Let $ V $ be a vector space with a decreasing filtration $$ V = F_0 V \supseteq F_1 V \supseteq F_2 V \supseteq\cdots $$ We define the associated graded of $ V $ to be $$ gr V := \oplus_{k=0}^\infty F_k V / F_{k+1} V $$ Of course $ gr V $ can also be regarded as a filtered vector space and we have a canonical isomorphism $gr (gr V) = gr V $.

We say that $ V $ ``admits an expansion'' if there is an isomorphism of filtered vector spaces between $ gr V $ and $ V $, which becomes the identity map after applying $ gr $ to both $ gr V $ and $ V $.

This condition is equivalent to the existence of subspaces $ W_k \subset F_k V $ such that $ F_k V = W_k \oplus F_{k+1} V $ and $ V = \oplus_k W_k $.

Note that not every filtered vector space admits an expansion. For example, the vector space $ V = \mathbb C[[x]] $ with the filtration $ F_k V = x^k \mathbb C[[x]]$ does not admit an expansion. On the other hand, $ V = \mathbb C[x] $ with the same filtration does admit an expansion.

Here are my questions:

  1. Does this property have a different name in the literature?
  2. Let $V, W $ be two filtered vector spaces which admit expansions. Suppose that I have a filtration-preserving map $ \phi : V \rightarrow W $ such that $ gr \phi : gr V \rightarrow gr W $ is an isomorphism. Can I conclude that $ \phi $ is an isomorphism?
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    $\begingroup$ 2. No, because look at the inclusion of the polynomial ring in the ring of power series (both filtered in your way). But I guess it is true if both $V $ and $W $ satisfy your condition, and maybe even if $W $ does. $\endgroup$ – darij grinberg Jun 29 '17 at 11:29
  • $\begingroup$ Actually, here's a slightly stronger claim: If the filtration on $W$ "admits an expansion", and if the filtration on $V$ has the property that $\bigcap\limits_{k\geq 0} F_k V = 0$, then any filtration-preserving linear map $\phi : V \to W$ whose associated graded map $\operatorname{gr} \phi : \operatorname{gr} V \to \operatorname{gr} W$ is an isomorphism must itself be an isomorphism. But both conditions are important; otherwise, the stupid filtration $V \supseteq V \supseteq V \supseteq \cdots$ would cause the zero map to the zero space to be an isomorphism. $\endgroup$ – darij grinberg Jun 29 '17 at 15:08
  • $\begingroup$ Sorry, I meant to demand that both V and W admit expansions. I am editing the question now. $\endgroup$ – Joel Kamnitzer Jun 29 '17 at 21:59
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    $\begingroup$ Ouch. Now I think the answer to 2. is "No", against everything I have written here so far. The $\mathbb{Q}$-algebra homomorphism $\mathbb{Q}\left[x\right] \to \mathbb{Q}\left[x\right]$ sending $x$ to $x^2 + x$ (where the filtration comes from the usual grading by degree, so $F_k V = x^k \mathbb{Q}\left[x\right]$) respects the filtration (right??) and its associated graded is the identity (right????), but it is not an isomorphism (this one at least I know even at this time of the night). Please check! (I'm extremely unused to decreasing filtrations.) $\endgroup$ – darij grinberg Jun 30 '17 at 0:30
  • $\begingroup$ Yes, I believe this is a good counterexample to 2. Thanks, now I can stop thinking about it! $\endgroup$ – Joel Kamnitzer Jun 30 '17 at 14:41

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