$\displaystyle\stackrel{{\scriptstyle\star}}{{\leq}}f_{Q}\left(\bar{\boldsymbol{\beta}},\bar{\boldsymbol{\gamma}},\mathbf{W}_{\rm
new}\right)$ (136c)
$\displaystyle\stackrel{{\scriptstyle\clubsuit}}{{\leq}}f_{Q}\left(\boldsymbol{\beta}^{\mathrm{F}}\left(\mathbf{W}_{\rm
new}\right),\boldsymbol{\gamma}^{\mathrm{F}}\left(\mathbf{W}_{\rm
new}\right),\mathbf{W}_{\rm new}\right)$ (136d)
$\displaystyle=\mathcal{R}_{\rm M}^{\mathrm{ssr}}\left(\bar{\mathbf{W}}_{\rm
new}\right)$ (136e)
$\displaystyle=\mathcal{R}_{\mathrm{q}}^{\mathrm{ssr}}\left(\bar{\mathbf{W}}_{\rm
new},\bar{{\boldsymbol{\phi}}},\bar{\mathbf{b}}\right)$ (136f)

where $\dagger$ follows (133), $\star$ comes from (135), and $\clubsuit$ is
concluded by (134b).

##  Appendix E Proof of Lemma 2

The proof follows the same approach as in Appendix D. Assuming the
initialization $\mathbf{W}_{0}=\bar{\mathbf{W}}$ and
$\mathbf{b}_{0}=\bar{\mathbf{b}}$, we start the proof by defining

$\displaystyle
f_{Q}\left(\mathbf{f},\boldsymbol{\varpi},{\boldsymbol{\phi}}\right)$
$\displaystyle=Q_{2}^{\mathrm{m}}\left({\boldsymbol{\phi}},\mathbf{f}\right)+Q_{2}^{\mathrm{e}}\left({\boldsymbol{\phi}},\boldsymbol{\varpi}\right)+\sum_{k=1}^{K}\omega_{k}\bar{b}_{k}\log
B_{2Uk}^{\mathrm{e}},$ (137a) $\displaystyle\mathcal{R}_{\rm
M}^{\mathrm{ssr}}\left({\boldsymbol{\phi}}\right)$
$\displaystyle=\mathcal{R}_{\mathrm{q}}^{\mathrm{ssr}}\left(\bar{\mathbf{W}},{\boldsymbol{\phi}},\bar{\mathbf{b}}\right)$
(137b)

with $Q_{2}^{\mathrm{m}}\left({\boldsymbol{\phi}},\mathbf{f}\right)$,
$Q_{2}^{\mathrm{e}}\left({\boldsymbol{\phi}},\boldsymbol{\varpi}\right)$ and
$B_{2Uk}^{\mathrm{e}}$ defined in Section V-B. We further define

$\displaystyle\mathbf{f}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$
$\displaystyle=\left[f_{1}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right),\ldots,f_{K}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)\right],$
(138a)
$\displaystyle\boldsymbol{\varpi}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$
$\displaystyle=\left[\varpi_{1}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right),\ldots,\varpi_{K}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)\right]$
(138b)

for a given ${\boldsymbol{\phi}}$, where

$\displaystyle f_{k}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$
$\displaystyle=\frac{\displaystyle\sqrt{\omega_{k}\bar{b}_{k}\left(1+\frac{A_{2k}^{\rm
m}\left({\boldsymbol{\phi}}\right)}{B_{2k}^{\rm
m}\left({\boldsymbol{\phi}}\right)}\right)}\left(\mathbf{h}^{\mathsf{H}}_{\mathrm{d},k}\bar{{\mathbf{w}}}_{k}+{\boldsymbol{\phi}}^{\mathsf{H}}\mathbf{H}^{\mathsf{H}}_{k}\bar{{\mathbf{w}}}_{k}\right)}{\left\lVert\mathbf{h}^{\mathsf{H}}_{\mathrm{d},k}\bar{\mathbf{W}}+{\boldsymbol{\phi}}^{\mathsf{H}}\mathbf{H}^{\mathsf{H}}_{k}\bar{\mathbf{W}}\right\rVert^{2}+\sigma_{k}^{2}},$
(139a) $\displaystyle\varpi_{k}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$
$\displaystyle=\sqrt{\frac{\displaystyle\omega_{k}\bar{b}_{k}\left(1+\frac{A_{2k}^{\rm
e}\left({\boldsymbol{\phi}}\right)}{B_{2k}^{\rm
e}\left({\boldsymbol{\phi}}\right)}\right)}{B_{2Uk}^{\rm e}}}\sqrt{A_{2k}^{\rm
e}\left({\boldsymbol{\phi}}\right)}.$ (139b)

By substitution, it is easily shown that for a given ${\boldsymbol{\phi}}$

$\displaystyle\mathcal{R}_{\rm
M}^{\mathrm{ssr}}\left({\boldsymbol{\phi}}\right)=f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),{\boldsymbol{\phi}}\right).$
(140)

Let $\bar{{\boldsymbol{\phi}}}$ and ${\boldsymbol{\phi}}_{\rm new}$ indicate
the phase-shift vectors updated at the end of iteration $t$ and $t+1$,
respectively. The convergence of the MM algorithm, i.e., Algorithm 1,
guarantees

$\displaystyle
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\bar{{\boldsymbol{\phi}}}\right)\leq
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),{\boldsymbol{\phi}}_{\rm
new}\right).$ (141)

This inequality holds as an identity, if Algorithm 1 iterates only for one
iteration. Following concavity of the marginal function
$f_{Q}\left(\mathbf{f},\boldsymbol{\varpi},{\boldsymbol{\phi}}\right)$ in
terms of $\mathbf{f}$ and $\boldsymbol{\varpi}$ with maximum being at
$\mathbf{f}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$ and
$\boldsymbol{\varpi}^{\mathrm{F}}\left({\boldsymbol{\phi}}\right)$, for a
given ${\boldsymbol{\phi}}$, we can write that

$\displaystyle
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),{\boldsymbol{\phi}}\right)\leq
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}\right),{\boldsymbol{\phi}}\right)$
(142)

for any feasible ${\boldsymbol{\phi}}$.

We hence can use the above inequalities and write

$\displaystyle\mathcal{R}_{\mathrm{q}}^{\mathrm{ssr}}\left(\bar{\mathbf{W}},\bar{{\boldsymbol{\phi}}},\bar{\mathbf{b}}\right)$
$\displaystyle=\mathcal{R}_{\rm
M}^{\mathrm{ssr}}\left(\bar{\boldsymbol{\phi}}\right)$ (143a)
$\displaystyle=f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\bar{\boldsymbol{\phi}}\right)$
(143b) $\displaystyle\leq
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left(\bar{{\boldsymbol{\phi}}}\right),{\boldsymbol{\phi}}_{\rm
new}\right)$ (143c) $\displaystyle\leq
f_{Q}\left(\mathbf{f}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}_{\rm
new}\right),\boldsymbol{\varpi}^{\mathrm{F}}\left({{\boldsymbol{\phi}}}_{\rm
new}\right),{\boldsymbol{\phi}}_{\rm new}\right)$ (143d)
$\displaystyle=\mathcal{R}_{\rm
M}^{\mathrm{ssr}}\left({\boldsymbol{\phi}}_{\rm new}\right)$ (143e)
$\displaystyle=\mathcal{R}_{\mathrm{q}}^{\mathrm{ssr}}\left(\bar{\mathbf{W}},{\boldsymbol{\phi}}_{\rm
new},\bar{\mathbf{b}}\right).$ (143f)

Noting that (141) is the only constraint we need for the convergence proof, in
terms of phase-tuning update algorithm, we conclude that the proof is valid
for any alternative of Algorithm 1 which converges to its fixed-point in a
non-decreasing fashion.

MIMO

    multiple-input multiple-output
mmW

    millimeter wave
TDD

    time division duplexing
SINR

    signal-to-interference-plus-noise ratio
CSI

    channel state information
AO

    alternating optimization
r.h.s.

    right hand side
l.h.s.

    left hand side
AWGN

    additive white Gaussian noise
i.i.d.

    independent and identically distributed
UT

    user terminal
TAS

    transmit antenna selection
RF

    radio frequency
SRZF

    secure regularized zero-forcing
IRS

    intelligent reflecting surface
MM

    majorization-maximization
MMSE

    minimum mean square error
FP

    fractional programming
BS

    base station
BCD

    block coordinate descent
QoS

    quality-of-service
MISO

    multiple-input single-output

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  *[MIMO]: multiple-input multiple-output
  *[mmW]: millimeter wave
  *[IRS]: intelligent reflecting surface
  *[IRSs]: intelligent reflecting surface
  *[BS]: base station
  *[RF]: radio frequency
  *[MISO]: multiple-input single-output
  *[UT]: user terminal
  *[UTs]: user terminal
  *[FP]: fractional programming
  *[AO]: alternating optimization
  *[TDD]: time division duplexing
  *[CSI]: channel state information
  *[i.i.d.]: independent and identically distributed
  *[SINR]: signal-to-interference-plus-noise ratio
  *[BCD]: block coordinate descent
  *[MM]: majorization-maximization
  *[SRZF]: secure regularized zero-forcing


$\hat{K}[[y_{1},\ldots,y_{n},\partial_{1}^{\prime},\ldots,\partial_{n-1}^{\prime},\partial_{n}^{-1}]]$).

The proof is by induction on $k$. Let $k=1$. Let $P_{0}(r)$ denotes the open
poly-cylinder in $\widehat{K}^{n+m}$ with the center at zero. Then for some
small $r$ we have for any $\lambda=(\lambda_{1},\ldots,\lambda_{n-1})\in
K^{n-1}$, $|\lambda|<r$ and
$(y_{1},\ldots,y_{n},\partial_{1}^{\prime}=t_{1},\ldots,\partial_{n-1}^{\prime}=t_{n-1},\partial_{n}^{-1})\in
P_{0}(r-|\lambda|)$ (see e.g. [55, Ch. 2])

$v_{1}^{\prime}(t_{1}+\lambda_{1},\ldots,t_{n-1}+\lambda_{n-1})=\frac{1}{\underline{j}!}\sum_{\underline{j}=(j_{1},\ldots,j_{n-1})\geq(\underline{0})}\underline{\partial}^{\underline{j}}(v_{1}^{\prime})(\lambda_{1},\ldots,\lambda_{n-1})t^{\underline{j}}.$

For any fixed $\lambda_{1},\ldots,\lambda_{n-1}\in K$, $|\lambda|<r$ we’ll
consider the function
$v_{1}^{\prime}(t_{1}+\lambda_{1},\ldots,t_{n-1}+\lambda_{n-1})$ as an element
of the ring
$\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[t_{1},\ldots,t_{n-1}]]$,
and we will use the same notation for the order functions $\mathop{\rm\bf
ord}$, $\mathop{\rm ord}\nolimits_{\Gamma}$ defined on elements of this ring
(setting $\mathop{\rm\bf ord}(t_{i})=0$, $\mathop{\rm
ord}\nolimits_{\Gamma}(t_{i})=(\underline{0})-(n-i)$).

Consider the function
$\tilde{v}_{1}:=\bar{v}_{1}^{\prime}|_{\partial_{n}^{-1}:=0}$. Since
$\mathop{\rm\bf ord}(v_{1}^{\prime})=0$ and
$\sigma(v_{1}^{\prime})=\sigma(\bar{v}_{1}^{\prime})$, we have
$\tilde{v}_{1}\neq 0$. Clearly, there is an open subset $0\in
U\subset\hat{K}^{n-1}$ of points $(\lambda_{1},\ldots,\lambda_{n-1})\in
K^{n-1}$, $|\lambda|<r$ such that
$\tilde{v}_{1}(\lambda_{1},\ldots,\lambda_{n-1})\neq 0$. Then

$\mathop{\rm
ord}\nolimits_{\Gamma}(v_{1}^{\prime}(t_{1}+\lambda_{1},\ldots,t_{n-1}+\lambda_{n-1}))=\mathop{\rm
ord}\nolimits_{\Gamma}(\bar{v}_{1}^{\prime}(t_{1}+\lambda_{1},\ldots,t_{n-1}+\lambda_{n-1}))=(\underline{0}).$

Without loss of generality we can assume
$\tilde{v}_{1}(\lambda_{1},\ldots,\lambda_{n-1})=1$. Note that the series
$f(y):=v_{1}^{\prime}|_{\partial_{1}^{\prime}=\ldots=\partial_{n-1}^{\prime}=\partial_{n}^{-1}=0}-1$
converges in the same poly-cylinder and therefore the series
$w_{1}:=v_{1}^{\prime}-f(y)v_{1}^{\prime}$ converges too; besides,
$\sigma(w_{1})$ is monic and $\sigma(w_{1})=\sigma(\bar{w}_{1})$, $\mathop{\rm
ord}\nolimits_{\Gamma}(\sigma(w_{1}))=\mathop{\rm
ord}\nolimits_{\Gamma}(w_{1})=(\underline{0})$.

