# A Deformation of Commutative Polynomial Algebras in Even Numbers of
Variables

Wenhua Zhao

###### Abstract.

We introduce and study a deformation of commutative polynomial algebras in
even numbers of variables. We also discuss some connections and applications
of this deformation to the generalized Laguerre orthogonal polynomials and the
interchanges of right and left total symbols of differential operators of
polynomial algebras. Furthermore, a more conceptual re-formulation for the
image conjecture [Z3] is also given in terms of the deformed algebras.
Consequently, the well-known Jacobian conjecture [Ke] is reduced to an open
problem on this deformation of polynomial algebras.

###### Key words and phrases:

The generalized Laguerre polynomials, total symbols of differential operators,
the image conjecture, the Jacobian conjecture.

###### 2000 Mathematics Subject Classification:

33C45, 32W99, 14R15

##  1\. Introduction

Let $\xi=(\xi_{1},\xi_{2},...,\xi_{n})$ and $z=(z_{1},z_{2},...,z_{n})$ be
$2n$ commutative free variables. Throughout this paper, we denote by
$\mathbb{C}[\xi,z]$, $\mathbb{C}[z]$ and $\mathbb{C}[\xi]$ the vector spaces
(without any algebra structures) over $\mathbb{C}$ of polynomials in
$({\xi,z})$, in $z$ and in $\xi$, respectively. The corresponding polynomial
algebras will be denoted respectively by ${\mathcal{A}}[\xi,z]$,
${\mathcal{A}}[z]$ and ${\mathcal{A}}[\xi]$.

For any $1\leq i\leq n$, we set $\partial_{i}\\!:=\partial_{z_{i}}$ and
$\delta_{i}=\partial_{\xi_{i}}$. Denote by
$\partial=(\partial_{1},\partial_{2},...,\partial_{n})$ and
$\delta=(\delta_{1},\delta_{2},...,\delta_{n})$. We also occasionally use
$\partial_{z}$ and $\partial_{\xi}$ to denote $\partial$ and $\delta$,
respectively.

Set
$\Omega\\!:=\sum_{i=1}^{n}(\partial_{i}\otimes\delta_{i}+\delta_{i}\otimes\partial_{i})$.
For any $t\in\mathbb{C}$, we define a new product $\ast_{t}$ for the vector
space $\mathbb{C}[\xi,z]$ by setting, for any $f,g\in\mathbb{C}[\xi,z]$,

(1.1) $\displaystyle f\ast_{t}g\\!:=\mu\left(e^{-t\,\Omega}(f\otimes
g)\right),$

where $\mu:\mathbb{C}[\xi,z]\otimes\mathbb{C}[\xi,z]\to\mathbb{C}[\xi,z]$
denotes the product map of the polynomial algebra ${\mathcal{A}}[{\xi,z}]$.

Denote by ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ the new algebra
$(\mathbb{C}[\xi,z],\ast_{t})$. For the case $t=1$, we also introduce the
following short notation:

(1.2) $\displaystyle\ast\\!:$ $\displaystyle=\ast_{t=1}.$ (1.3)
$\displaystyle{\mathcal{B}}[{\xi,z}]\\!:$
$\displaystyle={\mathcal{B}}_{t=1}[{\xi,z}].$

Note that, when $t=0$, the algebra ${\mathcal{B}}_{t=0}[{\xi,z}]$ coincides
with the usual polynomial algebra ${\mathcal{A}}[{\xi,z}]$.

In this paper, we first show that ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ gives a deformation of the polynomial algebra
${\mathcal{A}}[{\xi,z}]$. Actually, it is a trivial deformation in the sense
of deformation theory. To be more precise, set

(1.4) $\displaystyle\Lambda\\!:$
$\displaystyle=\sum_{i=1}^{n}\delta_{i}\partial_{i}.$ (1.5)
$\displaystyle\Phi_{t}\\!:$ $\displaystyle=e^{t\Lambda}=\sum_{m\geq
0}\frac{t^{m}\Lambda^{m}}{m!}.$ (1.6) $\displaystyle\Phi\\!:$
$\displaystyle=\Phi_{t=1}.$

Note that, $\Phi_{t}$ for any $t\in\mathbb{C}$ is a well-defined bijective
linear map from $\mathbb{C}[\xi,z]$ to $\mathbb{C}[\xi,z]$, whose inverse map
is given by $\Phi_{-t}=e^{-t\Lambda}$. This is because the differential
operator $\Lambda$ of $\mathbb{C}[\xi,z]$ is locally nilpotent, i.e. for any
$f(\xi,z)\in\mathbb{C}[\xi,z]$, $\Lambda^{m}f(\xi,z)=0$ when $m\gg 0$.

With the notation fixed above, we will show that, for any $t\in\mathbb{C}$,
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ actually is an
isomorphism of $\mathbb{C}$-algebras (See Proposition 2.1 and Corollary 2.2).

Note that, from the point view of deformation theory, the deformation
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ is not interesting at all.
But, surprisingly, as we will show in this paper, the algebra
${\mathcal{B}}_{t}[{\xi,z}]$ and the isomorphism $\Phi_{t}$ are actually
closely related with the generalized Laguerre polynomials (See [Sz], [PS] and
[AAR]) and the interchanges of right and left total symbols of differential
operators of polynomial algebras.

Furthermore, as we will show in Section 4, the algebras
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ and the isomorphism $\Phi_{t}$
via their connections with the image conjecture proposed in [Z3] are also
related with the Jacobian conjecture which was first proposed by O. H. Keller
[Ke] in $1939$ (See also [BCW] and [E]). Actually, the Jacobian conjecture can
be viewed as a conjecture which, in some sense, just claims that the algebra
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\neq 0)$ should not differ or change too much
from the polynomial algebra
${\mathcal{A}}[{\xi,z}]={\mathcal{B}}_{t=0}[{\xi,z}]$. Therefore, from this
point of view, the triviality of the deformation ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ (in the sense of deformation theory) can be viewed as a
fact in favor of the Jacobian conjecture. For another interesting application
of the isomorphism $\Phi$ to the Jacobian conjecture, see [Z5].

Considering the length of the paper, below we give a more detailed description
for the contents and the arrangement of the paper.

In Subsection 2.1, we prove some simple properties of the deformation
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ and the isomorphism
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$, which will be
needed for the rest of this paper. In particular, in this subsection the
triviality of the deformation ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$
in the sense of deformation theory is proved in Proposition 2.1. and Corollary
2.2.

In Subsection 2.2, we show that, for different $t\in\mathbb{C}$, the
$\ell$-adic topologies induced by ${\mathcal{B}}_{t}[{\xi,z}]$ on the common
base vector space $\mathbb{C}[\xi,z]$ are different. But they are all
homeomorphic to the $\ell$-adic topology induced by the polynomial algebra
${\mathcal{A}}[{\xi,z}]$ under the isomorphism
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ (viewed as an
automorphism of $\mathbb{C}[\xi,z]$). See Proposition 2.9 and also Corollary
2.10 for the precise statements.

In Subsection 2.3, we study the induced isomorphism $(\Phi_{t})_{*}$
$(t\in\mathbb{C})$ of $\Phi_{t}$ from the differential operator algebra, or
the Weyl algebra ${\mathcal{D}}_{t}[{\xi,z}]$ of ${\mathcal{B}}_{t}[{\xi,z}]$
to the Weyl algebra ${\mathcal{D}}[{\xi,z}]$ of ${\mathcal{A}}[{\xi,z}]$. The
main results of this subsection are Propositions 2.11 and 2.13. Proposition
2.11 says that the derivations $\partial_{z_{i}}$ and $\partial_{\xi_{i}}$
$(1\leq i\leq n)$ of ${\mathcal{A}}[{\xi,z}]$ are also derivations of
${\mathcal{B}}_{t}[{\xi,z}]$ for all $t\in\mathbb{C}$ and are fixed by the
isomorphism $(\Phi_{t})_{*}$. Proposition 2.13 gives explicitly the images
under $(\Phi_{t})_{*}$ of the multiplication operators with respect to the
product $\ast_{t}$ of ${\mathcal{B}}_{t}[{\xi,z}]$.

In Section 3, by using some results derived in Section 2, we show in Theorem
3.1 that $\Phi=\Phi_{t=1}$ (resp. $\Phi_{t=-1}$) as an automorphism of
$\mathbb{C}[\xi,z]$ actually coincides with the linear map which changes left
(resp. right) total symbols of differential operators of ${\mathcal{A}}[z]$ to
their right (resp. left) total symbols. Consequently, the products
$\ast_{t=\pm 1}$ appear naturally when one derives left or right total symbols
of certain differential operators of ${\mathcal{A}}[z]$ (See Corollary 3.2).
The results derived in this subsection also play some important roles in [Z5]
in which among some other results a more straightforward proof for the
equivalence of the Jacobian conjecture and the vanishing conjecture (See [Z1]
and [Z2]) will be given.

In Subsection 4.1, we study the Taylor series expansion of polynomials in
$\mathbb{C}[{\xi,z}]$ with respect to the new product $\ast_{t}$ and use it to
give a more conceptual proof for the expansion of polynomials given in Eq.
(4.6). This expansion was first proved in [Z3] and played a crucial role there
in the proof of the implication of the Jacobian conjecture from the image
conjecture (See Conjecture 4.3).

In Subsection 4.2, we first recall the notion of the so-called Mathieu
subspaces of commutative algebras (See Definition 4.2), which was first
introduced in [Z4], and also the image conjecture (See Conjecture 4.3) for the
differential operators $\xi_{i}-t\partial_{i}$ $(1\leq i\leq n)$ in terms of
the notion of Mathieu subspaces. We then give a re-formulation of Conjecture
4.3 in terms of the algebra ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$
(See Conjecture 4.5) and show in Theorem 4.6 that these two conjectures are
equivalent to each other. Since it has been shown in [Z3] that Conjecture 4.3
implies Jacobian conjecture, hence so does Conjecture 4.5.

Consequently, via its connections with Conjecture 4.5, the Jacobian conjecture
is reduced to an open problem on the deformation ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ of the polynomial algebra ${\mathcal{A}}[{\xi,z}]$. The
open problem asks if the ideal $\xi\mathbb{C}[{\xi,z}]$ of
${\mathcal{A}}[{\xi,z}]$ generated by $\xi$ will remain to be a Mathieu
subspace in the algebra ${\mathcal{B}}_{t}[{\xi,z}]$ for any $t\neq 0$. Note
that any ideal is automatically a Mathieu subspace, but not conversely.
Therefore, the triviality (in the sense of deformation theory) of the
deformation ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ proved in
Proposition 2.1 can be viewed as a fact in favor of the Jacobian conjecture.

Section 5 is mainly on a connection of the algebra ${\mathcal{B}}[{\xi,z}]$,
especially, its product $\ast$ with the multi-variable generalized Laguerre
polynomials, and also some of the applications of this connection to both
${\mathcal{B}}[{\xi,z}]$ and the generalized Laguerre polynomials.

In Subsection 5.1, we very briefly recall the definition of the (generalized)
Laguerre polynomials $L_{\alpha}^{[\bf k]}(z)$ $({\bf
k},\alpha\in{\mathbb{N}}^{n})$ (See Eqs. (5.1)–(5.3)) and also the orthogonal
property (See Theorem 5.1) of these polynomials.

In Subsection 5.2, we show in Theorem 5.2 that, for any ${\bf
k},\alpha\in{\mathbb{N}}^{n}$, we have

(1.7) $\displaystyle L_{\alpha}^{[{\bf k}]}(\xi
z)=\frac{(-1)^{|\alpha|}}{\alpha!}\,\xi^{-\bf k}(\xi^{\alpha+{\bf k}}\ast
z^{\alpha})=\frac{(-1)^{|\alpha|}}{\alpha!}\,z^{-{\bf k}}(\xi^{\alpha}\ast
z^{\alpha+{\bf k}}).$

Consequently, the generalized Laguerre polynomials $L_{\alpha}^{[\bf k]}(z)$
$({\bf k},\alpha\in{\mathbb{N}}^{n})$ can be obtained by evaluating the
polynomials $\xi^{-\bf k}(\xi^{\alpha+{\bf k}}\ast z^{\alpha})$ or $z^{-{\bf
k}}(\xi^{\alpha}\ast z^{\alpha+{\bf k}})$ at $\xi=(1,1,...,1)$. Note that the
evaluation map at $\xi=(1,1,...,1)$ is not an algebra homomorphism from
${\mathcal{B}}[{\xi,z}]$ to ${\mathcal{A}}[{\xi,z}]$. Otherwise, the
generalized Laguerre polynomials would be trivialized.

In the first part of Subsection 5.3, we use certain results of the generalized
Laguerre polynomials and the connection in Eq. (1.7) above to derive more
properties on the polynomials $\xi^{\alpha}\ast z^{\alpha}$ which, by
Proposition 2.6, $(c)$, are actually the monomials of $\xi$ and $z$ in the new
algebra ${\mathcal{B}}[{\xi,z}]$.

For example, by using the connection in Eq. (1.7) and I. Schur’s
irreducibility theorem [Sc1] of the Laguerre polynomials in one variable, we
immediately have that, when $n=1$, the monomials $\xi^{m}\ast z^{m}$ $(m\geq
2)$ of ${\mathcal{B}}[{\xi,z}]$ are actually irreducible over ${\mathbb{Q}}$
(See Theorem 5.9). Furthermore, by using I. Schur’s irreducibility theorem
[Sc2] and M. Filaseta and T.-Y. Lam’s irreducibility theorem [FL] on the
generalized Laguerre polynomials, we have that, all but finitely many of the
polynomials $\xi^{-k}(\xi^{m+k}\ast z^{m})$ and $z^{-k}(\xi^{m}\ast z^{m+k})$
$(m,k\in{\mathbb{N}})$ are irreducible over ${\mathbb{Q}}$ (See Theorem 5.10).

In the second part of Subsection 5.3, we use the connection given in Eq. (1.7)
and certain results of ${\mathcal{B}}[{\xi,z}]$ derived in Section 2 to give
new proofs, first, for some recurrent formulas of the generalized Laguerre
polynomials (See Proposition 5.11) and, second, for the fact that the
generalized Laguerre polynomials satisfy the so-called associate Laguerre
differential equation (See Theorem 5.12). At the end of this subsection, we
draw the reader’s attention to a conjecture, Conjecture 5.13, on the
generalized Laguerre polynomials, which is still open even for the classical
Laguerre polynomials in one variable.

