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. I want to think about $\text{Sh}(B\mathbf{G}_m)$, but I've found something disturbing.

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 $|u|=-1$. In this optic, the global sections functor 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 (=pulling back via $\pi:\text{pt}\to B\mathbf{G}_m$), the Gysin sequence gives $\pi^*(i^*j_*j^*k)=H^*(S^{2r-1})=k\oplus k[-2r+1]$. 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 with finite dimensional cohomology with $\pi^*(i^*j_*j^*k)=k\oplus k[-2r+1]$!

I'm not sure whether I've misunderstood something fundamental about $\text{Sh}(BG)$ or something else has happened.

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?