# What does the logarithm of the identity look like on a Lie group?

Let $G$ be a compact, connected Lie group with identity element $e$ and $\mathfrak g$ its Lie algebra. Consider the set $$L=\{A\in\mathfrak g\setminus\{0\};\exp(A)=e\}.$$ The most descriptive name for this set that I could come up with is the logarithm of the identity (or the set of those logarithms), whence the title of this question.

Question: What does the set $L$ look like in general? In particular, when is its radial projection to the unit sphere in $\mathfrak g$ surjective? And when it is not surjective, does it have dense image?

The surjectivity question can be reformulated more geometrically: Is every geodesic on $G$ periodic? (If you want an explicit metric on $G$, take any bi-invariant Riemannian one.) I have a faint feeling that the answer might be a somewhat standard result, but I couldn't find it. If anyone has encountered this set before or knows how to attach some useful structure to it, I would be happy to hear. The set $L$ is obviously closed under multiplication by nonzero integers, but that's not a lot of structure yet.

For two groups $G_1$ and $G_2$ and $G=G_1\times G_2$ we have $L=(L_1\cup\{0\})\times(L_2\cup\{0\})\setminus\{(0,0)\}$ and similarly for more groups. Since the case of tori is given below, analyzing all simple Lie groups should be enough.

Some wondered in the comments why I excluded zero from $L$. The reason for this is the application I have in mind (integral geometry on Lie groups, see my second comment below), but this is not all that important. If someone can give a good description of $L\cup\{0\}$, that obviously gives a good description of $L$ as well. And of course the radial projection is not meaningfully defined at the origin.

Examples:

• In the case of the torus $G=\mathbb R^n/\mathbb Z^n$ we have $L=\mathbb Z^n\setminus\{0\}$. (The projection has dense image but does not fill the unit sphere.)
• If $G=SU(2)=Sp(1)=S^3$, then by the geometry of the sphere $L=\{v\in\mathfrak g\setminus\{0\};|v|\in2\pi\mathbb N\}$. (The projection is surjective.)
• Let $G=U(n)$ and suppose $A\in L$. The matrix $A$ is antihermitean, so up to a unitary change of basis $A=iD$, where $D=\text{diag}(a_1,\dots,a_n)$ is a real diagonal matrix. Then $I=\exp(A)=\text{diag}(e^{ia_1},\dots,e^{ia_n})$ implies that $a_i\in2\pi\mathbb Z$ for all $i$. Therefore $L$ consists of the matrices with eigenvalues in $2\pi i\mathbb Z$. In the case of $SU(n)$ there is the additional constraint that the eigenvalues must sum to zero. (The projection has dense image but I'm not sure if it is surjective.)
• Do you have some specific reason for excluding $0$? It seems unnatural to me. (e.g., your set is currently closed under multiplication by integers, except when that integer is $0$). – Micah Oct 7 '14 at 4:52
• @Micah, I am considering geodesics on Lie groups. The zero elements corresponds to the constant curve, which does not qualify as a proper geodesic. – Joonas Ilmavirta Oct 7 '14 at 9:06
• Are finite groups not Lie groups because they are discrete? No. Is the constant curve a geodesic? Yes. They aren't as interesting, but I think the whole picture yields a clearer result. – Robin Goodfellow Oct 7 '14 at 17:52
• Also, I don't say this much, but...+1 for an awesome question. – Robin Goodfellow Oct 7 '14 at 17:55
• @RobinGoodfellow, I want to study questions of the following type: "If we know the integral of a function on a Lie group over all closed geodesics, can we recover the function?" Including constant curves as geodesics renders the problem trivial. And of course, if someone can describe the set $L\cup\{0\}$, that also explains the structure of $L$. Inclusion or exclusion of the zero is not that important, although I am not interested in the zero. And thanks! Let's hope there will be an awesome answer... – Joonas Ilmavirta Oct 7 '14 at 17:59

## 1 Answer

I claim that the radial projection of $L$ is always dense, but is surjective only for $S^1$, $SO(3)$ and $SU(2)$.

To see that the projection is dense, take a nonzero vector $v$ in $\mathfrak{g}$ and embed $v$ into a maximal Cartan $\mathfrak{h}$. Then the exponential map restricted to $\mathfrak{h}$ has kernel a lattice $\Lambda$ (the root lattice) of rank $\dim \mathfrak{h}$. We can approximate the spherical projection of $v$ arbitrarily well by spherical projections of lattice points, so the spherical projection of $L$ is dense.

You have already computed that the spherical projection is not surjective for $H = (\mathbb{R}/\mathbb{Z})^2$. Therefore, it is also not surjective if $G$ contains a copy of $H$. So you only care about groups of rank $\leq 1$. (Rank is the dimension of a maximal torus.) Looking through the classification of compact connected Lie groups, this is the trivial group, $S^1$, $SO(3)$ and $SU(2)$, and your conventions about the zero vector rule out the trivial group.

Finally, as to the question of what $L$ looks like. Fix a Cartan $\mathfrak{h}$, let $\Lambda$ be the root lattice (kernel of $\exp: \mathfrak{h} \to G$) and let $W$ be the Weyl group (which can be defined as $\{ g \in G: g \cdot \mathfrak{h} = \mathfrak{h} \}/\exp(\mathfrak{h})$). I claim that there is a connected component of $L$ for each element $\lambda$ of $(\Lambda \setminus \{ 0 \})/W$, and it is isomorphic to the homogenous space $G/P_{\lambda}$, where $P_{\lambda}$ is the stabilizer of $\lambda$.

For example, let $G=SU(2)$. Take $\mathfrak{h} = \mathbb{R} \left( \begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix} \right)$. Then $\Lambda = 2 \pi \mathbb{Z} \left( \begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix} \right)$ and $W = \pm 1$, acting on $\Lambda$ by $\pm 1$. For each $\lambda \in \Lambda$, the stabilizer $P_{\lambda}$ is an $S^1$ and $G/P_{\lambda} \cong S^2$. Try to visualize a nested chain of spheres in $\mathbb{R}^3$, of radius $2 \pi k \sqrt{2}$, and a line $\mathfrak{h}$ piercing each of them at the $2$ points $\pm 2 \pi k \left( \begin{smallmatrix} i & 0 \\ 0 & -i \end{smallmatrix} \right)$.

Proof sketch We write $\pi$ for the continuous map $\mathfrak{g} \to \mathfrak{h}/W$, whose fibers are the $G$ orbits on $\mathfrak{g}$. Since $L$ is $G$ invariant, we have $g \in L$ if and only if $\pi(g)$, lifted to an element of $\mathfrak{h}$, is in $L$. But $\mathfrak{h} \cap L = \Lambda$. So $L = \pi^{-1}(\Lambda/W) \setminus \{ 0 \}$. For $\lambda \in \Lambda/W$, the fiber $\pi^{-1}(\lambda)$ is the $G$-orbit through $\lambda$, which is $G/P_{\lambda}$. $\square$

• Thanks! This is a great answer, but it will take a while for me to digest it. I hope I can figure out the details myself. – Joonas Ilmavirta Oct 15 '14 at 10:21