For generic $k$ consider the $\hat{K}_{y}$-vector space
$H_{k}=\hat{K}_{y}\langle v_{1}^{\prime}(t),\ldots,v_{k}^{\prime}(t)\rangle$.
By induction, there exists a linear change of variables $\varphi$ such that
there is a basis $w_{1}(t),\ldots,w_{k-1}(t)$ in the space
$\hat{K}_{y}\langle\varphi(v_{1}^{\prime}(t)),\ldots,\varphi(v_{k-1}^{\prime}(t))\rangle$
with the desired properties. Then we can complete this basis to a basis
$w_{1}(t),\ldots,w_{k-1}(t),v_{k}(t)$ such that $\mathop{\rm
ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))\leq(\underline{0})-(k-1)$ (and all
other properties of $\sigma(v_{k}(t))$ remain valid, in particular it is
monic). If $\mathop{\rm
ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))=(\underline{0})-(k-1)$, then clearly
we can choose $v_{k}$ is such a way that $\mathop{\rm
ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))=\mathop{\rm
ord}\nolimits_{\Gamma}(v_{k}(t))$, and we are done, so we assume that
$\mathop{\rm ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))<(\underline{0})-(k-1)$.

If each entry of $\mathop{\rm ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))$ is not
more than one, then

$\sigma(v_{k}(t))=t_{j}+\mbox{terms of lower $\Gamma$-order},$

where $j<n-k$. Now apply the linear change
$\varphi:\partial_{j}\mapsto\partial_{j}+\epsilon\partial_{n-k-1}$,
$\partial_{i}\mapsto\partial_{i}$, $i\neq j$. It is not difficult to see that
$\varphi(w_{i})$, $\varphi(v_{k})$ are well defined for any $\epsilon$, belong
to
$K_{y}[[\partial_{1}^{\prime},\ldots,\partial_{n-1}^{\prime},\partial_{n}^{-1}]]$
and are analytic in the same poly-disc (clearly they are still algebraic;
besides, this linear change is a result of conjugating with the operator
$S:=\exp(\sum_{i=1}^{n-1}(\sum_{j=1}^{n-1}(c_{i,j}-\delta_{i,j})x_{j})*\partial_{i}))\in\hat{D}_{n}^{n}$
for appropriate constants $c_{i,j}$, cf. corollary 7.1 below). Obviously, they
satisfy all our conditions just for $\epsilon=1$.

Assume there is an entry of $\mathop{\rm
ord}\nolimits_{\Gamma}(\sigma(v_{k}(t)))$ which is more than one. Now we
represent the series $w_{i}(t+h)$, $v_{k}(t+h)$, where
$h=(y_{1}=0,\ldots,y_{n}=0,\lambda_{1},\ldots,\lambda_{n-1},\partial_{n}^{-1}=0)\in
P_{0}(r)$, $t=(y_{1},\ldots,y_{n},t_{1},\ldots,t_{n-1},\partial_{n}^{-1})\in
P_{0}(r-|h|)$, in powers of $t$ as vector-rows with entries

$\underline{\partial}^{\underline{j}=(j_{1},\ldots,j_{n-1})}(v_{i}^{\prime})(h):=\\\
\sum_{(i_{1},\ldots,i_{n+1})\geq(\underline{0})}\frac{1}{\underline{i}!}\partial_{y_{1}}^{i_{1}}\ldots\partial_{y_{n}}^{i_{n}}\partial_{\partial_{n}^{-1}}^{i_{n+1}}\frac{1}{\underline{j}!}\partial_{t_{1}}^{j_{1}}\ldots\partial_{t_{n-1}}^{j_{n-1}}(v_{i}^{\prime})(h)y_{1}^{i_{1}}\ldots
y_{n}^{i_{n}}\partial_{n}^{-i_{n+1}}\\\
\in\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[\lambda_{1},\ldots,\lambda_{n-1}]],$

each entry is labelled by the multi-index
$(j_{1},\ldots,j_{n-1})\geq(\underline{0})$, and these indices are ordered
w.r.t. the order mentioned before this lemma. Now we can reformulate the
problem of finding the needed basis in $H_{k}=\hat{K}_{y}\langle
w_{1}(t+h),\ldots,w_{k-1}(t+h),v_{k}(t+h)\rangle$: there is such a basis
$w_{i}$ iff the $k\times k$-minor of the matrix
$(\underline{\partial}^{\underline{j}}(v_{i}^{\prime})(h))$,
$\underline{j}<\underline{0}+k$, $i=1,\ldots,k$ is invertible, i.e. its
determinant is non-zero for some $h$.

To show that such $h$ exists we need to show that the determinant of this
matrix is not identically zero as a series in the ring
$\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[\lambda_{1},\ldots,\lambda_{n-1}]]$
(after, may be, some linear change of coordinates). For convenience, we’ll
present each entry of this minor as a series of the form
$f(\lambda_{1},\ldots,\lambda_{n-1})+({{\cal M}}^{\mathop{\rm\bf ord}f+1})$,
where $f(t_{1},\ldots,t_{n-1})$ is a homogeneous polynomial with respect to
the $\mathop{\rm\bf ord}$-order, and $({{\cal M}}^{\mathop{\rm\bf ord}f+1})$
denotes the sum of elements from the ideal ${{\cal M}}$ generated by
$y_{1},\ldots,y_{n},\partial_{n}^{-1},\lambda_{1},\ldots,\lambda_{n-1}$.
Clearly, we can assume that
$\underline{\partial}^{\underline{j}}(w_{i})(h)=({{\cal M}})$ for
$i=1,\ldots,k-1$,
$(\underline{0})-(k-2)\leq\underline{j}<(\underline{0})-(i-1)$. Making a
linear change
$\partial_{i}\mapsto\partial_{i}+\varepsilon_{i,k}\partial_{k}+\ldots+\varepsilon_{i,n}\partial_{n}$,
$i=1,\ldots,k-1$ for appropriate constants $\varepsilon_{i,j}\in K$ if
necessary we can also assume that
$\underline{\partial}^{\underline{j}}(w_{i})(h)=({{\cal M}})$ for
$i=k,\ldots,n$.

Note that

$\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[w_{1}(t),\ldots,w_{k-1}(t),t_{k},\ldots,t_{n}]]=\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[t_{1},\ldots,t_{n}]].$

Then, note that $\sigma(v_{k}(t))$ written as a series in the ring
$\widehat{K}[[y_{1},\ldots,y_{n},\partial_{n}^{-1}]][[w_{1},\ldots,w_{k-1},t_{k},\ldots,t_{n}]]$
depends on one of $t_{k},\ldots,t_{n}$ (it is not difficult to see that
$v_{k}(t)$ is analytical also as a series in new variables). Indeed, if it
does not depend on $t_{k},\ldots,t_{n}$, then it is algebraically independent
with $t_{k},\ldots,t_{n}$. It is not difficult to see that
$\sigma(v_{1}(t)),\ldots,\sigma(v_{k}(t))$ are algebraically independent over
$K$ iff $\varphi(\sigma(v_{1}(t))),\ldots,\varphi(\sigma(v_{k}(t)))$ are
algebraically independent for any linear change $\varphi$ from the
formulation, and also iff
$\sigma(w_{1}(t)),\ldots,\sigma(w_{k-1}(t)),\sigma(v_{k}(t))$ are
algebraically independent. So, $\sigma(v_{k}(t))$ is also algebraically
independent with $\sigma(w_{1}),\ldots,\sigma(w_{k-1})$ (as it is
algebraically independent with $\sigma(v_{1}),\ldots,\sigma(v_{k-1})$ by
assumption). Thus, applying the inverse linear change of variables to the
subspace generated by these elements, we get $n+1$ algebraically independent
elements in the ring $A$ \- a contradiction with lemma 6.5.

Making a linear change of variables if necessary we can assume that
$\sigma(v_{k}(t))$ depends on $t_{k}$. Then, making a linear change
$\partial_{i}\mapsto\partial_{i}+\lambda_{i}\partial_{n}$, $i=1,\ldots,k$ for
generic $\lambda_{i}$ from some open subset $0\in U\subset\hat{K}^{k}$ (as in
the first induction step) we’ll get a new series $v_{k}(\lambda)$ which
contains a monomial of the form $ct_{k}$, $0\neq c\in\hat{K}$. At the same
time, in the space $\hat{K}\langle
w_{1}(\lambda),\ldots,w_{k-1}(\lambda)\rangle$ we can still choose a basis
with the same properties as before. Then taking an appropriate linear
combination of this basis and new $v_{k}(\lambda)$, we can obtain a series of
the form $v_{k}(\lambda)=\lambda_{k}+({{\cal M}})$ with $\mathop{\rm
ord}\nolimits_{\Gamma}(v_{k})=\mathop{\rm
ord}\nolimits_{Gamma}(\sigma(v_{k}))=\mathop{\rm
ord}\nolimits_{\Gamma}(\lambda_{k})$.

With this notation the minor has the form

$\left(\begin{array}[]{ccccccc}1+({{\cal M}})&({{\cal M}})&\ldots&({{\cal
M}})&({{\cal M}})\\\ ({{\cal M}})&1+({{\cal M}})&\ldots&({{\cal M}})&({{\cal
M}})\\\ \vdots&\ldots&\ddots&({{\cal M}})&\vdots\\\ ({{\cal
M}})&\ldots&\ldots&1+({{\cal M}})&({{\cal M}})\\\ ({{\cal
M}})&\ldots&\ldots&({{\cal M}})&1+({{\cal M}})\\\ \end{array}\right)$

which means that its determinant is not equal to zero, and we are done.

∎

######  Definition 6.10.

Let $(A,W)$ be a 1-Schur pair. A linear change of variables $\varphi$ from
lemma 6.6 is called admissible if the spaces $\varphi(A),\varphi(W)$ are well
defined (i.e. all elements $\varphi(a),\varphi(w)$, $a\in A$, $w\in W$ are
well defined).

######  Remark 6.2.

1) If $K$ is a complete field and $(A,W)$ is a normalized 1-Schur pair
associated with a quasi-elliptic ring $B\subset\hat{D}_{n}$ of rank one, and
if there is a basis $\\{w_{i}\\}$ in $W$ such that for all $i$ $w_{i}\in
K_{y}[[\partial_{1}^{\prime},\ldots,\partial_{n-1}^{\prime},\partial_{n}^{-1}]]$
(the notation is from lemma 6.6)444This condition is trivial if $K_{y}=K$,
then there exist admissible changes. Indeed, by remark 6.2 $A\subset
W\subset{{\rm Quot\,}}(A)$. Then from the proof of lemma 6.6 we get that
elements from $W,A$ are analytic functions (after multiplication with an
appropriate power of $\partial_{n}$), and therefore there exist admissible
changes, because the intersection of countably many open neighbourhoods of
zero in $K^{n-1}$ is not empty for a complete field $K$.

If $A$ is finitely generated over $K$ and $W$ is a finitely generated
$A$-module, the same arguments show there exists an open subset $0\in U\subset
K^{n-1}$ such that the linear change $\varphi$ from lemma 6.6 is admissible
for any $(\lambda_{1},\ldots,\lambda_{n-1})\in U$.

2) Clearly, any invertible linear change of coordinates can be decomposed into
a composition of a linear change from lemma 6.6 and a linear change of the
form $\partial_{i}\mapsto\partial_{i}$, $i=1,\ldots,n-1$,
$\partial_{n}\mapsto\alpha\partial_{n}+\sum_{i=1}^{n-1}c_{i}\partial_{i}$,
$\alpha\neq 0$. Obviously, the last linear change is admissible, i.e. both
spaces are well defined after its application.

3) Note that if $B\subset\hat{D}_{n}$ is a quasi-elliptic ring and $\varphi$
is any linear change then $\varphi(B)$ is a well defined ring in
$\hat{D}_{n}^{sym}$.

4) Let $S$ be the Sato operator from construction 1 constructed by given
quasi-elliptic ring $B\subset\hat{D}_{n}$, let $(A,W)$ be the corresponding
1-Schur pair, let $\varphi$ be an admissible linear change. Then $\varphi(S)$
is also well defined operator from $\hat{\Pi}_{n}$. For, all slices of $S$ are
finite linear combinations of elements from $W$, cf. the proof of theorem 6.1.

5) Let $(A,W)$ be a 1-Schur pair. Then for any admissible $\varphi$ we have
$\varphi(a_{1}a_{2})=\varphi(a_{1})\varphi(a_{2})$,
$\varphi(aw)=\varphi(a)\varphi(w)$ for any $a_{i},a\in A$, $w\in W$ – this
follows easily from the Taylor series expansion.

