Homotopy theoretic description of homotopy fixed points (and obstructions) for an action of group $G$ on a groupoid $X$

There are several scattered statements about fixed points and obstructions which I'd very much like to see unified in some framework.

To state them let $G$ be a group acting on a connected (1-truncated) groupoid $X$. Firstly let's choose a point in $X$ and take $X=BAut(\pi_1(X))=:BA$. For the purpose of this question i'd like to find a unified homotopy theoretic approach for the following statements:

1. There's an obstruction in $H^2(G,A)$ for the existence of homotopy fixed points.

2. When the obstruction vanishes $\pi_0(X^{hG})\cong H^1(G,A)$

I'm hoping both of these statements can be explained using arguments about the following fiber sequence (and perhaps some additional close constructions):

$$BA \to E \to BG$$

Where $E$ is the total space of the $A$-gerbe corresponding to the $G$ action on $X$. Said differently we have by assumption a map $G \to Aut(BA)$ which we can $B(-)$ to get $BG \to BAut(BA)$ which we can use to pullback the universal fibration $BA \to BAut_*(BA) \to BAut(BA)$ to get the above fiber sequence.

Ideally both $H^2(G,A)$ and $H^1(G,A)$ will appear in the same long exact sequence (/spectral sequence) of homotopy groups for some fiber sequence.

• What do you mean by $H^2(G, A)$ if $A$ is nonabelian? – Qiaochu Yuan Sep 8 '17 at 0:50
• Also what do you mean by obstructing the 'existence' of homotopy fixed points? You said the group was acting on X so there's nothing in the way of taking homotopy fixed points for this action... – Dylan Wilson Sep 8 '17 at 0:52
• @DylanWilson Since taking homotopy fixed points is a limit in the category of unpointed spaces I assume it could be the empty space couldn't it? – Saal Hardali Sep 8 '17 at 2:03
• @QiaochuYuan I have no idea, I asked this question too in the comments to an answer for a different question but the discussion was to long so I posted this question: mathoverflow.net/questions/243312/… – Saal Hardali Sep 8 '17 at 2:03

I'm not sure exactly what kind of answer you're looking for, but I can try to give some context which may make things sound more reasonable. Let us think of of groupoids as $1$-truncated $\infty$-groupoids, or $1$-truncated spaces. Consider first the case where $A$ is abelian. In this case $BA$ will itself be an $\mathbb{E}_\infty$-group object in spaces, and will have its own classifying space $B^2A$ (which is no longer $1$-truncated, it's a $K(A,2)$). Since the formation of classifying spaces naturally lands in pointed spaces it follows that an arbitrary action of $G$ on $BA$ yields a base-point preserving action of $G$ on $B^2A$. Such a data can be geometrically realized as a fibration $p:F \to BG$ together with an identification $p^{-1}(x_0) \simeq B^2A$ of the fiber over the base point with $B^2A$ and equipped with a $0$-section $s_0:BG \to F$. Furthermore, $F$ will actually form an $\mathbb{E}_\infty$-group object in spaces over $BG$. We note that while we constructed $F$ out of the action of $G$ on $BA$, it actually only depends on the induced action of $G$ on $A$ (something that is well-defined when $A$ is abelian): indeed, as an $\mathbb{E}_\infty$-group object over $BG$, $F$ is just the double delooping of the local coefficients system $BG \to \mathcal{Ab}$ corresponding to the action of $G$ on $A$. The space of sections $\mathrm{Map}_{BG}(BG,F)$ of $F$ is then the space which "represents" the first three cohomology groups of $G$ with coefficients in $A$: namely, $\pi_i\mathrm{Map}_{BG}(BG,F) \cong H^{2-i}(G,A)$ for $i=0,1,2$.

We now turn to the fibration $E \to BG$ that you mention, which is associated to the action of $G$ on $BA$. What is the relation between $E$ and $F$? if the action of $G$ on $BA$ had a homotopy fixed point then $E \to BG$ would have a section and we could think of $F$ as a fiberwise delooping of $E$, or, alternatively, identify $E$ with the fiberwise looping of $F$, i.e., with the homotopy fiber product $BG \times^h_F BG$. In the general case (where we don't assume that there is a base point) one can show that a similar thing happens, only twistedly. Instead of taking the homotopy fiber product of $BG$ with itself over $F$, where the map from $BG$ to $F$ is the $0$-section in both cases, one can show that there actually exists (an essentially unique) other section $s: BG \to F$ such that $E$ is equivalent to the homotopy fiber product $BG \times^h_{F} BG$, where now one copy of $BG$ maps to $F$ via $s$ and the other copy via the $0$-section $s_0$. Roughly speaking, we may consider $E$ as the space of pairs $(x,\eta)$ where $x$ is a point of $BG$ and $\eta$ is path in the fiber $F_x$ from $s(x)$ to the base point $s_0(x) \in F_x$. The section $s$ then determines an element in $\pi_0(\mathrm{Map}_{BG}(BG,F)) \cong H^2(G,A)$, which vanishes if and only if $E \to BG$ admits a section (i.e., if and only if $BA$ has a homotopy fixed point). A choice of such a section is essentially the same as an identification of $E \to BG$ with the fiberwise looping of $F \to BG$, in which case the full space of sections satisfies $\pi_i(\mathrm{Map}_{BG}(BG,E)) \cong H^{1-i}(G,A)$ for $i=0,1$.

