10
$\begingroup$

$\DeclareMathOperator\gr{gr}$Let $ V $ be a vector space with a decreasing filtration $$ V = F_0 V \supseteq F_1 V \supseteq F_2 V \supseteq\dotsb .$$ We define the associated graded of $ V $ to be $$ \gr V := \bigoplus_{k=0}^\infty F_k V / F_{k+1} V. $$ 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 = \bigoplus_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?
$\endgroup$
6
  • 6
    $\begingroup$ 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. $\endgroup$ Jun 29, 2017 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. $\endgroup$ Jun 29, 2017 at 15:08
  • $\begingroup$ Sorry, I meant to demand that both V and W admit expansions. I am editing the question now. $\endgroup$ Jun 29, 2017 at 21:59
  • 4
    $\begingroup$ 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.) $\endgroup$ Jun 30, 2017 at 0:30
  • $\begingroup$ Yes, I believe this is a good counterexample to 2. Thanks, now I can stop thinking about it! $\endgroup$ Jun 30, 2017 at 14:41

2 Answers 2

2
$\begingroup$

Too long for a comment, I was wondering about the cost of completely unfolding Darij's excellent argument. In fact, we have a characterisation of that sort of perturbations of identity that are isomorphisms. The statement is as follows
Lemma. Let $k$ be a field and $x$ an indeterminate. For every $Q\in k[x]$, let $f_Q$ be the morphism of $k[x]$ sending $x$ to $x+x^2Q$. Then $f_Q$ is an isomorphism iff $Q=0$.
Proof. One way being obvious (if $Q=0$ $f_Q=Id$), it is sufficient to prove that, if $Q\not=0$ then $x\notin Im(f_Q)=f_Q(k[x])$.
It is not difficult to see that $Im(f_Q)$ is a subalgebra as follows $$Im(f_Q)\subset k\oplus (x+x^2Q)k[x]$$ which does not contain $x$ (indeed $P\in Im(f_Q)$ implies that $P(x)-P(0)$ can be divided by $(x+x^2Q)$ which is true for $P=x$ iff $Q=0$).

$\endgroup$
1
$\begingroup$
  1. NO. take V=W. If the map is identity on the grading, it only means the diagonal blocks are each identity. But the map only needs to be lower triangular. Restriction on the diagonal blocks leaves a lot of freedom for the map.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.