Humphreys page 63 - error? Looking at Humphreys linear algebraic groups, page 63.
Let $G$ be an affine algebraic group. Let $K[G]$ be the associated algebra. Let $\varphi:G\times G\to G$ be right translation $\varphi(g,x)=\varphi_g(x)=xg$ and let $\varphi^*_x:K[G]\to K[G]$ be its comorphism: $\varphi^*_x:f\mapsto f\circ \varphi_x$.
Humphreys writes generators $f_1,\dots,f_n\in K[G]$ and lets $F$ be the vector space given by the span of these elements. He then uses two propositions to say that $F$ is stable under translations of functions, and to show that $\varphi^* F\subset K[G]\otimes_K F$.
He then goes on to say with this we have $\varphi^*_x f_i=\sum_{j=1}^n m_{ij}(x)f_j$. Although technically correct here, I think this labeling is morally wrong, and should read: $$\varphi_x^* f_j = \sum_{i=1}^n m_{ij}(x)f_i$$
The objective was to assign $\Psi:G\to \text{GL}(F)$ so that we have a faithful representation of $G$. So I want to have $g\mapsto (m_{ij}(g))$, where $(m_{ij}(g))$ is the matrix corresponding to $\varphi^*_g|_E$, with respect to the basis $\{f_1,\dots,f_n\}$. In this case I would like for  $(m_{ij}(g))f_k = \sum_j m_{jk}f_j$. (Which works with my labeling)
For his to work, it seems I would take $(f_k)(m_{ij}(g))=\sum_j m_{kj}f_j$? I.e. the matrix multiplies from the right? Is this perhaps his intention?
 A: You want to have a representation $\Psi : G \to GL_d(F)$, that is, $\Psi(g) \Psi(h) = \Psi(gh)$ for all $g, h \in G$. Let us look at what happens when we use Humphrey's notation $\varphi^*_x f_i = \sum_{j} m_{i,j}(x) f_j $. Then you have
$$\begin{array}{rl}\sum_{k,j} m_{i,j}(g)m_{j,k}(h) f_k 
&= \sum_{j} m_{i,j}(g)  \varphi^*_h(f_j) \\ 
&= \varphi^*_h(\sum_{j} m_{i,j}(g) f_j) \\ 
&= \varphi^*_h(\varphi^*_g(f_i)) \\ 
&= \varphi^*_{gh}(f_i)  \\
&= \sum_{k} m_{i,k}(gh) f_k
\end{array} $$
And thus (Note that for this to work you have actually to take a basis of your finite dimensional vector space, not just a generating set) you obtain 
$$ m_{i,k}(gh) = \sum_{j} m_{i,j}(g) m_{j,k}(h) $$ which amounts to $\Psi(g) : = (m_{i,j}(g))$ having the desired property.
You might want to check for yourself that your convention $\varphi^*_x f_j = \sum_{j} m_{i,j}(x) f_i$ would lead to $\Psi(gh) = \Psi(h) \Psi(g)$ the other way around.
The main reason for this is that you have $\varphi^*_x \varphi^*_y = \varphi^*_{yx}$ and so your vector space becomes a right $G$-module and not a left one. Thus mulitplying from the right is more natural in this context.
