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

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 $\mathfrak{u}(n)$?

• (I just taught surface integrals to my vector calculus class today, and the question of the exponential map from a parallelogram in the tangent space of a surface to a curvilinear rectangle on the surface came up, and I couldn't remember this part of the derivation of the nomenclature.) May 6, 2020 at 1:45
• What is the definition of a geodesic map in this context? May 6, 2020 at 2:06
• Parallel transport of the velocity vector field along the curve is an isometry, IIRC? en.wikipedia.org/wiki/Geodesic#Affine_geodesics May 6, 2020 at 2:20
• Maybe the covariant derivative of the velocity vector field along the curve is 0? en.wikipedia.org/wiki/Geodesic#Affine_geodesics May 6, 2020 at 2:27
• (From the Wikipedia page - it's been a long time since I studies do Carmo) "[T]he equation ${\displaystyle \nabla _{\dot {\gamma }}{\dot {\gamma }}=0}$ means that the acceleration vector of the curve has no components in the direction of the surface (and therefore it is perpendicular to the tangent plane of the surface at each point of the curve). So, the motion is completely determined by the bending of the surface." May 7, 2020 at 1:43

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.

• This looks like it is exactly for what I was looking. I'll read it more carefully later today or tomorrow and accept it as the answer. (Sorry about the confusion about complex compact and compact real; I'm not really a Lie group/algebra theorist, more of a manifold topologist, and don't know the "lay of the land" well at all) May 6, 2020 at 12:45
• For a proof outline that the two exponential maps agree for compact Lie groups, see here: math.stackexchange.com/questions/3655754/…
– Max
May 6, 2020 at 16:13
• @Max Thanks so much! May 7, 2020 at 1:49
• It's not true that for a compact Lie group the exponential maps agree. What's true is that for bi-invariant metrics the exponential maps agree. And compact Lie groups turn out to possess such metrics (precisely a connected Lie group has such a metric iff it covers a compact Lie group). Beware that every nonabelian connected compact Lie group such as $\mathrm{SO}(3)$ possesses left-invariant non-bi-invariant metrics, for which the exponentials do not match. Precisely, for $S$ simple compact of dim $d$, the cone of left-invariant (resp. bi-invariant) metrics has dimension $d(d+1)/2$, resp. $1$.
– YCor
May 7, 2020 at 5:44
• @Dinisaur When I wrote $\mathrm{Spin}(n)$ I meant the compact real form, i.e. the double cover of $\mathrm{SO}(n, \mathbb{R})$. In particular, I meant $\mathrm{Spin}(3) \simeq \mathrm{SU}(2)$. Of course, the split real form of $\mathrm{Spin}(3, \mathbb{C})$ is isomorphic to $SL(2, \mathbb{R})$ but this is a non-compact group and hence one wouldn't expect it to have a bi-invariant metric. Feb 23, 2021 at 21:36