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Let $V$ be a vector space with an increasing filtration $V_j, j\in \mathbb Z$, we assume the filtration is Hausdorff $\bigcap_j V_j=0$ and exhaustive $\bigcup_j V_j=V$. Consider the associated graded space $\text{gr}V=\bigoplus_j V_j/{V_{j-1}}$, when is $\text{gr}V\cong V$ as a vector space?

In the case of $V_j$ is bounded below, say if $V_0=0$, then we do have $\text{gr}V\cong V$. But if it's bounded above, then $V=K^{\mathbb N}=\{(a_j):j\geq 0, a_j\in K\}$ with filtration $V_{-k}=\{(a_j)\in V:a_0=\cdots=a_k=0\}$ is a counterexample, since $\text{gr}V$ is direct sum of countable many $K$'s, while $V$ is a direct product of them.(this example can also be written as the fact that $\text{gr}(k[[t]])=k[t]$ with filtration given by the ideal $(t^n)$)

Is above arguments correct? Is there a general criterion of when is $\text{gr}V\cong V$?

Please correct me if I said anything wrong! Any comment or reference is welcome! It would also help me a lot to refer me to a detailed refrence on filtered space and algebra!

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    $\begingroup$ Two vector spaces are isomorphic if and only if they have the same dimensions. So the two are isomorphic if and only if $\sum_i \dim(V_i/V_{i+1})=\dim V$. A more natural question is when there exists a ``natural'' isomorphism, i.e. a basis $B=\bigsqcup_i B_i$ of $V$ such that $B_i$ is a basis of $V_i$ over $V_{i-1}$, but I don't think I can say much about this either. (I don't think there is a genuinely natural isomorphism in any nontrivial case, though.) $\endgroup$
    – tomasz
    Commented Mar 6 at 23:44
  • $\begingroup$ Do we have any natural map between $V$ and $\text{gr}(V)$? $\endgroup$
    – Eric Ley
    Commented Mar 6 at 23:56
  • $\begingroup$ I don't think so, that is my point in the parenthetical remark at the end. There is a "natural" one though: for each $i$, let $B_i\subseteq V_i$ be maximal linearly independent over $V_{i-1}$. Then let $B_\infty\subseteq V$ be maximal linearly independent over $\operatorname{Lin}\bigcup_i B_i$. Then $B=B_\infty\cup\bigcup B_i$ is a basis of $V$ and you can use it to define a "natural" map $V\to \operatorname{gr}(V)$, but this is heavily dependent on a lot of choices. I think whether or not $B_\infty=\emptyset$ does not depend on the choices though, but I'm not sure. $\endgroup$
    – tomasz
    Commented Mar 7 at 0:08
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    $\begingroup$ Equivalently, instead of bases, you can choose for each $i$ a $V_i'\leq V_i$ such that $V_i=V_{i-1}\oplus V_i'$, and then a $V_\infty\leq V$ such that $V_\infty\oplus \bigoplus_i V_i'=V$. Then $\bigoplus_i V_i'$ is naturally isomorphic to $\operatorname{gr}(V)$ (but the choice of $V_i'$ and $V_\infty$ themselves is anything but natural). $\endgroup$
    – tomasz
    Commented Mar 7 at 0:13
  • $\begingroup$ So after choosing a split, we do have a "natural" imbedding $\text{gr}(V)\to V$, at least we know $V$ is bigger right? :) $\endgroup$
    – Eric Ley
    Commented Mar 7 at 0:17

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