# Implementing Jastrow–Gutzwiller operators on a quantum computer using the
cascaded variational quantum eigensolver algorithm

John P. T. Stenger  U.S. Naval Research Laboratory, Washington, DC 20375,
United States  C. Stephen Hellberg  U.S. Naval Research Laboratory,
Washington, DC 20375, United States  Daniel Gunlycke  U.S. Naval Research
Laboratory, Washington, DC 20375, United States

###### Abstract

A Jastrow–Gutzwiller operator adds many-body correlations to a quantum state.
However, the operator is non-unitary, making it difficult to implement
directly on a quantum computer. We present a novel implementation of the
Jastrow–Gutzwiller operator using the cascaded variational quantum eigensolver
algorithm. We demonstrate the method on IBM Q Lagos for a Hubbard model.

##  I Introduction

Quantum computers access exponentially large Hilbert spaces that cannot be
efficiently simulated using classical computing [1, 2]. Significant effort has
been invested into using quantum computers to determine the ground state of
quantum systems [3, 4]. These algorithms can be categorized into two types:
those that time evolve the quantum system (e.g., the quantum phase estimation
algorithm [5] and probe qubit algorithms [6, 7, 8]) and those that evaluate
expectation values of the Hamiltonian with a variational ansatz state (e.g.,
the variational quantum eigensolver algorithm [9, 10, 11, 12, 13, 14, 15, 16,
17, 18]). Variational algorithms usually require shorter quantum circuits than
those required by time-evolution algorithms. Having short quantum circuits is
important on current and near-term quantum computers as these devices can only
perform a limited number of gate operations before the quantum state
decoheres. Therefore, variational algorithms are likely to be among the first
to show quantum advantage.

Variational algorithms require that a particular ansatz is chosen for the form
of the ground state. Examples include the unitary coupled cluster ansatz [9,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31], the quantum alternating
operator ansatz [32, 33, 34, 35, 36], the variational hamiltonian ansatz [4,
37, 38], and the hardware-efficient ansatz [10, 39, 40, 41, 42, 43, 44, 45,
46]. There is no ansatz which works best in every case because the space of
quantum states is exponentially large while the ansatz generally has a
subexponential number of parameters. The ansatz must be chosen such that its
manifold includes a path connecting the initial state to a good approximation
of the ground state of the particular Hamiltonian being considered. Therefore,
it is important to have a variety of ansatzes with some intuition as to when
each should be applied. A powerful approach for improving variational states
is to use a Jastrow–Gutzwiller operator to correlate the electrons in a trial
many-body state [47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61,
62, 10, 63]. The Jastrow operator rescales the quantum state using density-
density correlations [47]. The Gutzwiller operator is a specific form of the
Jastrow operator in which only doubly occupied orbitals are included [48].
Because the Jastrow–Gutzwiller operator is non-unitary, implementing it on a
quantum computer is challenging [64, 65, 66, 67].

In this paper, we apply the Jastrow–Gutzwiller operator using the cascaded
variational quantum eigensolver (CVQE) algorithm [68]. The ansatz in the CVQE
algorithm is a product of a diagonal operator dependent on the variational
parameters and a unitary operator. This includes the special case
$\hat{A}(\theta)=\hat{G}(\theta)\hat{U}$, where $\theta$ is a collection of
variational parameters, $\hat{G}(\theta)$ is the non-unitary
Jastrow–Gutzwiller operator, and $\hat{U}$ is a unitary Thouless operator. We
show how to use this ansatz to perform calculations for an arbitrary fermionic
system. As an illustration, we demonstrate our method for the four-site square
and triangular Hubbard models with periodic boundary conditions.

Section II presents specific forms of the operators $\hat{G}(\theta)$ and
$\hat{U}$, the derivation of the expectation values for a general Hamiltonian
$\hat{H}$, and the mapping to a quantum computer. In Sec. III, we demonstrate
the method for the Hubbard model. In Section IV presents results of the
demonstration obtained in part from calculations performed on IBM Q Lagos for
the Hubbard model on square and triangular lattices [69, 70, 71, 72].
Conclusions are shared in Section V.

##  II Method

Consider a general fermionic system and let $\mathcal{Q}=\\{0,1,...,Q-1\\}$ be
an index set for the basis states $\ket{\phi_{q}}$ for the $Q$-dimensional
one-body Hilbert space. A basis for the many-body Hilbert space can then be
formed by Fock states $\ket{\Phi_{n}}$ identified by fermionic configurations
$n=(n_{q})_{q\in\mathcal{Q}}$ of occupation numbers in $\\{0,1\\}$, where
$n_{q}$ equals $0$ and $1$ mean that the one-body state $\ket{\phi_{q}}$ is
unoccupied and occupied, respectively.

We prepare the initial state to be one of these Fock states, which we can
express as

$\ket{\Phi_{n}}=\prod_{q\in\mathcal{Q}}\big{(}c^{\dagger}_{q}\big{)}^{n_{q}}\ket{0},$
(1)

where $c_{q}^{\dagger}$ is the fermionic creation operator associated with $q$
and $\ket{0}$ is the vacuum state.

###  II.1 The Thouless operator

We are often interested in a state

$\ket{\Psi_{n}}=\hat{U}\ket{\Phi_{n}},$ (2)

which is not necessarily a Fock state based on the initial one-body basis
$\\{\ket{\phi_{q}}\\}$. In our example in Section IV, $\ket{\Psi_{n}}$ will be
the Fermi sea [73]. We consider states that can be produced by transforming
the one-body Hilbert space using the unitary change-of-basis operator
$\hat{f}$ defined by

$\ket{\psi_{q}}=\sum_{q^{\prime}\in\mathcal{Q}}f_{qq^{\prime}}\ket{\phi_{q^{\prime}}},$
(3)

where $f_{qq^{\prime}}=\braket{\phi_{q^{\prime}}}{\psi_{q}}$ are complex
coefficients. In Appendix A, we show that the effect of this transformation on
the many-body Fock space is captured by the Thouless operator [74, 67, 4, 30,
75, 76]

$\hat{U}=\exp\Big{[}\sum_{q,q^{\prime}\in\mathcal{Q}}\big{(}\log\hat{f}\big{)}_{qq^{\prime}}c^{\dagger}_{q^{\prime}}c_{q}\Big{]},$
(4)

where $\log\hat{f}$ is the operator logarithm of $\hat{f}$, which is
guaranteed to exist as $\hat{f}$ is invertible.

We implement the unitary operator $\hat{U}=\hat{U}(\hat{f})$ in a quantum
circuit by decomposing the one-body operator $\hat{f}$ into a set of
elementary operators $\hat{f}_{m}$ such that
$\hat{f}=\hat{f}_{1}\hat{f}_{2}\cdots\hat{f}_{M}$. The method for generating
these transformations and the associated quantum circuit is presented in
Appendix B. It then follows that

$\hat{U}(\hat{f})=\hat{U}(\hat{f}_{1})\,\hat{U}(\hat{f}_{2})\cdots\hat{U}(\hat{f}_{m}),$
(5)

where each $\hat{U}(\hat{f}_{m})$ is implemented using standard one- and two-
qubit gates.

###  II.2 The Jastrow–Gutzwiller operator

To address the effects of interactions, we use the Jastrow–Gutzwiller operator
[47, 48], which can be expressed as

$\hat{G}(\theta)=\exp\bigg{(}-\sum_{qq^{\prime}}\theta_{qq^{\prime}}\hat{n}_{q}\hat{n}_{q^{\prime}}\bigg{)},$
(6)

where $\hat{n}_{q}=c^{\dagger}_{q}c_{q}$ is the number operator for orbital
$q$, and $\theta_{qq^{\prime}}$ is a real-valued variational parameter in the
collection $\theta=(\theta_{qq^{\prime}})_{qq^{\prime}\in\mathcal{Q}}$.

The energy expectation value in the CVQE algorithm takes the form [68]

$E(\theta)=\frac{\bra{\Psi_{n}}\hat{G}(\theta)\hat{H}\hat{G}(\theta)\ket{\Psi_{n}}}{\bra{\Psi_{n}}\hat{G}^{2}(\theta)\ket{\Psi_{n}}},$
(7)

where $\hat{H}$ is the Hamiltonian of the system. The denominator is necessary
because $\hat{G}$ is a non-unitary operator. As we will show below, both the
numerator and the denominator can be computed efficiently.

The full ansatz $\hat{A}(\theta)=\hat{G}(\theta)\hat{U}$ is particularly
useful when we have a single-particle solution defined by $\hat{f}$ (e.g., the
eigenstates of the kinetic term of the Hamiltonian), and we want to turn on
certain interactions defined by a diagonal operator [68]. In Sec. III we
demonstrate the ansatz with a Hubbard model.

Let the Hamiltonian be a general linear combination of products of fermion
operators

$\hat{H}=\sum_{l=0}^{N_{H}-1}h_{l}\hat{N}_{\mathcal{Q}^{N}_{l}}C^{\dagger}_{\mathcal{Q}^{+}_{l}}C_{\mathcal{Q}^{-}_{l}}$
(8)

where $N_{H}$ is the number of terms in the Hamiltonian, $h_{l}$ are defining
parameters,
$\hat{N}_{\mathcal{Q}^{\prime}}=\prod_{q\in\mathcal{Q}^{\prime}}\hat{n}_{q}$,
$C_{\mathcal{Q}^{\prime}}^{\dagger}=\prod_{q\in\mathcal{Q}^{\prime}}c_{q}^{\dagger}$,
for any index set $\mathcal{Q}^{\prime}$, and $\mathcal{Q}^{N}_{l}$,
$\mathcal{Q}^{+}_{l}$ and $\mathcal{Q}^{-}_{l}$ are disjoint subsets of
$\mathcal{Q}$. Let us also define
$\mathcal{Q}_{l}=\mathcal{Q}_{l}^{+}\cup\mathcal{Q}_{l}^{-}$ and
$\bar{\mathcal{Q}}_{l}=\mathcal{Q}\backslash\mathcal{Q}_{l}$. Using this form
of the Hamiltonian, we obtain

$\displaystyle\hat{G}(\theta)\hat{H}\hat{G}(\theta)$
$\displaystyle=\sum_{l=0}^{N_{H}-1}h_{l}\hat{N}_{\mathcal{Q}^{N}_{l}}C^{\dagger}_{\mathcal{Q}^{+}_{l}}C_{\mathcal{Q}^{-}_{l}}$
$\displaystyle\times\exp\\!\bigg{(}\\!-\sum_{qq^{\prime}}\left[\epsilon_{lq}+\epsilon_{lq^{\prime}}+\zeta_{lqq^{\prime}}\right]\theta_{qq^{\prime}}\hat{n}_{q}^{\epsilon_{lq}}\hat{n}_{q^{\prime}}^{\epsilon_{lq^{\prime}}}\bigg{)}$
(9)

where the binary variables

$\epsilon_{lq}=\begin{cases}1&\text{if }q\in\bar{\mathcal{Q}}_{l},\\\
0&\text{if }q\in\mathcal{Q}_{l},\end{cases}$ (10)

and

$\zeta_{lqq^{\prime}}=\begin{cases}1&\text{if
}q,q^{\prime}\in\bar{\mathcal{Q}}^{+}_{l},\\\ 1&\text{if
}q,q^{\prime}\in\bar{\mathcal{Q}}^{-}_{l},\\\ 0&\text{ otherwise
}.\end{cases}$ (11)

The derivation is given in Appendix C. We see in Eq. (9) that the creation and
annihilation operators never act on the same orbital as the number operators.
Because the number operators are diagonal in the occupation basis, we do not
need separate measurements for each term in the exponential.

