# Itinerant spin polaron and metallic ferromagnetism in
semiconductor moiré superlattices

Margarita Davydova  Department of Physics, Massachusetts Institute of
Technology, Cambridge, MA 02139, USA  Yang Zhang  Department of Physics,
Massachusetts Institute of Technology, Cambridge, MA 02139, USA  Liang Fu
Department of Physics, Massachusetts Institute of Technology, Cambridge, MA
02139, USA

###### Abstract

Itinerant spin polaron and metallic ferromagnetism are theoretically predicted
in the Mott insulator in semiconductor moiré superlattices doped with holes
and electrons, respectively.

Recent experiments have discovered a plethora of novel electronic phases in
transition metal dichalcogenides (TMDs) heterostructures, including Mott-
Hubbard and charge transfer insulators [1, 2, 3, 4, 5, 6], generalized Wigner
crystals [3, 7, 8, 9, 10, 11], the quantum anomalous Hall state [12] and
light-induced ferromagnetism [13]. These remarkably rich phenomena result from
strong interaction effects in narrow moiré bands, which generally appear in
TMD heterostructures with large moiré wavelengths. Take the example of
WSe2/WS2: the lattice corrugation introduced by the moiré structure produces a
periodic spatial variation of the valence band edge, which acts as a
superlattice potential for holes in WSe2 layer. At large moiré wavelength,
moiré bands are formed by electron tunneling $t$ between adjacent potential
minima, which are well described by a simple tight-binding model on an
emergent lattice. Inclusion of the Coulomb interaction between electrons leads
to a Hubbard model description. As a hallmark of Hubbard model physics, Mott
insulating states are found in angle-aligned WSe2/WS2 [3, 4] and twisted AB-
homobilayer WSe2 [6] at the filling of $n=1$ hole per moiré unit cell.

One of the fundamental features of the Hubbard model is the local moment
formation driven by the on-site repulsion $U$. The presence of local moments
in WSe2/WS2 has been observed by measuring the dependence of optical response
on the magnetic field [4]. It is found that the exciton Zeeman splitting,
which is proportional dominantly to the magnetization, saturates above a
certain field where the spins are fully polarized. The saturation value of
Zeeman splitting depends on the filling factor $n$ and reaches the maximum at
$n=1$. This is expected from the Hubbard model where the fully spin polarized
state has a total spin $S_{\rm tot}=Nn/2$ for $n\leq 1$ and $N(1-n/2)$ for
$1<n\leq 2$ electrons per unit cell, where $N$ is the total number of unit
cells in the system.

In this work, we study the charge excitations of the Mott insulator in TMD
moiré superlattices in the presence of a magnetic field. By exactly solving
the problem of the Mott insulator with one doped hole, we find that as the
magnetic field is reduced, the fully polarized state becomes unstable to the
formation of a spin polaron—a bound state of the hole and a spin-flip. The
spin polaron has a kinetic origin due to the correlated hopping of the hole
and the spin-flip on the triangular lattice. Importantly, the binding energy
of spin polaron is on the order of the hole hopping amplitude $t$ and has a
strong dependence on the center-of-mass momentum $\bm{P}$, which we determine
exactly. Our work establishes spin polaron, a heavy-mass fermion of charge
$-e$ and spin $\frac{3}{2}$, as the low-energy charge carrier in hole-doped
Mott insulator over a wide range of magnetic fields. In contrast, the charge
carrier in electron-doped Mott insulator is the doublon with charge $e$ and
spin of $\frac{1}{2}$.

The dichotomy between the charge excitations of opposite signs leads to
distinct phases in electron- and hole-doped Mott insulators. At $n=1+\delta$,
metallic (Nagaoka) ferromagnetism is favored by the kinetic motion of
doublons. At $n=1-\delta$, a strange metallic state is formed by the dilute
Fermi gas of spin polarons, which has incomplete spin polarization and a gap
to adding or removing an electron. As a direct manifestation of the electron-
hole asymmetry, we predict a discontinuous jump of the saturation field at
zero doping. We further propose compressibility measurements for detecting the
spin polaron in TMD moiré materials directly.

Hubbard model description and the Mott insulator at $n=1$.— The starting point
for our analysis of a TMD moiré heterobilayer under a magnetic field is the
Hubbard model on a triangular lattice [1]:

$\displaystyle H=-t\sum_{\langle
i,j\rangle}(c^{\dagger}_{i}c_{j}+h.c.)+U\sum_{i}n_{i\uparrow}n_{i\downarrow}+\frac{h}{2}\sum_{i}\left(n_{i\uparrow}-n_{i\downarrow}\right).$

For WSe2/WS2, a DFT calculation found $t\sim 1$ meV [2], which is much smaller
than the on-site Coulomb repulsion $U$. Note that generalization of $H$ that
includes higher moiré bands is necessary to properly describe the charge-
transfer insulator (rather than Mott-Hubbard insulator) specific to some TMD
heterobilayers including WSe2/WS2 [2]. As we discuss later, the physics of
spin polarons in the single-band Hubbard model still applies to charge-
transfer insulators at hole doping. Our main conclusion is also unaffected by
the inclusion of longer-range interactions. Thus, below we focus on the
strong-coupling regime $U\gg t$ of the canonical triangular lattice Hubbard
model.

At half-filling ($n=1$), the Mott insulator is a quantum antiferromagnet
governed by the spin-$\frac{1}{2}$ Heisenberg model on the triangular lattice:
$H_{J}=J\sum_{\langle ij\rangle}{\bm{s}}_{i}\cdot{\bm{s}}_{j}$, with
$J=4t^{2}/U$ and ${\bm{s}}$ is the spin-$\frac{1}{2}$ operator. Since $t\ll
U$, the antiferromagnetic exchange interaction $J$ in TMD moiré superlattices
is generally weak. As a result, the antiferromagnetic Mott insulator becomes
fully polarized above a small saturation field $h_{s}^{0}$ whose value is set
by $J$ (see SM): $h_{s}^{0}=\frac{9}{2}J=18\frac{t^{2}}{U},$ where the
superscript ‘0’ refers to an undoped Mott insulator. For example, using $t=1$
meV and assuming $U\sim 50$ meV for angle-aligned WSe2/WS2, $J$ is only $0.08$
meV and the corresponding saturation field is $1$ T (using the $g$-factor
$6.7$ for holes in WSe2). This value is comparable to the saturation field
measured by MCD at $n=1$ [4].

Thanks to the narrow bandwidth, the full magnetization curve of TMD moiré
materials can be measured over the entire range of filling factors $0\leq
n\leq 2$, which is not possible elsewhere. The ability to achieve full spin
polarization in doped Mott insulators opens access to rich and previously
unexplored Hubbard model physics on the triangular lattice, as we shall show
below.

Charge exitations in Mott insulator.— As a first step towards the study of
doped Mott insulators, we consider charge excitations of the Mott insulator at
$n=1$ at full spin polarization induced by a magnetic field $h>h_{s}^{0}$.
Interestingly, we find that the charge $e$ and $-e$ excitations have very
different nature, as represented in Fig. 1.

Charge $e$ excitation is simply a doublon created by adding an electron with
minority spin, which consts a minimum energy

$\displaystyle E_{d}=U-\mu-6t+\frac{h}{2},$ (2)

where $\mu$ is the chemical potential, $-6t$ comes from the kinetic energy of
the added electron at the bottom of the band at ${\bm{k}}=0$, and $h/2$ comes
from the Zeeman energy of the added minority spin.

The nature of charge $-e$ excitation depends on the magnetic field $h$. When
$h$ is sufficiently large, the lowest energy excitation is simply a hole with
a minimum energy given by

$\displaystyle E_{h}=\mu-3t+\frac{h}{2},$ (3)

where $-3t$ comes from the kinetic energy of the hole at the band maxima $\bf
k=\pm K$.

Figure 1:  The gap edges of the Mott insulator as a function of the magnetic
field $h$. The upper and lower edges of the gap are defined by the energy cost
of adding a charge $e$ and $-e$ quasiparticle, respectively. The charge $e$
quasiparticle is a $s=\frac{1}{2}$ doublon. The charge $-e$ quasiparticle
transitions from a $s=\frac{1}{2}$ hole to a $s=\frac{3}{2}$ spin polaron at
$h=h^{*}\simeq 0.42t$, resulting in a change of slope in the lower gap edge
and the charge gap at $n=1$ (inset).

However, when the magnetic field is reduced below a certain value $h^{*}$
(with $h^{*}\gg h_{s}^{0}$ for $t\gg J$), we find that the lowest energy state
of the Mott insulator with one hole is no longer fully spin polarized, but
contains one spin flip that is bound to the hole. The bound state of the hole
and the spin-flip is a spin polaron, a composite quasiparticle carrying spin
$s=\frac{3}{2}$ instead of $\frac{1}{2}$, where $s$ denotes the component of
the spin along the field direction. $h^{*}$ is the saturation field for the
Mott insulator with one hole. Viewed from a complementary perspective, $h^{*}$
is the dividing line between the domains with two types of charge $-e$
excitations in the Mott insulator: the bare hole and the spin polaron.

As we show below, the origin of the spin polaron formation is purely kinetic.
This phenomenon already occurs in the $U\rightarrow\infty$ limit, where a spin
flip can only move by exchanging its position with an adjacent hole.
Remarkably, this highly restricted kinetic process is sufficient to bind them
together on the triangular lattice, but not on square or honeycomb lattices.
The result is an itinerant spin polaron whose binding energy depends on its
center-of-mass momentum $\bm{P}$. We find that at $h<h^{*}$ in the limit
$U=\infty$, the minimum energy cost of adding a hole is

$\displaystyle
E_{sp}=E_{h}-\varepsilon_{b}(0)+h\approx\mu-3.42t+\frac{3h}{2},$ (4)

where $\varepsilon_{b}(0)\approx 0.42t$ is the binding energy of the spin
polaron at $\bm{P}=0$, and the total Zeeman energy $\frac{3h}{2}$ comes from
the $s=\frac{3}{2}$ of the spin polaron. The wavefunction and full dispersion
of the spin polaron is determined exactly.

Comparing the expressions for $E_{sp}$ and $E_{h}$, we see that the spin
polaron has lower energy than a hole at $h<h^{*}$ with $h^{*}\sim t$
($h^{*}\approx 0.42t$ at $J=0$). Thus, in a wide range of fields
$h_{s}^{0}\sim J<h<h^{*}\sim t$, spin polarons are the lowest-energy charge
carriers upon hole doping of the Mott insulator. This is the first main result
of our work. Note that our spin polaron exists on top of the field-polarized
state of the Mott insulator, which is fundamentally different from the
magnetic polaron in quantum antiferromagnets at $h=0$ [14, 15, 16, 17, 18,
19]. Since the noncollinear antiferromagnetic state on the triangular lattice
spontaneously breaks spin rotational symmetry, the magnetic polaron at $h=0$
does not have a well-defined spin quantum number, in contrast with the
$s=\frac{3}{2}$ spin polaron we find in the ferromagnetic state.

The field-induced transition of the charged excitation between hole and spin
polaron is reflected in the charge gap of the Mott insulator, defined as
$\Delta=E_{+e}+E_{-e}$:

$\Delta(h)=\left\\{\begin{matrix}\Delta_{0}+h,\quad\quad\quad h>h^{*},\\\
\Delta_{0}-0.42t+2h,\ \ h_{s}^{0}<h<h^{*},\end{matrix}\right.$ (5)

where $\Delta_{0}=U-9t$ is field independent. Due to the different spin
quantum numbers of hole and spin polaron, $\Delta(h)$ shows a change of slope
at $h^{*}$, as illustrated in Fig. 1.

Figure 2:  (a) The energy spectrum of the Hubbard model at $n=1$ doped with
one hole as a function of the center of mass momentum $\bm{P}$ along
$\Gamma-K$ direction. The spectrum was obtained by exact diagonalization under
in the magnetization sector with a single spin flip. The dispersive bound
state (red line) is found below the continuum spectrum (blue), the bottom of
which is at $E=\frac{3}{2}h-3t$ and is shown by the dashed black line. (b-d)
The real space plots of the wavefunction at momenta
$\bm{P}=\Gamma,(K-\Gamma)/2,K$.

Spin polaron.— We now demonstrate the existence of a spin polaron
analytically. A single hole or doublon with momentum $\bm{k}$ will have energy
equal to $\varepsilon_{h/d}(\bm{k})=\frac{h}{2}\pm t\gamma_{\bm{k}}$ with
$\gamma_{\bm{k}}=\sum_{n}\left(e^{i\bm{k}\cdot\bm{t}_{n}}+e^{-i\bm{k}\cdot\bm{t}_{n}}\right)$,
where $\bm{t}_{n=1,2,3}$ are the three basis vectors on triangular lattice. To
examine the possibility of formation of a bound state between a hole and a
spin flip, we consider a general ansatz:

$\ket{\psi_{h}}=\sum_{n,m}\alpha_{nm}c_{n\downarrow}S^{+}_{m}\ket{FM_{n=1}}.$
(6)

Here, $S_{m}^{+}\equiv c^{\dagger}_{m\uparrow}c_{m\downarrow}$. The vacuum
state corresponds to a fully-polarized state with single occupancy at each
site $\ket{FM_{n=1}}=\prod_{i}c_{i\downarrow}^{\dagger}\ket{0}$. The
wavefunction must necessary vanish at the origin, reflecting the fact that a
spin flip and a hole cannot coincide.

