Two definitions of regular representation of a group

Question

A left action of a group $G$ on a set $X$ is a operation $G \times X \to X$ such that for $g,h \in G, x \in X$ we have $(gh)x = g(hx)$ and $1x = x$.

The following remarks from wikipedia makes me wonder, I cite:

For a finite group $G$, the left regular representation $\lambda$ (over a field $K$) is a linear representation on the $K$-vector space $V$ freely generated by the elements of $G$ [...] Given $g \in G$, $\lambda_g$ is the linear map determined by its action on the basis by left translation by $G$, i.e. $$ \lambda_g : h \mapsto gh $$ for all $h \in G$.

Then an alternate description is presented:

Alternatively, these representations can be defined on the $K$-vector space $W$ of all functions $G \to K$. [...] Given a function $f : G \to K$ and an element $g \in G$, $$ (\lambda_gf)(x) = f(\lambda_g^{-1}(x)) = f(g^{-1}x). $$

But with these definitions both actions are not "compatible" in the sense that for $f : G \to K$ and $x \in X$ we have $$ \lambda_{gh}f(x) = f((gh)^{-1}x) = f(h^{-1} g^{-1} x) = \lambda_h(\lambda_g f))(x) $$ but if $u \in G$ we have $$ \lambda_{gh}u = (gh)u = \lambda_g ( \lambda_h u ). $$ The last line is what should hold for a left action, but the application on functions reverses the order. But both should yield the same, for if I identify the functions $G \to K$ with formal sums (hence elements of the freely generated vector space), then $$ f = \sum_{u \in G} f(u) u $$ and linear extension of the first definition gives $$ \lambda_g f = \sum_{u\in G} f(u) gu = \sum_{g^{-1}u\in G} f(g^{-1}u) g(g^{-1}u) = \sum_{g^{-1}u\in G} f(g^{-1}u) u. $$ So on one side we have $$ \lambda_{gh} f = \sum_{u \in G} f(u) (gh)u = \lambda_g \left( \sum_{u \in G} f(u) hu \right) = \lambda_g (\lambda_h f ) $$ on the other side $$ \lambda_{gh} f = \sum_{(gh)^{-1}u\in G} f((gh)^{-1}u) u = \sum_{(gh)^{-1}u\in G} f(h^{-1} g^{-1}u) u = \lambda_h \left( \sum_{(gh)^{-1}u\in G} f( g^{-1}u) u \right) = \lambda_h (\lambda_g f ). $$ but in general $\lambda_g(\lambda_h f) \ne \lambda_h (\lambda_g f)$. What am I missing?

Answer 1

Take another look at your first computation: the $h$ action happens first, not second: $$ (\lambda_{gh}f)(x) = f((gh)^{-1}x) = f(h^{-1} g^{-1} x) = (\lambda_hf)(g^{-1}x) = (\lambda_g(\lambda_hf))(x) = ((\lambda_g\lambda_h)f)(x) $$

This is (I feel) one of the drawbacks to the "functional" definition of the groupring: this "twist"s (or perhaps "contravariant behavior") when you identify functions as actions can be a little disorienting.

If you look at the group ring as a set of formal linear combinations, then everything works like you think (the left action of $G$ on itself extends linearly to a left action of $R[G]$ on $R[G]$.)

Answer 2

For the first notation, as pointed out we have $$ (\lambda_h(\lambda_g f))(x) = (\lambda_g f)(h^{-1}x) = f(g^{-1}h^{-1}x) = f((hg)^{-1}) = \lambda_{hg} f. $$ The point with the other notation is that we can define $\lambda_g f = \sum_{u\in G} f(g^{-1}u)u$, but it is not the case that $$ \lambda_h (\lambda_g f) = \lambda_h \left( \sum_{u \in G} f(g^{-1}u) \right) = \sum_{u\in G} f(h^{-1} g^{-1}u) u. $$ With this a dependency between the variable $u$ and its coefficient is missed, in the sense that when it is not given as $f(u)u$ it must be handled differently, namely if we take $\hat f = \sum_{u\in G} f(g^{-1}u) u$ then $\hat f(u) = f(g^{-1}u)$, and writting $\hat f = \sum_{u\in G} \hat f(u)u$ gives $$ \lambda_h \hat f = \sum_{u \in G} \hat f (h^{-1}u)u $$ but $\hat f(h^{-1}u) = f(g^{-1}(h^{-1}u)$),i.e. when we put in $h^{-1}u$ for $u$ we must put it behind $g^{-1}$ to make a proper variable substitution, the "rewritting-rule" of putting in merely in front of the argument to the coefficient does not work as it "neglects the dependency" on the variable. So the correct "rewritting-rule" would be $$ f' = \sum_{u \in G} f(au) bu = \sum_{u \in G} f(a b^{-1} u) u $$ i.e. the $b$ goes in-between, as $f'(x) = f(au) \Leftrightarrow x = bu$, i.e. $u = b^{-1}x$. Said differently, the coefficient is the function "$f(g^{-1} "variable")$ and not the whole of $g^{-1}u$ is the variable.

Remark: This post seems to be related.