Geometry of Semi-Simple Lie Groups

Question

I wonder, what exactly does it mean for a Lie group to be semi-simple geometrically?

I know that any Lie group is a group and a smooth manifold simultaneously, so it can be considered merely as a vector field in dimension one. However, I don't get the geometrical philosophy behind the simplicity of a Lie group. What does the condition tell us about a vector field associated with the group itself?

How can one describe geometrically the condition of simplicity (or semi-simplicity) of a Lie group?

Answer 1

The sentence "so it can be considered merely as a vector field in dimension one" makes no sense as a manifold is not the same thing as a vector field. There is no single vector field canonically associated with a group. A Lie group is a smooth manifold $G$ equipped with a group operation which happens to be a smooth map $G\times G\to G$, don't read more into it.

For a connected Lie group, Cartan's second criterion says that $G$ is semi-simple if and only if its Killing form $B\colon \mathfrak{g}\times\mathfrak{g}\to\Bbb R$, given by $$B(X,Y) = {\rm tr}({\rm ad}(X){\rm ad}(Y)),$$is non-degenerate. If $G$ is assumed to be compact, then $-B$ is positive-definite and therefore defines a left-invariant Riemannian (Einstein) metric on $G$.

Answer 2

"What exactly does it mean for a Lie group to be semisimple geometrically?" - this can mean many different things depending on the geometric context. See for example this post (mentioning also Riemannian symmetric spaces)

What is the intuition behind simple Lie groups?

Another particular example is that a semisimple Lie group $G$ does not admit any left-invariant affine structure (whereas a non-semisimple Lie group usually admits one), because for its Lie algebra $\mathfrak{g}$ the first cohomology space $H^1(\mathfrak{g},M)$ with any finite-dimensional $\mathfrak{g}$-module $M$ vanishes, by Whitehead's first Lemma. This is equivalent to some fixed point theorem, for more see for example this MO-post.

Actually, a connected Lie group $G$ is semisimple if and only if $H^1(\mathfrak{g},M)=0$ for the Lie algebra $\mathfrak{g}$ of $G$, and for all finite-dimensional $\mathfrak{g}$-modules $M$. This is an algebraic description, but it is also related to Weyl's unitary trick (by Weyl's theorem, which implies the first Whitehead Lemma easily).