Inner Product on Lie Algebra So the Exponential Map Becomes the Matrix Exponential?

Question

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.

Answer 1

I may be misinterpreting your statement "for G a matrix group over C which is also a compact Lie group" but the only complex compact Lie groups are abelian varieties ("Chevalley's theorem"), while linear groups are affine, so very far from compact.. since you're interested in Riemannian metrics, I will assume that you want to consider compact real groups (such as U(n), SO(n), SU(n), ...)

In the case of a simple Lie group (SU(n), Spin(n), SO(n), Sp(n),...), there is a unique (up to scale) bi-invariant bilinear form (the Killing form), and it is negative definite if and only if the Lie group is compact. Hence, you can take the negative of this form, which lifts (thanks to bi-invariance) naturally to a Riemannian metric on G.

The general formula for the Killing form is $$B(X,Y) = \mathrm{tr} (ad(X) \circ ad(Y))$$ and so makes direct contact with the adjoint representation: The natural action of the Lie group on its Lie algebra. This is straightforward (yet perhaps a little annoying) to compute, and in the case of the classical Lie groups the form is just a constant multiple of $\mathrm{tr}(XY)$ (taken in the defining representation): e.g. for $\mathfrak{so}(n)$ it is $B(X,Y) = (n-2) \mathrm{tr}(XY)$ where you can think of $X$ and $Y$ just as special orthogonal $n \times n$-matrices. You can find more here: https://en.wikipedia.org/wiki/Killing_form#Matrix_elements

In any case, for a group with a bi-invariant metric like this, the matrix exponential and the Riemannian geometric exponential coincide, while for a group that does not permit a bi-invariant metric (like a non-compact group) these two will not agree.