Understanding the inverse in the definition $(\tilde{\Pi}_gf)(v)\equiv f(\Pi^{-1}_gv)$ I'm trying to understand the representation $\tilde{\Pi}$ induced from the fundamental representation $\Pi$, defined as $(\tilde{\Pi}_gf)(v)\equiv f(\Pi^{-1}_gv)$ for $g\in G,\hspace{1mm}f\in\mathcal{C}(V),\hspace{1mm}v\in V$, and $\tilde{\Pi}_g\equiv\tilde{\Pi}(g)$. Specifically, I'm trying to understand why it needs to be $\Pi^{-1}_g$ rather than $\Pi_g$. It seems like this would come out of requiring $\tilde{\Pi}$ to be a homomorphism, $\tilde{\Pi}(g_1g_2)=\tilde{\Pi}(g_1)\tilde{\Pi}(g_2)$, but that's the exact opposite of what I'm getting when I expand the two sides:
$$\text{LHS}:\hspace{2mm}(\tilde{\Pi}_{g_1g_2}f)(v)=f(\Pi_{g_1g_2}^{-1}v)=f((\Pi_{g_1}\Pi_{g_2})^{-1}v)=f(\Pi^{-1}_{g_2}\Pi^{-1}_{g_1}v)$$
$$\text{RHS}:\hspace{2mm}(\tilde{\Pi}_{g_1}\tilde{\Pi}_{g_2}f)(v)=\tilde{\Pi}_{g_1}f(\Pi^{-1}_{g_2}v)=f(\Pi^{-1}_{g_1}\Pi^{-1}_{g_2}v)$$
The fact that it's an inverse having $(\Pi_{g_1}\Pi_{g_2})^{-1}=\Pi^{-1}_{g_2}\Pi^{-1}_{g_1}$ is giving me opposite ordering, whereas $(\tilde{\Pi}_gf)(v)\equiv f(\Pi_gv)$ being the definition would give me agreement on the ordering. Where am I going wrong? 
 A: The presence of $^{-1}$ is only due to the fact that one wants
$$\tilde{\Pi}_g\tilde{\Pi}_h = \tilde{\Pi}_{gh}\:,$$
and not
$$\tilde{\Pi}_g\tilde{\Pi}_h = \tilde{\Pi}_{hg}\:.$$
This is wrong
$$"\tilde{\Pi}_{g_1}f(\Pi^{-1}_{g_2}v)=f(\Pi^{-1}_{g_1}\Pi^{-1}_{g_2}v)"$$
The left hand side is the action of $\tilde{\Pi}_{g_1}$ on the function $$g(v) := f(\Pi^{-1}_{g_2}v)$$
so that 
$$\left(\tilde{\Pi}_{g_1}f\right)(\Pi^{-1}_{g_2}v)=f(\Pi^{-1}_{g_2}(\Pi^{-1}_{g_1}v))$$
The point is that you are using a confusing notation as this one
$$"\tilde{\Pi}_{g_1}f(\Pi^{-1}_{g_2}v)=f(\Pi^{-1}_{g_1}\Pi^{-1}_{g_2}v)"$$
In summary, you have a group $G$ acting on the arguments $x\in X$ of your functions: if $g\in G$,  $x \mapsto gx$ is well defined and preserves the group structure: $(gg')x= g(g'(x))$ and $e(x)=x$ where $e\in G$ is the neutral element, i.e., the action of $G$ on $X$ is a group representation. 
This definition induces an action on the functions which, in turn is a representation of $G$ as well
$$f \mapsto \pi_g(f)$$ 
defined in this way
$$\left(\pi_g(f) \right)(x) := f(g^{-1}x) \quad \forall x \in X\tag{1}$$
With this definition you have to check that $$\pi_g\pi_h = \pi_{gh}\quad \forall g,h \in G\:.$$
Applying (1), you therefore have to check if
$$\left(\pi_g(\pi_h f)\right)(x) = f((gh)^{-1}x)\quad \forall x \in X \quad \mbox{and} \quad \forall g,h \in G\:.$$
The left-hand side, again applying (1), reads
$$\left(\pi_g s\right)(x)\quad \mbox{where}\quad s(x) := \left(\pi_hf\right)(x) = f(h^{-1}x)$$
so that 
$$\left(\pi_g s\right)(x) = s(g^{-1}x) = f(h^{-1}(g^{-1}x))= f((h^{-1}g^{-1})x)= f((gf)^{-1}x)\quad \forall x\in X$$
as wanted.
A: The inverse is needed to get a homomorphism,
let
$$
 \tilde f(x)= f((\pi(g_1)^{-1} x).
$$
Then 
$$ \pi(g_2) (\pi(g_1) f(x))\\
= \pi(g_2) f(\pi(g_1)^{-1} x)\\
= \pi(g_2) \tilde f(x)\\
= \tilde f(\pi(g_2)^{-1} x) \\
= f(\pi(g_1)^{-1}( \pi(g_2)^{-1}x))\\
=f(\pi(g_1)^{-1} \pi(g_2)^{-1}x)\\
= f((\pi(g_2)\pi(g_1))^{-1} x)\\
= (\pi(g_2)\pi(g_1))f(x)
$$
So one needs the inverse to get the product in the right order. 
