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Let $X$ be a topological space and $G$ be a topological group acting on $X$, both locally compact Hausdorff. Denote by $D^b(X)$ the derived category of sheaves (say of abelian groups) on $X$. We define a new category in the following way: An object is an object $M$ of $D^b(X)$, together with an isomorphism $$\pi^*M\leftrightarrow \rho^*M$$ that satisfies the cocycle condition, where $\pi,\rho\colon G\times X\to X$ are the projection and the action maps. A morphism is a morphism of objects in $D^b(X)$ which is compatible with the action isomorphisms.

Note that this category is a special case of the category of equivariant objects:

Is this category equivalent to the equivariant derived category? If not, why is the equivariant derived category a better construction in this case?

PS: Bernstein and Lunts give among others the following definition of the equivariant derived category in their book "Equivariant sheaves and functors" (p.32):

Denote by $[X/G]$ the "action simplex" $\Delta(n)=G^{n-1}\times X$.

By a simplicial sheaf we mean a collection of sheaves $F^n$ on $\Delta(n)$ together with maps $\alpha_h: h^*F^m \rightarrow F^n$ for each map $h$ in $[X/G]$, which satisfy the cocycle condition: $$\alpha_{h' h}=\alpha_h \circ h^* \alpha_{h'}$$ Denote by $Sh([X/G])$ the category of simplicial sheaves and by $Sh_{eq}([X/G])$ the full subcategory where all $\alpha_h$ are isomorphisms.

The derived equivariant category $D^b_G(X)$ is then the full subcategory of $D^b(Sh([X/G]))$ consisting of objects, which have cohomology in $Sh_{eq}([X/G])$.

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up vote 11 down vote accepted

The relation is easier to understand if $G$ is a discrete group. Then your definition is equivalent to an action of $G$ on an object $M$ in $D^b(X)$. This means that for each $g\in G$ we have a morphism $\phi_g\colon M\rightarrow M$ in $D^b(X)$ with, and this is the important part, $\phi_{gh}=\phi_g\phi_h$ in $D^b(X)$. Seen from the point of view of complexes (i.e., before we pass to $D^b(X)$) this means that we have a homotopy (or something slightly worse depending on what kind of complex $M$ is) $\phi_{gh}\sim\phi_g\phi_h$. Experience tells us that such actions up to homotopy is too weak a notion to be useful (things are a little bit tricky as there are non-trivial situations when such an action can in fact be replaced up to homotopy by a true action) and in general one should demand an actual action of $G$ on the complex $M$. The same goes (and in fact even more so) for morphisms, there may be too many morphisms if you just look at morphisms in $D^b(X)$ that commute in $D^b(X)$ with the action of $G$.

Addendum: Donu makes an excellent point and I just want to elaborate. Take an additive $G$-action on some $(\mathbb Z/p)^n$ which does not lift to an additive action on $(\mathbb Z/p^2)^n$. Then we may look at the morphism $(\mathbb Z/p)^n \to (\mathbb Z/p)^n[1]$ whose mapping cone is $(\mathbb Z/p^2)^n$. It will be a $G$-map $D^b(pt)$ but there is no compatible action of $G$ on a mapping cone.

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Nice. It's probably better to put brackets here $(\mathbb{Z}/p)^n$. – Donu Arapura Jun 10 '10 at 16:30
Thanks for the answer especially the Addendum! – Jan Weidner Jun 14 '10 at 19:23

There's a related discussion here. The idea that the equivariant version of the derived category consists simply of complexes with equivariance structure is correct, once you work in a refined enough setting such as differential graded, $A_\infty$ or stable $\infty$-categories. As was explained it's certainly not true just at the homotopy level. But at this enhanced level, the equivariant derived category is indeed just the derived invariants of the group acting on the derived category of X. The Bernstein-Lunts definition is the same, just writing down invariants using a standard resolution (ie the simplicial model for BG). A good reference for such things is chapter 7 of the manuscript of Beilinson-Drinfeld on Quantization of Hitchin's Hamiltonians.

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Thanks for this useful information! Although I must admit I find it quite hard to understand Beilinson-Drinfelds book. – Jan Weidner Jun 14 '10 at 19:22

In case you're not convinced by Torsten's answer, here's a question. Is it clear that your proposed category is even triangulated? Specifically, how would you prove that a morphism embeds into a triangle? If you had a map of equivariant complexes, you could take a mapping cone. However, you seem to have something weaker than that.

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In other words, you should do $C(X)$ equivariantly first and then mod out null homotopic things to get a triangulated category. On this category a notion of equivariant quasi-isomorphism should arise and then localize wrt to them. But of course, Bernstein and Lunts method has the merit of existence. – Leo Alonso Jun 10 '10 at 18:10
This is probably the wrong place to ask, but how have you been Leo? – Donu Arapura Jun 10 '10 at 19:41
Thanks for this answer as well as for your comment above! – Jan Weidner Jun 14 '10 at 19:24

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