Locally nilpotent derivation

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

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

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

The set of locally nilpotent derivations of a ring ${\displaystyle A}$ is denoted by ${\displaystyle \operatorname {LND} (A)}$. Note that this set has no obvious structure: it is neither closed under addition (e.g. if ${\displaystyle \partial _{1}=y{\tfrac {\partial }{\partial x}}}$, ${\displaystyle \partial _{2}=x{\tfrac {\partial }{\partial y}}}$ then ${\displaystyle \partial _{1},\partial _{2}\in \operatorname {LND} (k[x,y])}$ but ${\displaystyle (\partial _{1}+\partial _{2})^{2}(x)=x}$, so ${\displaystyle \partial _{1}+\partial _{2}\not \in \operatorname {LND} (k[x,y])}$) nor under multiplication by elements of ${\displaystyle A}$ (e.g. ${\displaystyle {\tfrac {\partial }{\partial x}}\in \operatorname {LND} (k[x])}$, but ${\displaystyle x{\tfrac {\partial }{\partial x}}\not \in \operatorname {LND} (k[x])}$). However, if ${\displaystyle [\partial _{1},\partial _{2}]=0}$ then ${\displaystyle \partial _{1},\partial _{2}\in \operatorname {LND} (A)}$ implies ${\displaystyle \partial _{1}+\partial _{2}\in \operatorname {LND} (A)}$[3] and if ${\displaystyle \partial \in \operatorname {LND} (A)}$, ${\displaystyle h\in \ker \partial }$ then ${\displaystyle h\partial \in \operatorname {LND} (A)}$.

Let ${\displaystyle A}$ be an algebra over a field ${\displaystyle k}$ of characteristic zero (e.g. ${\displaystyle k=\mathbb {C} }$). Then there is a one-to-one correspondence between the locally nilpotent ${\displaystyle k}$-derivations on ${\displaystyle A}$ and the actions of the additive group ${\displaystyle \mathbb {G} _{a}}$ of ${\displaystyle k}$ on the affine variety ${\displaystyle \operatorname {Spec} A}$, as follows.[3] A ${\displaystyle \mathbb {G} _{a}}$-action on ${\displaystyle \operatorname {Spec} A}$ corresponds to an ${\displaystyle k}$-algebra homomorphism ${\displaystyle \rho \colon A\to A[t]}$. Any such ${\displaystyle \rho }$ determines a locally nilpotent derivation ${\displaystyle \partial }$ of ${\displaystyle A}$ by taking its derivative at zero, namely ${\displaystyle \partial =\epsilon \circ {\tfrac {d}{dt}}\circ \rho ,}$ where ${\displaystyle \epsilon }$ denotes the evaluation at ${\displaystyle t=0}$. Conversely, any locally nilpotent derivation ${\displaystyle \partial }$ determines a homomorphism ${\displaystyle \rho \colon A\to A[t]}$ by ${\displaystyle \rho =\exp(t\partial )=\sum _{n=0}^{\infty }{\frac {t^{n}}{n!}}\partial ^{n}.}$

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

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

However, in practice it often happens that ${\displaystyle \ker \partial }$ is known to be finitely generated: notably, by the Maurer–Weitzenböck theorem,[6] 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 ${\displaystyle \ker \partial }$ is finitely generated. If ${\displaystyle A=k[g_{1},\dots ,g_{n}]}$ is a finitely generated algebra over a field of characteristic zero, then ${\displaystyle \ker \partial }$ can be computed using van den Essen's algorithm,[7] as follows. Choose a local slice, i.e. an element ${\displaystyle r\in \ker \partial ^{2}\setminus \ker \partial }$ and put ${\displaystyle f=\partial r\in \ker \partial }$. Let ${\displaystyle \pi _{r}\colon \,A\to (\ker \partial )_{f}}$ be the Dixmier map given by ${\displaystyle \pi _{r}(a)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\partial ^{n}(a){\frac {r^{n}}{f^{n}}}}$. Now for every ${\displaystyle i=1,\dots ,n}$, chose a minimal integer ${\displaystyle m_{i}}$ such that ${\displaystyle h_{i}\colon =f^{m_{i}}\pi _{r}(g_{i})\in \ker \partial }$, put ${\displaystyle B_{0}=k[h_{1},\dots ,h_{n},f]\subseteq \ker \partial }$, and define inductively ${\displaystyle B_{i}}$ to be the subring of ${\displaystyle A}$ generated by ${\displaystyle \{h\in A:fh\in B_{i-1}\}}$. By induction, one proves that ${\displaystyle B_{0}\subset B_{1}\subset \dots \subset \ker \partial }$ are finitely generated and if ${\displaystyle B_{i}=B_{i+1}}$ then ${\displaystyle B_{i}=\ker \partial }$, so ${\displaystyle B_{N}=\ker \partial }$ for some ${\displaystyle N}$. Finding the generators of each ${\displaystyle B_{i}}$ and checking whether ${\displaystyle B_{i}=B_{i+1}}$ is a standard computation using Gröbner bases.[7]

Assume that ${\displaystyle \partial \in \operatorname {LND} (A)}$ admits a slice, i.e. ${\displaystyle s\in A}$ such that ${\displaystyle \partial s=1}$. The slice theorem[3] asserts that ${\displaystyle A}$ is a polynomial algebra ${\displaystyle (\ker \partial )[s]}$ and ${\displaystyle \partial ={\tfrac {d}{ds}}}$.

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

However, in general it is not true that ${\displaystyle X\to X//\mathbb {G} _{a}}$ is locally trivial. For example,[8] let ${\displaystyle \partial =u{\tfrac {\partial }{\partial x}}+v{\tfrac {\partial }{\partial y}}+(1+uy^{2}){\tfrac {\partial }{\partial z}}\in \operatorname {LND} (\mathbb {C} [x,y,z,u,v])}$. Then ${\displaystyle \ker \partial }$ is a coordinate ring of a singular variety, and the fibers of the quotient map over singular points are two-dimensional.

If ${\displaystyle \dim X=3}$ then ${\displaystyle \Gamma =X\setminus U}$ is a curve. To describe the ${\displaystyle \mathbb {G} _{a}}$-action, it is important to understand the geometry ${\displaystyle \Gamma }$. Assume further that ${\displaystyle k=\mathbb {C} }$ and that ${\displaystyle X}$ is smooth and contractible (in which case ${\displaystyle S}$ is smooth and contractible as well[9]) and choose ${\displaystyle \Gamma }$ to be minimal (with respect to inclusion). Then Kaliman proved[10] that each irreducible component of ${\displaystyle \Gamma }$ is a polynomial curve, i.e. its normalization is isomorphic to ${\displaystyle \mathbb {C} ^{1}}$. The curve ${\displaystyle \Gamma }$ for the action given by Freudenburg's (2,5)-derivation (see below) is a union of two lines in ${\displaystyle \mathbb {C} ^{2}}$, so ${\displaystyle \Gamma }$ may not be irreducible. However, it is conjectured that ${\displaystyle \Gamma }$ is always contractible.[11]

Let ${\displaystyle k}$ be a field of characteristic zero (using Kambayashi's theorem one can reduce most results to the case ${\displaystyle k=\mathbb {C} }$[14]) and let ${\displaystyle A=k[x_{1},\dots ,x_{n}]}$ be a polynomial algebra.

The intersection of the kernels of all locally nilpotent derivations of the coordinate ring, or, equivalently, the ring of invariants of all ${\displaystyle \mathbb {G} _{a}}$-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 ${\displaystyle \mathbb {C} ^{3}}$, it is not.[25]

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