Locally nilpotent derivation

In mathematics, a derivation $\partial$ of a commutative ring $A$ is called a locally nilpotent derivation (LND) if every element of $A$ is annihilated by some power of $\partial$.

One motivation for the study of locally nilpotent derivations comes from the fact that some of the counterexamples to Hilbert's 14th problem are obtained as the kernels of a derivation on a polynomial ring.

Over a field $k$ of characteristic zero, to give a locally nilpotent derivation on the integral domain $A$, finitely generated over the field, is equivalent to giving an action of the additive group $(k,+)$ to the affine variety $X=\mathrm{Spec}(A)$. Roughly speaking, an affine variety admitting "plenty" of actions of the additive group is considered similar to an affine space.

Definition

Let $A$ be a ring. Recall that a derivation of $A$ is a map $\partial :A\to A$ satisfying the Leibniz rule $\partial (ab)=(\partial a)b+a(\partial b)$ for any $a,b\in A$. If $A$ is an algebra over a field $k$, we additionally require $\partial$ to be $k$-linear, so $k\subseteq \ker \partial$.

A derivation $\partial$ is called a locally nilpotent derivation (LND) if for every $a\in A$, there exists a positive integer $n$ such that ${\partial }^n(a)=0$.

If $A$ is graded, we say that a locally nilpotent derivation $\partial$ is homogeneous (of degree $d$) if $\mathrm{deg}\partial a=\mathrm{deg}a+d$ for every $a\in A$.

The set of locally nilpotent derivations of a ring $A$ is denoted by $\mathrm{LND}(A)$. Note that this set has no obvious structure: it is neither closed under addition (e.g. if ${\partial }_1=y\frac{\partial }{\partial x}$, ${\partial }_2=x\frac{\partial }{\partial y}$ then ${\partial }_1,{\partial }_2\in \mathrm{LND}(k[x,y])$ but $({\partial }_1+{\partial }_2{)}^2(x)=x$, so ${\partial }_1+{\partial }_2\in ̸\mathrm{LND}(k[x,y])$) nor under multiplication by elements of $A$ (e.g. $\frac{\partial }{\partial x}\in \mathrm{LND}(k[x])$, but $x\frac{\partial }{\partial x}\in ̸\mathrm{LND}(k[x])$). However, if $[{\partial }_1,{\partial }_2]=0$ then ${\partial }_1,{\partial }_2\in \mathrm{LND}(A)$ implies ${\partial }_1+{\partial }_2\in \mathrm{LND}(A)$ and if $\partial \in \mathrm{LND}(A)$, $h\in \ker \partial$ then $h\partial \in \mathrm{LND}(A)$.

Relation to Template:Math-actions

Let $A$ be an algebra over a field $k$ of characteristic zero (e.g. $k=ℂ$). Then there is a one-to-one correspondence between the locally nilpotent $k$-derivations on $A$ and the actions of the additive group ${𝔾}_a$ of $k$ on the affine variety $\mathrm{Spec}A$, as follows. A ${𝔾}_a$-action on $\mathrm{Spec}A$ corresponds to a $k$-algebra homomorphism $\rho :A\to A[t]$. Any such $\rho$ determines a locally nilpotent derivation $\partial$ of $A$ by taking its derivative at zero, namely $\partial =ϵ\circ \frac{d}{dt}\circ \rho ,$ where $ϵ$ denotes the evaluation at $t=0$. Conversely, any locally nilpotent derivation $\partial$ determines a homomorphism $\rho :A\to A[t]$ by $\rho =\exp (t\partial )=\sum _{n=0}^{\mathrm{\infty }}\frac{t^n}{n!}{\partial }^n.$

It is easy to see that the conjugate actions correspond to conjugate derivations, i.e. if $\alpha \in \mathrm{Aut}A$ and $\partial \in \mathrm{LND}(A)$ then $\alpha \circ \partial \circ {\alpha }^{-1}\in \mathrm{LND}(A)$ and $\exp (t\cdot \alpha \circ \partial \circ {\alpha }^{-1})=\alpha \circ \exp (t\partial )\circ {\alpha }^{-1}$

