# liftings of principal bundles

I would like to know what structure has the category of liftings of a principal bundle. Let me be more precise. Fix $k$ an algebraically closed field and $X$ a smooth projective variety over it (for eg. a curve) as well as a short exact sequence of smooth connected groups $1\to K\to G\stackrel{\pi}{\to} H\to 1$ (if one prefers one can work aver $\mathbb{C}$ and take the groups to be connected complex Lie groups). Since we're dealing with principal bundles pick your favourite topology, say $\tau$ (fpqc, etale, classical analytic - I hope that the answer will not depend in an essential way on this).

Now let us also fix $F_H$ an $H$-bundle on $X$. I'm interested in understading the category $\mathcal{K}_{F_H}$ of liftings of $F_H$ to $G$ (it is a stackover $X$, right?). This is defined as follows:

for an $X$-scheme $f:T\to X$ we have:

objects: pairs $(E,\alpha)$ where $E$ is a $G$-bundle on $T$ and $\alpha:\pi_*(E)\to f^*(F_H)$ an isomorphism of $H$-bundles.

morphisms: from a pair $(E,\alpha)$ to $(E',\alpha')$ are isomorphisms of $G$-bundles $u:E\to E'$ that are compatible with $\alpha$ and $\alpha'$, i.e. $\alpha' \pi_*(u) = \alpha$.

In case $K=\ker(\pi)$ is abelian $H$ acts naturally by conjugation on $K$ and one can form the following group scheme over $X$: $K_{F_H} = F_H\stackrel{H}{\times}K$.

One can see rather easily (for e.g. using local charts) that $\mathcal{K}_{F_H}$ is a gerbe over $X$ which is, locally on $X$, isomorphic to $BK_{F_H}\times X$.

The obstruction of $F_H$ to have a lift to a $G$-bundle is an element $\xi\in H^2_{\tau}(X,K_{F_H})$. Moreover, if $\xi=0$ the gerbe is trivial and hence the stack $\mathcal{K}_{F_H}$ is equivalent to the stack of $K_{F_H}$-bundles on $X$.

I hope I haven't messed things up yet.

My question is: what if $K$ is not abelian? then there's no natural action of $H$ on $K$ (even if the sequence is split the conjugation action depends on the splitting). Where lives the obstruction of lifting an $H$-bundle? And, what interests me more actually, what is the structure of the category of liftings of $F_H$ to $G$ once one knows it has a lift?

I'm pretty sure the answer should be in Giraud's book somewhere but I find it quite hard to read so I'm having troubles detecting the right place to look. Any references or comments would be helpful.

• It should be relevant somewhere if $K$ is central in $G$. If it is not central (e.g. $K$ not abelian), then $G$ acts on $K$ and forms a crossed module. Then your obstruction lives in Giraud's non-abelian cohomology with values in that crossed module. – Konrad Waldorf Feb 23 '15 at 20:27

The broad outlines of how this business works don't depend on the fact that you're working with varieties so let me work with spaces instead, by which I mean homotopy types.

Let $f : X \to BG$ be a principal bundle and let $BH \to BG$ be a map along which you'd like to lift. Then the space of lifts of $f$ to $BH$ (up to homotopy) is precisely the space of (homotopy) sections of the pullback $X \times_{BG} BH$, or equivalently the space of sections of the $G/H$-bundle associated to $f$. (Here by $G/H$ I mean the homotopy fiber of $BH \to BG$, which may or may not be the actual quotient $G/H$, whatever that means in your setting.)

The nicest case is when $BH \to BG$ is itself a homotopy fiber, or equivalently when it fits into a fiber sequence

$$\Omega Y \to BH \to BG \to Y$$

for some space $Y$ (which must necessarily deloop $G/H$; in this case the associated $G/H$-bundle is a "principal $\Omega Y$-bundle"). Then the space of lifts of $f$ to $BH$ is, by the universal property of the homotopy fiber, precisely the space of nullhomotopies of the composite map $X \to BG \to Y$. In the very nicest cases $Y$ is an Eilenberg-MacLane space.

In particular, if $G, H$ fit into a short exact sequence

$$K \to H \to G$$

then $BK, BH, BG$ fit into a fiber sequence

$$BK \to BH \to BG.$$

If $K$ is not only abelian but central in $H$, then this fiber sequence deloops to a fiber sequence

$$BK \to BH \to BG \to B^2 K$$

and so we can take $Y = B^2 K$, in which case the existence of a nullhomotopy of the composite $X \to BG \to Y$ is equivalent to the corresponding cohomology class in $H^2(X, K)$ vanishing. If $K$ is abelian but not central then this cohomology group needs to be taken with local coefficients. But sometimes $Y$ is more complicated than this. For example, if $G = \text{Spin}$ and $H = \text{String}$ then $Y = B^4 \mathbb{Z}$ and the existence of a lift is controlled by a cohomology class $\frac{p_1}{2} \in H^4(X, \mathbb{Z})$.

In general, though, $G/H$ just isn't a loop space, and so can't be delooped, which means that $Y$ doesn't exist and the obstruction theory gets harder. The obstructions are a sequence of classes in cohomology $H^{k+1}(X, \pi_k(G/H))$ with local coefficients, each of which is only well-defined provided that the previous one vanishes. A typical case here is $G = O(2n), H = U(n)$.

• Can you explain why local coefficients are required when $K$ is abelian but not central? – Chris Gerig Feb 23 '15 at 21:29
• @Chris: in that case the fiber sequence $BK \to BH \to BG$ just isn't classified by a map to $B^2 K$, but it is classified by cohomology with local coefficients. (Admittedly I'm most comfortable making statements of this form when $K$ is discrete.) – Qiaochu Yuan Feb 23 '15 at 21:40
• OK, what then makes the coefficient system constant when $K$ is central? Some sort of reduction map must be arising somewhere? – Chris Gerig Feb 24 '15 at 3:02
• @Chris: Isn't it because the conjugation action of G on K is trivial when K is central, so $K_F$ is just the constant group $K$? ($F$ is the $G$-fibration) – Dragos Fratila Feb 24 '15 at 5:11
• @Qiaochu: Thanks for your comment! If I understand correctly, if one doesn't have a delooping of BK the answer is not very nice. I don't know anything about looping/delooping stuff, so what happens, for example, when K=[H,H] and G=H/[H,H] for H a compact Lie group. Can you point to a reference where the "obstructions in $H^{k+1}(X,\pi_k(BK))$ are discussed (or used/computed, examples)? – Dragos Fratila Feb 24 '15 at 5:19