Acknowledgment The author would like to thank the anonymous referees for
pointing out many typos, minor errors of the previous version of this paper,
and also for suggesting the new proof of Lemma 5.8 without the condition that
the base filed $K$ has infinitely many elements.

##  2\. The Deformation ${\mathcal{B}}_{t}[{\xi,z}]$ of the Polynomial Algebra
${\mathcal{A}}[{\xi,z}]$

In this section, we first derive in Subsection 2.1 some properties and
identities for the algebra ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$. In
Subsection 2.2, we show that, for different $t\in\mathbb{C}$, the $\ell$-adic
topologies induced by the algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ on the common base vector space $\mathbb{C}[\xi,z]$ are
different. But they are all homeomorphic under the isomorphism
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ to the
$\ell$-adic topology on $\mathbb{C}[\xi,z]$ induced by the polynomial algebra
${\mathcal{A}}[{\xi,z}]$ (See Proposition 2.9 and also Corollary 2.10).

In Subsection 2.3, we study the isomorphism $(\Phi_{t})_{*}$ induced by
$\Phi_{t}$ from the Weyl algebra of ${\mathcal{B}}_{t}[{\xi,z}]$ to the Weyl
algebra of ${\mathcal{A}}[{\xi,z}]$. The main results in this subsection are
Propositions 2.11 and 2.13.

###  2.1. Some Properties of the Algebras ${\mathcal{B}}_{t}[{\xi,z}]$

First, one remark on notation and convention is that, we will freely use
throughout this paper some commonly used multi-index notations and
conventions. For instance, for $n$-tuples $\alpha=(k_{1},k_{2},...,k_{n})$ and
$\beta=(m_{1},m_{2},...,m_{n})$ of non-negative integers, we have

$\displaystyle|\alpha|$ $\displaystyle=\sum_{i=1}^{n}k_{i}.$
$\displaystyle\alpha!$ $\displaystyle=k_{1}!k_{2}!\cdots k_{n}!.$
$\displaystyle\binom{\alpha}{\beta}$
$\displaystyle=\begin{cases}\frac{\alpha!}{\beta!(\alpha-\beta)!}&\mbox{if
$k_{i}\geq m_{i}$ for all $1\leq i\leq n$};\\\
0,&\mbox{otherwise.}\end{cases}$

The notation and convention fixed in the previous section will also be used
throughout this paper.

The first main result of this section is the following proposition.

######  Proposition 2.1.

For any $t\in\mathbb{C}$ and $f,g\in\mathbb{C}[\xi,z]$, we have

(2.1) $\displaystyle\Phi_{t}(f\ast_{t}g)=\Phi_{t}(f)\Phi_{t}(g).$

Proof: We first set

(2.2) $\displaystyle
f\ast_{t}^{\prime}g\\!:=\Phi_{t}^{-1}\big{(}\Phi_{t}(f)\Phi_{t}(g)\big{)}=\Phi_{-t}\big{(}\Phi_{t}(f)\Phi_{t}(g)\big{)}.$

for any $f,g\in\mathbb{C}[{\xi,z}]$.

We view $t$ as a formal parameter which commutes with $\xi$ and $z$. Then, by
Eqs. (1.1), (2.2) and the fact that the differential operators $\Lambda$ and
$\Omega$ are locally nilpotent on $\mathbb{C}[\xi,z]$ and
$\mathbb{C}[\xi,z]\otimes\mathbb{C}[\xi,z]$, respectively, we see that
$f\ast_{t}g$ and $f\ast_{t}^{\prime}g$ are polynomials in $t$ with
coefficients in $\mathbb{C}[{\xi,z}]$. Furthermore, by setting $t=0$ in Eqs.
(1.1) and (2.2), we see that the constant terms (with respect to $t$) of
$f\ast_{t}g$ and $f\ast_{t}^{\prime}g$ are both $fg\in\mathbb{C}[{\xi,z}]$. In
other words, we have

(2.3) $\displaystyle
f\ast_{t}g\left.\right|_{t=0}=f\ast_{t}^{\prime}g\left.\right|_{t=0}=fg.$

From Eq. (1.1), we have,

(2.4) $\displaystyle\frac{\partial}{\partial t}(f\ast_{t}g)$
$\displaystyle=-\mu\left(\,e^{-t\Omega}(\Omega(f\otimes g))\,\right)$
$\displaystyle=-\sum_{i=1}^{n}\mu\left(e^{-t\Omega}\left((\delta_{i}f)\otimes(\partial_{i}g)+(\partial_{i}f)\otimes(\delta_{i}g)\right)\right)$
$\displaystyle=-\sum_{i=1}^{n}\left((\delta_{i}f)\ast_{t}(\partial_{i}g)+(\partial_{i}f)\ast_{t}(\delta_{i}g)\right).$

On the other hand, from Eq. (2.2), we have,

$\displaystyle\frac{\partial}{\partial t}(f$
$\displaystyle\ast_{t}^{\prime}g)=\frac{\partial}{\partial
t}\left(e^{-t\Lambda}((e^{t\Lambda}f)\,(e^{t\Lambda}g))\right)$
$\displaystyle=e^{-t\Lambda}\left(-\Lambda((e^{t\Lambda}f)\,(e^{t\Lambda}g))+(e^{t\Lambda}\Lambda
f)\,(e^{t\Lambda}g)+(e^{t\Lambda}f)\,(e^{t\Lambda}\Lambda g)\right).$

Note that, for any $u,v\in\mathbb{C}[\xi,z]$, it is easy to check that we have
the following identity:

$\displaystyle\Lambda(uv)=(\Lambda u)v+u(\Lambda
v)+\sum_{i=1}^{n}\left(\,(\delta_{i}u)(\partial_{i}v)+(\partial_{i}u)(\delta_{i}v)\,\right).$

By the last two equations above and also Eq. (2.2), we have

(2.5) $\displaystyle\frac{\partial}{\partial t}(f\ast_{t}^{\prime}g)$
$\displaystyle=-\sum_{i=1}^{n}e^{-t\Lambda}\left(\,((e^{t\Lambda}\delta_{i}f)\,(e^{t\Lambda}\partial_{i}g))+(e^{t\Lambda}\partial_{i}f)\,(e^{t\Lambda}\delta_{i}g)\,\right)$
$\displaystyle=-\sum_{i=1}^{n}\left(\,(\delta_{i}f)\ast_{t}^{\prime}(\partial_{i}g)+(\partial_{i}f)\ast_{t}^{\prime}(\delta_{i}g)\,\right).$

Next, we use the induction on $(\deg f+\deg g)$ to show Eq. (2.1). First, when
$\deg f+\deg g=0$, i.e. both $f$ and $g$ have degree zero, it is easy to see
from Eqs. (1.1) and (2.2) that $f\ast_{t}g=f\ast_{t}^{\prime}g=fg$ in this
case.

In general, by Eqs. (2.4), (2.5) and also the induction assumption, we have

(2.6) $\displaystyle\frac{\partial}{\partial
t}(f\ast_{t}g)=\frac{\partial}{\partial t}(f\ast_{t}^{\prime}g).$

Since $f\ast_{t}g$ and $f\ast_{t}^{\prime}g$ are polynomials in $t$ with
coefficients in $\mathbb{C}[{\xi,z}]$ and both satisfy Eqs. (2.3) and (2.6),
it is easy to see that they must be equal to each other. Hence, Eq. (2.1)
holds. $\Box$

######  Corollary 2.2.

For any $t\in\mathbb{C}$,
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ is an
isomorphism of algebras. Therefore, in the sense of deformation theory, the
deformation ${\mathcal{B}}_{t}[{\xi,z}]$ is a trivial deformation of the
commutative polynomial algebra ${\mathcal{A}}[{\xi,z}]$.

Next we derive some properties of the algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$, which will be needed for the rest of this paper.

######  Lemma 2.3.

For any $f,g\in\mathbb{C}[{\xi,z}]$, we have

(2.7) $\displaystyle
f\ast_{t}g=\sum_{\alpha,\beta\in{\mathbb{N}}^{n}}\frac{(-t)^{|\alpha|+|\beta|}}{\alpha!\beta!}\,\,(\delta^{\beta}\partial^{\alpha}f)(\partial^{\beta}\delta^{\alpha}g).$

Proof: Note first that, for any $1\leq i,j\leq n$,
$\partial_{i}\otimes\delta_{i}$ and $\delta_{j}\otimes\partial_{j}$ commute
with each other. So we have

(2.8) $\displaystyle e^{-t\,\Omega}$
$\displaystyle=e^{-t\sum_{i=1}^{n}\delta_{i}\otimes\partial_{i}}\,e^{-t\sum_{i=1}^{n}\partial_{i}\otimes\delta_{i}},$
(2.9) $\displaystyle e^{-t\sum_{i=1}^{n}\partial_{i}\otimes\delta_{i}}$
$\displaystyle=\prod_{i=1}^{n}e^{-t(\partial_{i}\otimes\delta_{i})}=\prod_{i=1}^{n}\sum_{k_{i}\geq
0}\frac{(-t)^{k_{i}}}{k_{i}!}\,(\partial_{i}^{k_{i}}\otimes\delta_{i}^{k_{i}})$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{(-t)^{|\alpha|}}{\alpha!}\,\,(\partial^{\alpha}\otimes\delta^{\alpha}).$

Similarly,

(2.10) $\displaystyle
e^{-t\sum_{i=1}^{n}\delta_{i}\otimes\partial_{i}}=\sum_{\beta\in{\mathbb{N}}^{n}}\frac{(-t)^{|\beta|}}{\beta!}\,\,(\delta^{\beta}\otimes\partial^{\beta}).$

Then it is easy to see that Eq. (2.7) follows directly from Eq. (1.1) and the
last three equations above. $\Box$

######  Proposition 2.4.

$(a)$ For any $\lambda_{i}(\xi)\in\mathbb{C}[\xi]$, $p_{i}(z)\in\mathbb{C}[z]$
$(i=1,2)$, we have

(2.11) $\displaystyle\lambda_{1}(\xi)\ast_{t}\lambda_{2}(\xi)$
$\displaystyle=\lambda_{1}(\xi)\lambda_{2}(\xi).$ (2.12) $\displaystyle
p_{1}(z)\ast_{t}p_{2}(z)$ $\displaystyle=p_{1}(z)p_{2}(z).$ (2.13)
$\displaystyle\lambda(\xi)\ast_{t}p(z)=\sum_{\alpha\in{\mathbb{N}}^{n}}$
$\displaystyle\frac{(-1)^{|\alpha|}t^{|\alpha|}}{\alpha!}(\delta^{\alpha}\lambda(\xi))(\partial^{\alpha}p(z)).$

$(b)$ For any $\lambda(\xi)\in\mathbb{C}[\xi]$, $p(z)\in\mathbb{C}[z]$ and
$g({\xi,z})\in\mathbb{C}[{\xi,z}]$, we have

(2.14) $\displaystyle\lambda(\xi)\ast_{t}g({\xi,z})$
$\displaystyle=\lambda(\xi-t\partial)g({\xi,z}).$ (2.15) $\displaystyle
p(z)\ast_{t}g({\xi,z})$ $\displaystyle=p(z-t\delta)g({\xi,z}).$

Note that the components $\xi_{i}-t\partial_{i}$ $(1\leq i\leq n)$ of the
$n$-tuple $\xi-t\partial$ in Eq. (2.14) commute with one another. So the
substitution $\lambda(\xi-t\partial)$ of $\xi-t\partial$ into the polynomial
$\lambda(\xi)$ is well-defined. Similarly, the substitution $p(z-t\delta)$ in
Eq. (2.15) is also well-defined.

Proof: Eqs. (2.11)–(2.13) follow directly from Eq. (2.7).

To show Eq. (2.14), first, by Eq. (2.7), we have

(2.16)
$\displaystyle\lambda(\xi)\ast_{t}g({\xi,z})=\sum_{\alpha\in{\mathbb{N}}^{n}}$
$\displaystyle\frac{(-1)^{|\alpha|}t^{|\alpha|}}{\alpha!}(\delta^{\alpha}\lambda(\xi))(\partial^{\alpha}g({\xi,z})).$

Second, note that the multiplication operators by $\xi_{i}$ $(1\leq i\leq n)$
and the derivations $\partial_{j}$ $(1\leq j\leq n)$ commute. By using the
Taylor series expansion of $\lambda(\xi-t\partial)$ at $\xi$, we have

(2.17) $\displaystyle\lambda(\xi-t\partial)g({\xi,z})$
$\displaystyle=\left(\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{(-1)^{|\alpha|}t^{|\alpha|}}{\alpha!}(\delta^{\alpha}\lambda)(\xi)\partial^{\alpha}\right)g({\xi,z})$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{(-1)^{|\alpha|}t^{|\alpha|}}{\alpha!}(\delta^{\alpha}\lambda(\xi))(\partial^{\alpha}g({\xi,z})).$

Hence, Eq. (2.14) follows from the last two equations. Eq. (2.15) can be
proved similarly. $\Box$

######  Lemma 2.5.

For any $t\in\mathbb{C}$, $\lambda(\xi)\in\mathbb{C}[\xi]$ and
$p(z)\in\mathbb{C}[z]$, we have

(2.18) $\displaystyle\Phi_{t}(\lambda(\xi))$ $\displaystyle=\lambda(\xi).$
(2.19) $\displaystyle\Phi_{t}(p(z))$ $\displaystyle=p(z).$ (2.20)
$\displaystyle\Phi_{t}(\lambda(\xi)p(z))$
$\displaystyle=\lambda(\xi)\ast_{-t}p(z).$

Proof: Since $\Lambda(\lambda(\xi))=\Lambda(p(z))=0$, $\Phi_{t}=e^{t\Lambda}$
fixes $\lambda(\xi)$ and $p(z)$. Hence we have Eqs. (2.18) and (2.19).

To show Eq. (2.20), by Eqs. (2.1), (2.18) and (2.19), we have

$\displaystyle\lambda(\xi)\ast_{t}p(z)$
$\displaystyle=\Phi_{-t}\left(\,\Phi_{t}(\lambda(\xi))\Phi_{t}(p(z))\,\right)$
$\displaystyle=\Phi_{-t}(\lambda(\xi)p(z)).$

Replacing $t$ be $-t$ in the equation above, we get Eq. (2.20). $\Box$

######  Proposition 2.6.

For any $t\in\mathbb{C}$, the following statements hold.

