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Write $\text{Sh}(X)$ for the triangulated/stable $\infty$ category of $\ell$-adic sheaves on $X$, and $k\in\text{Sh}(X)$ fo the unit object. In playing around with $\text{Sh}(B\mathbf{G}_m)$ I've found something disturbing I'd like to resolve.

Following 7.2 of [DG12], we can identify $\text{Sh}(B\mathbf{G}_m)$ with the modules in $\text{Sh}(\text{pt})$ for the algebra $B=H^*(\mathbf{G}_m,k)^\vee\in\text{Sh}(\text{pt})$. Note $B=k[u]/u^2$ where $|u|=-1$. Cohomology is $$H^*(B\mathbf{G}_m,-)\ :\ M \ \longrightarrow\ \text{Ext}_B(k,M).$$ For instance, $k\mapsto k[t]$ where $|t|=2$ and $B\mapsto k$.

Now let $V$ be a vector bundle over $B\mathbf{G}_m$ (=vector space with $\mathbf{G}_m$ action) of rank $r$. Write $i$ for the zero section and $j$ for the complementary open embedding. Applying $i^*$ to the distinguished triangle $i_!i^!k\to k\to j_*j^*k$ gives the Gysin sequence $$i^!k \ \stackrel{e}{\longrightarrow} \ i^*k \ \longrightarrow \ i^*j_*j^*k$$ Note that $i^!k\simeq k[-2r]$ and $e$ is multiplication by the Euler class. In particular, if $V$ has nonzero $\mathbf{G}_m$ weights, then on cohomology $e=t^{2r}:k[t]\to k[t]$ so $H^*(i^*j_*j^*k)$ is finite dimensional.

Question: however, what on earth is $i^*j_*j^*k$? Forgetting the $B$-module structure, $$i^*j_*j^*k\ =\ k\oplus k[-2r+1]$$ by pulling back the Gysin sequence to $\text{pt}$. In the rank $r=1$ case, this means that as a $B$ module it's $k\oplus k[-1]$ or $B$. Only the latter has finite dimensional cohomology it must be that: $$i^*j_*j^*k\ = \ B \ =\ H^*(\mathbf{G}_m)^\vee.$$ However, when the rank is greater than one there seems to be no $B$-module structure on $k\oplus k[-2r+1]$ which gives finite dimensional cohomology!

What on earth is going on here?

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    $\begingroup$ I think $du$ and $ud$ are both $0$ on the complex I wrote down, so they commute. This is because $d$ and $u$ go different ways because $d$ is a cohomological differential and $u$ is in degree $-1$. $\endgroup$
    – Will Sawin
    Commented Sep 22, 2020 at 13:17
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    $\begingroup$ I think there's lots of nice points of view here. (i) this is a prototypical case of Koszul duality; you could write down an explicit Koszul complex implementing the equivalence between (finite dimensional) B-modules and (perfect) k[t]-module, and then apply it k[t]/t^{2r}; (ii) if you do this you'll basically see the following play out: t corresponds to the extension with B in the middle and k's on each side, so then t^{2r} corresponds to splicing together 2r-copies of this extension, and this gives Will's answer; (contd) $\endgroup$
    – Dylan Wilson
    Commented Sep 22, 2020 at 13:24
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    $\begingroup$ (iii) maybe the confusion stems from the following line of thought: I want to define an algebra map B-->End(M) for some k-module, but in the example at hand it looks like nothing is in degree -1 (for large r), so how could there possibly be an interesting map here? The key is that we are in derived land! B is not free as an associative algebra: we have to choose a homotopy (or cycle) killing u^2, and then we have to choose something killing the resulting massey product <u,u,u>, etc. In the example at hand, it turns out that choosing an interesting Massey product at the 2rth-ish step $\endgroup$
    – Dylan Wilson
    Commented Sep 22, 2020 at 13:26
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    $\begingroup$ is what builds this module. Said slightly differently: on homotopy groups the action of B looks trivial, but there is a nontrivial massey product like <u,u, ..., u, x>=y which relates the two generators in $k\oplus k[-2r+1]$. $\endgroup$
    – Dylan Wilson
    Commented Sep 22, 2020 at 13:27
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    $\begingroup$ It seems my confusion was just down to confusing which direction $u$ went in! Normally I would delete this question in such a case, but I think Dylan and Will's comments make it worth keeping, thank you. $\endgroup$
    – Pulcinella
    Commented Sep 24, 2020 at 12:17

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