The Hubbard Hamiltonian at $U=\infty$, which forbids double occupancy, acting
on eq. (6) leads to a two-particle problem involving the hole and the spin-
flip. This problem can be separated into into center of mass and relative
motion, which in relative coordinates becomes a tight-binding model on the
triangular lattice. The process of exchanging the hole with spin flip
translates into a hopping from the origin’s nearest neighbor at coordinate
$\bm{r}=\pm\bm{t}_{n}$ to $-\bm{r}$ with an additional phase factor
$e^{i\bm{P}\cdot\bm{t}_{n}}$, where $\bm{P}$ is the center of mass momentum.
Importantly, this phase is caused by the shift of the center of mass and
therefore depends on ${\bm{P}}$.

For the center of mass momentum $\bm{P}=0$ the problem is analytically
tractable and all the details can be found in Supplemental Materials (SM). We
find that the bound state of a hole and a spin flip occurs for one of the
inversion-odd representations of the group $D_{6}$ and follows from the self-
consistency equation

$1+\sum_{\bm{q}}\frac{2t\sin\bm{q}\cdot\bm{t}_{1}\left(\sin\bm{q}\cdot\bm{t}_{1}+\sin\bm{q}\cdot\bm{t}_{2}+\sin\bm{q}\cdot\bm{t}_{3}\right)}{E-\frac{3}{2}h-t\gamma_{\bm{q}}}=0.$
(7)

This produces a bound state (spin polaron) with energy
$E_{sp}(\bm{P}=0)=E_{h}-\varepsilon_{b}^{(0)}+h$, where the binding energy is
found to be $\varepsilon_{b}^{(0)}\equiv\varepsilon_{b}(\bm{P}=0)\approx
0.42t$ and the $+h$ contribution is the energetic cost of a spin flip. The
fact that the binding energy is proportional to $t$ indicates the kinetic
origin of spin polaron formation as we discussed above. We also solve a
different two-particle problem with the spin flip held at a fixed position,
and find no bound state. This comparison further demonstrates that the
correlated hopping of the hole and spin flip is crucial for the bound state
formation, and the resulting spin polaron is necessarily an itinerant
quasiparticle.

Next, we numerically solve the tight-binding equation describing the relative
motion of the hole and the spin flip with a given center of mass momentum
using ansatz (6) in order to find the spin polaron dispersion (see SM for the
details). The spectrum for $\bm{P}$ along the $\Gamma-K$ direction obtained
from exact diagonalization on a lattice of 432 sites with periodic boundary
condition and is shown in Fig. 2(a). The dispersive bound state is found below
the band bottom ($E-\frac{3}{2}h=-3t$). We find that the mass of the bound
state is $m_{sp}\approx 13m_{h}$, where the mass of the bare hole is
$m_{h}=\frac{2}{3}\frac{1}{ta^{2}}$.

Fig. 2(b-d) shows the real-space wavefunction of the spin polaron in the
relative coordinates for several different momenta $\bm{P}$ obtained
numerically. At $\bm{P}=0$ the solution matches the analytical one. The spin
polaron is tightly bound on a lengthscale of the order of one lattice spacing
and the wavefunction realizes the one-dimensional antisymmetric irrep
$\Gamma_{3}$ of the dihedral group $D_{6}$ and vanishes exactly at the origin.
As seen both in Fig. 2(a) and (b-d), the state merges with continuum at
$\bm{P}=K$.

For a single doublon, we find, both analytically using the approach described
above and numerically (see SM), that the spin polaron does not form. It
similarly does not form on the square lattice, for either doping. To
understand this, we notice that the spin polaron wavefunction must vanish at
the origin in the relative coordinates, i.e. $\alpha_{nn}=\beta_{nn}=0$. The
single-particle spectrum of doped electron on the triangular lattice, or doped
electron/hole on a square lattice has only one band minimum. Therefore, the
low-energy states of such excitations cannot have a node at the origin that is
a necessary condition. In contrast, on the triangular lattice, the hole
dispersion has two band minima at $\pm K$ points. An antisymmetric
superposition of $\pm K$ states, shown in the inset in Fig. S3(b), lowers the
energy and satisfies the node condition, thus giving rise to a spin polaron.

In the case of charge-transfer insulator described by multi-band Hubbard
models [2], the dispersion of a hole in the fully polarized state at $n=1$,
which is determined by the noninteracting band structure, still has two band
minima at $\pm K$, which leads to the spin polaron formation. The situation
will be different for the electron doping, which we leave to a future study.

Figure 3:  The saturation field $h_{s}$ at different fillings obtained by
exact diagonalization of $t-J$ model with an odd number of doped
electrons/holes on $3\times L$ geometry with $L=9,12,15,18$ in the sector with
a single spin flip. Panel (a) represents the limit $U=\infty$ and panel (b)
corresponds to $J=0.1t$. The dashed black line shows the field $h^{*}$ for the
undoped Mott insulator at $n=1$.

Finite doping.— We now consider the case of finite doping density. The Hubbard
model on triangular lattice has been extensively studied in the absence of
magnetic field. It is known that, at $U\gg t$ and for the electron doping
($n>1$), the Nagaoka ferromagnetic state [20, 21, 22] arises due to the
kinetic energy gain of the doublons, which dominates over the weaker
antiferromagnetic exchange interaction between localized spins $J=4t^{2}/U\ll
t$. In contrast, for any amount of doped holes, the ferromagnetic state is
unstable at zero magnetic field [23, 24, 22, 25, 21], while the nature of the
true ground state is hard to determine.

Our results on charge excitations of the Mott insulator under a magnetic field
provide new insight. As we have shown, while the undoped Mott insulator is
already fully polarized at small magnetic fields above $h^{0}_{s}\sim J\propto
t^{2}/U$, the state with one hole can only achieve full polarization above a
much larger field $h^{*}\sim t\gg h_{s}^{0}$, at which the first spin flip
appears that is bound to the hole. Now consider a finite but small density
$\delta$ of holes. At high field, the fully polarized state is a dilute Fermi
gas of holes. As the field is reduced, provided that the hole density is
sufficiently low, the first spin flip to appear should also bind with one
hole. It follows from this argument that the saturation field $h_{s}$ at
finite hole density should approach $h^{*}$ as $\delta\rightarrow 0$. In
contrast, upon electron doping, Nagaoka mechanism leads to ferromagnetic
ground state in the limit $U=\infty$. Note that $h^{*}$ remains finite even
when $U=\infty$, whereas $h^{0}_{s}=0$ in this limit. Therefore, we conclude
that in the infinite-$U$ Hubbard model, the saturation field as a function of
doping shows a discontinuous jump from $h_{s}=h^{*}$ at $n=1^{+}$ to $0$ at
$n=1^{-}$.

This conclusion is supported by our calculation of the saturation field
$h_{s}$ as a function of doping, using exact diagonalization of the $t-J$
model shown Fig. 3. The calculations were performed at fixed number of holes
$N_{h}=1,3,5$ on $3$-leg ladders with periodic boundary conditions and various
lengths $L=9,12,15,18$. By comparing the energy of the state with a single
spin flip to that of a fully polarized state, we obtain a bound on the
saturation field from below. For $J=0$ (infinite-$U$ Hubbard model) shown in
Fig. 3(a), $h_{s}$ approaches $h^{*}\approx 0.66t$ in the limit $n\rightarrow
1^{-}$ and increases with hole doping. Its value agrees with $h^{*}$ we
independently obtain by calculating the binding energy of spin polaron on the
3-leg system at $L=\infty$. On the electron doping side, the ground state at
zero field is found to be fully polarized, as expected from Nagaoka
ferromagnetism.

We further consider the effect of small nonzero $J$. By first-order
perturbation theory, we find the correction to the energy of the spin polaron
bound state due to $H_{J}$: $\delta
E\approx\frac{\bra{\psi}H_{J}\ket{\psi}}{\bra{\psi}\ket{\psi}}\approx-3.94J.$
This yields the saturation field for the Mott insulator with one hole:
$h^{*}\approx\varepsilon_{b}+3.94J$. Indeed, our exact diagonalization study
finds that $h^{*}$ is larger at $J=0.1t$ (corresponding to $U/t\approx 40$)
than at $J=0$ by a value that agrees with our estimate, shown in Fig. 3(b).
The dependence of $h_{s}$ on doping is qualitatively similar to the case of
$J=0$: it changes discontinuously from $h^{*}$ to $h_{s}^{0}$ at $n=1$,
increases with hole doping and decreases with electron doping. In the presence
of the antiferromagnetic interaction $J$ between spins, metallic
ferromagnetism now onsets at finite electron doping density.

Experimental implications.— The particle-hole asymmetry of the saturation
field, especially its discontinuity at $n=1$, reflects the distinction between
the doublon and the spin polaron in doped Mott insulator on the triangular
lattice. In light of our theory, it is encouraging to note that the saturation
field in WSe2/WS2 measured at $T=1.7$ K [4] indeed decreases with doping at
$n>1$, increases with doping at $n<1$, and shows a large rapid change across
$n=1$, which we expect will sharpen into a discontinuity at $T=0$.

The presence of spin polaron can be established by the dependence of the lower
edge of the Mott gap at $n=1$ on the magnetic field, which can be obtained
from compressibility measurements. As shown in Fig. 1, our theory predicts a
linear dependence of the lower gap edge on the field with a change in the
slope by a factor of 3 at $h=h^{*}$, which shows the different spin quantum
numbers: $s=\frac{3}{2}$ for the spin polaron and $s=\frac{1}{2}$ for the bare
hole.

Finally, our theory predicts that at small hole doping, a Fermi liquid of
$s=\frac{3}{2}$ spin polarons can form in a range of magnetic fields below
$h^{*}$ and above $h_{s}^{0}$. This is a strange metallic state with heavy
fermion mass, incomplete magnetization, and a gap to adding an $s=\frac{1}{2}$
electron/hole. Its detailed study will appear shortly.