In order to perform measurements of $\hat{G}(\theta)\hat{H}\hat{G}(\theta)$ on
a quantum computer we need to map the fermion operators onto qubit operators.
Since we require that the number operators are diagonal, a natural map is the
Jordan–Wigner transformation

$\begin{split}c^{\dagger}_{q}=\frac{X_{q}-iY_{q}}{2}\prod_{q^{\prime}=0}^{q-1}Z_{q^{\prime}},\end{split}$
(12)

where $X_{q}$, $Y_{q}$, and $Z_{q}$ are the Pauli-x, Pauli-y, and Pauli-z
operators respectively. Once the fermion operators are replaced with Pauli
operators, we perform measurements of $\hat{G}(\theta)\hat{H}\hat{G}(\theta)$
on the quantum computer in the same way one would perform measurements of a
Hamiltonian, except that we do not expand the exponential in terms of Pauli
strings. Instead, we take measurements of the qubits in
$\bar{\mathcal{Q}}_{l}$ and insert those values directly into the exponential
function. This is possible because all of the Pauli terms in the exponential
can be measured in the same basis. See appendix C for more information on the
measurement procedure.

##  III The Hubbard Model with the Gutzwiller ansatz

We select the Hubbard model as a test case to demonstrate our method. The
Hubbard model is a particularly useful model for near-term quantum computing
as it is simple enough to have relatively few terms yet complex enough that it
has not been solved classically except for special cases. Furthermore, it is a
physically important model as it can be used to describe the most relevant
interactions in a variety of physical systems. The Hamiltonian for the Hubbard
model is

$\hat{H}=\mu\hat{M}+k\hat{K}+d\hat{D},$ (13)

where $\mu$ is the chemical potential,

$\hat{M}=\sum_{i\sigma}\hat{n}_{i\sigma},$ (14)

$k$ is a kinetic energy parameter,

$\hat{K}=\sum_{\langle
i,j\rangle}\sum_{\sigma}\left(c^{\dagger}_{i\sigma}c_{j\sigma}+c^{\dagger}_{j\sigma}c_{i\sigma}\right),$
(15)

$d$ is the interaction strength, and

$\hat{D}=\sum_{i}\hat{n}_{i\uparrow}\hat{n}_{i\downarrow},$ (16)

where $c^{\dagger}_{i\sigma}$ creates an electron on site
$i\in\\{0,1,...,N-1\\}$ with spin $\sigma\in\\{\uparrow,\downarrow\\}$,
$\hat{n}_{i\sigma}=c^{\dagger}_{i\sigma}c_{i\sigma}$ is the number operator,
$N$ is the number of sites, and $\langle i,j\rangle$ denotes neighboring sites
$i$ and $j$.

Writing each term as a summation over Pauli strings using the Jordan–Wigner
transformation in Eq. (12), the Hamiltonian operators become

$\hat{M}=\sum_{i\sigma}\frac{1-Z_{i\sigma}}{2},$ (17)
$\hat{D}=\sum_{i}\frac{1-Z_{i\uparrow}}{2}\frac{1-Z_{i\downarrow}}{2},$ (18)

and

$\hat{K}=\sum_{\langle
i,j\rangle}\sum_{\sigma}\bigg{(}X_{i\sigma}\prod_{l=i+1}^{j-1}Z_{l\sigma}X_{j\sigma}+Y_{i\sigma}\prod_{l=i+1}^{j-1}Z_{l\sigma}Y_{j\sigma}\bigg{)}.$
(19)

In this demonstration, we apply the Gutzwiller operator

$\hat{G}(\theta)=e^{-\theta\hat{D}}.$ (20)

for the non-unitary factor of the ansatz. Notice that there is only a single
optimization parameter in this case.

We next determine the Pauli operators that need to be computed with the
quantum computer. Both $\hat{M}$ and $\hat{D}$ contain only Pauli-$z$
operators and commute with $\hat{G}(\theta)$. Therefore, we have
$\hat{G}(\theta)\hat{M}\hat{G}(\theta)=\hat{M}\hat{G}^{2}(\theta)$ and
$\hat{G}(\theta)\hat{D}\hat{G}(\theta)=\hat{D}\hat{G}^{2}(\theta)$, which are
readily computable on the quantum computer. In contrast, $\hat{K}$ contains
Pauli-$x$ and Pauli-$y$ operators that need to be factored so that each term
in the expectation value can be easily diagonalized. The operator can be
written as

$\begin{split}&\hat{G}(\theta)\hat{K}\hat{G}(\theta)=\\\ &\sum_{\langle
i,j\rangle}\sum_{\sigma}\left(X_{i\sigma}\prod_{l=i+1}^{j-1}Z_{l\sigma}X_{j\sigma}+Y_{i\sigma}\prod_{l=i+1}^{j-1}Z_{l\sigma}Y_{j\sigma}\right)\\\
&\times
e^{-\frac{\theta}{2}(1-Z_{i\bar{\sigma}})}e^{-\frac{\theta}{2}(1-Z_{j\bar{\sigma}})}e^{-\frac{\theta}{2}\sum_{k\neq
i,j}(1-Z_{k\uparrow})(1-Z_{k\downarrow})}.\end{split}$ (21)

where $\bar{\sigma}$ is the complement of $\sigma$ (i.e., the opposite spin).
For each term in the summation, the Pauli-$x$ and Pauli-$y$ operators have
been factored such that there is at most one operator acting on any particular
spin-orbital. This allows us to easily diagonalize each term in the summation
so that they can be measured separately on the quantum computer. See Appendix
D for a derivation of Eq. (21).

All of the Pauli-$z$ operators can be measured simultaneously. Both
$\braket{\hat{G}(\theta)\hat{D}\hat{G}(\theta)}$ and
$\braket{\hat{G}(\theta)\hat{M}\hat{G}(\theta)}$ can be measured with a single
run of the quantum computer (where a run is composed of enough shots to
achieve statistically accurate measurements). The expectation value
$\braket{\hat{G}(\theta)\hat{K}\hat{G}(\theta)}$ requires a run for each term
in the summation, which is proportional to the number of lattice sites.

##  IV demonstration results

Figure 1: Lattices used to demonstrate the algorithm: (a) A four-site square
lattice with periodic boundary conditions. (b) A four-site triangular lattice
with periodic boundary conditions. The solid circles indicate the four sites
while the transparent circles are periodic images.  Figure 2: Ground-state
energy for the square and triangular lattices. The dashed-red line is the
exact energy calculated by diagonalizing the Hamiltonian. The dotted-blue
curve is an exact classical simulation of the Gutzwiller approach, the thick-
green curve is the Gutzwiller approach using the data from the IBM Q Lagos
quantum computer for spin-down electrons only, and the thin-orange curve uses
the quantum computer for both the spin-up and the spin-down expectation
values. (a) and (b) show the expectation value of the energy $E(\theta)$ as a
function of the Gutzwiller parameter $\theta$. (a) is for the square lattice
and (b) is for the triangular lattice. The optimal value $\theta^{*}$ of
$\theta$ is found by locating the minimum of $E(\theta)$. (c) and (d) show the
optimal expectation value $E(\theta^{*})$ as a function of the interaction
strength $d$ for the square lattice and the triangular lattice, respectively.
For information on the specifications of IMB Q Lagos, see Appendix E.

We demonstrate our algorithm on a four-site square lattice and a four-site
triangular lattice both with periodic boundary conditions. The calculations
are performed at the center point $\Gamma$ in the first Brillouin zone, and
thus, no phases are applied to the hopping terms across the periodic
boundaries [77]. The two lattices are represented in Fig. 1. The chemical
potential is set so that the ground state has half filling (two spin-up
electrons and two spin-down electrons). For the square and triangular
lattices, we set $\mu=-d/2$ and $\mu=1-2d/3$, respectively. In order to access
the half-filled state on the quantum computer we apply the two-particle Slater
determinant circuit as described in Sec. II.2.

We vary the single parameter $\theta$ to minimize the expectation value of the
energy. By moving the evaluation of $\hat{G}(\theta)$ to the classical
computer, we have not only removed the need to rerun the quantum circuits when
we vary the parameters, but we have also removed the coupling between the spin
species from the quantum computer. Evaluating the circuit for only a single
spin species reduces the number of required qubits by half [78].

Fig. 2(a,b) shows the energy expectation values for the square and triangular
lattices as a function of $\theta$ for $d=2k$. The dashed red-line shows the
exact ground-state energy which is plotted for reference. The dotted-blue
curve shows the energy expectation value for a noise-free simulation of our
method.

The two solid curves are computed using the quantum device. The thin-orange
curve is a direct calculation of all electrons on the quantum computer. The
thick-green curve is calculated using the quantum computer for the up
electrons and the noise-free simulator for the down electrons. The curves are
smooth even though the data is noisy because $\theta$ is varied using the
classical computer after the quantum measurement have been recorded. The
minimum energy expectation value can occur at infinite $\theta$ as in the
thin-orange curve in Fig. 2b.

Figs. 2(c,d) shows the optimized energy expectation values as a function of
$d$. At $d=0$ the Gutzwiller state is the exact ground state. Noise in the
quantum circuit generates error in the expectation values. The error is
greatly reduced when we run only a single spin species on the quantum
computer. Furthermore, the error in the energy expectation value stays
relatively constant as a function of $d$ and even slightly decreases when
$d\approx k$.

##  V conclusion

We have proposed a method for implementing a generalized Jastrow–Gutzwiller
ansatz on a quantum computer using the cascaded variational quantum
eigensolver algorithm. This allows us to implement the Jastrow–Gutzwiller
ansatz without adding auxiliary qubits or making approximations to the
operator. We demonstrate the method on the IBM Q Lagos device for a four-site
Hubbard model on both square and triangular lattices at half-filling.

###### Acknowledgements.

This work has been supported by the Office of Naval Research (ONR) through the
U.S. Naval Research Laboratory (NRL). J.P.T.S. thanks the National Research
Council Research Associateship Programs for support during his postdoctoral
tenure at NRL. We acknowledge quantum computing resources from IBM through a
collaboration with the Air Force Research Laboratory (AFRL).

