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In this question, representation means representation of a finite group $G$ on a finite-dimensional $\mathbb{C}$-vector space $V$.

When I studied this subject, I only learned about "left representations", i.e., when the action of $G$ on $V$ was a left action. What happens when we have a "right representation" (i.e., a group homomorphism $G^{op} \rightarrow GL(V)$)?

I know this is a bit vague but I find it confusing whenever I have to deal with these "right representations"...

One specific question would be: when people talk about a right representation, say $\rho: G^{op} \rightarrow GL(V)$, and then they mention the character $\chi_{\rho}$, do they mean $\chi_{\rho}(g) = trace(\rho(g))$ or $\chi_{\rho}(g) = trace(\rho(g^{-1}))$? What is the standard terminology when dealing with right representations?

Original question: I was reading a paper that mentioned a natural right representation of a given group $G$ and went on to say "irreducible representation of $G$". Should I assume he means a "right irreducible representation"? Or does he mean a standard "left irreducible representation"?

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$G\cong G^{op}$ via $g\mapsto g^{-1}$ –  user2035 Aug 3 '11 at 11:48
    
@a-fortiori: I added a specific question. –  expmat Aug 3 '11 at 11:53
    
I am pretty sure the character is $\chi_{\rho}\left(g\right) = \mathrm{trace}\left(\rho\left(g\right)\right)$, because when one speaks of right representations, one usually wants to think of them as being genuine right representations, rather than left representations in disguise. But I'll leave the definitive answer to experts. –  darij grinberg Aug 3 '11 at 12:36
    
Sorry for the vagueness and confusion. I just added the question that originally motivated this entry. Should I reformulate this whole entry? –  expmat Aug 3 '11 at 12:57

2 Answers 2

This left-right business is largely an accident of notation/language/writing, which makes it all the more difficult to distinguish genuine mathematical features from accidents of writing left-to-right horizonatally, etc.

Presumably a primary way in which $G^{op}$ arises is acting on the dual $V^\star$ of a given repn $V$ of $G$, by choosing to have $G$ act "on the right" on the dual, by $(\lambda g)(v)=\lambda(gv)$, that is, just moving the parentheses. The "problem", indeed, is the apparent left-right switch, if we write something like $\pi^\star(g)(\lambda)=\lambda\circ g$, namely, $\pi^\star(gh)=\pi^\star(h)\pi^\star(g)$.

It is a significant problem that $G^{op}$ [edit] cannot be isomorphic to $G$ non-abelian by $g\rightarrow g$... [as commented, my previous was messed-up...] Yes, it can be by $g\rightarrow g^{-1}$, but/and this amounts to converting everything back to left $G$-modules. My point would be that it is going the long way round to introduce the opposite group if we want to recover "left" $G$-modules in any case, and when the left-right business is syntactic, rather than semantic.

It is possibly better to have $G$ act "on the left" on the dual by $(g\lambda)(v)=\lambda(g^{-1}v)$.

Similarly, modules over rings can be "left" or "right", but, in fact, "right" modules are left modules over the opposite ring. Thus, a left $S$-module and right $R$-module is, as well, a left $S\otimes R^{op}$-module, where the left actions of $S$ and $R^{op}$ commute.

Some special classes of rings do have natural isomorphisms to their opposite rings, allowing identifications...

EDIT: yes, incorporating @Snark's comment, one common manner in which the left-right seeming-confusion arises is with a group $G$ acting on a set $X$, and then "inducing" a natural action of $G$ on functions $f$ on $X$. Much as taking dual reverses arrows, the action on functions on a set reverses arrows... thus the following. For $G$ acting on $X$ on the left, $G$ acts on functions "on the left" by $(gf)(x)=f(g^{-1}x)$. For $G$ acting on the right on $X$, $G$ acts on functions "on the left" by $(gf)(x)=f(xg)$. (Note that this avoids having to pretend that $G^{op}$ is a group, etc.) Once or twice in one's life, one should verify that the placement of inverse (or lack theoreof) makes the "left" action of $G$ on functions "associative", in the sense $(gh)f=g(hf)$ for $g,h\in G$.

Edit: ... and, in reference to the added "original question", operationally there is no "left/right" sense to a repn. One may choose different ways of writing, but the underlying things do not change. Many years ago, I. Herstein advocated writing functions on the right of their arguments, for the reason that then the order we'd read them would be the order of their application, but it mostly did not "catch on".