$(a)$ The subspaces $\mathbb{C}[\xi]$ and $\mathbb{C}[z]$ of
${\mathcal{B}}_{t}[{\xi,z}]$ are closed under the product $\ast_{t}$ and
hence, are actually subalgebras of ${\mathcal{B}}_{t}[{\xi,z}]$.

$(b)$ As associative algebras, $(\mathbb{C}[\xi],\ast_{t})$ and
$(\mathbb{C}[z],\ast_{t})$ are identical as the usual polynomial algebras
${\mathcal{A}}[\xi]$ and ${\mathcal{A}}[z]$ in $\xi$ and $z$, respectively.

$(c)$ ${\mathcal{B}}_{t}[{\xi,z}]$ is a commutative free algebra generated
freely by $\xi_{i}$ and $z_{i}$ $(1\leq i\leq n)$. The set of the monomials
generated by $\xi_{i}$ and $z_{i}$ $(1\leq i\leq n)$ in
${\mathcal{B}}_{t}[{\xi,z}]$ is given by
$\\{\xi^{\alpha}\ast_{t}z^{\beta}\,|\,\alpha,\beta\in{\mathbb{N}}^{n}\\}$.

Proof: Note that $(a)$ and $(b)$ follow immediately from Eqs. (2.11) and
(2.12).

To show $(c)$, first, by Eqs. (2.18) and (2.19), we know that the algebra
isomorphism
$\Phi_{-t}=\Phi_{t}^{-1}:{\mathcal{A}}[{\xi,z}]\to{\mathcal{B}}_{t}[{\xi,z}]$
as a linear map from $\mathbb{C}[\xi,z]$ to $\mathbb{C}[\xi,z]$ fixes
$\xi_{i}$ and $z_{i}$ $(1\leq i\leq n)$. Hence, ${\mathcal{B}}_{t}[{\xi,z}]$
is a commutative free algebra generated freely by $\xi_{i}$ and $z_{i}$
$(1\leq i\leq n)$.

The second part of $(c)$ follows from Eqs. (2.11), (2.12) and the fact that
the product $\ast_{t}$ is associative and commutative. $\Box$

The next two lemmas will be needed in Subsection 5.3.

######  Lemma 2.7.

For any $t\in\mathbb{C}$ and $\alpha,\beta\in{\mathbb{N}}$,

(2.21)
$\displaystyle(z\partial-\xi\delta)(\xi^{\alpha}\ast_{t}z^{\beta})=(|\beta|-|\alpha|)(\xi^{\alpha}\ast_{t}z^{\beta}),$

where
$z\partial-\xi\delta\\!:=\sum_{i=1}^{n}(z_{i}\partial_{i}-\xi_{i}\delta_{i})$.

Proof: First, by Euler’s lemma, we have

(2.22)
$\displaystyle(z\partial-\xi\delta)(\xi^{\alpha}z^{\beta})=(|\beta|-|\alpha|)(\xi^{\alpha}z^{\beta}).$

Second, note that $z\partial-\xi\delta$ commutes with $\Lambda$, hence also
with $\Phi_{t}$ for any $t\in\mathbb{C}$. Apply $\Phi_{-t}$ to Eq. (2.22), we
get

$\displaystyle(z\partial-\xi\delta)\Phi_{-t}(\xi^{\alpha}z^{\beta})=(|\beta|-|\alpha|)\Phi_{-t}(\xi^{\alpha}z^{\beta}).$

Then, by Eq. (2.20) with $t$ replaced by $-t$, Eq. (2.21) follows from the
equation above. $\Box$

######  Lemma 2.8.

For any $\lambda_{i}(\xi)\in\mathbb{C}[\xi]$ and $p_{i}(z)\in\mathbb{C}[z]$
$(i=1,2)$, we have

(2.23)
$\displaystyle(\lambda_{1}(\xi)p_{1}(z))\ast_{t}(\lambda_{2}(\xi)p_{2}(z))=(\lambda_{1}(\xi)\ast_{t}p_{2}(z))\,(\lambda_{2}(\xi)\ast_{t}p_{1}(z)).$

Proof: First, by Eq. (2.7), we have

$\displaystyle\quad(\lambda_{1}(\xi)p_{1}(z))\ast_{t}(\lambda_{2}(\xi)p_{2}(z))$
$\displaystyle=\sum_{\alpha,\beta\in{\mathbb{N}}^{n}}\frac{(-t)^{|\alpha|+|\beta|}}{\alpha!\beta!}\left((\delta^{\alpha}\lambda_{1}(\xi))(\partial^{\beta}p_{1}(z))\right)\left((\delta^{\beta}\lambda_{2}(\xi))(\partial^{\alpha}p_{2}(z))\right)$
Taking sum over $\alpha\in{\mathbb{N}}^{n}$ and applying Eq. (2.13):
$\displaystyle=(\lambda_{1}(\xi)\ast_{t}p_{2}(z))\sum_{\beta\in{\mathbb{N}}^{n}}\frac{(-t)^{|\beta|}}{\beta!}(\partial^{\beta}p_{1}(z))(\delta^{\beta}\lambda_{2}(\xi))$
Taking sum over $\beta\in{\mathbb{N}}^{n}$ and applying Eq. (2.13):
$\displaystyle=(\lambda_{1}(\xi)\ast_{t}p_{2}(z))(\lambda_{2}(\xi)\ast_{t}p_{1}(z)).$

Hence we get Eq. (2.23). $\Box$

###  2.2. The $\ell$-adic Topologies Induced by ${\mathcal{B}}_{t}[{\xi,z}]$
on $\mathbb{C}[\xi,z]$

We have seen that the algebras ${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$
share the same base vector space $\mathbb{C}[\xi,z]$ and, by Proposition 2.6,
$(c)$, they are all commutative free algebras generated freely by $\xi$ and
$z$. Therefore, we may talk about the $\ell$-adic topologies on
$\mathbb{C}[\xi,z]$ induced by the algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$, which are defined as follows.

For any $t\in\mathbb{C}[{\xi,z}]$ and $m\geq 0$, set $U_{t,m}$ to be the
subspace of $\mathbb{C}[\xi,z]$ spanned by the monomials
$\xi^{\alpha}\ast_{t}z^{\beta}$ of ${\mathcal{B}}_{t}[{\xi,z}]$ with
$\alpha,\beta\in{\mathbb{N}}^{n}$ and $|\alpha+\beta|\geq m$. The $\ell$-adic
topology induced from the algebra ${\mathcal{B}}_{t}[{\xi,z}]$ is the topology
whose open subsets are the subsets generated by $U_{t,m}$ $(m\in{\mathbb{N}})$
and their translations by elements of ${\mathcal{B}}_{t}[{\xi,z}]$. We denote
by ${\mathcal{T}}_{t}$ this topology on $\mathbb{C}[\xi,z]$.

The main result of this subsection is the following proposition.

######  Proposition 2.9.

$(a)$ For any $s\neq t\in\mathbb{C}$, we have
${\mathcal{T}}_{s}\neq{\mathcal{T}}_{t}$.

$(b)$ For any $t\in\mathbb{C}$, the algebra isomorphism
$\Phi_{t}:({\mathcal{B}}_{t}[{\xi,z}],{\mathcal{T}}_{t})\to({\mathcal{A}}[{\xi,z}],{\mathcal{T}}_{0})$
is also a homeomorphism of topological spaces. Consequently,
$({\mathcal{B}}_{t}[{\xi,z}],{\mathcal{T}}_{t})$ $(t\in\mathbb{C})$ as
topological spaces are all homeomorphic.

Proof: $(a)$ Let $\\{\alpha_{m}\in{\mathbb{N}}^{n}\,|\,m\geq 1\\}$ be any
sequence of elements of ${\mathbb{N}}^{n}$ such that $|\alpha_{m}|=m$ for any
$m\geq 1$.

Set $u_{m}\\!:=\xi^{\alpha_{m}}\ast_{t}z^{\alpha_{m}}$ for any $m\geq 1$.
Then, by the definition of ${\mathcal{T}}_{t}$, we see that the sequence
$\\{u_{m}\\}$ converges to $0\in\mathbb{C}[\xi,z]$ with respect to the
topology ${\mathcal{T}}_{t}$.

But, on the other hand, set $r:=s-t\neq 0$. Then, by Eq. (2.14), we have

$\displaystyle u_{m}$
$\displaystyle=\xi^{\alpha_{m}}\ast_{t}z^{\alpha_{m}}=(\xi-t\partial)^{\alpha_{m}}z^{\alpha_{m}}=((\xi-s\partial)+r\partial)^{\alpha_{m}}z^{\alpha_{m}}$
$\displaystyle=\sum_{\begin{subarray}{c}\beta,\gamma\in{\mathbb{N}}^{n}\\\
\beta+\gamma=\alpha_{m}\end{subarray}}\binom{\alpha_{m}}{\gamma}(\xi-s\partial)^{\gamma}(\partial^{\beta}z^{\alpha_{m}})$
$\displaystyle=\sum_{\begin{subarray}{c}\beta,\gamma\in{\mathbb{N}}^{n}\\\
\beta+\gamma=\alpha_{m}\end{subarray}}\binom{\alpha_{m}}{\gamma}\xi^{\gamma}\ast_{s}(\partial^{\beta}z^{\alpha_{m}})\equiv\alpha_{m}!\mod(U_{s,0}).$

From the equation above, we see that the sequence $\\{u_{m}\\}$ does not
converge to $0\in\mathbb{C}[\xi,z]$ with respect to the topology
${\mathcal{T}}_{s}$. Hence ${\mathcal{T}}_{s}\neq{\mathcal{T}}_{t}$.

$(b)$ Note that ${\mathcal{B}}_{t=0}[{\xi,z}]$ is the usual polynomial algebra
${\mathcal{A}}[{\xi,z}]$ and
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ is an algebra
isomorphism. Furthermore, from Eqs. (2.1), (2.18) and (2.19), we have

(2.24)
$\displaystyle\Phi_{t}(\xi^{\alpha}\ast_{t}z^{\beta})=\xi^{\alpha}z^{\beta}$

for any $\alpha,\beta\in{\mathbb{N}}^{n}$.

Therefore, for any $m\geq 0$, we have, $\Phi_{t}(U_{t,m})=U_{0,m}$ and
$\Phi_{t}^{-1}(U_{0,m})=U_{t,m}$. Hence, we have $(b)$. $\Box$

Actually, the proof above also shows that Proposition 2.9 also holds for the
following topologies on $\mathbb{C}[\xi,z]$ induced by the free algebras
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$.

For any $t\in\mathbb{C}[{\xi,z}]$ and $m\geq 0$, set

(2.25) $\displaystyle
U_{t,m}^{\prime}\\!:=\sum_{\begin{subarray}{c}\alpha\in{\mathbb{N}}^{n};\,\\\
|\alpha|\geq m\end{subarray}}\xi^{\alpha}\ast_{t}\mathbb{C}[z].$

Denote by ${\mathcal{T}}^{\prime}_{t}$ the topology on $\mathbb{C}[\xi,z]$
generated by $U_{t,m}^{\prime}$ and their translations (as open subsets).
Then, by a similar argument as in the proof of Proposition 2.9, it is easy to
see that the following corollary also holds.

######  Corollary 2.10.

$(a)$ For any $s\neq t\in\mathbb{C}$, we have
${\mathcal{T}}_{s}{}^{\prime}\neq{\mathcal{T}}_{t}{}^{\prime}$.

$(b)$ For any $t\in\mathbb{C}$, the algebra isomorphism
$\Phi_{t}:({\mathcal{B}}_{t}[{\xi,z}],{\mathcal{T}}_{t}{}^{\prime})\to({\mathcal{A}}[{\xi,z}],{\mathcal{T}}_{0}{}^{\prime})$
is also a homeomorphism of topological spaces.

Note that, due to the symmetric roles played by $\xi$ and $z$, the corollary
above also holds if $\xi$ in Eq. (2.25) is replaced by $z$.

###  2.3. The Induced Isomorphism $(\Phi_{t})_{\ast}$ on Differential Operator
Algebras

For any $t\in\mathbb{C}$, denote by ${\mathcal{D}}_{t}[{\xi,z}]$ the
differential operator algebra or the Weyl algebra of
${\mathcal{B}}_{t}[{\xi,z}]$, i.e. the associative algebra generated by the
$\mathbb{C}$-derivations and the multiplication operators of the algebra
${\mathcal{B}}_{t}[{\xi,z}]$. Since
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ is an algebra
isomorphism (See Corollary 2.2), it induces an algebra isomorphism, denoted by
$(\Phi_{t})_{*}:{\mathcal{D}}_{t}[{\xi,z}]\to{\mathcal{D}}[{\xi,z}]$, from the
Weyl algebra ${\mathcal{D}}_{t}[{\xi,z}]$ of ${\mathcal{B}}_{t}[{\xi,z}]$ to
the Weyl algebra ${\mathcal{D}}[{\xi,z}]$ of ${\mathcal{A}}[{\xi,z}]$.

Recall that the induced map $(\Phi_{t})_{*}$ is defined by setting

(2.26)
$\displaystyle(\Phi_{t})_{*}(\psi)=\Phi_{t}\circ\psi\circ\Phi_{t}^{-1}=\Phi_{t}\circ\psi\circ\Phi_{-t}$

for any $\psi\in{\mathcal{D}}_{t}[{\xi,z}]$.

The main result of this subsection are the following two propositions, even
though their proofs are very simple.

######  Proposition 2.11.

For any $t\in\mathbb{C}$, the following statements hold.

$(a)$ $\partial_{i}$ and $\delta_{i}$ $(1\leq i\leq n)$ are also derivations
of ${\mathcal{B}}_{t}[{\xi,z}]$.

$(b)$ For any $1\leq i\leq n$, we have

(2.27) $\displaystyle(\Phi_{t})_{*}(\partial_{i})$
$\displaystyle=\partial_{i},$ (2.28) $\displaystyle(\Phi_{t})_{*}(\delta_{i})$
$\displaystyle=\delta_{i},$

Proof: Note first that $\partial_{i}$ and $\delta_{i}$ $(1\leq i\leq n)$
commute with $\Lambda$, hence also with $\Phi_{t}$ for any $t\in\mathbb{C}$.
Then, Eqs. (2.27) and (2.28) follows immediately from this fact and the
definition of $(\Phi_{t})_{*}$ given in Eq. (2.26).

