# A note on GMP algebra, dipole symmetry, and Hohenberg-Merming-Wagner theorem
in the lowest Landau level.

Lev Spodyneiko<EMAIL_ADDRESS>Massachusetts Institute of Technology,
Cambridge, MA 02139, United States

###### Abstract

After projection to the lowest Landau level translational invariance and
particle conservation combine into dipole symmetry. We show that the new
symmetry forbids spontaneous $U(1)$ symmetry breaking at zero temperature. In
the case of the spatially inhomogeneous magnetic field, where the
translational invariance is absent, we show that the dipole symmetry
disappears and the constraint on the symmetry breaking is lifted. We pay
special attention to the fate of the Girvin-Macdonald-Platzman algebra in the
inhomogeneous magnetic field and show that a natural generalization of it is
still present even though the dipole symmetry is not.

##  I Introduction

In a strong magnetic field, the domination of the interactions over kinetic
energy leads to rich strongly correlated physics. The importance of many-body
effects complicates the theoretical understanding of the phase diagram of
these systems. It was proposed Kapustin and Spodyneiko (2022) recently that
some insight can be achieved by considering toy models with dipole symmetry
where both $U(1)$ charge and its dipole moment are conserved. The dipole
moment conservation leads to restricted mobility of charged excitations,
mimicking the quenching of kinetic energy in the magnetic field.

It was shown Kapustin and Spodyneiko (2022) that dipole symmetry leads to the
absence of spontaneous $U(1)$-symmetry breaking at zero temperature in two-
dimensional systems. This is an extension of the Hohenberg-Mermin-Wagner (HMW)
theorem Hohenberg (1967); Mermin and Wagner (1966) to this setting. An
important caveat is that the dipole symmetry algebra used in the proof did not
involve the effects of the non-trivial topology of the wave functions. In the
language of the band theory, this would mean that the Berry curvature and
magnetic field are zero. This is certainly not a good approximation even for
the simplest case of the spatially homogeneous magnetic field.

The primary purpose of the paper is to show that in the presence of a
homogeneous magnetic field the results of Kapustin and Spodyneiko (2022) still
hold. More precisely, we take the limit where the cyclotron energy is much
larger than the typical interaction scale and restrict the system to the
lowest Landau level (LLL). In this limit, a dipole symmetry emerges: not only
the projected charge is conserved but also its dipole moment. The additional
conservation law modifies the f-sum rule which in turn forbids the spontaneous
$U(1)$-symmetry breaking in the LLL at zero temperature. The same conclusion
was made in Pitaevskii and Stringari (1991) by L. Pitaevskii and S. Stringari.
In the present paper, we interpret their result using dipole symmetry and also
discuss how it generalizes to the inhomogeneous magnetic field.

In recent years, the question of whether the lowest Landau level physics and
fractional quantum Hall effect (FQHE) in particular can be reproduced in the
absence of a magnetic field attracted a lot of attention. For example, one can
expect to see FQHE state in a flat partially filled Chern band. Experimental
signatures of such fractional Chern insulating states were reported in Xie
_et al._ (2021). However, one still lacks a coherent theoretic understanding
of how these states can emerge. The LLL has a lot of fascinating features and
figuring out which of them are essential for the existence of FQHE phases is
an open problem.

In this paper, we will take an opportunity to shed some light on this problem
by focusing on the Girvin-Macdonald-Platzman (GMP) algebra Girvin  _et al._
(1986) which was discussed in this context in Roy (2014); Wang  _et al._
(2022) (see also Mera and Ozawa (2021); Ledwith  _et al._ (2022)). As the main
example, we will consider Landau levels in a spatially varying field, which is
a toy model for a flat band that satisfies the trace condition Mera and Ozawa
(2021), has non-trivial Chern class, and an inhomogeneous Berry curvature
across the Brillouin zone. A well-known fact in Kähler quantization literature
(see for example Proposition 4.1 in Bordemann  _et al._ (1994)) is that the
projected density operators form a closed algebra. Though the structure
constants of the resulting algebra are different from the homogeneous case, it
is still a natural generalization of the GMP algebra. Despite the presence of
the algebra, the physically important property of translational invariance
(and dipole symmetry) is absent leading to changes in the behavior such as the
inapplicability of the HMW theorem.

The plan of the paper is as follows. In Section II we review the lowest Landau
level projection using physical space instead of reciprocal space which is
more common in the literature. In section III we will prove that $U(1)$
symmetry cannot be broken in the LLL assuming that the magnetic field is
spatially homogeneous. In section IV, we will consider a simple model of
spatially varying magnetic field and compute its GMP algebra. We will explain
how it invalidates the proof of the HMW theorem of Section III. We will
discuss the limitations of the argument and possible future directions in the
last section.

We thank Anton Kapustin, Patrick J. Ledwith, Sergej Moroz, Dam Thanh Son, and
Senthil Todadri for discussions. The work was supported by the Simons
Collaboration on Ultra-Quantum Matter, which is a grant (651446) from the
Simons Foundation.

##  II LLL projection

In this section, we will introduce the lowest Landau level projection. The
procedure is rather standard by now but we will use a slightly different
perspective on it. We want to use the procedure in a spatially varying
magnetic field. In this case, the continuous translational symmetry is absent
and thus the use of position space notation instead of momentum space notation
is more appropriate. This is essentially a generalization of appendix A in
Read (1998) to the inhomogeneous magnetic field. The main point is that even
though the GMP algebra of density operators is still closed in an
inhomogeneous field, the dipole symmetry is lost. Since it is the dipole
symmetry that forbids the $U(1)$-symmetry breaking, the latter can be
spontaneously broken in an inhomogeneous field. More generally, one can
consider electors propagating on a curved manifold in the presence of an
inhomogeneous magnetic field Estienne  _et al._ (2023). After the transition
to isothermal coordinates, the problem essentially reduces to the flat space
and we will restrict ourselves to the latter for simplicity of presentation.
All results of this section can be generalized to a curved background with
suitable adjustments.

Consider a Hamiltonian of interacting fully spin-polarised fermions confined
to a plane in an inhomogeneous magnetic field

$\displaystyle H=\int d^{2}{\bf x}\Big{[}\,\Psi^{\dagger}({\bf x})$
$\displaystyle\frac{(\hat{\bf p}-{\bf A})^{2}}{2m}\Psi({\bf
x})-\frac{g}{4m}B({\bf x})\rho({\bf x})\Big{]}+\frac{1}{2}\int d^{2}{\bf
x}d^{2}{\bf y}\,V({\bf x}-{\bf y})\rho({\bf x})\rho({\bf y}),$

where $\rho({\bf x})=\Psi^{\dagger}({\bf x})\Psi({\bf x})$ is a density
operator, $\bf A$ is a vector potential of the magnetic field $B(\bf x)$
pointed perpendicular to the plane. We have included the Zeeman term and
ignored the other spin polarization since we will focus only on the LLL where
all particles are spin-polarized. The creation-annihilation operators satisfy
canonical anti-commutation relations

$\displaystyle\\{\Psi({\bf x}),\Psi^{\dagger}({\bf y})\\}=\delta({\bf x-y}),$
(1)

with other anti-commutators being zero.

By introducing the operator

$\displaystyle D=(\hat{p}_{1}-A_{1})+i(\hat{p}_{2}-A_{2}),$ (2)

we can rewrite the Hamiltonian as

$\displaystyle H=\int d^{2}{\bf x}\Big{[}\,\Psi^{\dagger}({\bf x})$
$\displaystyle\frac{D^{\dagger}D}{2m}\Psi({\bf x})-\frac{g-2}{4m}B({\bf
x})\rho({\bf x})\Big{]}+\frac{1}{2}\int d^{2}{\bf x}d^{2}{\bf y}\,V({\bf
x}-{\bf y})\rho({\bf x})\rho({\bf y}).$

Let us first ignore the interactions and focus on the single-particle part of
the Hamiltonian. After setting the $g$-factor111 When $g=2$ the Paul
Hamiltonian is equal to a square of 2d Dirac operator. In a magnetic field,
the Dirac operator conserves chirality and has chiral zero modes which are
protected by the Atiyah-Singer index theorem. Alternatively, one can motivate
$g=2$ by considering a slowly varying magnetic field. In this case, the first
term of the Hamiltonian will have zero-point energy (equal to half of the
cyclotron frequency) which will depend on the position. The Zeeman term with
$g=2$ exactly cancels this contribution. to $g=2$, the non-interacting
particle’s ground state is degenerate and is spanned by functions satisfying
$D\psi({\bf x})=0$. It is a natural generalization of the lowest Landau level
to the case of an inhomogeneous magnetic field. We will assume that the flux
of the magnetic field averaged over the whole space is positive. In this case,
the LLL is not empty and the number of states in it scales as the total flux
divided by $2\pi$. For a generic magnetic field configuration and positive
average flux, the first excited non-interacting level will be separated by a
gap proportional to inverse mass $\frac{1}{m}$. By taking the limit
$m\rightarrow 0$, we can neglect higher Landau levels and project the dynamics
to the lowest Landau level. On the physical grounds, it should be a good
approximation in the limit when typical interaction energy is less than the
cyclotron frequency. We will denote the full single-particle Hilbert space as
$\mathcal{H}$ and the ground state subspace as $\mathcal{H}_{\rm LLL}$.

Let $\psi_{n}({\bf x})$ be an orthonormal basis in the LLL enumerated by
$n\in\mathbb{N}$. Consider the following function

$\displaystyle K({\bf x},{\bf y})=\sum_{n\in\mathbb{N}}\psi_{n}({\bf
x})\psi_{n}^{\dagger}(\bf{y}).$ (3)

It is a kernel of the projection $\hat{K}$ to the LLL and satisfies a number
of useful properties

$\displaystyle K({\bf x},{\bf y})$ $\displaystyle=\int d^{2}{\bf z}\,K({\bf
x},{\bf z})K({\bf z},{\bf y}),$ $\displaystyle(\text{idempotnence})$ (4)
$\displaystyle K({\bf x},{\bf y})^{*}$ $\displaystyle=K({\bf y},{\bf x}),$
$\displaystyle(\text{hermicity})$ (5) $\displaystyle D_{\bf x}K({\bf x},{\bf
y})$ $\displaystyle=0,$ $\displaystyle(\text{holomorphicity in $\bf x$})$ (6)
$\displaystyle D_{\bf y}^{\dagger}K({\bf x},{\bf y})$ $\displaystyle=0,$
$\displaystyle(\text{antiholomorphicity in $\bf y$})$ (7)
$\displaystyle\psi({\bf x})$ $\displaystyle=\int d^{2}{\bf y}\,K({\bf x},{\bf
y})\psi({\bf y}),\quad\text{iff $\psi\in\mathcal{H}_{\rm LLL}$},$
$\displaystyle(\text{reproducing property})$ (8)

where $D_{\bf x}$ is the operator (2) acting only on $\bf x$. Note, that $D$
satisfies the integrability condition $[D,D]=0$ and therefore can be used to
define complex coordinate patches on the $U(1)$ bundle in which $\psi({\bf
x})$ lives. In this coordinate system, $D$ becomes an anti-holomorphic
derivative and the function $K({\bf x},{\bf y})$ becomes holomorphic in $\bf
x$ and antiholomorpohic in $\bf y$. In particular, it is uniquely determined
by its diagonal $K({\bf x},{\bf x})$. Note, we are not working in a specific
gauge (e.g. symmetric gauge), and in our approach, the complex charts are
defined by the connection $D$. In a symmetric gauge, our notion of
holomorphicity reduces to the usual one.

An important fact that we will use extensively in the following is that any
single-particle operator on the projected Hilbert space can be represented as
a projection of the operator of multiplication by a function. Denote, the
projected operators as $\hat{T}_{f}=\hat{K}\hat{f}\hat{K}$, where $\hat{f}$ is
an operator of multiplication by a function $\hat{f}\psi({\bf x})=f({\bf
x})\psi({\bf x})$ in the unprojected single-particle Hilbert space. We will
occasionally call operators of this form Toeplitz operators. Suppose that
there is an operator $\hat{A}$ which is orthogonal (in the Hilbert-Schmidt
sense) to $\hat{T}_{f}$ for all $f$, then

$\displaystyle 0$ $\displaystyle={\rm tr}(\hat{A}\hat{T}_{f})=\int d^{2}{\bf
x}\,A({\bf x},{\bf x})f({\bf x}),$ (9)

where we defined $A({\bf x},{\bf y})=\sum_{n,m}\psi_{n}({\bf x})\langle
n|\hat{A}|m\rangle\psi_{m}^{\dagger}({\bf y})$. This function is holomorphic
in $\bf x$ and antiholomorphic in $\bf y$. The equation above means222The
left-hand and right-hand sides of (9) must be well-defined for this argument
to work. This requires the operator $\hat{A}$ to be compact. One can refine
the argument by replacing the Hilbert-Schmidt topology with the strong
operator topology, i.e. by considering matrix elements instead of the trace.
The precise correspondence in strong operator topology is not between all
operators on $\mathcal{H}_{\rm LLL}$ and space of all functions on
$\mathbb{R}^{2}$ but between bounded operators and essentially bounded
functions. See Engliš (1992) for more details. that $A({\bf x},{\bf x})=0$. By
analytic continuation $A({\bf x},{\bf y})=0$ and so is $\hat{A}$.

On a plane, the equality of operators $\hat{T}_{f}=\hat{T}_{g}$ leads to the
equality of functions $f=g$ (see e.g. Proposition 2.95 in Folland (1989)). If
we compactify the system on a torus, this is non-longer true and the map
$f\rightarrow\hat{T}_{f}$ will have a non-trivial kernel. The representation
of operators by functions should be understood modulo the kernel in this case.