More edits! (Sorry for the delay... busy...) Certainly if one finds left-right distinctions/notations useful at least as a mnemonic, then they're good. For myself, somehow the variety of contexts relevant to my stuff makes me feel that left-right is insufficient. For one thing, different contexts would seem to impose conflicting requirements. For groups, many people do not want the contragredient to be a repn of the "opposite group", because "contragredient" is supposed to be a functor from the category of $G$-repns to itself. Similarly for special-but-important rings ("co-algebras", generally, I believe), such as group rings, universal enveloping algebras. "Oppositely", for rings without (relevant) involution, the bi-module, etc., notion is sometimes very useful. Nevertheless, it can get out of hand: for $R,S$-bimodule $M$ and $T,U$-bimodule $N$, $Hom_{\mathbb Z}(M,N)$ has a fairly obvious structure... until/unless we feel compelled to establish a once-and-for-all, "perfect" notation that allegedly suits all contexts.

I know that left-right is often used to suggest co/contra-variance, but, in fact, I claim that there are some technical advantages in not identifying the two things, one being innate, the other arguably less so.

Again, I'll claim that left-right should never be the pivotal issue, if the context is decently portrayed. Sure, something can be _tied_to_ notational left-right issues, but there's surely something underlying that is not so fragile as notation. This is not meant to "dis" anyone who finds left-right a marvelous device to remember distinctions!

Yet-another-edit: as with many things, the path by which one comes to a situation strongly affects one's attitude. Here, if one comes to "group repns" as a special and specially-equipped case of categories of modules over rings, yes, arguably, it makes complete sense to talk about the way in which categories over special types of rings have bonus properties. On the other hand, if one has followed the "opposite" route, starting with group repns as useful in various applications, and considered "imbedding" that story in the larger story of categories of modules over rings with-or-without various structures, one could just-as-easily take the viewpoint that the "general case" is a kind of "failure". That is, if categories of group repns are the "usual case" or "normal case", one might find no need to distinguish it from the "failing cases", and, likewise, not choose language that emphasizes how lucky we are to have "all these special features"... because they're not special?

Certainly some people prefer to proceed from the particular to the general, and others vice-versa, and such predilections color language and notation, obviously, and, often in manners nearly incomprehensible to people with "opposite" viewpoints.

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I'll just add to this explanation : both the left and right representations correspond to left actions. The thing is that going through a function transforms a left action to a right action (and vice versa), which explains the need to either write $f(xg)$ or $f(g^{-1}x)$ for a left action of $g$ on a function $f$. –  Julien Puydt Aug 3 '11 at 12:39
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$G^{op}$ is a group, for every group $G$. –  user2035 Aug 3 '11 at 12:58
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I disagree that "this left-right business is largely an accident of notation." We are talking about a special case of the distinction between covariant and contravariant functors, and this distinction is real, important, and far from an "accident." –  Qiaochu Yuan Aug 3 '11 at 14:47
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The issue is that group theorists like to fix which side the act on. Thus whenever one performs a contravariant operation, the immediately us inverses to remove the contravariance. Thus what many group theorists refer to the right regular representation for a certain left module. Semigroup theorists and ring theorists don't have this issue because dualizing a left module gives a right module pointe finale! –  Benjamin Steinberg Aug 3 '11 at 14:56
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It is quite a genuine mathematical feature of group rings that their categories of left and right representations are equivalent! It ca be pinned down precisely in several ways, but it is quite a feature :) –  Mariano Suárez-Alvarez Aug 4 '11 at 5:40

As has been pointed out in comments, any particular representation of a group can be realized as a representation with the "opposite handedness," but of another group, namely the opposite group. As has also been mentioned, every group is canonically isomorphic to its opposite group via the inversion map. One might be inclined to combine these two comments to conclude that the distinction between left and right representations is moot.

In my estimation: sometimes it is, and sometimes it isn't.

For example:

(1) When one defines the induction of a representation from a subgroup, one starts with a certain space of functions that transforms is a specified fashion on one side (depending on the original representation). The canonical representation of the original group (of the original handedness) on this space is on the other side, or equivalently, is a representation of the opposite group. However, we usually just trade it off for the same handedness representation of the same group through the inversion identification.

(2) On the other hand, sometimes one naturally is led to deal with both kinds of actions at once - like when studying functions (better: forms of some sort) on double coset spaces of a pair of subgroups in a group, and it can be useful to keep the right and left actions as such. Natural notation such as $H\setminus G/ K$ and the bi-chiral transformation formulas play more nicely together this way, for example.

I think that, in both cases the choice comes down to a convenience of one sort or another.

For that matter, I suppose I feel like this convenience extends to the covariant/contravariant commentary here. One can trade off one for the other by trading off a category for its opposite. This can certainly make a number of statements/constructions more awkward, but again this seems a matter of convenience.

All this aside, there is a lot to be said for convenience...

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