$(a)$ follows from the general fact that the induced map of any algebra
isomorphism maps derivations to derivations. It can also be checked directly
as follows.

For any $f,g\in{\mathcal{B}}_{t}[{\xi,z}]$, by Eq. (2.1) and the fact that
$\partial_{i}$ $(1\leq i\leq n)$ commute with $\Phi_{t}$ $(t\in\mathbb{C})$,
we have

$\displaystyle\partial_{i}(f\ast_{t}g)$
$\displaystyle=\partial_{i}\Big{(}\Phi_{-t}\big{(}\Phi_{t}(f)\Phi_{t}(g)\big{)}\Big{)}=\Phi_{-t}\Big{(}\partial_{i}\big{(}\Phi_{t}(f)\Phi_{t}(g)\big{)}\Big{)}$
$\displaystyle=\Phi_{-t}\big{(}(\partial_{i}\Phi_{t}(f))\Phi_{t}(g)\big{)}+\Phi_{-t}\big{(}\Phi_{t}(f)(\partial_{i}\Phi_{t}(g))\big{)}$
$\displaystyle=\Phi_{-t}\big{(}\Phi_{t}(\partial_{i}f)\Phi_{t}(g)\big{)}+\Phi_{-t}\big{(}\Phi_{t}(f)\Phi_{t}(\partial_{i}g)\big{)}$
$\displaystyle=(\partial_{i}f)\ast_{t}g+f\ast_{t}(\partial_{i}g).$

Similarly, we can show that $\delta_{i}$ $(1\leq i\leq n)$ are also
derivations of ${\mathcal{B}}_{t}[{\xi,z}]$. $\Box$

######  Corollary 2.12.

For any $\alpha,\beta,\gamma\in{\mathbb{N}}^{n}$, we have

(2.29) $\displaystyle\partial^{\gamma}(\xi^{\alpha}\ast_{t}z^{\beta})$
$\displaystyle=\gamma!\,\binom{\beta}{\gamma}\,(\xi^{\alpha}\ast_{t}z^{\beta-\gamma}),$
(2.30) $\displaystyle\delta^{\gamma}(\xi^{\alpha}\ast_{t}z^{\beta})$
$\displaystyle=\gamma!\,\binom{\alpha}{\gamma}\,(\xi^{\alpha-\gamma}\ast_{t}z^{\beta}).$

Proof: Note that, by Eqs. (2.11) and (2.12), we know that, for any
$\alpha,\beta\in{\mathbb{N}}^{n}$, $\xi^{\alpha}\ast_{t}z^{\beta}$ will remain
the same if we replace the (usual) product of ${\mathcal{A}}[{\xi,z}]$ in the
factors $\xi^{\alpha}$ and $z^{\beta}$ by the product $\ast_{t}$ of
${\mathcal{B}}_{t}[{\xi,z}]$. By Proposition 2.11, $(a)$, we know that
$\partial_{i}$ and $\delta_{i}$ $(1\leq i\leq n)$ are also the derivations of
${\mathcal{B}}_{t}[{\xi,z}]$. From these two facts, it is easy to see that
both equations in the corollary hold. $\Box$

######  Proposition 2.13.

For any $t\in\mathbb{C}$ and $f(\xi,z)\in\mathbb{C}[{\xi,z}]$,
$(\Phi_{t})_{*}$ maps the multiplication operator of
${\mathcal{B}}_{t}[{\xi,z}]$ by $f(\xi,z)$ $($with respect to the product
$\ast_{t}$$)$ to the multiplication operator of ${\mathcal{A}}[{\xi,z}]$ by
$\Phi_{t}(f(\xi,z))$ $($with respect to the product of
${\mathcal{A}}[{\xi,z}]$$)$.

Proof: We denote by $\psi_{f}$ the multiplication operator of
${\mathcal{B}}_{t}[{\xi,z}]$ by $f(\xi,z)$ $($with respect to the product
$\ast_{t}$$)$. Then for any $u({\xi,z})\in\mathbb{C}[{\xi,z}]$, by Eqs. (2.26)
and (2.1) we have

$\displaystyle(\Phi_{t})_{*}(\psi_{f})u({\xi,z})$
$\displaystyle=(\Phi_{t}\circ\psi_{f}\circ\Phi_{t}^{-1})u({\xi,z})=\Phi_{t}\big{(}f(\xi,z)\ast_{t}\Phi_{t}^{-1}(u({\xi,z}))\big{)}$
$\displaystyle=\Phi_{t}(f(\xi,z))\,\Phi_{t}\big{(}\Phi_{t}^{-1}(u(\xi,z))\big{)}=\Phi_{t}(f(\xi,z))\,u({\xi,z}).$

Hence, the proposition follows. $\Box$

By the proposition above and Eqs. (2.18) and (2.19), we also have the
following corollary.

######  Corollary 2.14.

For any $t\in\mathbb{C}$, $\lambda(\xi)\in\mathbb{C}[\xi]$ and
$p(z)\in\mathbb{C}[z]$, $(\Phi_{t})_{*}$ maps the multiplication operators of
${\mathcal{B}}_{t}[{\xi,z}]$ by $\lambda(\xi)$ and $p(z)$ $($with respect to
the product $\ast_{t}$$)$ to the multiplication operators of
${\mathcal{A}}[{\xi,z}]$ by $\lambda(\xi)$ and $p(z)$ $($with respect to the
product of ${\mathcal{A}}[{\xi,z}]$$)$, respectively.

Note that, as pointed out before, the algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ share the same base vectors space $\mathbb{C}[\xi,z]$.
Therefore, their Weyl algebras ${\mathcal{D}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$
are all subalgebras of the algebra of linear endomorphisms of
$\mathbb{C}[\xi,z]$. The following corollary says that all these subalgebras
turn out to be same, i.e. they do not depend on the parameter
$t\in\mathbb{C}$.

######  Corollary 2.15.

For any $t\in\mathbb{C}$, as subalgebras of the algebra of linear
endomorphisms of $\mathbb{C}[\xi,z]$,
${\mathcal{D}}_{t}[{\xi,z}]={\mathcal{D}}[{\xi,z}]$.

Proof: By Proposition 2.6, $(c)$, we know that ${\mathcal{B}}_{t}[{\xi,z}]$ is
a commuative free algebra generated freely by $\xi$ and $z$. By Proposition
2.11, $(a)$, we know that $\partial_{i}$ and $\delta_{i}$ $(1\leq i\leq n)$
are also derivations of ${\mathcal{B}}_{t}[{\xi,z}]$. Therefore, the Weyl
algebra ${\mathcal{D}}_{t}[{\xi,z}]$ as an associative algebra over
$\mathbb{C}$ is generated by the derivations $\partial_{i}$, $\delta_{i}$
$(1\leq i\leq n)$ and the multiplication operators (with respect to the
product $\ast_{t}$) by $\xi_{i},z_{i}\in{\mathcal{B}}_{t}[{\xi,z}]$ $(1\leq
i\leq n)$.

By Eqs. (2.14) and (2.15), we see that the multiplication operators by
$\xi_{i},z_{i}\in{\mathcal{B}}_{t}[{\xi,z}]$ $(1\leq i\leq n)$ are same as the
operators $\xi_{i}-t\partial_{i}$ and $z_{i}-t\delta_{i}$ which lie in
${\mathcal{D}}[{\xi,z}]$. Hence we have
${\mathcal{D}}_{t}[{\xi,z}]\subseteq{\mathcal{D}}[{\xi,z}]$.

To show ${\mathcal{D}}[{\xi,z}]\subseteq{\mathcal{D}}_{t}[{\xi,z}]$, by
Proposition 2.11, $(a)$, it will be enough to show that the multiplication
operators (with respect to the product of ${\mathcal{A}}[{\xi,z}]$) by
$\xi_{i},z_{i}\in{\mathcal{A}}[{\xi,z}]$ $(1\leq i\leq n)$ also belong to
${\mathcal{D}}_{t}[{\xi,z}]$.

But, for any $f({\xi,z})\in\mathbb{C}[{\xi,z}]$, by Eqs. (2.14) and (2.15), we
have

$\displaystyle\xi_{i}f({\xi,z})$
$\displaystyle=(\xi_{i}-t\partial_{i})f({\xi,z})+t\partial_{i}f({\xi,z})=\xi_{i}\ast_{t}f({\xi,z})+t\partial_{i}f({\xi,z}),$
$\displaystyle z_{i}f({\xi,z})$
$\displaystyle=(z_{i}-t\delta_{i})f({\xi,z})+t\delta_{i}f({\xi,z})=z_{i}\ast_{t}f({\xi,z})+t\delta_{i}f({\xi,z}).$

From the equations above, we see that the multiplication operators (with
respect to the product of ${\mathcal{A}}[{\xi,z}]$) by
$\xi_{i},z_{i}\in{\mathcal{A}}[{\xi,z}]$ $(1\leq i\leq n)$ do belong to
${\mathcal{D}}_{t}[{\xi,z}]$. $\Box$

##  3\. Connections with Interchanges of Right and Left Total Symbols of
Differential Operators

In this section, we show in Theorem 3.1 that the isomorphisms $\Phi_{t}$ with
$t=\pm 1$ coincide with the interchanges between total left and right symbols
of differential operators of the polynomial algebra ${\mathcal{A}}[z]$.

First, let us fix the following notation and convention for the differential
operators of ${\mathcal{A}}[z]$.

We denote by ${\mathcal{D}}[z]$ the differential operator algebra or the Weyl
algebra of ${\mathcal{A}}[z]$. For any differential operator
$\phi\in{\mathcal{D}}[z]$ and polynomial $u(z)\in{\mathcal{A}}[z]$, the
notation $\phi\,u(z)$ usually denotes the composition of $\phi$ and the
multiplication operator by $u(z)$. So $\phi\,u(z)$ is still a differential
operator of ${\mathcal{A}}[z]$. The polynomial obtained by applying $\phi$ to
$u(z)$ will be denoted by $\phi(u(z))$.

Next, let us recall the right and left total symbols of differential operators
of the polynomial algebra ${\mathcal{A}}[z]$.

For any $\phi\in{\mathcal{D}}[z]$, it is well-known (e.g. see Proposition 2.2
(pp. 4) in [B] or Theorem 3.1 (pp. 58) in [C]) that $\phi$ can be written
uniquely as the following two finite sums:

(3.1)
$\displaystyle\phi=\sum_{\alpha\in{\mathbb{N}}^{n}}a_{\alpha}(z)\partial^{\alpha}=\sum_{\beta\in{\mathbb{N}}^{n}}\partial^{\beta}b_{\beta}(z)$

where $a_{\alpha}(z),b_{\beta}(z)\in\mathbb{C}[z]$ but denote the
multiplication operators by $a_{\alpha}(z)$ and $b_{\beta}(z)$, respectively.

For the differential operator $\phi\in{\mathcal{D}}[z]$ in Eq. (3.1), the
right and left total symbols are defined to be the polynomials
$\sum_{\alpha\in{\mathbb{N}}^{n}}a_{\alpha}(z)\xi^{\alpha}\in\mathbb{C}[{\xi,z}]$
and
$\sum_{\beta\in{\mathbb{N}}^{n}}b_{\beta}(z)\xi^{\beta}\in\mathbb{C}[{\xi,z}]$,
respectively. We denote by
${\mathcal{R}}:{\mathcal{D}}[z]\to\mathbb{C}[\xi,z]$ (resp.
${\mathcal{L}}:{\mathcal{D}}[z]\to\mathbb{C}[\xi,z]$) the linear map which
maps any $\phi\in{\mathcal{D}}[z]$ to its right total symbol (resp. left total
symbol).

Note that, by the uniqueness of the expressions in Eq. (3.1), both
${\mathcal{R}}$ and ${\mathcal{L}}$ are isomorphisms of vector spaces over
$\mathbb{C}$. The interchange of the left (resp. right) total symbol of
differential operators to their right (resp. left) total symbols is given by
the isomorphism ${\mathcal{R}}\circ{\mathcal{L}}^{-1}$ (resp.
${\mathcal{L}}\circ{\mathcal{R}}^{-1}$) from $\mathbb{C}[\xi,z]$ to
$\mathbb{C}[\xi,z]$.

The main result of this section is the following theorem.

######  Theorem 3.1.

As linear maps from $\mathbb{C}[\xi,z]$ to $\mathbb{C}[\xi,z]$, we have

(3.2) $\displaystyle\Phi$
$\displaystyle={\mathcal{R}}\circ{\mathcal{L}}^{-1}.$ (3.3)
$\displaystyle\Phi_{t=-1}$
$\displaystyle={\mathcal{L}}\circ{\mathcal{R}}^{-1}.$

Proof: Note first that, Eq. (3.3) follows from Eq. (3.2) and the fact that
$\Phi_{t=-1}=\Phi_{t=1}^{-1}=\Phi^{-1}$.

To show Eq. (3.2), since both $\Phi$ and
${\mathcal{R}}\circ{\mathcal{L}}^{-1}$ are linear maps, it is enough to show
that, for any $\alpha,\beta\in{\mathbb{N}}^{n}$, we have

(3.4)
$\displaystyle\Phi(\xi^{\alpha}z^{\beta})=({\mathcal{R}}\circ{\mathcal{L}}^{-1})(\xi^{\alpha}z^{\beta}).$

Since

(3.5)
$\displaystyle({\mathcal{R}}\circ{\mathcal{L}}^{-1})(\xi^{\alpha}z^{\beta})={\mathcal{R}}(\partial^{\alpha}z^{\beta}),$

so we have to find the right total symbol of the differential operator
$\partial^{\alpha}z^{\beta}\in{\mathcal{D}}[z]$.

