# Short exact sequences of filtered vector spaces are split

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?

• 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 ... Jun 29, 2021 at 19:10
• ... 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. Jun 29, 2021 at 19:10

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.