# Is every graded subalgebra of gr(A) equal to gr(B) for some subalgebra B of A?

Let $A$ be a finitely generated reduced (i.e. affine) $\mathbb C$-algebra, and $D:A\longrightarrow A$ a locally nilpotent derivation (i.e. $D$ is $\mathbb C$-linear, satisfies Leibniz' rule and $\forall f\in A\,\exists n\in\mathbb N: D^n(f)=0$). Then there is a filtration on $A$ given by $A_i:=\ker D^{i+1}$, and an associated graded algebra $$gr_D(A):=\bigoplus_{i=0}^\infty A_i/A_{i-1}.$$ (it is understood that $A_{-1}=\{0\}$). Given a $D$-invariant subalgebra $B\subset A$, we get an induced filtration (by taking $B_i=B\cap A_i$), and a subalgebra $gr_D(B)\subset gr_D(A)$. I can show that $B$ is uniquely determined by $gr_D(B)$, i.e. if $B_1$ and $B_2$ are two $D$-invariant subalgebars of $A$, then $gr(B_1)=gr(B_2)$ implies $B_1=B_2$. My question is: which graded subalgebras $C\subset gr_D(A)$ are of the form $gr_D(B)$ for some $D$-invariant subalgebra $B\subset A$, and also, how should $B$ be chosen?

Motivation: I am trying to understand affine extensions of $\mathbb G_a$-principal bundles over $\mathbb A_* ^2$ as in Normality condition on graded algebra. In the question above, $spec(A)\rightarrow \mathbb A_* ^2$ is a non-trivial $\mathbb G_a$-principal bundle -- it is known that all non-trivial $\mathbb G_a$-bundles on $\mathbb A_* ^2$ are affine. Affine extensions of this principal bundle are going to correspond to subalgebras of $A$ (containing $\mathcal O(\mathbb A^2_*)=\mathbb C[x,y]$). Now the structural $\mathbb G_a$-action on $spec(A)$ corresponds to a locally nilpotent derivation on $A$, and my hope is that it will be fruitful in one way or another to consider the graded algebra corresponding to this derivation (locally nilpotent derivations on $A$ correspond to $\mathbb G_a$-actions on $spec(A)$ in a natural way).

PS: Thanks for the downvote (which helped me realize that the formulation of the question was not so great at first)! I hope it is better now.

Edit: Added "graded" in the title, as suggested.

Edit2: Paul is right in his comment below! And I think we have the following description of $gr_D(B)$ as a $B_0$-module: The sequence of ideals $\mathfrak b_n:=D^n(B_n)\hookrightarrow B_0$ is decreasing and satisifies $\mathfrak b_n\mathfrak b_m\subset \mathfrak b_{n+m}$, and furthermore $$gr_D(B)\cong B_0\oplus\bigoplus_{n=1}^\infty \mathfrak b_ns^n\hookrightarrow B_0[s].$$ The isomorphism is induced by $$gr_D(B)_n\rightarrow \mathfrak b_ns^n,\quad b+B_{n-1}\mapsto \frac {D^n(b)}{n!}s^n.$$ The question remains: Is every graded submodule of $gr_D(A)$ of this form for some subalgebra $B\subset A$?

Example: One algebra for which I am particularily interested in the answer to this question would be $A=\mathbb C[x,y,u,v]/(xv-yu-1)$, i.e. $\mathcal O(SL_2)$, with locally nilpotent derivation given by $D(x)=D(y)=0$ and $D(u)=x$, $D(v)=y$. Then $$gr_D(A)\cong \mathbb C[x,y]\oplus\bigoplus_{n=1}^\infty (x,y)^n s^n$$

Edit3: Finally I am not so sure that $B$ is uniquely determined by $gr_D(B)$ -- the argument I had in mind seems not to work.

-
To begin with, counter to the title, $gr(B)$ had better be a graded subalgebra of $gr(A)$. – Allen Knutson Mar 19 '13 at 11:47
Well I guess the first thing to note is that $D(A_i)\subset A_{i-1}$, so $D$ induces a degree $(-1)$ homogeneous map $gr\; D$ on $gr_D(A)$; then a necessary (but perhaps not sufficient?) condition on $C$ is that it is stable under $gr\; D$. – Paul Levy Mar 19 '13 at 11:49
Thanks Paul! You are right - that is a necessary condition! See Edit2 for what I think is "the precise" statement. – Isac Hedén Mar 21 '13 at 16:19