Note that, for any dummy $u(z)\in\mathbb{C}[z]$, by the Leibniz rule, we have

(3.6) $\displaystyle\partial^{\alpha}(z^{\beta}u(z))$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\alpha}{\gamma}(\partial^{\gamma}z^{\beta})(\partial^{\alpha-\gamma}u(z))$
$\displaystyle=\left(\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\alpha}{\gamma}(\partial^{\gamma}z^{\beta})\partial^{\alpha-\gamma}\right)u(z).$

Therefore, the right total symbol of the differential operator
$\partial^{\alpha}z^{\beta}\in{\mathcal{D}}[z]$ is given by

$\displaystyle{\mathcal{R}}(\partial^{\alpha}z^{\beta})$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\alpha}{\gamma}(\partial^{\gamma}z^{\beta})\xi^{\alpha-\gamma}=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\alpha}{\gamma}\xi^{\alpha-\gamma}(\partial^{\gamma}z^{\beta})$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\frac{1}{\gamma!}(\delta^{\gamma}\xi^{\alpha})(\partial^{\gamma}z^{\beta})$

Combining the equation above with Eqs. (2.13) and (2.20) with $t=-1$, we have

(3.7)
$\displaystyle{\mathcal{R}}(\partial^{\alpha}z^{\beta})=\xi^{\alpha}\ast_{t=-1}z^{\beta}=\Phi_{t=1}(\xi^{\alpha}z^{\beta})=\Phi(\xi^{\alpha}z^{\beta}).$

Hence, we have proved Eq. (3.4) and also the theorem. $\Box$

######  Corollary 3.2.

For any $\lambda(\xi)\in\mathbb{C}[\xi]$ and $p(z)\in\mathbb{C}[z]$, we have

(3.8) $\displaystyle{\mathcal{R}}(\lambda(\partial)p(z))$
$\displaystyle=\lambda(\xi)\ast_{t=-1}p(z).$ (3.9)
$\displaystyle{\mathcal{L}}(p(z)\lambda(\partial))$
$\displaystyle=\lambda(\xi)\ast p(z).$

Proof: By Eqs. (3.2) and (2.20) with $t=1$, we have

$\displaystyle{\mathcal{R}}(\lambda(\partial)p(z))$
$\displaystyle={\mathcal{R}}({\mathcal{L}}^{-1}(\lambda(\xi)p(z)))=({\mathcal{R}}\circ{\mathcal{L}}^{-1})(\lambda(\xi)p(z))$
$\displaystyle=\Phi_{t=1}(\lambda(\xi)p(z))=\lambda(\xi)\ast_{t=-1}p(z).$

So we have Eq. (3.8). Eq. (3.9) can be proved similarly by using Eqs. (3.3)
and (2.20) with $t=-1$. $\Box$

Finally, we end this section with the following one-variable example.

######  Example 3.3.

Let $n=1$ and $\phi=z^{2}\partial^{3}$. Then,

$\displaystyle{\mathcal{R}}(\phi)$
$\displaystyle={\mathcal{R}}(z^{2}\partial^{3})=\xi^{3}z^{2}.$
$\displaystyle{\mathcal{L}}(\phi)$
$\displaystyle={\mathcal{L}}(z^{2}\partial^{3})=\xi^{3}\ast
z^{2}=(z-\delta)^{2}\xi^{3}$
$\displaystyle=(z^{2}-2z\delta+\delta^{2})\xi^{3}=\xi^{3}z^{2}-6\xi^{2}z+6\xi.$

Therefore, we have

$\displaystyle\phi=z^{2}\partial^{3}=\partial^{3}z^{2}-6\partial^{2}z+6\partial.$

##  4\. A Re-formulation of the Image Conjecture on Commuting Differential
Operators of Order One with Constant Leading Coefficients

In this section, we show that the algebra ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ is closely related with a theorem (See Theorem 4.1) first
proved in [Z3] and also with the so-called image conjecture (See Conjecture
4.3) proposed in [Z3] on the differential operators $\xi-t\partial$
$(t\in\mathbb{C})$.

In Subsection 4.1, we use certain Taylor series expansion of elements of
${\mathcal{B}}_{t}[{\xi,z}]$ to give a new and more conceptual proof for
Theorem 4.1. In Subsection 4.2, we first give a new formulation (See
Conjecture 4.5) for Conjecture 4.3 in terms of the algebra
${\mathcal{B}}_{t}[{\xi,z}]$ and the notion of Mathieu subspaces (see
Definition 4.2) introduced in [Z4], and then show in Theorem 4.6 that the new
formulation is indeed equivalent to Conjecture 4.3.

###  4.1. The Taylor Series with Respect to the Product $\ast_{t}$

First, let us recall the following elementary fact on polynomials in $\xi$ and
$z$.

For any $f({\xi,z})\in{\mathcal{A}}[{\xi,z}]$, we may view $f({\xi,z})$ as a
polynomial in $\xi$ with coefficients in ${\mathcal{A}}[z]$. Then it has the
following Taylor series expansion

(4.1) $\displaystyle
f({\xi,z})=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{1}{\alpha!}\,\,\xi^{\alpha}c_{\alpha}(z)$

for some $c_{\alpha}(z)\in{\mathcal{A}}[z]$.

Let $ev_{{}_{0}}:{\mathcal{A}}[{\xi,z}]\to{\mathcal{A}}[z]$ be the evaluation
map of ${\mathcal{A}}[{\xi,z}]$ at $\xi=0$, i.e. for any
$u({\xi,z})\in{\mathcal{A}}[{\xi,z}]$, $ev_{{}_{0}}(u)\\!:=u(0,z)$. Then, the
$c_{\alpha}(z)$ $(\alpha\in{\mathbb{N}}^{n})$ in Eq. (4.1) are given by

(4.2) $\displaystyle c_{\alpha}(z)=ev_{{}_{0}}(\delta^{\alpha}f).$

Note that another characterization of the evaluation map $ev_{{}_{0}}$ is that
$ev_{{}_{0}}$ is the (unique) algebra homomorphism from
${\mathcal{A}}[{\xi,z}]$ to ${\mathcal{A}}[z]$ with $ev_{{}_{0}}(\xi_{i})=0$
and $ev_{{}_{0}}(z_{i})=z_{i}$ for any $1\leq i\leq n$.

Now, come back to our algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$. By Proposition 2.6, $(c)$, we know that
${\mathcal{B}}_{t}[{\xi,z}]$ is also a commutative free algebra generated
freely by $\xi$ and $z$ with the same base vector space $\mathbb{C}[\xi,z]$.
Hence, we should expect similar expansions as in Eq. (4.1) for polynomials
$f({\xi,z})\in\mathbb{C}[\xi,z]$ with respect to the product $\ast_{t}$.

But, in order to formulate the expected expansions precisely, we need first to
introduce the analogue of the evaluation map $ev_{{}_{0}}$ for the algebra
${\mathcal{B}}_{t}[{\xi,z}]$.

Note that, by Proposition 2.6, $(b)$, the subalgebra of
${\mathcal{B}}_{t}[{\xi,z}]$ generated by $z$ is also
${\mathcal{A}}[z]\subset\mathbb{C}[{\xi,z}]$. Parallel to the second
characterization of the evaluation map $ev_{{}_{0}}$ mentioned above, we let
${\mathcal{E}}_{t}$ be the unique algebra homomorphism from
${\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[z]$ such that
${\mathcal{E}}_{t}(\xi_{i})=0$ and ${\mathcal{E}}_{t}(z_{i})=z_{i}$ for any
$1\leq i\leq n$.

Note also that, by Eqs. (2.18) and (2.19), the algebra isomorphism
$\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$ maps $\xi_{i}$
(resp. $z_{i}$) to $\xi_{i}$ (resp. $z_{i}$) for any $1\leq i\leq n$. Hence
the composition
$ev_{{}_{0}}\circ\Phi_{t}:{\mathcal{B}}_{t}[{\xi,z}]\to{\mathcal{A}}[z]$ has
the same characterizing property of ${\mathcal{E}}_{t}$. Therefore, we have

(4.3) $\displaystyle{\mathcal{E}}_{t}=ev_{{}_{0}}\circ\Phi_{t}.$

Furthermore, we can also derive a more explicit formula for
${\mathcal{E}}_{t}$ as follows.

For any $\alpha\in{\mathbb{N}}^{n}$ and $p(z)\in\mathbb{C}[z]$, consider

(4.4) $\displaystyle{\mathcal{E}}_{t}(\xi^{\alpha}p(z))$
$\displaystyle=ev_{{}_{0}}(\Phi_{t}(\xi^{\alpha}p(z)))$ Applying Eq. (2.20)
and then Eq. (2.14) with $t$ replaced by $-t$:
$\displaystyle=ev_{{}_{0}}(\xi^{\alpha}\ast_{-t}p(z)))$
$\displaystyle=ev_{{}_{0}}((\xi+t\partial)^{\alpha}(p(z)))=t^{|\alpha|}\partial^{\alpha}(p(z)).$

From the formula above, we see that, for any $g(z,\xi)\in\mathbb{C}[z,\xi]$,
${\mathcal{E}}_{t}(g(z,\xi))\in\mathbb{C}[z]$ can be obtained by, first,
writing each monomial of $g(z,\xi)$ as $\xi^{\beta}z^{\gamma}$
$(\beta,\gamma\in{\mathbb{N}}^{n})$, i.e. putting the free variables
$\xi_{i}$’s to the most left in each monomial of $g(z,\xi)$, and then
replacing the part $\xi^{\beta}$ by the differential operator
$t^{|\beta|}\partial^{\beta}$ and applying it to the other part $z^{\gamma}$
of the monomial. For examples, we have

$\displaystyle{\mathcal{E}}_{t}(1)$ $\displaystyle=1;$
$\displaystyle{\mathcal{E}}_{t}(z^{\alpha})$
$\displaystyle=(t\partial)^{0}(z^{\alpha})=z^{\alpha}\qquad\quad\,\,\qquad\qquad\mbox{for
any }\alpha\in{\mathbb{N}}^{n};$
$\displaystyle{\mathcal{E}}_{t}(\xi^{\alpha})$
$\displaystyle=t^{|\alpha|}\partial^{\alpha}(1)=0\qquad\qquad\qquad\quad\quad\mbox{for
any }0\neq\alpha\in{\mathbb{N}}^{n};$
$\displaystyle{\mathcal{E}}_{t}(z_{1}^{m}\xi_{1}^{2})$
$\displaystyle=t^{2}\partial_{1}^{2}(z_{1}^{m})=m(m-1)t^{2}z_{1}^{m-2}\qquad\,\mbox{for
any }m\geq 2.$

Now we are ready to formulate and prove the expected expansion of polynomials
with respect to the new product $\ast_{t}$, which is parallel to the Taylor
expansion in Eq. (4.1).

######  Theorem 4.1.

For any $t\in\mathbb{C}$ and $f(\xi,z)\in\mathbb{C}[\xi,z]$, we have

(4.5) $\displaystyle f(\xi,z)$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{1}{\alpha!}\,\,\xi^{\alpha}\ast_{t}a_{\alpha}(z),$
(4.6) $\displaystyle f(\xi,z)$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{1}{\alpha!}\,(\xi-t\partial_{z})^{\alpha}a_{\alpha}(z),$

where, for any $\alpha\in{\mathbb{N}}^{n}$,

(4.7) $\displaystyle a_{\alpha}(z)={\mathcal{E}}_{t}(\delta^{\alpha}f).$

Furthermore, the expansions of the forms in Eqs. $(\ref{D_t-Taylor-e1})$ and
$(\ref{D_t-Taylor-e2})$ for $f({\xi,z})$ are unique.

Proof: Note first that, by Eq. (2.14) in Proposition 2.4, Eq. (4.5) and Eq.
(4.6) are actually equivalent. So we will focus only on Eq. (4.5).

The uniqueness of the expansion in Eq. (4.5) follows directly from Proposition
2.6, $(a)$-$(c)$.

To show that Eq. (4.5) with $a_{\alpha}(z)$ $(\alpha\in{\mathbb{N}}^{n})$
given in Eq. (4.7) does hold, we first write the Taylor series expansion of
$\Phi_{t}(f({\xi,z}))$ as in Eq. (4.1):

(4.8)
$\displaystyle\Phi_{t}(f({\xi,z}))=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{1}{\alpha!}\,\,\xi^{\alpha}a_{\alpha}(z)$

where $a_{\alpha}(z)\in\mathbb{C}[z]$ $(\alpha\in{\mathbb{N}}^{n})$ are given
by

(4.9) $\displaystyle a_{\alpha}(z)=ev_{{}_{0}}(\delta^{\alpha}\Phi_{t}(f)).$

Applying $\Phi_{-t}$ to Eq.(4.8) and, by Eq. (2.20) with $t$ replaced by $-t$,
we get Eq. (4.5).

Next, note that $\delta^{\alpha}$ $(\alpha\in{\mathbb{N}}^{n})$ commute with
$\Lambda$, hence they also commute with $\Phi_{t}=e^{t\Lambda}$. Then, by Eqs.
(4.9) and (4.3), we have

$\displaystyle
a_{\alpha}(z)=ev_{{}_{0}}(\Phi_{t}(\delta^{\alpha}f))=(ev_{{}_{0}}\circ\Phi_{t})(\delta^{\alpha}f)={\mathcal{E}}_{t}(\delta^{\alpha}f).$

Therefore, Eq. (4.7) also holds. $\Box$

Several remarks on Theorem 4.1 and the proof above are as follows.

First, Theorem 4.1 with $t=1$ was first proved in [Z3]. The proof in [Z3] is
more straightforward. It does not use the algebra ${\mathcal{B}}_{t}[{\xi,z}]$
and the product $\ast_{t}$. But the proof given here is more conceptual. For
example, the expansion in Eq. $(\ref{D_t-Taylor-e2})$ becomes much more
natural after we show here that it is just the usual Taylor series expansion
of polynomials as in Eq. $(\ref{Taylor-e1})$ but in the new context of the
algebra ${\mathcal{B}}_{t}[{\xi,z}]$.

Second, Eq. (4.7) can also be derived directly from Eq. (4.6) as in [Z3].
Namely, apply $\delta^{\alpha}$ to Eq. (4.6) and then replace $\xi$ by
$t\partial$ in both sides of the resulting equation.

Third, not all formal power series $f({\xi,z})\in{\mathcal{A}}[[{\xi,z}]]$ can
be expanded in the form of Eq. $(\ref{D_t-Taylor-e1})$ or $(\ref{D_t-
Taylor-e2})$. For example, let $n=1$ and $f({\xi,z})=e^{\xi z}$ and assume
that $(\ref{D_t-Taylor-e2})$ holds for $f({\xi,z})$. Then, by the argument in
the previous paragraph, we see that $a_{m}(z)$ $(m\geq 0)$ must be given by
Eq. (4.7). But, for the series $\delta^{m}f({\xi,z})=z^{m}\sum_{k\geq
0}\frac{(\xi z)^{k}}{k!}$, ${\mathcal{E}}_{t}$ is not well-defined, which is a
contradiction.

