
X^{\prime}=\emptyset$ and for any $u\in X$, $w\in X^{\prime}$, $u$ and $w$,
$u$ and $v_{3}$, $w$ and $v_{5}$ are not adjacent in $\Gamma^{*}$. Let
$|X|=\tau$ and $|X^{\prime}|=\tau^{\prime}$. Then
$|Y|=n-\tau-\tau^{\prime}-4$. By Lemma 2.6 and Claim 3.10, we see that
$\Gamma^{*}[X\cup\\{v_{5}\\}\cup Y]=(K_{n-\tau^{\prime}-3},+)$ and
$\Gamma^{*}[X^{\prime}\cup\\{v_{3}\\}\cup Y]=(K_{n-\tau-3},+)$. In particular,
$v_{5}$ is adjacent to all vertices in $X\cup Y$ and $v_{3}$ is adjacent to
all vertices in $X^{\prime}\cup Y$. The structure of $\Gamma^{*}$ is displayed
in Fig. 4.

$X$$Y$$X^{\prime}$$\underbrace{~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}}$$K_{n-\tau^{\prime}-4}$$\underbrace{~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}}$$K_{n-\tau-4}$$-$$v_{1}$$v_{5}$$v_{3}$$v_{2}$
Figure 4: The structure of $\Gamma^{*}$.

Note that

$\lambda_{1}(\Gamma^{*})x_{1}=-x_{2}+x_{5}+\sum\limits_{v_{i}\in X}x_{i}$

and

$\lambda_{1}(\Gamma^{*})x_{2}=-x_{1}+x_{3}+\sum\limits_{v_{j}\in
X^{\prime}}x_{j}.$

So

$(\lambda_{1}(\Gamma^{*})-1)(x_{1}-x_{2})+x_{3}-x_{5}=\sum\limits_{v_{i}\in
X}x_{i}-\sum\limits_{v_{j}\in X^{\prime}}x_{j}.$

Similarly, we have

$\lambda_{1}(\Gamma^{*})x_{5}=x_{1}+\sum\limits_{v_{i}\in
X}x_{i}+\sum\limits_{v_{k}\in Y}x_{k}$

and

$\lambda_{1}(\Gamma^{*})x_{3}=x_{2}+\sum\limits_{v_{j}\in
X^{\prime}}x_{j}+\sum\limits_{v_{k}\in Y}x_{k},$

so

$\lambda_{1}(\Gamma^{*})(x_{5}-x_{3})+x_{2}-x_{1}=\sum\limits_{v_{i}\in
X}x_{i}-\sum\limits_{v_{j}\in X^{\prime}}x_{j}.$

Thus

$\lambda_{1}(\Gamma^{*})(x_{1}-x_{2})+(\lambda_{1}(\Gamma^{*})+1)(x_{3}-x_{5})=0.$

Assume that $x_{1}\geq x_{2}$. Then $x_{3}\leq x_{5}$.

######  Claim 3.12.

$X^{\prime}=\emptyset$.

###### Proof.

Suppose that $X^{\prime}\neq\emptyset$. Let
$X^{\prime}=\\{v_{j_{1}},\dots,v_{j_{\tau^{\prime}}}\\}$ with
$\tau^{\prime}\geq 1$. Construct a signed graph $\Gamma^{\prime}$ by removing
the (positive) edges $v_{2}v_{j_{i}}$, $v_{3}v_{j_{i}}$ and adding the
positive edges $v_{1}v_{j_{i}}$ and $v_{5}v_{j_{i}}$ for all $1\leq
i\leq\tau^{\prime}$, which can be easily checked that is
$\mathcal{C}_{3,4}^{-}$-free and unbalanced. Then