Now, we can turn to the many-body Hamiltonian. We can introduce the projected
field as

$\displaystyle\overline{\Psi}({\bf x})$ $\displaystyle=\int d^{2}{\bf y}K({\bf
x},{\bf y})\Psi({\bf y}).$ (10)

In terms of the mode expansion $\Psi({\bf
x})=\sum_{\psi_{n}\in\mathcal{H}}\psi_{n}({\bf x})c_{n}$, the projected
operator $\overline{\Psi}({\bf x})=\sum_{\psi_{n}\in\mathcal{H}_{\rm
LLL}}\psi_{n}({\bf x})c_{n}$ contains only modes in the LLL. It satisfies

$\displaystyle\\{\overline{\Psi}({\bf x}),\overline{\Psi}^{\dagger}({\bf
y})\\}$ $\displaystyle=K({\bf x},{\bf y}),$ (11)
$\displaystyle\\{\overline{\Psi}({\bf x}),\overline{\Psi}({\bf y})\\}$
$\displaystyle=0,$ (12) $\displaystyle\\{\overline{\Psi}^{\dagger}({\bf
x}),\overline{\Psi}^{\dagger}({\bf y})\\}$ $\displaystyle=0,$ (13)
$\displaystyle D\overline{\Psi}({\bf x})=0.$ (14)

We will denote the projection of the many-body Hilbert space to the subspace
where particles occupy only the lowest Ladnau level by $P_{\rm LLL}$. After
these preliminaries the projection of the many-body Hamiltonian is
straightforward. One has to normal order it (with respect to
$\Psi^{\dagger},\Psi$), then replace $\Psi\rightarrow\overline{\Psi}$ and undo
the normal ordering. Normal ordering excludes processes where particles jump
to higher Landau levels and then return back to LLL. For example,

$\displaystyle P_{\rm LLL}\rho({\bf x)}\rho({\bf y})P_{\rm LLL}$
$\displaystyle=-P_{\rm LLL}\Psi^{\dagger}({\bf x)}\Psi^{\dagger}({\bf
y)}\Psi({\bf x)}\Psi({\bf y)}P_{\rm LLL}+\delta({\bf x-y})P_{\rm
LLL}\Psi^{\dagger}({\bf x})\Psi({\bf y})P_{\rm LLL}$
$\displaystyle=-\overline{\Psi}^{\dagger}({\bf
x)}\overline{\Psi}^{\dagger}({\bf y)}\overline{\Psi}({\bf
x)}\overline{\Psi}({\bf y)}+\delta({\bf x-y})\overline{\Psi}^{\dagger}({\bf
x})\overline{\Psi}({\bf y})$ $\displaystyle=\overline{\rho}({\bf
x})\overline{\rho}({\bf y})+(\delta({\bf x-y})-K({\bf x},{\bf
y}))\overline{\Psi}^{\dagger}({\bf x})\overline{\Psi}({\bf y}),$

where we have defined the projected density operator $\overline{\rho}({\bf
x})=\overline{\Psi}^{\dagger}({\bf x})\overline{\Psi}({\bf x})$. Due to the
holomorphicity (14) of $\overline{\Psi}$ the kinetic term projects to zero.

The resulting projected Hamiltonian is

$\displaystyle\overline{H}=\frac{1}{2}\int d^{2}{\bf x}d^{2}{\bf y}V({\bf
x}-{\bf y})\overline{\rho}({\bf x})\overline{\rho}({\bf y})$
$\displaystyle+\frac{1}{2}\int d^{2}{\bf x}V({\bf 0})\overline{\rho}({\bf x})$
$\displaystyle-\frac{1}{2}\int d^{2}{\bf x}d^{2}{\bf y}V({\bf x}-{\bf
y})K({\bf x},{\bf y})\overline{\Psi}^{\dagger}({\bf x})\overline{\Psi}({\bf
y}).$

According to the discussion above, the last term can be rewritten as a
projection of a single-particle potential

$\displaystyle-\frac{1}{2}\int d^{2}{\bf x}d^{2}{\bf y}V({\bf x}-{\bf
y})K({\bf x},{\bf y})\overline{\Psi}^{\dagger}({\bf x})\overline{\Psi}({\bf
y})=\int d^{2}{\bf x}\,\phi({\bf x})\overline{\rho}({\bf x}),$ (15)

where the potential $\phi({\bf x})$ depends on interactions $V({\bf x}-{\bf
y})$ and the kernel $K({\bf x},{\bf y})$. If the magnetic field is homogeneous
the resulting function $\phi({\bf x})$ will be constant which leads to a
simple shift of the chemical potential. However, for inhomogeneous field
$\phi({\bf x})$ will have a non-trivial spatial dependence. This may appear
counter-intuitive. Indeed consider a Landau level with just one particle. The
last term in the Hamiltonian suggests that the interaction will generate a
spatially non-trivial Hamiltonian even though the particle has nothing to
interact with. The resolution of this problem is that the first term is also
non-zero even for a single particle and it is exactly compensated by the last
term.

Let us define a projection of a density operator weighted by a function
$f({\bf x})$

$\overline{\rho}(f)=P_{\rm LLL}\int d^{2}{\bf x}\,f({\bf x})\rho({\bf
x})P_{\rm LLL}=\int d^{2}{\bf x}\,f({\bf x})\overline{\Psi}^{\dagger}({\bf
x})\overline{\Psi}({\bf x}).$ (16)

The commutator of two such operators is

$\displaystyle[\overline{\rho}(f),\overline{\rho}(g)]=\int d^{2}{\bf
x}d^{2}{\bf y}\,\overline{\Psi}^{\dagger}({\bf x})\big{[}f({\bf x})K({\bf
x},{\bf y})g({\bf y})-f({\bf y})K({\bf y},{\bf x})g({\bf
x})\big{]}\overline{\Psi}({\bf y}).$ (17)

The r.h.s. is a single particle operator. As explained above, all single-
particle operators are of the form $\overline{\rho}(h)$ for some function $h$.
Therefore we can express the r.h.s. as $h=f\star g-g\star f$, where the star
product is implicitly defined by the equation
$\hat{T}_{f}\hat{T}_{g}=i\hat{T}_{f\star g}$. We find,

$\displaystyle[\overline{\rho}(f),\overline{\rho}(g)]=i\overline{\rho}(f\star
g-g\star f).$ (18)

This is a natural generalization of GMP algebra to the case of the
inhomogeneous magnetic field. One can see that the projected Hamiltonian can
be expressed in terms of the density operators $\overline{\rho}$ which form a
closed algebra. Unfortunately, explicit formulas for the star product require
knowledge of the kernel $K({\bf x},{\bf y})$ which is only computable for
special cases. We will discuss an example with an inhomogeneous magnetic field
and compute the resulting GMP algebra in Section IV.

Suppose the Hamiltonian is invariant under a unitary transformation $U$

$UHU^{-1}=H.$ (19)

Commutativity with the Hamiltonian means that even after the projection the
operator $P_{\rm LLL}UP_{\rm LLL}$ stays unitary and commutes with the
projected Hamiltonian. We will assume that a single-particle operator
conjugated by $U$ stays a single-particle operator. In particular

$\displaystyle U\overline{\rho}(f)U^{-1}=\overline{\rho}(f^{U})$ (20)

for some function $f^{U}$. Then the star product is covariant under the
symmetry

$\displaystyle(f*g)^{U}=f^{U}\star g^{U}.$ (21)

Note, the action of symmetry $U$ on functions can be determined from the
action of $U$ on unprojected operators $\rho(f)$. For example, magnetic
translation Zak (1964) operators $\mathcal{T}_{\bf a}$ act as
$f^{\mathcal{T}_{\bf a}}({\bf x})=f({\bf x+a})$.

### Spatially homogeneous magnetic field

Let us now focus on the homogeneous magnetic field $B({\bf x})=B_{0}$. In this
case, the algebra of the projected densities $\overline{\rho}(f)$ is covariant
under the magnetic translation operators $\mathcal{T}_{\bf a}$. It is natural
to expand the projected densities in terms of eigenvectors of the operator
$\mathcal{T}_{\bf a}$ acting on $f$. We define

$\displaystyle\overline{\rho}_{\bf k}=\overline{\rho}(e^{-i{\bf k\cdot x}}),$
(22)

where minus on the right-hand side is to match the standard definition of the
Fourier transform.

Magnetic translations do not commute with each other and one cannot
diagonalize translations in different directions simultaneously. However, the
density operators transform under the adjoint representation (20) of magnetic
translation algebra in which the translations do commute. Thus one can expand
$\overline{\rho}$ in terms of usual plane waves.

The GMP algebra in momentum space333Note that the right-hand side contains
exponentially large structure constant. In position space, it will lead to an
algebra of $\overline{\rho}(f)$ that is well-defined only for analytic
functions $f$ on the whole plane. This does not lead to any problems since one
can replace arbitrary function $f$ with its convolution with sharp Gaussian
function $\widetilde{f}({\bf x})=\int{d^{2}{\bf
y}}\frac{1}{{2\pi\varepsilon}}\exp(-({\bf x}-{\bf y})^{2}/2\varepsilon)f({\bf
y})$. The resulting operator $\hat{T}_{\widetilde{f}}$ will converge to
$\hat{T}_{f}$ as $\varepsilon\rightarrow 0$. The GMP algebra can be computed
directly in the position space. Intermediate steps in the computation require
finding an inverse to a heat kernel, which can be found in Amour and Nourrigat
(2019).

$\displaystyle[\overline{\rho}_{\bf k},\overline{\rho}_{\bf
q}]=2i\,\exp\Big{(}{\dfrac{{\bf k}\cdot{\bf
q}}{2}l^{2}_{B}}\Big{)}\sin\Big{(}\frac{{\bf k}\wedge{\bf
q}}{2}l^{2}_{B}\Big{)}\,\overline{\rho}_{{\bf k}+{\bf q}},$ (23)

where $l_{B}=\sqrt{\frac{1}{B_{0}}}$ is the magnetic length, $(\wedge{\bf
k})_{i}=\epsilon_{ij}k_{j}$ with $\epsilon_{ij}$ being the Levi-Civita symbol,
${\bf k}\wedge{\bf p}={\bf k}\cdot(\wedge{\bf p})$, and summation over
repeating indices is assumed. By expanding the algebra at small $\bf k$, we
see that infinitesimal translations are generated by the operator
$P_{j}=il_{B}^{-2}\epsilon_{jm}\dfrac{\partial{\overline{\rho}_{\bf
k}}}{\partial k_{m}}\Big{|}_{{\bf
k}=0}=l_{B}^{-2}\epsilon_{jm}\overline{\rho}(x_{m})$

$[P_{i},\overline{\rho}_{\bf k}]=k_{i}\overline{\rho}_{\bf k},$ (24)

or in position space

$\displaystyle[P_{i},\overline{\rho}({\bf
x})]=\frac{\partial\overline{\rho}({\bf x})}{\partial x_{i}}.$ (25)

Let us define the total charge $Q=\overline{\rho}(1)=\int d^{2}{\bf
x}\,\overline{\rho}({\bf x})$ and the dipole charge
$D_{i}=\overline{\rho}(x_{i})=\int d^{2}{\bf x}\,x_{i}\overline{\rho}({\bf
x})$. Note, that dipole and momentum operators are related to each other
$D_{i}=-l_{B}^{2}\epsilon_{ij}P_{i}$. Importantly, the densities of $D_{i}$
and $Q$ are related as $d_{i}({\bf x})=x_{i}q({\bf x})$, i.e. $D_{i}$ can be
thought as a dipole moment of the charge $Q$. Therefore, the system projected
to the lowest Landau level conserves not only the charge but also the
associated dipole moment.

### Periodic magnetic field

Lastly, we will consider a periodic magnetic field $B({\bf x}+{\bf R})=B({\bf
x})$ for all $\bf R$ which belong to a lattice $\Lambda$. In this case, the
magnetic translations $\mathcal{T}_{\bf R}$ by the lattice vectors ${\bf
R}\in\Lambda$ act as a symmetry.

As in the homogeneous case, the projected densities $\overline{\rho}(\bf x)$
transform in the adjoint representation of the magnetic translation group
which is isomorphic to the adjoint representation of the translation group in
the absence of a magnetic field. Thus, it is natural to decompose an arbitrary
function $f$ as

$\displaystyle f({\bf r})=\sum_{n}\int_{{\bf K}\in{\rm BZ}}\frac{d^{2}{\bf
K}}{(2\pi)^{2}}f_{{\bf K},n}e^{-i{\bf K}\cdot{\bf r}}g_{n}({\bf r}),$ (26)

where $g_{n}({\bf r})$ is a basis of periodic functions $g_{n}({\bf r}+{\bf
R})=g_{n}({\bf r})$ for all ${\bf R}\in\Lambda$, and $\bf K$ is integrated
over first (non-magnetic) Brillouin zone. There is no a priori preferred
choice of the basis $g_{n}({\bf r})$.

Define

$\overline{\rho}_{{\bf K},n}=\overline{\rho}(e^{-i{\bf K\cdot{\bf
r}}}g_{n}({\bf r})).$ (27)

The GMP algebra will have the form

$\displaystyle[\overline{\rho}_{{\bf K},n},\overline{\rho}_{{\bf
Q},m}]=\sum_{l}C_{nm}^{l}({\bf K},{\bf Q})\,\overline{\rho}_{{\bf K}+{\bf
Q},l},$ (28)

where ${\bf K}+{\bf Q}$ is understood modulo reciprocal lattice vectors ${\bf
G}\in\Lambda^{\vee}$. The structure constants $C_{nm}^{l}({\bf K},{\bf Q})$
are complicated functionals of the magnetic field $B({\bf x})$ and its
computation is abstracted by the absence of explicit expression for the kernel
$K({\bf x},{\bf y})$.