Another way to look at the fact above is as follows. Even though
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\neq 0)$ and ${\mathcal{A}}[{\xi,z}]$ share
the same base vector space $\mathbb{C}[\xi,z]$, by Proposition 2.9, we know
that they induce different $\ell$-adic topologies on $\mathbb{C}[\xi,z]$.
Therefore, their completions with respect to the different $\ell$-adic
topologies will be different. In other words, the formal power series algebras
with respect to the product $\ast_{t}$ $(t\neq 0)$ and the usual formal power
series algebra ${\mathcal{A}}[[{\xi,z}]]$ do not share the same base vector
space anymore.

For the example $f({\xi,z})=e^{\xi z}$ above, we have
$f({\xi,z})\in{\mathcal{A}}[[{\xi,z}]]$. But, by the argument in the proof of
Proposition 2.9 with $\alpha_{m}$ $(m\geq 1)$ replaced by $m$, it is easy to
see that, for any $t\neq 0$, $f({\xi,z})=e^{\xi z}$ does not lie in the
completion of ${\mathcal{B}}_{t}[{\xi,z}]$ with respect to the $\ell$-adic
topology on $\mathbb{C}[{\xi,z}]$ induced by ${\mathcal{B}}_{t}[{\xi,z}]$.
Therefore, $f({\xi,z})=e^{\xi z}$ can not be written as a formal power series
with respect to the product $\ast_{t}$ as in Eq. (4.5).

###  4.2. Re-Formulation of the Image Conjecture in Terms of the Algebra
${\mathcal{B}}_{t}[{\xi,z}]$

First let us recall the following notion introduced recently in [Z4].

######  Definition 4.2.

Let $R$ be any commutative ring and ${\mathcal{A}}$ a commutative $R$-algebra.
We say that an $R$-subspace ${\mathcal{M}}$ of ${\mathcal{A}}$ is a Mathieu
subspace of ${\mathcal{A}}$ if the following property holds: if
$a\in{\mathcal{A}}$ satisfies $a^{m}\in{\mathcal{M}}$ for all $m\geq 1$, then,
for any $b\in{\mathcal{A}}$, we have $ba^{m}\in{\mathcal{M}}$ for all $m\gg
0$, i.e. there exists $N\geq 1$ $($depending on $a$ and $b$$)$ such that
$ba^{m}\in{\mathcal{M}}$ for all $m\geq N$.

From the definition above, it is easy to see that any ideal of ${\mathcal{A}}$
is automatically a Mathieu subspace of ${\mathcal{A}}$, but not conversely
(See [Z4] for some examples of Mathieu subspaces which are not ideals).
Therefore, the notion of Mathieu subspaces can be viewed as a generalization
of the notion of ideals.

Next, for any $t\in\mathbb{C}$, set

(4.10)
$\displaystyle\rm{Im\,}(\xi-t\partial)\\!:=\sum_{i=1}^{n}(\xi_{i}-t\partial_{i})\mathbb{C}[{\xi,z}].$

We call $\rm{Im\,}(\xi-t\partial)$ the image of the commuting differential
operators $(\xi_{i}-t\partial_{i})$ $(1\leq i\leq n)$.

With the notion and notation fixed above, the image conjecture proposed in
[Z4] for the commuting differential operators $(\xi-t\partial)$ can be re-
stated as follows.

######  Conjecture 4.3.

For any $t\in\mathbb{C}$, $\rm{Im\,}(\xi-t\partial)$ is a Mathieu subspace of
the polynomial algebra ${\mathcal{A}}[{\xi,z}]$.

One of the motivations of the conjecture above is the following theorem proved
in [Z3].

######  Theorem 4.4.

Conjecture 4.3 implies the Jacobian conjecture.

Actually, it has been shown in [Z3] that the Jacobian conjecture is equivalent
to some very special cases of Conjecture 4.3. For more detail, see [Z3].

The main result of this subsection is to show that the conjecture above can
actually be re-formulated as follows.

######  Conjecture 4.5.

Set $\xi\mathbb{C}[{\xi,z}]\\!:=\sum_{i=1}^{m}\xi_{i}\mathbb{C}[\xi,z]$. Then,
for any $t\in\mathbb{C}$, $\xi\mathbb{C}[{\xi,z}]$ as a subspace of
${\mathcal{B}}_{t}[{\xi,z}]$ is a Mathieu subspace of
${\mathcal{B}}_{t}[{\xi,z}]$.

######  Theorem 4.6.

Conjecture 4.5 is equivalent to Conjecture 4.3.

Proof: First, denote by $\xi\ast_{t}\mathbb{C}[{\xi,z}]$ the ideal of
${\mathcal{B}}_{t}[{\xi,z}]$ generated by $\xi_{i}$ $(1\leq i\leq n)$. View
$\xi\ast_{t}\mathbb{C}[{\xi,z}]$ as a subspace of ${\mathcal{A}}[{\xi,z}]$ and
apply Eqs. (4.10) and (2.14), we have

(4.11)
$\displaystyle\rm{Im\,}(\xi-t\partial)=\sum_{i=1}^{n}\xi_{i}\ast_{t}\mathbb{C}[{\xi,z}]=\xi\ast_{t}\mathbb{C}[{\xi,z}].$

Second, by Eqs. (2.1) and (2.18), we have

$\displaystyle\Phi_{t}(\xi\ast_{t}\mathbb{C}[{\xi,z}])=\Phi_{t}(\xi)\Phi_{t}(\mathbb{C}[{\xi,z}])=\xi\mathbb{C}[{\xi,z}]$

Hence, we also have

(4.12)
$\displaystyle\xi\ast_{t}\mathbb{C}[{\xi,z}]=\Phi_{t}^{-1}(\xi\mathbb{C}[{\xi,z}])=\Phi_{-t}(\xi\mathbb{C}[{\xi,z}]).$

Combine Eqs. (4.11) and (4.12), we get

(4.13)
$\displaystyle\Phi_{-t}(\xi\mathbb{C}[{\xi,z}])=\rm{Im\,}(\xi-t\partial).$

Third, by Proposition $4.9$ in [Z4], we know that pre-images of Mathieu
subspaces under algebra homomorphisms are still Mathieu subspaces, from which
it is easy to check that Mathieu subspaces are preserved by algebra
isomorphisms. By using this fact (on the algebra isomorphism
$\Phi_{-t}:{\mathcal{B}}_{-t}[{\xi,z}]\to{\mathcal{A}}[{\xi,z}]$) and also Eq.
(4.13), we see that, $\xi\mathbb{C}[{\xi,z}]$ is a Mathieu subspace of
${\mathcal{B}}_{-t}[{\xi,z}]$ iff $\rm{Im\,}(\xi-t\partial)$ is a Mathieu
subspace of ${\mathcal{A}}[{\xi,z}]$.

Replacing $t$ by $-t$ in the equivalence above, we have that,
$\xi\mathbb{C}[{\xi,z}]$ is a Mathieu subspace of ${\mathcal{B}}_{t}[{\xi,z}]$
for any $t\in\mathbb{C}$ iff $\rm{Im\,}(\xi+t\partial)$ is a Mathieu subspace
of ${\mathcal{A}}[{\xi,z}]$ for any $t\in\mathbb{C}$ iff
$\rm{Im\,}(\xi-t\partial)$ is a Mathieu subspace of ${\mathcal{A}}[{\xi,z}]$
for any $t\in\mathbb{C}$. Hence, we have proved the theorem. $\Box$

From Theorems 4.4 and 4.6, we immediately have the following corollary.

######  Corollary 4.7.

Conjecture 4.5 implies the Jacobian conjecture.

######  Remark 4.8.

Note that, when $t=0$, Conjecture 4.5 is trivial since $\xi\mathbb{C}[\xi,z]$
is an ideal of the algebra
${\mathcal{B}}_{t=0}[{\xi,z}]={\mathcal{A}}[{\xi,z}]$. In general, Conjecture
4.5 in some sense just claims that the algebras ${\mathcal{B}}_{t}[{\xi,z}]$
$(t\in\mathbb{C})$ do not differ or change too much from
${\mathcal{A}}[{\xi,z}]$ so that the vector subspace $\xi\mathbb{C}[\xi,z]$
still remains as a Mathieu subspace of ${\mathcal{B}}_{t}[{\xi,z}]$.

From this point of view, the triviality of the deformation
${\mathcal{B}}_{t}[{\xi,z}]$ $(t\in\mathbb{C})$ of the polynomial algebra
${\mathcal{A}}[{\xi,z}]$ given in Corollary 2.2 may be viewed as a fact in
favor of Conjecture 4.5, hence also to the Jacobian conjecture via the
implication in Corollary 4.7.

######  Remark 4.9.

Conjecture 4.5 and also the Jacobian conjecture can be viewed as problems
caused by the following fact. Namely, due to the change of the algebra
structure from ${\mathcal{A}}[{\xi,z}]$ to ${\mathcal{B}}_{t}[{\xi,z}]$, the
evaluation map at $\xi=0$, which is an algebra homomorphism from
${\mathcal{A}}[{\xi,z}]$ to ${\mathcal{A}}[z]$, is not an algebra homomorphism
from ${\mathcal{B}}_{t}[{\xi,z}]$ to ${\mathcal{A}}[z]$ if $t\neq 0$.
Therefore, its kernel $\xi\mathbb{C}[{\xi,z}]$ does not remain to be an ideal
of ${\mathcal{B}}_{t}[{\xi,z}]$ anymore.

But, on the other hand, as we will see later in Subsection 5.2 $($See Theorem
5.2 and Remark 5.4$)$, the same fact for the evaluation map at $\xi=1$, i.e.
$\xi_{i}=1$ $(1\leq i\leq n)$, in some sense also causes something truely
remarkable, namely, the generalized Laguerre polynomials.

##  5\. Connections with the Generalized Laguerre Polynomials

In this section, we study some connections and interactions of the monomials
of the algebra ${\mathcal{B}}[{\xi,z}]$ in $\xi$ and $z$ with the generalized
Laguerre polynomials in one or more variables.

In Subsection 5.1, we briefly recall the definition and the orthogonal
property of the generalized Laguerre polynomials. In Subsection 5.2, we show
that the generalized Laguerre polynomials can be obtained from certain
monomials of the algebra ${\mathcal{B}}[{\xi,z}]$ in $\xi$ and $z$ (See
Theorem 5.2 and Corollary 5.3).

In Subsection 5.3, we study some applications of the connection given in
Theorem 5.2. We first use certain properties of the generalized Laguerre
polynomials to derive some results on some monomials of
${\mathcal{B}}[{\xi,z}]$ in $\xi$ and $z$. We then use some results derived in
Section 2 on the algebra ${\mathcal{B}}[{\xi,z}]$ to give new proofs for some
important properties of the generalized Laguerre polynomials (see Proposition
5.11 and Theorem 5.12).

###  5.1. The Generalized Laguerre Orthogonal Polynomials

First, let us recall the generalized Laguerre orthogonal polynomials in one
variable.

For any $k\in{\mathbb{R}}$ and $m\in{\mathbb{N}}$, the generalized Laguerre
polynomial $L_{m}^{[k]}(z)$ in one variable is given by

(5.1) $\displaystyle
L_{m}^{[k]}(z)=\sum_{j=0}^{m}\binom{m+k}{m-j}\,\frac{(-z)^{j}}{j!}.$

Here we are only interested in the case that $k\in{\mathbb{N}}$. For any fixed
$k\in{\mathbb{N}}$, the generating function of the generalized Laguerre
polynomials $L_{m}^{[k]}(z)$ $(m\geq 0)$ is given by

(5.2)
$\displaystyle\frac{\exp(-\frac{zu}{1-u})}{(1-u)^{k+1}}=\sum_{m=0}^{+\infty}L_{m}^{[k]}(z)\,u^{m},$

where $u$ above denotes a formal variable which commutes with $z$.

The multi-variable generalized Laguerre polynomials are defined as follows.

Let ${\bf k}=(k_{1},k_{2},...,k_{n})\in{\mathbb{N}}^{n}$ and
${\alpha}=(a_{1},a_{2},...,a_{n})\in{\mathbb{N}}^{n}$. The generalized
Laguerre polynomials in $n$-variable $z=(z_{1},z_{2},...,z_{n})$ is defined by

(5.3) $\displaystyle L_{\alpha}^{[{\bf
k}]}(z)\\!:=L_{a_{1}}^{[k_{1}]}(z_{1})L_{a_{2}}^{[k_{2}]}(z_{2})\cdots
L_{a_{n}}^{[k_{n}]}(z_{n}).$

The polynomials $L_{\alpha}(z)\\!:=L_{\alpha}^{[0]}(z)$
$(\alpha\in{\mathbb{N}}^{n})$ are the so-called the (classical) Laguerre
polynomials. They were named after Edmond. N. Laguerre [L]. The generalized
Laguerre polynomials were introduced much later by G. Pólya and G. Szegö [PS]
in $1976$.

One of the most important properties of the generalized Laguerre polynomials
is the following theorem.

######  Theorem 5.1.

For any ${\bf k},\alpha,\beta\in{\mathbb{N}}^{n}$, we have

(5.4) $\displaystyle\int_{({\mathbb{R}}_{>0})^{n}}L_{\alpha}^{[{\bf
k}]}(z)L_{\beta}^{[{\bf
k}]}(z)\,w(z)\,dz=\delta_{\alpha,\beta}\,\frac{(\alpha+{\bf k})!}{\alpha!},$

where $\delta_{\alpha,\beta}$ is the Kronecker delta function and $w(z)$ given
by

(5.5) $\displaystyle w(z)\\!:=z^{\bf k}e^{-\sum_{i=1}^{n}z_{i}}.$

The function $w(z)$ above is called the weight function of the generalized
Laguerre polynomials $L_{\alpha}^{[\bf k]}(z)$ $(\alpha\in{\mathbb{N}}^{n})$.

Consequently, with any fixed ${\bf k}$, the generalized Laguerre polynomials
$L_{\alpha}^{[{\bf k}]}(z)$ $(\alpha\in{\mathbb{N}}^{n})$ form an orthogonal
basis of $\mathbb{C}[z]$ with respect to the Hermitian form defined by

(5.6)
$\displaystyle(f,g)=\int_{({\mathbb{R}}_{>0})^{n}}f(z)\bar{g}(z)\,w(z)\,dz,$

where $\bar{g}(z)$ denotes the complex conjugation of the polynomial
$g(z)\in\mathbb{C}[z]$.

