I'm familiar with the definition of the category of vector bundles, but I'm trying to derive it from some first principles about general fiber bundles. My intuition is that vector bundles should be understood as Cartan geometries (without a connection) of type $(\mathrm{Aff}(V),\mathrm{GL}(V))$. A $\mathrm{GL}$-structure on a fiber bundle obviously preserves the vector space structure of the model fiber, but I would think it should also force bundle morphisms to be fiberwise linear. Thus, starting with the definition of a bundle morphism I would like to understand how to properly constrain that definition in the context of a specific Cartan geometry $(G,H)$ in such a way that morphisms are represented, for example, by linear maps on fibers. I would like to avoid mentioning principal bundles in this process.

  1. The "default" structure group of a fiber bundle $E\overset{\pi}{\to}M$ with model fiber $F$ is $\mathrm{Diff}(F)$. In an atlas $\{(U_\alpha,\chi_\alpha)\}$ consisting of maps $\chi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times F$ we have transition functions $t_{\beta\alpha}:U_{\alpha\beta}\to \mathrm{Diff}(F)$ so that $\chi_\beta\circ\chi_\alpha^{-1}(p,f)=(p,t_{\beta\alpha}(p)\cdot f)$.
  2. A bundle morphism between two bundles $(E'\overset{\pi'}{\to}M',F)$ and $(E'\overset{\pi'}{\to}M',F')$ consists of two maps $\tilde{h}:E\to E'$ and $h:M\to M'$. Picking atlases $\{(U_\alpha,\chi_\alpha)\}$ and $\{(U_a',\chi_a')\}$ for these two bundles, and denoting $U_{\alpha a}=U_\alpha\cap h^{-1}(U_a')$, we can represent the morphism by local maps $\hat{h}_{\alpha a}:U_{\alpha a}\times F\to F'$ in the following way: $$\tilde{h}(\chi_\alpha^{-1}(p,f))=\chi_a^{\prime -1}(h(p),\hat{h}_{\alpha a}(p,f)).$$
  3. By combining these two definitions and writing out the definition of $\hat{h}_{\beta b}$ we get the transformation rule $$\boxed{\hat{h}_{\beta b}(p,t_{\beta\alpha}(p)f)=t_{ba}'(h(p))\hat{h}_{\alpha a}(p,f).}$$

Question: Now if I assume a choice of a $G$-structure on the first bundle and a $G'$-structure on the second, is there a way to simplify this transformation rule so as to turn it into a functor? More precisely, suppose we have two objects $(E\overset{\pi}{\to}M,G\curvearrowright F,\mathcal{G})$ and $(E'\overset{\pi'}{\to}M',G'\curvearrowright F',\mathcal{G}')$, where $\mathcal{G}$ is a $G$-structure on $E$ and $\mathcal{G}'$ is a $G'$-structure on $E'$. Furthermore, choose a map $h:M\to M'$, a map $\psi:F\to F'$, and a homomorphism $\lambda:G\to G'$. Is there a way of defining a map $\tilde{h}:E\to E'$ so that it is a bundle morphism which, in some sense, respects the given structures?

Attempt 1: Given the data that I provided, it would seem that we should be able to express any "reasonable" bundle morphism as $$\hat{h}_{\alpha a}(p,f)=\lambda(\bar{g}_a(p))\psi(g_\alpha(p)f)$$ using some maps $g_\alpha,\bar{g}_a:U_{\alpha a}\to G$. The transformation law then becomes $$\psi(f)=\lambda(\bar{g}_b)t_{ba}'(h(p))\lambda(\bar{g}_a)^{-1}\psi\left((g_\beta t_{\beta\alpha}g_\alpha^{-1})^{-1}f\right),$$ where all of the omitted arguments are $p$. Now, this turns out to quickly fail: consider for example isomorphisms of bundles, in which case $h=\mathrm{id}_M$, $\psi=\mathrm{id}_F$ and $\lambda=\mathrm{id}_G$, and we can also take $\alpha=a,\beta=b$ and $\bar{g}_a=e$. Then we get $$t_{\beta\alpha}'=g_\beta t_{\beta\alpha}g_\alpha^{-1}.$$ This is indeed what bundle isomorphisms generally look like, but it doesn't let us make any inferences about what $g_\alpha$'s must be.