## References

  * Wu  _et al._ [2018] F. Wu, T. Lovorn, E. Tutuc, and A. H. MacDonald, Hubbard Model Physics in Transition Metal Dichalcogenide Moiré Bands, Phys. Rev. Lett. 121, 026402 (2018).
  * Zhang  _et al._ [2020] Y. Zhang, N. F. Q. Yuan, and L. Fu, Moiré quantum chemistry: Charge transfer in transition metal dichalcogenide superlattices, Phys. Rev. B 102, 201115 (2020).
  * Regan  _et al._ [2020] E. C. Regan, D. Wang, C. Jin, M. I. Bakti Utama, B. Gao, X. Wei, S. Zhao, W. Zhao, Z. Zhang, K. Yumigeta, _et al._ , Mott and generalized Wigner crystal states in WSe2/WS2 moiré superlattices, Nature 579, 359 (2020).
  * Tang  _et al._ [2020] Y. Tang, L. Li, T. Li, Y. Xu, S. Liu, K. Barmak, K. Watanabe, T. Taniguchi, A. H. MacDonald, J. Shan, _et al._ , Simulation of Hubbard model physics in WSe2/WS2 moiré superlattices, Nature 579, 353 (2020).
  * Li  _et al._ [2021a] T. Li, S. Jiang, L. Li, Y. Zhang, K. Kang, J. Zhu, K. Watanabe, T. Taniguchi, D. Chowdhury, L. Fu, _et al._ , Continuous mott transition in semiconductor moiré superlattices, Nature 597, 350 (2021a).
  * Xu  _et al._ [2022] Y. Xu, K. Kang, K. Watanabe, T. Taniguchi, K. F. Mak, and J. Shan, Tunable bilayer Hubbard model physics in twisted WSe2 (2022).
  * Xu  _et al._ [2020] Y. Xu, S. Liu, D. A. Rhodes, K. Watanabe, T. Taniguchi, J. Hone, V. Elser, K. F. Mak, and J. Shan, Correlated insulating states at fractional fillings of moiré superlattices, Nature 587, 214 (2020).
  * Zhou  _et al._ [2021] Y. Zhou, J. Sung, E. Brutschea, I. Esterlis, Y. Wang, G. Scuri, R. J. Gelly, H. Heo, T. Taniguchi, K. Watanabe, _et al._ , Bilayer Wigner crystals in a transition metal dichalcogenide heterostructure, Nature 595, 48 (2021).
  * Jin  _et al._ [2021] C. Jin, Z. Tao, T. Li, Y. Xu, Y. Tang, J. Zhu, S. Liu, K. Watanabe, T. Taniguchi, J. C. Hone, _et al._ , Stripe phases in WSe2/WS2 moiré superlattices, Nature Materials 20, 940 (2021).
  * Li  _et al._ [2021b] H. Li, S. Li, E. C. Regan, D. Wang, W. Zhao, S. Kahn, K. Yumigeta, M. Blei, T. Taniguchi, K. Watanabe, _et al._ , Imaging two-dimensional generalized Wigner crystals, Nature 597, 650 (2021b).
  * Huang  _et al._ [2021] X. Huang, T. Wang, S. Miao, C. Wang, Z. Li, Z. Lian, T. Taniguchi, K. Watanabe, S. Okamoto, D. Xiao, _et al._ , Correlated insulating states at fractional fillings of the WS2/WSe2 moiré lattice, Nature Physics 17, 715 (2021).
  * Li  _et al._ [2021c] T. Li, S. Jiang, B. Shen, Y. Zhang, L. Li, Z. Tao, T. Devakul, K. Watanabe, T. Taniguchi, L. Fu, _et al._ , Quantum anomalous Hall effect from intertwined moiré bands, Nature 600, 641 (2021c).
  * Wang  _et al._ [2022] X. Wang, C. Xiao, H. Park, J. Zhu, C. Wang, T. Taniguchi, K. Watanabe, J. Yan, D. Xiao, D. R. Gamelin, _et al._ , Light-induced ferromagnetism in moiré superlattices, Nature 604, 468 (2022).
  * Kane  _et al._ [1989] C. L. Kane, P. A. Lee, and N. Read, Motion of a single hole in a quantum antiferromagnet, Phys. Rev. B 39, 6880 (1989).
  * Lee  _et al._ [2006] P. A. Lee, N. Nagaosa, and X.-G. Wen, Doping a mott insulator: Physics of high-temperature superconductivity, Rev. Mod. Phys. 78, 17 (2006).
  * Sachdev [1989] S. Sachdev, Hole motion in a quantum néel state, Phys. Rev. B 39, 12232 (1989).
  * Schmitt-Rink  _et al._ [1988] S. Schmitt-Rink, C. M. Varma, and A. E. Ruckenstein, Spectral function of holes in a quantum antiferromagnet, Phys. Rev. Lett. 60, 2793 (1988).
  * Su and Chen [1988] W. P. Su and X. Y. Chen, Magnetic polarons in the two-dimensional Hubbard model, Phys. Rev. B 38, 8879 (1988).
  * Schrieffer  _et al._ [1988] J. R. Schrieffer, X.-G. Wen, and S.-C. Zhang, Spin-bag mechanism of high-temperature superconductivity, Phys. Rev. Lett. 60, 944 (1988).
  * Nagaoka [1966] Y. Nagaoka, Ferromagnetism in a narrow, almost half-filled $s$ band, Phys. Rev. 147, 392 (1966).
  * Hanisch  _et al._ [1995] T. Hanisch, B. Kleine, A. Ritzl, and E. Müller-Hartmann, Ferromagnetism in the Hubbard model: instability of the Nagaoka state on the triangular, honeycomb and kagome lattices, Annalen der Physik 507, 303 (1995).
  * Shastry  _et al._ [1990] B. S. Shastry, H. R. Krishnamurthy, and P. W. Anderson, Instability of the Nagaoka ferromagnetic state of the U=$\infty$ Hubbard model, Phys. Rev. B 41, 2375 (1990).
  * Tasaki [1989] H. Tasaki, Extension of Nagaoka’s theorem on the large-U Hubbard model, Phys. Rev. B 40, 9192 (1989).
  * Richmond and Rickayzen [1969] P. Richmond and G. Rickayzen, Ferromagnetism in narrow non-degenerate energy bands-a variational principle, Journal of Physics C: Solid State Physics 2, 528 (1969).
  * Haerter and Shastry [2005] J. O. Haerter and B. S. Shastry, Kinetic antiferromagnetism in the triangular lattice, Phys. Rev. Lett. 95, 087202 (2005).
  * Weinberg and Bukov [2017] P. Weinberg and M. Bukov, QuSpin: a Python package for dynamics and exact diagonalisation of quantum many body systems part I: spin chains, SciPost Physics 2, 003 (2017).

Supplemental Materials

Our starting point is the Hubbard Hamiltonian in two dimensions in magnetic
field:

$H=U\sum_{i}n_{i\uparrow}n_{i\downarrow}+\frac{h}{2}\sum_{i}\left(n_{i\uparrow}-n_{i\downarrow}\right)-t\sum_{\langle
i,j\rangle}\left(c^{\dagger}_{i}c_{j}+h.c.\right)$ (S1)

[where]. Even though we are interested in the case of triangular lattice, we
do not specify the type of the lattice until later in order to make comparison
with the case of square lattice.

##  I Half filling

At half filling, the effective Hamiltonian is
$H_{eff}=\frac{t^{2}}{U}\sum_{\langle
i,j\rangle}\left(\bm{S}_{i}\cdot\bm{S}_{j}-1\right)+\frac{h}{2}\sum_{i}S_{zi}$,
which corrseponds to the 2D Heisenberg antiferromagnet in magnetic field (note
that we defined $S_{zi}=n_{i\uparrow}-n_{i\downarrow}$). Consider large field
limit first, $h>h_{c}$, where $h_{c}$ is the saturation field. Then the
excitations are gapped and we can assume that the spin-wave density is low. We
can expand the Holstein-Primakoff transformation to all orders and collect the
terms that only live in $0,1$-boson sectors:

$\begin{split}S_{i}^{+}&=a^{\dagger}_{i}\left(1-a_{i}^{\dagger}a_{i}\right)\\\
S_{i}^{-}&=\left(1-a_{i}^{\dagger}a_{i}\right)a_{i}\end{split}$ (S2)

The commutation relation gives

$\begin{split}[S^{+}_{i},S^{-}_{i}]=S_{zi}-3(a^{\dagger}_{i})^{2}(a_{i})^{2}\end{split}$
(S3)

where $S_{zi}=n_{i\uparrow}-n_{i\downarrow}=2a_{i}^{\dagger}a_{i}-1$.

The Hamiltonian is expressed through bosonic degrees of freedom:

$H_{eff}|_{\nu=\frac{1}{2}}=\frac{t^{2}}{U}\sum_{\langle
i,j\rangle}\left(2(a_{i}^{\dagger}a_{j}-a_{i}^{\dagger}a_{i})-2(a_{i}^{\dagger}a_{i}^{\dagger}a_{i}a_{j}+a_{i}^{\dagger}a_{j}^{\dagger}a_{j}a_{j}-a_{i}^{\dagger}a_{j}^{\dagger}a_{i}a_{j})\right)+\langle
i\Leftrightarrow j\rangle+\frac{h}{2}\sum_{i}(2a_{i}^{\dagger}a_{i}-1)$ (S4)

This is the Hamiltonian that we can use to study the tranisiton from the
‘ferromagnetic’ to the ferrimagnetic state the occurs when the gap closes and
the bosons condense. The single-particle spectrum is determined from

$\begin{split}H_{1pt}|_{\nu=\frac{1}{2}}&=\frac{2t^{2}}{U}\sum_{\langle
i,j\rangle}\left(a_{i}^{\dagger}a_{j}-a_{i}^{\dagger}a_{i}\right)+\langle
i\Leftrightarrow j\rangle+\frac{h}{2}\sum_{i}(2a_{i}^{\dagger}a_{i}-1)=\\\
&=\sum_{\bm{k}}\left(\frac{2t^{2}}{U}\sum_{n=1}^{z}(e^{i\bm{k}\cdot\bm{t}_{n}}-1)+h\right)a^{\dagger}_{\bm{k}}a_{\bm{k}}+const,\end{split}$
(S5)

where $\bm{t}_{n}$ are the vectors pointing from a site to the nearest
neighbors. Plugging that in, we obtain that the gap closes at
$h_{c}=2z\frac{t^{2}}{U}$ for bipartite lattices (at $\bm{k}_{0}=(\pi,\pi)$
for square lattice) and $h_{c}=18\frac{t^{2}}{U}$ for triangular lattice ( at
$K,K^{\prime}$ points). Below the critical value of the magnetic field
$h_{c}$, the interacting bosons condense.

##  II Bound state of a spin flip and a hole in a ferromagnetic background

First, we project the Hamiltonian onto the subspace forbidding double
occupancy. To the leading order in $t$, the effective Hamiltonian becomes:

$H_{eff}=\frac{h}{2}\sum_{i}\left(n_{i\uparrow}-n_{i\downarrow}\right)-t\sum_{\langle
i,j\rangle,\sigma}\left(c^{\dagger}_{j\sigma}c_{i\sigma}P_{j,\overline{\sigma}}P_{i,\overline{\sigma}}+h.c.\right)$
(S6)

For the purposes of this section, it is easier if we rewrite it more
explicitly:

$\begin{split}H_{eff}&=H_{h}+H_{t}^{\uparrow}+H_{t}^{\downarrow}=\\\
&=\frac{h}{2}\sum_{i}\left(n_{i\uparrow}-n_{i\downarrow}\right)-t\sum_{\langle
i,j\rangle}\left(c_{j\uparrow}^{\dagger}c_{i\uparrow}(1-n_{i\downarrow})(1-n_{j\downarrow})+h.c.\right)-t\sum_{\langle
i,j\rangle}\left(c_{j\downarrow}^{\dagger}c_{i\downarrow}(1-n_{i\uparrow})(1-n_{j\uparrow})+h.c.\right)\end{split}$
(S7)

Let us introduce the vacuum state:

$\ket{vac}=\prod_{i}c_{i\downarrow}^{\dagger}\ket{0}$ (S8)

Let us find the eigenstate of the Hamiltonian which has exactly one hole and
one spin-flip excitation in a half-filled ferromagnetic background. We use the
ansatz:

$\ket{\psi}=\sum_{n,m}\alpha_{nm}c_{n\downarrow}S^{+}_{m}\ket{vac}$ (S9)

Thus, $\alpha_{nm}$ is the amplitude of having a hole at position $n$ and a
spin-up at position $m$ in state $\ket{\psi}$. We must set $\alpha_{nn}=0$.

It is obvious that

$H_{h}\ket{\psi}=\frac{3}{2}h\ket{\psi}$ (S10)

After a straightforward but lengthy calculation, we find:

$H^{\uparrow}_{t}\ket{\psi}=t\sum_{\langle
i,j\rangle}\alpha_{ij}c_{j\downarrow}S^{+}_{i}+\alpha_{ji}c_{i\downarrow}S^{+}_{j}\ket{vac}$
(S11)

which has a clear physical meaning: this is the hopping that occurs only when
hole and spin-flip are neighboring.

Similarly,

$H^{\downarrow}_{t}\ket{\psi}=t\sum_{\langle
i,j\rangle}\left(c_{j\downarrow}\sum_{m}\alpha_{im}S^{+}_{m}\ket{vac}+c_{i\downarrow}\sum_{m}\alpha_{jm}S^{+}_{m}\ket{vac}\right)$
(S12)

which tells us that hole hops onto a spin-down place only if that place is not
occupied by a spin-flip, and the spin flip can be anywhere but on the
respective neighboring site.

Collecting everything, and projecting onto one of the orthogonal states in the
suprtposition, we find that the relation for the eigenstate is

$\frac{3}{2}h\alpha_{\ell
s}+t\sum_{n}\alpha_{s\ell}(\delta(\ell-s-\bm{t}_{n})+\delta(\ell-s+\bm{t}_{n}))+t\sum_{n}(\alpha_{\ell+\bm{t}_{n},s}+\alpha_{\ell-\bm{t}_{n},s})=E\alpha_{\ell
s}$ (S13)

Next, we separate the coordinates into the one of center of mass and the
relative one: $\alpha_{nm}=\alpha(\bm{R}+\bm{r},\bm{R})$, where
$\bm{R}=\bm{r}_{m}$ and $\bm{r}=\bm{r}_{n}-\bm{r}_{m}$.