There are many other interesting and important properties of the generalized
Laguerre polynomials. We refer the reader to [Sz], [PS], [AAR] and [DX] for
very thorough study on this family of orthogonal polynomials. See also the
Wolfram Research web sources [W1] and [W2] for over one hundred formulas and
identities on the (generalized) Laguerre polynomials.

###  5.2. The Generalized Laguerre Polynomials in Terms of the Product $\ast$

The main result of this subsection is the following theorem.

######  Theorem 5.2.

For any ${\bf k},\alpha\in{\mathbb{N}}^{n}$, we have

(5.7) $\displaystyle L_{\alpha}^{[{\bf k}]}(\xi z)$
$\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\,\xi^{-{\bf
k}}(\xi^{\alpha+{\bf k}}\ast z^{\alpha}).$ (5.8) $\displaystyle
L_{\alpha}^{[{\bf k}]}(\xi z)$
$\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\,z^{-{\bf k}}(\xi^{\alpha}\ast
z^{\alpha+{\bf k}}),$

where $\xi z\\!:=(\xi_{1}z_{1},\xi_{2}z_{2},...,\xi_{n}z_{n})$.

In particular, for the Laguerre polynomials, we have

(5.9) $\displaystyle L_{\alpha}(\xi
z)=\frac{(-1)^{|\alpha|}}{\alpha!}\,\,\xi^{\alpha}\ast z^{\alpha}.$

Proof: We first prove Eq. (5.9). Note first that, as pointed out in Subsection
$2.1$ [Z4], the Laguerre polynomials $L_{m}(z)$ $(m\in{\mathbb{N}})$ in one
variable can be obtained as

(5.10) $\displaystyle L_{m}(z)=\frac{1}{m!}(\partial-1)^{m}(z^{m}).$

Changing the variable $z\to\xi z$ in the equation above, we get

$\displaystyle L_{m}(\xi z)$
$\displaystyle=\frac{1}{m!}(\xi^{-1}\partial-1)^{m}(\xi^{m}z^{m})=\frac{1}{m!}\xi^{-m}(\partial-\xi)^{m}(\xi^{m}z^{m})$
$\displaystyle=\frac{1}{m!}(\partial-\xi)^{m}(z^{m})=\frac{(-1)^{m}}{m!}(\xi-\partial)^{m}(z^{m}).$

By Eq. (5.3) with ${\bf k}=0$ and the equation above, we see that the multi-
variable Laguerre polynomials $L_{\alpha}(z)$ $(\alpha\in{\mathbb{N}}^{n})$
can be given by

(5.11) $\displaystyle L_{\alpha}(\xi
z)=\frac{(-1)^{|\alpha|}}{\alpha!}(\xi-\partial)^{\alpha}(z^{\alpha}).$

Then, apply Eq. (2.14) with $\lambda(\xi)=\xi^{\alpha}$ and $t=1$, we get Eq.
(5.9).

To show Eq. (5.7), recall that we have the following well-known identity for
the one-variable generalized Laguerre polynomials, which can be easily derived
from the generating functions of the generalized Laguerre polynomials in Eq.
(5.2):

(5.12) $\displaystyle L_{m}^{[k]}(z)=(-1)^{k}\partial^{k}L_{m+k}(z).$

Now, by Eq. (5.3) and the equation above, we see that the multi-variable
generalized Laguerre polynomials can be given by

(5.13) $\displaystyle L_{\alpha}^{[{\bf k}]}(z)=(-1)^{|\bf k|}\partial^{\bf
k}L_{\alpha+{\bf k}}(z).$

Changing the variable $z\to\xi z$ in the equation above, we get

(5.14) $\displaystyle L_{\alpha}^{[{\bf k}]}(\xi z)$ $\displaystyle=(-1)^{|\bf
k|}(\partial^{\bf k}L_{\alpha+{\bf k}})(\xi z)$ $\displaystyle=(-1)^{|\bf
k|}\xi^{-{\bf k}}\partial^{\bf k}(L_{\alpha+{\bf k}}(\xi z))$ Applying Eq.
(5.9) and then Eq.(2.29): $\displaystyle=\frac{(-1)^{|\alpha|}}{(\alpha+{\bf
k})!}\xi^{-{\bf k}}\partial^{\bf k}(\xi^{\alpha+{\bf k}}\ast z^{\alpha+{\bf
k}}).$ $\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\xi^{-{\bf
k}}(\xi^{\alpha+{\bf k}}\ast z^{\alpha}).$

Hence, we get Eq. (5.7). Switching $\xi$ and $z$ in Eq. (5.7) and using the
commutativity of the product $\ast$, we get Eq. (5.8). $\Box$

######  Corollary 5.3.

For any ${\bf k},\alpha\in{\mathbb{N}}^{n}$, we have

$\displaystyle L_{\alpha}(z)$
$\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\,\,\left.(\xi^{\alpha}\ast
z^{\alpha})\right|_{\xi=1};$ $\displaystyle L_{\alpha}^{[{\bf k}]}(z)$
$\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\,\left.(\xi^{\alpha+{\bf
k}}\ast z^{\alpha})\right|_{\xi=1};$ $\displaystyle L_{\alpha}^{[{\bf k}]}(z)$
$\displaystyle=\frac{(-1)^{|\alpha|}}{\alpha!}\,\left.z^{-{\bf
k}}(\xi^{\alpha}\ast z^{\alpha+{\bf k}})\right|_{\xi=1},$

where $|_{{}_{\xi=1}}$ denotes the evaluation map from $\mathbb{C}[{\xi,z}]$
to $\mathbb{C}[z]$ by setting $\xi_{i}=1$ for any $1\leq i\leq n$.

######  Remark 5.4.

Note that, the evaluation map $|_{{}_{\xi=1}}$ viewed as a linear map from
${\mathcal{A}}[{\xi,z}]$ to ${\mathcal{A}}[z]$ is a homomorphism of algebras.
But, as a linear map from the algebra ${\mathcal{B}}[{\xi,z}]$ to the
polynomial algebra ${\mathcal{A}}[z]$, it is not a homomorphism of algebras
anymore. In particular, we have

$\displaystyle\left.(\xi^{\alpha}\ast z^{\alpha})\right|_{\xi=1}\neq 1\ast
z^{\alpha}=z^{\alpha}.$

Otherwise the generalized Laguerre polynomials would be trivialized.

Therefore, in some sense, the fact that the evaluation map
$|_{{}_{\xi=1}}:{\mathcal{B}}[{\xi,z}]\to{\mathcal{A}}[z]$ fails to be an
algebra homomorphism causes the non-trivial, actually truly remarkable,
generalized Laguerre polynomials. But, on the other hand, as we have discussed
in Subsection 4.2 $($See Remark 4.9$)$, the same fact for the evaluation map
at $\xi=0$ also causes some extremely difficult open problems such as
Conjecture 4.5 and the Jacobian conjecture.

Another immediate consequence of Theorem 5.2 is the following corollary.

######  Corollary 5.5.

For any $\alpha,\beta\in{\mathbb{N}}^{n}$, we have

(5.15) $\displaystyle\xi^{\beta}(\xi^{\alpha}\ast
z^{\alpha+\beta})=z^{\beta}(\xi^{\alpha+\beta}\ast z^{\alpha}).$

Note that the corollary follows immediately from Eqs. (5.7) and (5.8) with
${\bf k}=\beta$. But here we also give a more straightforward proof.

Proof: Consider

$\displaystyle\xi^{\beta}(\xi^{\alpha}\ast z^{\alpha+\beta})$
$\displaystyle=(\xi-\partial+\partial)^{\beta}(\xi^{\alpha}\ast
z^{\alpha+\beta})$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\beta}{\gamma}(\xi-\partial)^{\beta-\gamma}\partial^{\gamma}(\xi^{\alpha}\ast
z^{\alpha+\beta})$ Applying Eq. (2.29) and then Eq.(2.14):
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\beta}{\gamma}\frac{(\alpha+\beta)!}{(\alpha+\beta-\gamma)!}\,(\xi-\partial)^{\beta-\gamma}(\xi^{\alpha}\ast
z^{\alpha+\beta-\gamma})$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}\binom{\beta}{\gamma}\frac{(\alpha+\beta)!}{(\alpha+\beta-\gamma)!}\,\xi^{\beta-\gamma}\ast(\xi^{\alpha}\ast
z^{\alpha+\beta-\gamma})$
$\displaystyle=\sum_{\gamma\in{\mathbb{N}}^{n}}^{\beta}\binom{\beta}{\gamma}\frac{(\alpha+\beta)!}{(\alpha+\beta-\gamma)!}\,(\xi^{\alpha+\beta-\gamma}\ast
z^{\alpha+\beta-\gamma}).$

By switching $\xi\leftrightarrow z$ in the argument above and using the
commutativity of the product $\ast$, it is easy to see that we also have

$\displaystyle z^{\beta}(\xi^{\alpha+\beta}\ast
z^{\alpha})=\sum_{\gamma\in{\mathbb{N}}^{n}}^{\beta}\binom{\beta}{\gamma}\frac{(\alpha+\beta)!}{(\alpha+\beta-\gamma)!}\,(\xi^{\alpha+\beta-\gamma}\ast
z^{\alpha+\beta-\gamma}).$

Hence Eq. (5.15) follows. $\Box$

###  5.3. Some Applications of Theorem 5.2

First, let us derive some identities for the exponential series
$\exp_{*}(\cdot)=e_{\ast}^{\\{\cdot\\}}$ of the algebra
${\mathcal{B}}[{\xi,z}]$, i.e. the usual exponential series but with the
product replaced by $\ast$.

######  Proposition 5.6.

Let $u=(u_{1},u_{2},...,u_{n})$ be $n$ free commutative variables. Set
$\xi\ast z\\!:=(\xi_{1}\ast z_{1},\,\xi_{2}\ast z_{2},...,\,\xi_{n}\ast
z_{n})$ and $(\xi\ast z)u\\!:=\sum_{i=1}^{n}(\xi_{i}\ast z_{i})u_{i}$. Then,
for any ${\bf k}=(k_{1},k_{2},...,k_{n})\in{\mathbb{N}}^{n}$, we have

(5.16) $\displaystyle\xi^{-\bf k}\left(\xi^{\bf k}\ast e_{*}^{-(\xi\ast
z)u}\right)$
$\displaystyle=\prod_{i=1}^{n}\,\,\frac{\exp(-\,\frac{(\xi_{i}z_{i})\,u_{i}}{1-u_{i}})}{(1-u_{i})^{k_{i}+1}}.$
(5.17) $\displaystyle z^{-\bf k}\left(z^{\bf k}\ast e_{*}^{-(\xi\ast
z)u}\right)$
$\displaystyle=\prod_{i=1}^{n}\,\,\frac{\exp(-\,\frac{(\xi_{i}z_{i})\,u_{i}}{1-u_{i}})}{(1-u_{i})^{k_{i}+1}}.$

In particular, when ${\bf k}=0$, we have the following expression of the
exponential $\exp_{*}(-(\xi\ast z)u)$:

(5.18) $\displaystyle\exp_{*}(-(\xi\ast z)u)$
$\displaystyle=\prod_{i=1}^{n}\,\,\frac{\exp(-\,\frac{(\xi_{i}z_{i})\,u_{i}}{1-u_{i}})}{(1-u_{i})}.$

Proof: We give a proof for Eq. (5.16). The proof of Eq. (5.17) is similar.

First, by the commutativity and associativity of the product $\ast$ and also
by Proposition 2.6, $(b)$, it is easy to see that, for any
$\alpha,\beta\in{\mathbb{N}}^{n}$, we have

(5.19) $\displaystyle(\xi^{\alpha}\ast z^{\alpha})\ast(\xi^{\beta}\ast
z^{\beta})$ $\displaystyle=\xi^{\alpha+\beta}\ast z^{\alpha+\beta}.$ (5.20)
$\displaystyle(\xi\ast z)^{\ast\alpha}$ $\displaystyle=(\xi^{\alpha}\ast
z^{\alpha}),$

where $(\xi\ast z)^{\ast\alpha}$ denotes the “$\alpha^{\rm th}$” power of
$(\xi\ast z)$ with respect to the new product $\ast$.

By the last two equations above and Eq. (5.7), we have

(5.21) $\displaystyle\xi^{-{\bf k}}\left(\xi^{\bf k}\ast\exp_{*}^{-(\xi\ast
z)u}\right)$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}\frac{(-1)^{|\alpha|}}{\alpha!}\,\xi^{-\bf
k}(\xi^{\alpha+{\bf k}}\ast z^{\alpha})u^{\alpha}$
$\displaystyle=\sum_{\alpha\in{\mathbb{N}}^{n}}L_{\alpha}^{[\bf k]}(\xi
z)u^{\alpha}.$

On the other hand, by Eqs. (5.2) and (5.3), we see that the generating
function of the multi-variable generalized Laguerre polynomials
$L_{\alpha}^{[\bf k]}(z)$ $(\alpha\in{\mathbb{N}}^{n})$ is given by

(5.22)
$\displaystyle\prod_{i=1}^{n}\,\,\frac{\exp(-\,\frac{z_{i}\,u_{i}}{1-u_{i}})}{(1-u_{i})^{k_{i}+1}}=\sum_{\alpha\in{\mathbb{N}}^{n}}L_{\alpha}^{[\bf
k]}(z)u^{\alpha}.$

Replacing $z$ by $\xi z$ in the equation above, we get

(5.23)
$\displaystyle\prod_{i=1}^{n}\,\,\frac{\exp(-\,\frac{(\xi_{i}z_{i})\,u_{i}}{1-u_{i}})}{(1-u_{i})^{k_{i}+1}}=\sum_{\alpha\in{\mathbb{N}}^{n}}L_{\alpha}^{[\bf
k]}(\xi z)u^{\alpha}.$

Combining Eqs. (5.21) and (5.23), we get Eq. (5.16). $\Box$

Next we use the connection given in Theorem 5.2 to derive more properties on
the monomials in $\xi$ and $z$ with respect to the product $\ast$ from certain
results on the generalized Laguerre polynomials.