######  Proposition 6.1.

Let $K$ be a complete field. Let $B\subset\hat{D}_{n}$ be a quasi-elliptic
ring, let $(A,W)$ be a normalized 1-Schur pair associated with $B$ by
construction 1, let $\varphi$ be an admissible linear change of variables.

Then there exists an operator $U\in\hat{U}_{n}^{sym}$ such that
$B^{\prime}:=U^{-1}\varphi(B)U\subset\hat{D}_{n}$. Moreover, $B^{\prime}$ is
the ring associated with the 1-Schur pair $(\varphi(A),\varphi(W))$ by the
construction 2, and if this Schur pair is quasi-elliptic, then $B^{\prime}$ is
quasi-elliptic.

###### Proof.

Let $(A,W)$ be a normalized Schur pair associated with the ring $B$,
$A=S^{-1}BS$, $W=F\diamond S$, $S\in\hat{E}_{n}$, cf. remark 6.1. Then by
remark 6.2, for any $f\in A$ and corresponding operator $\mathtt{L}_{S}(f)$
the operators $\varphi(S)$, $\varphi(f)$, $\varphi(\mathtt{L}_{S}(f))$ are
well defined and moreover, $\varphi(Sf)=\varphi(S)\varphi(f)$,
$\varphi(\mathtt{L}_{S}(f)S)=\varphi(\mathtt{L}_{S}(f))\varphi(S)$. Indeed,
the products of operators on the right hand sides are well defined by their
definition, the left hand sides are well defined as all slices of $Sf$ and
$\mathtt{L}_{S}(f)S$ are finite sums of products of slices of $S$ and $f$ (for
the first product) and of $S$ and polynomials (for the second product). The
identities then follows from the Taylor series expansions of left and right
hand sides (alternatively one can compare
$\varphi(\underline{\partial}^{\underline{j}})\diamond l.h.s$ and
$\varphi(\underline{\partial}^{\underline{j}})\diamond r.h.s$ by induction on
$|\underline{j}|$). Finally we have

$\varphi(Sf)=\varphi(S)\varphi(f)=\varphi(\mathtt{L}_{S}(f)S)=\varphi(\mathtt{L}_{S}(f))\varphi(S),$

i.e. $\varphi(S)\varphi(A)=\varphi(B)\varphi(S)$. Besides,
$\varphi(W)=F\diamond\varphi(S)$ as $\varphi(F)=F$, i.e. $\varphi(S)$ is a
Sato operator for $\varphi(W)$. By theorem 6.1, item 4 there is a unique
factorisation $\varphi(S)=U\circ S_{0}$, where $U\in\hat{U}_{n}^{sym}$, hence
$S_{0}\varphi(A)S_{0}^{-1}=U^{-1}\varphi(B)U$. From construction 2 we know
(cf. also theorem 6.1, item 6) that
$B^{\prime}=S_{0}\varphi(A)S_{0}^{-1}\in\hat{D}_{n}$ and moreover it is quasi-
elliptic if $\varphi(A)$ is. ∎

##  7 Schur theory for $\hat{D}_{n}^{sym}$

There is an analogue of the Schur theory also for the ring
$\hat{D}_{n}^{sym}$. It is completely new and differs from the Schur theory
for $\hat{D}_{n}$.

###  7.1 Centralizers of operators with constant coefficients

Let’s introduce the following notation. For any $P\in\hat{D}_{n}^{sym}$ we put

$P_{[q]}:=\sum_{\underline{k}\in{\mbox{\sdbl
N}}_{0}^{n},|\underline{k}|=q}\frac{x^{\underline{k}}}{\underline{k}!}P_{(\underline{k})}$

(a partial slice decomposition), where we use the slice decomposition (6.1).
Thus, $P=\sum_{q\geq 0}P_{[q]}$. Put $\bar{P}:=P|_{y=0}$.

Let $k\in{\mbox{\dbl N}}$ be a number. Fix a primitive $k$-th root of unity
$\xi_{k}$ in an appropriate extension $\tilde{K}$ of the field $K$. For any
$i\in{\mbox{\dbl Z}}$ and $1\leq q\leq n$ we define operators

$A_{k;i,q}:=\exp((\xi_{k}^{i}-1)x_{q}*\partial_{q})\in\hat{D}_{n}^{sym}\otimes_{K}\tilde{K}$

(we use here the same notations as in theorem 2.1). Recall also the operators
of integration introduced in the proof of theorem 2.1:

$\int_{i}:=(1-\exp((-x_{i})*\partial_{i}))\cdot\partial_{i}^{-1}=\sum_{k=0}^{\infty}\frac{x^{k+1}}{(k+1)!}(-\partial)^{k}.$

Note that all these operators are homogeneous and $\mathop{\rm\bf
ord}(A_{k;i,q})=0$, $\mathop{\rm\bf ord}(\int_{i})=-1$. Besides,
$\partial_{i}\int_{i}=1$, but $\int_{i}\partial_{i}=1-\delta_{i}$.

######  Proposition 7.1.

The following properties hold:

  1. 1.

For any $i,j,q,p,l$ we have

$A_{k;i,q}A_{k;j,q}=A_{k;i+j,q},\quad A_{k;i,q}A_{k;j,l}=A_{k;j,l}A_{k;i,q},$
$\partial_{q}^{p}A_{k;i,q}=\xi_{k}^{pi}A_{k;i,q}\partial_{q}^{p},\quad
A_{k;i,q}x_{q}^{p}=\xi_{k}^{pi}x_{q}^{p}A_{k;i,q},\quad\int_{q}^{p}A_{k;i,q}=\xi_{k}^{-pi}A_{k;i,q}\int_{q}^{p},$
$[\partial_{q}^{p},A_{k;i,l}]=0,\quad[x_{q}^{p},A_{k;i,l}]=0,\quad[\int_{q}^{p},A_{k;i,l}]=0\mbox{
for $q\neq l$}$

In particular, $A_{k;i,q}A_{k;k-i,q}=1$.

  2. 2.

For a given $Q\in\hat{D}_{n}^{sym}$ assume that $[\partial_{q}^{k},Q]=0$ for
some $q\in\\{1,\ldots,n\\}$.

Then
$Q=c_{0}+c_{1}A_{k;1,q}+\ldots+c_{k-1}A_{k;k-1,q}\in\hat{D}_{n}^{sym}\otimes_{K}\tilde{K}$,
where $c_{i}\in\hat{D}_{n}^{sym}\otimes_{K}\tilde{K}$ are given by the
following formula:

$c_{i}=\sum_{m=0}^{\infty}c_{i,m}\partial_{q}^{m}+c_{i,-1}\int_{q}+\ldots+c_{i,-k+1}\int_{q}^{k-1},$

where $c_{i,j}$ don’t depend on $x_{q},\partial_{q}$ and $\mathop{\rm\bf
ord}(c_{i,j})\leq\mathop{\rm\bf ord}(Q)-j$ for all $i,j$ (so that
$\mathop{\rm\bf ord}(c_{i})\leq\mathop{\rm\bf ord}(Q)$).

  3. 3.

In particular, if $[\partial_{q}^{k_{j}},Q]=0$ for
$q=i_{1},\ldots,i_{l}\in\\{1,\ldots,n\\}$, $j=1,\ldots,l$, then

$Q=\sum_{j_{1}=0}^{k_{1}-1}\ldots\sum_{j_{l}=0}^{k_{l}-1}c_{j_{1},\ldots,j_{l}}A_{k_{1};j_{1},i_{1}}\ldots
A_{k_{l};j_{l},i_{l}},$

where $c_{j_{1},\ldots,j_{l}}\in\hat{D}_{n}^{sym}\otimes_{K}\tilde{K}$ are
given by analogous formulas and $\mathop{\rm\bf
ord}(c_{j_{1},\ldots,j_{l}})\leq\mathop{\rm\bf ord}(Q)$. Besides, the
coefficients $c_{j_{1},\ldots,j_{l}}$ are uniquely defined.

If $K_{y}=K$ and $\\{i_{1},\ldots,i_{l}\\}=\\{1,\ldots,n\\}$, then all
coefficients $c_{j_{1},\ldots,j_{l}}$ are polynomials in $\partial_{q}$,
$\int_{q}$, $q=1,\ldots,n$ with constant coefficients (from $\tilde{K}$) and
the degree of these polynomials with respect to $\partial_{q}$ is not greater
than $\mathop{\rm\bf ord}(Q)$ and the degree of these polynomials with respect
to $\int_{q}$ is not greater than $k-1$.

###### Proof.

1\. From general properties of operators (cf. theorem 2.1) we have

$A_{k;i,q}(f(x_{1},\ldots,x_{n}))=f(x_{1},\ldots,\xi_{k}^{i}x_{q},\ldots,x_{n})$

for any $f\in\tilde{K}[[x_{1},\ldots,x_{n}]]$. All identities in item 1
immediately follow from this observation.

2\. The identity $[\partial_{q}^{k},Q]=0$ can be rewritten as

$\sum_{i=1}^{k}C_{k}^{i}\partial_{q}^{i}(Q)\partial_{q}^{k-i}=0.$ (7.1)

Note that any solution $Q\in\hat{D}_{n}^{sym}$ of this equation gives a
solution
$Q^{\prime}\in\tilde{K}[[x_{1},\ldots,x_{n},\tilde{\partial}_{1},\ldots,\tilde{\partial}_{n}]]$
of the equation

$\sum_{i=1}^{k}C_{k}^{i}\partial_{q}^{i}(Q^{\prime})\tilde{\partial}_{q}^{k-i}=0,$

where $\tilde{\partial}_{q}$ means a new formal variable (commutative with all
$x_{i}$). Namely, we just replace $\partial_{i}$ by $\tilde{\partial}_{i}$ in
the series $Q$.

On the other hand, the last equation can be written in the form

$\prod_{i=1}^{k}(\partial_{q}+(1-\xi_{k}^{i})\tilde{\partial}_{q})(Q^{\prime})=0.$

Any solution of the last equation in the commutative ring
$\tilde{R}:=\tilde{K}[[x_{1},\ldots,x_{n},\tilde{\partial}_{1},\ldots,\widehat{\tilde{\partial}_{q}},\ldots,\tilde{\partial}_{n}]]((\tilde{\partial}_{q}))$
has the form

$Q^{\prime}=c_{0}+c_{1}\exp((\xi_{k}-1)x_{q}\tilde{\partial}_{q})+\ldots+c_{k-1}\exp((\xi_{k}^{k-1}-1)x_{q}\tilde{\partial}_{q}),$

where $c_{i}\in\tilde{R}$ don’t depend on $x_{q}$ (as it follows from
elementary differential algebra). If we choose a canonical representation form
of elements in $\tilde{R}$ such that each monomial has the form
$\underline{x}^{\underline{j}}\underline{\tilde{\partial}}^{\underline{k}}$
(with $x_{i}$ on the left and $\tilde{\partial}_{j}$ on the right), then the
right hand side of the last formula can be rewritten as

$Q^{\prime\prime}:=\sum_{\underline{j}=(j_{1},\ldots,\hat{j}_{q},\ldots,j_{n})\geq
0}\underline{x}^{\underline{j}}(c_{0,(\underline{j})}+\exp((\xi_{k}-1)x_{q}\tilde{\partial}_{q})c_{1,(\underline{j})}+\ldots+\exp((\xi_{k}^{k-1}-1)x_{q}\tilde{\partial}_{q})c_{k-1,(\underline{j})})=$
$\sum_{\underline{j}=(j_{1},\ldots,\hat{j}_{q},\ldots,j_{n})\geq
0}\underline{x}^{\underline{j}}(c_{0,(\underline{j})}+\tilde{A}_{k;1,q}c_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}c_{k-1,(\underline{j})}),$
(7.2)

where $c_{i,(\underline{j})}$ denotes the corresponding slice of the
coefficient $c_{i}$. Then $Q^{\prime}=Q^{\prime\prime}$ also as elements
written in this representation. Note that, since $Q^{\prime}$ contains only
non-negative powers of $\partial_{q}$, all brackets in (7.2) are series in
which $\partial_{q}$ appears only with non-negative powers (and if we replace
$\tilde{\partial}_{i}$, $i=1,\ldots,n$ by $\partial_{i}$ in all terms of
$Q^{\prime\prime}$, we get again the operator $Q$).

######  Lemma 7.1.

For any $i=0,\ldots,k-1$ we have $\mathop{\rm\bf
ord}(c_{i,(\underline{j})})\leq\mathop{\rm\bf ord}(Q)+|\underline{j}|$ in
formula (7.2), where the order $\mathop{\rm\bf ord}$ is defined on $\tilde{R}$
in the same way as on $\hat{{{\cal M}}}_{n}$.