##  Appendix A Generating Slater Determinants

In this appendix, we derive Eq. (2), where $\hat{U}$ is given by Eq. (4) and
$c^{\dagger}_{q}$ and $a^{\dagger}_{q}$ are related by

$a^{\dagger}_{q}=\sum_{q^{\prime}}\hat{f}_{qq^{\prime}}c^{\dagger}_{q^{\prime}}.$
(22)

Our derivation follows closely the one given in [30]. Specifically, we show
that

$\hat{U}c^{\dagger}_{q}\hat{U}^{\dagger}=a^{\dagger}_{q}.$ (23)

To do so, let us define
$\hat{\kappa}=\sum_{qq^{\prime}}\log(\hat{f})_{qq^{\prime}}c^{\dagger}_{q^{\prime}}c_{q}$.
Using the Baker–Campbell–Hausdorff formula we have

$\hat{U}c^{\dagger}_{q}\hat{U}^{\dagger}=e^{\hat{\kappa}}c^{\dagger}_{q}e^{-\hat{\kappa}}=c^{\dagger}_{q}+[\hat{\kappa},c^{\dagger}_{q}]+\frac{1}{2}[\hat{\kappa},[\hat{\kappa},c^{\dagger}_{q}]]+\ldots$
(24)

We need to evaluate the commutators
$\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}$,
$\big{[}\hat{\kappa},\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}\big{]}$,
$\ldots$, for which we get

$\displaystyle\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}$
$\displaystyle=\sum_{q^{\prime}q^{\prime\prime}}\big{(}\log\hat{f}\big{)}_{q^{\prime}q^{\prime\prime}}\big{[}c_{q^{\prime\prime}}^{\dagger}c_{q^{\prime}},c_{q}^{\dagger}\big{]}$
$\displaystyle=\sum_{q^{\prime}q^{\prime\prime}}\big{(}\log\hat{f}\big{)}_{q^{\prime}q^{\prime\prime}}\big{(}c_{q^{\prime\prime}}^{\dagger}c_{q^{\prime}}c_{q}^{\dagger}-c_{q}^{\dagger}c_{q^{\prime\prime}}^{\dagger}c_{q^{\prime}}\big{)}$
$\displaystyle=\sum_{q^{\prime}q^{\prime\prime}}\big{(}\log\hat{f}\big{)}_{q^{\prime}q^{\prime\prime}}\big{(}\delta_{q^{\prime}q}c_{q^{\prime\prime}}^{\dagger}-c_{q^{\prime\prime}}^{\dagger}c_{q}^{\dagger}c_{q^{\prime}}-c_{q}^{\dagger}c_{q^{\prime\prime}}^{\dagger}c_{q^{\prime}}\big{)}$
$\displaystyle=\sum_{q^{\prime}}\big{(}\log\hat{f}\big{)}_{qq^{\prime}}c_{q^{\prime}}^{\dagger}$
(25)

and

$\displaystyle\big{[}\hat{\kappa},\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}\big{]}$
$\displaystyle=\sum_{q^{\prime}}\big{[}\log\hat{f}\big{]}_{qq^{\prime}}\big{[}\hat{\kappa},c_{q^{\prime}}^{\dagger}\big{]}$
$\displaystyle=\sum_{q^{\prime}}\big{[}\log\hat{f}\big{]}_{qq^{\prime}}\sum_{q^{\prime\prime}}\big{[}\log\hat{f}\big{]}_{q^{\prime}q^{\prime\prime}}c_{q^{\prime\prime}}^{\dagger}$
$\displaystyle=\sum_{q^{\prime\prime}}\big{[}\big{(}\log\hat{f}\big{)}^{2}\big{]}_{qq^{\prime\prime}}c_{q^{\prime\prime}}^{\dagger},$
(26)

for the first two, where $\delta$ is the Kronecker delta. We see that each
commutator simply adds a power to $\log\hat{f}$. Returning to the
Baker–Campbell–Hausdorff formula, this gives

$\displaystyle\hat{U}$ $\displaystyle
c_{q}^{\dagger}\hat{U}^{\dagger}=e^{\hat{\kappa}}c_{q}^{\dagger}e^{-\hat{\kappa}}$
$\displaystyle=c_{q}^{\dagger}+\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}+\frac{1}{2!}\big{[}\hat{\kappa},\big{[}\hat{\kappa},c_{q}^{\dagger}\big{]}\big{]}+...$
$\displaystyle=\sum_{q^{\prime}}\big{[}\hat{I}+\log\hat{f}+\frac{1}{2!}\big{(}\log\hat{f}\big{)}^{2}+...\big{]}_{qq^{\prime}}c_{q^{\prime}}^{\dagger}$
$\displaystyle=\sum_{q^{\prime}}\big{[}e^{\log\hat{f}}\big{]}_{qq^{\prime}}c_{q^{\prime}}^{\dagger}$
$\displaystyle=\sum_{q^{\prime}}\hat{f}_{qq^{\prime}}c_{q^{\prime}}^{\dagger}$
$\displaystyle=a_{q}^{\dagger},$ (27)

where $\hat{I}$ is the identity operator.

Inserting identities into the string of creation operators we arrive at Eq.
(2):

$\displaystyle\hat{U}\ket{\Phi_{n}}$
$\displaystyle=\hat{U}\prod_{q}(c^{\dagger}_{q})^{n_{q}}\ket{0}$
$\displaystyle=\prod_{q}\hat{U}(c^{\dagger}_{q})^{n_{q}}\hat{U}^{\dagger}\ket{0}$
$\displaystyle=\prod_{q}(a^{\dagger}_{q})^{n_{q}}\ket{0}$
$\displaystyle=\ket{\Psi_{n}},$ (28)

where we have used $\hat{U}\ket{0}=\ket{0}$.

##  Appendix B Circuit model for the Thouless operator

In order to implement $\hat{U}(\hat{f})$ on the quantum computer, we use
single-particle rotations to diagonalize the one-body change-of-basis-operator
$\hat{f}$ and then find the Thouless operator for each of those rotations. We
use two basic single-particle rotations:

$\begin{split}[\hat{r}^{y}_{ij}(\phi_{y})]_{kl}&=\delta_{kl}+(\cos\phi_{y}-1)\delta_{ik}\delta_{il}+(\cos\phi_{y}-1)\delta_{jk}\delta_{jl}\\\
&+\sin\phi_{y}\delta_{ik}\delta_{jl}-\sin\phi_{y}\delta_{jk}\delta_{il}\end{split}$
(29)

and

$[\hat{r}^{z}_{i}(\phi_{z})]_{kl}=\delta_{kl}+(e^{i\phi_{z}}-1)\delta_{ik}\delta_{il},$
(30)

where the index $j$ defines the row in $\hat{f}$ that we are targeting and
index $i$ defines the row we are rotating into. In matrix form

$r^{y}_{ij}(\phi_{y})=\begin{pmatrix}\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots\\\
\ldots&1&0&0&\ldots&0&0&0&\ldots\\\
\ldots&0&\cos\phi_{y}&0&\ldots&0&\sin\phi_{y}&0&\ldots\\\
\ldots&0&0&1&\ldots&0&0&0&\ldots\\\
\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots\\\
\ldots&0&0&0&\ldots&1&0&0&\ldots\\\
\ldots&0&-\sin\phi_{y}&0&\ldots&0&\cos\phi_{y}&0&\ldots\\\
\ldots&0&0&0&\ldots&0&0&1&\ldots\\\
\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots\\\
\end{pmatrix}$ (31)

and

$r^{z}_{j}(\phi_{z})=\begin{pmatrix}\ddots&\vdots&\vdots&\vdots&\ddots\\\
\ldots&1&0&0&\ldots\\\ \ldots&0&e^{i\phi_{z}}&0&\ldots\\\
\ldots&0&0&1&\ldots\\\ \ddots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}.$ (32)

We use these rotations to rotate specific off-diagonal indices of $\hat{f}$ to
zero. The specific element in row $j$ of $\hat{f}$ that becomes zero depends
on the angles $\phi_{y}$ and $\phi_{z}$. Since $\hat{f}$ is unitary, we only
need to worry about the lower-left elements. The upper-right elements will be
automatically eliminated. In order to avoid undoing an element that has
already been eliminated, we start at the lower left corner and work up in the
following order:

$\begin{pmatrix}*&*&*&*&*&*&\\\ 5&*&*&*&*&*&\\\ 4&6&*&*&*&*&\\\
3&5&7&*&*&*&\\\ 2&4&6&8&*&*&\\\ 1&3&5&7&9&*&\\\ \end{pmatrix},$ (33)

where the asterisks indicate indices that do not need to be explicitly
targeted and indices with the same value can be eliminated in any order.

To avoid complications with fermion exchange, we restrict $r^{y}_{ij}(\phi)$
so that $j=i+1$. In this case, we can compute the Thouless operators for both
$r^{y}_{ij}(\phi)$ and $r^{z}_{j}(\phi)$ and write them in terms of quantum
gates

$U[r^{z}_{j}(\phi_{z})]=e^{i\phi_{z}c^{\dagger}_{j}c_{j}}=R^{Z}_{j}(\phi_{z})$
(34)

up to a global phase and

$\begin{split}U[r^{y}_{ii+1}(\phi_{y})]&=e^{\phi_{y}(c^{\dagger}_{i}c_{i+1}-c^{\dagger}_{i+1}c_{i})}=R^{XY}_{i,i+1}(\phi_{y})R^{YX}_{i,i+1}(-\phi_{y}),\end{split}$
(35)

where

$R^{Z}_{j}(\phi)=e^{-i\frac{\phi}{2}Z_{j}},$ (36)

is a rotation about the z-axis of the Bloch sphere of qubit $j$, and

$\begin{split}R^{XY}_{ij}(\phi)=e^{-i\frac{\phi}{2}X_{i}Y_{j}},\\\
R^{YX}_{ij}(\phi)=e^{-i\frac{\phi}{2}Y_{i}X_{j}},\end{split}$ (37)

are two qubit rotations. The two qubit rotations can be generated by single-
qubit rotations and C-not gates

$\begin{split}R^{XY}_{ij}(\phi)=R^{Y}_{i}(\pi/2)C^{X}_{ij}R^{Y}_{j}(\phi)C^{X}_{ij}R^{Y}_{i}(-\pi/2),\\\
R^{YX}_{ij}(\phi)=R^{Y}_{j}(\pi/2)C^{X}_{ji}R^{Y}_{i}(\phi)C^{X}_{ji}R^{Y}_{j}(-\pi/2),\end{split}$
(38)

where $R^{Y}_{i}(\phi)$ is a rotation about the y-axis of the Bloch sphere of
qubit $i$ and $C^{X}_{ij}$ is a controlled-not gate, targeting qubit $j$ and
controlled by qubit $i$.

Thus, we have a map from the single-particle rotations to quantum gates. Since
we know how to use the single-particle rotations to diagonalize $\hat{f}$, we
can reverse the rotations to generate $\hat{f}$ from the identity operator.
Transforming these reverse rotations onto quantum gates allows us to apply
$\hat{U}(\hat{f})$ on the quantum computer. See Appendix E for a depiction of
the circuit model.

A similar constructions of the quantum gates which generate a Thouless
operator can be found in [30, 75] where [75] provides an informative example.
For alternative constructions see [4, 76] .

