Say we have the intertwiner $T : \mathbb{C}[G] \rightarrow L^2(G)$ where the domain is the group ring over the complex numbers for $G$ and the codomain is the square integrable functions. I have the definition $$\left[ T \left( \sum_{g \in G} c_g e_g \right) \right](x) := c_x $$ where $c_g \in \mathbb{C}$ and $e_g$ are the basis vectors, and $x \in G$. One can show that this is indeed an intertwiner. But my confusion is as to whether above we are defining an action on the image of the intertwiner, or if we are defining the intertwiner itself? Somehow, I think it's the former since $T(\cdots)$ should take care of the "definition" of the linear map, and then by putting $x \in G$ after, we are saying what this image $T(\cdots)$ does on an element of $G$.
But then in the proof that $T$ is an intertwiner, we write $$\left[ T \left( h \sum_{g \in G} c_g e_g \right) \right](x) = \cdots = c_{h^{-1}x}$$ and here we introduce another element of $G$, namely $h$ to show the intertwinining relation $\rho_V(g)T(v) = T(\rho_W(g)v)$ where $\rho_*$ are the suitable representations. Basically, I'm unsure as to why the $x$ is relevant - I understand that the equation above (and another similar equation for the other side of the intertwining relation) shows that $T$ is an intertwiner, but since $h$ is already acting on an element of $\mathbb{C}[G]$ in the above, I'm not sure what $x$ is doing.
Thanks for any help.
Set $v=\sum_{g\in G} c_ge_g$. Then, $T(v)\in L^2(G)$ is a function $T(v):G\to \mathbb{C}$. The reason $x$ is relevant is because, in order to describe a function, you have to say what it does to every $x\in G$. You explain that this function is given by $$T(v)(x)=c_x.$$
Now, you want to show that this is an intertwiner. That means you need to show that for all $h\in G$, $T(h.v)=h.(T(v))$. Again, $T(h.v), h.T(v)\in L^2(G)$ are functions. In order to show they are equal, we need to show that $$T(h.v)(x)=(h.T(v))(x)$$ for every $x\in G$.
In order to describe what $T(h.v)$ is as a function, you need to say what it does to each $x\in G$. You have computed $$T(h.v)(x)=c_{h^{-1}x}.$$
Note that the group $G$ acts on a function $f:G\to\mathbb{C}$ by the formula $$(h.f)(x)=f(h^{-1}x).$$ Therefore, we compute $$c_{h^{-1}x}=T(v)(h^{-1}x)=(h.T(v))(x).$$ This shows that for every $x\in G$, $T(h.v)(x)=(h.T(v))(x)$. It follows that, as functions, $T(h.v)=h.T(v)$.