$\lambda_{1}(\Gamma^{\prime})-\lambda_{1}(\Gamma^{*})\geq\mathbf{x}^{T}(A(\Gamma^{\prime})-A(\Gamma^{*}))\mathbf{x}=\sum\limits_{1\leq
i\leq\tau^{\prime}}2x_{v_{j_{i}}}(x_{1}-x_{2}+x_{5}-x_{3})\geq 0,$

so $\lambda_{1}(\Gamma^{\prime})=\lambda_{1}(\Gamma^{*})$, implying that

$\lambda_{1}(\Gamma^{\prime})x_{1}=\lambda_{1}(\Gamma^{*})x_{1}+\sum\limits_{1\leq
i\leq\tau^{\prime}}x_{v_{j_{i}}},$

so $x_{v_{j_{i}}}=0$ for $1\leq i\leq\tau^{\prime}$, contradicting Claim 3.10.
∎

By Claim 3.12, $X^{\prime}=\emptyset$. Then $\Gamma^{*}$ is $\Gamma_{n,\tau}$
for some $\tau$ with $0\leq\tau\leq n-5$. By Lemma 2.2, $\tau=0,n-5$. As
$\Gamma_{n,n-5}$ is switching equivalent to $\Gamma_{n,0}$, $\Gamma^{*}$ is
switching equivalent to $\Gamma_{n,0}\cong\Gamma_{n}$. Therefore, $\Gamma$ is
switching isomorphic to $\Gamma_{n,0}$. By Lemma 2.2 (ii),
$\lambda_{1}(\Gamma_{n,0})$ is equal to the largest root of
$\lambda^{3}-(n-6)\lambda^{2}-3(n-4)\lambda-n+3=0$. ∎

Acknowledgement. This work was supported by National Natural Science
Foundation of China (No. 12071158).

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# Zero-field magnetic skyrmions in exchange-biased ferromagnetic-
antiferromagnetic bilayers

M. Pankratova  Department of Physics and Astronomy, Uppsala University, Box
516, SE-75120 Uppsala, Sweden<EMAIL_ADDRESS>O. Eriksson
Department of Physics and Astronomy, Uppsala University, Box 516, SE-75120
Uppsala, Sweden  Wallenberg Initiative Materials Science for Sustainability,
Uppsala University, 75121 Uppsala, Sweden  A. Bergman  Department of Physics
and Astronomy, Uppsala University, Box 516, SE-75120 Uppsala, Sweden

###### Abstract

We report on the stabilization of ferromagnetic skyrmions in zero external
magnetic fields, in exchange-biased systems composed of ferromagnetic-
antiferromagnetic (FM-AFM) bilayers. By performing atomistic spin dynamics
simulations, we study cases of compensated, uncompensated, and partly
uncompensated FM-AFM interfaces, and investigate the impact of important
parameters such as temperature, inter-plane exchange interaction,
Dzyaloshinskii-Moria interaction, and magnetic anisotropy on the skyrmions
appearance and stability. The model with an uncompensated FM-AFM interface
leads to the stabilization of individual skyrmions and skyrmion lattices in
the FM layer, caused by the effective field from the AFM instead of an
external magnetic field. Similarly, in the case of a fully compensated FM-AFM
interface, we show that FM skyrmions can be stabilized. We also demonstrate
that accounting for interface roughness leads to stabilization of skyrmions
both in compensated and uncompensated interface. Moreover, in bilayers with a
rough interface, skyrmions in the FM layer are observed for a wide range of
exchange interaction values through the FM-AFM interface, and the chirality of
the skyrmions depends critically on the exchange interaction.

##  I  Introduction

Modern society currently produces an ever-increasing amount of data, leading
to pressing challenges for data-storage technologies, one of which is storage
density, i.e. storing bigger amount of data in smaller space. Moreover, data
centers already use around one percent of global electricity demand, calling
for less energy-hungry data storage technologies. Magnetic skyrmions are
nanoscale topological spin structures [1, 2, 3], that are very promising
candidates for data-carrying or for logic elements in developing spintronic
devices [3]. Skyrmions are stable due to the combination of Heisenberg
exchange interactions, asymmetric exchange interactions, so-called
Dzyaloshinskii-Moriya (DMI) interaction, and an external magnetic field. DMI,
in contrast with the Heisenberg exchange interaction, tends to favour
perpendicular magnetic moment orientations and introduces chirality to the
system, which is central for the creation of skyrmions.

