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 $$X = \operatorname{Spec}(A)$$. 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 $$\partial\colon\, A\to A $$ 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 $$\deg \partial a=\deg a +d$$ for every $$a\in A$$.

The set of locally nilpotent derivations of a ring $$A$$ is denoted by $$\operatorname{LND}(A)$$. Note that this set has no obvious structure: it is neither closed under addition (e.g. if $$\partial_{1}=y\tfrac{\partial}{\partial x}$$, $$\partial_{2}=x\tfrac{\partial}{\partial y}$$ then $$\partial_{1},\partial_{2}\in \operatorname{LND}(k[x,y])$$ but $$(\partial_{1}+\partial_{2})^{2}(x)=x$$, so $$\partial_{1}+\partial_{2}\not\in \operatorname{LND}(k[x,y])$$) nor under multiplication by elements of $$A$$ (e.g. $$\tfrac{\partial}{\partial x}\in \operatorname{LND}(k[x])$$, but $$x\tfrac{\partial}{\partial x}\not\in\operatorname{LND}(k[x])$$). However, if $$[\partial_{1},\partial_{2}]=0$$ then $$\partial_{1},\partial_{2}\in \operatorname{LND}(A)$$ implies $$\partial_{1}+\partial_{2}\in \operatorname{LND}(A)$$ and if $$\partial\in \operatorname{LND}(A)$$, $$h\in\ker\partial$$ then $$h\partial\in \operatorname{LND}(A)$$.

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

It is easy to see that the conjugate actions correspond to conjugate derivations, i.e. if $$\alpha\in \operatorname{Aut} A$$ and $$\partial\in \operatorname{LND}(A)$$ then $$\alpha\circ\partial \circ \alpha^{-1}\in \operatorname{LND}(A)$$ 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 $$\mathbb{G}_{a}$$-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 $$X/\!/\mathbb{G}_{a}$$ is affine. By Zariski's finiteness theorem, it is true if $$\dim X\leq 3$$. On the other hand, this question is highly nontrivial even for $$X=\mathbb{C}^{n}$$, $$n\geq 4$$. For $$n\geq 5$$ 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 $$\pi_{r}\colon\, A\to (\ker \partial)_{f}$$ be the Dixmier map given by $$\pi_{r}(a)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\partial^{n}(a)\frac{r^{n}}{f^{n}}$$. Now for every $$i=1,\dots, n$$, chose a minimal integer $$m_{i}$$ such that $$h_{i}\colon = f^{m_{i}}\pi_{r}(g_{i})\in \ker\partial$$, 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 $$\partial\in\operatorname{LND}(A)$$ 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 $$\partial=\tfrac{d}{ds}$$.

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 $$\pi\colon\,X\to X//\mathbb{G}_{a}$$ is affine (e.g. when $$\dim X\leq 3$$ by the Zariski theorem), then it has a Zariski-open subset $$U$$ such that $$\pi^{-1}(U)$$ is isomorphic over $$U$$ to $$U\times \mathbb{A}^{1}$$, where $$\mathbb{G}_{a}$$ acts by translation on the second factor.

However, in general it is not true that $$X\to X//\mathbb{G}_{a}$$ is locally trivial. For example, let $$\partial=u\tfrac{\partial}{\partial x}+v\tfrac{\partial}{\partial y}+(1+uy^2)\tfrac{\partial}{\partial z}\in \operatorname{LND}(\mathbb{C}[x,y,z,u,v])$$. 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 $$\Gamma=X\setminus U$$ is a curve. To describe the $$\mathbb{G}_{a}$$-action, it is important to understand the geometry $$\Gamma$$. Assume further that $$k=\mathbb{C}$$ and that $$X$$ is smooth and contractible (in which case $$S$$ is smooth and contractible as well ) and choose $$\Gamma$$ to be minimal (with respect to inclusion). Then Kaliman proved that each irreducible component of $$\Gamma$$ is a polynomial curve, i.e. its normalization is isomorphic to $$\mathbb{C}^{1}$$. The curve $$\Gamma$$ for the action given by Freudenburg's (2,5)-derivation (see below) is a union of two lines in $$\mathbb{C}^{2}$$, so $$\Gamma$$ may not be irreducible. However, it is conjectured that $$\Gamma$$ is always contractible.

Example 1
The standard coordinate derivations $$\tfrac{\partial}{\partial x_i}$$ of a polynomial algebra $$k[x_1,\dots, x_n]$$ are locally nilpotent. The corresponding $$\mathbb{G}_a$$-actions are translations: $$t\cdot x_{i}=x_{i}+t$$, $$t\cdot x_{j}=x_{j}$$ for $$j\neq i$$.

=== Example 2 (Freudenburg's (2,5)-homogeneous derivation ) === Let $$f_1=x_1x_3-x_2^2$$, $$f_2=x_3f_1^2+2x_1^2x_2f_1+x^5$$, and let $$\partial$$ be the Jacobian derivation $\partial(f_{3})=\det \left[\tfrac{\partial f_{i}}{\partial x_{j}}\right]_{i,j=1,2,3}$. Then $$\partial\in \operatorname{LND}(k[x_1,x_2,x_3])$$ and $$\operatorname{rank}\partial=3$$ (see below); that is, $$\partial$$ annihilates no variable. The fixed point set of the corresponding $$\mathbb{G}_{a}$$-action equals $$\{x_1=x_2=0\}$$.

Example 3
Consider $$Sl_2(k)=\{ad-bc=1\}\subseteq k^{4}$$. The locally nilpotent derivation $$a\tfrac{\partial}{\partial b}+c\tfrac{\partial}{\partial d}$$ of its coordinate ring corresponds to a natural action of $$\mathbb{G}_a$$ on $$Sl_2(k)$$ via right multiplication of upper triangular matrices. This action gives a nontrivial $$\mathbb{G}_a$$-bundle over $$\mathbb{A}^{2}\setminus \{(0,0)\}$$. However, if $$k=\mathbb{C}$$ 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 $$k=\mathbb{C}$$ ) and let $$A=k[x_1,\dots, x_n]$$ be a polynomial algebra.

$n = 2$ ($G_{a}$-actions on an affine plane)
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$n = 3$ ($G_{a}$-actions on an affine 3-space)
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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 $$\leq 2$$ by Rentschler's theorem above, but it is not true for $$n\geq 3$$.

The derivation of $$k[x_1,x_2,x_3]$$ given by $$x_1\tfrac{\partial}{\partial x_2}+2x_2x_1\tfrac{\partial}{\partial x_3}$$ is not triangulable. Indeed, the fixed-point set of the corresponding $$\mathbb{G}_{a}$$-action is a quadric cone $$x_2x_3=x_2^2$$, while by the result of Popov, a fixed point set of a triangulable $$\mathbb{G}_{a}$$-action is isomorphic to $$Z\times \mathbb{A}^{1}$$ for some affine variety $$Z$$; and thus cannot have an isolated singularity.
 * Bass's example

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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 $$\mathbb{G}_{a}$$-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 $$\mathbb{C}^{3}$$, it is not.