Do we need transpose in the definition of a dual representation? Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. There is an action of $G$ on itself given by left multiplication: $G \times G \to G$, $(f,g) \mapsto fg$, $f, g \in G$. There is a corresponding action of $G$ on $\mathbb{C}[G]$ given by $(g, f)(h) = f(g^{-1}h)$, $g, h \in G, f\in \mathbb{C}[G]$. This is a left $G$-action on $\mathbb{C}[G]$. 
By wikipedia, a dual representation of $G$ is given by $\rho^*(g) = (\rho(g^{-1}))^T$. In $(g, f)(h) = f(g^{-1}h)$, we don't have transpose. But it seems that transpose is used implicitly in $(g, f)(h) = f(g^{-1}h)$. Does the transpose correspond to changing the right action of $G$ to the left action of $G$?
I think that the action of $\mathfrak{g}$ on $\mathbb{C}[G]$ is given by 
\begin{align}
(X.f)(h) = \frac{d}{dt}|_{t=0}(e^{-tX}h), X \in \mathfrak{g}, f \in \mathbb{C}[G], h \in G.
\end{align}
Is this formula correct? Any help will be greatly appreciated!
 A: Suppose $A:V\to W$ is a linear transformation between vector spaces. Define the dual map $A^*$ to be a linear transformation $A^*:W^*\to V^*$ between dual spaces $W^*$ and $V^*$ defined by $(A^*\phi)(v)=\phi(Av)$ for all functionals $\phi\in W^*$ (i.e. $\phi:W\to K$ where $K$ is the scalar field). Observe $A^*\phi$ takes $v\in V$ as input and outputs scalars, so indeed $A^*\phi\in V^*$.
IOW, when $A^*$ sees a map $W\to K$, it precomposes with $V\to W$, resulting in $V\to K$.
If we pick ordered bases (why does no one ever specify "ordered"?) of $V$ and $W$ then the transformation $A$ can be represented by an explicit matrix. The ordered bases of $V$ and $W$ induce ordered dual bases on the dual spaces $V^*$ and $W^*$, so the transformation $A^*$ may also be represented by a matrix with respect to them - it turns out the matrix of $A^*$ is the transpose of $A$'s original matrix.
But the concept of matrix transpose is not a requirement for defining dual maps. We could speak abstractly of vector spaces and linear transformations without mentioning bases, coordinates or matrices, which is what I did in the first paragraph.
Now, if $(V,\rho)$ is a linear representation of $G$, then for any $g\in G$, we know $\rho_V(g):V\to V$, so we can speak of corresponding dual maps $\rho_V(g)^*:V^*\to V^*$. This defines a linear action of $G$ on $V^*$. However, this group action is a right action of $G$, rather than a left action. 
Exercise: Suppose $G$ acts on a set $X$ from the left, and $Y$ is any other set. Verify $(g\cdot f)(x):=f(gx)$ is a right group action of $G$ on $\hom(X,Y)$, not a left action.
There is a canonical way to convert between right/left actions, though, which is to swap $g$ and $g^{-1}$:
Exercise: Suppose $\alpha:G\times X\to X$ is a left group action. Verify that the map $\beta:X\times G\to X$ defined by $\beta(x,g):=\alpha(g^{-1},x)$ is a right group action. State and verify the analogous claim for converting from right to left actions.
Note the transpose does not come into play here - this is true for any group actions, meaning barebones actions on mere sets without any regard for structure, like linearity. 
Anyway, this means $g\cdot \phi:=\rho_V(g^{-1})^*\phi$ is a linear left action of $G$ on $V^*$. This is the definition of the dual representation.
I remember my first reaction to the $g^{-1}$ in the definition of contragredient action was confusion. After learning that the "obvious" algebraic thing to do to create an action of $G$ on $\hom(X,Y)$ from an action of $G$ on $X$ (namely, just write down $f(gx)$) actualy creates a right instead of a left action, using $g^{-1}$ instead of $g$ still struck me as an artificial and inelegant fix. Eventually, I realized the counterintuitiveness of this phenomenon goes back all the way to transformations of graphs back in college algebra!
Suppose we have a graph $y=f(x)$ of some function $f:\mathbb{R}\to\mathbb{R}$. How does the graph of $y=f(x+h)$ relate to the original graph? (Let's assume $h>0$.) The student's gut reaction is to say it's the original graph translated $h$ to the right, but it's actually $h$ units translated to the left.
Here is the group action theory behind it. Suppose $G$ acts on $X$. If $Y$ is another set, then any function $f:X\to Y$ is formally encoded as a subset $\mathcal{G}\subset X\times Y$ s.t. $\forall x\in X~\exists !y\in Y:(x,y)\in\mathcal{G}$. Then we want $G$ to act on the graph by acting on the $x$-coordinates of individual points. (So $g\in G$ sends $(x,y)$ to the point $(gx,y)$. For $G=\mathbb{R}$ this means $g\in\mathbb{R}$ shifts any point $g$ units in the usual way.) Thus
$$g\mathcal{G}=g\{(x,f(x)):x\in X\}=\{(gx,f(x)):x\in X\}=\{(x,f(g^{-1}x):x\in X\}. $$
So actually the $g^{-1}$ is natural after all!
As a special case, if you want $S_n$ to permute the coordinates of $(x_1,\cdots,x_n)$ the "obvious" way (that is, if $\sigma:i\mapsto j$, then we want the value $x_i$ moved to the $j$th coordinate), we cannot write the naive definition $\sigma(x_1,\cdots,x_n)=(x_{\sigma(1)},\cdots,x_{\sigma(n)})$, as tempting as it may be. (Ditto for $S_n$ acting on the tensor power $V^{\otimes n}$, whose pure tensors look like $v_1\otimes\cdots\otimes v_n$.) I have seen multiple authoritative sources make this mistake!
One must think of the coordinate tuple as a function $x$ with domain $\{1,\cdots,n\}$ on which $S_n$ acts, and then have $S_n$ act on these functions contragrediently. Indeed, this makes sense, since we just said that if $\sigma:i\mapsto j$ then we want $(\sigma x)_j$ to be $x_i$, so $i=\sigma^{-1}(j)$ and $(\sigma x)_j=x_{\sigma^{-1}(j)}$.
(Sometimes it's better to have $S_n$ act on $X^n$ or $V^{\otimes n}$ from the right, because if some $G$ acts on a set $X$ or linearly on a space $V$ from the left then it acts on $X^n$ or $V^{\otimes n}$ too in a way that "commutes" with the action of $S_n$ from the right, which we may express symbolically as an "associative" property.)
A: For a representation $\rho : G \rightarrow GL(V)$ the dual representation is $\rho^{*} : G \rightarrow GL(V^{*})$ where the action is given in terms of that of G on $V$
 by  $\rho^*(g)(v^*)(v):=v^*({\rho(g^{-1})(v))}$. No transpose is implicit here. This definition is motivated by the requirement that the two representations should respect the natural pairing. Now if we see the elements of $V$ and $V^*$ as column vectors and the actions above as multiplications by the matrices $[\rho(g)]$ and $[\rho^*(g)]$ on the left then  ( natural pairing is obviously given by $<\phi,v>=\phi(v)=[\phi]^tv$( $[\phi]$ is the matrix of $ \phi $) ) the requirement that the reps respect the natural pairing implies $[\rho^*(g)(v^*)]^t [\rho(g)v]=[v^*]^tv$. So $[\rho^*(g)]=[\rho(g^{-1})]^t$.That is how we get the transpose !