Skyrmions are often stabilized using external magnetic fields [3], however,
for full exploration of spintronic devices involving skyrmions it is desirable
to stabilize skyrmions in the absence of a magnetic field. Many works have
been devoted to skyrmion stabilization in zero applied field, e.g. in bilayers
and multilayers [4, 5, 6, 7, 8, 9]. It was shown that the DMI interaction can
be transferred from an antiferromagnet (AFM) to a ferromagnet (FM) in FM-AFM
bilayers to observe skyrmions. Recently, AFM skyrmions in zero magnetic were
predicted in FM-AFM bilayers on triangular lattice by Monte Carlo simulations
[10].

Moreover, it was experimentally shown that skyrmions can be stabilized in zero
external magnetic field by using the exchange bias phenomena (EBP) in
ferromagnets [11, 12, 13] and antiferromagnets [7]. EBP [14] manifest itself
in the shift of magnetic hysteresis along the field axis, and appears in
systems with contacting FM-AFM, and other systems. The simplest model of
exchange bias involves an uncompensated antiferromagnetic interface [15, 16]
(non-zero magnetic moment of the AFM interface layer), therefore, an effective
field acts from the AFM to the FM layer. The resulting effective field can be
used to stabilize magnetic skyrmions in a FM, instead of an external applied
magnetic field. However, this uncompensated model is known to lead to an
overestimation of the shift of the hysteresis loop, when compared with
experimental studies. Therefore, other models assumed a compensated FM-AFM
interface (both AFM sublattices are presented at the interface and couple with
equal strength to the FM)[17]. Koon [18, 19] suggested that due to the
competition between couplings in AFM with a FM, the interface spins could
deviate from the easy axis, and form canted states, leading to the small
effective field observed in experiments, that is manifested by EBP with rather
small shifts of the hysteresis curve.

Another important factor to consider, while studying EBP, is the interface
roughness [19]. In most experiments some roughness or geometrical defects are
present at the interface. Also, it was shown that exchange interaction through
the interface is strongly impacted by the interface roughness [20]. It was
demonstrated experimentally in Ref.[21] that only about 7$\%$ of the spins at
the interface are uncompensated and contribute to the EBP. It was confirmed
theoretically in Ref.[22, 23] that only several percent of uncompensated
moments are enough to induce EBP. Also, it was shown that roughness and
defects impacts interfacial DMI [24] and therefore skyrmion stability. In
addition, roughness has a strong impact on skyrmion dynamics, and results for
rough and perfect interface cases can significantly differ [25], which makes
accounting of interface defects and roughness vital.

In this work, we study the possibility of stabilizing skyrmions in zero
external magnetic fields using the exchange bias phenomena. We investigate
various types of FM-AFM interfaces and the range of parameters, such as
exchange interaction, DMI, magnetic anisotropy, and temperature allowing
skyrmion stabilization. The paper is organized as follows: we start with
describing details of atomistic spin dynamics simulations, then we investigate
the case of an perfect interface. Finally, we study the impact of disorder or
(geometrical) roughness of the interface on the magnetic structure of FM and
AFM films.

##  II  Atomistic spin dynamics simulations

Atomistic spin dynamics (ASD) is governed by Landau-Lifshits-Gilbert equation:

$\displaystyle\frac{d\bm{m}_{i}}{dt}$
$\displaystyle=-\frac{\gamma}{(1+\alpha^{2})}\bm{m}_{i}\times(\bm{B}_{i}+\bm{B}_{i}^{\mathrm{fl}})$
(1)
$\displaystyle\quad-\frac{\gamma}{(1+\alpha^{2})}\frac{\alpha}{m_{i}}\bm{m}_{i}\times(\bm{m}_{i}\times[\bm{B}_{i}+\bm{B}_{i}^{\mathrm{fl}}]),$

where $\bm{m_{i}}$ represents an atomic magnetic moment, $m_{i}$ and $\gamma$
are the saturation magnetization and the gyromagnetic ratio correspondingly.
We obtain an effective exchange field $\bm{B}_{i}=-\partial
H_{\mathrm{SD}}/\partial\bm{m}_{i}$ from the spin Hamiltonian, $H_{SD}$. The
Hamiltonian used in this work includes FM, AFM parts and interaction between
FM and AFM subsystems:

