Proof. Set . Then it suffices to prove that for all $n\geq 1$ , there exists N such that for all $t\geq N$ . This follows since for all $i> n$ ,

Notation 2.6. For , consider , the commutator vector space of $\mathbb {C}\langle \kern -2.5pt\langle \mathsf {x}\rangle \kern -2.5pt\rangle $ . That is, elements of are finite sums

for elements $a_i,b_i\in \mathbb {C}\langle \kern -2.5pt\langle \mathsf {x}\rangle \kern -2.5pt\rangle $ and $\unicode{x3bb} _i\in \mathbb {C}$ . Write for the closure of the commutator vector space . Note that is only a vector space, not an ideal.

Proof. The proof follows the strategy used in [Reference Derksen, Weyman and ZelevinskyDWZ, 4.7], but as the axiomatics are different here, we give the full proof. By §2.4, there is an automorphism .

Since the depth of is $\geq i$ , by 2.2(1), differs from $\mathsf {f}_i$ only in degrees $> i$ . By (2), 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 . Since $f=\mathsf {f}_1$ , it is easy to see that

(8)

where 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 .

First, g exists, since by (2), $c_i\in \mathfrak {n}^{i+1}$ , and so since automorphisms preserve the maximal ideal, for all t. It follows easily that the sequence 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 has limit $F(g)$ follows, since for all $i>n$ ,

(by (7))

(since (φ ^t ⋯φ¹)⁻¹(c _t ) ∈ 𝔫 ^t+1)

(add zero)

(by (6))

Combining with (8) and taking limits, it follows that

(9) $$ \begin{align} F(f)=\lim \mathsf{f}_i + F(g). \end{align} $$

Now, it is easy to check that automorphisms preserve , so each term in the sequence belongs to . 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 . One final application of the fact that automorphisms preserve shows that , and so $F(f)\sim \lim \mathsf {f}_i $ .

Notation 2.10. When I is an ideal, write $(\kern -2pt( I)\kern -2pt)$ for its closure (in the sense of 2.5), which is again an ideal since the ring operations are continuous. Note that $(\kern -2pt( I)\kern -2pt)$ need not be finitely generated, even if I is.

Proof. (1) $(\kern -2pt( f_1,\dots ,f_s)\kern -2pt)$ is the smallest closed ideal containing all $f_i$ . Since $f_i=(f_iu_i)u_i^{-1}\in (\kern -2pt( f_1u_1,\dots ,f_su_s)\kern -2pt)$ for each i, by minimality, $(\kern -2pt( f_1,\dots ,f_s)\kern -2pt)\subseteq (\kern -2pt( f_1u_1,\dots ,f_su_s)\kern -2pt)$ . Repeating the same argument to $f_iu_i\in (\kern -2pt( f_1,\dots ,f_s)\kern -2pt)$ , the converse inclusion also holds.

(2) Certainly, for some ideal I, given it is an ideal of the quotient. This ideal I is closed by [Reference WarnerWa, 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 , and so .

However, is the smallest closed ideal containing . Setting $A=\mathbb {C}\langle \kern -2.5pt\langle \mathsf {x}\rangle \kern -2.5pt\rangle $ , , and , the third isomorphism theorem for topological rings [Reference WarnerWa, 5.13] asserts that there is a topological isomorphism

$$\begin{align*}(A/J)/(H/J)\cong A/H. \end{align*}$$

In particular, $A/H$ is Hausdorff since H is closed in A, by [Reference WarnerWa, 5.7(1)] applied to A. This being the case, $H/J$ is closed in $A/J$ , by [Reference WarnerWa, 5.7(1)] applied to $A/J$ . Hence, is a closed ideal, which clearly contains . By minimality, , and thus, . Combining inclusions, the required equality holds.

(3) Since $\unicode{x3c8} $ is a continuous isomorphism, the closed ideal generated by corresponds to the closed ideal generated by . Thus, there is a topological isomorphism

Now by (2), we have , and likewise for g. The statement follows by the third isomorphism theorem for topological rings [Reference WarnerWa, 5.13].

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, for example, [Reference LamL1, 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 $(\kern -2pt( \unicode{x3b4} f)\kern -2pt)$ is contained in $\mathfrak {n}$ , and so , and furthermore, for $n\ge 2$ .