Attempt 2: Now let's take some inspiration from Cartan geometry, where the model fiber is actually a homogeneous space $F=G/H=\{Hg\}_{g\in G}$ and the structure group is $H$. This means that there is a transitive left action of $G$ with stabilizers isomorphic to $H$: $G\curvearrowright F$. Then $t_{\beta\alpha}$'s can be assumed to take values in $H$, whereas $g_\alpha$'s can still take values in all of $G$. In this case the last formula obtained for bundle isomorphisms gets replaced with $$t'_{\beta\alpha}g_\alpha t_{\beta\alpha}^{-1}g_\beta^{-1}\in\mathrm{Stab}_G(f),\text{ for all }f\in F$$ but the intersection of all stabilizers is still trivial, so we get $t'_{\beta\alpha}g_\alpha t_{\beta\alpha}^{-1}g_\beta^{-1}=e$. This should somehow imply that $g_\alpha$'s actually commute with some subgroup of $G$ (the group of translations in the vector bundle case), but I don't see how that would happen.

Analysis: my approach so far gives me absolutely no information about what the values of $g$ should actually be, namely elements of a $\mathrm{GL}_k$ subgroup of $G=\mathrm{Aff}_k$. My goal is to come up with a definition of morphisms between bundles with $\mathrm{Aff}$-structures such that it implies that these morphisms must be equivariant with respect to the normal subgroup $\mathbb{R}^k$ of translations, i.e. actual fiberwise linear maps. Perhaps I need to be smart about requiring equivariance of $\psi$, but it's not clear how to do that considering that we need to be able to have different Lie groups acting on the two bundles (e.g. imagine embedding a vector bundle into another vector bundle of another rank, in which case $(G,H)\neq(G',H')$).

Attempt 3 (in response to Ben): every one of these maps $\hat{h}_{\alpha a}$ can be identified with a $G\times G'$-equivariant map $\Phi_{\alpha a}:U_{\alpha a}\times(G\times G')\to C^{\infty}(F,F')$ as follows: $$\Phi_{\alpha a}:\quad (p,g,g')\mapsto \left(f\mapsto g' \hat{h}_{\alpha a}\left(h(p),g^{-1}f\right)\right).$$ The action of $G\times G'$ on $C^{\infty}(F,F')$ is by composition $(g,g',\varphi)\mapsto g'\circ\varphi\circ g^{-1}$. The transformation rule becomes very simple: $$\boxed{\Phi_{\beta b}(p,g,g')=\Phi_{\alpha a}\left(p,t_{\beta\alpha}(p)g,t_{ba}'(h(p))g'\right).}$$ This clearly defines a global equivariant function on a principal $(G\times G')$-bundle but I want to omit that. I would like to set some restriction on these maps so that I could say that $h$ respects the $G$- and $G'$-structures in some sense. I can do it by analogy with the interpretation of a $G$-structure as a way of selecting an orbit inside the set of all isomorphisms: require all $\Phi_{\alpha a}$'s to take values in a single orbit of the action of $G\times G'$ on $C^\infty(F,F')$. This is a well-defined restriction because transition functions take values in $G\times G'$. It means that, up to a choice of local charts (compatible with the structures), the morphism "looks the same" in every fiber. Now, if $F$ and $F'$ have additional structure which selects in a natural way a subclass of maps $\mathrm{Hom}(F,F')\subset C^\infty(F,F')$ (and usually such that $G$ and $G'$ are the groups of automorphisms preserving those structures), then we also require those orbits to be orbits of maps from $\mathrm{Hom}(F,F')$.

Let's see what effect this has in the context of vector bundles. Here $G$ and $G'$ are some $\mathrm{GL}$-groups, say $G=\mathrm{GL}(F)$ and $G'=\mathrm{GL}(F')$. Then I'd like to say that a collection of maps $\varphi_p:F\to F'$ belonging to an orbit of a linear map under the action of $\mathrm{GL}(F)\times \mathrm{GL}(F')$ on $C^\infty(F,F')$ is equivalent to these maps being linear. This then leads to the standard definition of vector bundles. So the solution is that the additional structure of a vector space on the model fiber canonically selects of class of fiberwise maps, and these generate bundle morphisms.