This allows us to rewrite the last equation as

$\frac{3}{2}h\alpha(\bm{R}+\bm{r},\bm{R})+t\alpha(\bm{R},\bm{R}+\bm{r})\sum_{n}(\delta(\bm{r}-\bm{t}_{n})+\delta(\bm{r}+\bm{t}_{n}))+t\sum_{n}(\alpha(\bm{R}+\bm{r}+\bm{t}_{n},\bm{R})+\alpha(\bm{R}+\bm{r}-\bm{t}_{n},\bm{R}))=E\alpha(\bm{R}+\bm{r},\bm{R})$
(S14)

We use
$\alpha(\bm{R}+\bm{r},\bm{R})=\sum_{\bm{P}}\psi_{\bm{P}}(\bm{r})e^{i\bm{P}\cdot\bm{R}}$
in order to find solution with specific center of mass momentum:

$\frac{3}{2}h\psi_{\bm{P}}(\bm{r})+t\sum_{n}\psi_{\bm{P}}(-\bm{r})\left(e^{i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}-\bm{t}_{n})+e^{-i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}+\bm{t}_{n})\right)+t\sum_{n}\left(\psi_{\bm{P}}(\bm{r}+\bm{t}_{n})+\psi_{\bm{P}}(\bm{r}-\bm{t}_{n})\right)=E\psi_{\bm{P}}(\bm{r})$
(S15)

Which is more conveniently grouped as

$\left(E-\frac{3}{2}h\right)\psi_{\bm{P}}(\bm{r})-t\sum_{\pm
t_{n}}\psi_{\bm{P}}(\bm{r}+\bm{t}_{n})=t\sum_{\pm
t_{n}}\psi_{\bm{P}}(-\bm{r})e^{i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}-\bm{t}_{n})$
(S16)

And is solved together with the condition $\psi_{\bm{P}}(0)=0$. Let us perform
another Fourier transform
$\psi_{\bm{P}}(\bm{r})=\sum_{\bm{k}}\varphi_{\bm{P}}(\bm{k})e^{i\bm{k}\cdot\bm{r}}$,
which leads to

$\left(E-\frac{3}{2}h-t\gamma_{\bm{k}}\right)\varphi_{\bm{P}}(\bm{k})=\lambda_{\bm{P}}(\bm{k})$
(S17)

where
$\gamma_{\bm{k}}=\sum_{n}e^{i\bm{k}\cdot\bm{t}_{n}}+e^{-i\bm{k}\cdot\bm{t}_{n}}$
(where $\bm{t}_{n}$ are the two basis vectors of the square lattice or the
three for triangular one), and

$\lambda_{\bm{P}}(\bm{k})=t\sum_{\pm\bm{t}_{n}}\psi_{\bm{P}}(\bm{t}_{n})e^{i(\bm{k}-\bm{P})\cdot\bm{t}_{n}}$
(S18)

From here, there are two options:

(a) Either the condition

$\lambda_{\bm{P}}(\bm{k})=t\sum_{\pm\bm{t}_{n}}\psi_{\bm{P}}(\bm{t}_{n})e^{i(\bm{k}-\bm{P})\cdot\bm{t}_{n}}=0$
(S19)

is true. In this case,

$E(\bm{k})=\frac{3}{2}h+t\gamma_{\bm{k}}$ (S20)

And $\varphi_{\bm{P}}(\bm{k})$ are only restrained by the condition (S19) and
$\psi_{\bm{P}}(0)=0$. This corresponds to an independent motion of hole and
the spin wave, which are in this case not bound;

or (b):

$\lambda_{\bm{P}}(\bm{k})=t\sum_{\pm\bm{t}_{n}}\psi_{\bm{P}}(\bm{t}_{n})e^{i(\bm{k}-\bm{P})\cdot\bm{t}_{n}}\neq
0$ (S21)

In this case, the eigenstates states can be found as

$\varphi_{\bm{P}}(\bm{k})=\frac{\lambda_{\bm{P}}(\bm{k})}{E(\bm{P})-\frac{3}{2}h-t\gamma_{\bm{k}}}$
(S22)

and the energy $E=E(\bm{P})$ and is found from the self-consistent condition.
Let us derive the condition by plugging
$\psi_{\bm{P}}(-\bm{t}_{n})=\sum_{\bm{q}}\varphi_{\bm{P}}(\bm{q})e^{-i\bm{q}\cdot\bm{t}_{n}}$
in the following expression

$\sum_{\pm\bm{t}_{m}}\psi_{\bm{P}}(-\bm{t}_{m})e^{i(\bm{P}-\bm{k})\cdot\bm{t}}=\sum_{\pm\bm{t}_{m}}\sum_{\bm{q}}\varphi_{\bm{P}}(\bm{q})e^{i(\bm{P}-\bm{k}-\bm{q})\cdot\bm{t}_{m}}=\sum_{\pm\bm{t}_{m}}\sum_{\bm{q}}e^{i(\bm{P}-\bm{k}-\bm{q})\cdot\bm{t}_{m}}\frac{t\sum_{\pm\bm{t}_{n}}\psi_{\bm{P}}(-\bm{t}_{n})e^{i(\bm{P}-\bm{q})\cdot\bm{t}_{n}}}{E(\bm{P})-\frac{3}{2}h-t\gamma_{\bm{q}}}$
(S23)

where in the second equality, we used eq. (S22). Let us define
$\widetilde{\epsilon}_{\bm{P}}=\frac{E(\bm{P})-\frac{3}{2}h}{t}$, which brings
us to:

$\lambda_{\bm{P}}(\bm{k})=\sum_{\bm{q}}\frac{\gamma_{\bm{k}+\bm{q}-\bm{P}}\lambda_{\bm{P}}(\bm{q})}{\widetilde{\epsilon}_{\bm{P}}-\gamma_{\bm{q}}}$
(S24)

This integral equation can be rewritten as

$\sum_{\bm{q}}\lambda_{\bm{P}}(\bm{q})\left(\delta(\bm{q}-\bm{k})-\frac{\gamma_{\bm{k}+\bm{q}-\bm{P}}}{\widetilde{\epsilon}_{\bm{P}}-\gamma_{\bm{q}}}\right)=0$
(S25)

which only has nonzero solutions if the self-consistency condition

$\det\left(\delta(\bm{q}-\bm{k})-\frac{\gamma_{\bm{k}+\bm{q}-\bm{P}}}{\widetilde{\epsilon}_{\bm{P}}-\gamma_{\bm{q}}}\right)=0$
(S26)

is satisfied.

###  II.1 Square lattice

Before treating specific cases, let us review the symmetries of the solution
and the constraints imposed on it. We note that eq. (S16) has a symmetry under
simultaneous change $(\bm{r},\bm{P})\rightarrow(-\bm{r},-\bm{P})$ (global
inversion symmetry). Another condition $\alpha(R,R)=0$ translates into
$\psi_{\bm{P}}(0)=0$. Thus, we can search for solutions that are globally
inversion symmetric and asymmetric, i.e.
$\psi^{\pm}_{\bm{P}}(\bm{r})=\psi_{\bm{P}}(\bm{r})\pm\psi_{-\bm{P}}(-\bm{r})$,
respectively, such that
$\psi^{\pm}_{-\bm{P}}(-\bm{r})=\pm\psi^{\pm}_{\bm{P}}(\bm{r})$.

In Fourier space, similarly,
$\varphi^{\pm}_{-\bm{P}}(-\bm{k})=\pm\varphi^{\pm}_{\bm{P}}(\bm{k})$. This
especially simplifies at $\bm{P}=0$ and at $\bm{P}=(\pi,\pi)$, where the
function becomes even or odd of its argument.

####  II.1.1 $\bm{P}=0$

We search for odd solutions, and note that there is only one irrep of $C_{4v}$
group that is odd under in-plane inversion, which is two-dimensional. Denoting
$\psi_{1}=\psi(\bm{t}_{1})$ and $\psi_{2}=\psi(\bm{t}_{2})$, we can rewrite
(S24) as

$\psi_{1}\sin k_{x}+\psi_{2}\sin
k_{y}=\sum_{\bm{q}}\frac{\gamma_{\bm{k}+\bm{q}}\left(\psi_{1}\sin
q_{x}+\psi_{2}\sin
q_{y}\right)}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}=\sum_{\bm{q}}\frac{2t\left(\cos(k_{x}+q_{x})+\cos(k_{y}+q_{y})\right)\left(\psi_{1}\sin
q_{x}+\psi_{2}\sin
q_{y}\right)}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}$ (S27)

where we used $\gamma_{\bm{q}}=2t(\cos q_{x}+\cos q_{y})$. We expand the
cosines of sums into products and some of the integrals are trivially zero due
to symmetry constraints, and therefore we obtain

$\left(\psi_{1}\sin k_{x}+\psi_{2}\sin
k_{y}\right)\left(1+\sum_{\bm{q}}\frac{2t\sin^{2}q_{x}}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}\right)=0$
(S28)

where we used the symmetry of the integral under $x\Leftrightarrow y$. The
term in the second brackets present self-consistency condition; the integrand
is negative and maximum near BZ corners ($\bm{q}=(\pi,\pi)$), where its value
is finite because the numerator vanishes quadratically. We find that the self-
consistency equation is not satisfied for any $\epsilon_{\bm{0}}$ below the
band bottom (the integrals achieves its minimum value $\approx-0.36$ exactly
at the band bottom energy), and therefore, we conclude that on a square
lattice, there is no bound state at $\bm{P}=0$.

###  II.2 Triangular lattice

As before, we consider the solutions with
$\varphi^{\pm}_{-\bm{P}}(-\bm{k})=\pm\varphi^{\pm}_{\bm{P}}(\bm{k})$. This
especially simplifies at $\bm{P}=0$ the wavefunction becomes even or odd of
its argument.

####  II.2.1 $\bm{P}=0$

We search for odd solutions; there are two irreps (one-dimensional
$\Gamma_{3}$ and two-dimensional $\Gamma_{6}$) of $C_{3v}$ group that is odd
under in-plane inversion, which yield non-trivial function $\lambda(\bm{k})$.

For $\Gamma_{3}$, we can take
$\lambda(\bm{k})=\psi_{1}\left(\sin\bm{k}\cdot\bm{t}_{1}+\sin\bm{k}\cdot\bm{t}_{2}+\sin\bm{k}\cdot\bm{t}_{3}\right)$,
where $\bm{t}_{1}=(1,0)$,
$\bm{t}_{2}=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$, and
$\bm{t}_{3}=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$. Then eq. (S24)
turns into the self-consistency equation

$\sin\bm{k}\cdot\bm{t}_{1}+\sin\bm{k}\cdot\bm{t}_{2}+\sin\bm{k}\cdot\bm{t}_{3}=\sum_{\bm{q}}\frac{2t(\cos(\bm{k}+\bm{q})\cdot\bm{t}_{1}+\cos(\bm{k}+\bm{q})\cdot\bm{t}_{2}+\cos(\bm{k}+\bm{q})\cdot\bm{t}_{3})\left(\sin\bm{q}\cdot\bm{t}_{1}+\sin\bm{q}\cdot\bm{t}_{2}+\sin\bm{q}\cdot\bm{t}_{3}\right)}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}$
(S29)

where we used
$\gamma_{\bm{q}}=2t(\cos\bm{q}\cdot\bm{t}_{1}+\cos\bm{q}\cdot\bm{t}_{2}+\cos\bm{q}\cdot\bm{t}_{3})$.
We notice that the terms in the numerator that lead to nonvanishing integrals
are
$-\sin\bm{k}\cdot\bm{t}_{1}\sin\bm{q}\cdot\bm{t}_{1}\left(\sin\bm{q}\cdot\bm{t}_{1}+\sin\bm{q}\cdot\bm{t}_{2}+\sin\bm{q}\cdot\bm{t}_{3}\right)$
together with its cyclic permutation. Therefore, the self-consistency
condition can be simplified down to

$1+\sum_{\bm{q}}\frac{2t\sin\bm{q}\cdot\bm{t}_{1}\left(\sin\bm{q}\cdot\bm{t}_{1}+\sin\bm{q}\cdot\bm{t}_{2}+\sin\bm{q}\cdot\bm{t}_{3}\right)}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}=0$
(S30)

As the value of $\epsilon_{\bm{0}}$ approaches the band bottom ($-3t$), the
integrand diverges near the band minima, $K$ and $K^{\prime}$. Therefore, the
self-consistency condition can be easily saturated, which we find numerically
to occur at $\epsilon_{\bm{0}}\approx-3.55t$. Thus, we have found _a bound
state between a hole and a spin flip on a triangular lattice_ which we call
spin polaron, with binding energy $E_{b}=0.42t$, which is, remarkably,
commensurate with $t$.

For two-dimensional representation $\Gamma_{6}$, the general form of
$\lambda_{\bm{0}}(\bm{k})$ is
$\widetilde{\psi}_{1}\left(2\sin\bm{k}\cdot\bm{t}_{1}-\sin\bm{k}\cdot\bm{t}_{2}-\sin\bm{k}\cdot\bm{t}_{3}\right)+\widetilde{\psi}_{2}\left(\sin\bm{k}\cdot\bm{t}_{2}-\sin\bm{k}\cdot\bm{t}_{3}\right)$
for some independent constants $\widetilde{\psi}_{1,2}$ to be determined self-
consistently. We plug this ansantz in, perform similar simplifications as
before, and find the following self-consistency equation:

$1+\sum_{\bm{q}}\frac{2t\sin\bm{q}\cdot\bm{t}_{2}\left(\sin\bm{q}\cdot\bm{t}_{2}-\sin\bm{q}\cdot\bm{t}_{3}\right)}{\widetilde{\epsilon}_{\bm{0}}-\gamma_{\bm{q}}}=0$
(S31)

The integration-even part of the integrand is finite at $K,K^{\prime}$ points
because the numerator vanishes there, and the integral is bound from below by
approx. $-0.5$. Thus, as we find, there is no bound state solution. Thus, the
only solution that we find is for $\Gamma_{3}$ representation.

Figure S1:  (a) The $\bm{k}$-space and (b) the real-space plots of the
wavefunction (S22) for the bound state found from (S31). This bound state is
at $\bm{P}=0$, realizes the symmetry of $\Gamma_{3}$ irrep of $D_{6}$ group
and has a large binding energy $E_{b}\approx 0.42t$, which explains its small
localization length.  Figure S2:  Absolute value of the wavefunction from the
previous plot along $x$-axis, illustrating its quick decay on the lengthscale
of approximately one lattice spacing $\xi_{loc}\sim a$.

####  II.2.2 Dispersion $E(\bm{P})$

In order to fins the dispersion of the bound state, we solve the problem
numerically. For this, we recast the equation on the wavefunction:

$\sum_{n}\psi_{\bm{P}}(-\bm{r})\left(e^{i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}-\bm{t}_{n})+e^{-i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}+\bm{t}_{n})\right)+\sum_{n}\left(\psi_{\bm{P}}(\bm{r}+\bm{t}_{n})+\psi_{\bm{P}}(\bm{r}-\bm{t}_{n})\right)=\widetilde{\epsilon}_{\bm{P}}\psi_{\bm{P}}(\bm{r})$
(S32)

$\sum_{\bm{r}^{\prime}}\sum_{n}\left[\left(e^{i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}-\bm{t}_{n})+e^{-i\bm{P}\cdot\bm{t}_{n}}\delta(\bm{r}+\bm{t}_{n})\right)\delta(\bm{r}^{\prime}+\bm{r})+\delta(\bm{r}^{\prime}-\bm{r}+\bm{t}_{n})+\delta(\bm{r}^{\prime}-\bm{r}-\bm{t}_{n})\right]\psi_{\bm{P}}(\bm{r}^{\prime})=\widetilde{\epsilon}_{\bm{P}}\psi_{\bm{P}}(\bm{r})$
(S33)

We solve this equation numerically for a hexagon-shaped lattice with diameter
up to 25 sites and periodic boundary condition to find the dispersion
$E(\bm{P})$. We also check numerically that for electron doping for triangular
lattice, there is no bound state, and check that there is no bound state in
the case of square lattice.

The results of the numerical tight-binding calculations are shown in Fig. S3.
We find that the spin polaron has a mass $m_{p}\approx 13m_{h}$ near the
$\Gamma$-point, where $m_{h}=\frac{2}{3ta^{2}}$ is the mass of a hole.

Figure S3:  Color map of the bound state dispersion
$\widetilde{\epsilon}_{\bm{P}}$ in the Brillouin zone for $\bm{P}$.

##  III Bound state of a spin-flip and an electron in a ferromagnetic
background

The square lattice is bipartite, and therefore it has particle-hole symmetry
upon which $c_{A}\leftrightarrow c^{\dagger}_{A}$ and
$c_{B}\leftrightarrow-c^{\dagger}_{B}$ (where $A$ and $B$ are the two
sublattices), therefore the conclusions obtained for $\nu<\frac{1}{2}$ filling
can be applied for electron doping as well.

On triangular lattice, the attempt to perform particle-hole transoformation
will change the dispersion by taking $t\rightarrow-t$, which is important for
the bound state formation. Therefore, let us analytically consider the case of
$\bm{P}=0$ and electron doping on triangular lattice.

Let us similarly try to find the eigenstate of the Hamiltonian which has
exactly one electron and one spin-flip excitation in a half-filled
ferromagnetic background:

$\ket{\psi_{e}}=\sum_{n,m}\alpha_{nm}c_{n\uparrow}^{\dagger}S^{+}_{m}\ket{vac}$
(S34)

The equation determining the wavefunction is the same as eq. (S17), apart from
$t\rightarrow-t$ change. We use similar symmetry considerations and for
$\Gamma_{3}$ and $\Gamma_{6}$ irreps we obtain, respectively:

$1+\sum_{\bm{q}}\frac{2t\sin\bm{q}\cdot\bm{t}_{1}\left(\sin\bm{q}\cdot\bm{t}_{1}+\sin\bm{q}\cdot\bm{t}_{2}+\sin\bm{q}\cdot\bm{t}_{3}\right)}{\widetilde{\epsilon}_{\bm{0}}+\gamma_{\bm{q}}}=0$
(S35)

and

$1+\sum_{\bm{q}}\frac{2t\sin\bm{q}\cdot\bm{t}_{2}\left(\sin\bm{q}\cdot\bm{t}_{2}-\sin\bm{q}\cdot\bm{t}_{3}\right)}{\widetilde{\epsilon}_{\bm{0}}+\gamma_{\bm{q}}}=0$
(S36)

For both cases, we find that near the band bottom $(0,0)$ the integrand is
finite and the integral is not large enough for any value of
$\epsilon_{\bm{0}}$ in order to satisfy the self-consistency equation. Thus,
we, interestingly, find that due to the particle-hole asymmetry, the spin
polaron exists only for hole doping.

##  IV Metallic magnetism at finite doping

At finite doping density, we study the metallic magnetism and critical
magnetic field for fully polarized state in the infinite-U limit for
triangular lattice. Without double occupancy, the infinite-U Hubbard model is
reduced to the $t-J$ model with $J=0$ as shown in Eq. S7. We further consider
the case of finite $J=0.1t$ as following:

$\hat{H}=-t\sum_{\langle
ij\rangle,\sigma}\left(c_{i\sigma}^{\dagger}c_{j\sigma}+\text{ h.c.
}\right)+J\sum_{\langle
ij\rangle}\left(\vec{S}_{i}\cdot\vec{S}_{j}-\frac{n_{i}n_{j}}{4}\right)$ (S37)

Exact diagonalization is employed to calculate the magnetic ground state with
finite doping density up to system size $3\times 18$ for doping density less
than $1/3$. To reduce the dimension of Hilbert space, the full Hamiltonian is
divided into momentum and spin sectors by translation symmetry and spin
conserving. Comparing the energy difference between different spin sectors, we
can extract the critical magnetic field for fully polarized state. For the
case of one hole doping and one spin flip, the simulated system size goes up
to $24\times 24$. We cross check the results with QuSpin program, especially
for infinite U limit [26].

Figure S4:  The saturation field $h_{s}$ at different filling densities $n$
calculated by exact diagonalization on 3$\times$6, 3$\times$9, 3$\times$12,
3$\times$15, 3$\times$18 geometries in the limit $U=\infty$. For all
calculations, we only present the data points for odd number of electrons or
holes.

At the hole doping side, the ground state is antiferromagnetic, we only
consider the case of $0,1,2$ spin slip for the extrapolation of saturation
field. For $(L_{x},L_{y})=(3,6)$ and $(3,9)$ with two and three holes, the two
spin flip state is lower in energy than one spin flip when decreasing the
magnetic field below $h_{s}$. In Fig. 3, we plot the critical magnetic field
vs. doping density for system size
$(L_{x},L_{y})=(3,6),(3,9),(3,12),(3,15),(3,18)$ with odd number of electrons
and holes from n=1. It is found that critical field for fully polarized state
is increasing with the hole density up to $1/3$ doping. For system size
$(18,18)$, the saturation field for one hole doping is $0.423t$, approaching
the analytical value $h^{*}=0.42t$.


# The general position achievement game played on graphs

Sandi Klavžar $\ {}^{a,b,c}$ Email<EMAIL_ADDRESS>Neethu P. K.
$\ {}^{d}$ Email<EMAIL_ADDRESS>Ullas Chandran S. V. $\ {}^{d}$
Email<EMAIL_ADDRESS>
###### Abstract

A general position set of a graph $G$ is a set of vertices $S$ in $G$ such
that no three vertices from $S$ lie on a common shortest path. In this paper
we introduce and study the general position achievement game. The game is
played on a graph $G$ by players A and B who alternatively pick vertices of
$G$. A selection of a vertex is legal if has not been selected before and the
set of vertices selected so far forms a general position set of $G$. The
player who selects the last vertex wins the game. Playable vertices at each
step of the game are described, and sufficient conditions for each of the
players to win is given. The game is studied on Cartesian and lexicographic
products. Among other results it is proved that A wins the game on
$K_{n}\,\square\,K_{m}$ if and only if both $n$ and $m$ are odd, and that B
wins the game on $G\circ K_{n}$ if and only if either B wins on $G$ or $n$ is
even.

a Faculty of Mathematics and Physics, University of Ljubljana, Slovenia

b Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia

c Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia

d Department of Mathematics, Mahatma Gandhi College, University of Kerala,
Thiruvananthapuram-695004, Kerala, India

Key words: general position set; achievement game; Cartesian product of
graphs; lexicographic product of graphs

AMS Subj. Class: 05C12; 05C69

##  1 Introduction

The general position problem for graphs was independently introduced and
researched in [17, 27], but should be noted that in the case of hypercubes, it
has been studied much earlier by Körner [15]. Among motives for introducing
the problem is the more than a century old no-three-in-line problem of Dudeney
[5], see also [16, 19, 23]. For the related general position subset selection
problem in computational geometry see [6, 22].

A general position set of a graph $G=(V(G),E(G))$ is a set of vertices
$S\subseteq V(G)$ such that no three vertices from $S$ lie on a common
shortest path of $G$. The general position problem asks for the largest
possible size of a general position set of $G$; this number is denoted by
$\operatorname{gp}(G)$. Immediately after its introduction, the concept
received a great response [2, 7, 11, 12, 13, 18, 21, 24, 25, 26, 28].
Furthermore, in [14] general position sets have been extended to general
$d$-position sets, while in [10] the Steiner general position problem was
studied.

In this paper we study the achievement game associated with general position
sets. Achievement games have already been studied in different contexts. For
instance, in a finite group two players in turn select previously unselected
elements of the group, and the player who is the first to achieve a generating
set from the jointly selected elements wins the game [1, 3]. Similarly, and
closer to our game, two players in turn select vertices of a finite graph, and
the player who first plays such a vertex that the union of the intervals
between the vertices played contains all the vertices wins the game [4, 9,
20].

Let $G$ be a graph. Then the general position achievement game (gp achievement
game for short) is played by two players, player A and player B. The first
player A chooses a vertex $v_{1}$. The second player B then chooses a vertex
$v_{2}\neq v_{1}$. Next A picks a vertex $v_{3}\in
V(G)\setminus\\{v_{1},v_{2}\\}$ such that the set $\\{v_{1},v_{2},v_{3}\\}$ is
a general position set in $G$. The game then proceeds along the same way and
ends when there is no more vertex to be played, that is, there exists no
vertex such that the general position set consisting of the already played
vertices could be enlarged. The player who has played the last vertex wins the
game.

We proceed as follows. In the rest of this section, additional definitions and
notation needed are recalled. In the next section we give some general results
and provide several examples. Among other results we observe that A wins the
gp achievement game on a bipartite graph $G$ if and only if the number of
isolated vertices in $G$ is odd. In Section 3 we study the game on Cartesian
products, while in Section 4 we prove that B wins the game on the
lexicographic product $G\circ K_{n}$ if and only if either B wins on $G$ or
$n$ is even. At the end several concluding remarks are given, among which a
closely related avoidance game is commented.

All graphs considered are finite, simple, and without loops or multiple edges.
The distance $d_{G}(u,v)$ between vertices $u$ and $v$ of $G$ is the length of
a shortest $u,v$-path. A $u,v$-path of minimum length is also called an
$u,v$-geodesic. The interval $I_{G}[u,v]$ between $u$ and $v$ is the set of
vertices that lie on some $u,v$-geodesic of $G$. For $S\subseteq V(G)$, we set
$I_{G}[S]=\bigcup_{{}_{u,v\in S}}I_{G}[u,v]$. A subgraph $H$ of a graph $G$ is
convex if for every $u,v\in V(G)$, every $u,v$-geodesic in $G$ lies completely
in $H$.

##  2 Some general results and examples

The sequence of vertices played in the achievement game on a graph $G$ will be
denoted by $a_{1},b_{1},a_{2},b_{2},\ldots$, that is, the vertices played by A
are $a_{1},a_{2},\ldots$, and the vertices played by B are
$b_{1},b_{2},\ldots$ For instance, we may say that A starts the game by
playing $a_{1}=x$, where $x\in V(G)$. Suppose that $x_{1},\ldots,x_{j}$ are
vertices played so far on the graph $G$. Then we say that $y\in V(G)$ is a
playable vertex if $y\notin\\{x_{1},\ldots,x_{j}\\}$ and
$\\{x_{1},\ldots,x_{j}\\}\cup\\{y\\}$ is a general position set of $G$. Let
$\operatorname{Pl}_{G}(x_{1},\ldots,x_{j})$ be the set of all playable
vertices after the vertices $x_{1},\ldots,x_{j}$ have already been played; we
may sometimes simplify the notation
$\operatorname{Pl}_{G}(x_{1},\ldots,x_{j})$ to $\operatorname{Pl}_{G}(\ldots
x_{j})$. For instance, if $x$ and $y$ are arbitrary vertices of a path $P$,
then $\operatorname{Pl}_{P}(x)=V(P)\setminus\\{x\\}$ and
$\operatorname{Pl}_{P}(x,y)=\emptyset$. Denoting by $S$ the set of vertices
$\\{x_{1},\ldots,x_{j}\\}$ played so far, we may also write
$\operatorname{Pl}_{G}(S)$ for $\operatorname{Pl}_{G}(x_{1},\ldots,x_{j})$. In
the sequel we will implicitly but frequently use the following description of
playable vertices.

###### Lemma 2.1

Let $S$ be the sequence of played vertices so far in a gp achievement game on
a graph $G$. Then $x\in\operatorname{Pl}_{G}(S)$ if and only if the following
two conditions hold:

  1. (i)

if $u,v\in S$, then $x\notin I[u,v]$, and

  2. (ii)

if $u\in S$, then $I[x,u]\cap S=\\{u\\}$.

Proof. By definition, $x\in\operatorname{Pl}_{G}(S)$ if and only if
$S\cup\\{x\\}$ is a general position set. Since $S$ is a general position set
by the assumption of the game, deciding whether $S\cup\\{x\\}$ is a general
position set reduces to checking the conditions (i) and (ii). $\square$

Let us next look at some examples. Since in a complete graph every vertex
subset is a general position set, A wins the gp achievement game on the
complete graph $K_{n}$ if and only if $n$ is odd. In the course of the gp
achievement game on a graph $G$ of order at least $2$, at least two vertices
will be played. Hence B wins the game on graphs $G$ with
$\operatorname{gp}(G)=2$. As proved in [27], the only graphs with
$\operatorname{gp}(G)=2$ are paths and $C_{4}$. On the other hand, the class
of graphs $G$ with $\operatorname{gp}(G)=3$ has not yet been characterized. If
$\operatorname{gp}(G)=3$, then gp achievement game will take either two or
three moves. In fact, if $\operatorname{gp}(G)=3$ then B wins the gp
achievement game if only only if every vertex of $G$ lies in a maximal general
position set of order $2$. Applying this observation to cycles we infer that B
wins the gp achievement game on the cycle $C_{n}$, $n\geq 3$, if and only if
$n$ is even.

The following result is simple but at the same time quite useful.

###### Theorem 2.2

Let $G$ be a graph. Then the following holds.

(i) If A has a strategy such that after the vertex $a_{k}$, $k\geq 1$, is
played, the set $\operatorname{Pl}_{G}(\ldots
a_{k})\cup\\{a_{1},b_{1},\ldots,a_{k}\\}$ is a general position set and
$|\operatorname{Pl}_{G}(\ldots a_{k})|$ is even, then A wins the gp
achievement game.

(ii) If B has a strategy such that after the vertex $b_{k}$, $k\geq 1$, is
played, the set $\operatorname{Pl}_{G}(\ldots
b_{k})\cup\\{a_{1},b_{1},\ldots,b_{k}\\}$ is a general position set and
$|\operatorname{Pl}_{G}(\ldots b_{k})|$ is even, then B wins the gp
achievement game.

Proof. (i) Suppose that A has a strategy such that after A plays $a_{k}$, the
set $\operatorname{Pl}_{G}(\ldots a_{k})\cup\\{a_{1},b_{1},\ldots,a_{k}\\}$ is
a general position set. By definition, in the rest of the game only vertices
from $\operatorname{Pl}_{G}(\ldots a_{k})$ are playable. Moreover, each of
these vertices will actually be played because $\operatorname{Pl}_{G}(\ldots
a_{k})\cup\\{a_{1},b_{1},\ldots,a_{k}\\}$ is a general position set. Just
after this will be done, the game will be finished. Since
$|\operatorname{Pl}_{G}(\ldots a_{k})|$ is assumed to be even, this means that
A will be the last player to select a vertex.

(ii) Follows by a parallel argument. $\square$

For the first application of Theorem 2.2 consider the Petersen graph $P$.
Suppose that after A plays some vertex $a_{1}$ of $P$, B plays a vertex
$b_{1}$ adjacent to $a_{1}$. Then $\operatorname{Pl}_{P}(a_{1},b_{1})$
consists of four vertices which, together with $a_{1}$ and $b_{1}$, form a
general position set of $P$. Hence Theorem 2.2(ii) implies that B wins the gp
achievement game on the Petersen graph. As another application of Theorem 2.2
we have the following result.

###### Proposition 2.3

Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$, where
$k\geq 2$ and $n_{i}\geq 2$ for $i\in[k]$. Then A wins the gp achievement game
on $G$ if and only if $k$ is odd and at least one $n_{i}$ is odd.

Proof. Suppose first that $n_{i}$ is even for all $i\in[k]$. Let $X$ be the
partition set of $G$ in which the first move $a_{1}$ has been played by A.
Then B replies by playing a vertex $b_{1}\neq a_{1}$ from $X$. Note that
$\operatorname{Pl}_{G}(a_{1},b_{1})=X\setminus\\{a_{1},b_{1}\\}$. Since $X$ is
a general position set of $G$, Theorem 2.2(ii) applies and B wins the gp
achievement game on $G$.

Hence, the only possibility for A to win the game is that at least one $n_{i}$
is odd and that the first move $a_{1}$ is from an odd partition set $X$. Now,
if B would reply by playing a vertex in $X$, then by the argument of the
previous paragraph and with Theorem 2.2(i) in hand, A would win. So it is
better for B to play a vertex $b_{1}$ which lies in a partition set $Y\neq X$.
Since $\operatorname{Pl}_{G}(a_{1},b_{1})=V(G)\setminus(X\cup Y)$, the vertex
$a_{2}$ must lie in a partition set $Z$ different from both $X$ and $Y$.
Continuing in this manner, each of the subsequent played vertices belongs to
its private partition set. In conclusion, if some $n_{i}$ is odd, then A will
win if and only if $k$ is odd. $\square$

In view of Theorem 2.2(ii) we easily infer that B wins the gp achievement game
on an arbitrary connected, bipartite graph of order at least two. This
observation generalizes to arbitrary bipartite graphs as stated in the next
theorem, for which we need the following fact that was observed for the first
time in the proof of [2, Theorem 5.1].

###### Lemma 2.4

Let $G$ be a connected, bipartite graph. If $S$ is a general position set of
$G$ with $|S|\geq 3$, then $S$ is an independent set.

###### Theorem 2.5

Let $G$ be a bipartite graph. Then A wins the gp achievement game on $G$ if
and only if the number of isolated vertices in $G$ is odd.

Proof. Let $k$ be the number of isolated vertices of $G$.

First suppose that $k\geq 0$ is even and consider the following strategy of B.
Whenever A selects a vertex $v$ in some component of $G$ of order at least
$2$, B replies with a move on a neighbor of $v$. And whenever A plays an
isolated vertex, B replies by playing another isolated vertex. Note that after
two adjacent vertices of a component $H$ of $G$ are played, Lemma 2.4 implies
that no additional vertex from $H$ will be played in the rest of the game.
Moreover, since $k$ is even, whenever A plays an isolated vertex, there exists
at least one isolated vertex which was not played yet, hence B can follow the
described strategy. It follows that the game will finish when all the isolated
vertices and precisely two (adjacent) vertices from each component will be
played. So the number of played vertices will be even, hence B wins the gp
achievement game.

Second, let $k\geq 1$ be odd. Then A has the following strategy to win the gp
achievement game. The first vertex played will be an isolated vertex. After
that, the strategy of A is just as the described strategy of B in the above
paragraph: whenever B plays a vertex $v$ in some component of $G$ of order at
least $2$, A replies with a move on a neighbor of $v$, and if B plays an
isolated vertex, A replies by playing another isolated vertex. Using parallel
arguments as above, the total number of vertices played will be odd, which in
turn implies that A wins the game. $\square$

##  3 The game played on Cartesian products

The Cartesian product $G\,\square\,H$ of graphs $G$ and $H$ has the vertex set
$V(G)\times V(H)$, the vertices $(g_{1},h_{1}),(g_{2},h_{2})$ being adjacent
in $G\,\square\,H$ if either $g_{1}g_{2}\in E(G)$ and $h_{1}=h_{2}$, or
$g_{1}=g_{2}$ and $h_{1}h_{2}\in E(H)$. If $g\in V(G)$, then the subgraph of
$G\,\square\,H$ induced by the vertex set $\\{(g,h)$ $|$ $h\in v(H)\\}$ is an
$H$-layer ${}^{g}H$. $G$-layers $G^{h}$ are defined analogously. If
$S\subseteq V(G\,\square\,H)$, then the projection $\pi_{G}(S)$ of $S$ on $G$
is the set $\\{g\in V(G):\ (g,h)\in\ {\rm for\ some}\ h\in V(H)\\}$. The
projection $\pi_{H}(S)$ of $S$ on $H$ is defined analogously.

Throughout this section we will use the following basic fact about the
distance function in the Cartesian product. If $G$ and $H$ are connected
graphs and $(g,h),(g^{\prime},h^{\prime})\in V(G\,\square\,H)$, then the
distance formula holds:

$d_{G\,\square\,H}((g,h),(g^{\prime},h^{\prime}))=d_{G}(g,g^{\prime})+d_{H}(h,h^{\prime})\,.$
(1)

Moreover, if $P$ is a $(g,h),(g^{\prime},h^{\prime})$-geodesic in
$G\,\square\,H$, then $\pi_{G}(P)$ induces a $g,g^{\prime}$-geodesic in $G$
and $\pi_{H}(P)$ induces a $h,h^{\prime}$-geodesic in $H$. The distance
formula (1) implies that

$I_{G\,\square\,H}[(g,h),(g^{\prime},h^{\prime})]=I_{G}[g,g^{\prime}]\times
I_{H}[h,h^{\prime}]\,.$ (2)

For these results and more on the Cartesian product operation see the standard
book on product graphs [8]. We will also need the following known result.

###### Lemma 3.1

[26, Lemma 2.4] Let $G$ and $H$ be connected graphs and let $R$ be a general
position set of $G\,\square\,H$. If $u=(g,h)\in R$, then $V(^{g}{H})\cap
R=\\{u\\}$ or $V(G^{h})\cap R=\\{u\\}$.

We next prove two additional lemmas on general position sets in Cartesian
products.

###### Lemma 3.2

Let $G$ and $H$ be connected graphs and let $R\subseteq V(G\,\square\,H)$ has
the following two properties.

  1. (i)

If $(g,h)\in R$, then $V(^{g}{H})\cap R=\\{(g,h)\\}$ or $V(G^{h})\cap
R=\\{(g,h)\\}$.

  2. (ii)

$\pi_{G}(R)$ and $\pi_{H}(R)$ are general position sets of $G$ and $H$,
respectively.

Then $R$ is a general position set of $G\,\square\,H$.

Proof. Suppose on the contrary that $R$ contains three vertices
$x_{1}=(u_{1},v_{1})$, $x_{2}=(u_{2},v_{2})$, and $x_{3}=(u_{3},v_{3})$ such
that $x_{2}\in I_{G\,\square\,H}[x_{1},x_{3}]$. Applying the distance formula
(1) and the triangle inequality we can estimate as follows:

$\displaystyle d_{G\,\square\,H}(x_{1},x_{3})$
$\displaystyle=d_{G\,\square\,H}(x_{1},x_{2})+d_{G\,\square\,H}(x_{2},x_{3})$
$\displaystyle=(d_{G}(u_{1},u_{2})+d_{H}(v_{1},v_{2}))+(d_{G}(u_{2},u_{3})+d_{H}(v_{2},v_{3}))$
$\displaystyle=(d_{G}(u_{1},u_{2})+d_{G}(u_{2},u_{3}))+(d_{H}(v_{1},v_{2})+d_{H}(v_{2},v_{3}))$
$\displaystyle\geq d_{G}(u_{1},u_{3})+d_{H}(v_{1},v_{3})$
$\displaystyle=d_{G\,\square\,H}(x_{1},x_{3})\,.$

It follows that $d_{G}(u_{1},u_{3})=d_{G}(u_{1},u_{2})+d_{G}(u_{2},u_{3})$ and
$d_{H}(v_{1},v_{3})=d_{H}(v_{1},v_{2})+d_{H}(v_{2},v_{3})$.

Suppose first that $x_{1}$ and $x_{3}$ lie in a common $G$-layer or in a
common $H$-layer. By the commutativity of the Cartesian product we may without
loss of generality assume that they lie in a common $H$-layer, that is,
$u_{1}=u_{3}$. Since ${}^{u_{1}}H$ is a convex subgraph of $G\,\square\,H$
(see [8] again), it follows that $u_{1}=u_{2}=u_{3}$. Hence the vertices
$v_{1},v_{2},v_{3}$ are pairwise different, and so the fact
$d_{H}(v_{1},v_{3})=d_{H}(v_{1},v_{2})+d_{H}(v_{2},v_{3})$ yields a
contradiction with the assumption that $\pi_{H}(R)$ is a general position set
of $H$.

Assume second that $x_{1}$ and $x_{3}$ lie neither in a common $G$-layer nor
in a common $H$-layer. Then $u_{1}\neq u_{3}$ and $v_{1}\neq v_{3}$.
Assumption (i) then implies that
$(u_{2},v_{2})\notin\\{(u_{1},v_{3}),(u_{3},v_{1})\\}$. As a consequence, at
least one of the sets $\\{u_{1},u_{2},u_{3}\\}$ and $\\{v_{1},v_{2},v_{3}\\}$
is of cardinality $3$. But then we have a contradiction for one of these sets
just as in the previous paragraph. $\square$

The converse of Lemma 3.2 does not hold. As an example consider the path
$P_{3}$ on vertices $1,2,3$, and the Cartesian product
$P_{3}\,\square\,P_{3}$. Then $\\{(1,2),(2,1),(2,3),(3,2)\\}$ is a general
position set of $P_{3}\,\square\,P_{3}$, but neither its projection onto the
first factor not the projection onto the second factor is a general position
set.

If each $H$-layer contains at most one vertex from $R$, then the conditions of
Lemma 3.2 simplify as follows.

###### Lemma 3.3

Let $G$ and $H$ be connected graphs and let $R\subseteq V(G\,\square\,H)$. If
$\pi_{G}(R)$ is a general position set in $G$ and $\pi_{G}(R)=|R|$, then $R$
is a general position set of $G\,\square\,H$.

The proof of Lemma 3.3 proceeds along the same lines as the proof of Lemma 3.2
and is hence omitted. That the converse of Lemma 3.3 again does not hold,
consider again the Cartesian product $P_{3}\,\square\,P_{3}$. Then
$\\{(1,1),(2,2),(3,1)\\}$ is a general position set of $P_{3}\,\square\,P_{3}$
with exactly one vertex in each of the layers with respect to the first
factor, but its projection onto the first factor is not a general position
set.

###### Lemma 3.4

Let $G$ and $H$ be connected graphs. If for every $u\in V(G)$ there exists a
vertex $v$ such that $\operatorname{Pl}_{G}(u,v)\cup\\{u,v\\}$ is a clique of
even order, then B wins the gp achievement game on $G\,\square\,H$.

Proof. Consider the gp achievement game on $G\,\square\,H$. Let
$a_{1}=(u_{1},v_{1})$. Then there exists a vertex $u_{2}\in V(G)$ such that
$\operatorname{Pl}_{G}(u_{1},u_{2})\cup\\{u_{1},u_{2}\\}$ is a clique of even
order. The initial strategy of B is to play $b_{1}=(u_{2},v_{1})$. Suppose
that A next plays $a_{2}=(u_{3},v_{3})$. By Lemma 3.1,
$\operatorname{Pl}_{G\,\square\,H}(a_{1},b_{1})\subseteq
V(G\,\square\,H)\backslash(^{u_{1}}H\cup{}^{u_{2}}H)$, hence
$u_{3}\notin\\{u_{1},u_{2}\\}$. We claim that $\\{u_{1},u_{2},u_{3}\\}$ is a
general position set of $G$. If not, then, since $u_{1}u_{2}\in V(G)$, we may
without loss of generality assume that $u_{2}\in I_{G}[u_{1},u_{3}]$. But then
the distance formula implies that $(u_{2},v_{1})\in
I_{G\,\square\,H}[(u_{1},v_{1}),(u_{3},v_{3})]$. Hence the claim, which in
turn implies that $u_{3}\in\operatorname{Pl}_{G}(u_{1},u_{2})$. Since
$|\operatorname{Pl}_{G}(u_{1},u_{2})|$ is even and
$\operatorname{Pl}_{G}(u_{1},u_{2},u_{3})=\operatorname{Pl}_{G}(u_{1},u_{2})\backslash\\{u_{3}\\}$,
player B can continue the game by choosing the vertex $(u_{4},v_{3})$, where
$u_{4}\in\operatorname{Pl}_{G}(u_{1},u_{2},u_{3})$. By Lemma 3.3, the set
$S_{4}=\\{(u_{1},v_{1}),(u_{2},v_{1}),(u_{3},v_{3}),(u_{4},v_{3})\\}$ is a
general position set of $G\,\square\,H$. Player B then continues this strategy
and by repeatedly applying Lemma 3.3, we can see that each set $S_{n}$ is a
general position set of $G\,\square\,H$. Also since B wins on $G$, at each
stage of the game $|\operatorname{Pl}_{G\,\square\,H}(\ldots b_{k})|$ is even.
Hence by Theorem 2.2(ii), B wins on $G\,\square\,H$. $\square$

Since in a connected, bipartite graph, every pair of adjacent vertices is a
maximal general position set, the following theorem follows directly from
Lemma 3.4.

###### Theorem 3.5

Let $G$ be a connected graph and let $H$ be a connected bipartite graph with
at least one edge. Then B wins the gp achievement game on $G\,\square\,H$
after his first move.

Theorem 3.5 should be compared with the main result from [26] which asserts
that if $T$ and $T^{\prime}$ are trees, then
$\operatorname{gp}(T\,\square\,T^{\prime})=\ell(T)+\ell(T^{\prime})$, where
$\ell(G)$ is the number of leaves of a graph $G$.

We next resolve the gp achievement on Hamming graphs.

###### Theorem 3.6

If $n,m\geq 2$, then A wins the gp achievement game on $K_{n}\,\square\,K_{m}$
if and only if both $n$ and $m$ are odd.

Proof. Let $V(K_{n})=\\{u_{1},\ldots,u_{n}\\}$,
$V(K_{m})=\\{v_{1},\ldots,v_{m}\\}$, and set $G=K_{n}\,\square\,K_{m}$ for the
rest of the proof. If one of $n$ and $m$ is even, then B wins the gp
achievement game on $G$ by Lemma 3.4.

In the rest assume that both $n$ and $m$ are odd. We need to prove that in
this case A wins the gp achievement game. The strategy of A is to achieve the
following goal. After each move $a_{i}$, $i\geq 1$, we have that

$|\operatorname{Pl}_{G}(\ldots a_{i})\cap V(K_{n}^{v_{k}})|\ {\rm is\ even\
and}\ |\operatorname{Pl}_{G}(\ldots a_{i})\cap V(^{u_{j}}K_{m})|\ {\rm is\
even}$ (3)

for all layers $K_{n}^{v_{k}}$ and all layers ${}^{u_{j}}K_{m}$ in which at
least one vertex has already been played.

By the vertex-transitivity of $G$ we may assume that $a_{1}=(u_{1},v_{1})$.
Note that (3) holds true after this move. For the first move
$b_{1}=(u_{i},v_{j})$ of B we may, again using the symmetry of $G$, without
loss of generality assume that $i=2$ and $j\in[2]$.

Suppose first that $b_{1}=(u_{2},v_{1})$. Then A selects
$a_{2}=(u_{3},v_{1})$. Since by Lemma 3.1,
$\operatorname{Pl}_{G}(a_{1},b_{1},a_{2})\subseteq
V(G)\backslash(^{u_{1}}H\cup{}^{u_{2}}H\cup{}^{u_{3}}H)$, the condition (3) is
fulfilled after the move $a_{2}$. The next move of B must be in a new
$K_{m}$-layer, say $b_{2}=(u_{4},v_{j})$. Then A replies by the vertex
$a_{3}=(u_{5},v_{j})$. This is a legal move since $n$ is odd and because Lemma
3.3 guarantees that the so far selected vertices form a general position set
of $G$. The game then continues in this manner, that is, whenever it is B’s
turn, he must select a vertex $x$ from some new $K_{m}$-layer, and then A
replies with a playable neighbor of $x$ in the corresponding $K_{n}$-layer. As
$n$ is odd, A will play the last vertex.

Suppose second that $b_{1}=(u_{2},v_{2})$. In this case A replies by picking
$a_{2}=(u_{3},v_{3})$. Then
$\operatorname{Pl}_{G}(a_{1},b_{1},a_{2})=V(G)\setminus(\\{u_{1},u_{2},u_{3}\\}\times\\{v_{1},v_{2},v_{3}\\})$
and (3) is fulfilled after the move $a_{2}$. In the sequel of the game, if B
plays a vertex such that it is the first vertex played in the two layers in
which it lies, then A replies with another such vertex. Note that this is
possible as both $n$ and $m$ are odd. After each such move of A, the
conditions (3) remain fulfilled. Suppose now that at some point of the game, B
selects a vertex in a $K_{n}$-layer in which at least one vertex has been
played earlier. Because before this move (3) holds, $A$ can reply by playing a
vertex from the same $K_{n}$-layer. Now, in this $K_{n}$-layer exactly two
less vertices are playable, so the number of playable vertices in the layer is
even (possibly zero). Moreover, in the two $K_{m}$-layers, in which the last
two moves were played, no vertex is now playable, hence (3) holds also for
these two layers. In the case when at some point of the game, B selects a
vertex in a $K_{m}$-layer in which at least one vertex has been played
earlier, A proceeds analogously, that is, he plays next a vertex from the same
$K_{m}$-layer. Following this strategy, A wins the game. $\square$

Lemma 3.4 immediately implies the following.

###### Corollary 3.7

If $n$ is even and $G$ is a connected graph, then B wins the gp achievement
game on $K_{n}\,\square\,G$.

On the other hand, if $n$ is odd and $G$ is a connected graph, the the outcome
of the gp achievement game on $K_{n}\,\square\,G$ appears more involved. This
statement is in part justified by the following result.

###### Theorem 3.8

If $m\geq 3$, then A wins the gp achievement game on $K_{3}\,\square\,C_{m}$
if and only if $m\in\\{3,5\\}$.

Proof. Let $V(K_{3})=\\{u_{1},u_{2},u_{3}\\}$ and let
$V(C_{m})=\\{v_{1},\ldots,v_{2k+1}\\}$, where the edges are in natural order.
Set $G=K_{3}\,\square\,C_{m}$ for the rest of the proof.

If $m$ is even, then Theorem 3.5 implies that B wins the gp achievement game
on $G$. If $m=3$, then by Theorem 3.6 we know that A wins the game. Consider
next the case $m=5$. We are going to prove that A wins the game by considering
all possibilities (up to symmetry). Let A start the game with
$a_{1}=(u_{1},v_{1})$. Then, up to the symmetry of $G$ and having in mind that
each vertex of $G$ is at distance at most $3$ from $a_{1}$, we need to
consider the following replies of B: $(u_{1},v_{2})$, $(u_{1},v_{3})$,
$(u_{2},v_{1})$, $(u_{2},v_{2})$, and $(u_{2},v_{3})$. If
$b_{1}=(u_{1},v_{2})$, then A selects $a_{2}=(u_{1},v_{4})$ and wins the game.
If $b_{1}=(u_{1},v_{3})$, then A selects $a_{2}=(u_{1},v_{5})$ and wins the
game. If $b_{1}=(u_{2},v_{1})$, then the move $a_{2}=(u_{3},v_{1})$ finishes
the game. If $b_{1}=(u_{2},v_{2})$, then A replies by $a_{2}=(u_{3},v_{3})$.
Then we have three subcases: if $b_{2}=(u_{1},v_{4})$, then A plays
$a_{3}=(u_{2},v_{5})$; if $b_{2}=(u_{2},v_{4})$, then A plays
$a_{3}=(u_{3},v_{5})$; and if $b_{2}=(u_{3},v_{4})$, then A plays
$a_{3}=(u_{2},v_{5})$. In each of the subcases, A wins. Finally, if
$b_{1}=(u_{2},v_{3})$, then A selects $a_{2}=(u_{3},v_{2})$. Similarly as in
the second case we now see that B cannot win with the move $b_{2}$, while
afterwards A wins with his third move.

It remains to prove that B wins on $G$ when $m\geq 7$ is odd. By the vertex-
transitivity of $G$ we may assume that A starts with the vertex
$(u_{1},v_{1})$. Then B picks the vertex $b_{1}=(u_{2},v_{2})$. From here on,
we distinguish two cases.

Case 1: $a_{2}=(u_{3},v_{i})$, where $i>2$.
We may without loss of generality assume that $i\leq k+1$. By Lemma 2.1 and by
(2), the move $b_{2}=(u_{3},v_{s})$, where $s\geq k+2$, is a legal move of B.
Let $a_{3}=(u_{r},v_{l})$. If $r=1$ and $l>k+2$, then $a_{1}\in
I_{G}[a_{3},b_{1}]$, and if $r=2$ and $l\leq k+1$, then $b_{1}\in
I_{G}[a_{1},a_{3}]$. If $r=3$, then either $\\{a_{1},a_{2},a_{3}\\}$ or
$\\{a_{1},b_{2},a_{3}\\}$ is not a general position set of $G$. These cases
imply that

$\operatorname{Pl}_{G}(a_{1},b_{1},a_{2},b_{2})\subseteq(u_{1}\times\\{v_{1},\ldots,v_{k+2}\\})\cup(u_{2}\times\\{v_{k+2},\ldots,v_{2k+1}\\})\,.$
(4)

We can without loss of generality assume that A continues by playing
$a_{3}=(u_{2},v_{j})$, where $j>k+1$. Since $m\geq 7$, B can choose
$b_{3}=(u_{1},v_{t})$ with $2<t\leq k+2$ and $i\neq t\neq j$. Again by Lemma
2.1 and (2), $S=\\{a_{1},b_{1},a_{2},b_{2},a_{3},b_{3}\\}$ is a general
position set of $G$. We claim that $S$ is a maximal general position set.
Suppose on the contrary that $a_{4}=(u_{g},v_{h})$ is a legal move. Then by
(4), $g\neq 3$. If $g=1$, then applying (4) again, $h\leq k+2$. But then by
(2), $\\{b_{1},b_{3},a_{4}\\}$ is not be a general position set of $G$. And if
$g=2$, then again by (4), $h>k+1$. If $h=k+2$ or $j=k+2$, then
$\\{a_{1},a_{3},a_{4}\\}$ is not be a general position set. And if $h>k+2$ and
$j>k+2$, then $\\{b_{1},a_{3},a_{4}\\}$ is not be a general position set in
$G$. This proves the claim which in turn finishes th argument for Case 1.

