Filtered vector spaces

(1) Definition (Filtered vector spaces).

  • A filtered vector space \(S\) is a vector space with a decreasing filtration \((F^k S)_{k \in \IZ}\): \(\cdots \supset F^{-1} \supset F^0 \supset F^1 \supset \cdots\)

  • A morphism \(f: S \to T\) of filtered spaces is a linear map with \(f(F^k S) \subseteq F^k T\) for all \(k\).

  • The degree shift \(S[m]\) has \(F^k(S[m]) = F^{k+m} S\).

(2) Definition (Associated constructions).

  • The \(k\)-th graded piece is \(\gr^k(S) = F^k S / F^{k+1} S\).
  • The associated graded is \(\gr(S) = \bigoplus_k \gr^k(S)\).
  • The completion is \(\hat{S} = \varprojlim_k S / F^k S\).
  • The filtration is separated if \(\bigcap_k F^k S = \set{0}\), in which case \(S \hookrightarrow \hat{S}\).

(3) Definition (Tensor product and inner Hom).

  • The filtered tensor product \(S \otimes T\) has \(F^k(S \otimes T) = \sum_{a+b=k} F^a S \otimes F^b T\).

  • The inner Hom \(\Hom^k(S, T) = \set{f: S \to T \mid f(F^p S) \subseteq F^{p+k} T}\).

  • The filtered dual is \(S^\vee = \Hom^\bullet(S, \IR[0])\) with \((S^\vee)^k = \set{f: S \to \IR \mid f(F^{1-k} S) = 0}\) and total space \(S^\vee = \set{f: S \to \IR \mid \exists k: f(F^k S) = 0}\).

(4) Proposition (Completion).

  • \(\gr_k \hat{S} = \gr_k S\) (completion preserves graded pieces).
  • \(\hat{S}^\vee = S^\vee\) (filtered dual is insensitive to completion).

Proof. Ad graded pieces) The exact sequence \(0 \to F^{k+1}/F^n \to F^k/F^n \to F^k/F^{k+1} \to 0\) has surjective transition maps; passing to projective limits stays exact (Mittag-Leffler), giving \(\gr_k \hat{S} \cong F^k/F^{k+1} = \gr_k S\).

Ad dual) An element \(a \in \hat{S}^\vee\) vanishes on some \(F^k \hat{S}\); since \(F^k S \subseteq F^k \hat{S}\), it lies in \(S^\vee\). Conversely, \(a \in S^\vee\) factors through \(S/F^k\) and hence extends to \(\hat{S}\).

(5) Proposition (Filtered adjunction). For a filtered map \(f: S \to T\) there is a unique adjoint \(f^\vee: T^\vee \to S^\vee\) with \((f(a), b) = (a, f^\vee(b))\), and \(f^\vee\) is filtered of degree \(0\).

Proof. Set \(f^\vee(b)(a) := b(f(a))\). Since \(f\) preserves the filtration, \(f^\vee\) maps \((T^\vee)^m\) into \((S^\vee)^m\). Uniqueness by non-degeneracy of the pairing.


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