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 is induced by powers of the image of $\mathfrak {n}$ in the quotient [Reference WarnerWa, 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 is the radical-adic topology. Since $(\kern -2pt(\unicode{x3b4} f )\kern -2pt)$ is closed, is Hausdorff [Reference WarnerWa, 5.7(1)]. Under extra assumptions, it is also complete; see 8.4(3).

Proof. Certainly, 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 , and so . Otherwise, by Nakayama’s Lemma, $\mathfrak {J}^{n}=0$ for some n; hence, $\mathfrak {n}^{n}\subset (\kern -2pt( \unicode{x3b4} f)\kern -2pt)$ , and so and .

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 .

2. 2. If , then , where .

3. 3. Set $\mathsf {f}_1=f$ . If there exist and automorphisms such that

   1. (a) every is unitriangular of depth of $\geq i$ , and

   2. (b) , for all $i\geq 1$ ,

   then the sequence $(\mathsf {f}_i)_{i\ge 1}$ converges and where $g=\lim \mathsf {f}_i$ .

Proof. Writing $m=m_1m_2\ldots m_r$ , where each $m_i$ is a variable $x_{j(i)}$ , we have

$$ \begin{align*} rm - \operatorname{\mathrm{cyc}}(m) &= (r-1)m_1\ldots m_r - m_2\ldots m_rm_1 \\ &\qquad\qquad- m_3\ldots m_rm_1m_2 - \cdots - m_rm_1\ldots m_{r-1} \\ &= [m_1,m_2\ldots m_r] + [m_1m_2,m_3\ldots m_r] + \cdots + [m_1\ldots m_{r-1},m_r], \end{align*} $$

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

Proof. (1) At the level of ideals, $\mathfrak {n}^k\twoheadrightarrow \mathfrak {n}_{ \mathrm {ab} }^k$ for every $k\ge 0$ , since abelianisation is a ring homomorphism mapping each $x_i$ to $x_i$ .

(2) Since $(\unicode{x3b4} _{i}f)^{ \mathrm {ab} } = \partial f^{ \mathrm {ab} }/\partial x_i$ , where $\partial /\partial x_i$ 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.

Proof. This is [Reference Derksen, Weyman and ZelevinskyDWZ, 4.5] for the d-loop quiver. Since $\operatorname {\mathrm {ord}} g\ge 3$ , the derivatives of g impose no linear conditions, so necessarily, .

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

(13) $$ \begin{align} x\mapsto \unicode{x3b1} x+\unicode{x3b2} y,\qquad y\mapsto \unicode{x3b3} x + \unicode{x3b4} y \qquad\text{for } \unicode{x3b1},\unicode{x3b2},\unicode{x3b3},\unicode{x3b4}\in\mathbb{C} \end{align} $$

that maps $(f_3)^{ \mathrm {ab} }\in \mathbb {C}[x,y]$ 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 $p\ne 0$ 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 be the linear automorphism of $\mathbb {C}\langle \kern -2.5pt\langle x,y\rangle \kern -2.5pt\rangle $ defined by the same formula (13) 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 , although they are not equal, so . Thus , as claimed.

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

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

If q is zero, we are done, else after pulling out the lowest term, we can write $q = y^n \mathsf {u}$ for some 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, , as required.

Parts (1)–(2) are obvious since and , and uniqueness then follows since $\sum x_i^2+y^{n_1}\cong \sum x_i^2+y^{n_2}$ if and only if $n_1=n_2$ .

Proof. Consider and its unique crepant resolution . This has contraction algebra , realising the second case in 5.1. Moreover, by [Reference Donovan and WemyssDW1, 3.10], the Type $A$ m-Pagoda flop (with $m\geq 1$ ) has contraction algebra , which realises the infinite family in 5.1.

Proof. ( $\Leftarrow $ ) is clear from 5.1. For ( $\Rightarrow $ ), we prove the contrapositive. If , then $f_2=0$ , and we need to prove that 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

Write for the set of all noncommutative monomials of degree $3$ , and then factor by the ideal in . But in the four cases above, by differentiating then using the third isomorphism theorem, it follows that is one of

None of these factors is commutative, and so is not commutative.

Proof. Consider the unitriangular automorphism $\unicode{x3c8} $ which sends $x\mapsto x-\tfrac {1}3h_1$ , $y\mapsto y-\tfrac {1}{3}h_2$ . The result follows since

$$\begin{align*}\unicode{x3c8}(x^3+y^3) \sim x^3-h_1x^2+\tfrac{1}3h_1^2x-\tfrac{1}{27}h_1^3+y^3-h_2y^2+\tfrac{1}3h_2^2y-\tfrac{1}{27}h_2^3, \end{align*}$$

and $\unicode{x3c8} (m) \equiv m\ \mod \mathfrak {n}^{t+1}$ whenever $\deg (m) \ge 4$ .

