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\pgfsys@invoke{\lxSVG@closescope
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Since we are abusing notation, the higher squares commute simply since the
truncation of a truncation is itself a truncation. Thus all squares commute,
establishing (2.C).

Setting
$\upvartheta_{i}\colonequals\upvarphi^{i-1}\circ\cdots\circ\upvarphi^{1}\colon\mathbb{C}\langle\mathsf{x}\rangle/\mathfrak{m}^{i}\to\mathbb{C}\langle\mathsf{x}\rangle/\mathfrak{m}^{i}$,
again with the convention that $\upvartheta_{1}=\mathop{\rm{Id}}\nolimits$,
then each $\upvartheta_{i}$ is an automorphism since each $\upvarphi^{t}$ is.
Thus (2.A) induces, through (2.B), an automorphism of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ such
that for all $i\geq j$ the following diagram commutes.

$\begin{array}[]{c}\leavevmode\hbox to127.56pt{\vbox
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}{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}}
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}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\end{array}$ (2.D)

Write $\varprojlim\upvarphi^{n}\cdots\upvarphi^{1}$ for this induced
automorphism.

######  Lemma 2.4.

With notation and assumptions as directly above, for any
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
the sequence $(\upvarphi^{n}\cdots\upvarphi^{1}(f))_{n\geq 1}$ has limit
$\varprojlim\upvarphi^{n}\cdots\upvarphi^{1}(f)$.

###### Proof.

Set $F=\varprojlim\upvarphi^{n}\cdots\upvarphi^{1}$, then it suffices to prove
that for all $n\geq 1$, there exists $N$ such that
$\upvarphi^{t}\cdots\upvarphi^{1}(f)-F(f)\in\mathfrak{n}^{n}$ for all $t\geq
N$. This follows since for all $i>n$

$F(f)+\mathfrak{n}^{n}\stackrel{{\scriptstyle\scriptstyle\eqref{limit of hom
diagram}}}{{=}}\upvarphi^{n-1}\cdots\upvarphi^{1}(f)+\mathfrak{n}^{n}\stackrel{{\scriptstyle\scriptstyle\eqref{need
to comm 2}}}{{=}}\upvarphi^{i-1}\cdots\upvarphi^{1}(f)+\mathfrak{n}^{n}.\qed$

###  2.5. Closure and Cyclic Permutation

######  Definition 2.5.

For any subset
$\mathcal{S}\subset\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
its _closure_ is defined to be

$\overline{\mathcal{S}}=\bigcap_{i=0}^{\infty}(\mathcal{S}+\mathfrak{n}^{i}).$

That is, $b\in\overline{\mathcal{S}}$ if and only if for all $i\geq 0$, there
exists $s_{i}\in\mathcal{S}$ such that $b-s_{i}\in\mathfrak{n}^{i}$.

######  Notation 2.6.

For
$\mathcal{A}\colonequals\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
consider $\\{\mathcal{A},\mathcal{A}\\}$, the commutator vector space of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$. That
is, elements of $\\{\mathcal{A},\mathcal{A}\\}$ are finite sums

$\sum_{i=1}^{n}\uplambda_{i}(a_{i}b_{i}-b_{i}a_{i})$

for elements
$a_{i},b_{i}\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
and $\uplambda_{i}\in\mathbb{C}$. Write
$\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$ for the closure of
the commutator vector space $\\{\mathcal{A},\mathcal{A}\\}$. Note that
$\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$ is only a vector
space, not an ideal.

######  Definition 2.7.

Two elements
$f,g\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
are called cyclically equivalent, or $f$ is said to cyclically permute to $g$,
if $f-g\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$. We write
$f\sim g$ in this case.

######  Remark 2.8.

This notion of cyclic equivalence applied a pair of polynomials is finite and
elementary: it is generated over $\mathbb{C}$ by commutators $[m_{1},m_{2}]$
of monomials $m_{i}\in\mathbb{C}\langle\mathsf{x}\rangle$. With that in mind,
2.7 is then the natural notion for formal power series, as $f\sim g$ means
precisely that $f_{d}\sim g_{d}$ in every degree $d$, and no more: the closure
merely handles the possibility that $f$ and $g$ may differ by infinitely many
such operations.

###  2.6. Chasing into Higher Degrees

The following will be one of our main techniques for producing normal forms of
potentials in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$. The
basic idea is to start with a given $f$, then produce an infinite sequence of
automorphisms which chase terms into higher and higher degrees. Taking limits
then gives a single automorphism which takes $f$ to the desired normal form.
The subtle point is that at each stage the automorphisms in (2) below only
give the desired elements up to cyclic permutation. As such, the content in
the following is that, with care, limits interact well with cyclic
permutation.

######  Theorem 2.9.

Let
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
and set $\mathsf{f}_{1}=f$. Suppose that there exist elements
$\mathsf{f}_{2},\mathsf{f}_{3},\ldots$ and automorphisms
$\upvarphi^{1},\upvarphi^{2},\ldots$ such that

  1. (1)

Every $\upvarphi^{i}$ is unitriangular, of depth of $\geq i$, and

  2. (2)

$\upvarphi^{i}(\mathsf{f}_{i})-\mathsf{f}_{i+1}\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}\cap\mathfrak{n}^{i+1}$,
for all $i\geq 1$.

Then $\lim\mathsf{f}_{i}$ exists, and there exists an automorphism $F$ such
that $F(f)\sim\lim\mathsf{f}_{i}$.

###### Proof.

The proof follows the strategy used in [DWZ, 4.7], but as the axiomatics are
different here, we give the full proof. By §2.4 there is an automorphism
$F\colonequals\varprojlim\upvarphi^{n}\cdots\upvarphi^{1}$.

Since the depth of $\upvarphi^{i}$ is $\geq i$, by 2.2(1)
$\upvarphi^{i}(\mathsf{f}_{i})$ differs from $\mathsf{f}_{i}$ only in degrees
$>i$. By (2), $\upvarphi^{i}(\mathsf{f}_{i})$ differs from $\mathsf{f}_{i+1}$
only in degrees $>i$. Hence $\mathsf{f}_{i+1}$ differs from $\mathsf{f}_{i}$
only in degrees $>i$, from which it easily follows that $(\mathsf{f}_{n})$ is
a Cauchy sequence. Since Cauchy sequences converge in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$, the
limit $\lim\mathsf{f}_{i}$ exists.

Set
$c_{i}=\upvarphi^{i}(\mathsf{f}_{i})-\mathsf{f}_{i+1}\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}\cap\mathfrak{n}^{i+1}$.
Since $f=\mathsf{f}_{1}$, it is easy to see that

$\displaystyle\upvarphi^{n}\cdots\upvarphi^{1}(f)$
$\displaystyle=\mathsf{f}_{n+1}+\sum_{t=1}^{n}\upvarphi^{n}\cdots\upvarphi^{t+1}(c_{t})$
$\displaystyle=\mathsf{f}_{n+1}+\upvarphi^{n}\cdots\upvarphi^{1}\left(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})\right)$
(2.E)

where $\upvarphi^{n}\cdots\upvarphi^{t+1}$ is the identity when $t=n$. By 2.4
the left hand side has limit $F(f)$. The first part of the right hand side has
limit $\lim\mathsf{f}_{i}$, which exists by above. We next claim that the
rightmost term has limit $F(g)$, where $g$ is the limit of the sequence
$(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t}))_{n\geq 1}$.

First, $g$ exists, since by (2) $c_{i}\in\mathfrak{n}^{i+1}$, and so since
automorphisms preserve the maximal ideal,
$(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})\in\mathfrak{n}^{t+1}$ for all
$t$. It follows easily that the sequence
$\left(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})\right)_{n\geq
1}$ is Cauchy, and so its limit $g$ exists in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.
Given this, the fact that the sequence
$\left(\upvarphi^{n}\cdots\upvarphi^{1}(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t}))\right)_{n\geq
1}$ has limit $F(g)$ follows, since for all $i>n$

$\displaystyle F(g)+\mathfrak{n}^{n+1}$
$\displaystyle=\upvarphi^{n}\cdots\upvarphi^{1}(\uppi_{n+1}(g))+\mathfrak{n}^{n+1}$
(by (2.D))
$\displaystyle=\upvarphi^{n}\cdots\upvarphi^{1}\left(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})+\mathfrak{n}^{n+1}\right)+\mathfrak{n}^{n+1}$
(since $(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})\in\mathfrak{n}^{t+1}$)
$\displaystyle=\upvarphi^{n}\cdots\upvarphi^{1}\left(\sum_{t=1}^{i}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})+\mathfrak{n}^{n+1}\right)+\mathfrak{n}^{n+1}$
(add zero)
$\displaystyle=\upvarphi^{i}\cdots\upvarphi^{1}\left(\sum_{t=1}^{i}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t})\right)+\mathfrak{n}^{n+1}.$
(by (2.C))

Combining with (2.E) and taking limits it follows that

$F(f)=\lim\mathsf{f}_{i}+F(g).$ (2.F)

Now, it is easy to check that automorphisms preserve
$\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$, so each term in
the sequence
$\left(\upvarphi^{n}\cdots\upvarphi^{1}(\sum_{t=1}^{n}(\upvarphi^{t}\cdots\upvarphi^{1})^{-1}(c_{t}))\right)_{n\geq
1}$ belongs to $\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$.
But since
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ is
complete, every Cauchy sequence within a closed set has limit in that closed
set. It follows that the limit
$g\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$. One final
application of the fact that automorphisms preserve
$\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$ shows that
$F(g)\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}$, and so
$F(f)\sim\lim\mathsf{f}_{i}$. ∎

###  2.7. Elementary Properties of Closed Ideals

We finish this section with some technical results on closed ideals that are
used throughout §6–§8.

######  Notation 2.10.

When $I$ is an ideal, write $(\kern-2.5pt(I)\kern-2.5pt)$ for its closure (in
the sense of 2.5), which is again an ideal since the ring operations are
continuous. Note that $(\kern-2.5pt(I)\kern-2.5pt)$ need not be finitely
generated, even if $I$ is.

For a finite set of elements $\mathcal{S}$ in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
consider the closed ideal
$(\kern-2.5pt(\mathcal{S})\kern-2.5pt)=(\kern-2.5pt(s\mid
s\in\mathcal{S})\kern-2.5pt)$.

######  Lemma 2.11.

Let $\mathcal{S}$ be a finite subset of elements in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$, and
$f_{1},\dots,f_{s}\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.
Then the following statements hold.

  1. (1)

$(\kern-2.5pt(f_{1},\dots,f_{s})\kern-2.5pt)=(\kern-2.5pt(f_{1}u_{1},\dots,f_{s}u_{s})\kern-2.5pt)$
for any units
$u_{1},\dots,u_{s}\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.

  2. (2)

$(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\,)\kern-2.5pt)=(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$ in
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$.

  3. (3)

If
$\uppsi\colon\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\to\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$
is a topological isomorphism which sends
$f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\mapsto
g+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$, for two elements
$f,g\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
then there is an induced topological isomorphism

$\frac{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)}\buildrel\cong\over{\longrightarrow}\frac{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(g,s\mid
s\in\mathcal{S})\kern-2.5pt)}.$

###### Proof.

(1) $(\kern-2.5pt(f_{1},\dots,f_{s})\kern-2.5pt)$ is the smallest closed ideal
containing all $f_{i}$. Since
$f_{i}=(f_{i}u_{i})u_{i}^{-1}\in(\kern-2.5pt(f_{1}u_{1},\dots,f_{s}u_{s})\kern-2.5pt)$
for each $i$, by minimality
$(\kern-2.5pt(f_{1},\dots,f_{s})\kern-2.5pt)\subseteq(\kern-2.5pt(f_{1}u_{1},\dots,f_{s}u_{s})\kern-2.5pt)$.
Repeating the same argument to
$f_{i}u_{i}\in(\kern-2.5pt(f_{1},\dots,f_{s})\kern-2.5pt)$, the converse
inclusion also holds.

(2) Certainly
$(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\,)\kern-2.5pt)=I/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$
for some ideal $I$, given it is an ideal of the quotient. This ideal $I$ is
closed by [Wa, 5.2], since the map to the quotient is continuous, and hence
the inverse image of a closed set is closed. This closed ideal $I$ contains
both $f$ and $\mathcal{S}$, and so $(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)\subseteq I$.

On the other hand
$(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\,)\kern-2.5pt)$ is the
smallest closed ideal containing $f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$.
Setting
$A=\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
$J=(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$, and $H=(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)$, the third isomorphism theorem for topological
rings [Wa, 5.13] asserts that there is a topological isomorphism

$(A/J)/(H/J)\cong A/H.$

In particular, $A/H$ is Hausdorff, since $H$ is closed in $A$, by [Wa, 5.7(1)]
applied to $A$. This being the case, $H/J$ is closed in $A/J$, by [Wa, 5.7(1)]
applied to $A/J$. Hence $(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$ is a
closed ideal, which clearly contains
$f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$. By minimality
$(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\,)\kern-2.5pt)=I/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\subseteq(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$ and thus
$I\subseteq(\kern-2.5pt(f,s\mid s\in\mathcal{S})\kern-2.5pt)$. Combining
inclusions, the required equality holds.

(3) Since $\uppsi$ is a continuous isomorphism, the closed ideal generated by
$f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$ corresponds to the closed ideal
generated by $g+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$. Thus there is a
topological isomorphism

$\frac{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)}{(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\kern
1.0pt)\kern-2.5pt)}\longrightarrow\frac{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)}{(\kern-2.5pt(g+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\kern
1.0pt)\kern-2.5pt)}.$

Now by (2) we have
$(\kern-2.5pt(f+(\kern-2.5pt(\mathcal{S})\kern-2.5pt)\,)\kern-2.5pt)=(\kern-2.5pt(f,s\mid
s\in\mathcal{S})\kern-2.5pt)/(\kern-2.5pt(\mathcal{S})\kern-2.5pt)$, and
likewise for $g$. The statement follows by the third isomorphism theorem for
topological rings [Wa, 5.13]. ∎

##  3\. Jacobi Algebras

###  3.1. Differentiation

Consider the $\mathbb{C}$-linear maps
$\textnormal{\reflectbox{6}}_{i}\colon\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\to\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
which simply ‘strike off’ the leftmost $x_{i}$ of each monomial, in other
words act on monomials via the rule

$\textnormal{\reflectbox{6}}_{i}(m)=\begin{cases}n&\text{if }m=x_{i}n\\\
0&\text{otherwise.}\end{cases}$ (3.A)

The $\mathbb{C}$-linear cyclic symmetrisation map
$\operatorname{cyc}\colon\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\to\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
on monomials sends

$x_{i_{1}}\ldots x_{i_{t}}\mapsto\sum_{j=1}^{t}x_{i_{j}}x_{i_{j+1}}\ldots
x_{i_{t}}\cdot x_{i_{1}}\ldots x_{i_{j-1}}.$

Combining these two gives the cyclic derivatives. These are the
$\mathbb{C}$-linear maps
$\updelta_{i}\colon\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\to\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
which on monomials send

$x_{i_{1}}\ldots
x_{i_{t}}\mapsto\textnormal{\reflectbox{6}}_{i}\operatorname{cyc}(x_{i_{1}}\ldots
x_{i_{t}})=\sum_{j=1}^{t}\textnormal{\reflectbox{6}}_{i}(x_{i_{j}}x_{i_{j+1}}\ldots
x_{i_{t}}\cdot x_{i_{1}}\ldots x_{i_{j-1}}).$ (3.B)

######  Definition 3.1.

For
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
the Jacobi algebra is defined to be

$\mathcal{J}\mathrm{ac}(f)=\frac{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(\updelta_{1}f,\dots,\updelta_{d}f)\kern-2.5pt)}$

where
$(\kern-2.5pt(\updelta_{1}f,\dots,\updelta_{d}f)\kern-2.5pt)\,\reflectbox{\colonequals}\,(\kern-2.5pt(\updelta
f)\kern-2.5pt)$ is the closure of the two-sided ideal
$(\updelta_{1}f,\dots,\updelta_{d}f)$.

