Timeline for Deequivariantisation of indecomposable sheaves
Current License: CC BY-SA 4.0
14 events
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| Nov 22, 2021 at 14:15 | comment | added | rvk | @DavidBen-Zvi Ah! That clears it up. Somehow I missed the homology chains bit. I am way to used to working with the $H^*_G$ dg-modules. I posted an "explicit" version of what you are saying (that doesn't invoke Koszul duality). | |
| Nov 22, 2021 at 2:15 | comment | added | David Ben-Zvi | @rvk You and I are on different sides of Koszul duality - you're describing modules for the symmetric algebra $S=H^*(BG)$, and I'm discussing the equivalent (up to size issues) modules for $\Lambda=H_*(G)$. The regular module $S$ you discuss corresponds to the augmentation for $\Lambda$ (constant sheaf on point) - precisely by the functor $k\otimes -$ as you write. The functor to sheaves on a point is forgetful for $\Lambda$ but tensor with augmentation for $S$. | |
| Nov 22, 2021 at 1:25 | comment | added | rvk | Pedantic note (I’ll take my leave after this final comment): everywhere that I have written ‘$\otimes$’ above, strictly speaking it should say ‘$\otimes^L$. It doesnt matter in the context of $H^*(BG)$ as a dg-module over itself (since it’s free/flat over itself). But, point is that the forgetful functor $D_G(pt) \to D(pt)$ doesnt translate to something ‘simple’ under the Koszul dual description in terms of dg-algebras. Ok, I am out on this topic. | |
| Nov 22, 2021 at 0:27 | comment | added | rvk | Again the example is take the dg-module $H^*(BG)$ itself. It is non-zero in every even non-negative degree. When you apply the “forgetful” functor to this you get the degree 0 component. Back in the non-dg description/world, this is just the equivariant constant sheaf giving you the constant sheaf on forgetting equivariance. | |
| Nov 22, 2021 at 0:25 | comment | added | Peter McNamara | I'm happy this answer answers the question and kicking myself for not realising it before posting. Somehow I always forget the interesting objects in the G-equivariant derived category of a point exist, because I don't tend to come across them in the wild. | |
| Nov 22, 2021 at 0:23 | comment | added | rvk | Nope your edited answer doesnt address my concern. First, we dont need to complicate things with the full cochain algebra in char 0 - it’s formal so just $H^*(BG)$ with trivial differential will do. Now if $M$ is a dg-module for this, then under the equivalence $D_G(pt) \to dg-der(H^*(BG))$ the forgetful functor on the LHS corresponds to $k\otimes M$ on the RHS. I.e, the forgetful functor is not just forget the dg-algebra action. So just taking any non trivial module on the RHS wont do it. | |
| Nov 22, 2021 at 0:06 | comment | added | David Ben-Zvi | In particular for your last question, $For(M)$ will certainly be decomposable, so you just want to make sure $M$ is indecomposable. $M$ will be cone of the map you produced in the category of $H_*(G_m)$-modules, which is Koszul dual to $H^*(BG_m)$-modules. | |
| Nov 22, 2021 at 0:05 | comment | added | David Ben-Zvi | @rvk I'm not sure I follow exactly what you're worried about but hopefully the updated answer addresses this. | |
| Nov 22, 2021 at 0:04 | history | edited | David Ben-Zvi | CC BY-SA 4.0 |
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| Nov 21, 2021 at 22:50 | comment | added | rvk | In slightly more mundane terms: take a generator of $H^2(P^{\infty})$ and produce a complex $M$ in $D_{G_m}(pt)$ that corresponds to this map $k \to k[2]$ such that $M$ is indecomposable but $For(M)$ splits up. I don't know how to do this, and would be grateful if you explained how this is done. | |
| Nov 21, 2021 at 22:43 | comment | added | rvk | In general, under the equivalence with dg-modules the forget functor applied to a dg-$H^*(BG)$-module $M$ is given by $k\otimes M$ (tensor along augmentation). It is quite easy for $M$ to have plenty of stuff in non-zero degree as shown by the case of $H^*(BG)$ itself (all of which gets killed under the forgetful functor) be indecomposable and stay indecomposable under the forgetful functor. Note: I am ignoring boundedness issues. | |
| Nov 21, 2021 at 22:42 | comment | added | rvk | In fact, can you explain, "The augmentation module (trivial representation of $G$) has self Exts given by $H^*(BG_m)$". The trivial module for $G_m$ has no self-ext (say over the complex numbers) in any degree. Sure $H^*(BG_m)= H^*(P^{\infty})$ has plenty of non-zero terms, but I again maintain that conflating modules for $H^*(BG)$ with the dg-derived category of this dg-algebra will get you in trouble. For instance, the "constant sheaf" (i.e., monoidal unit) in $D_G(pt)$ corresponds to the dg-algebra $H^*(BG)$ itself. This stays indecomposable under the forgetful functor. | |
| Nov 21, 2021 at 21:18 | comment | added | rvk | While this example might work, you seem to be conflating the dg-derived category of the (formal) dg algebra $H^*_{G_m}$ with the derived category of its modules. I don’t see how you translate, derived Ext in the dg-derived category into Yoneda style extensions of complexes under the equivalence. | |
| Nov 21, 2021 at 19:56 | history | answered | David Ben-Zvi | CC BY-SA 4.0 |