This story is quite general, it will work essentially the same if we replace $A$ by any $\mathbb{E}_{\infty}$-group object, or even a spectrum, equipped with an action of $G$ (the cohomology of $G$ with coefficients in such a thing is exactly what you think it should be). Of course, everything works much less well when $A$ is not abelian. First of all, you might ask yourself what do we even mean by $H^i(G,A)$ when $A$ is not abelian. I have to say I don't really have a good answer for this. There exists an explicit definition using cocycles, which for $i=0,1$ requires an honest action of $G$ on $A$ and for $i=2$ only requires an outer action of $G$, namely, a homomorphism $G \to Out(A)$. In this definition $H^0(G,A)$ is a group (the fixed subgroup of $A$ under the action of $G$), $H^1(G,A)$ is a pointed set and $H^2(G,A)$ is a set with a collection of "neutral elements". This definition enjoys some basic properties: it reduces to the usual definition when $A$ is abelian, and admits some useful (not-so-long) long exact sequences. One can also show that, like in the abelian case, the elements of $H^2(G,A)$ are in one-to-one correspondence with extensions of $G$ by $A$ which are compatible with the given outer action, where the neutral elements correspond to the split extensions (note that there can be several split extensions of $G$ by $A$ with the same outer action). In particular, if we are given an action of $G$ on $BA$ then the corresponding fibration $E \to BG$ determines a class in $H^2(G,A)$ via the group extension $1 \to A \to \pi_1(E) \to G \to 1$. The corresponding element is neutral if and only if this sequence splits, which is equivalent to $E \to BG$ having a section. A choice of such a section/splitting determines in particular a lift of the outer action of $G$ on $A$ to an honest action, and hence determines a (non-abelian) local coefficient system $BG \to \mathcal{Grp}$. One may then identify $E \to BG$ with the fiberwise delooping of this local coefficient system and show that, as in the abelian case, $\pi_i\mathrm{Map}_{BG}(BG,E) \cong H^{1-i}(G,A)$ for $i=0,1$.

It is worth noting that even though the unabelian $H^2(G,A)$ is not an especially handsome invariant (which makes the obstruction theory above sounds a bit tautological at first), it can occasionally give useful and computable information, and more importantly, it can be shown to vanish under various assumptions. For example, if the map $Aut(A) \to Out(A)$ admits a section and the center $Z(A)$ vanishes (like if $A$ is the alternating group on at least 5 elements) then $H^2(G,A)$ always vanishes. One may then deduce that any action of $G$ on $BA$ admits a homotopy fixed point, a statement that is far from obvious at first sight.

Edit: My remark that everything works the same for $A$ an arbitrary $\mathbb{E}_{\infty}$-group should be reserved a bit. This is true if we consider actions of $G$ on $BA$ which are affine-linear, i.e., that factor through the group $\mathrm{Aff}(BA) = BA \rtimes \mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)$ of affine-linear transformations. Given such an action $\alpha: G \to \mathrm{Aff}(BA)$ the projected map $\overline{\alpha}:G \to \mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)$ determines the space $F$ by the homotopy fiber product $F = BG \times^h_{B\mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)} B\mathrm{Aff}(BA)$ while the lift $\alpha$ of $\overline{\alpha}$ determines the section $s: BG \to F$ mentioned above. In the case where $A$ is a discrete abelian group we have an equivalence $\mathrm{Aut}(BA) \simeq \mathrm{Aff}(BA)$, and so we can consider arbitrary actions of $G$ on $BA$, but I don't think this is true in general.

• Your claim that nonabelian $H^2$ recovers the usual one when $A$ is abelian contradicts ncatlab.org/nlab/show/nonabelian+group+cohomology. Does the nLab have the wrong definition of nonabelian $H^2$? – Marc Hoyois Sep 9 '17 at 4:41
• Yes, it seems to be a different definition than the one I'm familiar with (and the one that appears, for example, in Galois descent or in the Bousfield Kan spectral sequence). nLab defines it as the set of homotopy classes of morphisms from $BG \to \mathbf{B}Aut(A) \simeq BAut(BA)$, while the $H^2$ I know fixes in advance the outer action and looks only at maps to $BAut(BA)$ which induce a given action (maybe this can even be takes as a definition, I'm not sure). – Yonatan Harpaz Sep 9 '17 at 7:18
• I would like to add that Giraud's definition of nonabelian $H^2$ (from "Cohomologie non abélienne") agrees with the usual one when $A$ is abelian. At the time I found out about this I couldn't work out if this was a mistake in the nLab or due to a difference in definitions. – Robert Furber Sep 9 '17 at 9:14
• @YonatanHarpaz Is the following interpretation accurate? Lets work relatively in an (connected) $\infty$-topos $\mathfrak{X}$. Then the problem is basically obtaining an obstruction theory for determining when gerbes admit global sections (or just "points"). If $\mathcal{F}$ is any connected object of $\mathfrak{X}$ it must look locally like $BA$ for some $\infty$-group in $\mathfrak{X}$. But inside $\mathfrak{X}$ we can find a universal fibration $BA \to BAut_*(BA) \to BAut(BA)$. The long exact sequence of homotopy groups for this fibration only gets as far as... to be continued... – Saal Hardali Sep 10 '17 at 19:54
• $H^0(\mathfrak{X},BA) \to H^0(\mathfrak{X},BAut_*(BA)) \to H^0(\mathfrak{X},BAut(BA))$ which for the case $\mathfrak{X} = \text{Spaces}_{/ BG}$ gives $" H^1(G,A) \to H^1(G,Aut_*(BA)) \to H^1(G, Aut(BA)) "$. Then you claim there's an $H^2$ which one can define which extends this exact sequence to the right? If there was then $H^2(G,A)$ would by this definition have to be the obstructions to existence of sections of our original $\mathcal{F}$ which determined an element in $H^1(G, Aut(BA))$. – Saal Hardali Sep 10 '17 at 19:59