As was mentioned above, one is free to choose any basis for $g_{n}({\bf x})$.
A simple choice is $g_{\bf G}({\bf x})=\exp(-i{\bf G}\cdot{\bf x})$, where
index $n$ is replaced by a vector lying in the reciprocal lattice ${\bf
G}\in\Lambda^{\vee}$. We can combine $\bf K$ and $\bf G$ into a single vector
${\bf k}={\bf K}+{\bf G}$ which is not restricted to the first Brillouin zone

$\overline{\rho}_{{\bf k}}=\overline{\rho}_{{\bf K},{\bf
G}}=\overline{\rho}(e^{-i{\bf k\cdot{\bf r}}}).$ (29)

In this basis, the GMP algebra has the form

$\displaystyle[\overline{\rho}_{{\bf k}},\overline{\rho}_{{\bf q}}]=\sum_{{\bf
G}\in\Lambda^{\vee}}C^{{\bf G}}({\bf k},{\bf q})\overline{\rho}_{{\bf k}+{\bf
q}+{\bf G}}.$ (30)

One should note that the GMP algebra does not appear to be sensitive to the
central extension of the translational group to the magnetic translation
group. The above discussion work equally well for integer, rational, or even
irrational total flux through the unit cell. In contrast, the structure of the
wave functions and magnetic Brillouin zone strongly depends on the number
theoretic properties of the flux Novikov and Dubrovin (1980). It is an
interesting question whether something special happens to the structure
constants $C^{{\bf G}}({\bf k},{\bf q})$ at rational or integer flux through
unit cell.

##  III Hohenberg-Mermin-Wagner theorem

In this section, we prove the absence of spontaneous $U(1)$ symmetry breaking
in a system of particles in the LLL with density-density interaction and in
the presence of a spatially homogeneous magnetic field.

The Hamiltonian is

$\displaystyle\overline{H}=\frac{1}{2}\int d^{2}{\bf x}d^{2}{\bf y}\,V({\bf
x}-{\bf y})\overline{\rho}({\bf x})\overline{\rho}({\bf
y})=\frac{1}{2}\int\frac{d^{2}{\bf q}}{(2\pi)^{2}}V_{\bf
q}\overline{\rho}_{-\bf q}\overline{\rho}_{\bf q},$ (31)

where we have ignored the chemical potential shift, $V_{\bf q}$ is Fourier
transform of $V({\bf x}-{\bf y})$, and projected density operators satisfy the
standard GMP algebra (23).

The projected Hamiltonian is invariant under $U(1)$ and dipole symmetries
generated by

$\displaystyle Q=\overline{\rho}_{0},\quad
D_{i}=i\frac{\partial\overline{\rho}_{\bf k}}{\partial k_{i}}\Big{|}_{{\bf
k}=0}.$ (32)

These symmetries act on $\overline{\rho}_{\bf k}$ as

$\displaystyle[Q,\overline{\rho}_{\bf k}]=0,\quad[D_{i},\overline{\rho}_{\bf
k}]=-\epsilon_{ij}k_{j}\overline{\rho}_{\bf k}.$ (33)

The $U(1)$ symmetry is said to be broken if there exists a local operator
$\phi({\bf x})$ called the order parameter such that

$\langle[Q,\phi({\bf x})]\rangle=C\neq 0.$ (34)

We assume that continuous translational symmetry is unbroken and thus the
r.h.s. is $\bf x$-independent. After the Fourier transform we find

$\displaystyle C\delta({\bf k})=\frac{1}{(2\pi)^{2}}\int d^{2}{\bf
y}\exp(-i{\bf k}\cdot{\bf y})\langle[\overline{\rho}_{\bf k},\phi_{-\bf
k}]\rangle=\delta({\bf k})\langle[\overline{\rho}_{\bf k},\phi_{-\bf
k}]\rangle.$ (35)

Since in position space $\langle[\overline{\rho}({\bf x}),\phi({\bf
y})]\rangle$ decays at least exponentially for large separation $|\bf x-y|$,
the Fourier transform444A cautious reader may notice that this expression is
ill-defined in infinite volume. A more careful treatment is to work in real
space and replace the charge $Q$ of infinite space with a charge computed in a
large volume. The manipulations during the proof are almost the same as in
Kapustin and Spodyneiko (2022) and we will omit the technical complication for
sake of transparency. Alternatively, one can work in the momentum space of a
finite volume system as was done in the original Hohenberg’s paper Hohenberg
(1967) for example.  $\langle[\overline{\rho}_{\bf k},\phi_{-\bf k}]\rangle$
is a smooth function near ${\bf k}=0$ and we can divide this expression by
$\delta(\bf k)$. We find

$\displaystyle\langle[\overline{\rho}_{\bf k},\phi_{-\bf k}]\rangle=C+o(|{\bf
k}|).$ (36)

Since the l.h.s is smooth there exists a region $|{\bf k}|<k_{c}$ such that
$|\langle[\overline{\rho}_{\bf k},\phi_{-\bf k}]\rangle|>0$. We will restrict
ourselves to this region in the following.

We can apply to the l.h.s. the generalized uncertainty principle Pitaevskii
and Stringari (1991)

$|\langle[A_{1},A_{2}]\rangle|^{2}\leq\langle
A_{1}^{\dagger}A_{1}+A_{1}A_{1}^{\dagger}\rangle\langle
A_{2}^{\dagger}A_{2}+A_{2}A_{2}^{\dagger}\rangle,$

with $A_{1}=\phi_{-\bf k}$ and $A_{2}=\overline{\rho}_{{\bf k}}$. One finds

$0<|\langle[\overline{\rho}_{\bf k},\phi_{-\bf
k}]\rangle|^{2}\leq\langle\overline{\rho}_{\bf
k}^{\dagger}\overline{\rho}_{\bf k}+\overline{\rho}_{\bf
k}\overline{\rho}_{\bf k}^{\dagger}\rangle\langle\phi_{-\bf
k}^{\dagger}\phi_{-\bf k}+\phi_{-\bf k}\phi_{-\bf k}^{\dagger}\rangle.$ (37)

Using Cauchy-Schwartz inequality we find

$\displaystyle\begin{split}\langle\overline{\rho}_{\bf
k}^{\dagger}\overline{\rho}_{\bf
k}\rangle=\int_{0}^{\infty}\frac{d\omega}{2\pi}\langle\overline{\rho}_{\omega{\bf
k}}^{\dagger}\overline{\rho}_{\omega{\bf
k}}\rangle\leq\frac{1}{2}\sqrt{\int_{0}^{\infty}\frac{d\omega}{\pi}\omega\langle\overline{\rho}_{\omega{\bf
k}}^{\dagger}\overline{\rho}_{\omega{\bf
k}}\rangle\int_{0}^{\infty}\frac{d\omega}{\pi}\frac{1}{\omega}\langle\overline{\rho}_{\omega{\bf
k}}^{\dagger}\overline{\rho}_{\omega{\bf k}}\rangle}\\\
=\frac{1}{2}\sqrt{\langle[[\overline{\rho}^{\dagger}_{\bf
k},\overline{H}],\overline{\rho}_{\bf
k}]\rangle}\sqrt{\chi_{\overline{\rho}\overline{\rho}}({\bf k})},\end{split}$
(38)

where we introduced time Fourier transform $\overline{\rho}_{\bf
k}(t)=\int\frac{d\omega}{2\pi}\overline{\rho}_{\omega\bf k}e^{-i\omega t}$ and
static compressibility $\chi_{\overline{\rho}\overline{\rho}}({\bf
k})=\int_{0}^{\infty}\frac{d\omega}{\pi}\frac{1}{\omega}\langle\overline{\rho}_{\omega{\bf
k}}^{\dagger}\overline{\rho}_{\omega{\bf k}}\rangle$. Note that
compressibility with respect to the projected densities is the same as
compressibility with respect to unprojected ones. We will assume that
$\chi_{\overline{\rho}\overline{\rho}}({\bf k})$ is continuous function of
$\bf k$ and approximate it with its value at ${\bf k}=0$. Moreover, we will
assume that the static compressibility is finite
$\chi_{\overline{\rho}\overline{\rho}}(0)=\frac{\partial
n}{\partial\mu}<\infty$, where $n$ is particle density and $\mu$ is chemical
potential. For example, the FQHE state is incompressible $\frac{\partial
n}{\partial\mu}=0$ and satisfies this condition.

Putting everything together, we find

$\displaystyle 0<\langle|\phi_{-\bf
k}|^{2}\rangle\sqrt{\langle[[\overline{\rho}_{-\bf
k},[\overline{H},\overline{\rho}_{\bf k}]]\rangle}\sqrt{\frac{\partial
n}{\partial\mu}}.$

We assumed that all correlation functions are symmetric under reflection ${\bf
k}\rightarrow-{\bf k}$ for simplicity of presentation. The general case can be
done in the same fashion.

### Estimates

We demand that

$\displaystyle\int d^{2}{\bf k}\,\langle|\phi_{\bf k}|^{2}\rangle$ (39)

is finite. This condition is equivalent Kapustin and Spodyneiko (2022) to the
requirement of cluster decomposition for $\phi$. Physically, it means that
relative fluctuations of $\phi({\bf x})$ averaged over a region $U$ go to zero
as the area of the region $|U|$ increases

$\displaystyle\frac{1}{|U|^{2}}\int_{{\bf x},{\bf y}\in U}d^{2}{\bf
x}d^{2}{\bf y}\,\Big{(}\langle\phi^{\dagger}({\bf x})\phi({\bf
y})\rangle-|\langle\phi({\bf
x})\rangle|^{2}\big{)}\xrightarrow{|U|\rightarrow\infty}0.$ (40)

This is a natural requirement if we assume that the order parameter is a well-
defined quantity in a thermodynamical sense. Such assumption is implicit in
the original proofs of HMW Hohenberg (1967); Mermin and Wagner (1966).

In order for (39) to be finite, the $\langle|\phi_{\bf k}|^{2}\rangle$ cannot
diverge too fast as momentum decreases:

$\langle|\phi_{\bf k}|^{2}\rangle\sim|{\bf k}|^{\epsilon-2}\quad|{\bf
k}|\rightarrow 0,$ (41)

with $\epsilon>0$.

Next, we turn to computation of $\langle[[\overline{\rho}_{-\bf
k},[\overline{H},\overline{\rho}_{\bf k}]]\rangle$. At this point, it is
instructive to recall that before the LLL projection, one has the so-called
f-sum rule

$\int_{0}^{\infty}\frac{d\omega}{\pi}\omega\langle\rho_{\omega{\bf
k}}^{\dagger}\rho_{\omega{\bf k}}\rangle=\langle[[\rho_{-\bf k},[H,\rho_{\bf
k}]]\rangle=\frac{{\bf k}^{2}}{2m}$ (42)

After the projection, due to Kohn theorem Kohn (1961) ${\bf k}^{2}$ part of
the f-sum rule is saturated by cyclotron resonance which is projected out.
After going to LLL, the remainder can be found by direct computation Girvin
_et al._ (1986)

$\displaystyle\langle[[\overline{\rho}_{-\bf
k},[\overline{H},\overline{\rho}_{\bf k}]]\rangle\sim|{\bf k}|^{4},\qquad|{\bf
k}|\rightarrow 0,$ (43)

but a more insightful way is to use symmetries. The low momentum expansion of
density is

$\displaystyle\overline{\rho}_{\bf k}\sim Q-ik_{j}D_{j}+o(|{\bf
k}|^{2}),\quad|{\bf k}|\rightarrow 0,$ (44)

Since the Hamiltonian is invariant under charge and dipole symmetries the
terms up third order will be zero. Physically, the dipole symmetry forces
f-sum rule to start from quartic terms instead of the usual quadratic. A more
careful analysis shows that small ${\bf k}$ singularities in the second-order
derivatives of the Fourier transform $V_{\bf k}$ of the interaction energy can
contribute to the r.h.s. of (43). The second-order derivatives of $V_{\bf k}$
are non-singular at the origin if the interaction energy $V({\bf x}-{\bf y})$
decays faster then the fourth power of separation $\dfrac{1}{|{\bf x}-{\bf
y}|^{4}}$. Slower decays can invalidate our conclusions.

Combining the estimates and assuming that $U(1)$ symmetry is broken we find

$0<\langle|\phi_{\bf k}|^{2}\rangle\sqrt{\langle[[\overline{\rho}_{-\bf
k},[H,\overline{\rho}_{\bf k}]]\rangle}\sqrt{\chi_{BB}}\sim|{\bf
k}|^{\epsilon}$

which goes to $0$ as $|{\bf k}|\rightarrow 0$. This contradiction shows that
$U(1)$ symmetry cannot be broken at zero temperature in the lowest Landau
level.

##  IV Inhomogeneous magnetic field

As was indicated in section II, the GMP algebra is still present even in the
case of spatially varying magnetic field. However, the translational symmetry
is lost and, as a consequence, the dipole symmetry. The latter is responsible
for the absence of $k^{2}$ terms in the f-sum rule and the impossibility of
symmetry breaking. In this section, we will take a simple toy model as an
example and demonstrate this. In particular, we will explicitly compute the
resulting GMP algebra and show that f-sum rule contains quadratic terms.

We will consider a homogeneous magnetic field with a delta function insertion
of one negative $2\pi$-flux at the origin

$B({\bf x})=B_{0}-2\pi\delta({\bf x}).$ (45)

The ground state wave functions in the inhomogeneous magnetic field are of the
form Aharonov and Casher (1979)

$\psi({\bf x})=f(z)\exp(-\phi(\bf x)),$ (46)

where we introduced complex coordinate notation $z=x_{1}+ix_{2}$, $f(z)$ is a
holomorphic function which grows not too fast at infinity, $\phi({\bf x})$
satisfies

$\displaystyle\Delta\phi({\bf x})=B({\bf x}),$ (47)

and we used a symmetric gauge.

