Existence of $\mathbb{N}$-grading compatible with LNDs.

Question

Let $B$ be a finitely generated integral $\mathbb{C}$-domain. Let $\partial:B\to B$ be a LND, locally nilpotent derivation, i.e. a $\mathbb{C}$-linear map satisfying

1. Leibniz rule: $\partial(fg)=f\partial(g)+\partial(f)g$ for $f,g\in B$;
2. locally nilpotent: for all $f\in B$, there exists $n\in\mathbb{N}_+$ such that $\partial^n(f):=\partial\circ\cdots\circ \partial (f)=0$.

An $\mathbb{N}$-grading on $B$ is a decomposition into subspaces $$B=\bigoplus_{m\in\mathbb{N}}B_d$$ such that $B_{m_1}\cdot B_{m_2}\subseteq B_{m_1+m_2}$. We say this $\mathbb{N}$-grading grades $\partial$ with weight $d\in\mathbb{N}_+$ if $$\partial (B_m)\subseteq B_{m-d},\quad \forall m\in\mathbb{N}$$ where $B_{-1}=B_{-2}=\cdots=0$. In particular $B_0,\cdots,B_{d-1}\subseteq\ker(\partial)$. We say an $\mathbb{N}$-grading is $\partial$-compatible if for some $d\in\mathbb{N}_+$ the $\mathbb{N}$-grading grades $\partial$ with weight $d$.

My question is the following.

• Given $\partial:B\to B$ as above, does there always exist a $\partial$-compatible $\mathbb{N}$-grading?

Answer 1

We prove the following.

Lemma. If $\partial:B\to B$ is a LND, homogeneous of degree $-d<0$ with respect to an $\mathbb{N}$-grading on $B$, then $1\in\partial B\iff 1\in(\partial B).$

Proof. Assume $1\in(\partial B)$. Then for some $f_i,g_i\in B$ $$1=\sum_if_i\partial g_i. $$ Decompose by weight $$f_i=f_{i,0}+f_{i,>0},\quad f_{i,0}\in B_0,\;f_{i,>0}\in \bigoplus_{m>0}B_m$$ $$g_i=g_{i,<d}+g_{i,d}+g_{i,>d},\quad g_{i,<d}\in\bigoplus_{0\leq m<d}B_m,\;g_{i,d}\in B_d,\;g_{i,>d}\in\bigoplus_{m>d}B_m. $$

Then $$1=\sum_i f_{i,0}\partial g_{i,d}+\mathrm{err},\quad\mathrm{err}\in\bigoplus_{m>0}B_m.$$ In particular $s=\sum_if_{i.0}g_{i,d}\in B_d$ is a slice, i.e. $1\in\partial B$. QED.

Consider the action of $\mathbb{G}_a$ on $\mathrm{SL}_2$. Let $B=\mathcal{O}(\mathrm{SL}_2)$. The action is free, i.e. $1\in(\partial B)$ (section 1.5.2 of Freudenburg - Algebraic Theory of Locally Nilpotent Derivations (2017) or by direct calculation), and by the lemma above, there is a slice. By the Slice Theorem, we have $B\cong B^{\mathrm{G}_a}[t]\cong \mathbb{C}[a,c,t]$, a contradiction.