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In my algebraic topology course we have been studying covering spaces. There are two group actions on any fiber: the left action of the group of deck transformations and the right action of the fundamental group. There are various propositions proved about general group actions that will later be applied. In particular,

Let $X$ be a transitive left $G$-set. If a subgroup $H$ of $G$ is the stabilizer of a point in $X$, then $X$ is $G$-isomorphic to the left-coset space of $H$, i.e., the collection of all left-cosets of $H$, where $G$ acts on a coset by left-translation.

Here is my question: How do I translate this into a statement about right actions? Is it really so simple as flipping every occurence of 'left' to 'right'?

Here is what I know: Given a right action of a group $G$ on a set $X$, we can define a left action as $gx := xg^{-1}.$ Also, any right action of $G$ on $X$ gives a left action of $G^{\text{op}}$ on $X$ and every group is isomorphic to its opposite via inversion.

The Wiki article on group actions makes the claim that "...only left actions can be considered without any loss of generality." (Towards the bottom of the 'Definition' section.) This link seems to hint that the justification for such a claim lies in the fact that a group $G$ and its opposite are 'naturally' isomorphic. The problem is that I only have a vague idea as to what that means as I have only a very basic working knowledge of categories from my algebraic topology book. Though the 'rules for translation' that the latter link provides are helpful, I still feel like I don't have a very good understanding of what's going on.

I realize what I'm looking for is a sort of theorem about theorems, which is strange. But can anybody offer some insight? If possible, can somebody elaborate on what this natural isomorphism has to do with anything? Basically, is the translation really so simple?

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The concept of a natural isomorphism (and of natural transformations in general) is what drives the simplicity of the translation. I'll try and stay light on or refresh the necessary category theory, but essentially, a natural transformation is a morphism of functors. (refresher: functors take elements to elements, morphisms to morphism, identities and compositions are preserved - they're the morphisms of categories). With this machinery, the motivating "obviousness" of the translation can be given an accompanying rigorous justification.

I'm sorry if this is a bit pedantic, but setting up exactly what the concept of "a morphism of functors" entails is as good of a starting place as any. More concretely, if I have two categories $C$ and $D$, and functors $F$ and $G$ that map $C$ to $D$ , a natural transformation $\eta:F \rightarrow G$ is a collection of morphisms so that for every object $x$ in $C$, we have a morphism $\eta_x: F(x) \rightarrow G(x)$ so that the entire collection respects composition: that is, for all morphisms $f \in C$ taking $x \rightarrow y$, we have that the diagram below commutes. What this means is that for any objects $x, y \in C$ and a morphism $f \in C$ between them, taking either path from $F(x)$ to $G(y)$ will give the same answer in $D$.

$\begin{array}{ccccccccc} F(x)& \xrightarrow{F(f)} & F(y) \\ \downarrow{\eta_x} & & \downarrow{\eta_y} & \\ G(x) & \xrightarrow{G(f)} & G(y) \end{array}$

From here, a natural isomorphism can be defined as a natural transformation with an inverse (both left and right). Our statement about a group $G$ being naturally isomorphic with its opposite group can be phrased as a natural isomorphism between an identity functor from Grp to itself, and an "opposite functor" from Grp to itself that takes elements to their inverses and group homomorphisms to themselves. We can prove naturality by considering sqare diagrams above, and showing that they must all commute is a consequence of $f(x^{-1})$ = $f(x)^{-1}$ for any group homomorphism $f$ .

With these abstract tools set up, we can understand why we can talk about left and right actions without loss of generality. The naturality of the relation between $G$ and $G^{opp}$ tells us that our inversion commutes with group homomorphisms, including quotient maps and inclusion maps (key for group actions on cosets). Thanks to the relative abstraction of category theory, we can talk about things like group actions on left-coset spaces and "naturally" relate them to group actions on right-coset spaces exactly because we are sure that we're not harming any general structure by inverting.

In my opinion, a working knowledge of homological algebra and category theory is crucial for really getting a full understanding of algebraic topology - combining the abstraction with geometric reasoning and intuition will give you a great picture of what's really going on.

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Thanks so much for the help everybody. Jake, could you perhaps expand on the second to last paragraph a bit? In particular, how would one exhibit this "natural relation" between a group acting on left-coset spaces and a group acting on right-coset spaces? Are the G-Set categories involved here at all? –  John Myers Mar 8 '12 at 7:28
    
Sure: if a group G acts on left-coset spaces, we can "exhibit a natural relation" to a different group's action on right-coset spaces - namely, G's opposite group. The explicit correspondence looks like (g.x) <--> (x.g^-1), as Derek describes above. If I can prove something general about a left action (that is, as long as my statement makes sense in both the context of a left and a right action), the correspondence immediately lets us switch "left" to "right" and "G" to "G^opp", and the natural isomorphism lets us talk about G anyway, since this isomorphism commutes with related statements. –  JakeR Mar 8 '12 at 7:49
    
The category of G-sets are related, but not really directly involved. This result lets us talk about categories of G-sets more fluidly; our natural isomorphism commuting with group homomorphisms means that our statements involving the spaces a G-set acts on (e.g. cosets of stabilizers as with deck transformations/subgroups of the fundamental group) and general things about them. However, I would hesitate to say that categories of G-sets actively take a role here, other than that we're talking about group actions. We do our work in just the category of Groups in this case. –  JakeR Mar 8 '12 at 8:04
    
Great, thanks. I think I'm making some progress. I can see now how the natural isomorphism relates a right action of G and a left action of G^op: A right action of G on a set X induces a group morphism of G into Sym(X)^op where g is assigned the map x->xg. To say that inversion provides a natural isomorphism between G and G^op implies that the inversions, the morphism G->Sym(X)^op induced by the right action, and the left action G^op->Sym(X) all commute in the square that you drew. Correct? –  John Myers Mar 8 '12 at 9:01
    
Drawing the diagram, this interpretation seems to make sense. –  JakeR Mar 8 '12 at 14:27
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Given a left action $(g,x) \to g.x$ of $G$ on $X$, you can define a corresponding right action by defining $(x,g) \to g^{-1}.x$.

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Yes, simply translating that statement by inserting 'right' in for 'left' each time produces an equivalent, correct result. In essence, they're no different except for a notational thing.

Now that I've said that, there come times when it becomes more convenient to think of something as either acting on the left, or on the right. Left actions are nice because they line up with our ideas of function composition. For example, if $G$ is acting on $X$, $g_1g_2 x$ looks like $g_1 (g_2 (x))$, which it is. But $x g_1 g_2$ looks funny to me (unless I've been working out of a Herstein Algebra book, where often times functions compose on the left).

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I agree $x g_1 g_2$ looks funny, but when working with the functions themselves and not writing the $x$ explicitly, I like right actions because that makes $g_1 g_2$ mean "do $g_1$, then $g_2$", in left-to-right order. Particularly with permutations, I prefer using the right action. –  Ted Mar 8 '12 at 8:04
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