  • 5
    $\begingroup$ You have the wrong model, to my way of thinking. A vector space needs an origin. You are modelling vector bundle fibers on affine space $(G,H)=(\operatorname{Aff}_V,\operatorname{GL}_V)$ when you really want that $V-0=\operatorname{GL}_V/H$ where $H$ is the stabilizer of a vector. There is a notion of a bundle of affine spaces, which arises in particular in studying jet bundles. $\endgroup$
    – Ben McKay
    Commented Mar 30 at 10:34
  • $\begingroup$ @BenMcKay I agree that I need to fix an origin. But is that enough to derive the fiberwise linearity? $\endgroup$ Commented Mar 30 at 14:49

1 Answer 1


One obvious fact: Cartan geometries are rigid, i.e. the automorphisms of a Cartan geometry, on a manifold with finitely many connected components, form a finite dimensional Lie group acting smoothly. The automorphisms of a vector bundle form an infinite dimensional group, at least in some informal sense, and this can be made into a formal statement with suitable precise definitions. But clearly you will need to add a lot of extraneous structure to any vector bundle to get a Cartan geometry. But you seem to be happy with a Cartan geometry associated to each fiber, so let's try that.

I would look at the story a bit differently. To each vector bundle $E\xrightarrow{\pi}M$, associate its frame bundle $FE\xrightarrow{\pi}M$, which is the set of all pairs $(m,u)$ where $m\in M$ and $E_m\xrightarrow{u}V$ is a linear isomorphism to a fixed vector space $V$. Then each fiber $F_m E$, while not being itself a Lie group, has a flat Cartan geometry modelled on $\operatorname{GL}_V$ acted on by $\operatorname{GL}_V$. To see this, note that $FE$ is acted on by $\operatorname{GL}_V$ by $g,u\mapsto gu$. This action is simply transitive on each fiber $F_m E$. We can say that $F_m E$ is a $\operatorname{GL}_V$-torsor. Now suppose we take some closed subgroup $G\subseteq\operatorname{GL}_V$ and imagine that we have a principal left $G$-subbundle $F\subseteq FE$. For example, if $G=\operatorname{U}(n)$ is the unitary group, this $F$ is the orthonormal frame bundle of a unique Hermitian metric. This subbundle $F$ has thus a fiberwise Maurer--Cartan form $\omega$ which we will define. First, we differentiate the $G$-action by defining vector fields $\vec{A}$ for each $A\in\mathfrak{g}$ in the Lie algebra of $G$, by $$ \vec{A}(u):=\left.\frac{d}{dt}\right|_{t=0}e^{tA}u. $$ For any tangent vector $\dot{u}\in T_u F_m E$ by writing $\dot{u}$ as $\dot{u}=\vec{A}$ for a unique $A\in\mathfrak{g}$. Then define $\omega(\dot{u})=A$. A $G$-connection on the vector bundle is precisely an extension of $\omega$ to tangent vectors to $F$, instead of just tangent vectors to the fibers $F_m$. The problem with morphisms of these objects is simple: we want to allow vector bundle morphisms to have zeros, but that won't map the invertible linear maps on one vector space to those on the other. But if we have two vector bundles $E,E'\to M$, say with reductions $F\subseteq FE$, $F'\subseteq FE'$ to structure groups $G,G'$, we can consider on $F\times_M F'$ the $G\times G'$ equivariant maps $F\times_M F'\to V^*\otimes V'$. These maps are the vector bundle morphisms in this elaborate language. To be precise, if $E\xrightarrow{f}E'$ is a vector bundle morphism, so a linear map on each fiber, then associate to each $(u,u')\in F\times_M F'$ the llnear map $\varphi(u,u'):=u' \circ f \circ u^{-1}$. Conversely, you can check that any $G\times G'$-equivariant map $F\times_M F'\to V^*\otimes V'$ arises uniquely in this way.

  • $\begingroup$ ohh this is awesome, your last part about equivariant maps is what I was missing. I'm trying to avoid principal bundles for now, so I've added some notes at the end of my post in response to this. Do you mind taking a look and making sure that it's a reasonable definition of morphisms between bundles with G-structures? $\endgroup$ Commented Apr 1 at 18:25

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