###### Proof.

Since the elements $c_{i,(\underline{j})}$ are polynomials in
$\tilde{\partial}_{q}^{-1}$, any bracket in (7.2) can be written in the form

$(\tilde{c}_{0,(\underline{j})}+\tilde{A}_{k;1,q}\tilde{c}_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}\tilde{c}_{k-1,(\underline{j})})\tilde{\partial}_{q}^{-m_{\underline{j}}},$

where
$\tilde{c}_{i,(\underline{j})}\in\tilde{K}[[x_{1},\ldots,x_{n},\tilde{\partial}_{1},\ldots,\tilde{\partial}_{n}]]$,
and $m_{\underline{j}}\geq 0$, i.e. the series in brackets is divisible by
$\tilde{\partial}_{q}^{m_{\underline{j}}}$. We’ll additionally assume that
$m_{\underline{j}}$ is minimal, i.e. that
$GCD(\tilde{c}_{0,(\underline{j})},\ldots,\tilde{c}_{k-1,(\underline{j})})$ is
not divisible by $\tilde{\partial}_{q}$.

Obviously, the slice decomposition is unique also in the space $\tilde{R}$,
and therefore for any fixed $\underline{j}$ in (7.2) we have the unique
decomposition

$c_{0,(\underline{j})}+\tilde{A}_{k;1,q}c_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}c_{k-1,(\underline{j})}=\sum_{|\underline{l}|\geq-
m_{\underline{j}}}(c_{0,\underline{l}}+c_{1,\underline{l}}\tilde{A}_{k;1,q}+\ldots+c_{k-1,\underline{l}}\tilde{A}_{k;k-1,q})\underline{\tilde{\partial}}^{\underline{l}},$

where $c_{i,\underline{l}}\in\tilde{K}_{y}$. Since $\mathop{\rm\bf
ord}(Q)<\infty$ and $\mathop{\rm\bf ord}(A_{k;i,q})=0$, we should have

$\mathop{\rm\bf
ord}(c_{0,(\underline{j})}+\tilde{A}_{k;1,q}c_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}c_{k-1,(\underline{j})})\leq\mathop{\rm\bf
ord}(Q)+|\underline{j}|$

and therefore by lemma 7.2 below $c_{i,\underline{l}}=0$ for all
$\underline{l}$ with $|\underline{l}|\geq\mathop{\rm\bf
ord}(Q)+|\underline{j}|$ and all $i=0,\ldots,k-1$. But this exactly means that
$\mathop{\rm\bf ord}(c_{i,(\underline{j})})\leq\mathop{\rm\bf
ord}(Q)+|\underline{j}|$ for any $i$. ∎

######  Lemma 7.2.

The sum

$A:=c_{0,\underline{l}}+c_{1,\underline{l}}\tilde{A}_{k;1,q}+\ldots+c_{k-1,\underline{l}}\tilde{A}_{k;k-1,q},\quad
c_{i,\underline{l}}\in\tilde{K}_{y}$

is equal to zero iff $c_{i,\underline{l}}=0$, $i=0,\ldots,k-1$. If it is not
equal to zero, then it is of order zero.

###### Proof.

Clearly, $\mathop{\rm\bf ord}(A)\leq 0$. Note that the equality $A=0$ is
equivalent to the infinite system of linear equations

$c_{0,\underline{l}}+c_{1,\underline{l}}+\ldots+c_{k-1,\underline{l}}=0,\quad
c_{1,\underline{l}}(\xi_{k}-1)^{j}+\ldots+c_{k-1,\underline{l}}(\xi_{k}^{k-1}-1)^{j}=0,\quad
j\in{\mbox{\dbl N}}.$

By the well known property of the Vandermonde matrix this system has the
unique solution $c_{i,\underline{l}}=0$, $i=0,\ldots,k-1$. The second claim is
obvious. ∎

From lemma 7.1 it follows that the series $\tilde{c}_{i,(\underline{j})}$ will
belong to $\hat{D}_{n}^{sym}$ after replacing $\tilde{\partial}_{i}$ by
$\partial_{i}$ in all terms. Now note that in $\hat{D}_{n}^{sym}$ we have

$(\tilde{c}_{0,(\underline{j})}+\tilde{A}_{k;1,q}\tilde{c}_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}\tilde{c}_{k-1,(\underline{j})})\partial_{q}^{-m_{\underline{j}}}|_{\tilde{\partial}_{i}\mapsto\partial_{i}}=\\\
(\tilde{c}_{0,(\underline{j})}+\tilde{A}_{k;1,q}\tilde{c}_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}\tilde{c}_{k-1,(\underline{j})})|_{\tilde{\partial}_{i}\mapsto\partial_{i}}\int_{q}^{m_{\underline{j}}},$

i.e. $Q$ can be written in the form:

$Q=\sum_{\underline{j}=(j_{1},\ldots,\hat{j}_{q},\ldots,j_{n})\geq
0}\underline{x}^{\underline{j}}(\tilde{c}_{0,(\underline{j})}+\tilde{A}_{k;1,q}\tilde{c}_{1,(\underline{j})}+\ldots+\tilde{A}_{k;k-1,q}\tilde{c}_{k-1,(\underline{j})})|_{\tilde{\partial}_{i}\mapsto\partial_{i}}\int_{q}^{m_{\underline{j}}}.$
(7.3)

Besides, all summands in the sum (7.3) are well defined elements of
$\hat{D}_{n}^{sym}$ of order $\leq\mathop{\rm\bf ord}(Q)$ and their sum is
also well defined in $\hat{D}_{n}^{sym}$.

######  Lemma 7.3.

In the formula (7.3) we have $m_{\underline{j}}\leq k-1$.

###### Proof.

First note that all numbers $m_{\underline{j}}$ are bounded from above,
because the coefficients $c_{i}$ of $Q^{\prime}$ are polynomials in
$\tilde{\partial}_{q}^{-1}$. Now assume the converse: there exists
$m:=m_{\underline{j}}\geq k$. Assume that $m$ is a maximal such number. Let
$Q_{l}$ be a homogeneous component of $Q$ containing a summand with
$\int_{q}^{m}$. Note that each summand of $Q_{l}$ has the form
$c\partial_{q}^{i}$ or $c\int_{q}^{j}$, where $c$ is a linear combination of
$A_{k;s,q}$, $s=0,\ldots,k-1$ with coefficients not depending on
$x_{q},\partial_{q}$, and $j\leq m$, i.e.

$Q_{l}=\sum_{i=0}^{\infty}(a_{i,0}+a_{i,1}A_{k;1,q}+\ldots+a_{i,k-1}A_{k;k-1,q})\partial_{q}^{i}+\\\
(a_{-1,0}+\ldots+a_{-1,k-1}A_{k;k-1,q})\int_{q}+\ldots+(a_{-m,0}+a_{-m,1}A_{k;1,q}+\ldots+a_{-m,k-1}A_{k;k-1,q})\int_{q}^{m}.$
(7.4)

Since $Q_{l}$ is homogeneous, we have $\mathop{\rm\bf
ord}(a_{i,0}+a_{i,1}A_{k;1,q}+\ldots+a_{i,k-1}A_{k;k-1,q})=l-i$. Since
$(a_{-m,0}+a_{-m,1}A_{k;1,q}+\ldots+a_{-m,k-1}A_{k;k-1,q})\neq 0$, there is
$p\geq 0$ such that

$(a_{-m,0}+\frac{a_{-m,1}(\xi_{k}-1)^{p}x_{q}^{p}\partial_{q}^{p}}{p!}+\ldots+\frac{a_{-m,k-1}(\xi_{k}^{k-1}-1)^{p}x_{q}^{p}\partial_{q}^{p}}{p!})\neq
0.$

We’ll assume that $p$ is minimal such number. From lemma 7.2 it follows that
$p\leq k-1$.

Since $[\partial_{q}^{k},Q]=0$, it follows

$0=[\partial_{q}^{k},Q_{l}]=\partial_{q}^{k}(Q_{l})+k\partial_{q}^{k-1}(Q_{l})\partial_{q}+\ldots+k\partial_{q}(Q_{l})\partial_{q}^{k-1}.$

Now let’s expand all series in this sum and collect all summands containing
$x_{q}^{m-k+p}\partial_{q}^{p}$. Note that there are no such summands except
the expression

$m\ldots(m-k+1)(a_{-m,0}+\frac{a_{-m,1}(\xi_{k}-1)^{p}x_{q}^{p}\partial_{q}^{p}}{p!}+\ldots+\frac{a_{-m,k-1}(\xi_{k}^{k-1}-1)^{p}x_{q}^{p}\partial_{q}^{p}}{p!})x_{q}^{m-k+p}\partial_{q}^{p}$

(coming from
$(a_{-m,0}+a_{-m,1}A_{k;1,q}+\ldots+a_{-m,k-1}A_{k;k-1,q})\partial_{q}^{k}(\int_{q}^{m})$),
and since this expression is not zero, we get a contradiction. Indeed, for
$i\geq 0$ and any $0<l\leq k$ we have

$\partial_{q}^{l}((a_{i,0}+a_{i,1}A_{k;1,q}+\ldots+a_{i,k-1}A_{k;k-1,q})\partial_{q}^{i})\partial_{q}^{k-l}=(a_{i,1}^{\prime}A_{k;1,q}+\ldots+a_{i,k-1}^{\prime}A_{k;k-1,q})\partial_{q}^{i+k}$

(here $a_{i,r}^{\prime}=a_{i,j}\xi_{k}^{jr}$), and all monomials containing
$x_{q}^{m-k+p}$ from this series contain also $\partial_{q}^{m+p+i}$. For
other expressions from (7.4) we have

$\partial_{q}^{l}((a_{i,0}+a_{i,1}A_{k;1,q}+\ldots+a_{i,k-1}A_{k;k-1,q})\int_{q}^{j})\partial_{q}^{k-l}=\\\
\partial_{q}^{l}(\int_{q}^{j}(a_{i,0}+a_{i,1}^{\prime}A_{k;1,q}+\ldots+a_{i,k-1}^{\prime}A_{k;k-1,q}))\partial_{q}^{k-l}=\\\
\sum_{\mu+\nu=l}c_{\mu,\nu}\partial_{q}^{\nu}(\int_{q}^{i})\partial_{q}^{\mu}(a_{i,0}+a_{i,1}^{\prime}A_{k;1,q}+\ldots+a_{i,k-1}^{\prime}A_{k;k-1,q})\partial_{q}^{k-l}=\\\
\sum_{\mu+\nu=l}c_{\mu,\nu}\partial_{q}^{\nu}(\int_{q}^{i})(a_{i,1}^{\prime}(\xi_{k}-1)^{\mu}A_{k;1,q}+\ldots+a_{i,k-1}^{\prime}(\xi_{k}^{k-1}-1)^{\mu}A_{k;k-1,q})\partial_{q}^{k-\nu},$

where $c_{\mu,\nu}\in{\mbox{\dbl N}}$ and
$a_{i,r}^{\prime}=a_{i,j}\xi_{k}^{jr}$. Now consider any series

$\partial_{q}^{\nu}(\int_{q}^{i})(a_{i,1}^{\prime}(\xi_{k}-1)^{\mu}A_{k;1,q}+\ldots+a_{i,k-1}^{\prime}(\xi_{k}^{k-1}-1)^{\mu}A_{k;k-1,q})\partial_{q}^{k-\nu}$

from this formula and note that all monomials containing $x_{q}^{m-k+p}$ from
this series contain also
$\partial_{q}^{m-k+p-i+\nu+k-\nu}=\partial_{q}^{m+p-i}$. If $i<m$, then
$m+p-i>p$, and therefore there are no terms containing
$x_{q}^{m-k+p}\partial_{q}^{p}$.

If $i=m$, but $\nu<k$, then

$(a_{i,1}^{\prime}(\xi_{k}-1)^{\mu+s}x_{q}^{s}\partial_{q}^{s}+\ldots+a_{i,k-1}^{\prime}(\xi_{k}^{k-1}-1)^{\mu+s}x_{q}^{s}\partial_{q}^{s})=0$

for $s=0,\ldots,\nu-k+p$ by the condition on $p$ (its minimality), and
therefore there are no terms containing $x_{q}^{m-k+p}\partial_{q}^{p}$ again.
∎

Now, expanding all brackets in (7.3) and using the identities from item 1, we
can rewrite $Q$ in the form stated in item 2.

3\. This item immediately follows from item 2 by induction. Let’s show the
uniqueness of the coefficients.

