Let $G$ be a Lie group equipped with a bi-invariant metric. I have some questions concerning the totally geodesic and the flat (all sectional curvatures zero) submanifolds of $G$.

  1. I known that every Lie subgroup of $G$ is a totally geodesic submanifold. Can one say anything interesting about the other direction?
  2. Is there any relation between flat, totally geodesic submanifolds of $G$ and its Lie subgroups?

1 Answer 1


For (1): no. Think about $SU(2)$, which can be identified with the three sphere $\mathbb{S}^3$. The three sphere admits as its great "circle" a copy of $\mathbb{S}^2$ which is totally geodesic. But all compact Lie groups of dimensions $\leq 2$ are Abelian, and $\mathbb{S}^2$ cannot be represented as a Lie subgroup of $SU(2)$.

In the positive direction, the closest to the classification you are looking for can be stated in more generality for symmetric spaces, where one has a one-to-one correspondence between totally geodesic submanfiolds through a fixed point $p_0$ of a symmetric space, with "Lie triple systems" (a special type of subspace) of the corresponding Lie algebras. As the definitions take some time to write down, I'll just refer you to Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces (AMS Graduate Studies in Mathematics vol 34) Chapter IV, section 7.

Specifically in the case of a compact Lie group $G$ with Lie algebra $\mathfrak{g}$, this means that totally geodesic submanifolds are in correspondence with subspaces $\mathfrak{m}\subset \mathfrak{g}$ such that the property $$ X,Y,Z \in \mathfrak{m} \implies [X,[Y,Z]] \in \mathfrak{m} \tag{T}$$ holds.

In the case of $SU(2)$ letting the generators be $$ u_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \quad u_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad u_3 = \begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix} $$ you see that the linear span of any two out of the three forms a subspace for which (T) holds (i.e. the span is a Lie triple system). The submanifolds they generate are the 2 spheres.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.