##  Appendix C Derivation of
$\bra{\Psi_{0}}\hat{G}\hat{H}\hat{G}\ket{\Psi_{0}}$

Due to the idempotency of number operators ($\hat{n}_{q}^{2}=\hat{n}$), we can
write the Jastrow-Gutzwiller operator in Eq. (6) as

$\hat{G}(\theta)=\prod_{qq^{\prime}}(1+\gamma_{qq^{\prime}}\hat{n}_{q}\hat{n}_{q^{\prime}}),$
(39)

where $\gamma_{qq^{\prime}}=e^{-\theta_{qq^{\prime}}}-1$. To analyze
$\hat{G}(\theta)\hat{H}\hat{G}(\theta)$, we use the identities
$\hat{n}_{q}c^{\dagger}_{q}=c^{\dagger}_{q}$, $c_{q}\hat{n}_{q}=c_{q}$, and
$c^{\dagger}_{q}\hat{n}_{q}=n_{q}c_{q}=0$. Thus, we can write

$\begin{split}\hat{G}(\theta)\hat{H}\hat{G}(\theta)=&\sum_{l=0}^{N_{H}-1}h_{l}\hat{N}_{\mathcal{Q}^{N}_{l}}C^{\dagger}_{\mathcal{Q}^{+}_{l}}C_{\mathcal{Q}^{-}_{l}}\\\
\times&\prod_{qq^{\prime}\in\bar{\mathcal{Q}}_{l}}(1+\gamma_{qq^{\prime}}\hat{n}_{q}\hat{n}_{q^{\prime}})^{2}\\\
\times&\prod_{q\in\bar{\mathcal{Q}}_{l}}\prod_{q^{\prime}\in\mathcal{Q}_{l}}(1+\gamma_{qq^{\prime}}\hat{n}_{q})\\\
\times&\prod_{q\in\mathcal{Q}_{l}}\prod_{q^{\prime}\in\bar{\mathcal{Q}}_{l}}(1+\gamma_{qq^{\prime}}\hat{n}_{q^{\prime}})\\\
\times&\prod_{qq^{\prime}\in\mathcal{Q}^{+}_{l}}(1+\gamma_{qq^{\prime}})\\\
\times&\prod_{qq^{\prime}\in\mathcal{Q}^{-}_{l}}(1+\gamma_{qq^{\prime}}).\end{split}$
(40)

We then convert the number operator factors back into exponentials to arrive
at Eq. (9) in the main text.

To perform measurements of $\hat{G}(\theta)\hat{H}\hat{G}(\theta)$ on a
quantum computer, we map the fermion operators onto qubit operators using the
Jordan–Wigner transformation, Eq. (12). After the transformation, we have

$\begin{split}&\hat{G}(\theta)\hat{H}\hat{G}(\theta)=\sum_{l=0}^{\tilde{N}_{H}-1}\tilde{h}_{l}\hat{A}_{\mathcal{Q}_{l}}\hat{B}_{\bar{\mathcal{Q}}_{l}}\\\
&\times\exp\left[-\sum_{qq^{\prime}}\left(\epsilon_{lq}+\epsilon_{lq^{\prime}}+\zeta_{lqq^{\prime}}\right)\theta_{qq^{\prime}}\left(\frac{1-Z_{q}}{2}\right)^{\epsilon_{lq}}\left(\frac{1-Z_{q^{\prime}}}{2}\right)^{\epsilon_{lq^{\prime}}}\right]\end{split}$
(41)

where $\tilde{N}_{H}$ is the number of unique Pauli-strings that results from
the transformed Hamiltonian, $\hat{A}_{\mathcal{Q}_{l}}$ are Pauli-strings
containing only Pauli-x and Pauli-y operators acting on orbitals in
$\mathcal{Q}_{l}$, $\hat{B}_{\bar{\mathcal{Q}}_{l}}$ are Pauli-strings
containing only Pauli-z and identity operators acting on orbitals in
$\bar{\mathcal{Q}}_{l}$, and $\epsilon_{lq}$ and $\zeta_{lqq^{\prime}}$ are
defined in Eqs. (10) and (11).

All of the Pauli-z operators are diagonal in the $\\{\ket{\Phi_{n}}\\}$ basis.
In order to measure term $l$ in Eq. (41), we rotate the state so that each
Pauli-x and Pauli-y operator becomes diagonal. Let $\ket{\Phi^{l}_{n}}$ be the
state that has been rotated to measure term $l$ in Eq. (41). Because none of
the Pauli-z operators act on the same index as the Pauli-x or Pauli-y
operators, every operator in term $l$ is diagonal in the $\ket{\Phi^{l}_{n}}$
basis. Let
$\hat{A}_{\mathcal{Q}_{l}}\ket{\Phi^{l}_{m}}=\ket{\Phi^{l}_{m}}a_{lm}$,
$\hat{B}_{\bar{\mathcal{Q}}_{l}}\ket{\Phi^{l}_{m}}=\ket{\Phi^{l}_{m}}b_{lm}$,
and
$(1-Z_{q\in\bar{\mathcal{Q}}_{l}})/2\ket{\Phi^{l}_{m}}=\ket{\Phi^{l}_{m}}n_{qlm}$.
We decompose the final state Eq. (2) into linear combinations of the
measurement basis

$\ket{\Psi_{n}}=\sum_{m=0}^{N_{S}-1}u^{l}_{nm}\ket{\Phi^{l}_{m}},$ (42)

where $N_{S}=2^{Q}$ is the number of many-body basis states. This yields

$\begin{split}&\bra{\Psi_{n}}\hat{G}(\theta)\hat{H}\hat{G}(\theta)\ket{\Psi_{n}}=\sum_{l=0}^{\tilde{N}_{H}-1}\tilde{h}_{l}\sum_{m=0}^{N_{S}-1}|u^{l}_{nm}|^{2}a_{lm}b_{lm}\\\
&\times\exp\left(-\sum_{qq^{\prime}}\left[\epsilon_{lq}+\epsilon_{lq^{\prime}}+\zeta_{lqq^{\prime}}\right]\theta_{qq^{\prime}}n_{qlm}^{\epsilon_{lq}}n_{q^{\prime}lm}^{\epsilon_{lq^{\prime}}}\right).\end{split}$
(43)

Each time the quantum computer is measured it collapses into a state
$\ket{\Phi^{l}_{m}}$ with a random value of $m$ with a probability of
$|u^{l}_{nm}|^{2}$. For each state $\ket{\Phi^{l}_{m}}$, there are associated
values $a_{qlm}$, $b_{qlm}$, and $n_{qlm}$ that are obtained from the
measurement result. We can get an accurate estimate of the statistically
significant values of $|u^{l}_{nm}|^{2}$ with a subexponential number of
shots. The number of measurements scales only with the number of Pauli strings
in the Hamiltonian $\tilde{N}_{H}$. As there are at most $Q^{2}$ non-zero
values of $\theta_{qq^{\prime}}$. the full classical calculation scales as
$\tilde{N}_{H}\times Q^{2}$. Subexponential scaling is achieved by collecting
the eigenvalues $n_{qlm}$ into the exponential function Eq. (43).

##  Appendix D Derivation of $\hat{G}(\theta)\hat{K}\hat{G}(\theta)$

To derive Eq. (21), we expand $\hat{G}(\theta)$ as

$\begin{split}\hat{G}(\theta)&=e^{-\theta\sum_{i}n_{i\uparrow}n_{j\downarrow}}\\\
&=\prod_{i}e^{-\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}}\\\
&=\prod_{i}\left(1-\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}+\frac{1}{2}\theta^{2}\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}-\ldots\right)\\\
&=\prod_{i}\left[1-\left(\theta-\frac{1}{2}\theta^{2}+\ldots\right)\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}\right]\\\
&=\prod_{i}\left[1-(1-e^{-\theta})\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}\right],\end{split}$
(44)

where in the second line we used $\hat{n}_{i\sigma}^{2}=\hat{n}_{i\sigma}$.
From this expression, we can calculate
$\hat{G}(\theta)\hat{K}\hat{G}(\theta)$. First we examine how
$\hat{G}(\theta)$ acts on an arbitrary pair of creation and destruction
operators:

$\begin{split}&\hat{G}(\theta)c^{\dagger}_{j\uparrow}c_{k\uparrow}\hat{G}(\theta)\\\
&=\left[\prod_{i}(1-\lambda\hat{n}_{i\uparrow}\hat{n}_{i\downarrow})\right]c^{\dagger}_{j\uparrow}c_{k\uparrow}\prod_{i^{\prime}}(1-\lambda\hat{n}_{i^{\prime}\uparrow}\hat{n}_{i^{\prime}\downarrow})\\\
&=(1-\lambda\hat{n}_{j\uparrow}\hat{n}_{j\downarrow})(1-\lambda\hat{n}_{k\uparrow}\hat{n}_{k\downarrow})c^{\dagger}_{j\uparrow}c_{k\uparrow}\\\
&\quad\times(1-\lambda\hat{n}_{j\uparrow}\hat{n}_{j\downarrow})(1-\lambda\hat{n}_{k\uparrow}\hat{n}_{k\downarrow})\prod_{i\neq
j,k}e^{-2\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}}\\\
&=(1-\lambda\hat{n}_{j\downarrow})c^{\dagger}_{j\uparrow}c_{k\uparrow}(1-\lambda\hat{n}_{k\downarrow})\prod_{i\neq
j,k}e^{-2\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}}\\\
&=c^{\dagger}_{j\uparrow}c_{k\uparrow}e^{-\theta\hat{n}_{j\downarrow}}e^{-\theta\hat{n}_{k\downarrow}}\prod_{i\neq
j,k}e^{-2\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}},\end{split}$ (45)

where $\lambda=e^{-\theta}-1$ and we used
$\hat{n}_{i\sigma}c^{\dagger}_{i\sigma}=c^{\dagger}_{i\sigma}$,
$c_{i\sigma}\hat{n}_{i\sigma}=c_{i\sigma}$, and
$\hat{n}_{i\sigma}c_{i\sigma}=c^{\dagger}_{i\sigma}\hat{n}_{i\sigma}=0$. Using
Eq. (45) and an analogous one for spin down operators, it is straightforward
to calculate $\hat{G}(\theta)\hat{K}\hat{G}(\theta)$:

$\begin{split}&\hat{G}(\theta)\hat{K}\hat{G}(\theta)\\\
&=\sum_{\sigma}\sum_{\langle
i,j\rangle}\left[\hat{G}(\theta)c^{\dagger}_{j\sigma}c_{k\sigma}\hat{G}(\theta)+\hat{G}(\theta)c^{\dagger}_{k\sigma}c_{j\sigma}\hat{G}(\theta)\right],\\\
&=\sum_{\sigma}\sum_{\langle
i,j\rangle}\left(c^{\dagger}_{j\sigma}c_{k\sigma}+c^{\dagger}_{k\sigma}c_{j\sigma}\right)\\\
&\quad\times
e^{-\theta\hat{n}_{j\bar{\sigma}}}e^{-\theta\hat{n}_{k\bar{\sigma}}}\prod_{i\neq
j,k}e^{-2\theta\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}},\end{split}$ (46)

where $\bar{\sigma}$ is the complement of $\sigma$. With the Jordan–Wigner
transformation we arrive at the expression in Eq. (21).