It suffices to prove that the equality

$0=\sum_{j_{1}=0}^{k_{1}-1}\ldots\sum_{j_{l}=0}^{k_{l}-1}c_{j_{1},\ldots,j_{l}}A_{k_{1};j_{1},i_{1}}\ldots
A_{k_{l};j_{l},i_{l}}$

implies the equalities $c_{j_{1},\ldots,j_{l}}=0$. By lemma 6.2 it is
equivalent to the system of equalities
$\underline{\partial}^{\underline{q}}\diamond A=0$, where $A$ means our sum
above and $\underline{q}\in{\mbox{\dbl N}}_{0}^{n}$. Note that this system is
equivalent to the system $A_{(\underline{q})}=0$, $\underline{q}\in{\mbox{\dbl
N}}_{0}^{n}$.

Let $\underline{i}$ be the subset in $\\{1,\ldots,n\\}$ complement to the set
$\\{i_{1},\ldots,i_{l}\\}$. Then note that the last system is equivalent to
the system

$\sum_{j_{1}=0}^{k_{1}-1}\ldots\sum_{j_{l}=0}^{k_{l}-1}(c_{j_{1},\ldots,j_{l}})_{(\underline{i})}\xi_{k_{1}}^{q_{1}j_{1}}\ldots\xi_{k_{l}}^{q_{l}j_{l}}=0,\quad\underline{q}\in{\mbox{\dbl
N}}_{0}^{n}.$ (7.5)

Now note that system (7.5) is a system of linear equations with indeterminates
$(c_{j_{1},\ldots,j_{l}})_{(\underline{i})}$ and the matrix
$(\xi_{k_{1}}^{q_{1}j_{1}}\ldots\xi_{k_{l}}^{q_{l}j_{l}})$, which is just the
matrix of the tensor product of the Vandermonde matrices
$(\xi_{k_{s}}^{q_{s}j_{s}})$, $q_{s},j_{s}=0,\ldots,k_{s}$. Hence it is
invertible and the system has a unique (zero) solution.

∎

######  Definition 7.1.

An element $P\in\hat{D}_{n}^{sym}$ is called $p$-regular, where
$p\in{\mbox{\dbl N}}_{0}$, if the elements of the set
$\\{\underline{\partial}^{\underline{k}}\diamond\sigma(P)|\quad\underline{k}\in{\mbox{\dbl
N}}_{0}^{n}:|\underline{k}|\leq p\\}\subset F$ are linearly independent.

We start with the Schur theory for the case $n=1$.

###  7.2 Schur theory, $n=1$

######  Proposition 7.2.

Let $P\in\hat{D}_{1}^{sym}$ be a regular operator with $\mathop{\rm\bf
ord}(P)=\mathop{\rm\bf ord}(\bar{P})=k>0$ and such that $\bar{P}$ is regular.

Then there exists an invertible operator $S\in\hat{D}_{1}^{sym}$ with
$\mathop{\rm\bf ord}(S)=0$ such that
$P=S^{-1}\partial_{1}^{k}S$.

###### Proof.

We’ll find the operator $S$ recursively, looking for the partial slice
decompositions $S_{[q]}$, $S=\sum_{q\geq 0}S_{[q]}$.

For any $p\geq 0$ we need to solve the system

$(\partial_{1}^{k}S)_{[p]}=(SP)_{[p]}.$

Note that
$(\partial_{1}^{k}S)_{[p]}=\partial_{1}^{k}(S_{[p+k]})+\mbox{expression
depending on $S_{[q]}$, $q<p+k$}$, and
$(SP)_{[p]}=((S_{[0]}+\ldots+S_{[p]})P)_{[p]}$. Put $S_{[0]}:=1$, $S_{[p]}:=0$
for $1\leq p\leq k-1$. Then for $p=0$ we have

$(\partial_{1}^{k}S)_{[0]}=\partial_{1}^{k}(S_{[k]})+\partial_{1}^{k}=P_{[0]},$

and therefore the slice $S_{(k)}$ is uniquely determined; besides,
$\mathop{\rm\bf ord}(S_{[k]})\leq 0$. By the same reason all partial slice
decompositions $S_{[p]}$, $p>k$ will be uniquely determined by this system and
by induction $\mathop{\rm\bf ord}(S_{[p]})\leq 0$ as well. Since
$\mathop{\rm\bf ord}(S_{[0]})=0$, we get $\mathop{\rm\bf ord}(S)=0$ too. Note
that $(\partial_{1}^{k}\bar{S})_{[p]}=(\bar{S}\bar{P})_{[p]}$ automatically.

By corollary 6.1 $S$ is invertible iff $S$, $\bar{S}$ are regular. First note
that $S$, $\bar{S}$ are $k$-regular: for any $0\leq i\leq k$ we have

$\partial_{1}^{i}\diamond
S=\partial_{1}^{i}\diamond(S_{[0]}+\ldots+S_{[k]})=\left\\{\begin{array}[]{c}\partial_{1}^{i}\quad\mbox{if
$i<k$}\\\ P_{[0]}\quad\mbox{if $i=k$}\end{array}\right.,\quad$
$\partial_{1}^{i}\diamond\bar{S}=\partial_{1}^{i}\diamond(\bar{S}_{[0]}+\ldots+\bar{S}_{[k]})=\left\\{\begin{array}[]{c}\partial_{1}^{i}\quad\mbox{if
$i<k$}\\\ \bar{P}_{[0]}\quad\mbox{if $i=k$}\end{array}\right.$

Since $P$, $\bar{P}$ are regular, we have $\sigma(P)_{[0]}\neq 0$,
$\mathop{\rm\bf ord}(\sigma(P)_{[0]})=\mathop{\rm\bf ord}(P_{[0]})=k$ and
analogously $\overline{\sigma(P)}_{[0]}\neq 0$, $\mathop{\rm\bf
ord}(\overline{\sigma(P)}_{[0]})=\mathop{\rm\bf ord}(\bar{P}_{[0]})=k$,
therefore the elements
$1,\partial_{1},\ldots,\partial_{1}^{k-1},\overline{\sigma(P_{[0]})}$ are
linearly independent.

Now we can prove that $S$ is $p$-regular for $p>k$ by induction. For any $p>k$
we have

$\partial_{1}^{p}\diamond
S=\partial_{1}^{p-k}\diamond(\partial_{1}^{k}S)=\partial_{1}^{p-k}\diamond(SP)=\partial_{1}^{p-k}\diamond(S_{[0]}+\ldots+S_{[p-k]})P,$
$\partial_{1}^{p}\diamond\bar{S}=\partial_{1}^{p-k}\diamond(\partial_{1}^{k}\bar{S})=\partial_{1}^{p-k}\diamond(\bar{S}\bar{P})=\partial_{1}^{p-k}\diamond(\bar{S}_{[0]}+\ldots+\bar{S}_{[p-k]})\bar{P}.$

Since $P$, $\bar{P}$ are regular and $(S_{[0]}+\ldots+S_{[p-k]})$,
$(\bar{S}_{[0]}+\ldots+\bar{S}_{[p-k]})$ are $(p-k)$-regular by induction, the
symbols of the last elements in the equalities are not equal to zero (cf.
lemma 6.1), i.e. $S$, $\bar{S}$ are $p$-regular and therefore they are regular
by induction.

∎

######  Example 7.1.

Let’s illustrate propositions 7.1 and 7.2 on one well known Wallenber’s
example of two commuting ordinary differential operators of orders 2 and 3
with rational coefficients:

$L=\partial^{2}-\frac{2}{(x+1)^{2}},\quad
P=4\partial^{3}-\frac{12}{(x+1)^{2}}\partial+\frac{12}{(x+1)^{3}}.$

Then direct calculations with the help of these propositions show the
existence of an operator $S$ such that $SLS^{-1}=\partial^{2}$, and

$SPS^{-1}=(4\partial^{3}+2\int)-(4(1-\partial)-2\int)A_{2;1,1}.$

###  7.3 Schur theory, general case

######  Proposition 7.3.

Let $P\in\hat{D}_{n}^{sym}$ be a regular operator with $\mathop{\rm\bf
ord}(P)=\mathop{\rm\bf ord}(\bar{P})=k>0$ and such that $\bar{P}$ is regular.

Then for any $i=1,\ldots,n$ there exists an invertible operator
$S_{i}\in\hat{D}_{n}^{sym}$ with $\mathop{\rm\bf ord}(S_{i})=0$, $S_{i,0}=1$
such that $P=S_{i}^{-1}\partial_{i}^{k}S_{i}$.

###### Proof.

The proof is similar to proposition 7.2. First, note that
$\sigma(P)_{[0]}=\sigma(P_{[0]})\neq 0$ (since $P$ is regular) and therefore
$\mathop{\rm\bf ord}(P_{[0]})=\mathop{\rm\bf ord}(P)$. Analogously,
$\sigma(\bar{P})_{[0]}=\sigma(\bar{P}_{[0]})\neq 0$ and $\mathop{\rm\bf
ord}(\bar{P}_{[0]})=\mathop{\rm\bf ord}(\bar{P})$.

We’ll find the operator $S_{i}$ recursively, looking for the partial slice
decompositions $S_{i,[q]}$, $S_{i}=\sum_{q\geq 0}S_{i,[q]}$.

For any $p\geq 0$ we need to solve the system

$(\partial_{i}^{k}S_{i})_{[p]}=(S_{i}P)_{[p]}.$ (7.6)

Note that
$(\partial_{i}^{k}S_{i})_{[p]}=\partial_{i}^{k}(S_{i,[p+k]})+\mbox{expression
depending on $S_{i,[q]}$, $q<p+k$}$, and

$(S_{i}P)_{[p]}=((S_{i,[0]}+\ldots+S_{i,[p]})P)_{[p]}.$

In particular, $\partial_{i}^{k}(S_{i,[p+k]})$ is uniquely determined via
expressions depending on $S_{i,[q]}$, $q<p+k$ and $P$, i.e. all slices
$S_{i,(\underline{j})}$, where $\underline{j}=(j_{1},\ldots,j_{n})$ with
$j_{i}\geq k$ and $|\underline{j}|=k$, are uniquely determined.

Put $S_{i,[0]}:=1$, $S_{i,[p]}:=0$ for $1\leq p\leq k-1$. Then for $p=0$ we
have the system

$(\partial_{i}^{k}S_{i})_{[0]}=\partial_{i}^{k}(S_{i,[k]})+\partial_{i}^{k}=P_{[0]}.$

From this equation the slice $S_{i,(\underline{j})}$ with
$\underline{j}=(0,\ldots,0,k)$ is uniquely determined, and we define all other
slices $S_{i,(\underline{j})}$ with $|\underline{j}|=k$,
$\underline{j}\neq(0,\ldots,0,k)$ in such a way that the elements
$\underline{\partial}^{\underline{j}}+\sigma(S_{i,(\underline{j})})$ are
linearly independent with $\sigma(P)_{[0]}$ (and the elements
$\underline{\partial}^{\underline{j}}+\sigma(\bar{S}_{i,(\underline{j})})$ are
linearly independent with $\sigma(\bar{P})_{[0]}$ correspondingly). Then we
get $\mathop{\rm\bf ord}(S_{i,[k]})\leq 0$ (as $\mathop{\rm\bf
ord}(\partial_{i}^{k})=k$ and $\mathop{\rm\bf ord}(P_{[0]})=k$).

Note that the operators $(S_{i,[0]}+\ldots+S_{i,[k]})$,
$(\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[k]})$ are $k$-regular: for any
$\underline{j}\in{\mbox{\dbl N}}_{0}^{n}$ with $|\underline{j}|\leq k$ we have

$\underline{\partial}^{\underline{j}}\diamond(S_{i,[0]}+\ldots+S_{i,[k]})=\left\\{\begin{array}[]{c}\underline{\partial}^{\underline{j}}\quad\mbox{if
$|\underline{j}|<k$}\\\
\underline{\partial}^{\underline{j}}+S_{i,(\underline{j})}\quad\mbox{if
$|\underline{j}|=k$, $\underline{j}\neq(0,\ldots,0,k)$}\\\
P_{[0]}\quad\mbox{if $\underline{j}=(0,\ldots,0,k)$}\end{array}\right.,$
$\underline{\partial}^{\underline{j}}\diamond(\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[k]})=\left\\{\begin{array}[]{c}\underline{\partial}^{\underline{j}}\quad\mbox{if
$|\underline{j}|<k$}\\\
\underline{\partial}^{\underline{j}}+\bar{S}_{i,(\underline{j})}\quad\mbox{if
$|\underline{j}|=k$, $\underline{j}\neq(0,\ldots,0,k)$}\\\
\bar{P}_{[0]}\quad\mbox{if $\underline{j}=(0,\ldots,0,k)$}\end{array}\right.$

By construction of the slices $S_{i,(\underline{j})}$,
$\bar{S}_{i,(\underline{j})}$ all symbols of these elements are linearly
independent.

