Locally nilpotent derivations and triangularizability

Question

If $ k $ is a field of characteristic zero and $ \delta \in T_{\mathbb{A}^{n}_{k}/k} $, then $ \delta $ is triangular if $ \delta = \sum_{i=2}^{n} f_{i}(x_{1},\dots,x_{i-1}) \frac{\partial}{\partial x_{i}} $. A derivation $ \delta $ is triangularizable if there is an automorphism $ \phi: \mathbb{A}^{n}_{k} \to \mathbb{A}^{n}_{k} $ such that $ \phi \circ \delta \circ \phi^{-1} $ is triangular. A derivation $ \delta \in T_{\mathbb{A}^{n}_{k}/k} $ is locally nilpotent if for all $ f(X) \in k[X] $, there is an $ i \in \mathbb{N} $ such that $ \delta^{i}(f(X)) $ is equal to zero. An action of the linear algebraic group $ \mathbb{G}_{a} $ on a variety $ \operatorname{Spec}(A) $ over a field $ k $ of characteristic zero is determined by a locally nilpotent derivation.

Not all locally nilpotent derivations are triangular. For example if $ \delta $ is equal to $ x_{1} \frac{\partial}{\partial x_{2}}+x_{2} \frac{\partial}{\partial x_{3}} $, then the derivation $ \gamma $ equal to $ (x_{2}^{2}-2x_{1}x_{3}) \delta $ is locally nilpotent but not triangularizable. Because $ \delta $ is triangular, hence locally nilpotent, and $ \delta(x_{2}^{2}-2x_{1}x_{3}) $ is equal to zero, $ \gamma $ is locally nilpotent. In the following notes of Harm Derksen (see Locally nilpotent derivations), Derksen proves that $ \gamma $ is not triangularizable.

However, let us define a partial order $ \prec $ on derivations $ \delta \in \operatorname{Der}_{k}(k[X],k[X]) $ as follows. We say that $ \gamma \prec \delta $ if there exists $ f \in \ker(\delta) $ such that $ \gamma = f \delta $. We shall consider $ \delta $ as identical to $ c\delta $ for $ c \in k^{\ast} $. The order $ \prec $ is clearly antisymmetric, because if $ \gamma=f \delta $ and $ \delta =g \gamma $, then $ \gamma = gf \gamma $ and $ gf \in k^{\ast} $. It is transitive because if $ f \in k[X] $ and $ \gamma= f \delta $, and $ \delta = g \psi $, then $ \gamma = fg \psi $ amd $ f \in \ker(\psi) $.

For any $ \delta \in \operatorname{Der}_{k}(k[X],k[X]) $, there is a poset $ [\delta]_{\prec} $ which is equal to all derivations $ \gamma \in \operatorname{Der}_{k}(k[X],k[X]) $ such that $ \delta \prec \gamma $.

Because polynomials have a lower bound with respect to degree, there is a maximal element of any chain in this poset.

I claim that this maximal element is unique up to multiplication by an element of $ k^{\ast} $. If $ \gamma_{1},\gamma_{2} $ are two maximal elements then there exist polynomials $ f_{1}(X),f_{2}(X) $ such that $ \delta= f_{i}(X)\gamma_{i} $ for $ i =1,2 $. If $ \gamma_{i} $ is equal to $ \sum_{j=1}^{n} g_{i,j}(X) \frac{\partial}{\partial x_{i}} $, then $ f_{1}(X)g_{1,j}(X)=f_{2}(X)g_{2,j}(X) $ for $ 1 \le j \le n $.

Since $ k[X] $ is a UFD, any irreducible factor $ p(X) $ of $ f_{2}(X) $, either divides $ f_{1}(X) $ or divides $ g_{1,j}(X) $ for all $ 1 \le j \le n $. We may factor $ f_{2}(X) $ as $ h_{2}(X)r_{2}(X) $ where $ h_{2}(X) $ divides $ f_{1}(X) $ and $ r_{2}(X) $ divides $ g_{1,j}(X) $ for $ 1 \le j \le n $, but $ r_{2}(X) $ does not divide $ f_{1}(X) $. If $ r_{2}(X) \notin k^{\ast} $, then there is a derivation $ \psi $ equal to $ \sum_{j=1}^{n} g_{1,j}(X)/r_{2}(X) \frac{\partial}{\partial x_{j}} $ such that $ r_{2}(X)f_{1}(X) \psi = \delta $, and $ \gamma_{2} = r_{2}(X) \psi $. By maximality of $ \gamma_{2} $, the polynomial $ r_{2}(X) \in k^{\ast} $. Therefore $ f_{2}(X) $ divides $ f_{1}(X) $ and vice versa. Therefore $ \gamma_{1} = c \gamma_{2} $ for $ c \in k^{\ast} $.

In the case of the derivation $ (x_{2}^{2}-2x_{1}x_{3})\left(x_{1} \frac{\partial}{\partial x_{2}}+x_{2}\frac{\partial}{\partial x_{3}}\right) $, the maximal element is triangularizable.

Does anyone know if there is an example of a locally nilpotent derivation $ \delta \in \operatorname{Der}_{k}(k[X],k[X]) $ such that the maximal element $ \gamma $ of the poset $ [\delta]_{\prec} $ with respect to the partial order $ \prec $ is not triangularizable?

Answer 1

If $g\partial$ is a triangulable derivation, then $g$ does not have any isolated critical points (see Lemma 6.6. of your notes). In a similar vein, the set of fixed points $\operatorname{Fix}(\mathbb{G}_a)$ is equal to the zero set of $\operatorname{im}\partial$, and it was proved by Popov that if the action is triangulable then it is isomorphic to $\mathbb{A}^1 \times Z$ for some affine variety $Z$, which implies $\operatorname{Fix}(\mathbb{G}_a)$ can not contain any isolated singularities. An extension of this fact was proved by Freudenberg which states that any hypersurface $H \subseteq \operatorname{Fix}(\mathbb{G}_a)$ is isomorphic to $\mathbb{A}^{\operatorname{rank} \partial} \times Z$ for some affine variety $Z$, where $\operatorname{rank} \partial$ is a generalization of the usual linear algebraic rank, which is also defined by Freudenberg (these coincide for derivations linearized in the coordinates $x_1, \ldots, x_n$).

Freudenburg gives an example of a locally nilpotent derivation $\Delta$ of maximal rank (and hence non-triangulable) on $k[x,y,z]$ defined as follows: let $f = xz + y^2$ and $g = zf^2 + 2x^2yf - x^5$. Then $\Delta$ is given by the Jacobian $$ \Delta h = \frac{\partial(f, g, h)}{\partial(x,y,z)}. $$

The points at which $\Delta x$, $\Delta y$, and $\Delta z$ all vanish can be calcuated to be the line $x = y = 0$. From here it is not hard to see $\Delta$ is maximal (in your sense), since it can not be written in the form $p \gamma$ for any locally nilpotent derivation $\gamma$ and element $p$ of nonzero degree.

References

Popov, V. L., On actions of $\mathbb{G}_ a$ on $\mathbb{A}^n$, Algebraic groups, Proc. Symp., Utrecht/Neth. 1986, Lect. Notes Math. 1271, 237-242 (1987). ZBL0685.14027.

Freudenburg, Gene, Triangulability criteria for additive group actions on affine space, J. Pure Appl. Algebra 105, No. 3, 267-275 (1995). ZBL0857.14024.

Freudenburg, Gene, Actions of $G_a$ on $\mathbb{A}^3$ defined by homogeneous derivations, J. Pure Appl. Algebra 126, No. 1-3, 169-181 (1998). ZBL0904.14027.