Extending group actions to vector bundles

Question

Let $G$ be a group acting on a manifold $M$. Suppose $V$ is a rank $n$ vector bundle on $M$.

Is there any obstruction to extending the action of $G$ to $V$? In how many ways can the action be extended if obstructions vanish?

Given an $n$-dimensional real linear representation of $G$, when can an action of $G$ on $V$ be realized such that its action on $M$ and on each of the fibers of $V$ are the given ones?

I'd also like to know answers to these questions when (1) $G$ is a topological group acting continuously, or (2) $V$ is a principal $H$-bundle for a group $H$.

Thanks!

Answer 1

I'll discuss the case of complex bundles and a finite group $G$. If $V$ admits a compatible action of $G$ then there is a vector bundle $V_{hG}$ with total space $EV_{hG}=(EG\times EV)/G$ over the Borel construction $M_{hG}=(EG\times M)/G$. There is a canonical map $p\colon M\to M_{hG}$ with $p^*(V_{hG})\simeq V$, so the Chern classes satisfy $p^*(c_k(V_{hG}))=c_k(V)$. In particular, this implies that the Chern classes $c_k(V)$ are invariant under the action of $G$, so they appear in $E_2^{0,2k}$ of the spectral sequence $$ E_2^{pq}=H^p(G;H^q(M)) \Longrightarrow H^{p+q}(M_{hG}) $$ Moreover, they can support no differentials and must survive to $E_\infty$. In the real case you could do similar things with Stiefel-Whitney classes and Pontrjagin classes.

Note here that the symbol $H^p(G;H^q(M))$ refers to group cohomology, which is the same as the singular cohomology of the space $BG$ with a local coefficient system determined by the $G$-module $H^q(M)$. The same argument works for an infinite topological group $G$ except that we are forced to use cohomology of $BG$ with local coefficients in that case. If $G$ is connected then $BG$ will be simply connected so the local coefficient system must be constant and we just get a spectral sequence $$ E_2^{pq}=H^p(BG;H^q(M)) \Longrightarrow H^{p+q}(M_{hG}) $$ with ordinary untwisted coefficients.

Answer 2

$\DeclareMathOperator\Fr{Fr}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Sym{Sym}$Another, more geometric, approach to this question is as follows: let $G$ be a Lie group acting on $M$, and let $P$ be a principal $H$-bundle over $M$. (This addresses your question (2), but it includes (1) if you take $P = \Fr(V)$ to be the frame bundle of the vector bundle $V$.) Now, if the action of $g \in G$ is supposed to lift to $P$, the least we need is an isomorphism $\psi_g \colon P \to \Phi_g^*P$ of $H$-bundles over $M$. For these to actually provide an action, they must satisfy that $\Phi_g^*\psi_h \circ \psi_g$ agrees with $\psi_{hg}$ under the canonical isomorphism $\Phi_g^*\Phi_h^*P \cong (\Phi_h \circ \Phi_g)^*P \cong \Phi_{hg}^*P$.

This can be rephrased more globally as follows: let $\Aut(P)$ denote the group consisting of all diffeomorphisms $P \to P$ over $M$ (i.e. mapping fibres to fibres) that preserve the $H$-action. That is, these are bundle maps which cover any diffeomorphism $M \to M$. This comes with a projection $p \colon \Aut(P) \to \Diff(M)$. The map $p$ is a group homomorphism, and if we consider only diffeomorphisms isotopic to the identity on $M$, then it is even surjective (you can see this by a cohomology argument, or by endowing $P$ with a connection). Let $\mathcal{G}(P)$ be the fibre of $p$ over $1_M$ (this is the group of gauge transformations of $P$). We thus have a short exact sequence $$\mathcal{G}(P) \to \Aut(P) \to \Diff_0(M)\,,$$ where $\Diff_0(M)$ denotes the connected component of $1_M$ in $\Diff(M)$. This is even a short exact sequence of smooth groups (for instance in the world of diffeological spaces).

Our $G$-action is a smooth group homomorphism $\Phi \colon G \to \Diff(M)$, and I will assume that it factors through $\Diff_0(M)$. (Alternatively, assume that for every $g \in G$ a $\psi_g$ as above exists; otherwise there is no chance of finding a lift of the $G$-action $\Phi$ to $P$ anyway.) Then, we obtain a pullback sequence $$\mathcal{G}(P) \to \Sym(P) \to G\,,$$ where $\Sym(P) = \Phi^*\Aut(P)$.

A lifting of the $G$ action on $M$ to one on $P$ is now the same as a splitting of the above short exact sequence of smooth groups. The obstructions to the existence of lifts thus are the obstructions to splitting this sequence smoothly (I am not sure how well non-abelian group cohomology for diffeological groups is developed, but I would expect it to describe the obstructions; if $G$ is discrete, ordinary group cohomology should suffice). If you find lifts, then the sequence shows that these lifts are a torsor over $\mathcal{G}(P)$; this controls how many lifts there are in case there are any. You can also find a bit more about this in Bunk, Müller, and Szabo - Smooth 2-Group Extensions and Symmetries of Bundle Gerbes.

Answer 3

Let me point out that if $G$ is disconnected, then the action might not extend. The simplest example I can think of would be a real line bundle over a torus with $w_1$ equal to one of the cohomology generators. Then a ${\mathbb Z}_2$ action that switches the two generators of cohomology will not extend, because the bundle is not preserved. One can do the same with complex line bundles, e.g. on $S^2 \times S^2$.