Proof. We construct a sequence of power series $\mathsf {f}_1,\mathsf {f}_2,\dots $ and unitriangular automorphisms 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 .

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

for $\mathsf {h}_2, \mathsf {h}_2'\in \mathbb {C}\langle \kern -2.5pt\langle x,y\rangle \kern -2.5pt\rangle _2$ and $\unicode{x3bc} _4\in \mathbb {C}$ .

Hence, we begin the induction by setting

and note that $\mathsf {f}_1\sim g \cong f$ . Thus, $\mathsf {f}_1$ is in the desired form in degrees $\le 3$ 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

with $\mathsf {p}_3=0$ and by convention $\unicode{x3bc} _{t+3}=0$ for even t, where

1. 1. $(\mathsf {f}_{t})_{\leq t+2} = x^3+y^3 + \mathsf {p}_{t+2}(xy)$ , for some polynomial $\mathsf {p}_{t+2}\in \mathbb {C}[z]_{\ge 2}$ 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 $\mathsf {p}_{i+1} - \mathsf {p}_i=\unicode{x3bc} _{i+1}z^{(i+1)/2}$ for odd i, and

2. 2. $(\mathsf {f}_{t})_{t+3} = \mathsf {h}_{t+1}\cdot x^2 + \mathsf {h}_{t+1}'\cdot y^2 + \unicode{x3bc} _{t+3} (xy)^{\lfloor (t+3)/2\rfloor }$ for some homogeneous forms $\mathsf {h}_{t+1}$ , $\mathsf {h}_{t+1}'$ of degree $t+1$ .

Applying 6.2 with $h_1=\mathsf {h}_{t+1}$ and $h_2=\mathsf {h}_{t+1}'$ , there exists a unitriangular of depth $\geq t$ such that

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

for homogeneous forms $\mathsf {h}_{t+2}$ , $\mathsf {h}_{t+2}'$ of degree $t+2$ and some $\unicode{x3bc} _{t+4}\in \mathbb {C}$ , where again $\unicode{x3bc} _{t+4}=0$ for odd t. Thus, after setting $\mathsf {p}_{t+3}(xy)=\mathsf {p}_{t+2}(xy)+\unicode{x3bc} _{t+3}(xy)^{\lfloor (t+3)/2\rfloor }$ , define

Note that , and 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 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 $\mathsf {p}_{s+1}=\mathsf {p}_{s}+\unicode{x3bc} _{s+1}z^{(s+1)/2}$ ; thus, it is clear that . 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 , as required.

Proof. (1) Write $J_f = (\unicode{x3b4} _x f, \unicode{x3b4} _y f )$ , so that $(\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ is the closure of $J_f$ . Differentiating and pulling out the lowest terms, write

(14)

for some $q=\unicode{x3bb} _0 + \unicode{x3bb} _1z + \unicode{x3bb} _2z^2+\cdots \in \mathbb {C}[\![ z]\!]$ with $\unicode{x3bb} _0\ne 0$ . Writing $A\equiv B$ for $A-B \in J_f$ , then in particular,

(15) $$ \begin{align} x^2 \equiv -\tfrac{1}3 y(xy)^{s-1}q(xy) \qquad\text{and}\qquad y^2 \equiv -\tfrac{1}3 q(xy)(xy)^{s-1}x. \end{align} $$

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

At each substitution, the resulting power series has order $2s-3\ge 1$ 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, $yx^2\in \bigcap _{t\ge 2s}(J_f+\mathfrak {n}^t)$ , which is precisely the closure $(\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ . By symmetry in x and y, the analogous statement $xy^2\in (\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ also follows.

(2) This follows in an analogous way: start by writing

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 $A-B \in (\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ . Separating off the lowest term of $q(xy)$ in (14), we may write

and so $\unicode{x3bb} _0(xy)^{s-1}x \equiv -3y^2 - (q(xy)-\unicode{x3bb} _0)(xy)^{s-1}x$ where $q(xy)-\unicode{x3bb} _0\in \mathfrak {n}^2$ . Then

(by (15))

(yx ² ≡ 0 by(1))

The $\unicode{x3bb} _0(xy)^{s}x$ on each side cancel, showing that $q(xy)(xy)^{s}x\in (\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ . Since $q(xy)$ is a unit, it follows that $(xy)^{s}x\in (\kern -2pt( \unicode{x3b4} _x f, \unicode{x3b4} _y f )\kern -2pt)$ . Again, appealing to symmetry in x and y proves the final statement.

