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A well known fact (e.g. Moulinos - The geometry of filtrations) is that one can describe the category of filtered vector spaces as $\operatorname{FilVect} \simeq \operatorname{QCoh}(\mathbb{A}^1/\mathbb{G}_m)$. Writing $$\jmath: \mathbb{G}_m/\mathbb{G}_m \to \mathbb{A}^1/\mathbb{G}_m \leftarrow {\operatorname{pt}}/\mathbb{G}_m : \imath$$ for the generic and special points, and $p:\mathbb{A}^1/\mathbb{G}_m \to {\operatorname{pt}}/\mathbb{G}_m$ for the projection, we have the following operations:

  1. $\jmath^* : \operatorname{FilVect} \to \operatorname{Vect}$ corresponds to forgetting the filtration.
  2. $\imath^* : \operatorname{FilVect} \to \operatorname{QCoh}({\operatorname{pt}}/\mathbb{G}_m) \simeq \operatorname{Rep}(\mathbb{G}_m) \simeq \operatorname{GrVect}$ corresponds to taking the associated graded.
  3. $p^*: \operatorname{GrVect} \to \operatorname{FilVect}$ corresponds to the operation of taking a graded vector space $\bigoplus_n V_n$ to the filtration $F^nV = \bigoplus_{i \leq n}V_i$.

My questions:

  1. How can one describe the operation of taking a vector space $V$ to the filtered vector space $F^{\bullet}V$ with $F^{<0}V = 0$ and $F^{\geq 0}V = V \xrightarrow{\operatorname{id}} V \xrightarrow{\operatorname{id}} \dotsb$ in geometric terms?
  2. Is the pushforward $\imath_*$ defined? If so, is $\jmath^*\imath_* = 0$?
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1 Answer 1

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For question 1, by the Rees construction the corresponding quasi-coherent sheaf is the graded $k[t]$-vector space $V\otimes k[t]$. This is simply the quasi-coherent sheaf $V\otimes_k\mathcal O_{\mathbb A^1/\mathbb G_m}\in\mathrm{QCoh}(\mathbb A^1/\mathbb G_m)$. If you wish, this is the pullback of $V\in\mathrm{QCoh}(*)$ along the map $\mathbb A^1/\mathbb G_m\to *$.

For question 2, given a graded vector space $V_\bullet=\bigoplus_nV_n$, the filtered vector space $i_*V_\bullet$ is the filtered vector space given by $$\cdots \xrightarrow0V_{n-1}\xrightarrow0V_n\xrightarrow0V_{n+1}\xrightarrow0\cdots.$$ Indeed, $i_*$ is the right adjoint to $i^*$. Given a graded vector space $V_\bullet=\bigoplus_n V_n$ and a filtered vector space $F^\bullet W$, there is an isomorphism $$\mathrm{Hom}_{\mathrm{fil}}(F^\bullet W,i_*V_\bullet)\simeq \bigoplus_n\mathrm{Hom}(F^nW/F^{n-1}W,V_n).$$ Now the composition $j^*i_*$ is the colimit of the diagram, which is $0$.


Here is another way to think about Question 2: recall that the $X$-points of $\mathbb A^1/\mathbb G_m$ are virtual Cartier divisors of $X$, i.e., a line bundle $\mathcal L$ with a homomorphism $\mathcal L\to \mathcal O_X$. This is pulled back from the universal Cartier divisor $t\colon \mathcal O_{\mathbb A^1/\mathbb G_m}(-1)\to \mathcal O_{\mathbb A^1/\mathbb G_m}$, and $*/\mathbb G_m$ is the locus where $t=0$. Then the equivalence $$\mathcal D_{qc}(\mathbb A^1/\mathbb G_m)\simeq\mathcal D_{\mathrm{fil}}(k)$$ is given by tensoring an object $M\in \mathcal D_{qc}(\mathbb A^1/\mathbb G_m)$ with $$\cdots\xrightarrow t\mathcal O(-1)\xrightarrow t\mathcal O\xrightarrow t\mathcal O(1)\xrightarrow t\cdots$$ and taking global sections. The subcategory $i_*\colon\mathcal D_{qc}(*/\mathbb G_m)\hookrightarrow\mathcal D_{qc}(\mathbb A^1/\mathbb G_m)$ are those objects $M$ killed by $t$, so under the equivalence they correspond to the subcategory of filtered vector spaces $$\cdots\to F^{n-1}W\to F^nW\to F^{n+1}W\to\cdots$$ whose transition maps $F^nW\to F^{n+1}W$ are all zero.

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  • $\begingroup$ Thank you, that's very helpful! About the second part - in my question I meant the filtered vector space with 0 in negative degrees and identity maps in positive degrees (I'll edit my question to clarify). Is there a geometric operation that will produce that? $\endgroup$
    – E. KOW
    Commented Nov 20 at 18:05
  • $\begingroup$ Sorry I'm not sure what your question is: for Question 1, I claim the filtered vector space $\cdots0\to 0\to V\to V\to\cdots$ corresponds to $V\otimes_k\mathcal O_{\mathbb A^1/\mathbb G_m}$. $\endgroup$ Commented Nov 20 at 20:06
  • $\begingroup$ Oh sorry, I didn't notice the edit to the first part. Thank you! $\endgroup$
    – E. KOW
    Commented Nov 20 at 20:52

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