The kernel algorithm

The algebra $\ker \partial$ consists of the invariants of the corresponding ${𝔾}_a$-action. It is algebraically and factorially closed in $A$. A special case of Hilbert's 14th problem asks whether $\ker \partial$ is finitely generated, or, if $A=k[X]$, whether the quotient $X//{𝔾}_a$ is affine. By Zariski's finiteness theorem, it is true if $\dim X\le 3$. On the other hand, this question is highly nontrivial even for $X={ℂ}^n$, $n\ge 4$. For $n\ge 5$ the answer, in general, is negative. The case $n=4$ is open.

However, in practice it often happens that $\ker \partial$ is known to be finitely generated: notably, by the Maurer–Weitzenböck theorem, it is the case for linear LND's of the polynomial algebra over a field of characteristic zero (by linear we mean homogeneous of degree zero with respect to the standard grading).

Assume $\ker \partial$ is finitely generated. If $A=k[g_1,\dots ,g_n]$ is a finitely generated algebra over a field of characteristic zero, then $\ker \partial$ can be computed using van den Essen's algorithm, as follows. Choose a local slice, i.e. an element $r\in \ker {\partial }^2\setminus \ker \partial$ and put $f=\partial r\in \ker \partial$. Let ${\pi }_r:A\to (\ker \partial {)}_f$ be the Dixmier map given by ${\pi }_r(a)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}{\partial }^n(a)\frac{r^n}{f^n}$. Now for every $i=1,\dots ,n$, chose a minimal integer $m_i$ such that $h_i:=f^{m_i}{\pi }_r(g_i)\in \ker \partial$, put $B_0=k[h_1,\dots ,h_n,f]\subseteq \ker \partial$, and define inductively $B_i$ to be the subring of $A$ generated by $\{h\in A:fh\in B_{i-1}\}$. By induction, one proves that $B_0\subset B_1\subset \dots \subset \ker \partial$ are finitely generated and if $B_i=B_{i+1}$ then $B_i=\ker \partial$, so $B_N=\ker \partial$ for some $N$. Finding the generators of each $B_i$ and checking whether $B_i=B_{i+1}$ is a standard computation using Gröbner bases.

Slice theorem

Assume that $\partial \in \mathrm{LND}(A)$ admits a slice, i.e. $s\in A$ such that $\partial s=1$. The slice theorem asserts that $A$ is a polynomial algebra $(\ker \partial )[s]$ and $\partial =\frac{d}{ds}$.

For any local slice $r\in \ker \partial \setminus \ker {\partial }^2$ we can apply the slice theorem to the localization $A_{\partial r}$, and thus obtain that $A$ is locally a polynomial algebra with a standard derivation. In geometric terms, if a geometric quotient $\pi :X\to X//{𝔾}_a$ is affine (e.g. when $\dim X\le 3$ by the Zariski theorem), then it has a Zariski-open subset $U$ such that ${\pi }^{-1}(U)$ is isomorphic over $U$ to $U\times {𝔸}^1$, where ${𝔾}_a$ acts by translation on the second factor.

However, in general it is not true that $X\to X//{𝔾}_a$ is locally trivial. For example, let $\partial =u\frac{\partial }{\partial x}+v\frac{\partial }{\partial y}+(1+u{y}^2)\frac{\partial }{\partial z}\in \mathrm{LND}(ℂ[x,y,z,u,v])$. Then $\ker \partial$ is a coordinate ring of a singular variety, and the fibers of the quotient map over singular points are two-dimensional.