In our case, we find

$\displaystyle\phi({\bf x})=\frac{|{\bf x}|^{2}}{4l_{B}^{2}}-\log|{\bf x}|,$
(48)

and the normalized ground state wave functions are

$\displaystyle\psi_{n}({\bf
x})=\frac{z^{n-1}|z|}{\sqrt{2\pi(2l_{B}^{2})^{n}(n)!}}\exp\Big{(}-\frac{|z|^{2}}{4l_{B}^{2}}\Big{)},\qquad
n\geq 1$ (49)

where $l_{b}=\frac{1}{\sqrt{B_{0}}}$. All wave functions have a zero at
${z=0}$ and we can make a gauge transformation $\psi\rightarrow z\psi({\bf
x})/|z|$. The resulting wave functions in the new gauge are

$\displaystyle\psi_{n}({\bf
x})=\frac{z^{n}}{\sqrt{2\pi(2l_{B}^{2})^{n}(n)!}}\exp\Big{(}-\frac{|z|^{2}}{4l_{B}^{2}}\Big{)},\qquad
n\geq 1,$ (50)

which are the same as the usual wave functions except that the wave function
with $n=0$ is missing. It is reasonable to expect that since the Zeeman
interaction leads to an infinite potential at the origin $|{\bf x}|=0$.

The wave functions of the ground state are of the form
$\psi=zf(z)\exp\Big{(}-\frac{|z|^{2}}{4l_{B}^{2}}\Big{)}$ and one might be
tempted to consider $f(z)$ which has a pole at the origin since the resulting
wave function will be regular. However such functions should be excluded.
Heuristically, the number of LLL states in a region is proportional to the
total flux going through it. Since we have one unit of flux removed there
should be one less state. More formally, the delta function flux at the origin
should be physically thought of as a limit of insertion of a large negative
magnetic field in a small area around ${\bf x}=0$. In this setting, the
function $\phi({\bf x})$ does not have a logarithmic singularity which can
cancel a pole in $f(z)$. One can show that in the zero size limit of the flux
insertion area the wavefunctions reduce to (49).

One can see that the Hilbert space of a system with a negative flux insertion
can be thought of as a subspace of the homogeneous LLL with one state removed.
We use that in the following and we will express the operators in the former
Hilbert space as operators in the latter one which has zero matrix elements
whenever one of the particles is in $n=0$ state. The computations are
straightforward but cumbersome and we leave the details to the Appendix.

The Bergman kernel is

$\displaystyle\widetilde{K}({\bf x},{\bf y})=\frac{1}{2\pi
l_{B}^{2}}\exp\Big{(}-\frac{{\bf x}^{2}}{4l_{B}^{2}}-\frac{{\bf
y}^{2}}{4l_{B}^{2}}\Big{)}\left(\exp\Big{(}\frac{{\bf x}\cdot{\bf y}-i{\bf
x}\wedge{\bf y}}{2l_{B}^{2}}\Big{)}-1\right).$ (51)

Here and in the following, we will use a tilde to distinguish the case with a
flux inserted from a homogeneous magnetic field. The projected densities are

$\displaystyle\widetilde{\rho}_{\bf k}=\overline{\rho}_{\bf
k}-\exp\Big{(}-\frac{{\bf
k}^{2}l_{B}^{2}}{2}\Big{)}\int\frac{l_{B}^{2}d^{2}{\bf
q}}{2\pi}\left(2\exp\Big{(}\frac{l_{B}^{2}{\bf k}\cdot{\bf
q}}{2}\Big{)}\cos\Big{(}\frac{l_{B}^{2}{\bf k}\wedge{\bf
q}}{2}\Big{)}-1\right)\overline{\rho}_{\bf q},$ (52)

where again $\overline{\rho}_{\bf k},\widetilde{\rho}_{\bf k}$ are projected
densities for homogeneous and inhomogenous magnetic field respectively. It
satisfies

$\displaystyle\int\frac{d^{2}{\bf k}}{(2\pi)^{2}}\widetilde{\rho}_{\bf k}=0,$
(53)

which is to be expected since this is $\widetilde{\rho}({\bf x})$ at ${\bf
x}=0$. But $\widetilde{\rho}_{\bf k}$ obey even stronger condition

$\displaystyle\int\frac{d^{2}{\bf
k}}{(2\pi)^{2}}\exp\Big{(}\frac{l_{B}^{2}{\bf q}\cdot{\bf k}}{2}\pm
i\frac{l_{B}^{2}{\bf q}\wedge{\bf k}}{2}\Big{)}\widetilde{\rho}_{\bf k}=0,$
(54)

for any $\bf q$.

The GMP algebra is

$\displaystyle\begin{split}[\widetilde{\rho}_{\bf k},\widetilde{\rho}_{\bf
q}]&=2i\,e^{{\frac{{\bf k}\cdot{\bf q}}{2}}}\sin\Big{(}\frac{{\bf k}\wedge{\bf
q}}{2}\Big{)}\,\widetilde{\rho}_{{\bf k}+{\bf q}}\\\ &+2ie^{-\frac{{\bf
k}^{2}+{\bf q}^{2}+{\bf k}\cdot{\bf q}}{2}}\int\frac{d^{2}{\bf
p}}{2\pi}\,e^{\frac{{\bf p}\cdot({\bf k}+{\bf
q})}{2}}\sin\Bigg{(}\frac{1}{2}\Big{(}{\bf p}-\frac{{\bf k}+{\bf
q}}{2}\Big{)}\wedge({\bf k}-{\bf q})\Bigg{)}\widetilde{\rho}_{\bf
p},\end{split}$ (55)

where we set $l_{B}=1$ for simplicity. One can see that $\widetilde{\rho}$
forms a closed algebra (which is a subalgebra of homogeneous GMP algebra
generated by $\overline{\rho}$).

The dipole operator, defined as

$\widetilde{D}_{i}=i\frac{\partial\widetilde{\rho}_{\bf k}}{\partial
k_{i}}\Big{|}_{{\bf k}=0},$ (56)

acts on densities as

$\displaystyle[\widetilde{D}_{i},\widetilde{\rho}_{\bf
q}]=-\epsilon_{ij}q_{j}\widetilde{\rho}_{\bf q}-\int\frac{d^{2}{\bf
p}}{2\pi}e^{\frac{{\bf p}\cdot{\bf q}-{\bf
q}^{2}}{2}}\Bigg{(}p_{i}\sin\frac{{\bf q}\wedge{\bf
p}}{2}-\epsilon_{ij}p_{j}\cos\frac{{\bf q}\wedge{\bf
p}}{2}\Bigg{)}\widetilde{\rho}_{\bf p}.$ (57)

The latter term will lead to terms that are of the order ${\bf k}^{2}$ in the
sum rule (43). For small $|{\bf k}|$,

$[\widetilde{\rho}_{-\bf k},[\overline{H},\widetilde{\rho}_{\bf k}]]\sim
k_{i}k_{j}[D_{i},[\overline{H},D_{j}]]+\dots,\qquad|{\bf k}|\rightarrow 0$
(58)

and $[D_{i},[\overline{H},D_{j}]]$ is no longer equal zero on operator level.
The resulting expression for the order ${\bf k}^{2}$ term is rather
complicated, but we give a simple example for a translationally invariant
state in the appendix. The ${\bf k}^{2}$ term makes $U(1)$ symmetry breaking
possible in principle.

An important difference from the homogeneous case is that there are no such
vector-valued functions ${\bf h}({\bf x})$ that

$\displaystyle h_{i}\star f-f\star h_{i}=\partial_{i}f\quad\text{for any
$f$.}$

Indeed, if there were such a function one could translate the delta function
at the origin $\delta({\bf x})$ into $\delta({\bf x}-{\bf a})$ using
conjugation by a unitary which is exponentiation of the adjoint action of
${\bf a}\cdot{\bf h}$ (it is not equal to $\exp(i{\bf a}\cdot{\bf h})$).
However, $\widetilde{\rho}(0)=\int d^{2}{\bf x}\,\delta({\bf
x})\widetilde{\rho}({\bf x})$ is equal to 0 because of (53) while
$\widetilde{\rho}({\bf a})=\int d^{2}{\bf x}\,\delta({\bf x}-{\bf
a})\widetilde{\rho}({\bf x})$ is not zero for general ${\bf a}$ and thus these
operators cannot be related via a conjugation by a unitary.

In a constant magnetic field, the corresponding function $\bf h$ is a linear
function. One may wonder whether there is a remnant of dipole symmetry that
descends from the translational invariance of the interactions before the
projection. Or, in other words, whether the dipole symmetry gets deformed
instead of disappearing. An operator of a deformed symmetry should be of the
form $\int d^{2}{\bf x}\,\alpha_{i}({\bf x})\widetilde{\rho}(\bf x)$ (where
$\alpha_{i}({\bf x})$ is only linear on average) and should act as
infinitesimal translation on $\widetilde{\rho}(\bf x)$. The argument of the
previous paragraph shows that this is impossible and dipole symmetry is lost.

Let us conclude this section with a couple of straightforward generalizations.
First, one can add $N$ negative $2\pi$ fluxes at the origin instead of just
one flux. This will remove the $n=0,\dots N-1$ wave functions (50). It is an
interesting question what happens to GMP algebra in the limit when $N$ is
large but finite. In particular, can the algebra be approximated as a sum of
bulk and boundary components? Secondly, one can remove $2\pi$ fluxes forming
some lattice $\Lambda$ by subtracting $\sum_{{\bf R},{\bf
R^{\prime}}\in\Lambda}\psi_{\bf R}({\bf x})K_{\Lambda}^{-1}({\bf R},{\bf
R^{\prime}})\psi^{\dagger}_{\bf R^{\prime}}({\bf y})$ from the homogeneous
kernel. Here $\psi_{\bf R}({\bf x})=K({\bf x},{\bf R})$ is a coherent state
which corresponds to a wave function localized around $\bf R$ and
$K_{\Lambda}^{-1}({\bf R},{\bf R^{\prime}})$ is inverse to $K({\bf R},{\bf
R^{\prime}})$ understood as an infinite matrix labeled by discrete indices
${\bf R},{\bf R^{\prime}}\in\Lambda$, i.e. $\sum_{{\bf
R^{\prime\prime}}\in\Lambda}K({\bf R},{\bf
R^{\prime\prime}})K_{\Lambda}^{-1}({\bf R^{\prime\prime}},{\bf
R^{\prime}})=\delta_{{\bf R},{\bf R^{\prime}}}$. The inverse matrix
$K_{\Lambda}^{-1}({\bf R},{\bf R^{\prime}})$ is hard to compute, but one can
do it approximately by assuming that periods of the lattice $\Lambda$ are much
greater than the magnetic length. One can compute the resulting approximate
GMP algebra.

##  V Discussion

Let us discuss the limitation of the theorem. First, at zero temperature, the
condition of finite compressibility $\frac{\partial n}{\partial\mu}<\infty$ is
important with free boson theory being a counter-example. Another important
counter-example is the quantum Hall ferromagnet, where the relevant
susceptibility diverges since it will just reorient the vacuum magnetization.

Secondly, the argument does not forbid condensation of particles as long as
they are neutral under $U(1)$. In particular, the Moore-Read state Moore and
Read (1991) viewed as a superconducting state of composite fermions is not
forbidden. The composite fermion particle number is coupled to an emergent
gauge field which can undergo the Higgs transition and avoid the HMW argument.

A related issue is long-range interactions. One has to assume that the
interactions decay fast enough since otherwise there could be singularities in
the Fourier transform of the two-body interaction $V_{\bf k}$ which will
affect the sum rule.

Lastly, the higher Landau levels can lead to the breaking of $U(1)$ symmetry
but the relevant effect will be suppressed by the gap. One should note that
heating the system to a small non-zero temperature typically will lead to
enhanced fluctuations of the order parameter and further increase the tendency
to symmetry restoration.

Even though spontaneous symmetry breaking is absent and the existence of
massless excitations is not enforced by the Goldstone theorem, one can still
write down the would-be effective theory pretending that the symmetry is
broken. It can still be useful in understanding for example the
Berezinskii–Kosterlitz–Thouless transitions in this system. For the $U(1)$
symmetry breaking the relevant theory should be a non-commutative field theory
similar to the theory Tkachenko mode Du  _et al._ (2022) arising in the
rotating superfluid vortex lattice. It is an interesting problem what happens
in the effective theory after the introduction of long-range Coulomb
interaction. The long-range interaction could gap the would-be Goldstone modes
and prevent their fluctuations from restoring the $U(1)$ symmetry. The
resulting theory may have relevance to the superconductivity in magic-angle
twisted bilayer graphene.

## Appendix

In this appendix, we will set $l_{B}=1$ and use a tilde to distinguish a case
with a flux insertion from a homogeneous one. Everything without a tilde is
assumed to be computed for a homogeneous field. We will use the gauge where
(50) holds.

###  V.1 Homogeneous field

Let us first review the computation in the constant magnetic field. The
Bergman kernel is

$\displaystyle K({\bf x},{\bf y})=\sum_{n\geq 0}\psi_{n}({\bf
x})\psi_{n}^{\dagger}({\bf y})=\frac{1}{2\pi}\exp\Big{(}-\frac{{\bf
x}^{2}}{4}-\frac{{\bf y}^{2}}{4}\Big{)}\exp\Big{(}\frac{{\bf x}\cdot{\bf
y}-i{\bf x}\wedge{\bf y}}{2}\Big{)}$ (59)

where $\psi_{n}$ are given by (50).

