I have a stupid questions regarding Lie algebras and the exponential map: What inner product on $\mathfrak{g}$ (and, hence, what Riemannian metric on $G$, using left-multiplication as an isometry) makes the exponential map (as a map from the tangent space of a manifold to the manifold) the matrix exponential (as a map from the Lie algebra of a Lie group to the Lie group)?
So, to explain a little, if $(M,g)$ is a closed Riemannian manifold, it has a geodesic flow. Given $p \in M$ and $v \in T_p(m)$, there is a unique geodesic $\gamma_{p,v}(t)$ with $\gamma(0) = p$ and $\frac{d \gamma}{dt}(0) = v$. We call $\text{Exp}: T_p(M) \to M$ given by $\displaystyle \text{Exp}(v) = \gamma_{p,v}(1)$ the "according-to-Hoyle" exponential map, and, more generally, for $||v|| = 1$, $\text{Exp}(tv) = \gamma_{p,v}(t)$.
Now, for $G$ a matrix group over $\mathbb{C}$ [Edit: Per the answer below, this should be over $\mathbb{R}$/a real form?] which is also a compact Lie group, $I \in G$, and $\mathfrak{g} = T_I(G)$, if we have $A \in \mathfrak{g}$, we call $\text{Exp}: \mathfrak{g} \to G$ given by $\displaystyle \text{Exp}(A) = e^{A} = \sum\limits_{n=0}^{\infty} \frac{A^n}{n!} \in G$ the "according-to-Hoyle" exponential map, and, more generally, $\displaystyle \text{Exp}(tA) = e^{tA} = \sum\limits_{n=0}^{\infty} \frac{(tA)^n}{n!} \in G$.
My question (and, this probably a pretty basic/stupid one) is, is there an inner product $\langle\ |\ \rangle$ on $\mathfrak{g}$ that, together with the use of left-multiplication on $G$ to make a Riemannian metric $g$ on $G$ out of $\langle\ |\ \rangle$, so that the first sense of the exponential map agrees with the second sense? Is it
$$ \langle A\ |\ B\rangle = \Re\left\{\text{Tr}(B^*A)\right\} = \Re \left\{\sum\limits_{i=1}^n\sum\limits_{q=1}^n \bar{B}_{qi}A_{qi}\right\}$$
from this post Metric over a Lie algebra $\mathfrak{u}(n)$?
Thanks in advance.