##  Appendix E Device specifications

Figure 3: Circuit model specifications. a) layout of the IBM Lagos device.
Qubits are depicted as circles and the connectivity is shown by the lines. The
dashed boarder shows the qubits that were used in our demonstration. b)
Circuit model for the operator
$C(\phi_{y},\phi_{z})=U[r_{i,j}^{y}(\phi_{y})]U[r_{j}^{z}(\phi_{z})]$ as
defined in Eq. (34) and (35). The two qubit gates are $C^{X}$ gates, the
$R^{y}$ gates represent rotations around the $y$-axis of the block sphere, and
$R^{z}$ gates represent rotations around the $z$-axis. c) circuit model for
the full Thouless operator. From top to bottom the lines are applied to qubit
$q_{0}$, $q_{1}$, $q_{3}$, $q_{5}$ as shown if (a). The state is set to half
filling with the initial Pauli-X gates. The $R^{z}$ gates, in the next step,
set up the diagonalized $\hat{f}$ matrix. The $C$ gates are defined in (b).
All angles $\phi_{i}$, $\phi^{y}_{i}$, and $\phi^{z}_{i}$ are found during the
diagonalization process of $\hat{f}$ as described in Appendix B.

We demonstrate our method on IBM Q Lagos which is a seven-qubit quantum
computer. Lagos has the following qubit pairs connected

$\begin{matrix}&q_{0}&q_{1}&q_{2}&q_{3}&q_{4}&q_{5}&q_{6}\\\
q_{0}&0&1&0&0&0&0&0\\\ q_{1}&1&0&1&1&0&0&0\\\ q_{2}&0&1&0&0&0&0&0\\\
q_{3}&0&1&0&0&0&1&0\\\ q_{4}&0&0&0&0&0&1&0\\\ q_{5}&0&0&0&1&1&0&1\\\
q_{6}&0&0&0&0&0&1&0&,\\\ \end{matrix}$ (47)

where $q_{i}$ refers to qubit $i$ and $1$/$0$ specifies pairs which are
connected/disconnected. The connectivity map of Lagos is shown if Fig 3a. We
used only qubits $q_{0}$, $q_{1}$, $q_{3}$, and $q_{5}$. The pre-transpilation
circuit is shown in Fig 3(b,c). The circuit in Fig 3b represents a single step
during the diagonalization process of $\hat{f}$ as discussed in Appendix B.
This step is repeated several times with different angles and applied to
different qubits during the full implementation of the Thouless operator in
Fig. 3c. There are 24 CX gates in the circuit which is the largest
contributing factor to the error. After transpilation, all of the gates were
converted to z-rotations $R^{z}$, root-x gates $\sqrt{X}$, and controlled-not
gates CX using the qiskit transpile function. The specifications of the device
including the error rates of each gate can be found in Table 1. These
specifications are based on a calibration performed on May 5, 2022. The
$R^{z}$ gates have no error as they are implemented by changing the
measurement angles.

Table 1:  Specifications of IBM Q Lagos during the execution of our algorithm. | $q_{0}$ | $q_{1}$ | $q_{2}$ | $q_{3}$ | $q_{4}$ | $q_{5}$ | $q_{6}$
---|---|---|---|---|---|---|---
Frequency (GHz) | 5.24 | 5.1 | 5.19 | 4.99 | 5.29 | 5.18 | 5.06
T1 ($\mu$s) | 137.3 | 136.2 | 179.4 | 152.2 | 102.7 | 134.4 | 142.6
T2 ($\mu$s) | 49.8 | 114.6 | 130.1 | 76.4 | 44.9 | 98.2 | 194.0
Readout error (%) | 0.97 | 0.71 | 1.42 | 1.78 | 1.41 | 2.78 | 0.72
$\sqrt{X}$ error (%) | 0.018 | 0.023 | 0.022 | 0.018 | 0.036 | 0.063 | 0.028
$\text{CX}_{q_{0},q_{i}}$ error (%) |  | 0.707 |  |  |  |  |
$\text{CX}_{q_{1},q_{i}}$ error (%) | 0.707 |  | 0.661 | 0.675 |  |  |
$\text{CX}_{q_{2},q_{i}}$ error (%) |  | 0.661 |  |  |  |  |
$\text{CX}_{q_{3},q_{i}}$ error (%) |  | 0.675 |  |  |  | 1.163 |
$\text{CX}_{q_{4},q_{i}}$ error (%) |  |  |  |  |  | 3.406 |
$\text{CX}_{q_{5},q_{i}}$ error (%) |  |  |  | 1.163 | 3.406 |  | 0.908
$\text{CX}_{q_{6},q_{i}}$ error (%) |  |  |  |  |  | 0.908 |

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∎

11institutetext:  Arno van den Essen 22institutetext: Radboud University,
Nijmegen 33institutetext: Jan Schoone 44institutetext: Digital Security,
Radboud University, Nijmegen
44email<EMAIL_ADDRESS>
# Kernels of linear maps

A generalization of Duistermaat and Van der Kallens theorem

Arno van den Essen  Jan Schoone

(Received: date / Accepted: date)

###### Abstract

The theorem of Duistermaat and Van der Kallen from 1998 proved the first case
of the Mathieu conjecture. Using the theory of Mathieu-Zhao spaces, we can
reformulate this theorem as $\operatorname{Ker}L$ is a Mathieu-Zhao space
where $L$ is the linear map

$\displaystyle L\colon{\bf
C}[X_{1},\ldots,X_{n},X_{1},\ldots,X_{n}^{-1}]\to{\bf C},\ f\mapsto f_{0}.$

In this paper, we generalize this result (for $n=1$) to all non-trivial linear
maps $L\colon{\bf C}[X,X^{-1}]\to{\bf C}$ such that $\\{X^{n}\mid|n|\geq
N\\}\subset\operatorname{Ker}L$ for some $N\geq 1$.

###### Keywords:

algebraic geometry commutative algebra mathieu-zhao spaces mathieu conjecture

##  1 Introduction

In 1939, Ott-Heinrich Keller keller  formulated what is now known as the
Jacobian conjecture:

######  Conjecture 1 (Jacobian conjecture)

Let $k$ be a field of characteristic zero and $F\in k[X_{1},\ldots,X_{n}]^{n}$
a polynomial mapping such that its Jacobian determinant $J_{F}$ is a non-zero
constant. Then $F$ has an inverse function $G\in k[X_{1},\ldots,X_{n}]^{n}$.

Since then, this conjecture remained unsolved (apart from the case $n=1$). For
a historical treaty and a collection of results around the Jacobian
conjecture, one can read Ess0  and Ess2 . In 1997, Olivier Mathieu showed in
Mat  that the Jacobian conjecture follows from what is now known as the
Mathieu conjecture:

######  Conjecture 2 (Mathieu conjecture)

Let $G$ be a compact connected real Lie group with Haar measure $\sigma$. Let
$f$ be a complex valued $G$-finite function on $G$ such that
$\int_{G}f^{m}d\sigma=0$ for all $m\geq 1$. Then for every $G$-finite function
$g$ on $G$, also $\int_{G}gf^{m}d\sigma=0$ for all large $m$.

The Mathieu conjecture has been proved for the case of connected abelian
groups by Duistermaat and Van der Kallen in 1998 DuiKal .

From the formulation of the Mathieu conjecture, Wenhua Zhao formulated the
notion of Mathieu-Zhao spaces that are a generalization of ideals. For history
and details, see Zhao1  and Ess2 .

Together with this formulation, we manage to prove a generalization of the
theorem of Duistermaat and Van der Kallen in dimension one.

##  2 Notations and conventions

We write ${\bf Z}$ for the ring of integers, ${\bf C}$ for the field of
complex numbers, and $\mathbb{F}_{q}$ for a finite field of $q$ elements.
Furthermore, we always assume all rings to be commutative and to have an
identity. Similarly, we assume all algebras to be unital and associative.

##  3 Mathieu-Zhao spaces and partial fractions decomposition

Because of the shape of the statement in the Mathieu conjecture Mat , we can
define the following generalisation of ideals in associative algebras, which
we call Mathieu-Zhao spaces.

###### Definition 1

Let $R$ be any ring and $A$ an $R$-algebra. For any subset $M$ of $A$ we
define the _radical of $M$_ by

$\displaystyle r(M):=\\{a\in A\mid\forall m\gg 0:a^{m}\in M\\}$

with $m\gg 0$ meaning all $m\geq N$ for some $N\geq 1$.

###### Remark 1

For an ideal $I$ of $R$, this coincides with the usual definition of the
radical of $I$, being:

$\displaystyle\sqrt{I}:=\\{a\in R\mid\exists n:a^{n}\in I\\}.$

It is well-known that the radical of an ideal is an ideal itself, and hence
all $a^{n+i}$ are elements of $I$ for $i\geq 0$.

###### Definition 2

Let $R$ be any ring and $A$ an $R$-algebra. For any subset $M$ of $A$ we
define the _(left) strong radical of $M$_ by

$\displaystyle sr(M):=\\{a\in A\mid\forall b\in A\ \forall m\gg 0:ba^{m}\in
M\\}.$

###### Remark 2

Note that if $a\in sr(M)$, then $1\cdot a^{m}\in M$ for all $m\gg 0$, hence
$a\in r(M)$. So we always have $sr(M)\subset r(M)$.

###### Definition 3

Let $R$ be any ring and $A$ an $R$-algebra. We say that an $R$-submodule $M$
of $A$ is a Mathieu-Zhao space of $A$ (short: MZ-space) if and only if
$sr(M)=r(M)$.

A little less is needed in practice. Define $r^{\prime}(M)=\\{a\in
A\mid\forall m\geq 1:a^{m}\in M\\}$. If $r^{\prime}(M)\subset sr(M)$, then $M$
is an MZ-space of $A$.

######  Proposition 1 (Ess2 )

Let $R$ be a ring and $A$ an $R$-algebra. Let $M$ be an additive subset of $A$
and $I\subset A$ an ideal of $A$ such that $I\subset M$. Then $M$ is an MZ-
space of $A$ if and only if $M/I$ is an MZ-space of $A/I$.

###### Example 1

Let $R$ be a ring and $A$ an $R$-algebra. Let $I$ be an ideal of $A$. Then $I$
is an MZ-space.

_Proof._ We need to show that $r^{\prime}(M)\subset sr(M)$. Thus, let $a\in
r^{\prime}(I)$ be arbitrary. Then $a\in I$. Since $I$ is an ideal, we have for
any $b\in A$ that $ba^{n}a\in I$ for all $n\geq 0$. Therefore $a\in sr(I)$.
$\triangle$

###  3.1 Zhao’s Idempotency Theorem

One of the most important theorems regarding MZ-spaces is Zhao’s Idempotency
Theorem:

######  Theorem 3.1 (Zhao’s Idempotency Theorem Zhao1 )

Let $k$ be a field and $A$ a $k$-algebra. Let $M$ be a $k$-linear subspace of
$A$ such that all elements of $r(M)$ are algebraic over $k$. Then $M$ is an
MZ-space of $A$ if and only if $Ae\subset M$ for all idempotents $e$ which
belong to $M$.