$H_{\mathrm{SD}}=H_{\mathrm{FM}}+H_{\mathrm{AFM}}+H_{\mathrm{FM/AFM}}.$ (2)

Concretely:

$\displaystyle H_{\mathrm{SD}}$
$\displaystyle=-\frac{1}{2}\sum_{ij}J_{ij}^{\mathrm{FM}}\bm{m}_{i}\cdot\bm{m}_{j}-\frac{1}{2}\sum_{kl}J_{kl}^{\mathrm{AFM}}\bm{m}_{k}\cdot\bm{m}_{l}$
(3)
$\displaystyle\quad-\frac{1}{2}\sum_{ik}J_{ik}^{\mathrm{FM/AFM}}\bm{m}_{i}\cdot\bm{m}_{k}-\sum_{ij}D_{ij}^{\mathrm{FM}}\cdot(\bm{m}_{i}\times\bm{m}_{j})$
$\displaystyle\quad-
H_{\mathrm{ext}}\Bigl{(}\sum_{i}m_{i}^{z}+\sum_{k}m^{z}_{k}\Bigr{)}-K^{\mathrm{AFM}}\sum_{k}(m^{z}_{k})^{2},$

where $J_{ij}$, $J_{kl}$, and $J_{ik}$ are the exchange interaction in FM,
AFM, and through FM-AFM interface correspondingly. $H_{ext}$ denotes external
magnetic field. We take into account DMI interaction $D_{ij}^{\mathrm{FM}}$
and magnetocrystalline anisotropy $K^{\mathrm{AFM}}$ for AFM. The anisotropy
term is included mainly to ensure that the AFM layers stay pinned during the
simulations, and is thus larger than what is normally expected in real
systems.

In these Langevin-type simulations we employ stochastic fields,
$\bm{B}_{i}^{\mathrm{fl}}$ as white noise with properties $\langle
B_{i,\mu}^{\mathrm{fl}}(t)B_{j,\nu}^{\mathrm{fl}}(t^{\prime})\rangle=2D_{M}\delta_{ij}\delta_{\mu\nu}\delta(t-t^{\prime})$.
In our simulations, we use $D_{M}=\alpha k_{B}T/\gamma m$, $D_{L}=\nu Mk_{B}T$
, where $T$ and $k_{B}$ are temperature and Boltzmann constant respectively
(for details see e.g. Ref. [26]).

The atomistic spin dynamics simulations are complemented by Monte Carlo (MC)
calculations on the spin Hamiltonian. Since both LLG and MC simulations are
based on the same Hamiltonian, they exhibit the same energetics and give the
same ground state. However, the ground state search for the simulated systems
is often complicated by the large energy barriers present in the systems due
to the inclusion of strong DMI and anisotropies. To improve the efficiency of
the simulations we therefore consider several different starting states for
our simulations, including random, ferromagnetic, and spin-spiral states.

In our simulations we use the spin simulation package UppASD [26, 27] and
study the FM-AFM bilayer with square lattice consisting of three atomic layers
for AFM and one atomic layer for FM. The simulation cell size vary between
models, due to the reasons given above, and it is between $120\times 120\times
4$ and $160\times 160\times 4$. The value of magnetic moment in all
simulations is the same for FM and AFM, and equals 1 $\mu_{b}$/atom. The value
of exchange interactions in FM and AFM is fixed in all simulations and the
ratio is $|J^{\mathrm{AFM}}/J^{\mathrm{FM}}|=1$. Most simulations are
performed at zero temperatures, unless stated otherwise.

The topological charge $Q_{i}^{ab}$ is calculated using the method proposed in
[28], since the model is atomistic and the expression for the continuum limit
cannot be used. In particular, each unit cube of the lattice is divided into
12 triangles, two for each face. Then the chirality enclosed by the unit cube
is given by the expression:

$Q_{\mathrm{i}}^{ab}=\sum_{i=1}^{12}\bm{m_{i}}\cdot[\bm{m_{a}}\times\bm{m_{b}}],$
(4)

where $\bm{m_{i}}\cdot[\bm{m_{a}}\times\bm{m_{b}}]$ is the chirality of the
triangle formed by three neighboring spins. Spins are taken in in counter-
clockwise direction according to [28].