If $\dim X=3$ then $\mathrm{\Gamma }=X\setminus U$ is a curve. To describe the ${𝔾}_a$-action, it is important to understand the geometry $\mathrm{\Gamma }$. Assume further that $k=ℂ$ and that $X$ is smooth and contractible (in which case $S$ is smooth and contractible as well) and choose $\mathrm{\Gamma }$ to be minimal (with respect to inclusion). Then Kaliman proved that each irreducible component of $\mathrm{\Gamma }$ is a polynomial curve, i.e. its normalization is isomorphic to ${ℂ}^1$. The curve $\mathrm{\Gamma }$ for the action given by Freudenburg's (2,5)-derivation (see below) is a union of two lines in ${ℂ}^2$, so $\mathrm{\Gamma }$ may not be irreducible. However, it is conjectured that $\mathrm{\Gamma }$ is always contractible.

Examples

Example 1

The standard coordinate derivations $\frac{\partial }{\partial x_i}$ of a polynomial algebra $k[x_1,\dots ,x_n]$ are locally nilpotent. The corresponding ${𝔾}_a$-actions are translations: $t\cdot x_i=x_i+t$, $t\cdot x_j=x_j$ for $j\ne i$.

Example 2 (Freudenburg's (2,5)-homogeneous derivation)

Let $f_1=x_1{x}_3-x_2^2$, $f_2=x_3{f}_1^2+2x_1^2{x}_2{f}_1+x^5$, and let $\partial$ be the Jacobian derivation $\partial (f_3)=\det {\left[\frac{\partial f_i}{\partial x_j}\right]}_{i,j=1,2,3}$. Then $\partial \in \mathrm{LND}(k[x_1,x_2,x_3])$ and $\mathrm{rank}\partial =3$ (see below); that is, $\partial$ annihilates no variable. The fixed point set of the corresponding ${𝔾}_a$-action equals $\{x_1=x_2=0\}$.

Example 3

Consider $S{l}_2(k)=\{ad-bc=1\}\subseteq k^4$. The locally nilpotent derivation $a\frac{\partial }{\partial b}+c\frac{\partial }{\partial d}$ of its coordinate ring corresponds to a natural action of ${𝔾}_a$ on $S{l}_2(k)$ via right multiplication of upper triangular matrices. This action gives a nontrivial ${𝔾}_a$-bundle over ${𝔸}^2\setminus \{(0,0)\}$. However, if $k=ℂ$ then this bundle is trivial in the smooth category

LND's of the polynomial algebra

Let $k$ be a field of characteristic zero (using Kambayashi's theorem one can reduce most results to the case $k=ℂ$) and let $A=k[x_1,\dots ,x_n]$ be a polynomial algebra.

Triangular derivations

Let $f_1,\dots ,f_n$ be any system of variables of $A$; that is, $A=k[f_1,\dots ,f_n]$. A derivation of $A$ is called triangular with respect to this system of variables, if $\partial f_1\in k$ and $\partial f_i\in k[f_1,\dots ,f_{i-1}]$ for $i=2,\dots ,n$. A derivation is called triangulable if it is conjugate to a triangular one, or, equivalently, if it is triangular with respect to some system of variables. Every triangular derivation is locally nilpotent. The converse is true for $\le 2$ by Rentschler's theorem above, but it is not true for $n\ge 3$.

Bass's example

The derivation of $k[x_1,x_2,x_3]$ given by $x_1\frac{\partial }{\partial x_2}+2x_2{x}_1\frac{\partial }{\partial x_3}$ is not triangulable. Indeed, the fixed-point set of the corresponding ${𝔾}_a$-action is a quadric cone $x_2{x}_3=x_2^2$, while by the result of Popov, a fixed point set of a triangulable ${𝔾}_a$-action is isomorphic to $Z\times {𝔸}^1$ for some affine variety $Z$; and thus cannot have an isolated singularity.

Makar-Limanov invariant

The intersection of the kernels of all locally nilpotent derivations of the coordinate ring, or, equivalently, the ring of invariants of all ${𝔾}_a$-actions, is called "Makar-Limanov invariant" and is an important algebraic invariant of an affine variety. For example, it is trivial for an affine space; but for the Koras–Russell cubic threefold, which is diffeomorphic to ${ℂ}^3$, it is not.

Further reading

• A Nowicki, the fourteenth problem of hilbert for polynomial derivations