Why does the group act on the right on the principal bundle?

In many textbooks, in fact all textbooks I've seen, the fiberwise group action on the principal bundle is on the right. It seems to me that left actions and right actions are essentially the same. Then why do people keep acting on the right for the principal bundle, while acting on the left for many other cases? Is it because one might want to consider the possible additional left action on the bundle?

• The key example for a principal bundle is $G \rightarrow G/H$, where $G$ is a Lie group and $H$ is a Lie subgroup. If we use left cosets for the base, then the action on the fiber is a right action. – Deane Yang Dec 27 '10 at 5:56
• I think the underlying philosophy is usually this: When a group acts on some kind of object X, it also acts on various functions on X. But to make this second action compatible with the group product you usually have to pass from the given left action on X to a right action on functions. (Otherwise you have to use group inverses.) – Jim Humphreys Dec 28 '10 at 14:11
• I'm curious about this. I know nothing about differential geometry or algebraic topology, so my naive guess would be that this is a notation question. However the question doesn't have a "notation" tag, and the answers below look (from my lay perspective) to have some substance, so perhaps "right action" and "left action" refer to something more than the order in which your symbols appear? How exactly are those terms defined then? – Amit Kumar Gupta Dec 31 '10 at 1:05
• Nope, purely a notational question. What's subtle is figuring out how all the choices of left vs. right intertwine. A better question would be "what sensible seeming notational choice requires us to have the group act on the principal bundle on the right." – Ben Webster Dec 31 '10 at 6:19
• And this is exactly unknown's answer; if you think that left $G$-spaces are the default thing, and try to build a fiber bundle modeled on a left $G$-space $X$, with transition functions in $G$ acting on $X$, then that's the same data as a principal bundle with a right action. – Ben Webster Dec 31 '10 at 6:22

The answer to the first question is:

Because in local chart we want the action to commute with transition functions, and the latter are traditionally assumed to be acting on the left.

I'll explain better below. Assume we're working with a ringed space $(X, \mathcal{O})$ with structure sheaf $\mathcal{O}$ in your favourite geometry: $\mathcal{C}^{\; 0}$, $\mathcal{C}^{\infty}$, analytic, algebraic,...

First of all, what's a fiber bundle with fiber $F$ on a space $X$? It's essentially the datum of a covering {$U_i$} of $X$ and, for each double intersection $U_{ij}=U_i\cap U_j$ , some transition functions

$\varphi_{ij}:U_{ij}\times F\rightarrow U_{ij}\times F$

that verify some cocycle condition. The transition functions may be tautologically seen as functions

$g_{ij}:U_{ij}\rightarrow \mathrm{Aut}(F)$,

$x\mapsto g_{ij}(x)$

and the traditional convention is that $\mathrm{Aut}(F)$ acts on $F$ on the left.

Now, what's a principal bundle? It's a fiber bundle with $F=G$ and with transition functions $g_{ij}$ with values in the group $Left(G)\subset \mathrm{Aut}_{sp}(G)$ of left translations of $G$ (which is, btw, isomorphic to $G$ itself; and here of course we're considering automorphisms of $G$ as a space not as a group).

So, in local chart, we have: $\varphi_{ij}:(x,g)\mapsto (x,g_{ij}(x)\cdot g)$, where the dot is left group multiplication in $G$, and even actual left matrix multiplication in case $G$ is a matrix group.

Let's stick to the case $G=$ matrix group, for the sake of clearness (but the case of general $G$ is not different). Suppose we want to define an action of $G$ itself on the total space of the bundle.

The spontaneous idea is to write down things in local chart (like physicists usually do) and try the obvious matrix multiplication, say, on the left:

$h_{\cdot}:(x,g)\mapsto (x,h\cdot g)$ for $h\in G$. But... wait!! That doesn't glue, as that locally defined action doesn't commute with the action of the "gauge group" (i.e. transition functions), so it doesn't define an intrinsically defined global left action.

What about trying to do the same on the right?

$h:(x,g)\mapsto (x,g\cdot h)$ for $h\in G$

Well, now it works as the two actions clearly commute:

$h \circ g_{ij}=g_{ij}\circ h:(x,g)\mapsto (x, g_{ij}(x)\cdot g\cdot h)$

and we can glue and get a globally well defined right action.

Recall one way of defining a fibre bundle is a quintuple $(p,E,B,F,G)$, where $E$ is the total space, $B$ is the base, $p:E \rightarrow B$ is the projection, $F$ is the typical fibre, and $G$ is the structure group - that is, $G$ is a Lie group which acts on $F$ on the left.

Then a principal bundle is simply the special case where $G=F$ and the action is given by left translation. It is then an (easy) lemma that such a bundle $(p,E,B,G,G)$ carries a right action of $G$ on the total space $E$.

Left and right is mostly just a matter of convention: a left action by $G$ is precisely the same as a right action by the opposite group: the same object but with the multiplication in the reverse order.

In relation to you last question, you might want to consider the additional left action on the bundle, in a compatible way (acting left then right should be the same as acting right then left). This object is called a principal bibundle if the left action is also principal. These objects come up in two areas that I am aware of. They are used when defining non-abelian bundle gerbes, which are geometric objects representing classes in nonabelian cohomology $H^2(X,AUT(G))$ (here $AUT(G)$ is a coefficient object determined from a group $G$ - it is technically a 2-group or equivalently a crossed module).