For convenience, for any $\alpha\in{\mathbb{N}}^{n}$, we set

(5.24) $\displaystyle L_{\alpha}(z;\xi)\\!:=\xi^{\alpha}\ast z^{\alpha}.$

Note that, by Eqs. (5.1) and (5.9), the polynomials $L_{\alpha}(z;\xi)$
$(\alpha\in{\mathbb{N}}^{n})$ are polynomials with coefficients in
${\mathbb{Q}}$. In particular, for any fixed $\xi\in({\mathbb{R}}_{>0})^{n}$,
by Eqs. (5.1) and (5.9), it is easy to see that the polynomials
$L_{\alpha}(z;\xi)$ $(\alpha\in{\mathbb{N}}^{n})$ are polynomials in $z$ with
real coefficients and form a linear basis of $\mathbb{C}[z]$.

The next proposition says that this basis is also orthogonal with respect to
the following weight function:

(5.25) $\displaystyle
w_{\xi}(z)\\!:=e^{-\langle\xi,z\rangle}\prod_{i=1}^{n}\xi_{i},$

######  Proposition 5.7.

For any $\alpha,\beta\in{\mathbb{N}}^{n}$, we have

(5.26)
$\displaystyle\int_{({\mathbb{R}}_{>0})^{n}}L_{\alpha}(z;\xi)L_{\beta}(z;\xi)\,w_{\xi}(z)\,dz=(\alpha!)^{2}\delta_{\alpha,\beta}.$

Proof: Note that, under the change of variables $z_{i}\to\xi_{i}z_{i}$ $(1\leq
i\leq n)$, by Eqs. (5.9) and (5.24) the Laguerre polynomials $L_{\alpha}(z)$
will be changed to

(5.27) $\displaystyle L_{\alpha}(z)\to L_{\alpha}(\xi
z)=\frac{(-1)^{|\alpha|}}{\alpha!}L_{\alpha}(z;\xi).$

By Eq. (5.25) and also Eq. (5.5) with ${\bf k}=0$, the weight function $w(z)$
of the Laguerre polynomials is changed to

(5.28) $\displaystyle w(z)\to w_{\xi}(z)\prod_{i=1}^{n}\xi_{i}^{-1}.$

Now, apply the same changing of the variables to the integral in Eq. (5.4)
with ${\bf k}=0$, by the last two equations above, we get

(5.29) $\displaystyle\delta_{\alpha,\beta}$
$\displaystyle=\frac{(-1)^{|\alpha+\beta|}}{\alpha!\beta!}\int_{({\mathbb{R}}_{>0})^{n}}L_{\alpha}(z;\xi)L_{\beta}(z;\xi)\,w_{\xi}(z)\,dz$

Hence Eq. (5.26) follows. $\Box$

Denote by ${\mathcal{A}}_{\mathbb{Q}}[{\xi,z}]$ the polynomial algebra in
$\xi$ and $z$ over ${\mathbb{Q}}$. Next we assume $n=1$ and consider the
irreducibility of the polynomial $L_{\alpha}(z;\xi)$
$(\alpha\in{\mathbb{N}}^{n})$ as elements of
${\mathcal{A}}_{\mathbb{Q}}[{\xi,z}]$. But, first, we need to prove the
following lemma.

######  Lemma 5.8.

Let $\xi$ and $z$ be two commutative free variables and $K$ any field. Then,
for any $f(z)\in K[z]$ with $\deg f\geq 2$, $f(z)$ is irreducible over $K$ iff
$f(\xi z)\in K[\xi,z]$ $($as a polynomial in two variables$)$ is irreducible
over $K$.

Proof: The $(\Leftarrow)$ part of the lemma is trivial. We use the
contradiction method to show the $(\Rightarrow)$ part of the lemma.

Assume that $f(\xi z)$ is reducible in $K[{\xi,z}]$. Write

(5.30) $\displaystyle f(\xi z)=g({\xi,z})h({\xi,z})$

for some $g({\xi,z}),h({\xi,z})\in K[{\xi,z}]$ with $\deg g,\deg h\geq 1$.

Setting $\xi=1$ in the equation above, we also have

(5.31) $\displaystyle f(z)=g(1,z)h(1,z).$

Let $\bar{K}$ be the algebraic closure of $K$. Write
$f(z)=b\prod_{i=1}^{d}(z-a_{i})$ for some $b\in K\backslash\\{0\\}$ and
$a_{i}\in\bar{K}$ $(1\leq i\leq d)$. Then we have

(5.32) $\displaystyle f(\xi z)=b\prod_{i=1}^{d}(\xi z-a_{i}).$

Since $f(z)$ is irreducible over $K$ and $\deg f\geq 2$ by the assumption, we
have $a_{i}\neq 0$ $(1\leq i\leq d)$. Hence, for each $i$, $\xi z-a_{i}$ is
irreducible in $\bar{K}[{\xi,z}]$. Then by Eqs. (5.30) and (5.32), we have

(5.33) $\displaystyle g(\xi,z)=c\prod_{k=1}^{m}(\xi z-a_{i_{k}})$

for some $c\in\bar{K}\backslash\\{0\\}$, $1\leq m<d$ and $1\leq
i_{1}<i_{2}<\cdots<i_{m}\leq d$.

However, the equation above implies $g(1,z)=c\prod_{k=1}^{m}(z-a_{i_{k}})$.
Since $g(\xi,z)\in K[{\xi,z}]$, we also have $g(1,z)\in K[z]$. Then by Eq.
(5.31), $g(1,z)$ is a divisor of $f(z)$ in $K[z]$ with $1\leq\deg
g(1,z)=m<d=\deg f(z)$, which contradicts to the assumption that $f(z)$ is
irreducible in $K[z]$. $\Box$

######  Theorem 5.9.

Let $\xi$ and $z$ be two commutative free variables. For any $m\geq 2$,
$L_{m}(z;\xi)=\xi^{m}\ast z^{m}$ is irreducible in
${\mathcal{A}}_{\mathbb{Q}}[{\xi,z}]$.

Proof: By a theorem proved by I. Schur [Sc1], we know that, for any $m\geq 1$,
the Laguerre polynomials $L_{m}(z)$ in one variable is irreducible over
${\mathbb{Q}}$. Hence, by Eq. (5.9) and Lemma 5.8, the theorem holds. $\Box$

Note that I. Schur also proved in [Sc2] that the generalized Laguerre
polynomials $L_{m}^{[1]}(z)$ $(m\geq 0)$ in one variable are also irreducible
over ${\mathbb{Q}}$. Furthermore, M. Filaseta and T.-Y. Lam proved in [FL]
that, for any non-negative $k\in{\mathbb{Q}}$, all but finitely many of the
generalized Laguerre polynomials $L_{m}^{[k]}(z)$ $(m\geq 0)$ in one variable
are irreducible over ${\mathbb{Q}}$. Hence, by a similar argument as for
Theorem 5.9, we also have the following theorem.

######  Theorem 5.10.

Let $\xi$ and $z$ be two commutative free variables. Then, for any
$k\in{\mathbb{N}}$, all but only finitely many of the polynomials
$z^{-k}(\xi^{m}\ast z^{m+k})$ and $\xi^{-k}(\xi^{m+k}\ast z^{m})$
$(m\in{\mathbb{N}})$ are irreducible over ${\mathbb{Q}}$.

Next, we re-prove some important properties of the generalized Laguerre
polynomials by using their expressions given in Theorem 5.2. For simplicity,
we here only consider the one-variable case. Similar results for the multi-
variable generalized Laguerre polynomials can be simply derived from the one-
variable case via Eq. (5.3).

First, let us look at the following recurrent formulas of the Laguerre
polynomials in one variable.

######  Proposition 5.11.

For any $m\geq 1$, we have

(5.34) $\displaystyle(m+1)L_{m+1}(z)$
$\displaystyle=(2m+1-z)L_{m}(z)-mL_{m-1}(z),$ (5.35) $\displaystyle
zL^{\prime}_{m}(z)$ $\displaystyle=m(L_{m}(z)-L_{m-1}(z))$

Proof: Note first that, for any $m\geq 1$, by Eqs. (2.14) and (2.15), we have

$\displaystyle\xi\ast z^{m}$ $\displaystyle=(\xi-\partial)z^{m}=\xi
z^{m}-mz^{m-1},$ $\displaystyle z\ast\xi^{m}$
$\displaystyle=(z-\delta)\xi^{m}=z\xi^{m}-m\xi^{m-1}.$

Hence, we also have

$\displaystyle\xi\ast z$ $\displaystyle=\xi z-1,$ $\displaystyle\xi z^{m}$
$\displaystyle=\xi\ast z^{m}+mz^{m-1},$ $\displaystyle z\xi^{m}$
$\displaystyle=z\ast\xi^{m}+m\xi^{m-1}.$

By the last three equations above and also Eq. (2.23), we have

$\displaystyle(\xi z-1)(\xi^{m}\ast z^{m})$ $\displaystyle=(\xi\ast
z)(\xi^{m}\ast z^{m})=(z\xi^{m})\ast(\xi z^{m})$
$\displaystyle=(z\ast\xi^{m}+m\xi^{m-1})\ast(\xi\ast z^{m}+mz^{m-1})$
$\displaystyle=\xi^{m+1}\ast z^{m+1}+2m\,\xi^{m}\ast z^{m}+m^{2}\xi^{m-1}\ast
z^{m-1}.$

Multiply $(-1)^{m}/m!$ to the equation above and then apply Eq. (5.9), we get

$\displaystyle(\xi z-1)L_{m}(\xi z)=-(m+1)L_{m+1}(\xi z)+2mL_{m}(\xi
z)-mL_{m-1}(\xi z).$

Replace $\xi z$ by $z$ in the equation above, we get

$\displaystyle(z-1)L_{m}(z)=-(m+1)L_{m+1}(z)+2mL_{m}(z)-mL_{m-1}(z),$

Hence Eq. (5.34) follows.

To show Eq. (5.35), by Eqs. (2.15) and (2.30), we have,

$\displaystyle\xi^{m}\ast z^{m}$ $\displaystyle=z\ast(\xi^{m}\ast
z^{m-1})=(z-\delta)(\xi^{m}\ast z^{m-1})$ $\displaystyle=z(\xi^{m}\ast
z^{m-1})-m(\xi^{m-1}\ast z^{m-1})$
$\displaystyle=\frac{1}{m}z\partial(\xi^{m}\ast z^{m})-m(\xi^{m-1}\ast
z^{m-1}).$

Multiply $(-1)^{m}/m!$ to the equation above and then apply Eq. (5.9), we get

$\displaystyle L_{m}(\xi z)=\frac{1}{m}z\partial(L_{m}(\xi z))+L_{m-1}(\xi
z)=\frac{1}{m}\xi zL^{\prime}_{m}(\xi z)+L_{m-1}(\xi z).$

Replace $\xi z$ by $z$ in the equation above, we get

$\displaystyle L_{m}(z)=\frac{1}{m}zL^{\prime}_{m}(z)+L_{m-1}(z).$

Hence Eq. (5.35) follows. $\Box$

Next, we give a new proof for the following important property of the
generalized Laguerre polynomials in one variable.

######  Theorem 5.12.

For any $k,m\in{\mathbb{N}}$, $L_{m}^{[k]}(z)$ solves the following so-called
associated Laguerre differential equation:

(5.36) $\displaystyle zf^{\prime\prime}(z)+(k+1-z)f^{\prime}(z)+mf(z)=0.$

Proof: First, by Eq. (5.1), we have $L_{0}^{[k]}(z)=1$. It is easy to see that
the theorem holds for this case.

Assume $m\geq 1$. Then, by Eq. (2.15), we have

$\displaystyle\xi(\xi^{m+k}\ast z^{m})$ $\displaystyle=\xi(z\ast(\xi^{m+k}\ast
z^{m-1}))$ $\displaystyle=\xi(z-\delta)(\xi^{m+k}\ast z^{m-1})$
$\displaystyle=\xi z(\xi^{m+k}\ast z^{m-1})-\xi\delta(\xi^{m+k}\ast z^{m-1}).$

Add $z\partial(\xi^{m+k}\ast z^{m-1})$ to the equation above and apply Eq.
(2.21), we have

$\displaystyle\xi(\xi^{m+k}\ast z^{m})+z\partial(\xi^{m+k}\ast z^{m-1})$
$\displaystyle=\xi z(\xi^{m+k}\ast
z^{m-1})-(\xi\delta-z\partial)(\xi^{m+k}\ast z^{m-1})$ $\displaystyle=(\xi
z-k-1)(\xi^{m+k}\ast z^{m-1}).$

By Eq. (2.29), we may re-write the equation above as

$\displaystyle\xi(\xi^{m+k}\ast z^{m})+\frac{1}{m}z\partial^{2}(\xi^{m+k}\ast
z^{m})=\frac{1}{m}(\xi z-k-1)\partial(\xi^{m+k}\ast z^{m}).$

Multiply $\frac{(-1)^{m}\xi^{-k-1}}{(m-1)!}$ to both sides of the equation
above and then apply Eq. (5.7), we have

$\displaystyle mL_{m}^{[k]}(\xi z)+z\xi^{-1}\partial^{2}(L_{m}^{[k]}(\xi
z))=(\xi z-k-1)\xi^{-1}\partial(L_{m}^{[k]}(\xi z)).$

By the Chain rule, the equation above is same as

$\displaystyle mL_{m}^{[k]}(\xi z)+z\xi(\partial^{2}L_{m}^{[k]})(\xi z)=(\xi
z-k-1)(\partial L_{m}^{[k]})(\xi z).$

Replace $\xi z$ by $z$, or $z$ by $\xi^{-1}z$ in the equation above, we get

$\displaystyle mL_{m}^{[k]}(z)+z\partial^{2}L_{m}^{[k]}(z)=(z-k-1)\partial
L_{m}^{[k]}(z).$

Hence we have proved the theorem. $\Box$

Finally, let us point out the following conjecture on the generalized Laguerre
polynomials, which is a special case of Conjecture $3.5$ in [Z4] for all the
classical orthogonal polynomials.

######  Conjecture 5.13.

For any ${\bf k}\in{\mathbb{N}}^{n}$, the subspace ${\mathcal{M}}$ of the
polynomial algebra ${\mathcal{A}}[z]$ spanned by the generalized Laguerre
polynomials $L^{[\bf k]}_{\alpha}(z)$ $(0\neq\alpha\in{\mathbb{N}}^{n})$ is a
Mathieu subspace of ${\mathcal{A}}[z]$.

Despite the vast amount of known results on the generalized Laguerre
polynomials in the literature, the conjecture above is even still open for the
classical Laguerre polynomials, (i.e. the case with ${\bf k}=0$) in one
variable.

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Department of Mathematics, Illinois State University, Normal, IL 61790-4520.

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