Case 2: $a_{2}=(u_{i},v_{j})$, $i\in[2]$ or $j\in[2]$.
First suppose that $i\in[2]$, say $i=1$. Then clearly $j\neq 2$. If $j>k+2$,
then $a_{1}\in I_{G}[a_{2},b_{1}]$. Hence $2<j\leq k+2$. Set
$b_{2}=(u_{3},v_{2})$. By Lemma 2.1 and (2), the set
$\\{a_{1},b_{1},a_{2},b_{2}\\}$ is a general position set. Using Lemma 3.1, we
get $\operatorname{Pl}_{G}(a_{1},b_{1},a_{2},b_{2})\subseteq\,^{u_{1}}C_{m}$.
If A can choose $a_{3}=(u_{1},v_{r})$, then, as mentioned above, $r\leq k+2$.
But then by (2), $\\{b_{1},a_{2},a_{3}\\}$ is not be a general position set.
Hence B wins the game. Suppose second that $j\in[2]$, say $j=2$. Then clearly
$i=3$. Hence by Lemma 3.1, $\operatorname{Pl}(a_{1},b_{1},a_{2})\subseteq$
${}^{u_{1}}C_{m}$. Now B can choose $b_{2}=(u_{1},v_{s})$ with $2<s\leq k+2$.
Then as in the case $i=1$, B wins the gp achievement game on $G$. $\square$

##  4 The game played on lexicographic products

The lexicographic product $G\circ H$ of graphs $G$ and $H$ has the vertex set
$V(G)\times V(H)$, vertices $(g,h)$ and $(g^{\prime},h^{\prime})$ being
adjacent if either $gg^{\prime}\in E(G)$, or $g=g^{\prime}$ and
$hh^{\prime}\in E(H)$. Layers and projections are defined for the
lexicographic product in the same way as they are defined for the Cartesian
product. The distances in lexicographic products can be computed as follows,
see [8, Proposition 5.12].

###### Proposition 4.1

If $(g,h)$ and $(g^{\prime},h^{\prime})$ are two vertices of $G\circ H$, then

$d_{G\circ
H}\left((g,h),(g^{\prime},h^{\prime})\right)=\left\\{\begin{array}[]{ll}d_{G}(g,g^{\prime});&\mbox{$g\neq
g^{\prime}$}\,,\\\
d_{H}(h,h^{\prime});&g=g^{\prime},\operatorname{deg}_{G}(g)=0\,,\\\
\min\\{d_{H}(h,h^{\prime}),2\\};&g=g^{\prime},\operatorname{deg}_{G}(g)\neq
0\,.\\\ \end{array}\right.$

###### Lemma 4.2

If $G$ and $H$ are connected graphs, and $S$ is a general position set of
$G\circ H$, then $\pi_{G}(S)$ is a general position set of $G$.

Proof. Let $S$ be a general position set of $G\circ H$, and suppose on the
contrary that $\pi_{G}(S)$ is not a general position set of $G$. Then there
exist vertices $(u_{1},v_{1})$, $(u_{2},v_{2})$, and $(u_{3},v_{3})$ from $S$
such that $u_{2}\in I_{G}[u_{1},u_{3}]$. Since $u_{1}$,$u_{2}$, and $u_{3}$
are pairwise distinct, Proposition 4.1 yields

$\displaystyle d_{G\circ H}((u_{1},v_{1}),(u_{3},v_{3}))$
$\displaystyle=d_{G}(u_{1},u_{3})=d_{G}(u_{1},u_{2})+d_{G}(u_{2},u_{3})$
$\displaystyle=d_{G\circ H}((u_{1},v_{1}),(u_{2},v_{2}))+d_{G\circ
H}((u_{2},v_{2}),(u_{3},v_{3})),$

which is not possible as $S$ is a general position set of $G\circ H$.
$\square$

###### Theorem 4.3

If $G$ is a connected graph, then B wins the gp achievement game on $G\circ
K_{n}$ if and only if either B wins on $G$ or $n$ is even.

Proof. We first claim that $S\subseteq V(G\circ H)$ is a maximal general
position set of $G\circ K_{n}$ if and only if $S=S_{G}\times V(K_{n})$, where
$S_{G}$ is a maximal general position set of $G$. Proposition 4.1 and Lemma
4.2 imply that if $S_{G}$ is a maximal general position set of $G$, then
$S_{G}\times K_{n}$ is a maximal general position set of $G\circ K_{n}$. On
the other hand, let $S$ be a maximal general position set of $G\circ K_{n}$.
The maximality implies that if $(u,v)\in S$, then $V(^{u}K_{n})\subseteq S$.
Hence $S=\pi_{G}(S)\times V(K_{n})$, and clearly, $\pi_{G}(S)$ is a maximal
general position set in $G$. This proves the claim.

Suppose that $n$ is odd and that A wins the game on $G$. We will show that
then A wins also on $G\circ H$. The strategy of A is the following. First he
selects a vertex $(u,v)$, where $u$ is an optimal start vertex in the game
played on $G$ and $v$ is an arbitrary vertex of $K_{n}$. After that, A replies
to moves of B in the following way. Whenever B selects a vertex $b_{i}$ from a
$K_{n}$-layer, from which no vertex was played earlier, A replies with a
vertex $a_{i+1}$ such that $\pi_{G}(a_{i+1})$ is an optimal reply of A to the
move $\pi_{G}(b_{i})$ of B played in $G$. On the other hand, whenever $b_{i}$
belongs to a previously visited $K_{n}$-layer, A replies by choosing a vertex
$a_{i+1}$ such that $\pi_{G}(a_{i+1})=\pi_{G}(b_{i})$. Note that this is
possible since $n$ is odd. Because A wins on $G$, the described strategy
implies that vertices from an odd number of $K_{n}$-layers will be played
during the game. In addition, the above claim implies that all the vertices
from these $K_{n}$-layers will be played, hence in total odd number vertices
will be played. We conclude that A wins the game on $G\circ H$ when $n$ is odd
and A wins the game on $G$.

It remains to prove that in the other cases B has a winning strategy. If $n$
is even, then the claim implies that an even number of vertices will be played
during the game. This means that B wins. And if B has a winning strategy on
$G$, then B follows a similar strategy as A in the previous paragraph.
Whenever A plays in some new $K_{n}$-layer, B replies optimally (with respect
to the projected game on $G$) in a new $K_{n}$-layer, and whenever A plays in
an already visited $K_{n}$-layer, B plays in some other already visited
$K_{n}$-layer in which not all vertices has been played yet. In this way an
even number of $K_{n}$-layer will be visited during the game and then the
claim implies that B wins the game. $\square$

##  5 Concluding remarks

The main message of this paper is the following. If at an early stage of the
gp achievement game played on $G$, one of the players has a possibility to
play a vertex that significantly reduces the number of playable vertices
during the rest of the game, then it is often easier to analyze the gp
achievement game on $G$ than the general position number of $G$. On the other
hand, we have also seen that many challenging problems concerning the gp
achievement game remain open.

In parallel to achievement games one can also consider avoidance games. In our
particular case of the gp achievement game, the gp avoidance game is defined
analogously, the only difference is that in the general position avoidance
game the player who has played the last vertex loses the game. So the gp
achievement game and the gp avoidance game are much similar, however, they are
in general independent. For instance, one can check that player B wins both
games on the cycle $C_{6}$. Hence it would be interesting to investigate also
the gp avoidance game and to compare it with the gp achievement game.

## Acknowledgments

Sandi Klavžar acknowledges the financial support from the Slovenian Research
Agency (research core funding P1-0297, and projects N1-0095, J1-1693,
J1-2452). Neethu P K acknowledges the Council of Scientific and Industrial
Research(CSIR), Govt. of India for providing financial assistance in the form
of Junior Research Fellowship.

## References

  * [1] M. Anderson, F. Harary, Achievement and avoidance games for generating Abelian groups, Internat. J. Game Theory 16 (1987) 321–325.
  * [2] B.S. Anand, S.V. Ullas Chandran, M. Changat, S. Klavžar, E.J. Thomas, A characterization of general position sets in graphs, Appl. Math. Comput. 359 (2019) 84–89.
  * [3] B.J. Benesh, C.D. Ernst, N. Sieben, Impartial achievement games for generating nilpotent groups, J. Group Theory 22 (2019) 515–527.
  * [4] F. Buckley, F. Harary, Geodetic games for graphs, Quaest. Math. 8 (1987) 321–334.
  * [5] H.E. Dudeney, Amusements in Mathematics, Nelson, Edinburgh, 1917.
  * [6] V. Froese, I. Kanj, A. Nichterlein, R. Niedermeier, Finding points in general position, Internat. J. Comput. Geom. Appl. 27 (2017) 277–296.
  * [7] M. Ghorbani, S. Klavžar, H.R. Maimani, M. Momeni, F. Rahimi-Mahid, G. Rus, The general position problem on Kneser graphs and on some graph operations, Discuss. Math. Graph Theory 41 (2021) 1199–1213.
  * [8] R. Hammack, W. Imrich, S. Klavžar, Handbook of Product Graphs, Second Edition, CRC Press, Boca Raton, FL, 2011.
  * [9] T.W. Haynes, M.A. Henning, C. Tiller, Geodetic achievement and avoidance games for graphs, Quaest. Math. 26 (2003) 389–397.
  * [10] S. Klavžar, D. Kuziak, I. Peterin, I.G. Yero, A Steiner general position problem in graph theory, Comput. Appl. Math. 40 (2021) Article 223.
  * [11] S. Klavžar, B. Patkós, G. Rus, I.G. Yero, On general position sets in Cartesian products, Results Math. 76 (2021) Article 123.
  * [12] S. Klavžar, G. Rus, The general position number of integer lattices, Appl. Math. Comput. 390 (2021) Article 125664.
  * [13] S. Klavžar, I.G. Yero, The general position problem and strong resolving graphs, Open Math. 17 (2019) 1126–1135.
  * [14] S. Klavžar, D.F. Rall, I.G. Yero, General $d$-position sets, Ars Math. Contemp. 21 (2021) Article P1.03.
  * [15] J. Körner, On the extremal combinatorics of the Hamming space, J. Combin. Theory Ser. A 71 (1995) 112–126.
  * [16] C.Y. Ku, K.B. Wong, On no-three-in-line problem on $m$-dimensional torus, Graphs Combin. 34 (2018) 355–364.
  * [17] P. Manuel, S. Klavžar, A general position problem in graph theory, Bull. Aust. Math. Soc. 98 (2018) 177–187.
  * [18] P. Manuel, S. Klavžar, The graph theory general position problem on some interconnection networks, Fund. Inform. 163 (2018) 339–350.
  * [19] A. Misiak, Z. Stȩpień, A. Szymaszkiewicz, L. Szymaszkiewicz, M. Zwierzchowski, A note on the no-three-in-line problem on a torus, Discrete Math. 339 (2016) 217–221.
  * [20] M. Nečásková, A note on the achievement geodetic games, Quaest. Math. 12 (1988) 115–119.
  * [21] B. Patkós, On the general position problem on Kneser graphs, Ars Math. Contemp. 18 (2020) 273–280.
  * [22] M. Payne, D. R. Wood, On the general position subset selection problem, SIAM J. Discrete Math. 27 (2013) 1727–1733.
  * [23] M. Skotnica, No-three-in-line problem on a torus: periodicity, Discrete Math. 342 (2019) Article 111611.
  * [24] E.J. Thomas, S.V. Ullas Chandran, Characterization of classes of graphs with large general position number, AKCE Int. J. Graphs Comb. 17 (2020) 935–939.
  * [25] J. Tian, K. Xu, The general position number of Cartesian products involving a factor with small diameter, Appl. Math. Comput. 403 (2021) Article 126206.
  * [26] J. Tian, K. Xu, S. Klavžar, The general position number of the Cartesian product of two trees, Bull. Aust. Math. Soc. 104 (2021) 1–10.
  * [27] S.V. Ullas Chandran, G. Jaya Parthasarathy, The geodesic irredundant sets in graphs, Int. J. Math. Combin. 4 (2016) 135–143.
  * [28] Y. Yao, M. He, S. Ji, G. Li, On the general position set of two classes of graphs, arXiv:2002.03108 [math.CO].