The other place they turn up is via generalisation when $G$ is replaced by a (topological) groupoid (or Lie groupoid, if you are in a smooth setting), and you can in fact have two different groupoids, one acting on the left, and one acting on the right. These right principal bibundles (drop the principality of the left action here) are then generalised morphisms between topological/Lie groupoids which are used when presenting stacks by groupoids, especially in the Lie groupoid literature.

There is another possible reason for this convention: to make the notation for the Borel construction $EG \times_G X$ a little nicer. Writing $Y\times_G X$ generally implies that $Y$ has a right action and $X$ has a left action (although of course there really isn't any difference between right and left $G$-spaces). If we habitually work with left $G$-spaces $X$, then we end up wanting the universal principal bundle $EG \to BG$ to be defined by a right action.

I would propose this "explanation" : the archetypical principal bundles are tangent frame bundles of a given type on differentiable manifold (arbitrary or, say, orthonormal ones on a riemannian manifold, oriented ones on an oriented manifold, etc). Sticking to the bundle $P\to M$ of arbitrary tangent frames on $M$ for definiteness, they are naturally viewed as linear isomorphisms $p:\mathbb{R}^n\to T_x M$, $x\in M$.

Thus, using the "standard" (or not ?) convention for composition of maps ($g\circ f$ applies $f$ then $g$), the structure group $G=GL_n(\mathbb{R})$ naturally acts on the right.

Also, note that when you have a left action of $G$ on some space $F$, you naturally construct the associated bundle $P\times_G F \to M$ with fiber $F$, taking the quotient of $P\times F$ by $(p.g,\xi)\sim(p,g.\xi)$. If $F$ is a linear representation of $G$, this is very closely related to the local index notation of tensor fields on $M$ : local coordinates $(x_i)_i$ on $U\subset M$ define a local trivialization of $P$ (via the section $(\partial/\partial x_i)_i$), in which a tensor field of type $F$ is simply a map $U\to F$.

• Was it $(p.g,\xi)\sim(p,g.\xi)$ or $(p.g,\xi)\sim(p,g^{-1}.\xi)$ ? – Qfwfq Dec 29 '10 at 15:51
• The first is the right one, since $g$ acts on the right on $p$ and on the left on $\xi$ (think of $V=\mathbb{R}^n$, giving the tangent bundle). If you prefer, it is the quotient by the right action $(p,\xi).g=(p.g,g^{-1}\xi)$. – BS. Dec 30 '10 at 9:59
• ok ! – Qfwfq Dec 30 '10 at 22:04

You can look at principal fiber bundles as "half" of groupoids. And for a groupoid right and left actions have a more balanced and obvious meaning.

Consider a connected groupoid ${\bf K}$ (that is, between two objects there is always at least an arrow) and pick an object $x \in X = {\rm Obj}({\bf K})$. Let ${\bf K}(x)$ be the isotropy of $x$, that is the arrows with source and target $x$ (since the groupoid is connected, all the isotropies are isomorphic). This group acts by pre-composition on the arrows with source $x$ and by post-composition on arrows of target $x$. Let us make that clear, let us introduce $${\bf K}(x,-) \mbox{ the space of all arrows with source } x.$$ $${\bf K}(-,x) \mbox{ the space of all arrows with target } x.$$ We have two "actions" of $g \in {\bf K}(x)$, considering ${\bf K}(x,-)$ or ${\bf K}(-,x)$, that is:

\begin{align} \mbox{ For all $f \in {\bf K}(x,-)$,} & \mbox{ $g \cdot f \in {\bf K}(x,-)$.} \\ \mbox{ For all $f \in {\bf K}(-,x)$,} & \mbox{ $f \cdot g \in {\bf K}(x,-)$.} \\ \end{align}

The first one is an action of the group ${\bf K}(x)$, the second one is an anti-action. If we want also an action we have to consider $f \cdot g^{-1}$, but is not fundamental. So, ${\bf K}(x,-)$ and ${\bf K}(-,x)$ are two (equivalent) principal bundles over $X$, and the projection is given by the target map ${\rm trg} : {\rm Mor}({\bf K}) \to X$ in the first case, and the source map ${\rm src} : {\rm Mor}({\bf K}) \to X$ in the second case.

Conversely, if you have a principal bundle, taking the fiber square of the bundle and then the quotient by the diagonal action of the structure group, you get a groupoid, and these two constructions are inverse one from each other.

Now, if you want to consider not just algebraic principal bundles but smooth ones, you may equip the space of objects and the space of arrows of the groupoid with an accordingly smooth structure: manifold or whatever (I did it for diffeological spaces, actually it's this way diffeological fiber bundles are defined).

As David Roberts said (and you seem to already know), left vs right is just convention. In the same way that Riemannian geometry literature was influenced by the role it played in general relativity, perhaps the literature on principal bundles was influenced by the role they played in particle physics. In particular, Yang-Mills mills theory, the actions are on the right [Bleecker, D. Gauge theory and variational principles'' (1985) ]