User isac hedén - MathOverflow

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

2013-03-18T22:01:05Z

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 http://mathoverflow.net/questions/119702/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.

Normality condition on graded algebra

2013-01-23T22:28:46Z

Let $\mathbb G_a$ denote the additive group of complex numbers.

Definition: Let $V \subset Y$ be a dense open subset of the affine variety $Y$ and $\pi : P \longrightarrow V$ a $\mathbb G_a$-principal bundle. An affine extension is an affine $\mathbb G_a$-variety $\hat P$ together with a morphism $ \hat \pi: \hat P \longrightarrow Y$ and an equivariant open embedding $ \iota: P \hookrightarrow \hat P$, such that the diagram

$$ \begin{array}{ccc} P & \hookrightarrow & \hat P \\ \downarrow & & \downarrow \\ V & \hookrightarrow & Y \end{array} $$ is commutative and $\iota (P)= \hat \pi^{-1}(V).$

I am interested in affine extensions of the trivial $\mathbb G_a$-bundle over the affine plane punctured at the origin, i.e. $\mathbb A^2_*:=Sp(\mathbb C[x,y])\setminus{\mathbf o}$ with $\mathbf o:=(x,y)$, and I have the following description:

Proposition: If $P \longrightarrow \mathbb A_*^2$ is the trivial bundle, then any affine extension $\hat P \longrightarrow \mathbb A^2$ is of the form $$ \hat P = Sp (A), $$ where $$ A= \bigoplus_{\nu=0}^\infty \mathfrak m_\nu t^\nu \subset \mathbb C [x,y,t], $$ with a decreasing sequence $(\mathfrak m_\nu)$, $\nu \in \mathbb N$, of ideals $\mathfrak m_\nu \subset \mathbb C [x,y]$, such that

1. $ \mathfrak m_\nu \cdot \mathfrak m_\lambda \subset \mathfrak m_{\nu + \lambda} $ for all $\nu, \lambda\in \mathbb N$,

2. $\mathfrak m_0= \mathbb C [x,y]$, and

3. $V(\mathfrak m_\nu) \subset \mathbf o$ for $\nu > 0$.

On the other hand, every finitely generated $\mathbb C$-algebra of that form defines an affine extension of the trivial bundle.

Question: Does somebody know of a criterion on the sequence $(\mathfrak m_\nu)$$_{\nu \in \mathbb N}$ so that $A$ becomes normal?

Examples: a) If $\mathfrak m_\nu= \mathbb C [x,y]$ for all $\nu$ we have $\hat P \cong \mathbb A^2 \times \mathbb G_a$.

b) If $\mathfrak m_\nu =(x^m,y^n)^\nu$, we have $ A = \mathbb C [x,y,x^mt, y^nt]. $

In the second example, one can see that $A$ is normal for instance if $\mathfrak m_\nu=(x^2,y)^\nu$, but not if $\mathfrak m_\nu=(x^2,y^2)^\nu$ -- in the latter case I think the normalization would be defined by the sequence $\mathfrak m_\nu=(x^2,xy,y^2)^\nu$. Since $A = \mathbb C [x,y,x^mt, y^nt]\cong\mathbb C[x,y,u,v]/(x^mv-y^nu)$ in example b), $Sp(A)$ is a hypersurface in $\mathbb C^4$, so normality is equivalent to singularities being of codimension at least two.

Edit: If the question is difficult in general, I am also interested in the following special case: For which monomial ideals $\mathfrak m\subset\mathbb C[x,y]$, is the ring defined by the sequence $\mathfrak m^\nu$ normal? In this situation I would expect something like: $A$ is normal iff the support of $\mathfrak m^\nu$ consists of all lattice points in the convex hull of the support of $\mathfrak m^\nu$ in $\mathbb R^2$. Here the support of $\mathfrak m^\nu$ is the set of pairs $(k,l)\in\mathbb N^2$ such that $x^ky^l\in\mathfrak m^\nu$.

Answer by Isac Hedén for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications

2011-08-13T12:51:02Z

I've been asking around a little, and nobody seems to be able to tell whether or not Russell's hypersurface is analytically $\mathbb C^3$. Adrien Dubouloz shows in this article that the Makar-Limanov invariant of its product with $\mathbb C$ is trivial. Maybe that helps.

Comment by Isac Hedén

2013-03-21T16:19:18Z

Thanks Paul! You are right - that is a necessary condition! See Edit2 for what I think is "the precise" statement.

Comment by Isac Hedén

2013-01-26T18:08:48Z

Thanks for the answer! It looks like a good criterion for the case where $A$ is a Rees algebra. Unfortunately this is not always the case though, since we for instance could modify A by repeating every ideal once, i.e. take the ideal sequence $m_1,m_1,m_2, m_2,m_3,m_3,\ldots$ instead of $m_1,m_2,m_3,\ldots$ where all the $m_i$ are powers of a monomial ideal $m_1$ -- without violating the three conditions that the ideal sequence should satisfy. Could there be a way of associating a Rees algebra to $A$ which has the property of being normal exactly when $A$ is, e.g. by means of a Rees valuation?

Comment by Isac Hedén

2013-01-24T14:09:08Z

... Did you have a particular article by Kei-ichi Watanabe i mind? Thanks again for your answer!

Comment by Isac Hedén

2013-01-24T14:07:28Z

Thanks for the answer! The condition seems to be very strong, as formulated, and I don't really see how it is related to normality. Did you mean the condition should hold for all elements with $\nu(f)\geq 1$? Otherwise, if for instance we take $\mathfrak m$ to be a monomial ideal which contains a power of $x$, but not $x$ itself, and take the ideal sequence to be the powers of $\mathfrak m$, the condition fails (with $f=x$, since $\nu(f)=0$) although $A$ may very well be normal (as in example b)). What I'd really like is of course a condition which is both necessary and sufficient...