In general, the quotient of a complete topological ring by a closed ideal is
always separated, but it need not be complete.

######  Notation 3.2.

For any ring $R$, write $\mathfrak{J}(R)$ for its Jacobson radical. If $I$ is
any ideal of $R$ contained in $\mathfrak{J}(R)$, then
$\mathfrak{J}(R/I)=\mathfrak{J}(R)/I$ (see e.g. [L, 4.6]).

  1. (1)

It is clear that
$\mathfrak{J}(\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle)=\mathfrak{n}$.
If
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, then $(\kern-2.5pt(\updelta f)\kern-2.5pt)$ is contained in
$\mathfrak{n}$, and so
$\mathfrak{J}(\mathcal{J}\mathrm{ac}(f))=\mathfrak{n}/(\kern-2.5pt(\updelta
f)\kern-2.5pt)$, and furthermore
$\mathfrak{J(\mathcal{J}\mathrm{ac}(f))}^{n}=\big{(}\mathfrak{n}^{n}+(\kern-2.5pt(\updelta
f)\kern-2.5pt)\big{)}/(\kern-2.5pt(\updelta f)\kern-2.5pt)$ for $n\geq 2$.

  2. (2)

The topology on
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ is an
ideal topology generated by powers of $\mathfrak{n}$, so the natural quotient
topology on the quotient $\mathcal{J}\mathrm{ac}(f)$ is induced by powers of
the image of $\mathfrak{n}$ in the quotient [Wa, 5.5]. Thus, by (1), provided
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$ then the topology on both
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ and
$\mathcal{J}\mathrm{ac}(f)$ is the radical-adic topology. Since
$(\kern-2.5pt(\updelta f)\kern-2.5pt)$ is closed, $\mathcal{J}\mathrm{ac}(f)$
is Hausdorff [Wa, 5.7(1)]. Under extra assumptions it is also complete; see
8.4(3).

######  Remark 3.3.

A (polynomial or) power series $f=\sum
f_{i}\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
is called cyclically symmetric if $\operatorname{cyc}(f_{i})=if_{i}$ for each
graded piece $f_{i}\in\mathbb{C}\langle\mathsf{x}\rangle$. It is possible to
phrase the whole paper using only cyclically symmetric potentials, however
this becomes notationally unmanageable in §4–§6, since the property of being
cyclically symmetric is not preserved under change variables. Thus from the
viewpoint of NC singularity theory, it is much more natural to work with plain
old elements of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.
There are times when passing to cyclically symmetric potentials is convenient,
but this is confined entirely to §A.2.

###  3.2. Dimension

Being a quotient of formal noncommutative power series, determining which
dimension to use for $\mathcal{J}\mathrm{ac}(f)$ is a subtle point.

######  Definition 3.4.

For
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, we say that $\mathcal{J}\mathrm{ac}(f)$ has _polynomial growth_ (_with
respect to $\mathfrak{J}$_) if there exist $c,r\in\mathbb{R}$ such that
$\dim\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}^{n}\leq cn^{r}$ for all
$n\in\mathbb{N}$. In the case that $\mathcal{J}\mathrm{ac}(f)$ has polynomial
growth, then the _$\mathfrak{J}$ -dimension of $\mathcal{J}\mathrm{ac}(f)$_ is
the degree of that growth, precisely

$\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\colonequals\inf\left\\{r\in\mathbb{R}\mid\textrm{for
some $c\in\mathbb{R}$, }\dim\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}^{n}\leq
cn^{r}\text{ for every $n\in\mathbb{N}$}\right\\},$

and $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=\infty$ otherwise.

The $\mathfrak{J}$-dimension is analogous to the usual dimension of a
commutative noetherian local ring $(A,\mathfrak{m})$, defined as the degree of
the characteristic polynomial
$\chi_{\mathfrak{m}}(n)=\ell(A/\mathfrak{m}^{n})$, where, in that context, the
dimension is necessarily an integer [AM, 11.4, 11.14].

######  Lemma 3.5.

If $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\leq 1$, then either
$\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=0$ or $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=1$.

###### Proof.

Certainly
$\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}^{n+1}\twoheadrightarrow\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}^{n}$
for all $n\geq 1$, with equality if and only if
$\mathfrak{J}^{n+1}=\mathfrak{J}^{n}$. If each such map has nontrivial kernel,
then $\dim\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}^{n}\geq n$ and so
$\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\geq 1$. Otherwise, by
Nakayama’s Lemma, $\mathfrak{J}^{n}=0$ for some $n$, hence
$\mathfrak{n}^{n}\subset(\kern-2.5pt(\updelta f)\kern-2.5pt)$ and so
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)\leq\dim\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/\mathfrak{n}^{n}=2^{n}-1$
and $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=0$. ∎

######  Remark 3.6.

The $\mathfrak{J}$-dimension is used throughout, since it is better suited to
the complete local situation than the GK dimension [KL]. Indeed, it is well-
known that the GK dimension does not behave well with respect to completions.
For example, $\mathop{\rm GKdim}\nolimits\mathbb{C}[\\![x]\\!]=\infty$ whereas
$\mathop{\rm Jdim}\nolimits\mathbb{C}[\\![x]\\!]=1$. Compare [AB, §3.4], and
in particular [AB, §5.6]. Furthermore, $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=0$ if and only if
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)<\infty$, a property which does not
hold for GK dimension since $\mathcal{J}\mathrm{ac}(f)$ is not finitely
generated.

###  3.3. Equivalences and Isomorphisms

In what follows, recognising and producing isomorphisms of Jacobi algebras
will be key. The following techniques will be used extensively. The first is
trivial, but worth recording since it gives great flexibility in proofs; the
second two are more substantial with Part (2) being [DWZ, 3.7], and Part (3)
following from (2), together with 2.9. Recall the notation $f\sim g$ from 2.7.

######  Summary 3.7.

Suppose that
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.

  1. (1)

If $f$ cyclically permutes to $g$, so $f\sim g$, then
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(g)$.

  2. (2)

If
$\upvarphi\in\operatorname{Aut}\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
then $\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(g)$, where
$g=\upvarphi(f)$.

  3. (3)

Set $\mathsf{f}_{1}=f$. If there exist $\mathsf{f}_{2},\mathsf{f}_{3},\ldots$
and automorphisms $\upvarphi^{1},\upvarphi^{2},\ldots$ such that

    1. (a)

every $\upvarphi^{i}$ is unitriangular of depth of $\geq i$, and

    2. (b)

$\upvarphi^{i}(\mathsf{f}_{i})-\mathsf{f}_{i+1}\in\\{\kern-3.0pt\\{\mathcal{A},\mathcal{A}\\}\kern-3.0pt\\}\cap\mathfrak{n}^{i+1}$,
for all $i\geq 1$,

then the sequence $(\mathsf{f}_{i})_{i\geq 1}$ converges and
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(g)$ where
$g=\lim\mathsf{f}_{i}$.

######  Lemma 3.8.

Let
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
and
$m\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
be a monomial. Then the following hold.

  1. (1)

$\operatorname{cyc}(m)\sim\deg(m)m$.

  2. (2)

If $f$ contains $\uplambda m$, then $f\sim f+\uplambda\left(\frac{1}{\deg
m}\operatorname{cyc}(m)-m\right)$.

  3. (3)

Let $h$ be the sum of terms of $f$ whose monomials appear in
$\operatorname{cyc}(m)$. Then

$f\sim f-h+\upalpha\operatorname{cyc}(m)\sim f-h+\upalpha\deg(m)\,m,$

for some $\upalpha\in\mathbb{C}$.

###### Proof.

Writing $m=m_{1}m_{2}\ldots m_{r}$, where each $m_{i}$ is a variable
$x_{j(i)}$, we have

$\displaystyle rm-\operatorname{cyc}(m)$ $\displaystyle=(r-1)m_{1}\ldots
m_{r}-m_{2}\ldots m_{r}m_{1}$ $\displaystyle\qquad\qquad-m_{3}\ldots
m_{r}m_{1}m_{2}-\cdots-m_{r}m_{1}\ldots m_{r-1}$
$\displaystyle=[m_{1},m_{2}\ldots m_{r}]+[m_{1}m_{2},m_{3}\ldots
m_{r}]+\cdots+[m_{1}\ldots m_{r-1},m_{r}]$

and (1) follows. (2) follows at once from (1). The final claim (3) follows by
applying (2) to each monomial of $h$ in turn. ∎

Below it will be convenient to work with the following three equivalence
relations.

######  Definition 3.9.

For elements
$f,g\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
(recall and) define

  1. (1)

$f\sim g$ if
$f-g\in\\{\kern-3.0pt\\{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle,\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\\}\kern-3.0pt\\}$
(see 2.7).

  2. (2)

$f\simeq g$ if there is an equality of ideals
$(\kern-2.5pt(\updelta_{1}f,\dots,\updelta_{d}f)\kern-2.5pt)=(\kern-2.5pt(\updelta_{1}g,\dots,\updelta_{d}g)\kern-2.5pt)$.

  3. (3)

$f\cong g$ if there is an isomorphism of algebras
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(g)$.

Clearly $f\sim g$ implies $f\simeq g$ implies $f\cong g$, but the converse
implications do not hold. The relation $\sim$ is additive by definition, but
$\simeq$ is not: $x^{2}+y^{3}\simeq x^{2}+2y^{3}$ but $x^{2}\not\simeq
x^{2}+y^{3}$.

The Jacobi isomorphism relation $\cong$ is the equivalence relation that we
will classify up to, but the others help understand the structure of the
various arguments. For example, by 2.9, the symmetrisation relation $\sim$
behaves well in limits. It appears to permit creation from the void, in the
sense that $0\sim xy-yx$, but of course this form has all derivatives zero, so
does not contribute to Jacobi ideals. The relation $\simeq$ is useful for
cancelling high order terms in potentials (see e.g. the proof of 6.5), whereas
$\cong$ is most suited to, and is often a by-product of, analytic changes in
coordinates.

##  4\. NC Singularity Theory 101

###  4.1. Corank and the Splitting Lemma

The closed vector subspace of commutators
$\\{\kern-3.0pt\\{\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle,\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\\}\kern-3.0pt\\}$
generates the much larger closed ideal of commutators, and the quotient of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ by
this ideal is the ring of commutative power series
$\mathbb{C}[\\![x_{1},\dots,x_{d}]\\!]$. The quotient, or ‘abelianisation’,
map
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\to\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]$
written $g\mapsto g^{\mathop{\rm ab}\nolimits}$ simply takes the expression
for $g$ to the same expression in the commutative ring.

######  Lemma 4.1.

With notation as above, the following hold:

  1. (1)

The abelianisation map
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle\to\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]$
is continuous and surjective.

  2. (2)

For any
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$,
the map $f\mapsto f^{\mathop{\rm ab}\nolimits}$ descends to a surjection

$\mathcal{J}\mathrm{ac}(f)\twoheadrightarrow\frac{\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]}{(\kern-2.5pt(\textnormal{\reflectbox{6}}f^{\mathop{\rm
ab}\nolimits}/\textnormal{\reflectbox{6}}x_{i}\mid
i=1,\ldots,n)\kern-2.5pt)}.$ (4.A)

###### Proof.

(1) At the level of ideals,
$\mathfrak{n}^{k}\twoheadrightarrow\mathfrak{n}_{\mathop{\rm
ab}\nolimits}^{k}$ for every $k\geq 0$, since abelianisation is a ring
homomorphism mapping each $x_{i}$ to $x_{i}$.

(2) Since $(\updelta_{i}f)^{\mathop{\rm
ab}\nolimits}=\textnormal{\reflectbox{6}}f^{\mathop{\rm
ab}\nolimits}/\textnormal{\reflectbox{6}}x_{i}$, where
$\textnormal{\reflectbox{6}}/\textnormal{\reflectbox{6}}x_{i}$ is the usual
differentiation of commutative functions, surjectivity at the level of
(unclosed) Jacobian ideals follows. Since the abelianisation map is continuous
and surjective by (1), this passes to their closures, as claimed. ∎

Below we will consider

$\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm
ab}\nolimits}\colonequals\frac{\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]}{(\kern-2.5pt(\textnormal{\reflectbox{6}}f^{\mathop{\rm
ab}\nolimits}/\textnormal{\reflectbox{6}}x_{i}\mid
i=1,\ldots,n)\kern-2.5pt)}=\frac{\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]}{(\textnormal{\reflectbox{6}}f^{\mathop{\rm
ab}\nolimits}/\textnormal{\reflectbox{6}}x_{i}\mid i=1,\ldots,n)}$

where, since $\mathbb{C}[\\![x_{1},\ldots,x_{d}]\\!]$ is commutative
noetherian, all ideals are closed [M1, 8.1(1)].

######  Remark 4.2.

In classical singularity theory, for
$g\in\mathbb{C}[\\![x_{1},\dots,x_{d}]\\!]$ both the Milnor algebra
$\mathbb{C}[\\![x_{1},\dots,x_{d}]\\!]/(\updelta_{1}g,\dots,\updelta_{d}g)$
and the Tjurina algebra
$\mathbb{C}[\\![x_{1},\dots,x_{d}]\\!]/(g,\updelta_{1}g,\dots,\updelta_{d}g)$
are defined, and play a major role. In the noncommutative setting, the
analogous Tjurina algebra is not well defined on $\sim$ classes. For example,
the potentials $0\sim xy-yx$ determine the same Jacobi algebra, but their
naively-defined Tjurina algebras are $\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ and $\mathbb{C}[\\![x,y]\\!]$ respectively. To
have any hope of classifying elements in the completed free algebra, some
identification is required, and for us identifying $\sim$ classes is essential
for applications. Compare [HZ], where the lack of a noncommutative Tjurina
algebra motivates the use of Hochschild classes to generalise Saito’s theorem
on homogeneous potentials.

######  Definition 4.3.

For
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, the corank of $f$ is defined to be

$\mathcal{C}\mathrm{rk}(f)=\dim_{\mathbb{C}}\left(\frac{\mathfrak{J}}{\mathfrak{J}^{2}}\right)$

where $\mathfrak{J}$ is the Jacobson radical of $\mathcal{J}\mathrm{ac}(f)$.

######  Remark 4.4.

Clearly $0\leq\mathcal{C}\mathrm{rk}(f)\leq d$. Since
$\mathfrak{J}/\mathfrak{J^{2}}\cong(\mathfrak{n}+I)/(\mathfrak{n}^{2}+I)$,
where $I=(\kern-2.5pt(\updelta f)\kern-2.5pt)$, the exactness of the sequence
of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/\mathfrak{n}=\mathbb{C}$-vector
spaces

$0\longrightarrow\frac{\mathfrak{n}^{2}+I}{\mathfrak{n}^{2}}\longrightarrow\frac{\mathfrak{n}}{\mathfrak{n}^{2}}\longrightarrow\frac{\mathfrak{n}+I}{\mathfrak{n}^{2}+I}\longrightarrow
0$

shows that
$\mathcal{C}\mathrm{rk}(f)=d-\dim_{\mathbb{C}}\left(\frac{\mathfrak{n}^{2}+I}{\mathfrak{n}^{2}}\right)$,
so that the corank is determined by the linear conditions imposed by
derivatives, and is therefore uniquely determined by $f_{2}$.