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

(16)

for some $q=\unicode{x3bb} _0 + \unicode{x3bb} _1z + \unicode{x3bb} _2z^2+\cdots \in \mathbb {C}[\![ z]\!]$ with $\unicode{x3bb} _0\ne 0$ . Set $g=x^3+y^3+(\unicode{x3bb} _0/s)(xy)^s$ . Now 6.4 applies equally well to both f and g; hence, both $(xy)^{s}x$ and $(yx)^sy$ belong to both the Jacobi ideals associated to f and g. Consequently,

(cancel higher terms from δ _x f and δ _y f)

It follows that $f\simeq g$ . The coordinate change $x\mapsto ax$ , $y\mapsto ay$ for $a=\sqrt [2s-3]{s/\unicode{x3bb} _0}$ then normalises the constant factor $\unicode{x3bb} _0/s\ne 0$ , as required.

(1) Consider the case $p=0$ . As in 3.2, if $\mathfrak {J}$ is the Jacobson radical of , then

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

This is always a two-dimensional vector space since if d is even, it has basis $(xy)^{d/2}$ and $(yx)^{d/2}$ , while if d is odd, it has basis $x(yx)^{(d-1)/2}$ and $y(xy)^{(d-1)/2}$ . It follows that is an infinite-dimensional $\mathbb {C}$ -algebra, with .

When $p\ne 0$ , 6.4 shows at once that is finite dimensional: indeed, any monomial of degree $t\ge 2s+1$ either contains one of the monomials listed in 6.4(1–3) or is $x^{t}$ or $y^{t}$ . But by (16), $x^t=x^{t-4}\cdot x^2\cdot x^2$ and $y^t=y^2\cdot y^2\cdot y^{t-4}$ is equivalent, modulo $(\kern -2pt( \unicode{x3b4} _xf,\unicode{x3b4} _yf)\kern -2pt)$ , 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 is finite dimensional.

(2) We compute a standard basis of $(\kern -2pt( \unicode{x3b4} _xf,\unicode{x3b4} _yf)\kern -2pt)$ with respect to a local graded monomial order, where we refer the reader to 6.6 below for references to the formal theory and its properties in this case. In particular, leading terms have lowest degree, and lexicographical order selects the leading term when there is more than one of lowest degree.

The proof of (1) above introduces a simplifying factor: since all monomials of degree $t=2s+1$ lie in the closed Jacobian ideal, it is sufficient to work in the quotient

where we write for the set of all noncommutative monomials of degree t, and closure is no longer an issue.

Following [Reference Gerritzen and HoltkampGH, 3.6], we compile a standard basis $\{g_1,g_2,\dots \}$ of starting with the normalised derivatives

which have leading terms $x^2$ and $y^2$ , respectively, since $s\ge 2$ . The computation proceeds by resolving nontrivial overlaps among leading terms. The leading term of $g_1$ overlaps nontrivially with itself to produce (after scaling by $\tfrac {3}s$ to normalise coefficients)

This does not reduce further modulo existing leading terms. The analogous overlap $yg_2-g_2y$ gives the same $g_3$ , and all other overlaps have order $\ge t$ , so reduce to zero modulo . Hence, the standard basis is $\{g_1,g_2,g_3\}$ . Note that we can then dispense with since those monomials all reduce to zero under $g_1,g_2,g_3$ – that is exactly what 6.4 demonstrates – and so they do not appear in the standard basis.

As in the commutative case, the set of monomials not divisible by the leading term of any of $g_1$ , $g_2$ , $g_3$ descends to give a monomial $\mathbb {C}$ -vector space basis for the quotient, by, for example, [Reference Gerritzen and HoltkampGH, 3.5 and 3.1–2]. Thus, working in increasing degree, $1,x,y$ are in the basis. Then, in each pair of degrees $2e,2e+1$ for $1\le e\le s-1$ , the basis consists of the four monomials

$$\begin{align*}(xy)^{e},(yx)^{e},\ (xy)^{e}x, (yx)^{e}y, \end{align*}$$

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, , which is four-dimensional. When $s=2$ , it is also easy to verify that , so in all cases, the dimension is four.

