# 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?

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.

• Kind of confused. If we call the transformed vector $w\equiv\Pi^{-1}_{g_2}v$ and have no knowledge that it is the result of a transformation, then the definition would lead us to $\tilde{\Pi}_{g_1}f(w)=f(\Pi^{-1}_{g_1}w)$. How is it pre-acting on the $v$ before the $\Pi^{-1}_{g_2}$ has already been shifted inside, solely based on us knowing this first transformation has already happened?
– dsm
Feb 10, 2020 at 13:25
• You are using bad notations. This is the source of the confusion. Feb 10, 2020 at 13:31
• this $\tilde{\Pi}_{g_1}f(\Pi^{-1}_{g_2}v)$ is very ambiguous. Feb 10, 2020 at 13:33
• If the notation is bad I apologize, just using the notation from my text (this is my first foray into representation theory). My above comment is still my confusion for your answer. So, if I'm understanding you right, this definition of $\tilde{\Pi}$ acting on elements of $\mathcal{C}(V)$ requires any consecutive transformation to first act on the original element $v\in V$. Just seems like we have to have knowledge about the history of $v$ to know that it didn't first come from a different transformation $\tilde{\Pi}_{g^*}$. Hopefully I'm making sense.
– dsm
Feb 10, 2020 at 13:42
• I extended my answer, please have a look at it... Feb 10, 2020 at 13:52

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.