######  Theorem 4.5 (Splitting Lemma).

Let
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.
Then $f\cong x_{1}^{2}+\cdots+x_{r}^{2}+g$ for some
$g\in\mathbb{C}\langle\kern-2.5pt\langle
x_{r+1},\ldots,x_{d}\rangle\kern-2.5pt\rangle_{\geq 3}$, where
$d-r=\mathcal{C}\mathrm{rk}(f)$. In particular,

$\mathcal{J}\mathrm{ac}(f)\cong\frac{\mathbb{C}\langle\kern-2.5pt\langle
x_{r+1},\ldots,x_{d}\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(\updelta_{x_{r+1}}g,\ldots,\updelta_{x_{d}}g)\kern-2.5pt)}.$

###### Proof.

This is [DWZ, 4.5] for the $d$-loop quiver. Since $\operatorname{ord}g\geq 3$,
the derivatives of $g$ impose no linear conditions, so necessarily
$d-r=\mathcal{C}\mathrm{rk}(f)$. ∎

###  4.2. Golod–Shafarevich–Vinberg

The classical approach to growth of algebras comes from the Golod–Shafarevich
theorem [GS], adapted by Vinberg [V3] to power series; see also [E2]. This
result constrains $f$ to achieve $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)<\infty$, and we develop a stronger
version in 4.7 adapted to Jacobi algebras.

######  Theorem 4.6 (Golod–Shafarevich, Vinberg).

Let
$I=(g_{1},\dots,g_{s})\subset\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
be an ideal, set $r_{i}=\operatorname{ord}g_{i}$ for each $i=1,\dots,s$, and
write $h=1-dt+t^{r_{1}}+\cdots+t^{r_{s}}\in\mathbb{R}[\\![t]\\!]$. If the
coefficients of $(1-t)/h$ are non negative, then
$\dim_{\mathbb{C}}\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(I)\kern-2.5pt)=\infty$.

In most cases where the result applies, one can in fact show exponential
growth. The Golod–Shafarevich–Vinberg estimates readily show that $\mathop{\rm
Jdim}\nolimits\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/(\kern-2.5pt(g_{1},\dots,g_{d})\kern-2.5pt)=\infty$
in the following cases:

  1. (1)

$d=2$ with either $r_{1}\geq 3$, $r_{2}\geq 8$, or $r_{1}\geq 4$, $r_{2}\geq
5$.

  2. (2)

$d=3$ with either $r_{1}\geq 2$, $r_{2},r_{3}\geq 3$, or $r_{1}=r_{2}=2$,
$r_{3}\geq 5$.

  3. (3)

$d\geq 4$ with $r_{i}\geq 2$ for every $i$.

For example, in the case $d=4$, it is sufficient to observe the exponential
growth of

$(1-4t+4t^{2})^{-1}=1+3t+8t^{2}+20t^{3}+48t^{4}+\cdots+(1+k)2^{k}t^{k}+\cdots$

as this bounds the growth of the algebra from below in the case of an order
$3$ potential with four order $2$ derivatives.

Setting aside quadratic terms by the Splitting Lemma, this then puts
constraints on the motivating problem 1.1. Indeed, if
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
3}$ and $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)<\infty$, then
either

  1. (1)

$d=2$, $\operatorname{ord}f\leq 5$ and $f_{\leq 5}\not\sim\ell^{5}$ for a
linear form $\ell=\ell(x_{1},x_{2})$, or

  2. (2)

$d=3$, $\operatorname{ord}f=3$, and $f_{3}\not\sim\ell^{3}$ for a linear form
$\ell=\ell(x_{1},x_{2},x_{3})$.

It turns out that these estimates can be substantially improved, but this
requires much more work. Iyudu and collaborators [ISm, IS2] introduce several
new ideas that exploit the Jacobi structure; in Appendix A we extend their
techniques into the power series context, and establish the following. Recall
that $\mathsf{x}=x_{1},\ldots,x_{d}$.

######  Theorem 4.7 (A.13).

Suppose that $d=2$ and $k\geq 4$, or $d\geq 3$ and $k\geq 3$. If
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
has order $k$, then $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\geq
3$.

######  Remark 4.8.

Together with the Splitting Lemma, the above 4.7 reduces the classification of
those $f$ satisfying $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\leq
1$ to the case of two variables ($d=2$).

###  4.3. Higher Coranks

Higher-degree versions of the corank exist, and contain more detailed
information about Jacobi algebras.

######  Definition 4.9.

Let
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$. For $m\geq 1$, the $m$th corank of $f$ is defined to be

$\mathcal{C}\mathrm{rk}_{m}(f)=\dim_{\mathbb{C}}\left(\frac{\mathfrak{J}^{m}}{\mathfrak{J}^{m+1}}\right),$

where $\mathfrak{J}$ is the Jacobson radical of $\mathcal{J}\mathrm{ac}(f)$.
We also define
$\mathcal{C}\mathrm{rk}_{0}(f)=\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)/\mathfrak{J}=1$
and note that $\mathcal{C}\mathrm{rk}_{1}(f)=\mathcal{C}\mathrm{rk}(f)$.

######  Remark 4.10.

Since
$\mathfrak{J}^{m}/\mathfrak{J}^{m+1}\cong(\mathfrak{n}^{m}+I)/(\mathfrak{n}^{m+1}+I)$,
the exactness of the sequence of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle/\mathfrak{n}=\mathbb{C}$-vector
spaces

$0\to\frac{\mathfrak{n}^{m}\cap I}{\mathfrak{n}^{m+1}\cap
I}\cong\frac{(\mathfrak{n}^{m}\cap
I)+\mathfrak{n}^{m+1}}{\mathfrak{n}^{m+1}}\to\frac{\mathfrak{n}^{m}}{\mathfrak{n}^{m+1}}\to\frac{\mathfrak{n}^{m}+I}{\mathfrak{n}^{m+1}+I}\to
0$

shows that
$\mathcal{C}\mathrm{rk}_{m}(f)=d^{m}-\dim_{\mathbb{C}}\left(\frac{\mathfrak{n}^{m}\cap
I}{\mathfrak{n}^{m+1}\cap I}\right)$; compare [V3, (4)]. Thus the $m$th corank
is determined by the conditions imposed on the leading terms of elements of
the Jacobian ideal of order exactly $m$. In particular,
$0\leq\mathcal{C}\mathrm{rk}_{m}(f)\leq d^{m}$. If $\operatorname{ord}(f)\geq
m+1$, then

$\frac{(\mathfrak{n}^{m}\cap
I)+\mathfrak{n}^{m+1}}{\mathfrak{n}^{m+1}}\cong\frac{\mathfrak{n}^{m+1}+I}{\mathfrak{n}^{m+1}}$

matching (1.A), (1.B), and 4.4.

By definition, the $\mathfrak{J}$-dimension is the growth of the sum of
coranks. Calculating the $m$th corank is not necessarily straightforward:
essentially it amounts to calculating a Gröbner basis of the Jacobian ideal
with a local monomial order to at least order $m$.

To study Jacobi algebras $\mathcal{J}\mathrm{ac}(f)$ of
$\mathfrak{J}$-dimension $\leq 1$, 4.7 constrains the number of variables to
$d\leq 2$ and $k=\operatorname{ord}(f)\leq 3$. The corank controls the rank of
$f_{2}$. The main case is when $f_{2}=0$, when it is clear that
$2\leq\mathcal{C}\mathrm{rk}_{2}(f)\leq 4$. The two derivatives
$\updelta_{x}f_{3}$ and $\updelta_{y}f_{3}$ are linearly independent when
$\mathcal{C}\mathrm{rk}_{2}(f)=2$ and they are dependent when
$\mathcal{C}\mathrm{rk}_{2}(f)=3$. The case $\mathcal{C}\mathrm{rk}_{2}(f)=4$
holds only when $f_{3}=0$, which is ruled out by 4.7.

This provides a numerical characterisation of the ADE types. The first Type A
case is when $\mathcal{C}\mathrm{rk}(f)=0$, in which case
$\mathcal{J}\mathrm{ac}(f)\cong\mathbb{C}$. In addition to this, if
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$
has $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\leq 1$, we say $f$
has Type A, D or E according to the following table.

Type | $\mathcal{C}\mathrm{rk}(f)$ | $\mathcal{C}\mathrm{rk}_{2}(f)$
---|---|---
A | 1 | 1
D | 2 | 2
E | 2 | 3

The higher coranks provide much more detail. In Type A they provide enough
information to classify up to isomorphism, however in Type D this is not true.

######  Example 4.11.

Consider the families $D_{n,\infty}$ and $D_{n,m}$ from the introduction. The
higher coranks are given by the following table.

Type | $f$ |  Conditions |  $\mathcal{C}\mathrm{rk}_{i}(f)$, $i=0,1,\dots$
---|---|---|---
$D_{n,\infty}$ | $xy^{2}+x^{2n}$ |  $n\geq 2$ |  $1,\overbrace{2,\dots,2}^{2n-1},\overbrace{1,\dots,1}^{2n-2}$
$D_{n,m}$ | $xy^{2}+x^{2n}+x^{2m+1}$ |  $2n-2\geq m\geq n\geq 2$
$D_{n,m}$ | $xy^{2}+x^{2m+1}+x^{2n}$ |  $n>m\geq 1$ |  $1,\overbrace{2,\dots,2}^{2n-1},\overbrace{1,\dots,1}^{2m-1}$

In particular $\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=6n-3$ in the first
families, which is independent of $m$, whilst
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=4n+2m-2$ in the final family.

###  4.4. Linear Changes in Coordinates and Discriminants

In light of 4.8, from §6 onwards we work in two non-commuting variables $x$
and $y$.

The following is an immediate consequence of the Splitting Lemma and
abelianisation.

######  Lemma 4.12.

Let $f\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 2}$ with
$f_{2}=ax^{2}+b_{1}xy+b_{2}yx+cy^{2}\not\sim 0$. Set $b=b_{1}+b_{2}$ and
consider the discriminant $\Delta=b^{2}-4ac$. Then $f\cong g$, for some
$g\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
2}$ with

$g_{2}=\begin{cases}x^{2}+y^{2}&\text{if }\Delta\neq 0\\\ x^{2}&\text{if
}\Delta=0.\end{cases}$

As for the quadratic forms above, up to $\sim$ we may commute variables
appearing in cubic forms in $\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$, and we use this to simplify the statement of
the following lemma, writing $bx^{2}y$ rather than
$b_{1}x^{2}y+b_{2}xyx+\dots$, and so on. Note that, in general, cyclic
equivalence no longer simulates commutativity in higher degree, as $xyxy\nsim
x^{2}y^{2}$.

######  Lemma 4.13.

[I] Let $f\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$ with $f_{3}\sim
ax^{3}+bx^{2}y+cxy^{2}+dy^{3}$ for some $a,b,c,d\in\mathbb{C}$, not all zero.
Let $\Delta=-27a^{2}d^{2}+18abcd-4ac^{3}-4b^{3}d+b^{2}c^{2}\in\mathbb{C}$ be
the cubic discriminant. Then $f\cong g$, for some
$g\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
3}$ with

$g_{3}=\begin{cases}x^{3}+y^{3}&\text{if }\Delta\neq 0\\\ x^{2}y&\text{if
}\Delta=0\text{ and }\begin{cases}a(b^{2}-3ac)\neq 0\text{ or }\\\
(c^{2}-3bd)d\neq 0\text{ or }\\\ a=d=0\end{cases}\\\
x^{3}&\text{otherwise}.\\\ \end{cases}$

Thus these three leading cubic normal forms are characterised by whether
$f_{3}^{\mathop{\rm ab}\nolimits}$ has three, two or one distinct factors
respectively.

###### Proof.

Consider the linear automorphism of $\mathbb{C}[x,y]$

$x\mapsto\upalpha x+\upbeta y,\qquad y\mapsto\upgamma x+\updelta
y\qquad\text{for $\upalpha,\upbeta,\upgamma,\updelta\in\mathbb{C}$}$ (4.B)

that maps $(f_{3})^{\mathop{\rm ab}\nolimits}\in\mathbb{C}[x,y]$ to one of the
normal forms $x^{3}+y^{3}$, $xy^{2}$ or $x^{3}$. The choice of normal form is
determined by the cubic determinant. The additional conditions on the
coefficients in the statement are simply that $p\neq 0$ in the depressed form
after completing the cube $x^{3}+pxy^{2}+qy^{3}$, in which
$\Delta=-4p^{3}-27q^{2}$, and accounting for the fact that $a=0$ or $d=0$ or
both are possible.

Let $\upvarphi$ be the linear automorphism of
$\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle$ defined by
the same formula (4.B) and $g_{3}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ be the corresponding cubic normal form. Then
$\upvarphi(f_{3})\sim g_{3}$, although they are not equal, so
$\upvarphi(f)\sim g_{3}+\upvarphi(f_{\geq 4})$. Thus $f\cong\upvarphi(f)\cong
g_{3}+\upvarphi(f_{\geq 4})$, as claimed. ∎

##  5\. Type $A$ and Commutativity

This section considers the most elementary situation, namely
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$ with large quadratic part. Normal forms are established in §5.1. Together
with the linear coordinate changes from §4.4, this proves in §5.2 that for any
$f\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
2}$, the algebra $\mathcal{J}\mathrm{ac}(f)$ is commutative if and only if $f$
has corank at most $1$. This fact is used in later sections.

###  5.1. Normal Forms of Type A

It is notationally convenient to identify $y=x_{d}$ and work in the ring
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle=\mathbb{C}\langle\kern-2.5pt\langle
x_{1},\dots,x_{d-1},y\rangle\kern-2.5pt\rangle$.

######  Theorem 5.1.

If
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$ with $\mathcal{C}\mathrm{rk}(f)\leq 1$, then there is a unique polynomial
$g$ of the form $g=x_{1}^{2}+\cdots+x_{d-1}^{2}+\upvarepsilon y^{n}$ for some
$n\geq 2$ and $\upvarepsilon\in\\{0,1\\}$ such that $f\cong g$.

  1. (1)

If $\upvarepsilon=1$, then $\mathcal{J}\mathrm{ac}(f)$ is commutative with
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=n-1$.

  2. (2)

If $\upvarepsilon=0$, then $\mathcal{J}\mathrm{ac}(f)$ is commutative with
$\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=1$.

###### Proof.

By the Splitting Lemma 4.5, there is
$\mathsf{f}\in\mathbb{C}\langle\kern-2.5pt\langle
x_{1},\dots,x_{d-1},y\rangle\kern-2.5pt\rangle$ with $f\cong\mathsf{f}$ and
either

$\mathsf{f}=\begin{cases}x_{1}^{2}+\cdots+x_{d-1}^{2}+y^{2}&\text{if
}\mathcal{C}\mathrm{rk}(f)=0\\\ x_{1}^{2}+\cdots+x_{d-1}^{2}+q(y)&\text{if
}\mathcal{C}\mathrm{rk}(f)=1\end{cases}$

for some $q\in\mathbb{C}[\\![y]\\!]$ with $\operatorname{ord}(q)\geq 3$.

If $q$ is zero we are done, else after pulling out the lowest term, we can
write $q=y^{n}\mathsf{u}$ for some
$\mathsf{u}=c_{n}+c_{n+1}y+\ldots\in\mathbb{C}[\\![y]\\!]$ with $c_{n}\neq 0$.
The homomorphism $\mathbb{C}\langle\kern-2.5pt\langle
x_{1},\dots,x_{d-1},y\rangle\kern-2.5pt\rangle\to\mathbb{C}\langle\kern-2.5pt\langle
x_{1},\dots,x_{d-1},y\rangle\kern-2.5pt\rangle$ which sends $x_{k}\mapsto
x_{k}$ and $y\mapsto y\sqrt[n]{\mathsf{u}}$ is an automorphism. Since
$\sqrt[n]{\mathsf{u}}$ is a power series only in $y$, it commutes with $y$,
and so this automorphism sends $\sum x_{i}^{2}+y^{n}$ to $\sum
x_{i}^{2}+y^{n}\mathsf{u}=\mathsf{f}$. Hence $\mathcal{J}\mathrm{ac}(\sum
x_{i}^{2}+y^{n})\cong\mathcal{J}\mathrm{ac}(\mathsf{f})\cong\mathcal{J}\mathrm{ac}(f)$,
as required.