It is convenient to write single-particle operators in the LLL in terms of
their kernels

$A({\bf x},{\bf y})=\sum_{n,m\geq 0}\psi_{n}({\bf x})\langle
n|\hat{A}|m\rangle\psi_{m}^{\dagger}({\bf y}).$ (60)

It is straightforward to show that

$\displaystyle(\hat{A}\psi)({\bf x})$ $\displaystyle=\int d^{2}{\bf y}\,A({\bf
x},{\bf y})\psi({\bf y}),$ (61) $\displaystyle\int d^{2}{\bf z}\,K({\bf
x},{\bf z})A({\bf z},{\bf y})$ $\displaystyle=\int d^{2}{\bf z}\,A({\bf
x},{\bf z})K({\bf z},{\bf y})=A({\bf x},{\bf y}),$ (62)
$\displaystyle(\hat{A}\hat{B})({\bf x},{\bf y})$ $\displaystyle=\int d^{2}{\bf
z}\,A({\bf x},{\bf z})B({\bf z},{\bf y}),$ (63) $\displaystyle{\rm
Tr}\,\hat{A}$ $\displaystyle=\int d^{2}{\bf x}\,A({\bf x},{\bf x}),$ (64)
$\displaystyle T_{f}({\bf x},{\bf y})$ $\displaystyle=\int d^{2}{\bf
z}\,K({\bf x},{\bf z})f({\bf z})K({\bf z},{\bf y}),$ (65)

where $\hat{T}_{f}$ is a projection of multiplication by function $f({\bf x})$
to the LLL.

In the homogeneous magnetic field, the magnetic translation operators Zak
(1964) act on wavefunctions as

$\displaystyle\hat{\mathcal{T}}_{\bf a}\psi({\bf
x})=\exp\Big{(}\frac{i}{2}{\bf x}\wedge{\bf a}\Big{)}\psi({\bf x}+{\bf a}).$
(66)

They commute with the Hamiltonian for all ${\bf a}$ and thus with the
projection

$\displaystyle\hat{\mathcal{T}}_{\bf a}\hat{K}=\hat{K}\hat{\mathcal{T}}_{\bf
a},$ (67)

or in terms of kernel

$\displaystyle\exp\Big{(}\frac{i}{2}{\bf x}\wedge{\bf a}\Big{)}K({\bf
x+a},{\bf y})=\exp\Big{(}\frac{i}{2}{\bf y}\wedge{\bf a}\Big{)}K({\bf x},{\bf
y-a}).$ (68)

The magnetic translations satisfy the magnetic translation algebra before and
after the projection

$\hat{\mathcal{T}}_{\bf a}\hat{\mathcal{T}}_{\bf b}=\exp\left(i{{\bf
a}\wedge{\bf b}}\right)\hat{\mathcal{T}}_{\bf b}\hat{\mathcal{T}}_{\bf a}.$
(69)

They act on Toeplitz operators $\hat{T}_{f}$ as

$\displaystyle\hat{\mathcal{T}}_{\bf a}\hat{T}_{f}\hat{\mathcal{T}}_{-\bf
a}=\hat{T}_{f^{\hat{\mathcal{T}}_{\bf a}}},\qquad\text{where
$f^{\hat{\mathcal{T}}_{\bf a}}({\bf x})=f({\bf x}+{\bf a})$}.$ (70)

The projection of the magnetic translation operator is expressible as Toeplitz
operator $\hat{T}_{f_{\bf a}}$ for some function $f_{\bf a}({\bf x})$. From
the magnetic translation algebra (69) we find

$\hat{\mathcal{T}}_{\bf b}\hat{\mathcal{T}}_{\bf a}\hat{\mathcal{T}}_{-\bf
b}=\exp\left(-i{{\bf a}\wedge{\bf b}}\right)\hat{\mathcal{T}}_{\bf a}.$ (71)

By comparing it with (70), we find that up to $\bf x$-independent prefactor
$f_{\bf a}({\bf x})\sim\exp\left(i{\bf x}\wedge{\bf a}\right)$. The prefactor
can be found by projecting $\exp\left(i{\bf x}\wedge{\bf a}\right)$ and
comparing it with the magnetic translation. One finds for the exponential
functions $e_{\bf k}({\bf x})=\exp(-i{\bf k}\cdot{\bf x})$

$\displaystyle T_{e_{\bf k}}({\bf x},{\bf y})=\exp\Big{(}-\frac{{\bf
k}^{2}}{2}+\frac{{\bf k}\wedge({\bf x}-{\bf y})}{2}-\frac{i{\bf k}\cdot({\bf
x}+{\bf y})}{2}\Big{)}K({\bf x},{\bf y}).$ (72)

And we find

$f_{\bf a}({\bf x})=\exp\left(\frac{{\bf a}^{2}}{4}+i{\bf x}\wedge{\bf
a}\right).$ (73)

Thus magnetic translation operators $\hat{\mathcal{T}}_{\bf a}$ are

$\displaystyle\hat{\mathcal{T}}_{\bf a}=\hat{T}_{f_{\bf a}}.$ (74)

We can find a nice basis of unitary operators

$\displaystyle\hat{\tau}_{\bf k}=\hat{\mathcal{T}}_{-\wedge{\bf
k}}=\exp\Big{(}\frac{{\bf k}^{2}}{4}\Big{)}\hat{T}_{e_{-\bf k}}.$ (75)

Labeling the basis in terms of $\bf k$ instead of ${\bf a}=-{\wedge{\bf k}}$
is conventional and chosen here in order to reduce to the standard Fourier
basis $\exp(i{\bf k}\cdot{\bf x})$ in $l_{B}\rightarrow 0$ limit. The kernels
of these operators are

$\displaystyle\tau_{\bf k}({\bf x},{\bf y})=\exp\Big{(}-\frac{{\bf
k}^{2}}{4}-\frac{{\bf k}\wedge({\bf x}-{\bf y})}{2}+\frac{i{\bf k}\cdot({\bf
x}+{\bf y})}{2}\Big{)}K({\bf x},{\bf y}),$ (76)

and they satisfy the identity

$\displaystyle\hat{\tau}_{\bf k}\hat{\tau}_{\bf q}=\exp\Big{(}i\frac{{\bf
k}\wedge{\bf q}}{2}\Big{)}\tau_{{\bf k}+{\bf q}}.$ (77)

They form a basis for the space of operators

$\displaystyle\begin{split}\hat{\tau}^{\dagger}_{\bf k}&=\hat{\tau}_{-\bf
k},\\\ {\rm Tr}\,\Big{(}\hat{\tau}_{\bf k}\hat{\tau}_{\bf
p}\Big{)}&=2\pi\delta({\bf k}+{\bf p}),\\\ \int\frac{d^{2}{\bf
k}}{2\pi}\,\langle n|\hat{\tau}_{\bf k}|m\rangle\langle l|\hat{\tau}_{-\bf
k}|s\rangle&=\delta_{ns}\delta_{ml},\end{split}$ (78)

where $|n\rangle$ is ket notation for the orthonormal basis $\psi_{n}$. Any
operator can be expanded in this basis

$\displaystyle\hat{A}=\int\frac{d^{2}{\bf k}}{2\pi}\,A_{{\bf
k}}\hat{\tau}_{\bf k},$ (79)

where the Fourier components $A_{{\bf k}}$ are defined as

$\displaystyle A_{\bf k}={\rm Tr}(\hat{A}\hat{\tau}_{-{\bf k}})=\int d^{2}{\bf
x}d^{2}{\bf y}\,A({\bf x},{\bf y})\tau_{-\bf k}({\bf y},{\bf x}).$ (80)

Note a useful identity

$\displaystyle\hat{A}=\hat{T}_{f_{\hat{A}}},\qquad\text{where
$f_{\hat{A}}({\bf x})=\int\frac{d^{2}{\bf k}}{2\pi}A_{\bf
k}\exp\Big{(}{\frac{{\bf k}^{2}}{4}+i{\bf k}\cdot{\bf x}}\Big{)}$}.$ (81)

This is one of the main formulas we were aiming for. It allows the expression
of an arbitrary operator in the lowest Landau level in terms of projected
multiplication by a function $\hat{T}_{f}$.

For example, we find

$\displaystyle{\rm Tr}\big{(}[\hat{T}_{e_{\bf k}},\hat{T}_{e_{\bf
q}}]\hat{\tau}_{-{\bf p}}\big{)}=4\pi i\,\exp\Big{(}-{\dfrac{{\bf k}^{2}+{\bf
q}^{2}}{4}}\Big{)}\sin\Big{(}\frac{{\bf k}\wedge{\bf q}}{2}\Big{)}\delta({\bf
p}+{\bf k}+{\bf q}),$ (82)

which leads to

$\displaystyle[\hat{T}_{e_{\bf k}},\hat{T}_{e_{\bf
q}}]=2i\,\exp\Big{(}{\dfrac{{\bf k}\cdot{\bf
q}}{2}}\Big{)}\sin\Big{(}\frac{{\bf k}\wedge{\bf
q}}{2}\Big{)}\,\hat{T}_{e_{{\bf k}+{\bf q}}}.$ (83)

Using the identity

$\displaystyle\overline{\rho}(f)=\int d^{2}{\bf x}d^{2}{\bf
y}\,\overline{\Psi}^{\dagger}({\bf x})T_{f}({\bf x},{\bf
y})\overline{\Psi}^{\dagger}({\bf y})$ (84)

together with (17) and (83), we find the standard GMP algebra (23).

In this subsection, we took a long path that extensively utilizes the
continuous magnetic translation symmetry in order to find Toeplitz operators
corresponding to any projected operator and compute their algebra. This way
once again emphasizes the importance of symmetry in the computability of the
standard GMP algebra. One can use a similar method for any symmetric space. In
the case of an inhomogeneous system, the symmetry disappears along with the
nice basis $\hat{\tau}_{\bf k}$. One can still project the plane waves and
find $\hat{T}_{e_{\bf k}}$, but they won’t be orthogonal to each other in the
sense of eqs. (78).

This machinery allows a straightforward computation of GMP algebra in the case
of a homogeneous field with a negative flux insertion.

###  V.2 Homogeneous field with a negative flux

The kernel can be found as

$\displaystyle\begin{split}\widetilde{K}({\bf x},{\bf y})=\sum_{n\geq
1}\psi_{n}({\bf x})\psi_{n}^{\dagger}({\bf y})&=K({\bf x},{\bf
y})-\psi_{0}({\bf x})\psi_{0}^{\dagger}({\bf y})\\\
&=\frac{1}{2\pi}\exp\Big{(}-\frac{{\bf x}^{2}}{4}-\frac{{\bf
y}^{2}}{4}\Big{)}\left(\exp\Big{(}\frac{{\bf x}\cdot{\bf y}-i{\bf x}\wedge{\bf
y}}{2}\Big{)}-1\right).\end{split}$ (85)

As obvious from the second equality, we can think about projection to the
system with a negative flux as first projecting to the homogeneous LLL and
then projecting out a single state $n=0$. We will use that in the following
and express the operators acting in the $\widetilde{\mathcal{H}}_{LLL}$ as
operators acting in the $\mathcal{H}_{LLL}$ with zero matrix elements with
$n=0$ state.

We can express the kernel in terms of the basis using

$\displaystyle{\rm Tr}\big{(}\widehat{\widetilde{K}}\hat{\tau}_{-{\bf
q}}\big{)}=2\pi\delta({\bf q})-\exp\left({-\frac{\bf q^{2}}{4}}\right).$ (86)

By expressing $\widetilde{K}$ in terms of $\tau_{\bf q}$ the computation of
the GMP algebra becomes a straightforward exercise.

The projection of exponential functions gives

$\displaystyle{\rm Tr}\big{(}\widehat{\widetilde{K}}\hat{T}_{e_{\bf
k}}\widehat{\widetilde{K}}\hat{\tau}_{-{\bf
q}}\big{)}=2\pi\exp\Big{(}-\frac{k^{2}}{4}\Big{)}\delta({\bf k}+{\bf
q})-\exp\Big{(}-\frac{{\bf k}^{2}}{2}-\frac{{\bf
q}^{2}}{4}\Big{)}\left(2\exp\Big{(}-\frac{{\bf k}\cdot{\bf
q}}{2}\Big{)}\cos\Big{(}\frac{{\bf k}\wedge{\bf q}}{2}\Big{)}-1\right),$

or in terms of density operators

$\displaystyle\widetilde{\rho}_{\bf k}=\overline{\rho}_{\bf
k}-\exp\Big{(}-\frac{{\bf k}^{2}}{2}\Big{)}\int\frac{d^{2}{\bf
q}}{2\pi}\left(2\exp\Big{(}\frac{{\bf k}\cdot{\bf
q}}{2}\Big{)}\cos\Big{(}\frac{{\bf k}\wedge{\bf
q}}{2}\Big{)}-1\right)\overline{\rho}_{\bf q}.$ (87)

Note, if we act on both sides of (87) with projections $\widetilde{P}_{LLL}$
it must become a trivial identity. In order for this to be true
$\widetilde{\rho}_{\bf q}$ must satisfy

$\displaystyle\int\frac{d^{2}{\bf q}}{2\pi}\exp\Big{(}\frac{{\bf k}\cdot{\bf
q}}{2}\Big{)}\cos\Big{(}\frac{l_{B}^{2}{\bf k}\wedge{\bf
q}}{2}\Big{)}\widetilde{\rho}_{\bf q}=0,$ (88)

which can be checked directly. Actually, a stronger condition holds. It
originates from the fact that similar self-consistency conditions must hold
when one expresses $\widetilde{\rho}_{\bf k}$ in terms of operators
$\widetilde{\rho}_{\bf k}^{L}$ and $\widetilde{\rho}_{\bf k}^{R}$ which are
$\overline{\rho}_{\bf k}$ projected by $\widetilde{P}_{LLL}$ only on the left
or on the right respectively. The resulting conditions are

$\displaystyle\int\frac{d^{2}{\bf q}}{2\pi}\exp\Big{(}\frac{{\bf q}\cdot{\bf
k}}{2}\pm i\frac{{\bf q}\wedge{\bf k}}{2}\Big{)}\widetilde{\rho}_{\bf
q}=0,\qquad\text{for any $\bf k$}$ (89)

whose again can be checked directly. As clear from the above discussions,
these conditions enforce matrix elements of $\widetilde{\rho}$ to be zero
whenever one of the particles is in $n=0$ state.

