Let $k$ be a field. By a filtered vector space over $k$, I mean a pair $(V,F)$ where $V$ is a finite dimensional $k$-vector space and $F=(F^pV)_{p\in \mathbb{Z}}$ is an increasing filtration of $V$ by $k$-subspaces. A morphism of filtered vector spaces $(V,F)\to (W,G)$ is a morphism of the underlying vector spaces $f:V\to W$ which preserves the filtration, that is, $f(F^pV)\subset G^pW$ for all $p\in \mathbb{Z}$. The morphism $f$ is said to be strict if, for all $p\in \mathbb{Z}$, we have $f(F^pV)=f(V)\cap G^pW$.

It is claimed somewhere in the literature that any short exact sequences of filtered vector spaces with strict morphisms is split. Why is this true?

Let $(V_i,F_i)$ $(i\in \{1,2,3\})$ be filtered vector spaces and let $0\to V_1\stackrel{i}{\to} V_2\stackrel{p}{\to} V_3\to 0$ be a short exact sequence of $k$-vector spaces such that $i$ and $p$ are strict for the respective filtrations. The question is equivalent to: can we chose a splitting $s:V_3\to V_2$ which preserves the filtration?

  • $\begingroup$ Would it suffice to find a splitting $V_2 \to V_1$ that preserves the filtration and is strict? I think it can be done fairly directly: Rename $V_1$ and $V_2$ as $U$ and $V$, and rename the filtered parts $F^p V_1$ and $F^p V_2$ as $U_p$ and $V_p$. Then, the strictness of $i$ says that $U_p = V_p \cap U$ for all $p$. Thus, for each $p$, we have $V_p \cap U_{p+1} = U_p$. Thus, any linear map $V_p \to U_p$ that splits the canonical inclusion $i_p : U_p \to V_p$ can be extended to a linear map $V_{p+1} \to U_{p+1}$ that splits the canonical inclusion $i_{p+1}$ (just project ... $\endgroup$ Jun 29, 2021 at 19:10
  • $\begingroup$ ... from $V_{p+1}$ to the subspace $V_p + U_{p+1}$ somehow, and then use the surjection $V_p + U_{p+1} \to U_{p+1}$ obtained from the surjection $V_p \to U_p$ by adding $U_{p+1}$ to both sides). It shouldn't be hard to show that this is strict. $\endgroup$ Jun 29, 2021 at 19:10

1 Answer 1


Since $i$ and $p$ are strict, we know two things:

  • $p(F^j_2(V_2)) = F^j_3(V_3)$ for each $j$.
  • We can identify $(V_1,F_1)$ with a filtration of a subspace $V_1\subseteq V_2$ such that $F_1^jV_1=V_1\cap F_2^jV_2$ for each $j$.

In particular, since $V_1=\ker p$, we know that $\ker(p|F_2^j(V_2))=F_2^j(V_2)\cap \ker p$ and thus $$F_3^j(V_3)\cong F_2^j(V_2)/(F_2^j(V_2)\cap \ker p)$$ for each $j$. By repeatedly applying basis extension, we can choose a basis $v_1,\dots,v_n$ for $\ker p$ that respects the filtration, in that for each $j$ there is a $n_j$ such that a basis for $F^j_2(V_2)\cap\ker p$ is given by $v_1,\dots,v_{n_j}$. By repeatedly applying basis extension again, we can choose additional vectors $w_1,\dots,w_m$ such that for each $j$ there is an $m_j$ such that $v_1,\dots,v_{n_j},w_1,\dots,w_{m_j}$ is a basis for $F_2^j(V_2)$. By construction, for each $j$ the vectors $p(w_1),\dots,p(w_{m_j})$ form a basis for $F^j_3(V_3)$, and in particular $p(w_1),\dots,p(w_m)$ form a basis for $V_3$. Thus, we can use these to define an injective linear map $s:V_3\to V_2$ with $s(p(w_i))=w_i$ for each $i$.

Evidently, $s$ is a splitting of $p$ as a linear map that respects the filtration. What's more, $s(F_3^j(V_3))=s(V_3)\cap F_2^j(V_2)$ for each $j$, so it is strict.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .