Definition of action

Question

Consider the following definition of "action":

Definition 1. Let $\Omega$ and $E$ be two sets. A mapping of $\Omega$ into the set $E^E$ of mappings of into itself is called an action of $\Omega$ on $E$.

Let $\alpha\mapsto f_{\alpha}$ be an action of $\Omega$ on $E$. The mapping $(\alpha,x)\mapsto f_{\alpha}(x)$ (resp. $(x,\alpha)\mapsto f_{\alpha}(x)$) is called the law of left (resp. right) action of $\Omega$ on E associated with the given action of $\Omega$ on E.

Now, let $f:E\times E\rightarrow E$ be a law of composition on a set $E$. We have the following pair of derived actions: \begin{align*} \gamma&:E\rightarrow E^E,\, x\mapsto (y\mapsto f(x,y)),\\ \delta&:E\rightarrow E^E,\, x\mapsto (y\mapsto f(y,x)). \end{align*}

According to the above definition, we have the following mappings \begin{align*} l_{\gamma}&:E\times E\rightarrow E,\,(x,y)\mapsto\gamma_{x}(y)=f(x,y),\\ r_{\gamma}&:E\times E\rightarrow E,\,(y,x)\mapsto\gamma_{x}(y)=f(x,y); \end{align*}

But these are the same mappings—no? What is the difference between $l_{\gamma}(x,y)$ and $r_{\gamma}(x,y)$ for $(x,y)\in E\times E$? Does merely switching the components of the pair in the specification of $r_{\gamma}$ give us a map different from $l_{\gamma}$?

When $\Omega$ and $E$ are not equal, one actually obtains two different mappings by the procedure referred to above (since in that case the domains are different). In this case, however, it is not clear how one can differentiate between a law of left and a law of right action.

The book claims that $l_{\gamma}=f$ and $r_{\gamma}=f^{op}$, where $f^{op}$ denotes the opposite law of $f$. If $l_{\gamma}=r_{\gamma}$, then that implies that $f=f^{op}$ when $E$ is not commutative. But then again I don't see how $l_{\gamma}\ne r_{\gamma}$...

Answer 1

But these are the same mappings—no?

No. Note that your \begin{align*} l_{\gamma}&:E\times E\rightarrow E,\,(x,y)\mapsto\gamma_{x}(y)=f(x,y),\\ r_{\gamma}&:E\times E\rightarrow E,\,(y,x)\mapsto\gamma_{x}(y)=f(x,y); \end{align*} has "$(x,y)\mapsto \dots$" and "$(y,x)\mapsto \dots$". These are not both images of $(x,y)$. Specifically, to your question

What is the difference between $l_{\gamma}(x,y)$ and $r_{\gamma}(x,y)$ for $(x,y)\in E\times E$? Does merely switching the components of the pair in the specification of $r_{\gamma}$ give us a map different from $l_{\gamma}$?

\begin{align} l_\gamma(x,y) &= f(x,y) \text{ and} \\ r_\gamma(x,y) &= f(y,x) \end{align}

Unless $f$ is symmetric in its arguments, these are not the same image, so $l_\gamma$ and $r_\gamma$ are not the same map. Not all laws of composition are symmetric.