Parts (1)–(2) are obvious, since $\mathcal{J}\mathrm{ac}(\sum
x_{i}^{2}+y^{n})\cong\mathbb{C}[\\![y]\\!]/(y^{n-1})$ and
$\mathcal{J}\mathrm{ac}(\sum x_{i}^{2})\cong\mathbb{C}[\\![y]\\!]$, and
uniqueness then follows since $\sum x_{i}^{2}+y^{n_{1}}\cong\sum
x_{i}^{2}+y^{n_{2}}$ if and only if $n_{1}=n_{2}$. ∎

Recall from the introduction our geometric applications in the setting of cDV
singularities. These correspond to Jacobi algebras in two non-commutating
variables, so we set $d=2$ and write the variables as $x,y$.

######  Corollary 5.2.

Every $\mathcal{J}\mathrm{ac}(f)$, where
$f\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
2}$ with $f_{2}\not=0$, is geometric.

###### Proof.

Consider $\mathcal{R}=\mathbb{C}[\\![x,y,z]\\!]^{\frac{1}{2}(1,1,0)}$, and its
unique crepant resolution $\mathcal{X}\to\mathop{\rm
Spec}\nolimits\mathcal{R}$. This has contraction algebra
$\mathbb{C}[\\![y]\\!]\cong\mathcal{J}\mathrm{ac}(x^{2})$, realising the
second case in 5.1. On the other hand, by [DW1, 3.10] the Type $A$ $m$-Pagoda
flop (with $m\geq 1$) has contraction algebra
$\mathbb{C}[\\![y]\\!]/y^{m}\cong\mathcal{J}\mathrm{ac}(x^{2}+y^{m+1})$, which
realises the infinite family in 5.1. ∎

######  Example 5.3.

Consider
$f=x^{2}+\tfrac{2}{3}(xy^{2}+yxy+y^{2}x)\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$. This has $3$-dimensional Jacobi algebra

$\mathcal{J}\mathrm{ac}(f)\cong\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle/(\kern-2.5pt(x+y^{2},xy+yx)\kern-2.5pt)\cong\mathbb{C}[y]/y^{3}$

so, by 5.1 or Reid’s Pagoda [R2], $f$ gives the same Jacobi algebra as
$g=x^{2}+y^{4}$. Commutatively, one would see this by completing the square,
but that automorphism does not work directly in the noncommutative context:
$x\mapsto x-\frac{2}{3}y^{2}$ gives $f\mapsto
x^{2}+\frac{2}{3}yxy-\frac{8}{9}y^{4}$, and we cannot attack the $yxy$ term by
coordinate changes that preserve $f_{2}=x^{2}$. But $f\sim
x^{2}+xy^{2}+y^{2}x$, which then allows us to complete the square (and a
scalar on $y$) to conclude. This exemplifies the way $\sim$ helps to navigate
the Jacobi isomorphism classes.

###  5.2. Commutativity

The following characterisation of commutative Jacobi algebras in $d=2$
variables will be used later.

######  Proposition 5.4.

$f\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
2}$, then $\mathcal{J}\mathrm{ac}(f)$ is commutative if and only if
$\mathcal{C}\mathrm{rk}(f)\leq 1$.

###### Proof.

($\Leftarrow$) is clear from 5.1. For ($\Rightarrow$), we prove the
contrapositive. If $\mathcal{C}\mathrm{rk}(f)\not\leq 1$, then $f_{2}=0$, and
we need to prove that $\mathcal{J}\mathrm{ac}(f)$ is not commutative. For
this, it suffices to exhibit a factor that is not commutative. By 4.13 without
loss of generality we can assume that $f$ equals

$x^{3}+y^{3}+\mathcal{O}_{4},\quad xy^{2}+\mathcal{O}_{4},\quad
x^{3}+\mathcal{O}_{4}\quad\mbox{or}\quad\mathcal{O}_{4}.$

Write $\mathcal{M}_{3}$ for the set of all noncommutative monomials of degree
$3$, then factor by the ideal
$(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f,\mathcal{M}_{3})\kern-2.5pt)/(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$
in $\mathcal{J}\mathrm{ac}(f)$. But in the four cases above, by
differentiating then using the third isomorphism theorem it follows that
$\mathcal{J}\mathrm{ac}(f)$ is one of

$\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(x^{2},y^{2},\mathcal{M}_{3})\kern-2.5pt)},\quad\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(y^{2},xy+yx,\mathcal{M}_{3})\kern-2.5pt)},\quad\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(x^{2},\mathcal{M}_{3})\kern-2.5pt)},\quad\mbox{or}\quad\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(\mathcal{M}_{3})\kern-2.5pt)}.$

None of these factors is commutative, and so $\mathcal{J}\mathrm{ac}(f)$ is
not commutative. ∎

##  6\. Type $D$ normal forms

This section considers the next case, namely those
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$ with $\mathcal{C}\mathrm{rk}(f)=2$ and $\mathcal{C}\mathrm{rk}_{2}(f)=2$.
Reducing to two variables by the Splitting Lemma 4.5, the assumption
$\mathcal{C}\mathrm{rk}_{2}(f)=2$ is then equivalent to the first two cases in
4.13, namely those $f\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$ with $f_{3}\neq 0$ for which
$f_{3}^{\mathop{\rm ab}\nolimits}$ has either two or three distinct linear
factors. Full normal forms are obtained in both situations, and are then
merged into a unified form in §6.4. These are the Type $D$ normal forms in the
tables in §1.2.

Throughout this section, it will be convenient to adopt the following
language.

######  Definition 6.1.

We say that a monomial $m\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ contains $x^{2}$ if $m\sim nx^{2}$ for some
monomial $n$, else $m$ does not contain $x^{2}$. Similarly, an element
$f\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle$
contains $x^{2}$ if for some (nonzero) term $\uplambda m$ of $f$, the monomial
$m$ contains $x^{2}$, else $f$ does not contain $x^{2}$. We also use the
analogous expressions for $y^{2}$.

###  6.1. Abelianized Cubic with Three Factors

This subsection considers the case of 4.13 where $f_{3}^{\mathop{\rm
ab}\nolimits}$ has three distinct factors, that is $f\cong
x^{3}+y^{3}+\mathcal{O}_{4}$, and in 6.5 and 6.6 provides two different, but
equivalent, normal forms.

Recall from 2.2(1) that a unitriangular automorphism $\upvarphi$ of
$\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$ has
depth $e\geq 1$ if and only if $\upvarphi(f)_{\leq e}=f_{\leq e}$ for all
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle$.

######  Lemma 6.2.

Fix $t\geq 4$, and let $f=x^{3}+y^{3}+f_{4}+\cdots+f_{t}+\mathcal{O}_{t+1}$.
For any $h_{1},h_{2}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ with $\operatorname{ord}(h_{i})\geq t-2$, there
is a unitriangular automorphism $\uppsi$ of depth $\geq t-3$ such that

$\uppsi(f)\sim
x^{3}+y^{3}+f_{4}+\cdots+f_{t-1}+\left(f_{t}-h_{1}x^{2}-h_{2}y^{2}\right)+\mathcal{O}_{t+1}$

and $\uppsi(f)-(x^{3}+y^{3}+f_{4}+\cdots+f_{t-1})\in\mathfrak{n}^{t}$.

###### Proof.

Consider the unitriangular automorphism $\uppsi$ which sends $x\mapsto
x-\tfrac{1}{3}h_{1}$, $y\mapsto y-\tfrac{1}{3}h_{2}$. The result follows since

$\uppsi(x^{3}+y^{3})\sim
x^{3}-h_{1}x^{2}+\tfrac{1}{3}h_{1}^{2}x-\tfrac{1}{27}h_{1}^{3}+y^{3}-h_{2}y^{2}+\tfrac{1}{3}h_{2}^{2}y-\tfrac{1}{27}h_{2}^{3},$

and $\uppsi(m)\equiv m\mod\mathfrak{n}^{t+1}$ whenever $\deg(m)\geq 4$. ∎

With the preparatory lemma in place, the strategy is to first find a standard
power series form of each potential, and then distill that down to a
polynomial normal form.

######  Proposition 6.3.

Suppose that $f=f_{3}+\mathcal{O}_{4}$ where $f_{3}^{\mathop{\rm
ab}\nolimits}$ has three distinct factors. Then $f\cong x^{3}+y^{3}+p(xy)$ for
some power series $p(z)\in\mathbb{C}[\\![z]\\!]$ with
$\operatorname{ord}(p)\geq 2$.

Recall the Conventions 1.7 on denoting graded pieces of sequence elements: we
denote sequence elements $\mathsf{f}_{n}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ in Greek font, and we write
$(\mathsf{f}_{n})_{t}$ for its degree $t$ piece, and $(\mathsf{f}_{n})_{<t}$
and $(\mathsf{f}_{n})_{>t}$ for sub- and super-degree $t$ portions
respectively.

###### Proof.

We construct a sequence of power series $\mathsf{f}_{1},\mathsf{f}_{2},\dots$
and unitriangular automorphisms $\upvarphi^{1},\upvarphi^{2},\dots$
inductively, with each $\mathsf{f}_{t}$ having the form of the target power
series $x^{3}+y^{3}+p(xy)$ in small degree. Summary 3.7(3) then constructs
$\mathsf{f}=\lim\mathsf{f}_{i}$ of the required form with
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(\mathsf{f})$.

By 4.13, $f\cong g$ where $g_{3}=x^{3}+y^{3}$. After grouping together terms
containing $x^{2}$ or $y^{2}$ and cyclically permuting, we may write

$g\sim x^{3}+y^{3}+\mathsf{h}_{2}\cdot x^{2}+\mathsf{h}_{2}^{\prime}\cdot
y^{2}+\upmu_{4}(xy)^{2}+\mathcal{O}_{5}$

for
$\mathsf{h}_{2},\mathsf{h}_{2}^{\prime}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{2}$ and $\upmu_{4}\in\mathbb{C}$.

Hence we begin the induction by setting

$\mathsf{f}_{1}=x^{3}+y^{3}+\mathsf{h}_{2}\cdot
x^{2}+\mathsf{h}_{2}^{\prime}\cdot y^{2}+\upmu_{4}(xy)^{2}+g_{\geq 5}$

and note that $\mathsf{f}_{1}\sim g\cong f$. Thus $\mathsf{f}_{1}$ is in the
desired form in degrees $\leq 3$ and has its degree 4 piece prepared in
standard form for further analysis.

For the inductive step more generally, we may suppose that
$\mathsf{f}_{t}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ has been constructed of the form

$\mathsf{f}_{t}=\big{(}x^{3}+y^{3}+\mathsf{p}_{t+2}(xy)\big{)}+\big{(}\mathsf{h}_{t+1}\cdot
x^{2}+\mathsf{h}_{t+1}^{\prime}\cdot
y^{2}+\upmu_{t+3}(xy)^{\lfloor(t+3)/2\rfloor}\big{)}+\mathcal{O}_{t+4}$

with $\mathsf{p}_{3}=0$ and by convention $\upmu_{t+3}=0$ for even $t$, where

  1. (1)

$(\mathsf{f}_{t})_{\leq t+2}=x^{3}+y^{3}+\mathsf{p}_{t+2}(xy)$, for some
polynomial $\mathsf{p}_{t+2}\in\mathbb{C}[z]_{\geq 2}$ of degree
$\leq(t+2)/2$, where the polynomials $\mathsf{p}_{3},\dots,\mathsf{p}_{t+2}$
satisfy $\mathsf{p}_{i+1}=\mathsf{p}_{i}$ for even $i$ and
$\mathsf{p}_{i+1}-\mathsf{p}_{i}=\upmu_{i+1}z^{(i+1)/2}$ for odd $i$, and

  2. (2)

$(\mathsf{f}_{t})_{t+3}=\mathsf{h}_{t+1}\cdot
x^{2}+\mathsf{h}_{t+1}^{\prime}\cdot
y^{2}+\upmu_{t+3}(xy)^{\lfloor(t+3)/2\rfloor}$ for some homogeneous forms
$\mathsf{h}_{t+1}$, $\mathsf{h}_{t+1}^{\prime}$ of degree $t+1$.

Applying 6.2 with $h_{1}=\mathsf{h}_{t+1}$ and
$h_{2}=\mathsf{h}_{t+1}^{\prime}$, there exists a unitriangular
$\upvarphi^{t}$ of depth $\geq t$ such that

$\upvarphi^{t}(\mathsf{f}_{t})\sim\big{(}x^{3}+y^{3}+\mathsf{p}_{t+2}(xy)\big{)}+\upmu_{t+3}(xy)^{\lfloor(t+3)/2\rfloor}+\mathcal{O}_{t+4}.$

In degree $t+4$, again grouping together the terms containing $x^{2}$ or
$y^{2}$ and cyclically permuting, we may write

$\upvarphi^{t}(\mathsf{f}_{t})_{t+4}\sim\mathsf{h}_{t+2}\cdot
x^{2}+\mathsf{h}_{t+2}^{\prime}\cdot
y^{2}+\upmu_{t+4}(xy)^{\lfloor(t+4)/2\rfloor}$

for homogeneous forms $\mathsf{h}_{t+2}$, $\mathsf{h}_{t+2}^{\prime}$ of
degree $t+2$ and some $\upmu_{t+4}\in\mathbb{C}$, where again $\upmu_{t+4}=0$
for odd $t$. Thus, after setting
$\mathsf{p}_{t+3}(xy)=\mathsf{p}_{t+2}(xy)+\upmu_{t+3}(xy)^{\lfloor(t+3)/2\rfloor}$,
define

$\mathsf{f}_{t+1}=x^{3}+y^{3}+\mathsf{p}_{t+3}(xy)+\big{(}\mathsf{h}_{t+2}\cdot
x^{2}+\mathsf{h}_{t+2}^{\prime}\cdot
y^{2}+\upmu_{t+4}(xy)^{\lfloor(t+4)/2\rfloor}\big{)}+\upvarphi^{t}(\mathsf{f}_{t})_{\geq
t+5}.$

Note that $\upvarphi^{t}(\mathsf{f}_{t})\sim\mathsf{f}_{t+1}$, and
$\upvarphi^{t}(\mathsf{f}_{t})-\mathsf{f}_{t+1}\in\mathfrak{n}^{t+3}\subset\mathfrak{n}^{t+1}$
using the last statement of 6.2.

Thus we have constructed a sequence of power series
$\mathsf{f}_{1},\mathsf{f}_{2},\dots$ and unitriangular automorphisms
$\upvarphi^{1},\upvarphi^{2},\dots$ to which 3.7(3) applies. For $s\geq 3$
either $s$ is even, in which case $\mathsf{p}_{s+1}=\mathsf{p}_{s}$, or $s$ is
odd, in which case $\mathsf{p}_{s+1}=\mathsf{p}_{s}+\upmu_{s+1}z^{(s+1)/2}$,
thus it is clear that
$p\colonequals\lim\mathsf{p}_{s}=\sum_{s=2}^{\infty}\upmu_{2s}z^{s}$. Further,
$\mathsf{f}=\lim\mathsf{f}_{i}=x^{3}+y^{3}+p(xy)$, since the difference
$(\mathsf{f}_{i}-(x^{3}+y^{3}+p(xy)))_{i\geq 1}$ converges to zero.

It follows from 3.7(3) that
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(x^{3}+y^{3}+p(xy))$, as
required. ∎

The next step is to replace the power series $p(xy)$ by its leading term,
without changing the Jacobi algebra.

