# Fundamental group of a component of $GL_n({\bf R})$

Let $G$ be a component of $GL_n({\bf R})$ such that element has a positive determenant.

(1) Since it contains $SO(n)$, $\pi_1(SO(n))$ ? What is a fundamental group of $G$ ?

(2) It has a curvature bound ? That is to say, we can have bound $-1$ below or $-\infty$ ?

• The subgroup $SO(n)$ is a deformation retract of $G$ (we have discussed this several times here: if you search a bit you should find it), so $SO(n)$ and $G$ have the same $\pi_1$. On the other hand, for (2) you need to tell us what metric you want to consider, for otherwise we simply cannot answer; being a group, it is natural to consider left invariant metrics on $G$, and then the curvature will be constant, so you question becomes slightly uninteresting! Commented Oct 18, 2013 at 7:36
• (Here I wrote the complex case and here Quiaochu did the real one) Commented Oct 18, 2013 at 7:40
• $GL_n^+({\bf R})$ can have nonnegative curvature ? Or it is a bundle ? For instance, vector bundle over $SO(n)$ ? Commented Oct 18, 2013 at 7:50
• Milnor mentions a theorem of Wallace to the point that «If the universal covering of G is not homeomorphic to Euclidean space (or equivalently if G contains a compact non-commutative subgroup), then G admits a left invariant metric of strictly positive scalar curvature» in his amazing review Curvatures of left invariant metrics on lie groups. Commented Oct 18, 2013 at 7:54

$(1)$ Yes, like $SO(n)$, the group $GL_+(n, \mathbb{R})$ is not simply connected for $n\ge 2$, but rather has a fundamental group isomorphic to $\mathbb{Z}$ for $n=2$ or $\mathbb{Z}/2\mathbb{Z}$ for $n>2$.