##  III  Results

###  III.1 Uncompensated FM-AFM interface

We start with an uncompensated model of the AFM interface. This means that the
magnetic moments in each AFM layer are parallel to each other with
$J^{\mathrm{AFM}}_{\mathrm{inlayer}}>0$ with an antiparallel coupling between
layers, $J^{\mathrm{AFM}}_{\mathrm{interlayer}}<0$ (the so-called A-type AFM).
Therefore, the total magnetic moment of the AFM interface layer in contact to
the FM layer is non-zero and there is an effective exchange field that acts on
FM film (see Fig.1a).

(a) | (b) | (c)
---|---|---
|  |
Figure 1: The scheme of uncompensated (a), compensated (b), and partly-
uncompensated (c) AFM interface models. In all our calculations we use one FM
layer and three AFM layers.

Moreover, we assume a sizeable magnetic anisotropy $K^{\mathrm{AFM}}$ in each
AFM layer, so AFM magnetic moments are almost ”frozen” at their easy axis
positions during the reversal of an external magnetic field [16, 15]. Due to
the effective field acting on the FM layer, we observe an exchange bias
phenomenon with the shift of the magnetic hysteresis depending on the strength
of exchange interaction $J^{\mathrm{FM/AFM}}$ through the FM-AFM interface and
on the magnetic anisotropy in the AFM layer. The appearance and stabilization
of skyrmions will also depend on the strengths of this effective field, and
therefore, the exchange coupling $J^{\mathrm{FM/AFM}}$. In Fig.2 we show the
magnetic structure of the FM layer for various values of DMI, and exchange
interaction through the FM-AFM interface $J^{\mathrm{FM/AFM}}$ from
simulations without an external magnetic field. One can see that in the
absence of exchange interaction through the FM-AFM interface we observe stripe
domains in FM (Fig.2, first column). However, when the exchange interaction,
$J^{\mathrm{FM/AFM}}$, is increased, single skyrmions, then a skyrmion
lattice, appears, as can be seen in Fig.2 (for example, the second figure in
the second row). A further increase of the exchange interaction leads to a
collapse of parts of the skyrmions (Fig.2). A continued increase of the
exchange interaction, $J^{\mathrm{FM/AFM}}$, causes the effective field to be
sufficiently strong to overcome the DMI and the magnetization of FM layer will
become uniform.

Another important aspect to consider, while studying skyrmions is the impact
of the DMI on the magnetic structure of the FM layer. We should note that the
skyrmion lattice presented in Fig.2 is very sensitive to the DMI value. As can
be seen from Fig.2, the increase of DMI value in FM leads to skyrmions
observation for larger values of exchange interactions through the interface,
since a stronger effective field is required to overcome DMI.

In Fig.3 the magnetic hysteresis loops are shown for two values of interplane
exchange interactions, $J^{\mathrm{FM/AFM}}/J^{FM}$= –0.4 and –0.8. It can be
seen that in both cases the hysteresis loops are shifted and the shift is
increasing with the increase of the exchange interaction
$J^{\mathrm{FM/AFM}}$. We notice that hysteresis loops consist of several
loops, something that was also observed experimentally in Ref. [6]. At zero
fields the inclined part of the hysteresis loop corresponds to non-collinear
magnetization structure in the film.

The results in Fig.2 are obtained for very low temperatures. However, it is
known that the stability and inner structure of skyrmions depend on the
temperature, which is especially pronounced near the Curie temperature (please
see Ref.[29] and reference therein). Therefore, we have studied the impact of
temperature on the skyrmion lattice. For low temperatures the magnetic
structure remains similar to the one presented in Fig.2, however, with the
rise of the temperature the magnetic structure changes (see, for example,
Fig.4a) and skyrmions disappear for sufficiently high temperatures. The
explicit temperature dependence of the skyrmion number of the simulation cell
is presented in Fig.4c, showing that the topological magnetic order disappears
at a temperature just above 2.5 kB/JFM.