######  Lemma 6.4.

If $f=x^{3}+y^{3}+p(xy)\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ for some $0\neq p(z)\in\mathbb{C}[\\![z]\\!]$
for which $s=\operatorname{ord}(p)\geq 2$, then the following statements hold.

  1. (1)

$yx^{2},xy^{2}\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.

  2. (2)

$x^{2}y,y^{2}x\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.

  3. (3)

$(xy)^{s}x,(yx)^{s}y\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.

###### Proof.

(1) Write $J_{f}=(\updelta_{x}f,\updelta_{y}f)$, so that
$(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$ is the closure of
$J_{f}$. Differentiating and pulling out the lowest terms, write

$\updelta_{x}f=3x^{2}+y(xy)^{s-1}q(xy)\qquad\textrm{and}\qquad\updelta_{y}f=3y^{2}+q(xy)(xy)^{s-1}x$
(6.A)

for some
$q=\uplambda_{0}+\uplambda_{1}z+\uplambda_{2}z^{2}+\cdots\in\mathbb{C}[\\![z]\\!]$
with $\uplambda_{0}\neq 0$. Writing $A\equiv B$ for $A-B\in J_{f}$, then in
particular

$x^{2}\equiv-\tfrac{1}{3}y(xy)^{s-1}q(xy)\qquad\textrm{and}\qquad
y^{2}\equiv-\tfrac{1}{3}q(xy)(xy)^{s-1}x.$ (6.B)

Substituting for $x^{2}$ or $y^{2}$ at each step, we see that

$\displaystyle y\cdot x^{2}$
$\displaystyle\equiv\tfrac{-1}{3}y^{2}\cdot(xy)^{s-1}q(xy)$
$\displaystyle\in\mathfrak{n}^{2s\phantom{+2}}$
$\displaystyle\equiv\tfrac{+1}{3^{2}}q(xy)(xy)^{s-1}x\cdot(xy)^{s-1}q(xy)$
$\displaystyle=\tfrac{+1}{3^{2}}q(xy)(xy)^{s-1}\cdot x^{2}\cdot
y(xy)^{s-2}q(xy)$ $\displaystyle\in\mathfrak{n}^{4s-3}$
$\displaystyle\equiv\tfrac{-1}{3^{3}}q(xy)(xy)^{s-1}\cdot
y(xy)^{s-1}q(xy)\cdot y(xy)^{s-2}q(xy)$
$\displaystyle=\tfrac{-1}{3^{3}}q(xy)(xy)^{s-2}x\cdot
y^{2}\cdot(xy)^{s-1}q(xy)y(xy)^{s-2}q(xy)$
$\displaystyle\in\mathfrak{n}^{6s-6}$ $\displaystyle=\ldots$

At each substitution, the resulting power series has order $2s-3\geq 1$ higher
than the previous one. It follows from the above that for all $t\geq 2s$ there
exists $n_{t}\in\mathfrak{n}^{t}$ such that $yx^{2}-n_{t}\in J_{f}$. Hence
$yx^{2}\in\bigcap_{t\geq 2s}(J_{f}+\mathfrak{n}^{t})$, which is precisely the
closure $(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$. By symmetry
in $x$ and $y$, the analogous statement
$xy^{2}\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$ also follows.
(2) This follows in an analogous way: start by writing

$\updelta_{x}f=3x^{2}+r(yx)(yx)^{s-1}y\qquad\textrm{and}\qquad\updelta_{y}f=3y^{2}+x(yx)^{s-1}r(yx)$

then consider $x^{2}\cdot y\equiv\tfrac{-1}{3}r(yx)(yx)^{s-1}y^{2}$, etc.
(3) Now write $A\equiv B$ for
$A-B\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$. Separating off
the lowest term of $q(xy)$ in (6.A), we may write

$\updelta_{y}f=3y^{2}+\uplambda_{0}(xy)^{s-1}x+(q(xy)-\uplambda_{0})(xy)^{s-1}x$

and so
$\uplambda_{0}(xy)^{s-1}x\equiv-3y^{2}-(q(xy)-\uplambda_{0})(xy)^{s-1}x$ where
$q(xy)-\uplambda_{0}\in\mathfrak{n}^{2}$. Then

$\displaystyle\uplambda_{0}(xy)^{s}x$
$\displaystyle=\left(\uplambda_{0}(xy)^{s-1}x\right)yx$
$\displaystyle\equiv\left(-3y^{2}-(q(xy)-\uplambda_{0})(xy)^{s-1}x\right)(yx)$
$\displaystyle=-3y\cdot y^{2}\cdot x-(q(xy)-\uplambda_{0})(xy)^{s}x$
$\displaystyle\equiv y\cdot q(xy)(xy)^{s-1}x\cdot
x-(q(xy)-\uplambda_{0})(xy)^{s}x$ (by (6.B))
$\displaystyle\equiv-(q(xy)-\uplambda_{0})(xy)^{s}x$ ($yx^{2}\equiv 0$ by (1))

The $\uplambda_{0}(xy)^{s}x$ on each side cancel, showing that
$q(xy)(xy)^{s}x\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.
Since $q(xy)$ is a unit, it follows that
$(xy)^{s}x\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$. Again,
appealing to symmetry in $x$ and $y$ proves the final statement. ∎

######  Proposition 6.5.

Suppose that $f=x^{3}+y^{3}+p(xy)$ where $p(z)\in\mathbb{C}[\\![z]\\!]$ with
$s=\operatorname{ord}(p)\geq 2$. Then

$f\cong\begin{cases}x^{3}+y^{3}&\text{when $p=0$}\\\
x^{3}+y^{3}+(xy)^{s}&\text{when $p\neq 0$}.\\\ \end{cases}$

Furthermore

  1. (1)

$\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)\leq 1$, with equality if
and only if $p=0$.

  2. (2)

If $p\neq 0$, then $\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=4s$, and
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm ab}\nolimits}=4$.

Therefore the expressions $x^{3}+y^{3}+(xy)^{s}$ with $s\in\mathbb{Z}_{\geq
2}\cup\\{\infty\\}$ form a set of normal forms.

###### Proof.

For the first statement, if $p=0$ we are done, so suppose $p\neq 0$.
Continuing the notation in the proof of 6.4 above, after differentiating and
pulling out the lowest terms, we may write

$\updelta_{x}f=3x^{2}+y(xy)^{s-1}q(xy)\qquad\textrm{and}\qquad\updelta_{y}f=3y^{2}+q(xy)(xy)^{s-1}x$
(6.C)

for some
$q=\uplambda_{0}+\uplambda_{1}z+\uplambda_{2}z^{2}+\cdots\in\mathbb{C}[\\![z]\\!]$
with $\uplambda_{0}\neq 0$. Set $g=x^{3}+y^{3}+(\uplambda_{0}/s)(xy)^{s}$. Now
6.4 applies equally well to both $f$ and $g$, hence both $(xy)^{s}x$ and
$(yx)^{s}y$ belong to _both_ the Jacobi ideals associated to $f$ and $g$.
Consequently

$\displaystyle(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$
$\displaystyle=(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f,(xy)^{s}x,(yx)^{s}y)\kern-2.5pt)$
$\displaystyle=(\kern-2.5pt(\updelta_{x}g,\updelta_{y}g,(xy)^{s}x,(yx)^{s}y)\kern-2.5pt)$
(cancel higher terms from $\updelta_{x}f$ and $\updelta_{y}f$)
$\displaystyle=(\kern-2.5pt(\updelta_{x}g,\updelta_{y}g)\kern-2.5pt).$

It follows that $f\simeq g$. The coordinate change $x\mapsto ax$, $y\mapsto
ay$ for $a=\sqrt[2s-3]{s/\uplambda_{0}}$ then normalises the constant factor
$\uplambda_{0}/s\neq 0$, as required.
(1) Consider the case $p=0$. As in 3.2, if $\mathfrak{J}$ is the Jacobson
radical of $\mathcal{J}\mathrm{ac}(f)$, then

$\frac{\mathfrak{J}^{d}}{\mathfrak{J}^{d+1}}=\frac{\mathfrak{n}^{d}+(\kern-2.5pt(x^{2},y^{2})\kern-2.5pt)}{\mathfrak{n}^{d+1}+(\kern-2.5pt(x^{2},y^{2})\kern-2.5pt)}.$

This is always a two-dimensional vector space, since if $d$ is even it has
basis $(xy)^{d/2}$ and $(yx)^{d/2}$, whilst if $d$ is odd it has basis
$x(yx)^{(d-1)/2}$ and $y(xy)^{(d-1)/2}$. It follows that
$\mathcal{J}\mathrm{ac}(f)$ is an infinite-dimensional $\mathbb{C}$-algebra,
with $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=1$.

When $p\neq 0$, 6.4 shows at once that $\mathcal{J}\mathrm{ac}(f)$ is finite
dimensional: indeed any monomial of degree $t\geq 2s+1$ either contains one of
the monomials listed in 6.4(1–3) or is $x^{t}$ or $y^{t}$. But by (6.C)
$x^{t}=x^{t-4}\cdot x^{2}\cdot x^{2}$ and $y^{t}=y^{2}\cdot y^{2}\cdot
y^{t-4}$ is equivalent, modulo
$(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$, to a monomial that
contains one of those listed, and thus is equivalent to zero. Consequently,
the entire graded piece of degree $t$ is zero, and so
$\mathcal{J}\mathrm{ac}(f)$ is finite dimensional.
(2) We compute a suitable Gröbner basis with a local graded monomial order,
that is, the leading terms are those of lowest order (there is no need here
for lexicographical considerations to break ties). The proof of part (1) above
introduces a key simplifying factor: since all monomials of degree $t=2s+1$
lie in the closed Jacobian ideal, it is sufficient to work in the quotient

$\mathcal{J}\mathrm{ac}(f)\cong\frac{\mathbb{C}\langle
x,y\rangle}{(\updelta_{x}f,\updelta_{y}f,\mathcal{M}_{t})}$

where $\mathcal{M}_{t}$ for the set of all noncommutative monomials of degree
$t$, where now closure is no longer an issue.

We compile a Gröbner basis $\\{g_{1},g_{2},\dots\\}$ of
$(\updelta_{x}f,\updelta_{y}f,\mathcal{M}^{t})$ starting with

$g_{1}=3x^{2}+sy(xy)^{s-1}\quad\text{and}\quad g_{2}=3y^{2}+sx(yx)^{s-1}$

which have leading terms $3x^{2}$ and $3y^{2}$ respectively, as $s\geq 2$. The
computation proceeds by resolving non-trivial overlaps among leading terms.
The leading term of $g_{1}$ overlaps non-trivially with itself to produce

$g_{3}=xg_{1}-g_{1}x=s(xy)^{s}-s(yx)^{s}$

which does not reduce further modulo existing leading terms. The analogous
overlap $yg_{2}-g_{2}y$ gives the same $g_{3}$, and all other overlaps have
order $\geq t$, so are zero in the ideal. Hence the Gröbner basis is
$\\{g_{1},g_{2},g_{3}\\}$.

This Gröbner basis provides a monomial basis for the quotient as follows.
Among all monomials of degree $<t$, any of the form $m_{1}x^{2}m_{2}$, for
monomials $m_{1},m_{2}$, reduce modulo $g_{1}$ to zero at once, and similarly
with $y^{2}$. Thus a basis consists of all other monomials, with the single
relation $g_{3}$. That is, in each pair of degrees $2e,2e+1$ for $1\leq e\leq
s-1$ there are 4 monomials

$(xy)^{e},(yx)^{e},\ (xy)^{e}x,(yx)^{e}y,$

and then $1,x,y$ in degrees $\leq 1$ and finally $(xy)^{s}\equiv(yx)^{s}$ in
degree $2s$. Summing up, this basis has size $4s$, as claimed.

In the abelianisation, if $s>2$ we may rewrite the derivatives as
$x^{2}(\mathrm{unit})$ and $y^{2}(\mathrm{unit})$. Hence
$\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm
ab}\nolimits}\cong\mathbb{C}[\\![x,y]\\!]/(x^{2},y^{2})$, which is four
dimensional. When $s=2$, it is also easy to verify that
$\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm
ab}\nolimits}\cong\mathbb{C}[\\![x,y]\\!]/(x^{2},y^{2})$, so in all cases the
dimension is four. ∎

In order to state a unified theorem with the $xy^{2}$ case in §6.4 below, it
is convenient to mildly change basis. This is rather cheap, largely because
there are no moduli.

######  Corollary 6.6.

Suppose that $f\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$ where $f_{3}^{\mathop{\rm
ab}\nolimits}$ has three roots. Then either

$f\cong\begin{cases}xy^{2}+x^{3}&\\\ xy^{2}+x^{3}+x^{2n}&\mbox{for some }n\geq
2.\\\ \end{cases}$

Furthermore, the above are normal forms.

###### Proof.

Each of the forms $f$ listed satisfies the condition that $f_{3}^{\mathop{\rm
ab}\nolimits}$ has three roots. Thus there exists some $g$ from the list in
6.5 with $g\cong h$. Furthermore,
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(xy^{2}+x^{3})=\infty$, whereas
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(xy^{2}+x^{3}+x^{2n})=4n$ (see e.g.
[vG, §5] and [Ka, §5], or 4.11), and so all options are uniquely covered.
Since the $g$ listed in 6.5 are normal forms, it follows that the $f$ listed
here are normal forms. ∎

###  6.2. Isomorphisms on the Quantum Plane

The following, which may be of independent interest, is one of the key
reduction steps that will be used in §6.3.

######  Lemma 6.7.

For any units $v,w\in\mathbb{C}[\\![x^{2}]\\!]$, the unitriangular
automorphism $\upvarphi$ of $\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ sending $x\mapsto xv$, $y\mapsto yw$ descends to
a topological isomorphism

$\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}\buildrel\cong\over{\longrightarrow}\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}.$

###### Proof.

The inverse of $\upvarphi$, as an automorphism of
$\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle$, is clearly
given by the unitriangular automorphism $\uppsi\colon x\mapsto xv^{-1}$,
$y\mapsto yw^{-1}$. Set $I=(\kern-2.5pt(xy+yx)\kern-2.5pt)$, then since
$\upvarphi$ is a topological isomorphism by 2.3, we just need to prove that
$\upvarphi(I)=I$.

Now $x$ commutes with $v$ and $w$, being power series in $x^{2}$, and also
$vw=wv$. But, modulo $I=(\kern-2.5pt(xy+yx)\kern-2.5pt)$, $y$ commutes with
$x^{2}$, thus since the ideal is closed $y$ commutes with both $v$ and $w$. It
follows that

$\upvarphi(xy+yx)=xvyw+ywxv\equiv(xy+yx)vw\equiv 0\quad\textnormal{mod }I.$
(6.D)

Since $\upvarphi$ is a continuous isomorphism, and $I$ is the smallest closed
ideal containing $xy+yx$, $\upvarphi(I)$ is the smallest closed ideal
containing $\upvarphi(xy+yx)$. But by (6.D) $\upvarphi(xy+yx)$ also belongs to
the closed ideal $I$, so by minimality $\upvarphi(I)\subseteq I$.

Since $v^{-1}$ and $w^{-1}$ are also units in $\mathbb{C}[\\![x^{2}]\\!]$,
exactly the same logic applied to $\uppsi$ shows that $\uppsi(I)\subseteq I$.
Applying $\upvarphi$ to this inclusion, we see that
$I=\upvarphi\uppsi(I)\subseteq\upvarphi(I)$. Combining inclusions gives
$\upvarphi(I)=I$. ∎

###  6.3. Abelianized Cubic with Two Factors

This subsection considers the case of 4.13 where $f_{3}^{\mathop{\rm
ab}\nolimits}$ has two distinct factors, that is $f\cong
x^{2}y+\mathcal{O}_{4}$, and in 6.17 provides normal forms. This is
substantially harder than in §6.1.