By the same reason all partial slice decompositions $S_{i,[p+k]}$, $p>0$ can
be determined by the equation

$(\partial_{i}^{k}S_{i})_{[p]}=(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[p+k]}))_{[p]}=(S_{i}P)_{[p]}=((S_{i,[0]}+\ldots+S_{i,[p]})P)_{[p]},$

where all slices $S_{i,(\underline{j})}$, where $|\underline{j}|=p+k$,
$\underline{j}=(j_{1},\ldots,j_{n})$ with $j_{i}\geq k$, are uniquely
determined by the equation, and to define the slices $S_{i,(\underline{j})}$,
where $|\underline{j}|=p+k$, $\underline{j}=(j_{1},\ldots,j_{n})$ with
$j_{i}<k$, let’s consider first elements
$v_{k,\underline{j}}:=S_{i,(\underline{j})}+\underline{\partial}^{\underline{j}}\diamond(S_{i,[0]}+\ldots+S_{i,[p+k-1]})$,
where $\underline{j}=(j_{1},\ldots,j_{n})$ with $j_{i}\geq k$,
$|\underline{j}|=p+k$. Observe that for all $q<p$ we have

$(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[p+k]}))_{[q]}=(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[q+k]}))_{[q]}=\\\
((S_{i,[0]}+\ldots+S_{i,[q]})P)_{[q]}=((S_{i,[0]}+\ldots+S_{i,[p]})P)_{[q]}$

by induction, and therefore

$v_{k,\underline{j}}=\underline{\partial}^{(j_{1},\ldots,j_{i}-k,\ldots,j_{n})}\diamond(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[p+k-1]}+\frac{x^{\underline{j}}}{\underline{j}!}S_{i,(\underline{j})}))=\\\
\underline{\partial}^{(j_{1},\ldots,j_{i}-k,\ldots,j_{n})}\diamond(\sum_{q=0}^{p}(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[p+k-1]}+\frac{x^{\underline{j}}}{\underline{j}!}S_{i,(\underline{j})}))_{[q]})=\\\
\underline{\partial}^{(j_{1},\ldots,j_{i}-k,\ldots,j_{n})}\diamond(\sum_{q=0}^{p}((S_{i,[0]}+\ldots+S_{i,[p]})P)_{[q]})=\underline{\partial}^{(j_{1},\ldots,j_{i}-k,\ldots,j_{n})}\diamond((S_{i,[0]}+\ldots+S_{i,[p]})P),$

where from the elements $v_{k,\underline{j}}$ are linearly independent
(moreover, their symbols are linearly independent), since $P$ is regular and
$(S_{i,[0]}+\ldots+S_{i,[p]})$ is $p$-regular by induction. Analogously
defined elements
$v_{\underline{j}}:=\underline{\partial}^{\underline{j}}\diamond(S_{i,[0]}+\ldots+S_{i,[p+k-1]})$
with $|\underline{j}|<p+k$ and their symbols are also linearly independent by
induction, and symbols of the elements $v_{\underline{j}}$,
$v_{k,\underline{j}}$ are linearly independent because elements from two
different groups (with $|\underline{j}|<p$ and with $|\underline{j}|=p$) have
different $\mathop{\rm\bf ord}$\- orders.

Now we define the slices $S_{i,(\underline{j})}$, where $|\underline{j}|=p+k$,
$\underline{j}=(j_{1},\ldots,j_{n})$ with $j_{i}<k$, in such a way that
symbols of the elements
$v_{p,\underline{j}}:=S_{i,(\underline{j})}+\underline{\partial}^{\underline{j}}\diamond(S_{i,[0]}+\ldots+S_{i,[p+k-1]})$
are linearly independent with the symbols of elements $v_{k,\underline{j}}$
and $v_{\underline{j}}$, $|\underline{j}|<p+k$ (and analogous properties hold
for $\bar{v}_{p,\underline{j}}$).

As a result, the operators $(S_{i,[0]}+\ldots+S_{i,[p+k]})$,
$(\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[p+k]})$ are $(p+k)$-regular and satisfy
the equations

$(\partial_{i}^{k}(S_{i,[0]}+\ldots+S_{i,[p+k]}))_{[p]}=((S_{i,[0]}+\ldots+S_{i,[p]})P)_{[p]}=((S_{i,[0]}+\ldots+S_{i,[p+k]})P)_{[p]},$
$(\partial_{i}^{k}(\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[p+k]}))_{[p]}=((\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[p]})\bar{P})_{[p]}=((\bar{S}_{i,[0]}+\ldots+\bar{S}_{i,[p+k]})\bar{P})_{[p]}.$

By induction and construction $\mathop{\rm\bf ord}(S_{i,[p+k]})\leq 0$ as
well.

Now consider the operator $S_{i}:=\sum_{q\geq 0}S_{i,[q]}$. Since
$\mathop{\rm\bf ord}(S_{i,[0]})=0$, we get $\mathop{\rm\bf ord}(S_{i})=0$ too.
Moreover, $S$, $\bar{S}$ are regular (as they are $p$-regular for all $p>0$)
and satisfy equation (7.6) and analogous equation for $\bar{S}$ (for all
partial slice decompositions). By corollary 6.1 $S_{i}$ is invertible, and we
are done. ∎

######  Theorem 7.1.

Let $P_{1},\ldots,P_{n}\in\hat{D}_{n}^{sym}$ be commuting operators with
$\mathop{\rm\bf ord}(P_{i})=\mathop{\rm\bf ord}(\bar{P}_{i})=k$ for all
$i=1,\ldots,n$. Assume that the spectral module $F$ of the ring
$K[\sigma(\bar{P}_{1}),\ldots,\sigma(\bar{P}_{n})]$ is finitely generated and
Cohen-Macaulay.

Then there exists an invertible operator $S\in\hat{D}_{n}^{sym}$ with
$\mathop{\rm\bf ord}(S)=0$ such that

$S^{-1}\partial_{i}^{k_{i}}S=P_{i},\quad i=1,\ldots,n.$

######  Remark 7.1.

The condition on the spectral module is satisfied for many examples arising in
quantum integrable systems, e.g. for rings of commuting partial differential
operators, cf. [32, Th. 4], [62, Cor. 1], [61, Th. 4.1]. Further examples will
appear in a subsequent paper.

###### Proof.

The proof is similar to proposition 7.3.

Construction. Namely, for any $p\geq 0$ we need to solve the system

$(\partial_{n}^{k_{n}}S)_{[p]}=(SP_{n})_{[p]},\quad,\ldots,\quad(\partial_{1}^{k_{1}}S)_{[p]}=(SP_{1})_{[p]}$
(7.7)

We have again

$(\partial_{i}^{k_{i}}S)_{[p]}=\partial_{i}^{k_{i}}(S_{[p+k_{n}]})+\mbox{expression
depending on $S_{[q]}$, $q<p+k_{i}$},$

and $(SP_{i})_{[p]}=((S_{[0]}+\ldots+S_{[p]})P_{i})_{[p]}$. In particular, if
this system is compatible, the expressions
$\partial_{i}^{k_{i}}(S_{[p+k_{i}]})$ are uniquely determined via expressions
depending on $S_{[q]}$, $q<p+k_{i}$ and $P_{i}$.

Put $S_{[0]}:=1$, $S_{(\underline{j})}:=0$ for
$\underline{j}=(j_{1},\ldots,j_{n})\neq\underline{0}$ with $j_{i}<k_{i}$,
$i=1,\ldots,n$. Then for $p=0$ we have

$(\partial_{n}^{k_{n}}S)_{[0]}=\partial_{n}^{k_{n}}(S_{[k_{n}]})+\partial_{n}^{k_{n}}=P_{n,[0]},\quad,\ldots,\quad(\partial_{1}^{k_{1}}S)_{[0]}=\partial_{1}^{k_{1}}(S_{[k_{1}]})+\partial_{1}^{k_{1}}=P_{1,[0]}$

From this system the slices $S_{(\underline{j})}$ with
$\underline{j}=(0,\ldots,0,k_{i},0,\ldots,0)$, $i=1,\ldots,n$ are uniquely
determined.

Note that we have defined the partial slice decompositions $S_{[q]}$ with
$q\leq k_{1}$. We will define slices $S_{(\underline{j})}$ with
$q:=|\underline{j}|>k_{1}$ by induction on $q$. Assume we have defined all
slices $S_{(\underline{j})}$ with $|\underline{j}|<q$. We first determine the
slices $S_{(\underline{j})}$ with $j_{1}\geq k_{1}$ from the equation

$(\partial_{1}^{k_{1}}S)_{[q-k_{1}]}=(SP_{1})_{[q-k_{1}]}$

of the system (7.7). We claim that other equations of this system remain
compatible, i.e. for all $\underline{j}$ with $j_{1}\geq k_{1}$ as above we
have

$(\partial_{i}^{k_{i}}S)_{(\underline{j})}=(SP_{i})_{(\underline{j})},\quad
i=2,\ldots,n.$

Indeed, we have on the one side

$(\partial_{i}^{k_{i}}S)_{(\underline{j})}=\underline{\partial}^{\underline{j}}\diamond(\partial_{i}^{k_{i}}S)=(\underline{\partial}^{\underline{j}}\partial_{i}^{k_{i}}S)_{[0]}=(\partial_{i}^{k_{i}}\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}SP_{1})_{[0]}=(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}\partial_{i}^{k_{i}}S)_{[0]}\diamond
P_{1},$

and on the other side

$(SP_{i})_{(\underline{j})}=\underline{\partial}^{\underline{j}}\diamond(SP_{i})=(\underline{\partial}^{\underline{j}}SP_{i})_{[0]}=(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}SP_{1}P_{i})_{[0]}=\\\
(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}SP_{i}P_{1})_{[0]}=(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}SP_{i})_{[0]}\diamond
P_{1}.$

By induction, we have

$(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}\partial_{i}^{k_{i}}S)_{[0]}=\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}\diamond(\partial_{i}^{k_{i}}S)=(\partial_{i}^{k_{i}}S)_{(j_{1}-k_{1},j_{2},\ldots,j_{n})}=\\\
(SP_{i})_{(j_{1}-k_{1},j_{2},\ldots,j_{n})}=(\underline{\partial}^{\underline{j}-(k_{1},0,\ldots,0)}SP_{i})_{[0]},$

thus $(\partial_{i}^{k_{i}}S)_{(\underline{j})}=(SP_{i})_{(\underline{j})}$.

Therefore, continuing this line of reasoning, we can determine recursively the
slices $S_{(\underline{j})}$ with $j_{i}\geq k_{i}$, $j_{l}<k_{l}$ for $l<i$
from the equation

$(\partial_{i}^{k_{i}}S)_{[q-k_{i}]}=(SP_{i})_{[q-k_{i}]}$

for each $i=2,\ldots,n$.

At the end of the induction procedure we’ll get the operator $S$. Note that
$\mathop{\rm\bf ord}(S)=0$ by construction. Automatically we get also the
analogue of the equation (7.7) for $\bar{S}$, $\bar{P}$.

Regularity. By corollary 6.1 $S$ is invertible iff $\bar{S}$ is regular. Note
that for any
$\underline{j}=(l_{1}+q_{1}k_{1},l_{2}+q_{2}k_{2},\ldots,l_{n}+q_{n}k_{n})$,
where $q_{i}\in{\mbox{\dbl Z}}_{+}$, $l_{i}<k_{i}$ for $i=1,\ldots,n$, we have

$\underline{\partial}^{\underline{j}}\diamond\sigma(S)=\sigma(\underline{\partial}^{\underline{j}}\diamond\bar{S})=\sigma(\underline{\partial}^{\underline{l}}\diamond\bar{S}\bar{P}_{1}^{q_{1}}\ldots\bar{P}_{n}^{q_{n}})=\sigma(\underline{\partial}^{\underline{l}}\diamond\bar{P}_{1}^{q_{1}}\ldots\bar{P}_{n}^{q_{n}})=\underline{\partial}^{\underline{l}}\diamond\sigma(\bar{P}_{1}^{q_{1}}\ldots\bar{P}_{n}^{q_{n}}).$

Now note that the elements $\underline{\partial}^{\underline{l}}$ belong to
the set of generators of the free
$K[\sigma(\bar{P}_{1}),\ldots,\sigma(\bar{P}_{n})]$-module $F$ ($F$ is free
since it is Cohen-Macaulay, see [7]). Indeed, for any fixed system of
generators $e_{1},\ldots,e_{m}$ the $\mathop{\rm\bf ord}$-order of any linear
combination $\alpha_{1}e_{1}+\ldots+\alpha_{m}e_{m}$, where $\alpha_{i}$ are
polynomials without zero terms, is greater than $k$. Thus, each monomial
$\underline{\partial}^{\underline{l}}$ must be a linear combination of
$e_{i}$’s with constant coefficients. Thus, making a linear change, we can
replace some of generators by these monomials. On the other hand, all such
monomials are obviously linear independent over $K$.