Figure 2: Ferromagnetic skyrmions in FM/AFM bilayer in zero external magnetic
field for various values of exchange interaction through the FM/AFM interface
and the strength of the DMI for interaction. Note that in the figure these two
interactions are divided by the strength of the exchange interaction of the
ferromagnetic layer. Magnetic anisotropy use din calculations is
$K^{\mathrm{AFM}}/J^{\mathrm{FM}}=-0.5$.  Figure 3: Exchange-biased magnetic
hysteresis of a FM film for $J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}=-0.4$ (blue),
and $J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}=-0.8$ (red).
---
Figure 4: Ferromagnetic skyrmions in FM-AFM bilayer in zero external magnetic
field for different temperatures (a) $Tk_{B}/J^{FM}=0.8$, (b)
$Tk_{B}/J^{FM}=1.4$. Skyrmion number [28] versus temperature (c). The value of
$J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}=-0.4$.

###  III.2 Compensated FM-AFM interface

Next we study the case of afully compensated AFM interface. The structure of
the model can be seen in Fig.1(b). In this case, the total magnetic moment of
the AFM interface atoms in contact with the FM layer is zero and therefore
there is no effective exchange field acting on the magnetic moments in the FM
layer that stems from atoms in the AMF layer, at least if the AFM moments are
frozen in perfect AFM order with vanishing fluctuations. This is expected to
not result in the formation of skyrmions. However, fluctuations of magnetic
moments will generate an effective exchange field across the AFM-FM interface,
with the possibility to induce skyrmions. To allow for this we have for the
simulations presented in this section used a magnetic anisotropy in the AFM
layers that is half of that used in the uncompensated model. We present the
magnetic structure of FM and AFM layers in zero magnetic field, for and lower
magnetic anisotropy in the AFM layer, see Fig.5. In the absence of exchange
through the FM-AFM interface, the magnetic structure of FM film is naturally
similar to uncompensated model (see Fig.2 for
$J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}=0$). It can be seen from Fig.5(a), that
there are skyrmions in the FM layer. We would like to note that no skyrmions
would be observed for high values of the magnetic anisotropy, for the same
values of other parameters. Moreover, if we study closely the magnetic
structure of the AFM layers, as displayed in Fig.5(b), an imprint of FM
magnetic structure, in this case, of skyrmions in the AFM layer can be
observed. The transfer of the magnetic structure of FM to AFM in exchange-
biased bilayers was shown experimentally in Ref.[7]. Such random topological
magnetic textures could be useful for e.g. reservoir computing [30]. Finally,
it was shown experimentally in [7], that induction of skyrmions in AFM by
transferring FM magnetic order to AFM by exchange bias, is a very promising
technique. Our findings support these experimental observations.
Antiferromagnetic skyrmions are very promising candidates for memory devices
because they are expected to move without a skyrmion Hall effect.

Figure 5: Magnetic structure of FM layer (a) and first (b) AFM layer
correspondingly for $K^{\mathrm{AFM}}/J^{\mathrm{FM}}=-0.25$ and
$|J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}|=0.6$.

###  III.3 Partly uncompensated interface. Impact of interface roughness

In this section, we study the case of a partly uncompensated interface, where
uncompensated spins appear due to, for example, interface roughness or
disorder. It was shown in Ref.[30] that skyrmions pinned by inhomogeneities
can be useful for the implementation of reservoir computing. The scheme of the
model is presented in Fig.1(c). We start with a perfectly compensated AFM
interface (checkerboard AFM pattern) and then assume that one out of four AFM
magnetic moments on the interface is uncompensated (the orange one in
Fig.1(c)). We study cases when coupling ferromagnetic
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}>0$, or antiferromagnetic
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}<0$ to the FM layer and to its neighbors
in the layer. The other AFM layers are ordered as perfect AFM. Therefore, in
this system, the EBP, and, therefore, the effective field will depend on the
ratio between coupling in an AFM interface layer, and through the FM-AFM
interface.