In particular, if $M$ is an MZ-space of $A$ and $1\in M$, then $M=A$.

###### Corollary 1

Let $A$ be a $k$-algebra and suppose that all elements of $r(M)$ are algebraic
over $k$. Then $M$ is an MZ-space of $A$ if and only if $r(M)$ is an ideal of
$A$.

Not all MZ-spaces are ideals, as the following example shows:

###### Example 2

The finite field $\mathbb{F}_{4}$ has the following four MZ-spaces:

$\displaystyle\mathbb{F}_{4}\qquad\\{0\\}\qquad\\{0,x\\}\qquad\\{0,x+1\\}.$

_Proof._ The sets $\\{0\\}$, $\\{0,x\\}$ and $\\{0,x+1\\}$ are the only ${\bf
Z}$-linear subspaces of $\mathbb{F}_{4}$ that do not contain $1$. Since
$\mathbb{F}_{4}^{*}$ is cyclic, we find that $r^{\prime}(M)=\\{0\\}$ when $M$
is a ${\bf Z}$-linear subspace that does not contain $1$. It is clear that
$r^{\prime}(M)\subset sr(M)$, thus $M$ is an MZ-space. $\triangle$

###  3.2 Partial fractions decomposition

We have the partial fractions decomposition of rational maps:

###### Theorem 3.2

_(Partial Fractions Decomposition)_ Let $k$ be a field, $a_{1},\ldots,a_{n}\in
k$ disinct and $\alpha\in k^{*}$. Define $U(X)=\alpha(X-a_{1})\cdots(X-a_{n})$
and let $V(X)\in k[X]$ with $\deg_{X}V(X)<n$ be arbitrary. Then

$\displaystyle\frac{V(X)}{U(X)}=\sum_{i=1}^{n}\frac{A_{i}}{X-a_{i}}$

with $A_{i}=\frac{V(a_{i})}{U^{\prime}(a_{i})}$.

###### Proof

Note first that $U^{\prime}(a_{i})\neq 0$ for all $i=1,\ldots,n$. This is
since all $a_{i}$ are distinct and we know that if both $U(a_{i})=0$ and
$U^{\prime}(a_{i})=0$, then $a_{i}$ is a double root of $U$. Since this is not
the case, and we know that $U(a_{i})=0$, we must have $U^{\prime}(a_{i})\neq
0$ for all $i=1,\ldots,n$.

Then consider the set $\mathcal{V}_{U(X)}:=\left\\{\frac{V(X)}{U(X)}\mid\deg
V(X)<n\right\\}$ as a subset of $k(X)$, the field of rational functions. Since
$\deg V(X)<n$, we see that $\mathcal{V}_{U(X)}$ is spanned by the set
$\left\\{\frac{1}{U(X)},\frac{X}{U(X)},\ldots,\frac{X^{n-1}}{U(X)}\right\\}$.
It is clearly a $k$-linear subspace of $k(X)$. Indeed, the elements
$\frac{1}{U(X)},\frac{X}{U(X)},\ldots,\frac{X^{n-1}}{U(X)}$ are $k$-linearly
independent as well. Hence $\dim_{k}\mathcal{V}_{U(X)}=n$.

Now consider the elements $\frac{1}{X-a_{1}},\ldots,\frac{1}{X-a_{n}}$. One
easily verifies that they belong to $\mathcal{V}_{U(X)}$. We show that they
are linearly independent, and hence form a basis. Therefore, suppose that
there exist $\lambda_{1},\ldots,\lambda_{n}\in k$ such that

$\displaystyle\sum_{i=1}^{n}\frac{\lambda_{i}}{X-a_{i}}=0.$

Then also

$\displaystyle\lambda_{1}+\sum_{i=2}^{n}\frac{\lambda_{i}(X-a_{1})}{X-a_{i}}=0.$

By substituting $a_{1}$ for $X$, we find $\lambda_{1}=0$.

Then $\sum_{i=2}^{n}\frac{\lambda_{i}}{X-a_{i}}=0$ and we repeat this
procedure. We find that indeed $\frac{1}{X-a_{1}},\ldots,\frac{1}{X-a_{n}}$
are linearly independent.

Since now these elements constitute a basis for $\mathcal{V}_{U(X)}$, given
any $V(X)$ with $\deg V(X)<n$, we can write

$\displaystyle\frac{V(X)}{U(X)}=\sum_{i=1}^{n}\frac{A_{i}}{X-a_{i}}$

for some $A_{1},\ldots,A_{n}\in k$. We now determine those $A_{i}$. By
multiplying both sides with $(X-a_{j})$ we get

$\displaystyle\frac{V(X)}{U(X)}\cdot(X-a_{j})=A_{j}+\sum_{i\neq
j}\frac{A_{i}(X-a_{j})}{X-a_{i}}.$

Note that the $X-a_{j}$ cancels with the $X-a_{j}$ that occurs in $U(X)$. So
we can substitute $a_{j}$ for $X$ and obtain

$\displaystyle\frac{V(a_{j})}{U^{\prime}(a_{j})}=A_{j},$

as required. Note that we used that $U^{\prime}(a_{j})=\alpha\prod_{i\neq
j}(a_{j}-a_{i})$. ∎

##  4 The Theorem of Duistermaat and Van der Kallen

In this section we state the Theorem of Duistermaat and Van der Kallen in one
dimension. We shall prove a more general theorem later on, adapting a
technique by Monsky. Before we do that, we need to discuss some theorems, all
of which are self-contained and needed for our proof. At the end, we shall
state the Theorem of Duistermaat and Van der Kallen. The first necessary
theorem is the Newton-Puiseux Theorem. (See: New , Pui , Pui2 )

######  Theorem 4.1 (Newton-Puiseux Theorem New , Pui , Pui2 )

Let $k$ be an algebraically closed field of characteristic zero and
$f(X)=\sum_{i=0}^{n}a_{i}(t)X^{i}\in k((t))[X]$ with $n:=\deg_{X}f(X)\geq 1$.
Then there exists some $p\geq 1$ such that $f(X)$ splits completely in linear
factors over $k((t^{1/p}))$.

Next, we need a special ring for our proof:

###### Definition 4

Let $k$ be a field and $v$ a valuation on $k$ such that $k$ is complete with
respect to $v$. Define $k[[X,X^{-1}]]$ to be the ring of formal series
$\sum_{n=-\infty}^{\infty}c_{n}X^{n}$ such that $\lim_{|n|\to\infty}c_{n}=0$.

###### Proposition 2

Let $k={\bf C}((z^{1/p}))$ and $v$ the valuation of $k$ such that
$v(z^{1/p})=1/p$. Then ${\bf C}[X,X^{-1}][[z]]$ is a subring of
$k[[X,X^{-1}]]$.

###### Proof

Note that ${\bf C}((z^{1/p}))$ is complete with respect to $v$, one could
write $z^{1/p}=t$, then $k={\bf C}((t))$ and $v(t)=1/p$. Hence $k[[X,X^{-1}]]$
is defined.

Let $f(z)=a_{0}(X)+a_{1}(X)z+a_{2}(X)z^{2}+\ldots$ be an arbitrary element of
${\bf C}[X,X^{-1}][[z]]$ and let $\varepsilon>0$ be arbitrary. Write this
element $f(z)=\sum_{n=-\infty}^{\infty}c_{n}X^{n}$.

We want to determine $N$ such that $|c_{n}|<\varepsilon$ for all $|n|\geq N$.
First, observe that $v(z^{i})=iv(z)=i/p$. So
$|z^{i}|=2^{-v(z^{i})}=2^{-i/p}<\varepsilon$ if
$i>p\log_{2}(\frac{1}{\varepsilon})$.

Write $N(\varepsilon):=\lceil p\log_{2}(\frac{1}{\varepsilon})\rceil$. Then
determine

$\displaystyle N:=\max_{0\leq i\leq N(\varepsilon)}\\{\deg a_{i}(X),\deg
a_{i}(X^{-1})\\}.$

Then, if $|n|>N$, all $z^{i}$ contributing to $c_{n}$ come from $a_{i}(X)$
with $i>N(\varepsilon)$. Since $|z^{i}|<\varepsilon$ if $i>N(\varepsilon)$, it
follows that $|c_{n}|<\varepsilon$. Hence $|c_{n}|<\varepsilon$ if $|n|>N$.

Hence indeed, $f(z)\in k[[X,X^{-1}]]$, as required. ∎

The theorem of Duistermaat and Van der Kallen is as follows:

######  Theorem 4.2 (Duistermaat - Van der Kallen DuiKal )

Let $X_{1},\ldots,X_{n}$ be $n$ commutative variables and let $M$ be the
subspace of the Laurent polynomial algebra ${\bf
C}[X_{1},\ldots,X_{n},X_{1}^{-1},\ldots,X_{n}^{-1}]$ consisting of those
Laurent polynomials with no constant term:

$\displaystyle M=\\{f\in{\bf
C}[X_{1},\ldots,X_{n},X_{1}^{-1},\ldots,X_{n}^{-1}]\mid f_{0}=0\\}.$

Then $M$ is an MZ-space of ${\bf
C}[X_{1},\ldots,X_{n},X_{1}^{-1},\ldots,X_{n}^{-1}]$.

We omit the proof, and prove a more general theorem, for dimension $n=1$, in
the next section, see Theorem 5.1.

##  5 Kernels of linear maps: A generalisation of Duistermaat - Van der Kallen
in dimension one

In this section we will discuss a generalisation of Duistermaat - Van der
Kallen. Note that

$\displaystyle\varphi\colon{\bf C}[X,X^{-1}]\to{\bf C},f\mapsto f_{0}$

is a linear map of ${\bf C}$-vector spaces for which $\varphi(X^{N})=0$ for
all $|N|\geq 1$. The theorem of Duistermaat and Van der Kallen can thus be
phrased as: $\operatorname{Ker}\varphi$ is an MZ-space.

We now investigate arbitrary ${\bf C}$-linear maps of a form, $L\colon{\bf
C}[X,X^{-1}]\to{\bf C}$ or $L\colon{\bf C}[X]\to{\bf C}$.

###### Remark 3

  1. 1.

If $L$ is injective, i.e., $\operatorname{Ker}L=0$, then $\operatorname{Ker}L$
is an MZ-space.

  2. 2.

If $L=0$, i.e., the trivial linear map, then $\operatorname{Ker}L={\bf
C}[X,X^{-1}]$ (or indeed $\operatorname{Ker}L={\bf C}[X]$), an MZ-space.

From now on, when we write $L$ is a linear map, we mean a non-trivial non-
injective linear map.

###### Lemma 1

Let $L\colon{\bf C}[X]\to{\bf C}$ be a ${\bf C}$-linear map for which there
exists an $N\geq 1$ such that $L(X^{n})=0$ for all $n\in{\bf Z}_{\geq N}$.
Then $\operatorname{Ker}L$ is an MZ-space of ${\bf C}[X]$ if and only if
$L(1)\neq 0$.