######  Lemma 6.8.

Fix $t\geq 4$, and let $f=xy^{2}+f_{4}+\cdots+f_{t}+\mathcal{O}_{t+1}$. For
any $h\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle$
with $\operatorname{ord}(h)=t-2$, the unitriangular automorphism $x\mapsto
x-h$, $y\mapsto y$ sends

$f\mapsto
xy^{2}+f_{4}+\cdots+f_{t-1}+\left(f_{t}-hy^{2}\right)+\mathcal{O}_{t+1}.$

###### Proof.

Write $\uppsi$ for the stated automorphism. The result follows since
$\uppsi(xy^{2})=xy^{2}-hy^{2}$ and $\uppsi(m)\equiv m\mod\mathfrak{n}^{t+1}$
whenever $\deg(m)\geq 4$. ∎

The next lemma is much less elementary.

######  Lemma 6.9.

Fix $t\geq 4$, and let $f=xy^{2}+f_{4}+\cdots+f_{t}+\mathcal{O}_{t+1}$, where
furthermore

$f_{t}=\mathsf{h}_{t-2}\cdot y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y+\upalpha x^{t}$ (6.E)

for some homogeneous form $\mathsf{h}_{t-2}$ of degree $t-2$, each $a_{i}\geq
1$, $r\geq 1$ and $r+\sum a_{i}=t$, $\upalpha\in\mathbb{C}$ and each
$\uplambda_{a}=\uplambda_{a_{1}\cdots a_{r}}\in\mathbb{C}$. Then there exists
a unitriangular automorphism $\upvarphi$ of depth $\geq t-3$ such that

$\upvarphi(f)=xy^{2}+f_{4}+\cdots+f_{t-1}+(g_{t}+\upalpha
x^{t})+\mathcal{O}_{t+1}.$

where $g_{t}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{t}$ satisfies $g_{t}\sim 0$.

######  Example 6.10.

It is worth considering an example to make the notation of both the statement
and proof more transparent. Consider

$f=xy^{2}+\Big{(}\uplambda_{51}x^{5}yxy+\uplambda_{42}x^{4}yx^{2}y+\uplambda_{33}x^{3}yx^{3}y\Big{)}+f_{\geq
9}$

which has $f_{8}$ of the form (6.E). Applying $\upvarphi_{1}\colon x\mapsto x$
and $y\mapsto y-\uplambda_{33}x^{3}yx^{2}$, where we cancelled $xy$ from the
right of the target $\uplambda_{33}$ term to obtain the subtracted term, gives

$\upvarphi_{1}(f)=xy^{2}+\uplambda_{51}x^{5}yxy+(\uplambda_{42}-\uplambda_{33})x^{4}yx^{2}y+g_{1}+\mathcal{O}_{9}$

where $g_{1}=\uplambda_{33}(x^{3}yx^{3}y-xyx^{3}yx^{2})\sim 0$. Ignoring
$g_{1}$, the summation in degree 8 symbolically now has only two terms, which
is progress.

An analogous automorphism $\upvarphi_{2}$ sending $x\mapsto x$ and
$y\mapsto-(\uplambda_{42}-\uplambda_{33})x^{4}yx$, where we cancelled $xy$
from the right of the next target term, gives

$\upvarphi_{2}\upvarphi_{1}(f)=xy^{2}+(\uplambda_{51}-\uplambda_{42}+\uplambda_{33})x^{5}yxy+g_{2}+\mathcal{O}_{9}$

for some $g_{2}\sim 0$, and again the number of terms in degree 8 (outside
$g_{2}$) has not increased. Repeating again with an analogous automorphism
$\upvarphi_{3}$ gives

$\upvarphi_{3}\upvarphi_{2}\upvarphi_{1}(f)=xy^{2}-(\uplambda_{51}-\uplambda_{42}+\uplambda_{33})x^{6}y^{2}+g_{3}+\mathcal{O}_{9}$

with $g_{3}\sim 0$. We are now in a position to apply 6.8 to leave only
$g_{3}$ in degree 8.

The proof below confirms that this inductive idea works more generally.

###### Proof.

If the middle sum in the expression for $f_{t}$ is zero, we are done by 6.8
(with $g_{t}=0$), so we may assume that the sum is nonzero.

Suppose that the middle sum contains a term
$\mathsf{t}_{1}=\uplambda_{b}x^{b_{1}}y\cdots x^{b_{r}}y$ with $r>1$. In this
case, consider the unitriangular automorphism $\upvarphi$ defined by $x\mapsto
x$, $y\mapsto y-\uplambda_{b}x^{b_{1}}y\cdots yx^{b_{r}-1}$, where we have
simply cancelled $xy$ from the right-hand side of the target term
$\mathsf{t}_{1}$. As in 6.8, $\upvarphi(m)\equiv m\mod\mathfrak{n}^{t+1}$
whenever $\deg(m)\geq 4$, so any change in degree $\leq t$ comes from
$\upvarphi(xy^{2})$, and thus

$\displaystyle\upvarphi(f)=xy^{2}$ $\displaystyle+f_{4}+\cdots+f_{t-1}$
$\displaystyle+(f_{t}-\uplambda_{b}xyx^{b_{1}}y\cdots
yx^{b_{r}-1}-\uplambda_{b}x^{b_{1}+1}y\cdots yx^{b_{r}-1}y)+\mathcal{O}_{t+1}$
(6.F)

Writing $\mathsf{g}_{1}=\mathsf{t}_{1}-\uplambda_{b}xyx^{b_{1}}y\cdots
yx^{b_{r}-1}\sim 0$, then the degree $t$ term of (6.F) equals

$\mathsf{g}_{1}+h\cdot y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y+\upalpha x^{t}$

where, under the summand, the target term $\mathsf{t}_{1}$ has been replaced
by a term of the form $\mathsf{t}_{2}=-\uplambda_{b}x^{b_{1}+1}y\cdots
yx^{b_{r}-1}y$, so the sum has the same number of terms or fewer (depending on
whether $\mathsf{t}_{2}$ cancels with existing terms or not).

If $b_{r}=1$, then the new term $\mathsf{t}_{2}$ equals $hy^{2}$ for
$h=-\uplambda_{b}x^{b_{1}+1}y\cdots yx^{b_{r-1}}$, and we so may apply 6.8 to
(6.F) to find $\uppsi$ such that

$\uppsi\upvarphi(f)=xy^{2}+f_{4}+\cdots+f_{t-1}+(f_{t}-\uplambda_{b}xyx^{b_{1}}y\cdots
yx^{b_{r}-1})+\mathcal{O}_{t+1}$

where the degree $t$ term is equal to

$\displaystyle f_{t}-\uplambda_{b}xyx^{b_{1}}y\cdots yx^{b_{r}-1}$
$\displaystyle=\mathsf{g}_{1}+f_{t}-\uplambda_{b}x^{b_{1}}y\cdots x^{b_{r}}y$
$\displaystyle=\mathsf{g}_{1}+h\cdot y^{2}+\sum_{a\neq
b}\uplambda_{a}x^{a_{1}}y\ldots x^{a_{r}}y+\upalpha x^{t}$

and the number of terms under the summand is now strictly reduced.

Otherwise, $b_{r}>1$. Set $\upvarphi_{1}=\upvarphi$, and repeating the
original construction of a unitriangular automorphism by cancelling $xy$ from
the right, we can construct $\upvarphi_{2}$ such that

$\displaystyle\upvarphi_{2}\upvarphi_{1}(f)=xy^{2}$
$\displaystyle+f_{4}+\cdots+f_{t-1}$
$\displaystyle+\big{(}\mathsf{g}_{2}+h\cdot
y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots x^{a_{r}}y+\upalpha
x^{t}\big{)}+\mathcal{O}_{t+1}$ (6.G)

where $\mathsf{g}_{2}\sim 0$ is the sum of $\mathsf{g}_{1}$ and another
binomial $\sim 0$, and in the sum we have replaced the term
$-\uplambda_{b}x^{b_{1}+1}y\cdots yx^{b_{r}-1}y$ by
$\uplambda_{b}x^{b_{1}+2}y\cdots yx^{b_{r}-2}y$. Repeating this, we find
unitriangular automorphisms $\upvarphi_{1},\dots,\upvarphi_{b_{r}-1}$ so that
$\upvarphi_{b_{r}-1}\cdots\upvarphi_{2}\upvarphi_{1}(f)$ has the form of
(6.G), and the sum has the same number of terms or fewer, but in which the
target monomial we are focussing on has become
$x^{b_{1}+b_{r}-1}yx^{b_{2}}\cdots yxy$. A further repetition with a
unitriangular automorphism $\upvarphi_{b_{r}}$ replaces that term by one that
contains $y^{2}$, and once again we may apply 6.8 to find a unitriangular
automorphism $\uppsi$ that moves this term into higher degree. Thus after
applying the single unitriangular automorphism
$\uppsi\upvarphi_{b_{r}}\cdots\upvarphi_{1}$ to $f$, the number of terms in
the summation when parsed in the form (6.E) has strictly reduced.

We repeat this process inductively, and it will terminate when there are no
terms under the summation sign of the form $\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y$ with $r>1$. Each step was achieved by a single unitriangular
automorphism (itself built as a composition of unitriangular automorphisms),
and composing each of these gives a single unitriangular automorphism
$\upvartheta$ such that

$\upvartheta(f)=xy^{2}+f_{4}+\cdots+f_{t-1}+(\mathsf{g}+h\cdot y^{2}+\uplambda
x^{t-1}y+\upalpha x^{t})+\mathcal{O}_{t+1}$

for some $\mathsf{g}$ with $\mathsf{g}\sim 0$.

To conclude, the unitriangular automorphism $\upvarphi$ defined by $x\mapsto
x-h$, $y\mapsto y-\frac{\uplambda}{2}x^{a_{1}-1}=y-\frac{\uplambda}{2}x^{t-2}$
has depth $t-3$, so again $\upvarphi(m)\equiv m\mod\mathfrak{n}^{t+1}$
whenever $\deg(m)\geq 4$ and thus

$\upvarphi\upvartheta(f)=xy^{2}+f_{4}+\cdots+f_{t-1}+(\mathsf{g}+\underbrace{\uplambda
x^{t-1}y-\tfrac{\uplambda}{2}xyx^{t-2}-\tfrac{\uplambda}{2}x^{t-2}y}_{\mathsf{h}}+\upalpha
x^{t})+\mathcal{O}_{t+1}.$

Set $g_{t}=\mathsf{g}+\mathsf{h}$, then since both $\mathsf{g}\sim 0$ and
$\mathsf{h}\sim 0$, we are done. ∎

From here, the strategy of §6.1 remains: first find a standard power series
form of each potential, then simplify into polynomial normal form.

######  Proposition 6.11.

Suppose that $f=f_{3}+\mathcal{O}_{4}$ where $f_{3}^{\mathop{\rm
ab}\nolimits}$ has two distinct linear factors. Then $f\cong xy^{2}+q(x)$ for
some power series $q(x)\in\mathbb{C}[\\![x]\\!]$ with
$\operatorname{ord}(q)\geq 4$.

Recall the Conventions 1.7, used in 6.3, on graded pieces of sequence
elements: namely sequence elements
$\mathsf{f}_{n}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ are in Greek font, whilst $(\mathsf{f}_{n})_{t}$
is the degree $t$ piece of $\mathsf{f}_{n}$.

###### Proof.

We construct a sequence of power series $\mathsf{f}_{1},\mathsf{f}_{2},\dots$
and unitriangular automorphisms $\upvarphi^{1},\upvarphi^{2},\dots$
inductively, with each $\mathsf{f}_{t}$ having the form of the target power
series $xy^{2}+q(x)$ in low degree. Summary 3.7(3) will then construct
$\mathsf{f}=\lim\mathsf{f}_{i}$ of the required form with
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(\mathsf{f})$.

By 4.13, $f\cong g$ where $g_{3}=xy^{2}$. After grouping together the terms
containing $y^{2}$, then the terms that contain $y$ but not $y^{2}$, and
cyclically permuting, we may write

$g_{4}\sim\mathsf{h}_{2}\cdot y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y+\upmu_{4}x^{4}$

for $\mathsf{h}_{2}\in\mathbb{C}\langle x,y\rangle_{2}$, $r\geq 1$ and each
$a_{i}\geq 1$ and $\upmu_{4}\in\mathbb{C}$ and where we use the abbreviated
notation $\uplambda_{a}:=\uplambda_{a_{1}\cdots a_{r}}\in\mathbb{C}$. It is
convenient to write the sum as $\sum\uplambda_{a}x^{a_{1}}y\ldots x^{a_{r}}y$
by analogy with the general case, noting that here it is nothing more than
$\uplambda_{11}xyxy+\uplambda_{3}x^{3}y$.

Hence we begin the induction by setting

$\mathsf{f}_{1}=xy^{2}+(\mathsf{h}_{2}\cdot y^{2}+\sum\uplambda
x^{a_{1}}y\ldots x^{a_{r}}y+\upmu_{4}x^{4})+g_{\geq 5}$

and note that $\mathsf{f}_{1}\sim g\cong f$. Thus $\mathsf{f}_{1}$ is in the
desired form in degrees $\leq 3$ and has its degree $4$ piece prepared in
standard form for further analysis.

For the inductive step more generally, we may suppose that
$\mathsf{f}_{t}\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ has been constructed of the form

$\displaystyle\mathsf{f}_{t}=\big{(}xy^{2}+\mathsf{q}_{t+2}(x)\big{)}+\big{(}{\mathsf{h}}_{t+1}\cdot
y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y+\upmu_{t+3}x^{t+3}\big{)}+\mathcal{O}_{t+4}$

with $\mathsf{q}_{3}=0$ and

  1. (1)

$(\mathsf{f}_{t})_{\leq t+2}=xy^{2}+\mathsf{q}_{t+2}(x)$, for some polynomial
$\mathsf{q}_{t+2}\in\mathbb{C}[x]_{\geq 4}$ of degree $\leq t+2$, where the
polynomials $\mathsf{q}_{3},\dots,\mathsf{q}_{t+2}$ satisfy
$\mathsf{q}_{i+1}-\mathsf{q}_{i}=\upmu_{i+1}x^{i+1}$ for
$\upmu_{i+1}\in\mathbb{C}$, and

  2. (2)

$(\mathsf{f}_{t})_{t+3}=\mathsf{h}_{t+1}\cdot
y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots x^{a_{r}}y+\upmu_{t+3}x^{t+3}$ for
some homogeneous form $\mathsf{h}_{t+1}$ of degree $t+1$, each $a_{i}\geq 1$,
$r\geq 1$ and $r+\sum a_{i}=t+3$, and $\upmu_{t+3}\in\mathbb{C}$.

By 6.9 there exists a unitriangular $\upvarphi^{t}$ of depth $t$ such that

$\upvarphi^{t}(\mathsf{f}_{t})=xy^{2}+q_{t+2}(x)+(k_{t+3}+\upmu_{t+3}x^{t+3})+\mathcal{O}_{t+4}.$

where $k_{t+3}\sim 0$. In degree $t+4$, again grouping together the terms
containing $y^{2}$, then the terms that contain $y$ but not $y^{2}$, and
cyclically permuting, we may write

$\upvarphi^{t}(\mathsf{f}_{t})_{t+4}\sim\mathsf{h}_{t+2}\cdot
y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots x^{a_{r}}y+\upmu_{t+4}x^{t+4}.$

Thus after setting
$\mathsf{q}_{t+3}(x)=\mathsf{q}_{t+2}(x)+\upmu_{t+3}x^{t+3}$, define

$\mathsf{f}_{t+1}=xy^{2}+\mathsf{q}_{t+3}(x)+\big{(}\mathsf{h}_{t+2}\cdot
y^{2}+\sum\uplambda_{a}x^{a_{1}}y\ldots
x^{a_{r}}y+\upmu_{t+4}x^{t+4}\big{)}+\upvarphi^{t}(\mathsf{f}_{t})_{\geq
t+5}.$

Note that $\upvarphi^{t}(\mathsf{f}_{t})\sim\mathsf{f}_{t+1}$, and that
$\upvarphi^{t}(\mathsf{f}_{t})-\mathsf{f}_{t+1}\in\mathfrak{n}^{t+3}\subset\mathfrak{n}^{t+1}$.