Proof. Each of the forms f listed satisfies the condition that $f_3^{ \mathrm {ab} }$ has three roots. Thus, there exists some g from the list in 6.5 with $g\cong f$ . Furthermore, , whereas (see, for example, [Reference van GarderenvG, §5] and [Reference KawamataKa, §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.

Proof. The inverse of , 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 $\unicode{x3c8} \colon x\mapsto xv^{-1}$ , $y\mapsto yw^{-1}$ . Set $I=(\kern -2pt( xy+yx)\kern -2pt)$ . Then since is a topological isomorphism by 2.3, we just need to prove that .

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

(17)

Since is a continuous isomorphism, and I is the smallest closed ideal containing $xy+yx$ , is the smallest closed ideal containing . But by (17), also belongs to the closed ideal I, so by minimality, .

Since $v^{-1}$ and $w^{-1}$ are also units in $\mathbb {C}[\![ x^2]\!]$ , exactly the same logic applied to $\unicode{x3c8} $ shows that $\unicode{x3c8} (I)\subseteq I$ . Applying to this inclusion, we see that . Combining inclusions gives .

Proof. Write $\unicode{x3c8} $ for the stated automorphism. The result follows since $\unicode{x3c8} (xy^2) = xy^2 - hy^2$ and $\unicode{x3c8} (m) \equiv m \mod \mathfrak {n}^{t+1}$ whenever $\deg (m) \ge 4$ .

Proof. If the middle sum in the expression for $f_t$ is zero, we are done by 6.9 (with $g_t=0$ ), so we may assume that the sum is nonzero.

Suppose that the middle sum contains a term $\mathsf {t}_1 = \unicode{x3bb} _b x^{b_1}y\cdots x^{b_r}y$ with $r>1$ . In this case, consider the unitriangular automorphism defined by $x\mapsto x$ , $y\mapsto y-\unicode{x3bb} _b x^{b_1}y\cdots yx^{b_r-1}$ , where we have simply cancelled $xy$ from the right-hand side of the target term $\mathsf {t}_1$ . As in 6.9, whenever $\deg (m) \ge 4$ , so any change in degree $\le t$ comes from , and thus,

(19)

Writing $\mathsf {g}_1=\mathsf {t}_1-\unicode{x3bb} _b xyx^{b_1}y\cdots yx^{b_r-1}\sim 0$ , then the degree t term of (19) equals

where, under the summand, the target term $\mathsf {t}_1$ has been replaced by a term of the form $\mathsf {t}_2 = -\unicode{x3bb} _b x^{b_1+1}y\cdots yx^{b_r-1}y$ , 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 $h=-\unicode{x3bb} _b x^{b_1+1}y\cdots yx^{b_{r-1}}$ , and we so may apply 6.9 to (19) to find $\unicode{x3c8} $ such that

where the degree t term is equal to

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

Otherwise, $b_r>1$ . Set , and repeating the original construction of a unitriangular automorphism by cancelling $xy$ from the right, we can construct such that

(20)

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 $-\unicode{x3bb} _b x^{b_1+1}y\cdots yx^{b_r-1}y$ by $\unicode{x3bb} _b x^{b_1+2}y\cdots yx^{b_r-2}y$ . Repeating this, we find unitriangular automorphisms so that has the form of (20), 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 replaces that term by one that contains $y^2$ , and once again, we may apply 6.9 to find a unitriangular automorphism $\unicode{x3c8} $ that moves this term into higher degree. Thus, after applying the single unitriangular automorphism to f, the number of terms in the summation when parsed in the form (18) has strictly reduced.

We repeat this process inductively, and it will terminate when there are no terms under the summation sign of the form 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 such that

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

To conclude, the unitriangular automorphism defined by $x\mapsto x-h$ , $y\mapsto y-\frac {\unicode{x3bb} }{2}x^{a_1-1}=y-\frac {\unicode{x3bb} }{2}x^{t-2}$ has depth $t-3$ , so again, whenever $\deg (m) \ge 4$ , and thus,

Set $g_t=\mathsf {g}+\mathsf {h}$ . Then since both $\mathsf {g}\sim 0$ and $\mathsf {h}\sim 0$ , we are done.

Proof. We construct a sequence of power series $\mathsf {f}_1,\mathsf {f}_2,\dots $ and unitriangular automorphisms 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 .

