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

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

2010-12-27T05:53:39Z

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?

Answer by David Roberts for Why does the group act on the right on the principal bundle?

2010-12-27T08:14:46Z

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.

Answer by Will Merry for Why does the group act on the right on the principal bundle?

2010-12-27T18:28:40Z

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 $E$.

Answer by Qfwfq for Why does the group act on the right on the principal bundle?

2010-12-28T18:57:19Z

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.

Answer by Jeffrey Giansiracusa for Why does the group act on the right on the principal bundle?

2010-12-29T09:23:46Z

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.

Answer by BS for Why does the group act on the right on the principal bundle?

2010-12-29T14:32:32Z

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 $U\to F$.

Answer by Patrick I-Z for Why does the group act on the right on the principal bundle?

2010-12-29T17:37:23Z

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).