Thus we have constructed a sequence of power series
$\mathsf{f}_{1},\mathsf{f}_{2},\dots$ and unitriangular automorphisms
$\upvarphi^{1},\upvarphi^{2},\dots$ to which 3.7(3) applies. Since at each
stage $\mathsf{q}_{s}=\mathsf{q}_{s-1}+\upmu_{s}x^{s}$, it is clear that
$q\colonequals\lim\mathsf{q}_{s}=\sum_{s=4}^{\infty}\upmu_{s}x^{s}$, and that
$\mathsf{f}=\lim\mathsf{f}_{i}=xy^{2}+q$, since the difference
$(\mathsf{f}_{i}-(xy^{2}+q))_{i\geq 1}$ converges to zero. Hence
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(xy^{2}+q)$, as required.
∎

The next step is to reduce the options for $q(x)$, using the following
preliminary lemma.

######  Lemma 6.12.

Let $u\in\mathbb{C}[\\![x]\\!]$ be an even power series: that is, $u$ is a
power series in $x^{2}$.

  1. (1)

If $u$ is a unit, then $u^{-1}$ and $\sqrt[n]{u}$ are also even power series
for any $n\geq 2$.

  2. (2)

Let $U\in\mathbb{C}[\\![x]\\!]$ be a unit and $n\in\mathbb{Z}$ a nonzero
integer. Then there is a unit $t\in\mathbb{C}[\\![x]\\!]$ with $t^{n}=U(xt)$.
Furthermore, if $U$ is even then $t$ is even.

###### Proof.

(1) Consider $v\in\mathbb{C}[\\![z]\\!]$ with $u(x)=v(x^{2})$. If $u$ is a
unit, then $v$ is a unit and $v^{-1}$ and
$\sqrt[n]{v}\in\mathbb{C}[\\![z]\\!]$ for all $n\geq 2$. Then
$u^{-1}(x)=v^{-1}(x^{2})$ and $\sqrt[n]{u}(x)=\sqrt[n]{v}(x^{2})$.

(2) Write $U=a_{0}+a_{1}x+a_{2}x^{2}+\cdots$ with $a_{0}\not=0$. Consider the
case $n>0$. We show that we may solve inductively for the coefficients $b_{d}$
of the expansion $t=b_{0}+b_{1}x+b_{2}x^{2}+\cdots$ in the equation
$t^{n}=U(xt)$.

It is clear that the coefficient of $x^{d}$ in $t^{n}$ is a sum of
$nb_{0}^{n-1}b_{d}$ with terms involving only coefficients $b_{i}$ with $i<d$.
On the other side of the equation, the coefficient of $x^{d}$ in $U(xt)$ is a
sum of terms involving $a_{i}$ and $b_{j}$ with $i\leq d$ and $j<d$. Putting
these together, $b_{d}$ does not appear in the coefficient of $x^{i}$ for any
$i<d$, and it appears linearly with nonzero coefficient for the first time in
the coefficient of $x^{d}$, and so we may solve for it. Working inductively in
increasing $d\geq 0$, and taking the limit, determines $t$ as claimed. For
$n<0$, the same argument proves the existence of the unit $t^{-1}$, which is
equivalent.

Suppose that $U$ is even, and let $b_{2n+1}$ be the smallest nonzero odd-
degree coefficient of $t$. Then the odd-degree term with smallest degree in
$t^{n}$ is $nb_{0}^{n-1}b_{2n+1}x^{2n+1}$, while in $U(xt)$ it is
$2a_{2}b_{0}b_{2n+1}x^{2n+3}$ which appears in the summand $a_{2}(xt)^{2}$, a
contradiction. So $t$ must be even. ∎

The point is now simple: $y$ is under control, and so there is a relation
$xy+yx$ in the Jacobi algebra. Then 6.7 yields the following key preparation
result.

######  Proposition 6.13.

If $f=xy^{2}+p(x)$ where $p(x)\in\mathbb{C}[\\![x]\\!]$ with
$\operatorname{ord}(p)\geq 4$, then

$f\cong xy^{2}+\upalpha\,x^{2n}+\upbeta\,x^{2m+1}$

for some $n,m\geq 2$, and some $\upalpha,\upbeta\in\\{0,1\\}$. Furthermore,
$\upalpha=1$ if and only if $p$ has a nonzero even-degree term, in which case
$2n$ is the least even degree appearing in $p$, and similarly the analogous
criterion for $\upbeta=1$ and least odd degree term in $p$.

###### Proof.

First note that

$\mathcal{J}\mathrm{ac}(f)=\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle/(\kern-2.5pt(xy+yx,y^{2}+\updelta_{x}p)\kern-2.5pt).$

We exhibit an automorphism of $\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle$ that takes the two generators of the Jacobian
ideal to $(xy+yx)\text{(unit)}$ and $\left(y^{2}+\upbeta x^{2m}+\upalpha
x^{2n-1}\right)\text{(unit)}$, respectively, where either $\upalpha=0$ or
$2n\geq 4$ is the least even degree appearing in $p$, and either $\upbeta=0$
or $2m+1\geq 5$ is the least odd degree appearing in $p$. This proves all the
claims.

Parsing $\updelta_{x}p$ into even and odd terms, write the Jacobi algebra
relations as

$xy+yx\quad\text{and}\quad y^{2}+ax^{2N}u+bx^{2M-1}v$

where $u,v\in\mathbb{C}[\\![x^{2}]\\!]$ are each either a unit or zero,
$N,M\geq 2$, and $a,b\in\mathbb{C}$ are any nonzero numbers that carry through
the calculation undisturbed; we choose $a=2N+1$ and $b=2M$ at the end.

Suppose in the first place that $u\neq 0$, then fix a square root
$s=\sqrt{u}\in\mathbb{C}[\\![x^{2}]\\!]$ and consider the unitriangular
automorphism $\upvarphi$ sending $x\mapsto x$, $y\mapsto ys$. By 6.7 this
induces a topological isomorphism

$\bar{\upvarphi}\colon\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}\xrightarrow{\sim}\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}.$

In the codomain of this map, $y$ commutes with $x^{2}$ and thus commutes with
$s\in\mathbb{C}[\\![x^{2}]\\!]$. It follows that
$\bar{\upvarphi}(y^{2})=ysys=y^{2}s^{2}=yu$ and thus

$\bar{\upvarphi}(y^{2}+\updelta_{x}p)=(y^{2}+ax^{2N})u+bx^{2M-1}v.$

By 2.11(1)(3) after right multiplying by the unit $u^{-1}$, we obtain an
isomorphism

$\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx,y^{2}+\updelta_{x}p)\kern-2.5pt)}\xrightarrow{\sim}\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx,y^{2}+ax^{2N}+bx^{2M-1}\tfrac{v}{u})\kern-2.5pt)}.$
(6.H)

If $v=0$, then (6.H) asserts that
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(xy^{2}+x^{2N+1})$, and
so we are done. Hence we may assume that $v\neq 0$.

As $u$ and $v$ are both unit power series in $x^{2}$, so is $\frac{v}{u}$. By
6.12(2), since $2N-2M+1$ is nonzero, we may choose a unit
$t\in\mathbb{C}[\\![x^{2}]\\!]$ such that $t^{2N}=t^{2M-1}v(xt)/u(xt)$.
Consider the unitriangular automorphism $\uppsi$ sending $x\mapsto xt$,
$y\mapsto yt^{N}$. Again by 6.7 there is an induced topological isomorphism

$\bar{\uppsi}\colon\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}\xrightarrow{\sim}\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx)\kern-2.5pt)}.$

Clearly $x$ commutes with $t,t^{N}\in\mathbb{C}[\\![x^{2}]\\!]$, and further
in the codomain of $\bar{\uppsi}$ the element $y$ commutes with $x^{2}$ and
thus commutes with $t,t^{N}\in\mathbb{C}[\\![x^{2}]\\!]$. Thus

$\displaystyle\bar{\uppsi}(y^{2}+ax^{2N}+bx^{2M-1}\tfrac{v}{u})$
$\displaystyle=y^{2}t^{2N}+ax^{2N}t^{2N}+bx^{2M-1}t^{2M-1}u(xt)/v(xt)$
$\displaystyle=(y^{2}+ax^{2N}+bx^{2M-1})t^{2N}.$

Again 2.11(1)(3) then induces an isomorphism

$\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx,y^{2}+ax^{2N}+bx^{2M-1}\tfrac{v}{u})\kern-2.5pt)}\xrightarrow{\sim}\frac{\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle}{(\kern-2.5pt(xy+yx,y^{2}+ax^{2N}+bx^{2M-1})\kern-2.5pt)}.$
(6.I)

Setting $a=2N+1$ and $b=2M$, the right hand side is
$\mathcal{J}\mathrm{ac}(xy^{2}+x^{2N+1}+x^{2M})$. Hence composing (6.H) with
(6.I) gives an isomorphism
$\mathcal{J}\mathrm{ac}(f)\cong\mathcal{J}\mathrm{ac}(xy^{2}+x^{2N+1}+x^{2M})$.

The case $u=0$ and $v\neq 0$ works in exactly the same way as the case $u\neq
0$ and $v=0$ above, applying an automorphism with $s=\sqrt{v}$, while the case
$u=v=0$ is trivial. ∎

The above is not quite yet in normal form, since some of the polynomial
potentials in 6.13 have isomorphic Jacobi algebras. The next step is to
discard cases where the odd term in $p(x)$ has significantly greater degree
than the even term.

######  Lemma 6.14.

If $f=xy^{2}+x^{2n}+\upvarepsilon x^{2m+1}$ where $\upvarepsilon\in\\{0,1\\}$
and $m\geq n$, then $y^{3}\in(\updelta_{x}f,\updelta_{y}f)$. In particular
$y^{3}\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.

###### Proof.

Set
$I=(\updelta_{x}f,\updelta_{y}f)=(y^{2}+2nx^{2n-1}+\upvarepsilon(2m+1)x^{2m},xy+yx)$,
and below write $\equiv$ for an equality mod $I$. Since
$y^{2}\equiv-2nx^{2n-1}-\upvarepsilon(2m+1)x^{2m}$, multiplying on the left by
$y$ and on the right by $y$ gives

$\displaystyle-2nx^{2n-1}y-\upvarepsilon(2m+1)x^{2m}y\equiv y^{3}$
$\displaystyle\equiv-2nyx^{2n-1}-\upvarepsilon(2m+1)yx^{2m}$
$\displaystyle\equiv 2nx^{2n-1}y-\upvarepsilon(2m+1)x^{2m}y$

where the last line holds since $xy\equiv-yx$ using the second generator of
$I$. Inspecting the right and lefthand sides, the $x^{2m}y$ terms cancel, and
so $4nx^{2n-1}y\equiv 0$, thus $x^{2n-1}y\equiv 0$. Finally, since $m\geq n$,
taking out the common factor we see that

$y^{3}\equiv(2n-\upvarepsilon(2m+1)x^{2m-2n-1})x^{2n-1}y\equiv 0.$

Thus $y^{3}\in I$. The final statement follows immediately. ∎

######  Corollary 6.15.

If $f=xy^{2}+x^{2n}+\upvarepsilon x^{2m+1}$ where $m\geq 2n-1$, then
$x^{4n-2}\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$.

###### Proof.

Continue to write $\equiv$ for an equality mod
$(\updelta_{x}f,\updelta_{y}f)$. Then

$\displaystyle x^{4n-2}=(-x^{2n-1})^{2}$
$\displaystyle\equiv\tfrac{1}{(2n)^{2}}(y^{2}+\upvarepsilon(2m+1)x^{2m})^{2}$
(since $\updelta_{x}f\equiv 0$)
$\displaystyle\equiv\tfrac{\upvarepsilon(2m+1)}{(2n)^{2}}(y^{2}x^{2m}+x^{2m}y^{2}+\upvarepsilon(2m+1)x^{4m})$
($y^{3}\equiv 0$ by 6.14)
$\displaystyle\equiv\tfrac{\upvarepsilon(2m+1)}{(2n)^{2}}(2x^{2m}y^{2}+\upvarepsilon(2m+1)x^{4m}).$
($xy\equiv-yx$)

Taking out the $x^{2m}$ common factor from the front, we may write
$x^{4n-2}\equiv x^{2m}g$ for some $g$ with no constant term. Then, since
$2m\geq 4n-2$ by assumption, we see that $x^{4n-2}\equiv
x^{4n-2}(x^{2m-(4n-2)}g)$, and so $x^{4n-2}(1-x^{2m-(4n-2)}g)\equiv 0$.

Given this statement holds mod $(\updelta_{x}f,\updelta_{y}f)$, it also holds
mod $(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$, and hence
$x^{4n-2}(1-x^{2m-(4n-2)}g)=0$ in $\mathcal{J}\mathrm{ac}(f)$. But there,
$1-x^{2m-(4n-2)}g$ is a unit, and so it follows that $x^{4n-2}=0$ in
$\mathcal{J}\mathrm{ac}(f)$, as required. ∎

The above two results combine to remove the case when the odd-degree $x$ term
is sufficiently larger than the even-degree $x$ term, as follows.

######  Corollary 6.16.

If $f=xy^{2}+x^{2n}+x^{2m+1}$ where $m\geq 2n-1$, then $f\cong xy^{2}+x^{2n}$.

###### Proof.

By 6.15 we have
$x^{4n-2}\in(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$ and
$x^{4n-2}\in(\kern-2.5pt(y^{2}+2nx^{2n-1},xy+yx)\kern-2.5pt)$. Since $2m\geq
4n-2$, it follows that $x^{2m}$ belongs to _both_ of the ideals above, and
thus

$\displaystyle(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f)\kern-2.5pt)$
$\displaystyle\colonequals(\kern-2.5pt(y^{2}+2nx^{2n-1}+(2m+1)x^{2m},xy+yx,x^{2m})\kern-2.5pt)$
$\displaystyle=(\kern-2.5pt(y^{2}+2nx^{2n-1},xy+yx,x^{2m})\kern-2.5pt)$
$\displaystyle=(\kern-2.5pt(y^{2}+2nx^{2n-1},xy+yx)\kern-2.5pt).$

As this final ideal is obtained from $xy^{2}+x^{2n}$ by differentiation, the
result follows. ∎

Summarising the above gives the following, which is the main result of this
subsection.

######  Corollary 6.17.

Suppose that $f\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$ where $(f_{3})^{\mathop{\rm
ab}\nolimits}$ has two roots. Then either

$f\cong\begin{cases}xy^{2}&\\\ xy^{2}+x^{2m+1}&m\geq 2\\\ xy^{2}+x^{2n}&n\geq
2\\\ xy^{2}+x^{2m+1}+x^{2n}&2\leq m\leq n-1\\\ xy^{2}+x^{2n}+x^{2m+1}&2\leq
n\leq m\leq 2(n-1)\end{cases}$

All of the above are mutually non-isomorphic.

###### Proof.

The fact that the stated list covers all cases follows from 6.13, using 6.16
to discount the case when the odd-degree $x$ term is sufficiently larger than
the even-degree $x$ term. We now claim that the potentials listed give
pairwise non-isomorphic Jacobi algebras.