In particular, in Fig.6 we present the case when the coupling of uncompensated
AFM magnetic moments (orange ones in Fig.1c) through the interface is much
stronger than the coupling of compensated AFM spins (green ones in Fig.1c)
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}>J^{\mathrm{FM/AFM}}_{\mathrm{comp}}$.
At zero exchange through the interface, similarly, to compensated or
uncompensated model, we observe stripe domains in FM layer (see Fig.6(b)).
Then, with an increase of $J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}$ exchange
coupling we can observe the formation of a skyrmion lattice in the FM layer
(please see Fig.6(c)). Skyrmions become smaller with increasing exchange
interaction Fig.6(d) and then, skyrmions lattice collapse.

Figure 6: Ferromagnetic skyrmions in FM-AFM bilayer with partly uncompensated
interface in zero external magnetic field for various values of exchange
interaction $J^{\mathrm{FM/AFM}}$ through the FM-AFM interface. In particular,
(a)
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}/J^{\mathrm{FM/AFM}}_{\mathrm{comp}}=-10.5$,
(b)
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}/J^{\mathrm{FM/AFM}}_{\mathrm{comp}}=0.0$,
(c)
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}/J^{\mathrm{FM/AFM}}_{\mathrm{comp}}=30.5$,
(d)
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}/J^{\mathrm{FM/AFM}}_{\mathrm{comp}}=100.0$.

Above, we have studied the case of positively coupled through the FM-AFM
interface uncompensated AFM spins $J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}>0$,
while were $J^{\mathrm{FM/AFM}}_{\mathrm{comp}}<0$. However, one can observe
skyrmions in this model even for negative values of exchange interaction
through the interface of uncompensated AFM spins
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}<0$. For small negative values values of
$J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}$ the magnetization of FM layer is
similar to Fig.6, however, with the rise of exchange interaction value, we can
observe skyrmion lattice, as presented in Fig.6(a). Importantly, for negative
values of coupling $J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}$ skyrmions change
chirality, and one can observe a lattice of skyrmions (Fig.6(a)), with
negative chirality. With further increase of coupling number of skyrmions
decrease (Fig.6b) and then they disappear. The dependence of skyrmions number
as a function of exchange interaction $J^{\mathrm{FM/AFM}}_{\mathrm{uncomp}}$
is presented in Fig.7. One can observe, that there is the critical value of
exchange interaction allowing skyrmions formation, however, skyrmions are
observed in a wide range of exchange interactions through the interface, both
positive and negative ones. We can conclude, that with only 25$\%$ of spins
uncompensated at the AFM interface, it takes enormous couplings to saturate
magnetization in an FM film, and therefore, skyrmions exist in a much wider
range of exchange interaction values than in the case of fully uncompensated
interface considered above.

Figure 7: Skyrmion number depends on the exchange interaction ratio between
compensated and uncompensated spins in the AFM interface. The minus/plus sign
in the skyrmion number denotes skyrmion chirality.

###  III.4 Random interface. Impact of interface roughness

In the previous section, we studied a model when a perfectly compensated
interface became partly uncompensated due to, for example, disorder. In that
case, the uncompensated spins were distributed periodically at the interface.
However, in real systems with rough/disordered interfaces, these defects would
be distributed randomly.

In this section, we demonstrate the impact of interface roughness/disorder by
introducing randomly distributed magnetic moments with positive and negative
exchange interaction through the interface. The case when all of the spins
have $J^{\mathrm{FM/AFM}}>0$ (or $J^{\mathrm{FM/AFM}}<0$) would correspond to
the uncompensated interface, studied above. In turn, the situation when the
number of spins with $J^{\mathrm{FM/AFM}}>0$ and $J^{\mathrm{FM/AFM}}<0$ is
the same corresponds to a compensated interface. However, unlike for the
perfectly compensated interface above, here, spins from two AFM sublattices
are distributed randomly, and therefore, effective fields can arise and lead
to skyrmions stabilization. The other ratio between two AFM sublattices will
lead to cases close to the partly uncompensated interface.