Lastly, we want to compute the GMP algebra. There are two ways to do it. One
is to compute ${\rm Tr}\big{(}[\widehat{\widetilde{T}}_{e_{\bf
k}},\widehat{\widetilde{T}}_{e_{\bf q}}]\hat{\tau}_{-{\bf p}}\big{)}$, where
$\widehat{\widetilde{T}}_{e_{\bf k}}=\widehat{\widetilde{K}}\hat{T}_{e_{\bf
k}}\widehat{\widetilde{K}}$. The other way is to use (87) together with
standard GMP algebra relations (23) and constraints (89). Either computation
is lengthy but straightforward. The result is (55).

Consider a translationally and rotationally invariant state (even though the
Hamiltonian is not invariant)

$\displaystyle\langle\widetilde{\rho}_{\bf p}\widetilde{\rho}_{\bf
q}\rangle=(2\pi)^{2}s_{\bf q}\delta({\bf p}+{\bf q}),$ (90)

where we introduced the structure factor $s_{\bf q}$ which depends only on
$|\bf q|$. In this state one finds

$\sum_{i}\langle[D_{i},[\overline{H},D_{i}]]\rangle=\int\frac{d^{2}{\bf
k}}{\pi}4\pi{\bf k}^{2}e^{\frac{-{\bf k}^{2}}{2}}V_{\bf k}s_{\bf
k}+\int\frac{d^{2}{\bf k}d^{2}{\bf p}}{(2\pi)^{4}}{\bf k}^{2}e^{{\bf
p}\cdot{\bf k}{-{\bf p}^{2}}}V_{\bf k}s_{\bf p},$ (91)

which is not zero for general interaction.

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4^{k}{(1+\varepsilon_{1})^{2}}(1-2\xi)^{-2k},\quad K=2\ln(2M/\delta_{2}).$
(153)

Then Lemma 4 combined with the union bound gives that the following holds for
all $i$ with probability at least $1-\delta_{2}$:

$\left|\hat{o}_{i}-\mathbb{E}(\hat{o}_{i}^{(r)})\right|\leq\varepsilon_{2}.$
(154)

Combining Eq. (150) and Eq. (154) using the triangular inequality gives

$|\hat{o}_{i}-\operatorname{Tr}(O_{i}\rho)|\leq\varepsilon_{1}+\varepsilon_{2},$
(155)

which holds with probability at least $1-\delta_{1}-\delta_{2}$. This
completes the proof. ∎

Specifically, if $\xi\ll\frac{1}{2}$ then
$(1-2\xi)^{-2k}=\left((1-2\xi)^{-\frac{1}{2\xi}}\right)^{4k\xi}\approx
e^{4k\xi}$. That is how we get the bound in Theorem 4.

##  Appendix D The effect of state preparation noise

In this section, we will prove Theorem 5 and Theorem 6 in the main text
establishing the robustness of RShadow against state preparation noise in the
calibration procedure. Let’s first fix the notations: We assume $\ket{\bm{0}}$
is experimentally prepared as some other state $\rho_{\bm{0}}$ which is fixed
over time, and we will use a subscript “SP” to denote the state-preparation
noisy version of our estimators. For example,
$\widehat{}\mathcal{M}_{\mathrm{SP}}=\sum_{\lambda\in
R_{\mathbb{G}}}\hat{f}_{\lambda,\mathrm{SP}}\Pi_{\lambda}$ is our estimation
for the physical channel
$\widetilde{}\mathcal{M}\mathrel{\mathop{:}}\nobreak\mkern-1.2mu=\sum_{\lambda\in
R_{\mathbb{G}}}f_{\lambda}\Pi_{\lambda}$ when the calibration process suffers
from state preparation error.

###  D.1 Robustness of RShadow with Global Clifford Group

######  Lemma 10.

For RShadow using ${\sf Cl}(2^{n})$, if the state-preparation fidelity
satisfies

$F(|\bm{0}\rangle\langle\bm{0}|,\rho_{\bm{0}})\geq
1-\varepsilon_{\mathrm{SP}},$ (156)

then the SP-noisy single-round estimator $\hat{f}_{SP}^{(r)}$ satisfies

$\displaystyle
f\geq\mathbb{E}(\hat{f}_{\mathrm{SP}}^{(r)})\geq(1-2\varepsilon_{\mathrm{SP}})f,$
(157)
$\displaystyle\mathrm{Var}(\hat{f}_{\mathrm{SP}}^{(r)})\leq\frac{6d}{(d-1)^{3}}.$

###### Proof.

According to the calibration procedure described in Algorithm 2 or Protocol 1
of App. B , we have

$\displaystyle\mathbb{E}(\hat{F}_{\mathrm{SP}}^{(r)})$
$\displaystyle=\mathbb{E}_{U\sim{\sf
Cl}(2^{n})}\sum_{b}\langle\\!\langle\bm{0}|{\mathcal{U}}^{\dagger}|b\rangle\\!\rangle\langle\\!\langle
b|\Lambda{\mathcal{U}}|\rho_{0}\rangle\\!\rangle$ (158)
$\displaystyle=\langle\\!\langle\bm{0}|~{}\left[|\sigma_{\bm{0}}\rangle\\!\rangle\langle\\!\langle\sigma_{\bm{0}}|+f(I-|\sigma_{\bm{0}}\rangle\\!\rangle\langle\\!\langle\sigma_{\bm{0}}|)\right]~{}|\rho_{0}\rangle\\!\rangle$
$\displaystyle=\frac{1}{d}+f(\bra{\bm{0}}\rho_{\bm{0}}\ket{\bm{0}}-\frac{1}{d}).$
$\displaystyle\mathbb{E}(\hat{f}_{\mathrm{SP}}^{(r)})$
$\displaystyle=\frac{d\mathbb{E}(\hat{F}_{\mathrm{SP}}^{(r)})-1}{d-1}$
$\displaystyle=\frac{d\bra{\bm{0}}\rho_{\bm{0}}\ket{\bm{0}}-1}{d-1}f$
$\displaystyle\geq(1-\varepsilon_{\mathrm{SP}}\frac{d}{d-1})f.$

One can immediately conclude that
$f\geq\mathbb{E}(\hat{f}^{(r)}_{SP})\geq(1-2\varepsilon_{\mathrm{SP}})f$.

The second moment of $\hat{F}^{(r)}_{SP}$ can be written as (see Eq. (78))

$\displaystyle\mathbb{E}(\hat{F}_{\mathrm{SP}}^{(r)^{2}})$
$\displaystyle=\sum_{b\in\\{0,1\\}^{n}}\operatorname{Tr}\left[\mathbb{E}_{U\sim{\sf
Cl}(2^{n})}\left(U\rho_{0}U^{\dagger}\otimes
U\ket{\bm{0}}\bra{\bm{0}}U^{\dagger}\otimes
U\ket{\bm{0}}\bra{\bm{0}}U^{\dagger}\right)~{}\left(\Lambda^{\dagger}(\ket{b}\bra{b})\otimes\ket{b}\bra{b}\otimes\ket{b}\bra{b}\right)\right]$
(159)
$\displaystyle=\sum_{b\in\\{0,1\\}^{n}}\operatorname{Tr}\left[\Phi_{\mathrm{Haar}}^{(3)}(\rho_{0}\otimes|\bm{0}\rangle\langle\bm{0}|\otimes|\bm{0}\rangle\langle\bm{0}|)~{}\left(\Lambda^{\dagger}(\ket{b}\bra{b})\otimes\ket{b}\bra{b}\otimes\ket{b}\bra{b}\right)\right]$
$\displaystyle=\sum_{b\in\\{0,1\\}^{n}}\sum_{\pi,\sigma\in
S_{3}}c_{\pi,\sigma}\operatorname{Tr}\left[W_{\pi}(\rho_{0}\otimes|\bm{0}\rangle\langle\bm{0}|\otimes|\bm{0}\rangle\langle\bm{0}|)\right]\operatorname{Tr}\left[W_{\sigma}\left(\Lambda^{\dagger}(\ket{b}\bra{b})\otimes\ket{b}\bra{b}\otimes\ket{b}\bra{b}\right)\right]$
$\displaystyle=\frac{2(d-2F_{Z}-2F_{0}+2dF_{Z}F_{0})}{(d^{2}-1)(d+2)}$
$\displaystyle\leq\frac{6d}{(d^{2}-1)(d+2)},$

where we define
$F_{0}\mathrel{\mathop{:}}\nobreak\mkern-1.2mu=\bra{\bm{0}}\rho_{0}\ket{\bm{0}}$
and $F_{Z}\mathrel{\mathop{:}}\nobreak\mkern-1.2mu=F_{Z}(\Lambda)$. Therefore,

$\mathrm{Var}(\hat{f}_{\mathrm{SP}}^{(r)})=\frac{d^{2}}{(d-1)^{2}}\mathrm{Var}(\hat{F}_{\mathrm{SP}}^{(r)})\leq\frac{d^{2}}{(d-1)^{2}}\mathbb{E}(\hat{F}_{\mathrm{SP}}^{(r)^{2}})\leq\frac{6d}{(d-1)^{3}}.$
(160)

∎

The following theorem is a more detailed formalisation of Theorem 5 in the
main text.

######  Theorem 11.

For RShadow using $\sf{Cl}(2^{n})$, if the state-preparation fidelity
satisfies

$F(\ket{\bm{0}}\bra{\bm{0}},\rho_{\bm{0}})\geq 1-\varepsilon_{\mathrm{SP}},$
(161)

then with $R=\tilde{}\mathcal{O}(\varepsilon^{-2}F_{Z}^{-2})$ calibration
samples, the subsequent estimation procedure with high probability satisfies

$\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|\leq(\varepsilon+2\varepsilon_{\mathrm{SP}})\|O\|_{\infty}.$
(162)

up to the first order of $\varepsilon$ and $\varepsilon_{\mathrm{SP}}$ for any
observable $O$. We have assumed
$F_{Z}\mathrel{\mathop{:}}\nobreak\mkern-1.2mu=F_{Z}(\Lambda)\gg 1/d$.

###### Proof.

First notice that the target function can be upper bounded as

$\displaystyle\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|$
$\displaystyle=\left|\langle\\!\langle
O|\widehat{}\mathcal{M}_{\mathrm{SP}}^{-1}\widetilde{}\mathcal{M}-1|\rho\rangle\\!\rangle\right|$
(163) $\displaystyle=\left|\langle\\!\langle
O_{0}|\widehat{}\mathcal{M}_{\mathrm{SP}}^{-1}\widetilde{}\mathcal{M}-1|\rho\rangle\\!\rangle\right|$
$\displaystyle\leq\left|\langle\\!\langle
O_{0}|\rho\rangle\\!\rangle\right|\cdot\left|\hat{f}_{\mathrm{SP}}^{-1}f-1\right|$
$\displaystyle\leq\left\|O\right\|_{\infty}\cdot\left|\hat{f}_{\mathrm{SP}}^{-1}f-1\right|.$

According to Lemma 4, by taking the parameters of the median of mean
estimators as

$\displaystyle N$
$\displaystyle=34\mathrm{Var}(\hat{f}^{(r)}_{\mathrm{SP}})\varepsilon^{-2}f^{-2},$
(164) $\displaystyle K$ $\displaystyle=2\ln(2\delta^{-1}),$

the following holds with probability at least $1-\delta$,

$\left|\hat{f}_{\mathrm{SP}}-\mathbb{E}(\hat{f}_{\mathrm{SP}}^{(r)})\right|\leq\varepsilon
f.$ (165)

We also have, from Lemma 10, that

$\left|\mathbb{E}(\hat{f}_{\mathrm{SP}}^{(r)})-f\right|\leq
2\varepsilon_{\mathrm{SP}}f.$ (166)

Therefore, our final bound is as claimed

$\displaystyle\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|$
$\displaystyle\leq\|O\|_{\infty}\cdot\frac{|f-\hat{f}_{\mathrm{SP}}|}{|\hat{f}_{\mathrm{SP}}|}$
(167)
$\displaystyle\leq\|O\|_{\infty}\cdot\frac{\varepsilon+2\varepsilon_{\mathrm{SP}}}{1-\varepsilon-2\varepsilon_{\mathrm{SP}}}$
$\displaystyle=\|O\|_{\infty}\cdot(\varepsilon+2\varepsilon_{\mathrm{SP}}+o(\varepsilon+2\varepsilon_{\mathrm{SP}})).$

The sample complexity is

$R=NK\leq 2\ln(2\delta^{-1})\cdot
204\varepsilon^{-2}(F_{Z}-1/d)^{-2}\frac{(d+1)^{2}}{d(d-1)}=\tilde{}\mathcal{O}(\varepsilon^{-2}F_{Z}^{-2}),$
(168)

for $F_{Z}\mathrel{\mathop{:}}\nobreak\mkern-1.2mu=F_{Z}(\Lambda)\gg 1/d$.
Here we have used Lemma 10 and Proposition 1 to bound
$\mathrm{Var}(\hat{f}^{(r)}_{\mathrm{SP}})$ and $f$, respectively. ∎

###  D.2 Robustness of RShadow with Local Clifford Group

Note that, we consider a local state-preparation noise model for the results
in this section, i.e., no cross-talk between qubits.