_Proof._ $\Rightarrow:)$ Suppose that $L(1)=0$, then
$1\in\operatorname{Ker}L$. Since $\operatorname{Ker}L$ is an MZ-space of ${\bf
C}[X]$, we find that $\operatorname{Ker}L={\bf C}[X]$, i.e., $L=0$, a
contradiction. So $L(1)\neq 0$.

$\Leftarrow:)$ Note that $L(X^{N})=0$. So for any $f=\sum_{i=0}^{n}a_{i}X^{i}$
we have

$\displaystyle
L(fX^{N})=L\left(\sum_{i=0}^{n}a_{i}X^{i+N}\right)=\sum_{i=0}^{n}a_{i}L(X^{i+N})=0,$

hence $(X^{N})\subset\operatorname{Ker}L$. Then by Corollary 1 it suffices to
show that $\operatorname{Ker}L/(X^{N})$ is an MZ-space of ${\bf
C}[X]/(X^{N})$. We will also use Zhao’s Theorem. We therefore need to show
that all elements in $r(\operatorname{Ker}L/(X^{N}))$ are algebraic over ${\bf
C}$ and that ${\bf C}[X]/(X^{N})e\subset\operatorname{Ker}L/(X^{N})$ for all
idempotents $e$ in $\operatorname{Ker}L/(X^{N})$.

Note first that $\operatorname{Ker}L/(X^{N})\subset{\bf C}[X]/(X^{N})$, hence
any idempotent of $\operatorname{Ker}L/(X^{N})$ is an idempotent of ${\bf
C}[X]/(X^{N})$. We start by determining the idempotents of ${\bf
C}[X]/(X^{N})$. Since $(X)\subset{\bf C}[X]$ is a maximal ideal, we find that
${\bf C}[X]/(X^{N})$ is a local ring. A local ring has only idempotents $0,1$.
Since $L(1)\neq 0$, we see that $0$ is the only idempotent in
$\operatorname{Ker}L/(X^{N})$, and we trivially have
$0\subset\operatorname{Ker}L/(X^{N})$.

Hence it remains to show that every element in
$r(\operatorname{Ker}L/(X^{N}))$ is algebraic over ${\bf C}$. Note that

$\displaystyle r(\operatorname{Ker}L/(X^{N}))\subset{\bf C}[X]/(X^{N}),$

and every element of ${\bf C}[X]/(X^{N})$ is algebraic over ${\bf C}$ due to
its dimension being finite.

Since we now satisfy the hypotheses of Zhao’s Theorem, we find that
$\operatorname{Ker}L/(X^{N})$ is an MZ-space of ${\bf C}[X]/(X^{N})$ and hence
$\operatorname{Ker}L$ is an MZ-space of ${\bf C}[X]$. $\triangle$

Note that the above lemma can be generalized to an arbitrary field $k$, with
an identical proof. The following is the generalization of the $1$-dimensional
version of the theorem of Duistermaat and Van der Kallen.

###### Theorem 5.1

Let $L\colon{\bf C}[X,X^{-1}]\to{\bf C}$ be a ${\bf C}$-linear map for which
there exists an $N\geq 1$ such that $L(X^{n})=0$ for all $n\in{\bf Z}_{\geq
N}$ and all $n\in{\bf Z}_{\leq-N}$. Then $\operatorname{Ker}L$ is an MZ-space
of ${\bf C}[X,X^{-1}]$ if and only if $L(1)\neq 0$.

The proof of this theorem relies on the following lemmas and is based on the
proof of the $1$-dimensional Duistermaat-Van der Kallen theorem by Monsky
Monskyproof , monskylemma  (Lemmas 3.6 and 3.7).

###### Lemma 2

Let $L\colon{\bf C}[X,X^{-1}]\to{\bf C}$ be a ${\bf C}$-linear map for which
there exists an $N\geq 1$ such that $L(X^{n})=0$ for all $n\in{\bf Z}_{\geq
N}$ and all $n\in{\bf Z}_{\leq-N}$. If $L(1)\neq 0$ and $f\in
r^{\prime}(\operatorname{Ker}L)$, then $f\in{\bf C}[X]$ or $f\in{\bf
C}[X^{-1}]$.

_Proof._ Suppose $f\not\in{\bf C}[X]$ and $f\not\in{\bf C}[X^{-1}]$, i.e.
$f=\alpha X^{-s}+\ldots+\beta X^{r}$ with $\alpha,\beta\neq 0$ and $s,r\geq
1$. Then consider the power series

$\displaystyle W(z)=\sum_{j=0}^{\infty}L(f^{j})z^{j}.$

We will show that $W(z)\neq 1$. That is, $L(f^{m})\neq 0$ for some $m\geq 1$.
This contradicts $f\in r^{\prime}(\operatorname{Ker}L)$.

Write $U(X)=X^{s}(1-zf(X))\in{\bf C}(z)[X]\subset{\bf C}((z))[X]$ and
$n:=r+s$. Then by the Newton-Puiseux Theorem (see Theorem 4.1) there exists
some $p\geq 1$ such that

$\displaystyle U(X)=(-\beta z)(X-a_{1})\cdots(X-a_{n}),\ \text{with all
}a_{i}\in{\bf C}((z^{\frac{1}{p}})).$

Write $k:={\bf C}((z^{\frac{1}{p}}))$. There we have the valuation $\nu$
defined by $\nu(z^{\frac{1}{p}})=\frac{1}{p}$ and $k$ is complete with respect
to this valuation. Furthermore $k[[X,X^{-1}]]$ is a ring, by Theorem 4. Extend
$L$ to $k[[X,X^{-1}]]$ as

$\displaystyle L\colon k[[X,X^{-1}]]\to k$

by

$\displaystyle
L\left(\sum_{n=-\infty}^{\infty}c_{n}X^{n}\right)=\sum_{n=-(N-1)}^{N-1}c_{n}L(X^{n}).$

Since ${\bf C}[X,X^{-1}][[z]]$ is a subring of $k[[X,X^{-1}]]$ (see
Proposition 2), we have

$\displaystyle\sum_{m\geq 0}f^{m}z^{m}\in k[[X,X^{-1}]].$

Then $1-zf$ is invertible in $k[[X,X^{-1}]]$ and so is $U(X)$. Write

$\displaystyle
W(z)=L\left(\frac{1}{1-zf}\right)=L\left(\frac{X^{s}}{U(X)}\right).$

We will now use the partial fractions decomposition of $X^{s}/U(X)$.
Therefore, note that since we have $(-1)^{n}a_{1}\cdots a_{n}=\alpha/\beta$,
we find that $a_{i}\neq 0$ for all $i=1,\ldots,n$. From $U(a_{i})=0$ we find
then that $f(a_{i})=1/z$ for all $i=1,\ldots,n$.

So $f^{\prime}(a_{i})a_{i}^{\prime}=-z^{-2}$. Also
$U^{\prime}(X)=sX^{s-1}(1-zf(X))+X^{s}(-z)f^{\prime}(X)$ and
$U^{\prime}(a_{i})=-za_{i}^{s}f^{\prime}(a_{i})\neq 0$. Hence all $a_{i}$ are
distinct. Then by Theorem 3.2 we get a partial fractions decomposition of the
form

$\displaystyle\frac{X^{s}}{U(X)}=\sum_{i=1}^{n}\frac{A_{i}}{X-a_{i}}$

with
$A_{i}=\frac{a_{i}^{s}}{U^{\prime}(a_{i})}=-\frac{1}{zf^{\prime}(a_{i})}$.
Therefore

$\displaystyle\frac{1}{1-zf(X)}=\frac{X^{s}}{U(X)}=\sum_{i=1}^{n}-\frac{1}{zf^{\prime}(a_{i})(X-a_{i})}.$
(1)

We now compute the inverse of each factor $X-a_{i}$ in $k[[X,X^{-1}]]$.
Observe that $f(a_{i})=1/z$ implies that $\nu(a_{i})\neq 0$. Then
$\nu(a_{i})>0$ or $\nu(a_{i})<0$.

If $\nu(a_{i})>0$, then we have

$\displaystyle(X-a_{i})^{-1}=X^{-1}(1-a_{i}X^{-1})^{-1}=X^{-1}\sum_{m=0}^{\infty}a_{i}^{m}X^{-m}\in
k[[X,X^{-1}]]$

while if $\nu(a_{i})<0$, then

$\displaystyle(X-a_{i})^{-1}=-a_{i}^{-1}(1-a_{i}^{-1}X)^{-1}=-a_{i}^{-1}\sum_{m=0}^{\infty}(a_{i}^{-1}X)^{m}\in
k[[X,X^{-1}]].$

So we find that

$\displaystyle\frac{1}{1-zf(X)}$ $\displaystyle=-\left(\sum_{i\in
S^{+}}\frac{1}{zf^{\prime}(a_{i})(X-a_{i})}+\sum_{i\in
S^{-}}\frac{1}{zf^{\prime}(a_{i})(X-a_{i})}\right)$
$\displaystyle=-\left(\sum_{i\in
S^{+}}\left(\frac{1}{zf^{\prime}(a_{i})}\right)\left(\sum_{m=0}^{\infty}a_{i}^{m}X^{-m}\right)X^{-1}-\sum_{i\in
S^{-}}\left(\frac{1}{zf^{\prime}(a_{i})}\right)\left(\sum_{m=0}^{\infty}(a_{i}^{-1}X)^{m}\right)a_{i}^{-1}\right)$

where $S^{+}=\\{i\mid v(a_{i})>0\\}$ en $S^{-}=\\{i\mid v(a_{i})<0\\}$.