Therefore, all elements
$\underline{\partial}^{\underline{j}}\diamond\sigma(\bar{S})$ are linearly
independent, and $\bar{S}$ is regular.

∎

######  Corollary 7.1.

Any invertible linear change of variables
$\partial_{i}\mapsto\sum_{j=1}^{n}c_{i,j}\partial_{j}$, $c_{i,j}\in K$,
$i=1,\ldots,n$ can be obtained by conjugation with an operator
$S\in\hat{U}_{n}^{sym}$. More precisely,

$S=c_{0}\exp(\sum_{i=1}^{n}(\sum_{j=1}^{n}(c_{i,j}-\delta_{i,j})x_{j})*\partial_{i}))\in\hat{D}_{n}^{sym},$
(7.8)

where $c_{0},c_{i,j}\in K$, $c_{0}\neq 0$, and $\delta_{i,j}$ is the Kronecker
delta. In particular, $\mathop{\rm\bf ord}(S)=\mathop{\rm\bf ord}(S^{-1})=0$,
and

$S^{-1}\partial_{i}S=\sum_{j=1}^{n}c_{i,j}\partial_{j},\mbox{\quad}x_{i}\mapsto\sum_{j=1}^{n}c_{i,j}^{\prime}x_{j},$

where $(c_{i,j}^{\prime})$ denotes the inverse matrix of $(c_{i,j})$.

###### Proof.

The proof immediately follows from theorem 7.1. If $(c_{i,j}^{\prime})$
denotes the inverse matrix of $(c_{i,j})$, then it sends
$x_{i}\mapsto\sum_{j=1}^{n-1}c_{i,j}^{\prime}x_{j}$, $1\leq i\leq n-1$.

By theorem 2.1, item 4) the operator $S$ gives the automorphism
$x_{i}\mapsto\sum_{j=1}^{n-1}c_{i,j}x_{j}$, $1\leq i\leq n-1$ of the ring
$\hat{R}$. In particular, it is invertible with the inverse
$S^{-1}=\exp(\sum_{i=1}^{n}(\sum_{j=1}^{n}(c_{i,j}^{\prime}-\delta_{i,j})x_{j})*\partial_{i}))$.
Note that $\mathop{\rm\bf ord}(S)=\mathop{\rm\bf ord}(S^{-1})=0$. Direct
calculations show that $S^{-1}\partial_{i}S=\sum_{j=1}^{n}c_{i,j}\partial_{j}$
for $1\leq i\leq n$. ∎

##  8 Classification theorems

Now we are ready to prove a first classification theorem - a classification of
certain equivalent classes of quasi-elliptic rings in terms of Schur pairs.

######  Definition 8.1.

(cf. [61, Def.3.4]) The commutative quasi-elliptic rings $B_{1}$,
$B_{2}\subset\hat{D}_{n}$ are said to be equivalent in $\hat{D}_{n}$ if there
is an invertible operator $S\in\hat{D}_{n}^{n}$ such that
$B_{1}=SB_{2}S^{-1}$.

######  Proposition 8.1.

Any 1-admissible operator $T$ can be written in the form

$T=S_{2}S_{1},$

where $S_{2}\in\hat{E}_{n}$ is monic, satisfies $A_{1}$, $\mathop{\rm\bf
ord}(S_{2})=\mathop{\rm\bf ord}(S_{2}^{-1})=\mathop{\rm
ord}\nolimits_{n}(S_{2}^{-1})=\mathop{\rm ord}\nolimits_{n}(S_{2})=0$,
$\mathop{\rm ord}\nolimits_{\Gamma}(S_{2})=\mathop{\rm
ord}\nolimits_{\Gamma}(S_{2}^{-1})=(0,\ldots,0)$, and
$S_{1}\in\hat{D}_{n}^{n}$ is invertible and $\mathop{\rm\bf ord}(S_{1})=0$.

###### Proof.

Put $q_{i}^{\prime}=T^{-1}\partial_{i}T$. From lemma 5.3 it follows that for
$i=1,\ldots,n-1$ we have $q_{i}^{\prime}=\sum_{j=1}^{n-1}\sum_{k\geq
0}c_{i;j,k}\partial_{j}^{k}+l.o.t.$, where l.o.t. mean terms with $\mathop{\rm
ord}\nolimits_{n}$-order less than zero and $c_{i;j,k}\in K_{y}$, and
$q_{n}^{\prime}=\partial_{n}+l.o.t.$, where l.o.t. mean terms with
$\mathop{\rm ord}\nolimits_{n}$-order less than one. Moreover, $\mathop{\rm\bf
ord}(\sum_{j=1}^{n-1}\sum_{k\geq 0}c_{i;j,k}\partial_{j}^{k})=1$.

Note that ${HT}_{n}(T\cdot{HT}_{n}(q_{i}^{\prime})\cdot T^{-1})=\partial_{i}$,
i.e. in particular the map

$\partial_{i}\mapsto\partial_{i}^{\prime}:=\sum_{j=1}^{n-1}\sum_{k\geq
1}c_{i;j,k}\partial_{j}^{k},\quad 1\leq i\leq n-1$

is invertible. Since $\mathop{\rm\bf ord}(\partial_{i}^{\prime})=1$, and
${HT}_{n}(T\partial_{i}^{\prime}T^{-1})=\partial_{i}\notin\mathfrak{m}$, we
must have $\overline{\partial_{i}^{\prime}}\neq 0$, and the map

$\partial_{i}\mapsto\overline{\partial_{i}^{\prime}}:=\sum_{j=1}^{n-1}c_{i;j}\partial_{j},\quad
1\leq i\leq n-1\quad c_{i;j}\in K$

is invertible too. In particular, the spectral module
$F_{n-1}=K[\partial_{1},\ldots,\partial_{n-1}]$ of the ring
$K[\overline{\partial_{1}^{\prime}},\ldots,\overline{\partial_{n-1}^{\prime}}]$
is Cohen-Macaulay. Then by theorem 7.1 there exists an invertible operator
$S_{0}\in\hat{D}_{n-1}^{sym}\subset\hat{D}_{n}^{n}$ with $\mathop{\rm\bf
ord}(S_{0})=0$ such that $S_{0}\partial_{i}^{\prime}S_{0}^{-1}=\partial_{i}$
for all $i=1,\ldots,n-1$.

Put $T_{1}=TS_{0}$. Then the operators $T_{1}^{-1}\partial_{i}T_{1}$ are monic
formally quasi-elliptic. Then by lemma 5.2 there exists an operator
$S_{1}=c_{0}\exp(\sum_{i=1}^{n}e_{i}x_{i})\exp(\sum_{i=1}^{n-1}d_{i,n}x_{n}\partial_{i})$
(see the proof of lemma 5.2) such that the operators
$S_{1}^{-1}T_{1}^{-1}\partial_{i}T_{1}S_{1}$ are normalized. Note that again
$\mathop{\rm\bf ord}(S_{1})=\mathop{\rm\bf ord}(S_{1}^{-1})=0$. And by theorem
5.1 there exists an operator $S_{2}\in\hat{E}_{n}$ such that $\mathop{\rm\bf
ord}(S_{2})=|\mathop{\rm ord}\nolimits_{\Gamma}(S_{2})|=\mathop{\rm\bf
ord}(S_{2}^{-1})=|\mathop{\rm ord}\nolimits_{\Gamma}(S_{2}^{-1})|=0$ ,
$S_{2},S_{2}^{-1}$ satisfies $A_{1}$ and
$S_{2}^{-1}S_{1}^{-1}T_{1}^{-1}\partial_{i}T_{1}S_{1}S_{2}=\partial_{i}$ for
$1\leq i\leq n$.

Therefore, $T^{\prime}:=T_{1}S_{1}S_{2}\in
K_{y}[[\partial_{1},\ldots,\partial_{n-1}]]((\partial_{n}^{-1}))\cap\Pi_{n}$.
Since all operators in this product are of order zero and $\mathop{\rm
ord}\nolimits_{n}(T_{1}S_{1}S_{2})=\mathop{\rm
ord}\nolimits_{n}(T)+\mathop{\rm ord}\nolimits_{n}(S_{0})+\mathop{\rm
ord}\nolimits_{n}(S_{1})+\mathop{\rm ord}\nolimits_{n}(S_{2})=0$, it follows
that $\mathop{\rm\bf ord}(T^{\prime})=0$, where from $\mathop{\rm\bf
ord}((T^{\prime})^{-1})=\mathop{\rm ord}\nolimits_{n}((T^{\prime})^{-1})=0$.
Thus, $T=T^{\prime}S_{2}^{-1}S_{1}^{-1}S_{0}^{-1}$ and therefore
$\mathop{\rm\bf ord}(T)=\mathop{\rm\bf ord}(T^{-1})=0$, $\mathop{\rm
ord}\nolimits_{n}(T)=\mathop{\rm ord}\nolimits_{n}(T^{-1})=0$. The rest of the
proof is straightforward. ∎

######  Theorem 8.1.

(cf. [61, Th.3.2]) There is a one to one correspondence between the classes of
equivalent quasi-elliptic 1-Schur pairs $(A,W)$ in $V_{n}$ of rank $r$ with
$\mathop{\rm Supp}(W)=F$ and the classes of equivalent in $\hat{D}_{n}$ quasi-
elliptic rings $B\subset\hat{D}_{n}^{sym}$ of rank $r$.

######  Remark 8.1.

In [61, Th.3.2] (see also [32, Sec. 2]) a correspondence between slightly
narrower classes was established: namely, the 1-admissible operators were
assumed to satisfy an extra condition. In fact, the extended equivalence
classes are not much more, as it can be seen from the proof.

###### Proof.

In one way we use the construction 1 from section 6.3: given a 1-quasi
elliptic ring $B$ we construct a pair $(A,W)$ with the required properties. If
we choose any equivalent 1-quasi elliptic ring $B^{\prime}$, then obviously
the pair constructed by $B^{\prime}$ will be equivalent to $(A,W)$.

In another way we need to use the analogue of the Sato theory for subspaces in
$V_{n}$ from section 6. Namely, given a quasi-elliptic 1-Schur pair $(A,W)$,
we construct the corresponding ring $B$ by construction 2 from section 6.3
with the help of a Sato operator $S$.

If $(A^{\prime},W^{\prime})$ is an equivalent 1-Schur pair, then
$A^{\prime}=T^{-1}AT$, $W^{\prime}=WT$ for some 1-admissible operator $T$,
which can be written (see proposition 8.1) in the form $T=S_{2}S_{1}$. Then it
is easy to see that

$W^{\prime}=W\diamond(S_{2}S_{1})=(F\diamond
S_{1})\diamond(S_{1}^{-1}SS_{2}S_{1})=F\diamond(S_{1}^{-1}SS_{2}S_{1}),$

i.e. the corresponding Sato operator for the space $W^{\prime}$ from theorem
6.1 is
$S^{\prime}=S_{1}^{-1}SS_{2}S_{1}$. So, the corresponding ring
$B^{\prime}=S^{\prime}A^{\prime}(S^{\prime})^{-1}=S_{1}^{-1}BS_{1}$, i.e. it
is equivalent to $B$.

At last, note that the composition of two constructions
$B\rightsquigarrow(A,W)$, $(A,W)\rightsquigarrow B$ leads to equivalent
objects. ∎

The theorem above can be improved for quasi-elliptic rings of rank one: as we
have noted in remark 6.2 the set of admissible linear changes is not empty for
such rings.

######  Definition 8.2.

Two quasi-elliptic rings $B_{1},B_{2}\subset\hat{D}_{n}$ are said to be weakly
equivalent if there exits a unit $U\in\hat{U}_{n}^{sym}$ such that
$B_{1}=UB_{2}U^{-1}$.

Two quasi-elliptic 1-Schur pairs $(A,W)$ and $(A^{\prime},W^{\prime})$ are
said to be weakly equivalent if $A^{\prime}=\varphi_{m}\circ
Ad(T_{m})\circ\ldots\varphi_{1}\circ Ad(T_{1})(A)$ and
$W^{\prime}=\varphi_{m}\circ Ad(T_{m})\circ\ldots\varphi_{1}\circ
Ad(T_{1})(W)$ for some $m\in{\mbox{\dbl N}}$, where $T_{i}$ are 1-admissible
operators and $\varphi_{i}$ are admissible linear changes of variables555Of
course, $\varphi_{i}$ is an admissible change for the pair
$(Ad(T_{i-1})\circ\ldots\varphi_{1}\circ
Ad(T_{1})(A),Ad(T_{i-1})\circ\ldots\varphi_{1}\circ Ad(T_{1})(A)(W))$,
$Ad(T)(A):=T^{-1}AT$, $Ad(T)(W):=W\diamond T$.