######  Lemma 11.

For RShadow using ${\sf Cl}_{2}^{\otimes n}$, if the prepared state is in a
product form, i.e., $\rho_{\bm{0}}=\bigotimes_{i=1}^{n}\rho_{0,i}$, and the
single-qubit state-preparation fidelity satisfies

$F(|0\rangle\langle 0|,\rho_{0,i})\geq
1-\xi_{\mathrm{SP}},\quad\forall~{}i\in[n],$ (169)

for some $\xi_{\mathrm{SP}}<1/2$, then the SP-noisy single-round estimator
$\hat{f}^{(r)}_{z,\mathrm{SP}}$ satisfies

$\displaystyle
f_{z}\geq\mathbb{E}(\hat{f}^{(r)}_{z,\mathrm{SP}})\geq(1-2\xi_{\mathrm{SP}}|z|)f_{z},$
(170) $\displaystyle\mathrm{Var}(\hat{f}^{(r)}_{z,\mathrm{SP}})\leq
3^{-|z|},\quad\forall z\in\\{0,1\\}^{n}.$

###### Proof.

According to the calibration procedure described in Algorithm 2 or Protocol 2
of App. C , we have

$\displaystyle\mathbb{E}(\hat{f}^{(r)}_{z,\mathrm{SP}})$
$\displaystyle=\mathbb{E}_{U\sim{\sf Cl}_{2}^{\otimes
n}}\sum_{b}\langle\\!\langle
P_{z}|{\mathcal{U}}^{\dagger}|b\rangle\\!\rangle\langle\\!\langle
b|\Lambda{\mathcal{U}}|\rho_{\bm{0}}\rangle\\!\rangle$ (171)
$\displaystyle=\langle\\!\langle
P_{z}|\sum_{m\in\\{0,1\\}^{n}}f_{m}\Pi_{m}|\rho_{\bm{0}}\rangle\\!\rangle$
$\displaystyle=f_{z}\langle\\!\langle P_{z}|\rho_{\bm{0}}\rangle\\!\rangle$
$\displaystyle=f_{z}\prod_{i:z_{i}=1}(2\bra{0}\rho_{0,i}\ket{0}-1)$
$\displaystyle\geq(1-2|z|\,\xi_{\mathrm{SP}})f_{z}.$

One can immediately conclude that
$f_{z}\geq\mathbb{E}(\hat{f}^{(r)}_{z,\mathrm{SP}})\geq(1-2|z|\,\xi_{\mathrm{SP}})f_{z}$.

To calculate the second moment,

$\mathbb{E}(\hat{f}^{(r)^{2}}_{z,\mathrm{SP}})=\sum_{b}\operatorname{Tr}\left[\mathbb{E}_{U\sim{\sf
Cl}_{2}^{\otimes n}}\left(U\rho_{\bm{0}}U^{\dagger}\otimes
UP_{z}U^{\dagger}\otimes
UP_{z}U^{\dagger}\right)\left(\Lambda^{\dagger}(\ket{b}\bra{b})\otimes\ket{b}\bra{b}\otimes\ket{b}\bra{b}\right)\right],$
(172)

we can first investigate the single-qubit case:

$\displaystyle\mathbb{E}_{U\sim{\sf
Cl}_{2}}\left(U\rho_{0,i}U^{\dagger}\otimes UP_{I}U^{\dagger}\otimes
UP_{I}U^{\dagger}\right)$ $\displaystyle=\frac{1}{2}I_{2}^{\otimes 3},$ (173)
$\displaystyle\mathbb{E}_{U\sim{\sf
Cl}_{2}}\left(U\rho_{0,i}U^{\dagger}\otimes UP_{Z}U^{\dagger}\otimes
UP_{Z}U^{\dagger}\right)$
$\displaystyle=\Phi^{(3)}_{\mathrm{Haar}}(\rho_{0,i}\otimes P_{Z}\otimes
P_{Z}),$

To further simplify the second expressions, one can verify that

$\operatorname{Tr}(\vec{W}(\rho_{0,i}\otimes P_{Z}\otimes
P_{Z}))=\begin{bmatrix}0&0&0&2&1&1\end{bmatrix},$ (174)

where $\vec{W}$ is defined in Eq. (59). Calculating the Haar integral using
Eq. (56), one immediately notice that the form of $\rho_{0,i}$ has nothing to
do with the result. So we can safely replace all $\rho_{0,i}$ with
$\ket{0}\bra{0}$ and retrieve the result with no state preparation error:
$\mathbb{E}(\hat{f}^{(r)^{2}}_{z,\mathrm{SP}})=\mathbb{E}(\hat{f}_{z}^{(r)^{2}})=3^{-|z|}$
and hence $\mathrm{Var}(\hat{f}^{(r)}_{z,\mathrm{SP}})\leq 3^{-|z|}$. ∎

The following theorem is a more detailed formalisation of Theorem 6 in the
main text.

######  Theorem 12.

For RShadow using $\sf{Cl}_{2}^{\otimes n}$, if the state is prepared as some
product state $\rho_{\bm{0}}=\bigotimes_{i=1}^{n}\rho_{0,i}$ and the single-
qubit state-preparation fidelity satisfies

$F(\ket{0}\bra{0},\rho_{0,i})\geq 1-\xi_{\mathrm{SP}},\quad\forall i\in[n],$
(175)

then with $R=\tilde{}\mathcal{O}(3^{k}\varepsilon^{-2}F_{Z}^{-2})$ calibration
samples, the subsequent estimation procedure with high probability satisfies

$\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|\leq(\varepsilon+2k\xi_{SP})2^{k}\|O\|_{\infty}.$
(176)

up to the first order of $\varepsilon$ and $k\xi_{SP}$, for any $k$-local
observable $O$.

###### Proof.

Suppose $O$ is a $k$-local observable for some $k$. Following exactly the same
procedure as in the proof of Theorem 9 (see Eq. (129)), we can bound our
target function as follows,

$\displaystyle\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|$
$\displaystyle\leq 2^{k}\|O\|_{\infty}\cdot\max_{|z|\leq
k}\left|\hat{f}^{-1}_{z,\mathrm{SP}}f_{z}-1\right|.$ (177)

According to Lemma 4, by taking the parameters of the median of mean
estimators as

$\displaystyle N$ $\displaystyle=\max_{|z|\leq
k}34\mathrm{Var}(\hat{f}^{(r)}_{z,\mathrm{SP}})\varepsilon^{-2}f_{z}^{-2},$
(178) $\displaystyle K$ $\displaystyle=2\ln(2(\delta/n^{k})^{-1}),$

the following holds with probability at least $1-\delta/n^{k}$ for any $z$
whose weight is no larger than $k$, hence simultaneously holds for all such
$z$ with probability at least $1-\delta$ by the union bound:

$\left|\hat{f}_{z,\mathrm{SP}}-\mathbb{E}(\hat{f}^{(r)}_{z,\mathrm{SP}})\right|\leq\varepsilon
f_{z},\quad\forall~{}z\in\\{0,1\\}^{n}:|z|\leq k.$ (179)

We also have, from Lemma 11, that

$\left|\mathbb{E}(\hat{f}_{z,\mathrm{SP}}^{(r)})-f_{z}\right|\leq
2\xi_{\mathrm{SP}}|z|f_{z},\quad\forall~{}z\in\\{0,1\\}^{n}.$ (180)

Therefore, our final bound is as claimed

$\displaystyle\left|\mathbb{E}(\hat{o}^{(r)})-\operatorname{Tr}(O\rho)\right|$
$\displaystyle\leq 2^{k}\|O\|_{\infty}\cdot\max_{|z|\leq
k}\frac{|f_{z}-\hat{f}_{\mathrm{z,SP}}|}{|\hat{f}_{\mathrm{z,SP}}|}$ (181)
$\displaystyle\leq
2^{k}\|O\|_{\infty}\cdot\frac{\varepsilon+2k\xi_{\mathrm{SP}}}{1-\varepsilon-2k\xi_{\mathrm{SP}}}$
$\displaystyle=2^{k}\|O\|_{\infty}\cdot(\varepsilon+2k\xi_{\mathrm{SP}}+o(\varepsilon+2k\xi_{\mathrm{SP}})).$

The sample complexity is

$\displaystyle R=NK$ $\displaystyle\leq 2\ln(2\delta^{-1}n^{k})\cdot 34\cdot
3^{k}\varepsilon^{-2}\Gamma_{z}(\Lambda)^{-2}$ (182) $\displaystyle\leq
2\ln(2\delta^{-1}n^{k})\cdot 34\cdot 3^{k}\varepsilon^{-2}F_{Z}(\Lambda)^{-2}$
$\displaystyle=\tilde{}\mathcal{O}(3^{k}\varepsilon^{-2}F_{Z}^{-2}),$

where we have used Lemma 11 and Proposition 2 to bound
$\mathrm{Var}(\hat{f}^{(r)}_{z,\mathrm{SP}})$ and $f_{z}$, respectively. The
second inequality is by Lemma 7. We remark that one can alternatively use a
stronger bound given in Lemma 8 when the noise model is assumed to be local. ∎

##  Appendix E More numerical results

Here, we present more numerical results to show the performance of RShadow in
the task of estimating the 2-point correlation function of the $5$-qubit GHZ
state using local Clifford gates. We perform the estimation task under two
different noise models: single-qubit $X$-axis rotation and two-qubit $XX$
cross-talk noises. When the single-qubit $X$-axis rotation error happens, each
qubit will experience a coherent rotation after the implementation of the
random unitary gate $R_{X}(\theta)=e^{-i\theta X}$, where
$\theta=\frac{k\pi}{40}$, $k=0,1,2,3,4,5$. When the two-qubit $XX$ cross-talk
noise happens, each two adjacent qubits will experience a coherent rotation
after the implementation of the random unitary gate
$R_{XX}(\theta)=e^{-i\theta XX}$, where $\theta=\frac{3k\pi}{100}$,
$k=0,1,2,3,4,5$. For clarity, we estimate the 2-point correlation function of
the $5$-th qubit with every other qubit and plot the average values.

$0$$5$$10$$15$$20$$25$$0.4$$0.6$$0.8$$1$$1.2$$\text{Rotation angle
}\theta\text{ * 200/}\pi$Estimated valueSingle-qubit X-axis rotation errorTrue
ValueStd-
ShadowRShadow$0$$3$$6$$9$$12$$15$$0.2$$0.4$$0.6$$0.8$$1$$1.2$$\text{Rotation
angle }\theta\text{ * 100/}\pi$Estimated valueTwo-qubit coupled XX rotation
errorTrue ValueStd-ShadowRShadow Figure 7: $5$-qubit GHZ 2-point correlation
function estimation under the coherent noise models, including single-qubit
$X$-axis rotation and two-qubit $XX$-coupling noise.


1 0 :
1 2022 Machine Learning for Health (ML4H) 2022

# Full Title of ArticleThis Title Has A Line Break††thanks: sample footnote

###### Abstract

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###### keywords:

List of keywords

††editors: List of editors’ names

##  1 Introduction

This is a sample article that uses the jmlr class with the pmlr class option.
Please follow the guidelines in this sample document as it can help to reduce
complications when combining the articles into a book. Please avoid using
obsolete commands, such as `\rm`, and obsolete packages, such as epsfig.111See
http://www.ctan.org/pkg/l2tabu

Please also ensure that your document will compile with PDFLaTeX. If you have
an error message that’s puzzling you, first check for it at the UK TUG FAQ
https://texfaq.org/FAQ-man-latex. If that doesn’t help, create a minimal
working example (see https://www.dickimaw-books.com/latex/minexample) and post
to somewhere like TeX on StackExchange (https://tex.stackexchange.com/) or the
LaTeX Community Forum (https://latex.org/forum/).

###### Note

This is an numbered theorem-like environment that was defined in this
document’s preamble.

###  1.1 Sub-sections

Sub-sections are produced using `\subsection`.

####  1.1.1 Sub-sub-sections

Sub-sub-sections are produced using `\subsubsection`.

##### Sub-sub-sub-sections

Sub-sub-sub-sections are produced using `\paragraph`. These are unnumbered
with a running head.

###### Sub-sub-sub-sub-sections

Sub-sub-sub-sub-sections are produced using `\subparagraph`. These are
unnumbered with a running head.

##  2 Cross-Referencing

Always use `\label` and `\ref` (or one of the commands described below) when
cross-referencing. For example, the next section is Section 3. The jmlr class
provides some convenient cross-referencing commands: `\sectionref`,
`\equationref`, `\tableref`, `\figureref`, `\algorithmref`, `\theoremref`,
`\lemmaref`, `\remarkref`, `\corollaryref`, `\definitionref`,
`\conjectureref`, `\axiomref`, `\exampleref` and `\appendixref`. The argument
of these commands may either be a single label or a comma-separated list of
labels. Examples:

Referencing sections: sec:math or sec:intro,sec:math or
sec:intro,sec:math,sec:tables,sec:figures.

Referencing equations: eq:trigrule or eq:trigrule,eq:df or
eq:trigrule,eq:f,eq:df,eq:y.

Referencing tables: tab:operatornames or tab:operatornames,tab:example or
tab:operatornames,tab:example,tab:example-booktabs.

Referencing figures: fig:image or fig:image,fig:teximage or
fig:image,fig:teximage,fig:subfigex or fig:image-a,fig:image-b.

Referencing algorithms: alg:gauss or alg:gauss,alg:moore or
alg:gauss,alg:moore,alg:net.

Referencing theorem-like environments: thm:eigenpow, lem:sample, rem:sample,
cor:sample, def:sample, con:sample, ax:sample and ex:sample.

Referencing appendices: apd:first or apd:first,apd:second.

##  3 Equations

The jmlr class loads the amsmath package, so you can use any of the commands
and environments defined there. (See the amsmath documentation for further
details.222Either texdoc amsmath or http://www.ctan.org/pkg/amsmath)

Unnumbered single-lined equations should be displayed using `\[` and `\]`. For
example:

$E=mc^{2}$

Numbered single-line equations should be displayed using the equation
environment. For example:

$\cos^{2}\theta+\sin^{2}\theta\equiv 1$ (1)

This can be referenced using `\label` and `\equationref`. For example,
eq:trigrule.

