I am reading a paper DRINFELD DOUBLE OF GLn AND GENERALIZED CLUSTER STRUCTURES. I have a question about $\nabla^R f$ in Section 2 of page 7.
Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. Let $\langle \cdot, \cdot \rangle$ be a non-degenerate bilinear form on $\mathfrak{g}$. Let $f \in C^{\infty}(G)$. $\nabla^R f$ is defined as follows \begin{align} \langle \nabla^R f(X), \xi \rangle = \frac{d}{dt}|_{t=0} f(Xe^{t \xi}), \end{align} where $X \in G, \xi \in \mathfrak{g}$. I think that $\nabla^R f(X)$ is in $\mathfrak{g}$. Is this correct? Where does $\nabla^R f$ belong to? Thank you very much.
Fix $X\in G$. Then the mapping $\alpha : \mathfrak g \to \mathbb R$, $$ \xi \mapsto \frac{d}{dt}\bigg|_{t=0}f(Xe^{t\xi})$$ is linear and thus $\alpha$ is an element in $\mathfrak g^*$.
Now you are given a bilinear form $\langle \cdot, \cdot\rangle$ on $\mathfrak g$. So you can define $A: \mathfrak g \to \mathfrak g^*$ by
$$ A(\xi) (\psi) = \langle \xi, \psi\rangle.$$
That the bilinear form is non-degenerate means that $A$ is invertible. Thus $\nabla^R f(X)$ is just $A^{-1}\alpha$. This is true for all $X\in G$, so you can think of $\nabla^R f$ as a map $\nabla^R f : G\to \mathfrak g$.