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

for $\mathsf {h}_2\in \mathbb {C}\langle x,y\rangle _2$ , $r\ge 1$ and each $a_i\ge 1$ and $\unicode{x3bc} _4\in \mathbb {C}$ and where we use the abbreviated notation $\unicode{x3bb} _a := \unicode{x3bb} _{a_1\cdots a_r}\in \mathbb {C}$ . It is convenient to write the sum as by analogy with the general case, noting that here it is nothing more than $\unicode{x3bb} _{11}xyxy+\unicode{x3bb} _3x^3y$ .

Hence, we begin the induction by setting

and note that $\mathsf {f}_1\sim g\cong f$ . Thus, $\mathsf {f}_1$ is in the desired form in degrees $\le 3$ 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

with $\mathsf {q}_3=0$ and

1. 1. $(\mathsf {f}_{t})_{\leq t+2} = xy^2 + \mathsf {q}_{t+2}(x)$ , for some polynomial $\mathsf {q}_{t+2}\in \mathbb {C}[x]_{\ge 4}$ of degree $\leq t+2$ , where the polynomials $\mathsf {q}_3,\dots , \mathsf {q}_{t+2}$ satisfy $\mathsf {q}_{i+1} - \mathsf {q}_i=\unicode{x3bc} _{i+1}x^{i+1}$ for $\unicode{x3bc} _{i+1}\in \mathbb {C}$ , and

2. 2. for some homogeneous form $\mathsf {h}_{t+1}$ of degree $t+1$ , each $a_i\ge 1$ , $r\ge 1$ and $r+\sum a_i=t+3$ , and $\unicode{x3bc} _{t+3}\in \mathbb {C}$ .

By 6.10, there exists a unitriangular of depth t such that

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

Thus, after setting $\mathsf {q}_{t+3}(x)=\mathsf {q}_{t+2}(x)+\unicode{x3bc} _{t+3}x^{t+3}$ , define

Note that , and that .

Thus, we have constructed a sequence of power series $\mathsf {f}_1,\mathsf {f}_2,\dots $ and unitriangular automorphisms to which 3.7(3) applies. Since at each stage $\mathsf {q}_s=\mathsf {q}_{s-1}+\unicode{x3bc} _sx^s$ , it is clear that , 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, , as required.

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 $n\ge 2$ . Then $u^{-1}(x) = v^{-1}(x^2)$ and $\sqrt [n]{u}(x)=\sqrt [n]{v}(x^2)$ .

(2) Write $U=a_0+a_1x+a_2x^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_1x+b_2x^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 $i\le d$ 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 $d\ge 0$ , 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 $2a_2b_0b_{2n+1}x^{2n+3}$ which appears in the summand $a_2(xt)^2$ , a contradiction. So t must be even.

Proof. First note that

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 $\left (y^2 + \unicode{x3b2} x^{2m} + \unicode{x3b1} x^{2n-1}\right )\text {(unit)}$ , respectively, where either $\unicode{x3b1} =0$ or $2n\ge 4$ is the least even degree appearing in p, and either $\unicode{x3b2} =0$ or $2m+1\ge 5$ is the least odd degree appearing in p. This proves all the claims.

Parsing $\unicode{x3b4} _xp$ into even and odd terms, write the Jacobi algebra relations as

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

where $u,v\in \mathbb {C}[\![ x^2]\!]$ are each either a unit or zero, $N,M\ge 2$ , 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 $u\neq 0$ , and then fix a square root $s=\sqrt {u}\in \mathbb {C}[\![ x^2]\!]$ and consider the unitriangular automorphism sending $x\mapsto x$ , $y\mapsto ys$ . By 6.8, this induces a topological isomorphism

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

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

(21)

If $v=0$ , then (21) asserts that , 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 6.13(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 $\unicode{x3c8} $ sending $x\mapsto xt$ , $y\mapsto yt^N$ . Again by 6.8, there is an induced topological isomorphism

$$\begin{align*}\bar{\unicode{x3c8}}\colon \frac{\mathbb{C}\langle\kern -2.5pt\langle x,y\rangle\kern -2.5pt\rangle}{(\kern -2pt( xy+yx)\kern -2pt)} \xrightarrow{\sim} \frac{\mathbb{C}\langle\kern -2.5pt\langle x,y\rangle\kern -2.5pt\rangle}{(\kern -2pt( xy+yx )\kern -2pt)}. \end{align*}$$

