⇤Math diary
May 27, 2023:
Today I saw something I already knew in a new light. A while ago I learned about the morse theory proof that $\pi_2$ of all Lie groups is trivial. The idea is, $\pi_2(G) = \pi_1(\Omega G)$ where $\Omega G$ is the based loop space. This consists of paths $\gamma:S^1 \to G$ such that $\gamma(1)=1$. We study the topology $\Omega G$ using the energy function, $E(\gamma) = \int_{S^1} \vert \gamma’\vert ^2$. The critical points are exactly the closed geodesics on $G$, which are one-parameter subgroups, or homomorphisms form $S^1\to G$. Morse theory provides a cell decomposition of $\Omega G$ indexed by the critical points of $E$, consisting of all points which flow to the given critical point under the gradient flow of $E$. We can compute the index of each critical point using lie algebra theory, and we find they are all even. Thus, the cell structure of $\Omega G$ is purely made of even dimensional cells, so it must be simply connected. I summarized this proof in this tweet thread.
Turns out, these cells are the infinite-dimensional Burhat decomposition of $LG$. $\Omega G \cong LG_\mathbb{C}/LG_{\mathbb{C}}^+$ is a sort of grassmanian, and these cells now are the schubert cells of this grassmanian. Also, every one-parameter subgroup of $G$, up to conjugation, lives in some maximal torus. In the maximal torus $T$, the one-parameter subgroups are indexed by the set of maps from $S^1$ to $T$, which is the cocharecter lattice $\check{T}$. Therefore, the cells of $\Omega G$ modulo congugacy are indexed by cocharecters modulo the weyl group.
It turns out, $\Omega G$ is not just a riemannian manifold, but a Kahler one. It carries a symplectic structure which I talked about here, but in this case is nondegenerate. The energy function has Hamiltonain vector field which is a rigid rotation of loops. (Note that this does NOT act on $\Omega G$ as a subgroup of the free loop group $LG$, only on the coset space $LG/G$.) This provides a symplectic perspective to things.