Then

$\displaystyle W(z)$ $\displaystyle=L\left(\frac{1}{1-zf(X)}\right)$
$\displaystyle=L\left(-\sum_{i\in
S^{+}}\left(\frac{1}{zf^{\prime}(a_{i})}\right)\left(\sum_{m=0}^{\infty}a_{i}^{m}X^{-m}\right)X^{-1}\right)+L\left(\sum_{i\in
S^{-}}\left(\frac{1}{zf^{\prime}(a_{i})}\right)\left(\sum_{m=0}^{\infty}(a_{i}^{-1}X)^{m}\right)a_{i}^{-1}\right)$
$\displaystyle=-\sum_{i\in
S^{+}}\frac{1}{zf^{\prime}(a_{i})}L\left(X^{-1}\sum_{m=0}^{\infty}a_{i}^{m}X^{-m}\right)+\sum_{i\in
S^{-}}\frac{1}{zf^{\prime}(a_{i})}L\left(a_{i}^{-1}\sum_{m=0}^{\infty}(a_{i}^{-1}X)^{m}\right)$
$\displaystyle=-\sum_{i\in
S^{+}}\frac{1}{zf^{\prime}(a_{i})}\sum_{m=0}^{N-1}a_{i}^{m}L(X^{-(m+1)})+\sum_{i\in
S^{-}}\frac{1}{zf^{\prime}(a_{i})}\sum_{m=0}^{N-1}a_{i}^{-(m+1)}L(X^{m})$

Since $f^{\prime}(a_{i})a_{i}^{\prime}=-z^{-2}$ we have

$\displaystyle W(z)=\sum_{i\in
S^{+}}\sum_{m=0}^{N-1}za_{i}^{\prime}a_{i}^{m}L(X^{-(m+1)})-\sum_{i\in
S^{-}}\sum_{m=0}^{N-1}za_{i}^{\prime}a_{i}^{-(m+1)}L(X^{m})$

We want to show $W(z)\neq 1$, i.e. we need to show that

$\displaystyle\sum_{i\in
S^{+}}\sum_{m=0}^{N-1}a_{i}^{\prime}a_{i}^{m}L(X^{-(m+1)})-\sum_{i\in
S^{-}}\sum_{m=0}^{N-1}a_{i}^{\prime}a_{i}^{-(m+1)}L(X^{m})\neq\frac{1}{z}$ (2)

To prove this inequality, we study the $a_{i}$ at infinity, i.e. we set
$t=\frac{1}{z}$. Now fix an $i$. Then $f(a_{i})=t$. Since ${\bf C}(z)={\bf
C}(t)$ we find that $a_{i}$ is algebraic over ${\bf C}(t)$ and hence also over
${\bf C}((t))$. Then again by the Newton-Puiseux theorem we can regard $a_{i}$
inside ${\bf C}((t^{\frac{1}{p}}))$ for some $p\geq 1$. Since $a_{i}\neq 0$ we
can write $a_{i}=\sum_{n=m}^{\infty}c_{n}t^{n/p}$ for some $c_{i}\in{\bf C}$
with $c_{m}\neq 0$. Write $w$ for the valuation on ${\bf C}((t^{1/p}))$
defined by $w(t^{1/p})=1/p$. Then $w(a_{i})=m/p$.

Suppose that $w(a_{i})>0$, then

$\displaystyle w(f(a_{i}))=w(\alpha c_{m}^{-s}t^{-ms/p})=-ms/p=-sw(a_{i})<0$

since $s\geq 1$. But since $f(a_{i})=t$ we also have $w(f(a_{i}))=w(t)=1$, a
contradiction.

Similarly, if $w(a_{i})<0$, then $w(f(a_{i}))=w(\beta
c_{m}^{r}t^{mr/p})=mr/p=w(a_{i})r<0$, a contradiction.

Hence $w(a_{i})=0$ and $a_{i}=\sum_{n=0}^{\infty}c_{n}t^{n/p}$ with
$c_{0}\in{\bf C}^{*}$.

Note that $t=\frac{1}{z}$, and $w(t)=1$. So if we show that
$w(a_{i}^{\prime}a_{i}^{m})>1$ for all $m\in{\bf Z}$ we have shown our
inequality 2, and hence that $W(z)\neq 1$. Note that since $f(a_{i})=t$ we
have $a_{i}\not\in{\bf C}$. Hence there exists some $j>0$ with $c_{j}\neq 0$.
Choose such $j$ minimal and write $a_{i}=c_{0}+c_{j}t^{j/p}+R$ with
$w(R)>j/p$. Then $w\left(\frac{da_{i}}{dt}\right)=\frac{j}{p}-1$.

Also
$\frac{a_{i}^{\prime}(z)}{a_{i}(z)}=-\frac{\frac{da_{i}}{dt}}{a_{i}}t^{2}$ and
$w(a_{i})=0$, therefore we have

$\displaystyle
w(a_{i}^{\prime}a_{i}^{m})=w(-\frac{da_{i}}{dt}t^{2}a_{i}^{m})=\frac{j}{p}-1+2>1,$

which concludes the proof. $\triangle$

###### Lemma 3

Let $L\colon{\bf C}[X,X^{-1}]\to{\bf C}$ be a ${\bf C}$-linear map for which
there exists an $N\geq 1$ such that $L(X^{n})=0$ for all $n\in{\bf Z}_{\geq
N}$ and all $n\in{\bf Z}_{\leq-N}$. If $L(1)\neq 0$ and $f\in{\bf C}[X]\cap
r(\operatorname{Ker}L)$, then $f\in X{\bf C}[X]$.

_Proof._ Let $f\in{\bf C}[X]\cap r(\operatorname{Ker}L)$, then $f\in
r(\operatorname{Ker}L_{\mid{\bf C}[X]})$. By Corollary 1 and Lemma 1 we know
that $r(\operatorname{Ker}L_{\mid{\bf C}[X]})$ is an ideal of ${\bf C}[X]$.
Since $L(1)\neq 0$ we find that $\operatorname{Ker}L_{\mid{\bf C}[X]}\neq{\bf
C}[X]$, and hence $r(\operatorname{Ker}L_{\mid{\bf C}[X]})\neq{\bf C}[X]$. As
$r(\operatorname{Ker}L_{\mid{\bf C}[X]})$ is a principal ideal (${\bf C}[X]$
is Euclidean), it is generated by some monic $g\in{\bf C}[X]\setminus{\bf C}$.
Since it is clear that $X\in r(\operatorname{Ker}L_{\mid{\bf C}[X]})$ we find
$(X)\subset r(\operatorname{Ker}L_{\mid{\bf C}[X]})$. Hence $X$ is a multiple
of this $g$. Hence $g=X$. So $r(\operatorname{Ker}L_{\mid{\bf C}[X]})=(X)$ and
$f\in(X)=X{\bf C}[X]$. $\triangle$

We will use the following definition(s):

###### Definition 5

For $f\in{\bf C}[X,X^{-1}]\setminus\\{0\\}$ we have the following quantities:

$\displaystyle\deg_{+}f=\max\\{i:f_{i}\neq 0\\}$

and

$\displaystyle\deg_{-}f=\min\\{i:f_{i}\neq 0\\}.$

###### Proof

_(of Theorem 5.1)_ $\Rightarrow:)$ Identical to the proof of Lemma 1.

$\Leftarrow:)$ Suppose that $L(1)\neq 0$ and $f\in r(\operatorname{Ker}L)$. We
want to show that $gf^{m}\in\operatorname{Ker}L$ for all $m\gg 0$. By Lemma 2
we find that $f^{N}\in{\bf C}[X]$ or $f^{N}\in{\bf C}[X^{-1}]$ for some $N\geq
1$, but then also $f\in{\bf C}[X]$ or $f\in{\bf C}[X]$.

Assume that $f\in{\bf C}[X]$, the argument is similar when $f\in{\bf
C}[X^{-1}]$. Then $f\in{\bf C}[X]\cap r(\operatorname{Ker}L)$. By Lemma 3 we
then find $f\in X{\bf C}[X]$. Hence $L(f^{m})=0$ for all $m\geq N$.

Then $L(gf^{m})=0$ for all $m\geq N-\deg_{-}g$. ∎

Theorem 5.1 is false in characteristic $p>0$, as was shown in Wil .

##  6 Encore: Generalizations

In this section we list several generalizations of various results we have
seen in this paper. The proofs can be found in Thesis .

###### Proposition 3

_(Generalization of Lemma 1)_ Let $L\colon{\bf C}[X_{1},\ldots,X_{n}]\to{\bf
C}$ be a ${\bf C}$-linear map for which there exists an $N\geq 1$ such that
$(X_{1},\ldots,X_{n})^{l}\subset\operatorname{Ker}L$ for all $l\geq N$. Then
$\operatorname{Ker}L$ is an MZ-space of ${\bf C}[X_{1},\ldots,X_{n}]$ if and
only if $L(1)\neq 0$.

We cannot prove

Let $L\colon{\bf C}[X_{1},\ldots,X_{n}]\to{\bf C}$ be a ${\bf C}$-linear map
for which there exists an $N\geq 1$ such that $L(X_{i}^{m})=0$ for all
$m\in{\bf Z}_{\geq N}$ and all $i\in\\{1,\ldots,n\\}$. Then
$\operatorname{Ker}L$ is an MZ-space of ${\bf C}[X_{1},\ldots,X_{n}]$ if and
only if $L(1)\neq 0$.

For this we have the following counterexample:

###### Example 3

Take $n=2$ and $N=2$ in the following form. Let $L\colon{\bf
C}[X_{1},X_{2}]\to{\bf C}$ be the ${\bf C}$-linear map given by

$\displaystyle L(X_{1}^{i}X_{2}^{j})=\begin{cases}1&\mbox{if }i=1\mbox{ or
}i=j=0\\\ 0&\mbox{else }\end{cases}$

Then we have $L(1)=1$, and $L(X_{i}^{m})=0$ for all $m\geq 2$. It is then
clear that $X_{2}\in r(\operatorname{Ker}L)$ by construction, but
$X_{1}X_{2}^{m}\not\in\operatorname{Ker}L$ for all $m\geq 1$. Therefore
$\operatorname{Ker}L$ is not an MZ-space.

While it is technically not a generalization, we also have the following
lemma, which uses the same method-of-proof:

###### Proposition 4

Let $k$ be a field, $A$ a _finite-dimensional_ $k$-algebra and $\mathfrak{m}$
a maximal ideal of $A$. Let $L\colon A\to k$ be a $k$-linear map for which
there exists an $N\geq 1$ such that
$\mathfrak{m}^{l}\subset\operatorname{Ker}L$ for all $l\geq N$. Then
$\operatorname{Ker}L$ is an MZ-space of $A$ if and only if $L(1)\neq 0$.

In this section we discuss a generalization of Zhao’s Idempotency Theorem
(Theorem 3.1), that is in the exercises of Ess2 .

###### Definition 6

Let $R$ be a ring, $S$ a multiplicatively closed subset of $R$ and $A$ an
$R$-algebra. An $R$-submodule $M$ of $A$ is called _$S$ -saturated_ if we have
$a\in M$ for all $a\in A$ for which there exists some $s\in S$ such that
$sa\in M$, i.e.,

$\displaystyle\\{a\in A\mid\exists s\in S[sa\in M]\\}\subset M.$

###### Definition 7

Let $R$ be a ring, $S$ a multiplicatively closed subset of $R$ and $A$ an
$R$-algebra. Then $e\in A$ is an _$S$ -idempotent_ if there exists some $s\in
S\setminus\\{0\\}$ such that $e^{2}=se$.

######  Theorem 6.1 (Generalized Zhao Theorem Ess2 )

Let $R$ be a domain, $S=R\setminus\\{0\\}$ and $A$ an $R$-algebra. Let $M$ be
an $S$-saturated $R$-submodule of $A$ such that all elements of $r(M)$ are
algebraic over $R$. Then $M$ is an MZ-space of $A$ if and only if for every
$S$-idempotent $e$ of $A$ which belongs to $M$ we have that for every $a\in A$
there exists an $s\in S$ such that $sae\in M$.

Using this Generalized Zhao Theorem, we can prove the following generalization
of Lemma 1:

###### Proposition 5

Let $R$ be a domain and $L\colon R[Y]\to R$ be a linear map such that there
exists an $N\geq 1$ for which $L(Y^{n})=0$ for all $n\geq N$. Then
$\operatorname{Ker}L$ is an MZ-space of $R[Y]$ if and only if $L(1)\neq 0$.

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