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Suppose that $A = \bigcup_{n=0}^{\infty} A_n$ is a filtered algebra over a field $k$. The associated graded algebra is $\mathrm{gr} A = \bigoplus_{n=0}^{\infty} A_n/A_{n-1}$, where we define $A_{-1} = (0)$. There is no canonical algebra map from $A$ to $\mathrm{gr} A$, but there is a well-defined function $\gamma : A \to \mathrm{gr} A$ given by $$ \gamma(x) = x + A_{n-1} \in A_n/A_{n-1}, $$ where $n$ is the unique natural number (or 0) such that $x \in A_n$ but $x \notin A_{n-1}$. (To forestall nitpicking, let's say that $\gamma(0) = 0 \in A_0$.) Of course, this map fails to be even additive, but it does exist.


Given a set of generators $\{x_i\}$ for $A$, when is it the case that the set $\{\gamma(x_i)\}$ generates $\mathrm{gr} A$?

Here is an easy example where this fails to happen. Let $\mathfrak{h}$ be the 3-dimensional Heisenberg Lie algebra (over $\mathbb{C}$, say), spanned by three elements $X,Y,Z$ with $[X,Y] = Z$ and $Z$ central. Let $A = U(\mathfrak{h})$ be its universal enveloping algebra with the usual filtration. Since $XY - YX = Z$, it follows that $A$ can be generated just by $X$ and $Y$. But Poincare-Birkhoff-Witt tells us that $\mathrm{gr} A \cong \mathbb{C}[X,Y,Z]$, which is certainly not generated just by $X$ and $Y$.

The problem here is with the relation $XY-YX=Z$: since $Z$ has lower degree than $XY$ and $YX$, it drops out of the relation in the associated graded.

Can anything be said about this, in general? Are there any nice criteria on the filtration and the generating set (and the relations, obviously) that ensure things don't go wrong in this way? Also I am amenable to making assumptions on the algebra $A$, for example that it is finitely generated or Noetherian (or ...?), if that helps. I do not want to increase the size of the generating set.

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It's hard for me to think about this question, because the associated-graded construction requires the structure of an additive category, but your "$\gamma$" isn't even a morphism of vector spaces. You can probably say the following without conditions (although I don't know how): the question of generating sets and relations can be encoded homologically, and there will be a spectral sequence connecting the filtered and graded versions. For example, you can "see" that $X,Y$ generate $\mathfrak h$ by looking at the Chevalley–Eilenberg complex, which has homology $1,2,2,1$ not $1,3,3,1$. –  Theo Johnson-Freyd Aug 26 '12 at 4:33
@Theo: to encode generators and relations homologically, you need appropriate nilpotency assumptions. For instance, if you replace $\mathfrak{h}$ by $sl_2$, you surely will not see generators and relations that way. –  Vladimir Dotsenko Aug 26 '12 at 9:24
Note that, with degree(Z)=2, we seem to repair the setting. –  Duchamp Gérard H. E. Aug 26 '12 at 10:38
OK, it is in view of chasing "gradation change" –  Duchamp Gérard H. E. Aug 26 '12 at 10:41
That's very near to the question I'm asking myself at the moment! In my case I already know that the $\gamma(x_i)$ generate $gr(A)$. Instead I would like know if the $\gamma(R_j)$ are a set of defining relations $gr(A)$ and this generating set if $R_j$ is a set of defining relations for $A$ and $x_i$. I would like to add that question as an addendum to the OP's question. –  Johannes Hahn Aug 26 '12 at 20:16

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