All Questions

What is the easiest way to show that a compact hausdorff topological group is a closed subgroup of a product of finite dimensional Lie groups? Here are the relevant definitions: Definition: (compact ...

Basically what the title says — if a Lie group is infinite-dimensional, is it necessarily noncompact?

Let $G$ be a simple (i.e. every proper normal subgroup is discrete) simply connected compact Lie group. Define the degree of $k$-nilpotence of $G$ to be the Haar measure of the set $$\{(x_1,\dotsc,x_{...

I know that $G_2$ can be described as the subgroup of $SO(7)$ preserving a specific element of $\Lambda^3(\mathbb{R}^7)^*$. It can thus be realized as a matrix group. Prof. Robert Bryant did describe ...

Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$. Of course any ...