Does there exist an isotropic left invariant Riemannian metric on $GL_n^+$? Let $GL_n^+$ be the group of $n \times n$ real invertible matrices with positive determinant
Does there exist a left invariant isotropic Riemannian metric on $GL_n^+$?
(By "isotropic" I mean that the sectional curvature $k(p,\sigma)$ for $\sigma \subseteq T_pM$ does not deped on $\sigma$. By left invariance it does not depend on $p$ as well).

I present a proof below, but I suspect that there are simpler ones:
Suppose such a metric $g$ exists. Then the left invariance implies $(GL_n^+,g)$ is complete, hence it's universal covering space is complete, simply connected, and with constant curvature, so it must be diffeomorphic to $\mathbb{R}^{n^2}$ or   $\, \mathbb{S}^{n^2}$. 
Since compact spaces can cover only compact spaces, this rules out $\mathbb{S}^{n^2}$, so the universal cover of $GL_n^+$ must be $\mathbb{R}^{n^2}$.
However, Milnor states here that the universal cover of a Lie group $G$ is not homeomorphic to Euclidean space iff $G$ contains a compact non-abelian subgroup*. Since $GL_n^+$ contains $SO(n)$ we have a contradiction, at least for $n \ge 3$.

*I actually don't know how to prove this. Any hints or references  would be welcome. Milnor states this in theorem 3.4.
 A: Here is a topological argument ruling out covering by $\mathbb R^{n^2}$. However, it might be an overkill for this statement.
Assume $n>2$. If $\mathbb R^{n^2}$ were the universal cover of $GL^+_n$, we have $GL^+_n=K(\pi, 1)$, where $\pi=\pi_1(GL_n^+)$. At the same time, $\pi_1(GL_n^+)=\pi_1(SO(n))=\mathbb Z_2$, so $K(\pi, 1)=\mathbb R P^{\infty}$, which is clearly not homotopy equivalent to a finite dimensional space.
A: $GL_n^{+}$ deformation retracts onto $SO(n)$. The universal cover of this (for $n \ge 3$) is a double cover, since $\pi_1(SO(n)) = \mathbb{Z}_2$. It's known as the Spin group $Spin(n)$, and in particular, because it's a double cover it's compact. Hence it's certainly not homotopy equivalent to a Euclidean space (e.g. because its top $\mathbb{Z}_2$ homology is nontrivial), which means the universal cover of $GL_n^{+}$ can't be either.
Various other arguments are possible. For example, we also know that (again, for $n \ge 3$) $\pi_3(Spin(n)) \cong H_3(Spin(n)) \cong \mathbb{Z}$, so the same is true of the universal cover of $GL_n^{+}$. 