Figure 8: Ferromagnetic skyrmions in FM-AFM bilayer for rough interface with
various percentage of spins from two AFM sublattices. In particular, (a)
10$\%$ of $J^{\mathrm{FM/AFM}}>0$, 90$\%$ of $J^{\mathrm{FM/AFM}}<0$ (b)
50$\%$ of $J^{\mathrm{FM/AFM}}>0$, 50$\%$ of$J^{\mathrm{FM/AFM}}<0$ (c) 80$\%$
of $J^{\mathrm{FM/AFM}}>0$, 20$\%$ of $J^{\mathrm{FM/AFM}}<0$ (d) 90$\%$ of
$J^{\mathrm{FM/AFM}}>0$, 10$\%$ of $J^{\mathrm{FM/AFM}}<0$. The values of
$|J^{\mathrm{FM/AFM}}/J^{\mathrm{FM}}|=0.4$ for both AFM sublattices.

The magnetic structure of a FM layer is presented in Fig.8. We study various
percentage of spins from two AFM sublattices presented at the AFM interface
layer, which corresponds to various percentage of positively of negatively
coupled spins at the interface. In Fig.8a,c,d we illustrate the cases when
most of the spins either negative or positively coupled. It can be seen from
the figure that the behavior of the system is not symmetric, concerning the
amount of positive (negative) coupling through the AFM-AFM interface. In
particular, one can observe the coexistence of labyrinth domains and skyrmions
for case Fig.8(a), however, in case Fig.8(d) we obtain stable skyrmions
lattice. It is due to the sign of DMI, and also to the exchange bias effect
itself. In Fig.8(b) we present the stabilization of skyrmions in FM-AFM
bilayer, when the amount of positively and negatively coupled spins at the
interface is the same. In all presented cases, one can observe the imprint of
FM magnetic structure in the AFM interface layer, for example, skyrmions
lattice from Fig.8(d) will be imprinted in the AFM interface layer, leading to
the similar periodic pattern, with a bigger deviation of AFM magnetic moments
in places corresponding to skyrmions positions.

##  IV Conclusions

We have performed atomistic spin dynamic simulations to study the
stabilization of magnetic skyrmions in an exchange-biased FM-AFM bilayer in a
zero external magnetic field. We have investigated the cases of compensated,
uncompensated FM-AFM interfaces, and the impact of geometrical roughness at
the interface. We have shown that, in the case of an uncompensated interface,
the effective field acting from AFM to FM film leads to the appearance of a
skyrmions and even a skyrmion lattice. However, either too weak or too strong
an exchange through the FM-AFM interface destroys skyrmions. The fully
compensated AFM interface results in the appearance of skyrmions, however,
only for low values of magnetic anisotropy in AFM. Moreover, due to the
transfer by exchange bias of the magnetic structure of a FM to an AFM, one can
observe imprint of FM skyrmions in AFM, which was also observed
experimentally. In the case of a partly uncompensated interface, where only
25$\%$ of the spins at the AFM interface are uncompensated, due to competition
between compensated and uncompensated spins, skyrmions can be observed both
for positive and negative exchange interaction through the FM-AFM interface in
the wide range of exchange values and skyrmions chirality depend of the
exchange interactions through the FM-AFM interface. We show that the interface
roughness can lead to skyrmions lattice stabilization in the FM layer for the
cases of both compensated and uncompensated interfaces. The results of this
work could be useful in reducing the energy consumption of perspective memory
devices based on skyrmions and for skyrmion-based reservoir computing
applications.

##  V Data availability

The data used to produce results presented in paper, are obtained using UppASD
software, available to download, from the UppASD web-page [27].

##  VI ACKNOWLEDGEMENTS

This work was financially supported by Olle Engkvist foundation and Knut och
Alice Wallenbergs Stiftelse, contract 2018.0060. Computations were enabled by
resources provided by the Swedish National Infrastructure for Computing (SNIC)
at NSC, partially funded by the Swedish Research Council. O.E. acknowledge
support from the Wallenberg Initiative Materials Science for Sustainability
(WISE) funded by the Knut and Alice Wallenberg Foundation (KAW), the Swedish
Research Council (VR), and STandUPP. A.B. and O.E. acknowledge support from
eSSENCE. M.P. acknowledge support from WISE-WASP initiative.

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