Multi-lined numbered equations should be displayed using the align
environment.333For reasons why you shouldn’t use the obsolete eqnarray
environment, see Lars Madsen, _Avoid eqnarray!_ TUGboat 33(1):21–25, 2012. For
example:

$\displaystyle f(x)$ $\displaystyle=x^{2}+x$ (2) $\displaystyle f^{\prime}(x)$
$\displaystyle=2x+1$ (3)

Unnumbered multi-lined equations should be displayed using the align*
environment. For example:

$\displaystyle f(x)$ $\displaystyle=(x+1)(x-1)$ $\displaystyle=x^{2}-1$

If you want to mix numbered with unnumbered lines use the align environment
and suppress unwanted line numbers with `\nonumber`. For example:

$\displaystyle y$ $\displaystyle=x^{2}+3x-2x+1$ $\displaystyle=x^{2}+x+1$ (4)

An equation that is too long to fit on a single line can be displayed using
the split environment. Text can be embedded in an equation using `\text` or
`\intertext` (as used in thm:eigenpow). See the amsmath documentation for
further details.

###  3.1 Operator Names

Predefined operator names are listed in tab:operatornames. For additional
operators, either use `\operatorname`, for example $\operatorname{var}(X)$ or
declare it with `\DeclareMathOperator`, for example

    \DeclareMathOperator{\var}{var}

and then use this new command. If you want limits that go above and below the
operator (like `\sum`) use the starred versions (`\operatorname*` or
`\DeclareMathOperator*`).

tab:operatornames  \arccos $\arccos$ \deg $\deg$ \lg $\lg$ \projlim $\projlim$
\arcsin $\arcsin$ \det $\det$ \lim $\lim$ \sec $\sec$ \arctan $\arctan$ \dim
$\dim$ \liminf $\liminf$ \sin $\sin$ \arg $\arg$ \exp $\exp$ \limsup $\limsup$
\sinh $\sinh$ \cos $\cos$ \gcd $\gcd$ \ln $\ln$ \sup $\sup$ \cosh $\cosh$ \hom
$\hom$ \log $\log$ \tan $\tan$ \cot $\cot$ \inf $\inf$ \max $\max$ \tanh
$\tanh$ \coth $\coth$ \injlim $\injlim$ \min $\min$ \csc $\csc$ \ker $\ker$
\Pr $\Pr$

Table 1: Predefined Operator Names (taken from amsmath documentation)

\varlimsup | $\varlimsup$ | \varinjlim | $\varinjlim$
---|---|---|---
\varliminf | $\varliminf$ | \varprojlim | $\varprojlim$

##  4 Vectors and Sets

Vectors should be typeset using \vec. For example $\vec{x}$. The jmlr class
also provides \set to typeset a set. For example $\set{S}$.

##  5 Floats

Floats, such as figures, tables and algorithms, are moving objects and are
supposed to float to the nearest convenient location. Please don’t force them
to go in a particular place. In general it’s best to use the htbp specifier
and don’t put the figure or table in the middle of a paragraph (that is make
sure there’s a paragraph break above and below the float). Floats are supposed
to have a little extra space above and below them to make them stand out from
the rest of the text. This extra spacing is put in automatically and shouldn’t
need modifying.

To ensure consistency, please _don’t_ try changing the format of the caption
by doing something like:

    \caption{\textit{A Sample Caption.}}

or

    \caption{\em A Sample Caption.}

You can, of course, change the font for individual words or phrases, for
example:

    \caption{A Sample Caption With
    Some \emph{Emphasized Words}.}

###  5.1 Tables

Tables should go in the table environment. Within this environment use
`\floatconts` (defined by jmlr) to set the caption correctly and center the
table contents.

tab:example  Dataset Result Data1 0.12345 Data2 0.67890 Data3 0.54321 Data4
0.09876

Table 2: An Example Table

If you want horizontal rules you can use the booktabs package which provides
the commands `\toprule`, `\midrule` and `\bottomrule`. For example, see
tab:example-booktabs.

tab:example-booktabs  Dataset Result Data1 0.12345 Data2 0.67890 Data3 0.54321
Data4 0.09876

Table 3: A Table With Horizontal Lines

If you want vertical lines as well, you can’t use the booktabs commands as
there’ll be some unwanted gaps. Instead you can use LaTeX’s `\hline`, but the
rows may appear a bit cramped. You can add extra space above or below a row
using `\abovestrut` and `\belowstrut`. For example, see tab:example-hline.

tab:example-hline  2.2exDataset Result 2.2exData1 0.12345 Data2 0.67890 Data3
0.54321 0.2exData4 0.09876

Table 4: A Table With Horizontal and Vertical Lines

If you want to align numbers on their decimal point, you can use the siunitx
package. For example, see tab:example-siunitx. For further details see the
siunitx documentation444Either texdoc siunitx or
http://www.ctan.org/pkg/siunitx.

tab:example-siunitx  Dataset Result Data1 $0.123\,45$ Data2 $10.6789$ Data3
$50.543$ Data4 $200.098\,76$

Table 5: A Table With Numbers Aligned on the Decimal Point

If the table is too wide, you can adjust the inter-column spacing by changing
the value of `\tabcolsep`. For example:

    \setlength{\tabcolsep}{3pt}

If the table is very wide but not very long, you can use the sidewaystable
environment defined in the rotating package (so use `\usepackage{rotating}`).
If the table is too long to fit on a page, you should use the longtable
environment defined in the longtable package (so use
`\usepackage{longtable}`).

###  5.2 Figures

Figures should go in the figure environment. Within this environment, use
`\floatconts` to correctly position the caption and center the image. Use
`\includegraphics` for external graphics files but omit the file extension. Do
not use `\epsfig` or `\psfig`. If you want to scale the image, it’s better to
use a fraction of the line width rather than an explicit length. For example,
see fig:image.

fig:image

Figure 1: Example Image

If your image is made up of LaTeX code (for example, commands provided by the
pgf package) you can include it using \includeteximage (defined by the jmlr
class). This can be scaled and rotated in the same way as \includegraphics.
For example, see fig:teximage.

fig:teximage [angle=45]example-teximage

Figure 2: Image Created Using LaTeX Code

If the figure is too wide to fit on the page, you can use the sidewaysfigure
environment defined in the rotating package.

It’s best not to use `\graphicspath` with the jmlr class as it can cause
problems with the production editing process. If the images are contained in a
subdirectory, specify this when you include the image, for example
`\includegraphics{figures/mypic}`.

###  5.3 Sub-Figures

Sub-figures can be created using `\subfigure`, which is defined by the jmlr
class. The optional argument allows you to provide a subcaption. The label
should be placed in the mandatory argument of `\subfigure`. You can reference
the entire figure, for example fig:subfigex, or you can reference part of the
figure using `\figureref`, for example fig:image-a. Alternatively you can
reference the subfigure using `\subfigref`, for example
fig:image-a,fig:image-b in fig:subfigex.

fig:subfigex [Image A] [Image B]

Figure 3: An Example With Sub-Figures.

By default, the sub-figures are aligned on the baseline. This can be changed
using the second optional argument of `\subfigure`. This may be t (top), c
(centered) or b (bottom). For example, the subfigures
fig:image-a2,fig:image-b2 in fig:subfigex2 both have `[c]` as the second
optional argument.

fig:subfigex2 [Image A][c] [Image B][c]

Figure 4: Another Example With Sub-Figures (with a Cramped Sub-Caption).

Note that a very narrow sub-float will leave little space for the sub-caption,
which will likely cause Underfull/Overfull hbox warnings.

You can set the length `\jmlrminsubcaptionwidth` to indicate the minimum width
to be made available for the sub-captions, as in fig:subfigex3.

fig:subfigex3 [Image A][c] [Image B][c]

Figure 5: Another Example With Sub-Figures (with a Less Cramped Sub-Caption).

###  5.4 Sub-Tables

There is an analogous command `\subtable` for sub-tables. It has the same
syntax as `\subfigure` described above. You can reference the table using
`\tableref`, for example tab:subtabex or you can reference part of the table,
for example tab:ab. Alternatively you can reference the subtable using
`\subtabref`, for example tab:ab,tab:cd in tab:subtabex.

tab:subtabex  A B 1 2 C D 3 4 5 6

Table 6: An Example With Sub-Tables

By default, the sub-tables are aligned on the top. This can be changed using
the second optional argument of `\subtable`. This may be t (top), c (centered)
or b (bottom). For example, the sub-tables tab:ab2,tab:cd2 in tab:subtabex2
both have `[c]` as the second optional argument.

tab:subtabex2 [][c]  A B 1 2 [][c]  C D 3 4 5 6

Table 7: Another Example With Sub-Tables

###  5.5 Algorithms

Enumerated textual algorithms can be displayed using the algorithm
environment. Within this environment, use use an enumerate or nested enumerate
environments. For example, see alg:gauss. Note that algorithms float like
figures and tables.

alg:gauss

  1. 1.

For $k=1$ to maximum number of iterations

    1. (a)

For $i=1$ to $n$

      1. i.

$x_{i}^{(k)}=\frac{b_{i}-\sum_{j=1}^{i-1}a_{ij}x_{j}^{(k)}-\sum_{j=i+1}^{n}a_{ij}x_{j}^{(k-1)}}{a_{ii}}$

      2. ii.

If $\|\vec{x}^{(k)}-\vec{x}^{(k-1)}<\epsilon\|$, where $\epsilon$ is a
specified stopping criteria, stop.

Algorithm 1 The Gauss-Seidel Algorithm

If you’d rather have the same numbering throughout the algorithm but still
want the convenient indentation of nested enumerate environments, you can use
the enumerate* environment provided by the jmlr class. For example, see
alg:moore.

alg:mooreGiven a connected graph $G$, where the length of each edge is 1:

Set the label of vertex $s$ to 0

Set $i=0$

Locate all unlabelled vertices adjacent to a vertex labelled $i$ and label
them $i+1$

If vertex $t$ has been labelled,

the shortest path can be found by backtracking, and the length is given by the
label of $t$. otherwise

increment $i$ and return to step 2

Algorithm 2 Moore’s Shortest Path

Pseudo code can be displayed using the algorithm2e environment. This is
defined by the algorithm2e package (which is automatically loaded) so check
the algorithm2e documentation for further details.555Either texdoc algorithm2e
or http://www.ctan.org/pkg/algorithm2e For an example, see alg:net.

Input: $x_{1},\ldots,x_{n},w_{1},\ldots,w_{n}$

Output: $y$, the net activation

$y\leftarrow 0$;

for  _$i\leftarrow 1$ to $n$_ do

$y\leftarrow y+w_{i}*x_{i}$;

end for

Algorithm 3 Computing Net Activation

##  6 Description Lists

The jmlr class also provides a description-like environment called
altdescription. This has an argument that should be the widest label in the
list. Compare:

add

A method that adds two variables.

differentiate

A method that differentiates a function.

with differentiate

A method that adds two variables.

A method that differentiates a function.

##  7 Theorems, Lemmas etc

The following theorem-like environments are predefined by the jmlr class:
theorem, example, lemma, proposition, remark, corollary, definition,
conjecture and axiom. You can use the proof environment to display the proof
if need be, as in thm:eigenpow.

######  Theorem 7.1 (Eigenvalue Powers).

If $\lambda$ is an eigenvalue of $\vec{B}$ with eigenvector $\vec{\xi}$, then
$\lambda^{n}$ is an eigenvalue of $\vec{B}^{n}$ with eigenvector $\vec{\xi}$.

######  Proof 7.2.

Let $\lambda$ be an eigenvalue of $\vec{B}$ with eigenvector $\xi$, then

$\displaystyle\vec{B}\vec{\xi}$ $\displaystyle=\lambda\vec{\xi}$ premultiply
by $\vec{B}$: $\displaystyle\vec{B}\vec{B}\vec{\xi}$
$\displaystyle=\vec{B}\lambda\vec{\xi}$
$\displaystyle\Rightarrow\vec{B}^{2}\vec{\xi}$
$\displaystyle=\lambda\vec{B}\vec{\xi}$
$\displaystyle=\lambda\lambda\vec{\xi}\qquad\text{since
}\vec{B}\vec{\xi}=\lambda\vec{\xi}$ $\displaystyle=\lambda^{2}\vec{\xi}$

Therefore true for $n=2$. Now assume true for $n=k$:

$\displaystyle\vec{B}^{k}\vec{\xi}$ $\displaystyle=\lambda^{k}\vec{\xi}$
premultiply by $\vec{B}$: $\displaystyle\vec{B}\vec{B}^{k}\vec{\xi}$
$\displaystyle=\vec{B}\lambda^{k}\vec{\xi}$
$\displaystyle\Rightarrow\vec{B}^{k+1}\vec{\xi}$
$\displaystyle=\lambda^{k}\vec{B}\vec{\xi}$
$\displaystyle=\lambda^{k}\lambda\vec{\xi}\qquad\text{since
}\vec{B}\vec{\xi}=\lambda\vec{\xi}$ $\displaystyle=\lambda^{k+1}\vec{\xi}$

Therefore true for $n=k+1$. Therefore, by induction, true for all $n$.

######  Lemma 7.3 (A Sample Lemma).

This is a lemma.

######  Remark 7.4 (A Sample Remark).

This is a remark.

######  Corollary 7.5 (A Sample Corollary).

This is a corollary.

######  Definition 7.6 (A Sample Definition).

This is a definition.

######  Conjecture 7.7 (A Sample Conjecture).

This is a conjecture.

######  Axiom 1 (A Sample Axiom).

This is an axiom.

######  Example 7.8 (An Example).

This is an example.

##  8 Citations and Bibliography

The jmlr class automatically loads natbib. This sample file has the citations
defined in the accompanying BibTeX file pmlr-sample.bib. For a parenthetical
citation use `\citep`. For example (guyon-elisseeff-03). For a textual
citation use `\citet`. For example guyon2007causalreport. Both commands may
take a comma-separated list, for example guyon-elisseeff-03;
guyon2007causalreport.

These commands have optional arguments and have a starred version. See the
natbib documentation for further details.666Either texdoc natbib or
http://www.ctan.org/pkg/natbib

The bibliography is displayed using `\bibliography`.

Acknowledgements go here.

##  Appendix A First Appendix

This is the first appendix.

##  Appendix B Second Appendix

This is the second appendix.