Clearly, x commutes with $t,t^N\in \mathbb {C}[\![ x^2]\!]$ , and further in the codomain of $\bar {\unicode{x3c8} }$ , the element y commutes with $x^2$ and thus commutes with $t,t^N\in \mathbb {C}[\![ x^2]\!]$ . Thus,

$$ \begin{align*} \bar{\unicode{x3c8}}(y^2+ax^{2N}+bx^{2M-1}\tfrac{v}{u})&= y^2t^{2N}+ax^{2N}t^{2N}+bx^{2M-1}t^{2M-1}u(xt)/v(xt) \\ &=(y^2+ax^{2N}+bx^{2M-1})t^{2N}. \end{align*} $$

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

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

Setting $a=2N+1$ and $b=2M$ , the right-hand side is . Hence, composing (21) with (22) gives an isomorphism .

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.

Proof. Set , and below write $\equiv $ for an equality mod I. Since , multiplying on the left by y and on the right by y gives

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

Thus, $y^3\in I$ . The final statement follows immediately.

Proof. Continue to write $\equiv $ for an equality mod $(\unicode{x3b4} _xf,\unicode{x3b4} _yf )$ . Then

(since δ _x f ≡ 0)

(y ³ ≡ 0 by 6.15)

(xy ≡−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 $2m\ge 4n-2$ 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 $(\unicode{x3b4} _xf,\unicode{x3b4} _yf)$ , it also holds mod $(\kern -2pt(\unicode{x3b4} _xf,\unicode{x3b4} _yf )\kern -2pt)$ , and hence, $x^{4n-2}(1-x^{2m-(4n-2)}g)=0$ in . But there, $1-x^{2m-(4n-2)}g$ is a unit, and so it follows that $x^{4n-2}=0$ in , as required.

Proof. By 6.16, we have $x^{4n-2}\in (\kern -2pt(\unicode{x3b4} _xf,\unicode{x3b4} _yf )\kern -2pt)$ and $x^{4n-2}\in (\kern -2pt( y^2+2nx^{2n-1},xy+yx)\kern -2pt)$ . Since $2m\geq 4n-2$ , it follows that $x^{2m}$ belongs to both of the ideals above, and thus,

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

Proof. The fact that the stated list covers all cases follows from 6.14, using 6.17 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 , whereas , 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 [Reference Brown and WemyssBW1, 4.7], or by using [Reference KawamataKa, 5.10] directly. Further, all members of family five are also mutually non-isomorphic for dimension reasons since for f in family five and by either [Reference van GarderenvG, §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 for f in family five is even, and the dimension of for g in families three and four is odd [Reference KawamataKa, 5.10], so there can be no such isomorphisms.

Proof. By the Splitting Lemma 4.5, the condition implies that for some $g\in \mathbb {C}\langle \kern -2.5pt\langle x,y\rangle \kern -2.5pt\rangle _{\geq 3}$ . The condition 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 $g_3^{ \mathrm {ab} }$ has either two or three distinct linear factors. The options for all such g thus follow from combining 6.7 and 6.18

The fact that follows since by linear change in coordinates, and by 6.5(1). The statements that for all $m\geq 2$ and 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 8.16) 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 6.7 and 6.18, 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 6.7 to the normal forms of 6.18, 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.

Proof. When is a central vertex (which covers all cases except the awkward one in 7.1), the isomorphism is a direct application of [Reference MellitM2, Theorem 1]; see also [Reference Crawley-BoeveyCB, p53]. The final remaining case, when e is the vertex marked $5$ in $E_8$ , can be proved using Auslander–Reiten theory; alternatively, we may simply observe that is the factor of the algebra in [Reference KarmazynK, 1.3(5)] by the ideal generated (in the notation there) by . The result [Reference KarmazynK, 1.3(5)] then gives a presentation

which is clearly isomorphic to the algebra in (24).

Proof. Set . Then it is clear that . This set is clearly a two-sided ideal of , and further, consists of units in . These two properties imply that equals the Jacobson radical of ; see, for example, [Reference LamL1, 4.5]. The fact that is clear.

Proof. (1) In all cases, $\unicode{x3b4} _yf=xy+yx$ , and so certainly $x^2$ commutes with y in . Obviously, $x^2$ commutes with x; thus, since we are considering closed ideals, it follows that $x^2$ is central in . Similarly, $y^2$ is central.