######  Theorem 8.2.

Let $K$ be a complete field. Assume $K_{y}=K$.

There is a one to one correspondence between the classes of weakly equivalent
quasi-elliptic 1-Schur pairs $(A,W)$ in $V_{n}$ of rank one with $\mathop{\rm
Supp}(W)=F$ and the classes of weakly equivalent quasi-elliptic rings of
commuting operators $B\subset\hat{D}_{n}^{sym}$ of rank one.

###### Proof.

The proof is essentially the same as the proof of theorem 8.1. Given two
weakly equivalent rings $B_{1},B_{2}$ we need to show that the corresponding
quasi-elliptic 1-Schur pairs are weakly equivalent.

Assume $B_{1}=UB_{2}U^{-1}$, denote by $(A_{i},W_{i})$ normalized 1-Schur
pairs of rings $B_{i}$ constructed by the construction 1 of section 6.3.

First, taking appropriate powers of formally quasi-elliptic elements in
$A_{1}$ and making an appropriate linear change of variables
$\varphi^{\prime}$ from lemma 6.6 we can replace the 1-Schur pair
$(A_{1},W_{1})$ with the quasi-elliptic 1-Schur pair
$(A_{1}^{\prime},W_{1}^{\prime}):=(\varphi^{\prime}(A_{1}),\varphi^{\prime}(W_{1}))$
such that $\varphi^{\prime}(A_{1})$ contains formally quasi-elliptic operators
of the same $\mathop{\rm\bf ord}$-order (we can find an admissible such
$\varphi^{\prime}$ by remark 6.2 since the Schur pairs are of rank one). By
proposition 6.1 the quasi-elliptic ring $B_{1}^{\prime}$ corresponding to this
Schur pair by construction 2 from section 6.3 is weakly equivalent to $B_{1}$.
i.e. $B_{1}^{\prime}$ is weakly equivalent to $B_{2}$. Obviously, it suffices
to prove that $(A_{1}^{\prime},W_{1}^{\prime})$ is weakly equivalent to
$(A_{2},W_{2})$, and we’ll assume below that $(A_{1},W_{1})$ satisfies the
properties of the pair $(A_{1}^{\prime},W_{1}^{\prime})$.

Let $B_{i}=S_{i}A_{i}S_{i}^{-1}$, $W_{i}=F\diamond S_{i}$. Denote by
$\varphi_{i}:W_{i}\rightarrow F$, $w_{i}\mapsto w_{i}\diamond S_{i}^{-1}$ the
isomorphisms of $A_{i}$-modules $W_{i}$ with $B_{i}$-modules $F$
($a_{i}\mapsto S_{i}a_{i}S_{i}^{-1}$). Then $\varphi_{i}^{-1}$ are given as
$\varphi_{i}:F\rightarrow W_{i}$, $w\mapsto w\diamond S_{i}$. At last, denote
by $\psi:F\rightarrow F$, $w\mapsto w\diamond U^{-1}$ the isomorphism of
$B_{i}$-modules $F$ ($b_{1}\mapsto Ub_{1}U^{-1}$). Then the composition
$\tilde{\varphi}:=\varphi_{2}^{-1}\circ\psi\circ\varphi_{1}$ determines an
isomorphism of $A_{i}$-modules $W_{i}$. By construction, it preserves the
$\mathop{\rm\bf ord}$-order filtration on $W_{i}$ and on $A_{i}$. Let
$a_{1},\ldots,a_{n}\in A_{1}$ be formally quasi-elliptic elements of the same
$\mathop{\rm\bf ord}$-order, with $\mathop{\rm
ord}\nolimits_{\Gamma}(a_{1})=(0,\ldots,0,p)$, $\mathop{\rm
ord}\nolimits_{\Gamma}(a_{i})=(\underline{0})-(i-1)$. Let $\tilde{a}_{1}$ be
the monic $p$-th root of $a_{1}$: $\tilde{a}_{1}^{p}=a_{1}$. Applying lemma
6.6 and remark 6.2, we can find an admissible linear change of variables
$\varphi$ such that $\mathop{\rm
ord}\nolimits_{\Gamma}(\varphi(\tilde{\varphi}(a_{i})))=\mathop{\rm
ord}\nolimits_{\Gamma}(a_{i})$, $i=1,\ldots,n$. Applying theorem 5.1 and lemma
5.2, we can find a 1-admissible operator $T$ such that
$T\varphi(\tilde{\varphi}(a_{i}))T^{-1}=a_{i}$, $i=2,\ldots,n$ and
$T\varphi(\tilde{\varphi}(\tilde{a}_{1}))T^{-1}=\tilde{a}_{1}$. Then
$T\varphi(\tilde{\varphi}(A_{1}))T^{-1}=A_{1}$. For, arguing as in lemma 6.6,
for any $a\in A_{1}$ we have $a\tilde{a}_{1}^{-\mathop{\rm\bf ord}(a)}\in
K[[a_{2}\tilde{a}_{1}^{-\mathop{\rm\bf
ord}(a_{2})},\ldots,a_{n}\tilde{a}_{1}^{-\mathop{\rm\bf
ord}(a_{n})},\tilde{a}_{1}^{-1}]]$. Thus, $\tilde{\varphi}=\varphi^{-1}\circ
Ad(T)$, i.e. $\varphi(A_{2})=T^{-1}A_{1}T$ and $(A_{2},W_{2})$ is weakly
equivalent to $(A_{1},W_{1})$.

Conversely, two weakly equivalent quasi-elliptic 1-Schur pairs of rank one
give, via the construction 2 and proposition 6.1, two weakly equivalent quasi-
elliptic rings in $\hat{D}_{n}^{sym}$. ∎

######  Remark 8.2.

If $K_{y}\neq K$, the theorem holds if an additional assumption is added:
namely, in each equivalence class of Schur pairs corresponding to quasi-
elliptic rings via construction 1, there should exist a representative pair
satisfying the extra condition of remark 6.2, item 1).

######  Example 8.1.

Let’s illustrate theorem 8.2 as well as some other theorems of these paper on
the explicit example calculated in [9, Sec. 6].

Recall that there was a family of 1-Schur pairs $(A,W_{\beta})$ defined, where
$A,W_{\beta}\in{\mbox{\dbl C}}[z_{1},z_{2}]$, $A={\mbox{\dbl
C}}[z_{1}^{2},z_{1}^{3},z_{2}^{2},z_{2}^{3}]$, where $z_{i}=\partial_{i}$, and

$W_{\beta}={\mbox{\dbl C}}\cdot w+(\xi_{2}+\xi_{2}^{2}z_{2}+\beta
z_{1})z_{1}^{2}{\mbox{\dbl C}}[z_{1}]+(\xi_{1}+\xi_{1}^{2}z_{1}+\beta
z_{2})z_{2}^{2}{\mbox{\dbl C}}[z_{2}]+z_{1}^{2}z_{2}^{2}{\mbox{\dbl
C}}[z_{1},z_{2}],$

where
$w=1+\xi_{1}z_{1}+\xi_{2}z_{2}+(\xi_{1}\xi_{2}-\beta)\cdot\left(z_{1}z_{2}+\left(\dfrac{z_{1}^{2}}{\xi_{2}^{2}}+\dfrac{z_{2}^{2}}{\xi_{1}^{2}}\right)\right).$
In [9, Th. 6.5] corresponding Sato operators $S_{\beta}$ were found,
$S_{\beta}=S_{0}+\beta T$, where

$S_{0}:=\partial_{1}\partial_{2}+\frac{1}{\xi_{2}-x_{2}}\partial_{1}+\frac{1}{\xi_{1}-x_{1}}\partial_{2}+\frac{1}{(\xi_{1}-x_{1})(\xi_{2}-x_{2})}$

and

$\begin{split}T=&\frac{1}{(\xi_{1}-x_{1})(\xi_{2}-x_{2})}\left(\frac{1}{\xi_{2}}\left(\delta_{2}\partial_{1}+(\xi_{1}-x_{1})\delta_{2}\partial_{1}^{2}\right)+\frac{1}{\xi_{1}}\left(\delta_{1}\partial_{2}+(\xi_{2}-x_{2})\delta_{1}\partial_{2}^{2}\right)\right)+\\\
&\frac{1}{(\xi_{1}\xi_{2}-\beta)(\xi_{1}-x_{1})(\xi_{2}-x_{2})}\delta_{1}\delta_{2}\left(1+\beta\left(\frac{\partial_{1}}{\xi_{2}}+\frac{\partial_{2}}{\xi_{1}}\right)\right),\end{split}$

It is not difficult to see that $A$ is not quasi-elliptic, but for a generic
linear change of variables (obviously, all linear changes are admissible in
our case) it becomes quasi-elliptic. Besides, for a generic linear change
$\varphi$ we have $\mathop{\rm Supp}(\varphi(W_{\beta})z_{2}^{-2})=F$. By
definition, all such quasi-elliptic Schur pairs
$(\varphi(A),\varphi(W_{\beta})z_{2}^{-2})$ are weakly equivalent. By theorem
6.1, item 3 there exists another Sato operator $S_{n}$ (invertible in
$\hat{E}_{n}$) for the space $\varphi(W_{\beta})z_{2}^{-2}$. By theorem 6.1,
item 4 we have $\varphi(S_{\beta})\partial_{2}^{-2}=U\circ S_{n}$ for some
unity $U\in\hat{U}_{n}^{sym}$.

By theorem 7.1 the linear change $\varphi$ can be defined by conjugation with
a Schur operator $S_{\varphi}\in\hat{D}_{2}^{sym}$ of order zero. Then the
corresponding rings of commuting operators are just conjugated rings
$S_{\varphi}\mathtt{L}(A)S_{\varphi}^{-1}$. By theorem 8.2 they are weakly
equivalent as well.

Besides, such pairs $(\varphi(A),\varphi(W_{\beta})z_{2}^{-2})$ fit definition
of a Schur pair (and even a pre-Schur pair of rank one) from papers [61], [62,
Def. 14], and therefore determine an algebraic-geometric pre-spectral data,
see e.g. [62, Th. 4] (in these constructions the Sato operator $S_{n}$ above
was essentially used). By [62, Th. 3] the last data can be extended to a
(reduced) geometric datum $(X,C,p,{{\cal F}}_{\beta})$ of rank one, where $X$
is a projective surface (a natural compactification of ${{\rm Spec\,}}(A)$),
$C$ is an integral ample Cartier divisor (by [62, Th. 2]), ${{\cal
F}}_{\beta}={{\rm Proj\,}}(\widetilde{W_{\beta}})$ is a torsion free spectral
sheaf with fixed Hilbert polynomial (see [9, Sec. 6] or [62, Def. 13, 14]),
and $p$ is a regular point on $C$.

From [9, Sec. 6] we know that the normalisation of the surface ${{\rm
Spec\,}}(A)$ is just the affine plane ${\mbox{\dbl A}}^{2}$. By [31, Th.2.1.]
we have $C^{2}=1$ and $C$ is a rational curve, and therefore by [32, Th. 3.2]
the normalisation of $X$ is isomorphic to the projective plane ${\mbox{\dbl
P}}^{2}$. From the explicit description of the spaces $W_{\beta}$ and from
definition of the spectral sheaf ${{\cal F}}_{\beta}$ it is not difficult to
see that ${{\cal F}}_{\beta}|_{C}\simeq n_{*}({{\cal O}}_{{\mbox{\sdbl
P}}^{1}})$, where $n:{\mbox{\dbl P}}^{1}\rightarrow C$ is the normalisation
morphism, iff $\beta=0$. In this case the operator $S_{0}$ is just a
differential operator and $\varphi(S_{\beta})\partial_{2}^{-2}=cS_{n}$,
$c\in{\mbox{\dbl C}}$. In view of [63, Th. 7.9] this confirms the conjecture
7.11 about characterisation of commuting PDOs between all quasi-elliptic rings
from [63].

It is not difficult to see that the corresponding to such a geometric data
Schur pairs from [62, Th. 1] will be weakly equivalent for different choices
of $p$. Thus, the isomorphism class of weakly equivalent Schur pairs (or an
isomorphism class $[S_{\varphi}\mathtt{L}(A)S_{\varphi}^{-1}]$ of weakly
equivalent quasi-elliptic rings) determines an isomorphism class of pre-
spectral datum $(X,C,{{\cal F}})$. A detailed description of this
correspondence (in a more general situation) will appear in [64].

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can be found e.g. at https://www.researchgate.net/publication/
340952902AlgebraGeometryandAnalysisofCommutingOrdinaryDifferentialOperators

A. Zheglov, Lomonosov Moscow State University, faculty of mechanics and
mathematics, department of differential geometry and applications, Leninskie
gory, GSP, Moscow, 119899, Russia
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