The first two families both have infinite dimensional Jacobi algebras, whereas
the bottom three are all finite dimensional. As such, the only possibilities
for isomorphisms are between members in families one and two, or between
members in families three, four and five. But
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(xy^{2})^{\mathop{\rm
ab}\nolimits}=\infty$, whereas
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(xy^{2}+x^{2m+1})^{\mathop{\rm
ab}\nolimits}=2m+2$, and so all members of families one and two are mutually
non-isomorphic.

For the final three families, all members of families three and four and
mutually non-isomorphic, as can be seen by extending the method of [BW, 4.7],
or by using [Ka, 5.10] directly. Further, all members of family five are also
mutually non-isomorphic for dimension reasons, since for $f$ in family five
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm ab}\nolimits}=2m+2$
and $\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=(2m+2)+4(n-1)$ by either [vG,
§5] or §4.3, and thus we can distinguish between all different $m$ and $n$.
The only remaining possibility is an isomorphism between a member of family
five, and a member of family three or four. But by above the dimension of
$\mathcal{J}\mathrm{ac}(f)$ for $f$ in family five is even, and the dimension
of $\mathcal{J}\mathrm{ac}(g)$ for $g$ in families three and four is odd [Ka,
5.10], so there can be no such isomorphisms. ∎

###  6.4. Overview of Type D normal forms

The previous subsections combine to give the following, which is the main
result of this section.

######  Theorem 6.18.

Let
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$ with $\mathcal{C}\mathrm{rk}(f)=2$ and $\mathcal{C}\mathrm{rk}_{2}(f)=2$.
Then either

$f\cong\begin{cases}z_{1}^{2}+\cdots+z_{d-2}^{2}+xy^{2}&\\\
z_{1}^{2}+\cdots+z_{d-2}^{2}+xy^{2}+x^{2m+1}&\mbox{with }m\geq 1\\\
z_{1}^{2}+\cdots+z_{d-2}^{2}+xy^{2}+x^{2n}&\mbox{with }n\geq 2\\\
z_{1}^{2}+\cdots+z_{d-2}^{2}+xy^{2}+x^{2n}+x^{2m+1}&\mbox{with }n\geq
2,\,\,n\leq m\leq 2n-2\\\
z_{1}^{2}+\cdots+z_{d-2}^{2}+xy^{2}+x^{2m+1}+x^{2n}&\mbox{with }m\geq
1,\,\,n\geq m+1.\end{cases}$

The Jacobi algebras of these potentials are all mutually non-isomorphic, and
furthermore the following statements hold.

  1. (1)

Every $f$ in the first two families satisfies $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=1$.

  2. (2)

Every $f$ in the last three families satisfies $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(f)=0$.

    1. (a)

For any fixed $n\geq 2$, the algebras in families three and four combine to
give $n-1$ non-isomorphic Jacobi algebras, all of which satisfy
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm ab}\nolimits}=2n+1$
and $\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=(2n+1)+4(n-1)=6n-3$.

    2. (b)

In the fifth family, $\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)^{\mathop{\rm
ab}\nolimits}=2m+2$ and
$\dim_{\mathbb{C}}\mathcal{J}\mathrm{ac}(f)=(2m+2)+4(n-1)$.

###### Proof.

By the Splitting Lemma 4.5 the condition $\mathcal{C}\mathrm{rk}(f)=2$ implies
that $f\cong z_{1}^{2}+\ldots+z_{d-2}^{2}+g$ for some
$g\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
3}$. The condition $\mathcal{C}\mathrm{rk}_{2}(f)=2$ is then equivalent to the
first two cases in 4.13, namely those $g\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$ with $g_{3}\neq 0$ for which
$g_{3}^{\mathop{\rm ab}\nolimits}$ has either two or three distinct linear
factors. The options for all such $g$ thus follow from combining 6.6 and 6.17

The fact that $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(xy^{2}+x^{3})=1$ follows since
$\mathcal{J}\mathrm{ac}(xy^{2}+x^{3})\cong\mathcal{J}\mathrm{ac}(x^{3}+y^{3})$
by linear change in coordinates, and $\mathop{\rm
Jdim}\nolimits\mathcal{J}\mathrm{ac}(x^{3}+y^{3})=1$ by 6.5(1). The statements
that $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(xy^{2}+x^{2m-1})=1$ for
all $m\geq 2$ and $\mathop{\rm Jdim}\nolimits\mathcal{J}\mathrm{ac}(xy^{2})=1$
can be shown by a very similar explicit method as in the proof of 6.5(1), or
alternatively by using 8.5 below, once we know (in 8.12) that all such Jacobi
algebras are contraction algebras. The stated vector space dimensions of the
Jacobi algebras in all remaining cases have already been justified in the
proofs of 6.6 and 6.17 respectively.

The fact that the above are all mutually non-isomorphic, and thus a list of
normal forms, then follows. Indeed, by inspecting $\mathfrak{J}$-dimension,
the only possible isomorphisms are between members of families one and two, or
between members of families three, four and five. Given we have just added the
normal forms of 6.6 to the normal forms of 6.17, the only remaining possible
isomorphisms are between these two cases. But again, either the dimension of
the abelianisation, or the dimension of the contraction algebra itself,
distinguishes in all cases. ∎

##  7\. Central Elements and General Elephants

This section algebraically extracts ADE information from the normal forms in
§1.2, using generic central elements and contracted preprojective algebras.

###  7.1. Generic Central Sections

For any
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, set $\mathcal{Z}=\mathcal{Z}(\mathcal{J}\mathrm{ac}(f))$ to be the centre
of $\mathcal{J}\mathrm{ac}(f)$, and write $\mathfrak{m}_{\mathcal{Z}}$ for the
Jacobson radical of $\mathcal{Z}$. Recall that $\mathfrak{J}$ denotes the
Jacobson radical of the local ring $\mathcal{J}\mathrm{ac}(f)$.

######  Lemma 7.1.

We have $\mathfrak{m}_{\mathcal{Z}}=\mathfrak{J}\cap\mathcal{Z}$ and
$\mathcal{Z}/\mathfrak{m}_{\mathcal{Z}}\cong\mathbb{C}$. Thus $\mathcal{Z}$ is
also a local ring.

###### Proof.

Set $J_{f}=(\kern-2.5pt(\updelta_{1}f,\ldots,\updelta_{d}f)\kern-2.5pt)$, then
it is clear that $\mathfrak{J}\cap\mathcal{Z}=\\{g+J_{f}\in\mathcal{Z}\mid
g\in\mathfrak{n}\\}$. This set is clearly a two-sided ideal of $\mathcal{Z}$,
and further $1+\mathfrak{J}\cap\mathcal{Z}$ consists of units in
$\mathcal{Z}$. These two properties imply that $\mathfrak{J}\cap\mathcal{Z}$
equals the Jacobson radical $\mathfrak{m}_{\mathcal{Z}}$ of $\mathcal{Z}$, see
e.g. [L, 4.5]. The fact that
$\mathcal{Z}/\mathfrak{m}_{\mathcal{Z}}\cong\mathbb{C}$ is clear. ∎

Generic elements of the centre $\mathcal{Z}$ will be used to intrinsically
extract ADE information.

######  Definition 7.2.

Given
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, we say that $\mathcal{J}\mathrm{ac}(f)$ has Type $X$ if for all finite
dimensional vector spaces $V\subset\mathfrak{m}_{\mathcal{Z}}$ such that
$V\twoheadrightarrow\mathfrak{m}_{\mathcal{Z}}^{\phantom{2}}/\mathfrak{m}_{\mathcal{Z}}^{2}$,
there exists a Zariski open subset $U$ of $V$ such that
$\mathcal{J}\mathrm{ac}(f)/(u)\cong e\Uppi e$ for all $u\in U$, where $\Uppi$
is the preprojective algebra of Type $X$, and $e$ is an idempotent marked  in
(1.C).

Equivalently, in the language of [R1, 2.5], $\mathcal{J}\mathrm{ac}(f)$ has
Type $X$ provided that a general hyperplane section $u$ of $\mathcal{Z}$
satisfies $\mathcal{J}\mathrm{ac}(f)/(u)\cong e\Uppi e$ where $\Uppi$ is the
preprojective algebra of Type $X$, and $e$ is an idempotent marked  in (1.C).
We also remark that there are two different Type $E_{8}$’s in 7.2,
corresponding to the two different choices of  in $E_{8}$ in (1.C). This
feature matches the two different $E_{8}$ cases in the length classification
of flops [KM].

Much like the definition of cDV singularities, 7.2 is only designed to be
useful in specific situations. Indeed, for general
$f\in\mathbb{C}\langle\kern-2.5pt\langle\mathsf{x}\rangle\kern-2.5pt\rangle_{\geq
2}$, it is not clear whether the centre $\mathcal{Z}$ of
$\mathcal{J}\mathrm{ac}(f)$ is noetherian, nor whether its maximal ideal
$\mathfrak{m}_{\mathcal{Z}}$ is finitely generated as a $\mathcal{Z}$-module.
Consequently, work is required to establish that
$\mathfrak{m}_{\mathcal{Z}}^{\phantom{2}}/\mathfrak{m}_{\mathcal{Z}}^{2}$ is
finite dimensional, which is needed for there to exist a finite dimensional
vector space $V$ surjecting onto it.

When $\mathcal{J}\mathrm{ac}(f)$ is finite dimensional, these difficulties
disappear, since $\mathcal{J}\mathrm{ac}(f)$ and thus $\mathcal{Z}$,
$\mathfrak{m}_{\mathcal{Z}}$ and
$\mathfrak{m}_{\mathcal{Z}}^{\phantom{2}}/\mathfrak{m}_{\mathcal{Z}}^{2}$ are
all finite dimensional vector spaces. Other cases are more tricky, but for our
purposes the following suffices.

######  Lemma 7.3.

If $f$ is a normal form from 6.18, then the following statements hold.

  1. (1)

If $u\in\mathfrak{m}_{\mathcal{Z}}$, then $u\equiv\uplambda x^{2}+h$ in
$\mathcal{J}\mathrm{ac}(f)$ for some $h\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 3}$.

  2. (2)

If further $f$ has Type $D_{\infty,\infty}$ or $D_{\infty,m}$, then
$\mathcal{Z}\cong\mathbb{C}[\\![x^{2}]\\!]$.

In particular, in all cases
$\dim_{\mathbb{C}}(\mathfrak{m}_{\mathcal{Z}}^{\phantom{2}}/\mathfrak{m}_{\mathcal{Z}}^{2})<\infty$.

###### Proof.

(1) In all cases $\updelta_{y}f=xy+yx$, and so certainly $x^{2}$ commutes with
$y$ in $\mathcal{J}\mathrm{ac}(f)$. Obviously $x^{2}$ commutes with $x$, thus
since we are considering closed ideals, it follows that $x^{2}$ is central in
$\mathcal{J}\mathrm{ac}(f)$. Similarly $y^{2}$ is central.

We next claim that there are no elements in $\mathfrak{m}_{\mathcal{Z}}$ that
contain linear terms. Write $u\in\mathfrak{m}_{\mathcal{Z}}$ as $u=k+g$ for
some $k=\uplambda_{1}x+\uplambda_{2}y$ and
$g\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
2}$. Now $u$ remains central after factoring by $\mathfrak{J}^{3}$, so set
$I=(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f,\mathfrak{n}^{3})\kern-2.5pt)$,
and observe that

$0+I=[x+I,u+I]=2\uplambda_{2}xy+I$

For any (finite-dimensional) $\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle/(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f,\mathfrak{n}^{r})\kern-2.5pt)$,
$r\geq 3$, we may choose a basis of the form
$1,x,\ldots,x^{i},xy,x^{2}y,\ldots,x^{j}y$ for some $i,j\geq 2$. Since $xy$
forms part of this basis it follows that $\uplambda_{2}=0$. Repeating using
the commutator $[y+I,u+I]$ shows that $\uplambda_{1}=0$.

Thus $u=g$ for some $g\in\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle_{\geq 2}$. Using the relation $xy+yx$ to move
$x$’s to the left, and the other relation to move $y^{2}$ either to zero, to
$x^{2}$, or into higher degree, write

$u\equiv\uplambda_{1}x^{2}+\uplambda_{2}xy+h$

in $\mathcal{J}\mathrm{ac}(f)$, for some
$h\in\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle_{\geq
3}$. We next claim that $\uplambda_{2}=0$. Since $u$ is central, and $x^{2}$
is central, it follows that $v\colonequals\uplambda_{2}xy+h$ is also central
in $\mathcal{J}\mathrm{ac}(f)$. In particular it is still central after
factoring by $\mathfrak{J}^{4}$. Set
$I=(\kern-2.5pt(\updelta_{x}f,\updelta_{y}f,\mathfrak{n}^{4})\kern-2.5pt)$, so
that

$0+I=[x+I,v+I]=2\uplambda_{2}x^{2}y+I$

and thus $2\uplambda_{2}x^{2}y\in I$. But $x^{2}y$ forms part of a basis of
$\mathbb{C}\langle\kern-2.5pt\langle x,y\rangle\kern-2.5pt\rangle/I$, so
$\uplambda_{2}=0$.

(2) Either set $I=(xy+yx,y^{2})$, or $(xy+yx,y^{2}+x^{2m})$, so by assumption
$\mathcal{J}\mathrm{ac}(f)\cong\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle/(\kern-2.5pt(I)\kern-2.5pt)$. Consider an
arbitrary element $u\in\mathcal{Z}$. By using the first relation to move all
the $x$’s to the left, and the second relation to either move $y^{2}$ to zero
or to higher powers of $x$, we may write $u\equiv p+qy$ in
$\mathcal{J}\mathrm{ac}(f)$, where by (1) $p\in\mathbb{C}[\\![x]\\!]_{\geq 2}$
and $q\in\mathbb{C}[\\![x]\\!]_{\geq 2}$. Observe that in
$\mathcal{J}\mathrm{ac}(f)$

$0\equiv[x,u]\equiv[x,p+qy]=2xqy$

and so $xqy\in(\kern-2.5pt(I)\kern-2.5pt)$. Thus $xq_{\leq t-3}y\in
I+\mathfrak{n}^{t}$ for all $t\geq 3$. Now
$\mathbb{C}\langle\kern-2.5pt\langle
x,y\rangle\kern-2.5pt\rangle/(I+\mathfrak{n}^{t})$ has basis
$1,x,\ldots,x^{t-1},y,xy,\ldots,x^{t-2}y$. Write $q_{\leq
t-3}=\sum_{i=2}^{t-3}\uplambda_{i}x^{i}$, then the second part of this basis
on the equation $xq_{\leq t-3}y\in I+\mathfrak{n}^{t}$ shows that
$\uplambda_{2}=\ldots=\uplambda_{t-3}=0$. This holds for all $t$, and so
$q=0$.

Thus the central element $u\equiv p$. Splitting into even and odd terms, write
$u\equiv P(x^{2})+xQ(x^{2})$ in $\mathcal{J}\mathrm{ac}(f)$ for some
$P,Q\in\mathbb{C}[\\![x]\\!]_{\geq 1}$. Then, in $\mathcal{J}\mathrm{ac}(f)$,
since $x^{2}$ is central

$0\equiv[y,u]\equiv[y,P+xQ]\equiv-2(xQ)y.$

Using the same argument as above, $Q=0$, and so $u\equiv P(x^{2})$, as
claimed. This shows that $\mathcal{Z}\subseteq\mathbb{C}[\\![x^{2}]\\!]$, but
since $x^{2}$ is central by (1), equality holds, proving (2).

For the very last statement, all the finite-dimensional
$\mathcal{J}\mathrm{ac}(f)$ satisfy
$\dim_{\mathbb{C}}(\mathfrak{m}_{\mathcal{Z}}^{\phantom{2}}/\mathfrak{m}_{\mathcal{Z}}^{2})<\infty$.
Since the only other potentials in 6.18 are those in (2), where visibly